tag:blogger.com,1999:blog-22574588387273157112024-03-13T20:28:36.083-07:00K6JCA—•— —•••• •——— —•—• •—Jeffhttp://www.blogger.com/profile/04853314106806116765noreply@blogger.comBlogger178125tag:blogger.com,1999:blog-2257458838727315711.post-86336216278932334722023-11-28T16:38:00.000-08:002023-12-19T16:20:08.129-08:00Bringing an ARC-5 Receiver Back to Life, Part 2 (An AC Power Supply)<div class="separator" style="clear: both; text-align: center;"><br /></div><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjrHMePeW6z4TG709r2HQJGK1kFWQ5cq1WHPkpUd49Q7cTHW187pxuivjumjs2Ta84VTOFhi7uGVHYbrQt5DhKCtK5Ke-WfGN4fQpDcaSu8uEKbNMANc4I4MZhtw21gP0f2cJYTbipIwe-GmxND3VZWOI3kxR5_lmKm2SgB7FF1vSznlw8Ht5hmdrkIYm8/s640/IMG_9922.jpg" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="480" data-original-width="640" height="240" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjrHMePeW6z4TG709r2HQJGK1kFWQ5cq1WHPkpUd49Q7cTHW187pxuivjumjs2Ta84VTOFhi7uGVHYbrQt5DhKCtK5Ke-WfGN4fQpDcaSu8uEKbNMANc4I4MZhtw21gP0f2cJYTbipIwe-GmxND3VZWOI3kxR5_lmKm2SgB7FF1vSznlw8Ht5hmdrkIYm8/s320/IMG_9922.jpg" width="320" /></a></div><p>This post describes an AC power supply I designed and built for powering old military ARC-5 receivers that are either in unmodified condition (i.e. designed for 28VDC operation) or modified for either a 12.6 VAC or 6.3VAC filament voltage)..</p><p><u><b>Power Supply Requirements:</b></u></p><p></p><ul style="text-align: left;"><li>+HV (High Voltage): Must be between 200 and 250 VDC</li><li>LV (Low Voltage): This is a receiver's filament voltage. Three different filament voltages (25.2 VAC, 12.6 VAC, and 6.3 VAC) should be available for powering ARC-5 receivers.</li><li>Audio: RCA jack for attaching an 8-ohm speaker, and an internal Impedance Matching transformer to transform the speaker's 8-ohm impedance to 600 ohms as load for the receiver's audio output signal.</li><li>BFO ON/OFF switch (front-panel mount).</li><li>GAIN Control (front-panel mount).</li><li>All output voltages and control signals available both on a female Octal connector and on an eight-terminal terminal strip.</li><li>Power ON/OFF switch (controls both +HV and LV) and Indicator Lamp (ON when power is ON) both mounted on the front panel.</li></ul><p></p><p><u><b><br /></b></u></p><p><u><b>Schematic:</b></u></p><p>The schematic is shown below:</p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhR5nNUVUkTMj6Sys62T2sYk5e6XYnkwFna83-K4JmSBe0qiHgEPhIGBjU9mo3Fk5OpHQpQTZ_iIGGTySt_b0v5dyWnt9R7EJpjyFYpBTfkgRjreH8O_04Mta934ABZzo0cs4EZ4RVUjnFxsESEWBRwytQCAAi8KWQTMEJijYCboWImip8_yIv1kPpAI6s/s3109/231128_schematic.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="2349" data-original-width="3109" height="242" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhR5nNUVUkTMj6Sys62T2sYk5e6XYnkwFna83-K4JmSBe0qiHgEPhIGBjU9mo3Fk5OpHQpQTZ_iIGGTySt_b0v5dyWnt9R7EJpjyFYpBTfkgRjreH8O_04Mta934ABZzo0cs4EZ4RVUjnFxsESEWBRwytQCAAi8KWQTMEJijYCboWImip8_yIv1kPpAI6s/s320/231128_schematic.png" width="320" /></a></div><div><br /></div><b>Schematic Notes:</b><br /><p></p><ul style="text-align: left;"><li>The Transformer's secondary specification is 300 Vrms, center-tapped (Vprimary = 115 VAC, Secondary Load = 86 mA). Assuming full-wave rectification (i.e. center-tap tied to ground) and a capacitor-only filter (which acts as a peak-detector) the peak DC voltage (and ideally the output voltage, if the capacitor is large enough) should be the peak of one-half of the secondary's Vrms spec, which calculates to be 212 VDC.</li><li>If the secondary's peak voltage is 212 VDC, the diode Peak Reverse Voltage (PRV, or PIV) should be at least 2x 212 VDC, or 424 VDC, plus a good margin to account for AC Line voltage variation, etc. 1N4005 diodes (PRV = 600V) should suffice, but I went with higher-rated diodes (1N4006, PRV = 800V) from my junk box.</li><li>The higher the capacitance attached to the full-wave rectifier, the lower will be the AC ripple on the DC line. My BC-454-B receiver's frequency varies if +HV changes (the frequency change is *very* roughly 60 Hz per 10V change in +HV), so if +HV had significant ripple, there could be FM'ing of the audio signal.</li><li>The 68K bleeder resistor has a time-constant of 32 seconds in conjunction of 470 uF (and especially important, it was in my junkbox). And its power dissipation across 212 VDC is about 0.7 watts, so the resistor's 2 watt rating provides a nice margin.</li><li>The Audio transformer's primary will have an impedance of 1.2K ohms if terminated with an 8-ohm load. The receiver requires a load of 600 ohms, which is available at the primary's center-tap (i.e. its impedance will be one-half of the total primary impedance of 1200 ohms). Note that the selection of this transformer (a Xicon 42TM003-RC transformer) is discussed further in my <a href="https://k6jca.blogspot.com/2023/11/bringing-arc-5-receiver-back-to-life.html">BC-454-B post</a>.</li><li>I chose a lamp power-on indicator (with a #44 bulb), instead of an LED to match the era of the ARC-5 receivers.</li></ul><p></p><p><br /></p><p><u><b>Front View:</b></u></p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjOxgisByUKloVnAipivE0451qFkbNx7Od2YljsdbHTE-OgWCIEelnE6wLvNVpM8cIVzYNuaw3cL80xSkupV3ONcjlmVyCcc8MRbnmzHEXlWY-aFn9gRaRMHWnCwEhBH32uD7I_9p4tk8biIkx8VDQEviAcMHP2Q0hd3aiCSn9eE0tZwckH4jkQiRW_dL0/s640/IMG_9921.jpg" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="480" data-original-width="640" height="240" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjOxgisByUKloVnAipivE0451qFkbNx7Od2YljsdbHTE-OgWCIEelnE6wLvNVpM8cIVzYNuaw3cL80xSkupV3ONcjlmVyCcc8MRbnmzHEXlWY-aFn9gRaRMHWnCwEhBH32uD7I_9p4tk8biIkx8VDQEviAcMHP2Q0hd3aiCSn9eE0tZwckH4jkQiRW_dL0/s320/IMG_9921.jpg" width="320" /></a></div><p>On the left is the POWER ON/OFF switch and above it a panel-lamp (with a #44 bulb) to indicate when power is ON.</p><p>The middle switch is BFO ON/OFF.</p><p>And the right control is the GAIN potentiometer.</p><p><br /></p><p><u><b>Back View:</b></u></p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjsYQ_mbgGtUgOCCxt3CaLHePJ3h8wo8AXAgnA_nV-s11jvZRwHgaiqaamY4s7WOgRiL5QrdR3qFUkRFJ2N3KkPrj1hyphenhyphenapIsU6ox0PNeSItKb3IpTW3WrIpumrs5_ZYHSY9VQ2Gwi5wSDEJJzDg4Xx32Og0wQcv59ETuEHU3VKGQ7pExAMXvUJxAy8kipA/s640/IMG_9919.jpg" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="480" data-original-width="640" height="240" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjsYQ_mbgGtUgOCCxt3CaLHePJ3h8wo8AXAgnA_nV-s11jvZRwHgaiqaamY4s7WOgRiL5QrdR3qFUkRFJ2N3KkPrj1hyphenhyphenapIsU6ox0PNeSItKb3IpTW3WrIpumrs5_ZYHSY9VQ2Gwi5wSDEJJzDg4Xx32Og0wQcv59ETuEHU3VKGQ7pExAMXvUJxAy8kipA/s320/IMG_9919.jpg" width="320" /></a></div><p>Across the top of the back panel is a Terminal Strip consisting of 8 screw terminals.</p><p>Below it, from left to right, are:</p><p></p><ul style="text-align: left;"><li>Octal socket (with same signals assigned to the same pin numbers as those on the terminal strip)</li><li>Fuse Holder</li><li>RCA jack (connect to an 8-ohm speaker)</li><li>AC Power Connector.</li></ul><p></p><p><u><br /></u></p><p><u><b>Bottom View, Looking Towards Rear Panel.</b></u></p><p>(For reference)</p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEijLiOp4aNDdScFMEc0KBwxkrth5UX9lpsTLGMu2X6xUF-SXC9M0v8owJvuhxvIyEVM2ZHnXVOQJyhOQgsBofADSs1F1_UKkQP2pSFpRRek9RN7K1NA-gcSyY36PLZIns7ngnFkVgDO_-wPdU6t1n_dW7w2jgu20pa4taZr-tkfOKYnrOdv6qhngcNvZ5g/s640/IMG_9916.jpg" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="640" data-original-width="480" height="320" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEijLiOp4aNDdScFMEc0KBwxkrth5UX9lpsTLGMu2X6xUF-SXC9M0v8owJvuhxvIyEVM2ZHnXVOQJyhOQgsBofADSs1F1_UKkQP2pSFpRRek9RN7K1NA-gcSyY36PLZIns7ngnFkVgDO_-wPdU6t1n_dW7w2jgu20pa4taZr-tkfOKYnrOdv6qhngcNvZ5g/s320/IMG_9916.jpg" width="240" /></a></div><p>At the upper right is the audio impedance-matching transformer (mounted on a piece of perf-board).</p><p>Just below it is the 470 uF capacitor, the 68K bleeder resistor, and the two diodes.</p><p><br /></p><p><u><b>Bottom View, Looking Towards Front Panel:</b></u></p><p>(For reference)</p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhaQgATa8Lf2dVfAQyojtJq7OvqsSXK7qY3JVpqUGFXJzObRVRaXC0l2qryADZb3piBU3DlVkEpzTUoAi3DdQyDtmgKLw9M6RQnqn5TBs-2MZcvirdPfO966Wt3F15ix7aFnWIVZVjINK4UX2w_OnBOVNUyShgUDBS6cjTF9V2SJLI2ONjK2McUSuooL54/s640/IMG_9917.jpg" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="640" data-original-width="480" height="320" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhaQgATa8Lf2dVfAQyojtJq7OvqsSXK7qY3JVpqUGFXJzObRVRaXC0l2qryADZb3piBU3DlVkEpzTUoAi3DdQyDtmgKLw9M6RQnqn5TBs-2MZcvirdPfO966Wt3F15ix7aFnWIVZVjINK4UX2w_OnBOVNUyShgUDBS6cjTF9V2SJLI2ONjK2McUSuooL54/s320/IMG_9917.jpg" width="240" /></a></div><p><br /></p><p><u><b>Terminal Strip, Identification of Terminals:</b></u></p><p>Note the color coding under each screw-terminal's name -- I used the color-code to identify terminals for cabling purposes -- the associated spade terminals (that connect at the power supply side of the cable) are color coded to match their screw terminals. </p><p>For example, the first terminal on the left (Terminal One) is the +HV terminal, and its associated color is BROWN (this might not be obvious from the photograph, but there is a brown line drawn under the "+HV" text. The associated +HV wire (in the cable bundle) from the receiver that attaches to this screw terminal has its spade-terminal color coded with a matching BROWN.</p><p><br /></p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhHo6t2LPEfHV1heBuwF7xvTSkCYpuFA_gw_NJnQs69-w_ppKnGhdnE0Px6n013mnGNfbBS3-U1Jfk1UdTRtn2hXJs0S5JxekXPx83xqmxKurG5e8O_lAPYuOomNbnep6SjHFIhO768rTthJmFL9q6ILWmtAEXc3RAP22qgsMYL8PT-uXa_WmblKgraJHY/s640/IMG_9918.jpg" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="480" data-original-width="640" height="240" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhHo6t2LPEfHV1heBuwF7xvTSkCYpuFA_gw_NJnQs69-w_ppKnGhdnE0Px6n013mnGNfbBS3-U1Jfk1UdTRtn2hXJs0S5JxekXPx83xqmxKurG5e8O_lAPYuOomNbnep6SjHFIhO768rTthJmFL9q6ILWmtAEXc3RAP22qgsMYL8PT-uXa_WmblKgraJHY/s320/IMG_9918.jpg" width="320" /></a></div><p>The next terminal to the right is Terminal Two, and its color is RED. It is the GND terminal.</p><p>The Terminal Numbers increase as we move to the right and the color code increases to match the numbers. At the far right is the eighth terminal (BFO On/Off). Per the color code, this should be grey, but I did not have a grey Sharpie, which would have been the appropriate color per the color-code, so I instead colored it with alternating black and white lines.</p><p>Note that the numbers assigned to the terminal strip match the pin number of the octal socket. For example, Terminal One (+HV) of the Terminal Strip connects to the octal socket's pin 1, Terminal Two (GND) to the octal socket's pin 2, etc.</p><div><br /></div><p><u><b>Interconnect Cable (Power Supply Terminal Strip to Receiver J3):</b></u></p><p>Both ends of the cables that interconnect the Power Supply's Terminal Strip to the Receiver's J3 connector (on the back panel) are color coded. For example, in the image below, the spade terminal that attaches to the Terminal Strip has a red band, identifying that it should attach to the second terminal (from the left) of the Power Supply's terminal strip (i.e. the GND terminal).</p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEihsi2A6lhSSp0-CDyytJ6f4LkBcmuYDMDiHlvsIolkNRfbDcOvRdQiyiTBBMRCSdrALG4GBX0_D4rlWQqqNGSD-jZcrjIoxL0nv70CwLasYLoBOwKlKuacmQX9aFdvivRsDms2uBe3vJqBFU_iIUOGELdIhTnhw_pqkuZvf3bC63vVwnK2Z4FQ3d4jTOA/s1093/cable%20ends,%20cropped.jpg" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="947" data-original-width="1093" height="277" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEihsi2A6lhSSp0-CDyytJ6f4LkBcmuYDMDiHlvsIolkNRfbDcOvRdQiyiTBBMRCSdrALG4GBX0_D4rlWQqqNGSD-jZcrjIoxL0nv70CwLasYLoBOwKlKuacmQX9aFdvivRsDms2uBe3vJqBFU_iIUOGELdIhTnhw_pqkuZvf3bC63vVwnK2Z4FQ3d4jTOA/s320/cable%20ends,%20cropped.jpg" width="320" /></a></div><div><br /></div><div>On the other end of the wire is a mini Banana Plug (2.5 mm diameter), which will insert into Pin 1 of the Receiver's rear panel connector, J3. Note that it has a brown band, identifying that it should be inserted into pin 1 of J3 (i.e. the receiver's GND pin).</div><div><br /></div><div>I sourced these miniature Banana Plugs, from Amazon:</div><div><br /></div><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhDGGpjkDaEUXbVihUTTMXlvAFDNypXLTiAQMlbceFyeDUD1x9_i6Z8rzajgwrMBK31F0m91rWgYv4sb-evnszrXR6a9j2BCSVkLDdLpTUhSCzGjnDJF0jNWvGTJIyiE8pAb_D6WAXtlJAqZxtSBQQJTaVnL_BV7qAC4frcpse5UA5jwEFBYUaejdRLXV8/s1106/banana%20plugs.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="620" data-original-width="1106" height="179" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhDGGpjkDaEUXbVihUTTMXlvAFDNypXLTiAQMlbceFyeDUD1x9_i6Z8rzajgwrMBK31F0m91rWgYv4sb-evnszrXR6a9j2BCSVkLDdLpTUhSCzGjnDJF0jNWvGTJIyiE8pAb_D6WAXtlJAqZxtSBQQJTaVnL_BV7qAC4frcpse5UA5jwEFBYUaejdRLXV8/s320/banana%20plugs.png" width="320" /></a></div><div><br /></div>Note: not all 2.5mm Banana Plugs are compatible with the receiver's jack. I purchased a different set (that were only metal -- there was no plastic covers) and they did not fit. So you might need to do some experimentation.<div><br /></div><div><br /></div><div><p><u><b>Interconnect Cable (Power Supply Octal Socket to Receiver J3):</b></u></p><div>Below is an interconnect cable that uses the power supply's Octal socket in lieu of its terminal strip. This specific cable is for radios that require a filament voltage of 12.6 V.</div><div><br /></div><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjfrh3A1amP_dZtrxfTEvEYd56v3jI2Jum8uyw-wNE_kPn3Ylyjw2CEzFPZUu0ay1qIOYqm6FSS4tC_vTpd-8nsoCZwJKyOoJ4G1kHSI_sYTDwtieVY-7KLimqkz2kaC9mdrbPhBKIiRpuFldlJNiRcK81eCT0GVceDXKEa9p78KUezkso4yAZkLpyDGIQ/s640/12v6%20cable.jpg" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="474" data-original-width="640" height="237" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjfrh3A1amP_dZtrxfTEvEYd56v3jI2Jum8uyw-wNE_kPn3Ylyjw2CEzFPZUu0ay1qIOYqm6FSS4tC_vTpd-8nsoCZwJKyOoJ4G1kHSI_sYTDwtieVY-7KLimqkz2kaC9mdrbPhBKIiRpuFldlJNiRcK81eCT0GVceDXKEa9p78KUezkso4yAZkLpyDGIQ/s320/12v6%20cable.jpg" width="320" /></a></div><br /><div><br /></div><div><u><b>Banana Plugs Inserted into a Receiver's J3 connector:</b></u></div><p>The image below shows six miniature Banana Plugs inserted into a receiver's J3, which is a seven-pin jack. Note that only J3 pin 5 (+screen grid voltage) is not used, and therefore it does not have a Banana Plug inserted into it.</p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh4opbS2g7UXXarms00gJljTX1q05VevZhA1yDEBePv1vLJEntWcLuRHlrMZxGrrR-C7pU87I9zik74PDM9LRm6rAMjxCvST5vsUepWuNxWSquvJHSuynciCs7bBy09Sv92bEWEcM6Y9ZIAWdUJK1tdCAQ0mhsQKMvmyX2SuV4k5CZXhjOJrkPxRaTd1po/s640/IMG_9923.jpg" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="480" data-original-width="640" height="240" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh4opbS2g7UXXarms00gJljTX1q05VevZhA1yDEBePv1vLJEntWcLuRHlrMZxGrrR-C7pU87I9zik74PDM9LRm6rAMjxCvST5vsUepWuNxWSquvJHSuynciCs7bBy09Sv92bEWEcM6Y9ZIAWdUJK1tdCAQ0mhsQKMvmyX2SuV4k5CZXhjOJrkPxRaTd1po/s320/IMG_9923.jpg" width="320" /></a></div><div><br /></div><div><br /></div><div><u><b>Receiver J3 Pinout:</b></u></div><div><br /></div><div>The partial schematic, below, identifies the signals at each pin of an ARC-5 receiver's J3 rear panel connector:</div><div><br /></div><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh6xWijw3d84yyfq75R6p_YAucJWNOE4DdGGlkgijfFcymXnB17YPo1VKKNaou062GzB0QXQgH_NOCNUzE6yBbxUgXRcfoMI2MBtuCeb_R3idGN5S6BnZwwMh11IlPqSZZmAKGrqnIVImqL3DyW7UmkEGbpoYptMMRHVgtdQqLL3JQNkI3Dqz098W5tw6Q/s921/rear%20panel%20conn.jpg" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="674" data-original-width="921" height="234" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh6xWijw3d84yyfq75R6p_YAucJWNOE4DdGGlkgijfFcymXnB17YPo1VKKNaou062GzB0QXQgH_NOCNUzE6yBbxUgXRcfoMI2MBtuCeb_R3idGN5S6BnZwwMh11IlPqSZZmAKGrqnIVImqL3DyW7UmkEGbpoYptMMRHVgtdQqLL3JQNkI3Dqz098W5tw6Q/s320/rear%20panel%20conn.jpg" width="320" /></a></div><div><br /></div>Note that pin 5 is not used -- therefore only six miniature Banana Plugs are required.<div><br /><div><br /></div><div><u><b>Measurements:</b></u></div><div><br /></div><div>Given a measured AC line voltage of 118 VAC, I measured the following power supply output voltages:</div><div><br /></div><div>Connected to a BC-454-B, powered ON, with filaments wired for 28V:</div><div><ul style="text-align: left;"><li>+HV: measured (loaded) 212 VDC with 0.15 VAC ripple.</li><li>25.2 VAC: measured (loaded) 28.2 VAC.</li><li>12.6 VAC: measured (unloaded) 14.1 VAC (13.6 VAC when loaded with an R-27 Receiver).</li><li>6.3 VAC: measured (loaded with panel lamp, but no receiver): 7.2 VAC.</li></ul></div><div><br /></div><div>Connected to an R-27 receiver, powered ON, whose filaments have been rewired for 12V:</div><div><div><ul><li>+HV: measured (loaded) 212 VDC with 0.15 VAC ripple.</li><li>12.6 VAC: measured (loaded) 13.6 VAC.</li></ul><div>Notes on the measurements:</div></div></div><div><ul style="text-align: left;"><li>The loaded 25.2 VAC filament voltage (when attached to the BC-454-B) measures 28.2 VAC -- almost 10 percent higher than the recommended filament voltage of 25.2 VAC (for two equal filaments, wired in series). However, this RMS voltage is essentially the same as the DC voltage used to power the receiver when a Dynamotor is used (28 VDC). Note, too, that it is close to <a href="https://frank.pocnet.net/sheets/049/suppinfo/rca_hb3_gen.pdf">RCA's recommended voltage tolerance of +/- 10%</a>. If the AC line voltage were to increase above 118 volts, this 10% specification would be exceeded.</li></ul><ul style="text-align: left;"><li>The loaded 12.6 VAC filament voltage (when attached to the R-27 receiver) measures 13.6 VAC, or 8% above the recommended filament voltage. If line voltage were to increase, then RCA's 10% maximum variation spec could be exceeded. </li></ul></div><div><ul><li>If it is important to a user that the filament voltages, when loaded, measure to be 25.2, 12.6, and 6.3 VAC when the AC input line voltage is at its nominal value (typically 120 VAC), resistors could be added, within the power-supply, in series with each filament line to provide the necessary voltage drop. The resistors would probably only be a few ohms each, and probably spec'd at a watt, or less.</li></ul></div><div><br /></div><div><br /></div><div><u><b>Other Notes:</b></u></div><div><ul style="text-align: left;"><li>L14 is in series with the LV (filament) voltage -- does it have an effect on filament voltage? This part is an RF choke whose inductance is 112 uH. At 120 Hz its reactance is only 0.084 ohms -- in other words, negligible.</li></ul><ul style="text-align: left;"><li>I strongly recommend covering pin 2 (LV) and pin 3 (+HV) on the Receiver's J2 connector (the Dynamotor connector) to prevent accidentally shorting these pins to ground (or shocking oneself). Perhaps use heat-shrink tubing?</li></ul><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhtm2xoFggO9C-ErojaokxKgXu9yLLlw06YKvYIPJ1ktqexBopQ4P9M8Ev9UKEcSnewVJh7n99jN_OgWIzh4VbnFiXtrnm9UMKn1idIdvSZydW_XVcXJo98Zy9Nq0s88dek1iKKMrQZuN3GtCearG0qRlO584jbbvTZxCVJQavIhcCrdpNChYPkMa2NP1Q/s390/dynamotor%20conn%20pinout.jpg" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="320" data-original-width="390" height="263" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhtm2xoFggO9C-ErojaokxKgXu9yLLlw06YKvYIPJ1ktqexBopQ4P9M8Ev9UKEcSnewVJh7n99jN_OgWIzh4VbnFiXtrnm9UMKn1idIdvSZydW_XVcXJo98Zy9Nq0s88dek1iKKMrQZuN3GtCearG0qRlO584jbbvTZxCVJQavIhcCrdpNChYPkMa2NP1Q/s320/dynamotor%20conn%20pinout.jpg" width="320" /></a></div><br /><div><br /></div></div><div><br /></div><div><p><b><u>Standard Caveat:</u></b></p><p>As always, I might have made a mistake in my equations, assumptions, drawings, or interpretations. If you see anything you believe to be in error or if anything is confusing, please feel free to contact me or comment below.<br /><br />And so I should add -- this information is distributed in the hope that it will be useful, but WITHOUT ANY WARRANTY; without even the implied warranty of MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.</p></div></div></div>Jeffhttp://www.blogger.com/profile/04853314106806116765noreply@blogger.com0tag:blogger.com,1999:blog-2257458838727315711.post-3069050365677487502023-11-16T14:17:00.000-08:002023-11-23T15:10:48.188-08:00Bringing an ARC-5 Receiver Back to Life, Part 1 (BC-454-B)<p></p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg9rDuetPCSGWej0uW_JB63BzycD4yOZEFNFnK_KE-4XtdvWlFy7bBRPETImwDarNXhc3t-ZnF2E1y-ZL6LIsdx-KMjKbJ1O_raGCtvjx6eHmQ20Sz_mAdlxQolZoTzflWWqMy36lPiiEX_gsteGz8VxbfSV7xzHZd1M-UYmNhhLCrQLYhpkKul4ShSB2c/s533/IMG_9895.jpg" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="426" data-original-width="533" height="256" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg9rDuetPCSGWej0uW_JB63BzycD4yOZEFNFnK_KE-4XtdvWlFy7bBRPETImwDarNXhc3t-ZnF2E1y-ZL6LIsdx-KMjKbJ1O_raGCtvjx6eHmQ20Sz_mAdlxQolZoTzflWWqMy36lPiiEX_gsteGz8VxbfSV7xzHZd1M-UYmNhhLCrQLYhpkKul4ShSB2c/s320/IMG_9895.jpg" width="320" /></a></div><br />Back when I was a young ham, I was given an ARC-5 receiver, with power supply, that covered the 80-meter ham band.<p></p><p>That receiver provided hours of listening enjoyment, and it, along with its power supply, were small enough that I could take them on family vacations for listening during the evenings.</p><p>Somewhere along the way the receiver went one way and I another. But I've always had fond memories of it.</p><p>In more recent years I've picked up a few ARC-5 receivers at various swap-meets, and I've finally decided to see if I could get any of them back on the air (some are more heavily modified by previous owners than others, and those might not be worth the trouble).</p><p>My ARC-5 receiver stable currently consists of the following six radios:</p><p></p><ul style="text-align: left;"><li><u>CCT-46129</u> (0.19 - 0.55 MHz). One of the ARA series of receivers.</li><li><u>CCT-46104</u> (1.5 - 3 MHz). One of the ARA series of receivers.</li><li><u>BC-454-B</u> (3 - 6 MHz). One of the SCR-274-N series of receivers.</li><li><u>R-26</u> (3-6 MHz). One of the AN/ARC-5 series of receivers.</li><li><u>CBY-46106</u> (6 - 9.1 MHz). One of the ARA series of receivers.</li><li><u>R-27</u> (6-9.1 MHz). One of the AN/ARC-5 series of receivers. (Somehow this one wound up with a cover from an R-23 receiver).</li></ul><p></p><p>In this blog post I will:</p><p></p><ul style="text-align: left;"><li>Introduce the three receiver series that comprise the ARC-5 receiver family.</li><li>Discuss the differences in design between these three series.</li><li>Describe how to recognize which series a receiver belongs to.</li><li>Describe how to power receiver.</li><li>Describe the receiver's connectors and their signals.</li><li>Describe controlling the receiver.</li><li>Describe how to interface with an 8-ohm speaker.</li><li>List the steps to take when first checking out a receiver.</li><li>And finally, describe of my efforts in bringing my BC-454-B back to life.</li></ul><p></p><p><b><u><br /></u></b></p><p><b><u>Introduction, the ARC-5 Receiver Series:</u></b></p><p>Per <a href="https://en.wikipedia.org/wiki/AN/ARC-5">Wikipedia</a>, the term ARC-5 is used to describe three similar (yet different) series of receivers deployed during World War Two. Of these three series, only one series has the "official" ARC-5 nomenclature (the AN/ARC-5series), which was actually the last of the three series to be introduced (in 1943). </p><p>The other two series, unofficially, are also referred to as "ARC-5" receivers. The first series, developed for the Navy prior to World War Two, was the "ARA" series. This series was followed in 1941 by the "SCR-274-N" series, which was initially adopted for the US Army Air Corps. </p><p>The table below (from the <a href="https://en.wikipedia.org/wiki/AN/ARC-5">AN/ARC-5</a> Wikipedia entry) lists the receivers of each of the three series that cover the medium through high-frequency range of 0.19 to 9.1 MHz. This frequency span is typically divided between five different receivers (note that the SCR-274-N series does not have a receiver covering 1.5-3.0 MHz).</p><table class="wikitable" style="background-color: #f8f9fa; border-collapse: collapse; border: 1px solid rgb(162, 169, 177); color: #202122; font-family: sans-serif; font-size: 14px; margin: 1em 0px;"><tbody><tr><th style="background-color: #eaecf0; border: 1px solid rgb(162, 169, 177); padding: 0.2em 0.4em; text-align: center;"></th><th style="background-color: #eaecf0; border: 1px solid rgb(162, 169, 177); padding: 0.2em 0.4em; text-align: center;">MHz</th><th style="background-color: #eaecf0; border: 1px solid rgb(162, 169, 177); padding: 0.2em 0.4em; text-align: center;">ARA-ATA</th><th style="background-color: #eaecf0; border: 1px solid rgb(162, 169, 177); padding: 0.2em 0.4em; text-align: center;">SCR-274-N</th><th style="background-color: #eaecf0; border: 1px solid rgb(162, 169, 177); padding: 0.2em 0.4em; text-align: center;">AN/ARC-5</th></tr><tr><td style="border: 1px solid rgb(162, 169, 177); padding: 0.2em 0.4em;">Navigation Receiver (Beacon Band)</td><td style="border: 1px solid rgb(162, 169, 177); padding: 0.2em 0.4em;">0.19 - 0.55</td><td style="border: 1px solid rgb(162, 169, 177); padding: 0.2em 0.4em;">CBY-46129</td><td style="border: 1px solid rgb(162, 169, 177); padding: 0.2em 0.4em;">BC-453-(*)</td><td style="border: 1px solid rgb(162, 169, 177); padding: 0.2em 0.4em;">R-23(#)/ARC-5</td></tr><tr><td style="border: 1px solid rgb(162, 169, 177); padding: 0.2em 0.4em;">Navigation Receiver (Broadcast Band)</td><td style="border: 1px solid rgb(162, 169, 177); padding: 0.2em 0.4em;">0.52 - 1.5</td><td style="border: 1px solid rgb(162, 169, 177); padding: 0.2em 0.4em;">CBY-46145</td><td style="border: 1px solid rgb(162, 169, 177); padding: 0.2em 0.4em;">BC-946-(*)</td><td style="border: 1px solid rgb(162, 169, 177); padding: 0.2em 0.4em;">R-24/ARC-5</td></tr><tr><td style="border: 1px solid rgb(162, 169, 177); padding: 0.2em 0.4em;">Communication Receiver</td><td style="border: 1px solid rgb(162, 169, 177); padding: 0.2em 0.4em;">1.5 - 3.0</td><td style="border: 1px solid rgb(162, 169, 177); padding: 0.2em 0.4em;">CBY-46104</td><td style="border: 1px solid rgb(162, 169, 177); padding: 0.2em 0.4em;"></td><td style="border: 1px solid rgb(162, 169, 177); padding: 0.2em 0.4em;">R-25/ARC-5</td></tr><tr><td style="border: 1px solid rgb(162, 169, 177); padding: 0.2em 0.4em;">Communication Receiver</td><td style="border: 1px solid rgb(162, 169, 177); padding: 0.2em 0.4em;">3.0 - 6.0</td><td style="border: 1px solid rgb(162, 169, 177); padding: 0.2em 0.4em;">CBY-46105</td><td style="border: 1px solid rgb(162, 169, 177); padding: 0.2em 0.4em;">BC-454-(*)</td><td style="border: 1px solid rgb(162, 169, 177); padding: 0.2em 0.4em;">R-26/ARC-5</td></tr><tr><td style="border: 1px solid rgb(162, 169, 177); padding: 0.2em 0.4em;">Communication Receiver</td><td style="border: 1px solid rgb(162, 169, 177); padding: 0.2em 0.4em;">6.0 - 9.1</td><td style="border: 1px solid rgb(162, 169, 177); padding: 0.2em 0.4em;">CBY-46106</td><td style="border: 1px solid rgb(162, 169, 177); padding: 0.2em 0.4em;">BC-455-(*)</td><td style="border: 1px solid rgb(162, 169, 177); padding: 0.2em 0.4em;">R-27/ARC-5</td></tr></tbody></table><p><span face="sans-serif" style="background-color: white; color: #202122; font-size: 14px;">(*) A (basic model) or B (1st revision). (#) No letter (basic model) or A (revision). </span></p><p><b><u><br /></u></b></p><p><b><u>Differences between the ARC-5 Receiver Series:</u></b></p><p>(I will summarize the information below at the end of this section)</p><p><u>1. Front Panel Plug wiring:</u></p><p>Both the ARA series and the SCR-274-N series bring the headset audio output signal to pin 4 of the front-panel connector (audio also appears on the rear panel connector). The AN/ARC-5 series <u><i>has no audio connection to the front panel</i></u>! Audio for this series only appears on the rear panel connector.</p><p><u>2. Automatic Volume Control (AVC):</u></p><p>Of the three series, only the AN/ARC-5 series (i.e. R-23 through R-27 receivers) has <i>Automatic Volume Control </i>(AVC) circuitry. You should be able to quickly identify if a radio has this functionality (and thus is part of the AN/ARC-5 series) by looking for the following differences:</p><p></p><ul style="text-align: left;"><li>12SF7 replacing the 12SK7 tube in the Second IF Amplifier stage.</li></ul><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh59NaL3EFFW-T0ZI2u56I81brk9UK8zYEh3CUgFXB5x2nyqv2QOgf0JvrfzgxQIlyjZvRKqe0VOkvq1IqKgqhvSFw5gOX3i1E7zUyR1Cj76IRhc9hsMVJOYJVfg97G_U9HaCmLsJkF1KPuZc-iCwPVKlX0wLPMl6KkQJAoP2w2W1UQeordCX2d1peSJKM/s1267/12sf7.jpg" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="979" data-original-width="1267" height="247" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh59NaL3EFFW-T0ZI2u56I81brk9UK8zYEh3CUgFXB5x2nyqv2QOgf0JvrfzgxQIlyjZvRKqe0VOkvq1IqKgqhvSFw5gOX3i1E7zUyR1Cj76IRhc9hsMVJOYJVfg97G_U9HaCmLsJkF1KPuZc-iCwPVKlX0wLPMl6KkQJAoP2w2W1UQeordCX2d1peSJKM/s320/12sf7.jpg" width="320" /></a></div><div><br /></div><ul style="text-align: left;"><li>C26 (a capacitor connected between pin 2 of the 12SR7 (middle tube, last row) and the middle terminal of Z-4, i.e. L12) does <i><b>not</b></i> have a resistor attached in parallel across it (R14).</li></ul><p></p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj5yG0HwvW4UJL4LXOm2JZE5jGF_hWvss6E6TzAoGl_0sJRomirfA2l3O3x5V1nb-zzyBmKAN9zECh-x_SPhd5WQd4UARa7Aot8quQtfzO6HXnMfe8mf-4FM2qPSKQJo9bOksm3JkIpQWfzrRIOb_RIU_hLC6oO517oJeQTqZY-J-jeAINVlsTco0XrXvs/s803/C26%20no%20R.jpg" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="590" data-original-width="803" height="235" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj5yG0HwvW4UJL4LXOm2JZE5jGF_hWvss6E6TzAoGl_0sJRomirfA2l3O3x5V1nb-zzyBmKAN9zECh-x_SPhd5WQd4UARa7Aot8quQtfzO6HXnMfe8mf-4FM2qPSKQJo9bOksm3JkIpQWfzrRIOb_RIU_hLC6oO517oJeQTqZY-J-jeAINVlsTco0XrXvs/s320/C26%20no%20R.jpg" width="320" /></a></div><blockquote style="border: none; margin: 0px 0px 0px 40px; padding: 0px;"><p style="text-align: left;">For comparison, here is an example of the wiring for non AN/ARC-5 receivers:</p></blockquote><p></p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEguxVwPC51aeGlo5GXDEJP0NSbJQbUviKZFJeTRY_djLJI629I8uWrYgwQ_xiCVzFTthkZesbyiY6biBYLuyv-hl6h43s-7kUH4T8p3UNJ0ro8dGn4YfoqrHWkRq-LiL9f6FkBYeLZ9_NreEjbT1iwBREUnQdJ6BuJqa7j7NRoY_xkTVP_uAuYs90uULbU/s812/c26%20paralleled%20with%20R14.jpg" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="669" data-original-width="812" height="264" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEguxVwPC51aeGlo5GXDEJP0NSbJQbUviKZFJeTRY_djLJI629I8uWrYgwQ_xiCVzFTthkZesbyiY6biBYLuyv-hl6h43s-7kUH4T8p3UNJ0ro8dGn4YfoqrHWkRq-LiL9f6FkBYeLZ9_NreEjbT1iwBREUnQdJ6BuJqa7j7NRoY_xkTVP_uAuYs90uULbU/s320/c26%20paralleled%20with%20R14.jpg" width="320" /></a></div><br /><p><u>3. Filament Voltage:</u></p><p>Per my reading of the ARA, SCR-274-N, and AN/ARC-5 manuals, only one receiver was designed for 14 V operation, and that was the R-148/ARC-5X receiver, covering 0.19 to 0.55 MHz.</p><p>In this receiver, 12V filament tubes were used, and all filaments were <u>wired in parallel</u>.</p><p>All other receivers used 28 volts for their filament voltage. Given that the tubes are 12V filament tubes, the filaments of <i>pairs</i> of tubes were first connected in series (the RF Amp & Mixer pair, the 1st IF and 2nd IF pair, and the Det./Osc. & Audio Amp pair) and then each pair was fed with 28V.</p><p><b>Never assume that a receiver has been wired for 28V filament voltage! I have made this mistake! Always verify filament circuitry before applying power to a newly acquired receiver. Previous owners might have modified it.</b></p><p>For example, my R-27 AN/ARC-5 receiver should be wired for 28 V filament voltage, but all filaments are wired in parallel, while my CBY-46106 ARA receiver is also wired with all filaments in parallel and the original 12V tubes were replaced with 6V tubes!</p><p><u>4. Audio Output Transformer and Impedance:</u></p><p>Of the three receiver series, only the SCR-274-N series was designed for connection to high-impedance headphones (e.g. 8000 ohms). Of this series, the "-A" version was designed for use <i>only</i> with high impedance headphones, while the "-B" version could be used with either high-impedance or low-impedance (e.g. 600 ohm) headphones, depending upon how the Audio Output Transformer T1 was wired.</p><p>The other two receiver series (ARA and AN/ARC-5) were designed for use only with low impedance (e.g. 600 ohms) headphones.</p><p><u>Audio Transformer (T1) Summary:</u></p><blockquote style="border: none; margin: 0px 0px 0px 40px; padding: 0px; text-align: left;"><p><u>ARA receiver:</u> The Audio Transformer should have a "5613" part number. Its primary to secondary turns ratio is 8:1, and spec'd for low impedance headsets (e.g. 600 ohms). (See Table 19 of the <a href="https://www.tuberadio.com/robinson/Manuals/ATA_ARA_part3.pdf">Instruction Manual</a>).</p><p><u>SCR-274-N receiver:</u> Audio Transformers can have two different part numbers:</p></blockquote><p style="text-align: left;"></p><ul style="text-align: left;"><ul><li>"6308": Used only with high impedance headsets (e.g. 8000 ohms). This transformer was used in the "-A" revision of the receiver.</li><li>"ES-691027": Provides connection to either a high impedance headset (e.g. 8000 ohms, via terminal 3) or a low impedance headset (e.g. 600 ohms, via terminal 6). This transformer was used in the "-B" revision of the receiver. Refer to Table 20 of the <a href="https://hangarthirteen.org/wp-content/uploads/2020/08/16-40SCR274-5-Maintenance-Instructions-for-SCR-274-N.pdf">Maintenance Instructions</a>. (Note, the turns listed for the secondary's tap were most likely counted from the "hot" end of the secondary (terminal 3), rather than its ground end, meaning that the tapped secondary would actually be 475 turns rather than 1325 turns).</li></ul></ul><p></p><blockquote style="border: none; margin: 0px 0px 0px 40px; padding: 0px;"><p style="text-align: left;">The illustration below shows both the schematic and the wiring diagram for ES-691027 transformer in the "-B" revision of the SCR-274-N receivers: </p></blockquote><p></p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhI7zxjK84T8FYQz1IgDBQeVEL0U_PFNGAc5G2SClCwrlqBHDlqfLRaiR2DXuDGei63Jr_bJ0jXVbIsEthXN2C49_7XQD3AjB4UQ3G5zIkdpD1cmDRD8snW1DOYwi3mOPu_sACeo0I9c2L6mD06lQADO_dw2CwfWkLYXH0lxCq9elcLVhVwwcKItfCS6q4/s1190/BC%20B-version%20audio%20transformer.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="763" data-original-width="1190" height="205" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhI7zxjK84T8FYQz1IgDBQeVEL0U_PFNGAc5G2SClCwrlqBHDlqfLRaiR2DXuDGei63Jr_bJ0jXVbIsEthXN2C49_7XQD3AjB4UQ3G5zIkdpD1cmDRD8snW1DOYwi3mOPu_sACeo0I9c2L6mD06lQADO_dw2CwfWkLYXH0lxCq9elcLVhVwwcKItfCS6q4/s320/BC%20B-version%20audio%20transformer.png" width="320" /></a></div><p></p><blockquote style="border: none; margin: 0px 0px 0px 40px; padding: 0px; text-align: left;"><p><u>AN/ARC-5 receiver:</u> Audio Transformers support low impedance headsets (e.g. 600 ohms) and, per Table 20 of the <a href="https://hangarthirteen.org/wp-content/uploads/2020/08/16-40SCR274-5-Maintenance-Instructions-for-SCR-274-N.pdf">Maintenance Instructions</a>, can have one of two part numbers:</p></blockquote><p style="text-align: left;"></p><ul style="text-align: left;"><ul><li>"5631"</li><li>"640268"</li></ul></ul><p></p><blockquote style="border: none; margin: 0px 0px 0px 40px; padding: 0px;"><p style="text-align: left;">An example of the different T1 part numbers in different receivers:</p></blockquote><p></p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi-UoTZgnZ00mTeYoxgWg4ochZOB75HkdhYs9Fyt4zzxN7Vhkh-ChArYG6hWaO119EE5_kyyXKqc14kgnAZNvBYKc-vdjIceovTck2C8X7-XtCw3uscFjubjd7dD_yDcWDRaScUCfuztFJ8Jd1KWTQiJdyKzfEo73wS1e6CwrAJmqaxlTZlZpstrWo5sF4/s1199/Audio%20Transformers.jpg" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="1017" data-original-width="1199" height="271" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi-UoTZgnZ00mTeYoxgWg4ochZOB75HkdhYs9Fyt4zzxN7Vhkh-ChArYG6hWaO119EE5_kyyXKqc14kgnAZNvBYKc-vdjIceovTck2C8X7-XtCw3uscFjubjd7dD_yDcWDRaScUCfuztFJ8Jd1KWTQiJdyKzfEo73wS1e6CwrAJmqaxlTZlZpstrWo5sF4/s320/Audio%20Transformers.jpg" width="320" /></a></div><p>(By the way, for an interesting discussion of military headsets circa the 1940's, go <a href="https://aafradio.org/flightdeck/Peripherals-headsets.html">here</a>).</p><div><br /></div><div><u><b>Summary of the Differences between Command-set Receiver series:</b></u></div><div><br /></div><blockquote style="border: none; margin: 0px 0px 0px 40px; padding: 0px; text-align: left;"><div><u><b>ARA Series:</b></u></div></blockquote><ul style="text-align: left;"><ul><li><u>Audio Line at Front Connector (pin 4):</u> Yes</li><li><u>AVC Circuitry:</u> No </li><li><span style="text-decoration-line: underline;"><u>Dynamotor (Filament) Voltage:</u></span> 28 VDC</li><li><u style="text-decoration-line: underline;">Headset Impedance</u><u>:</u> Low Impedance (e.g. 600 ohms<u>)</u></li></ul></ul><p> </p><blockquote style="border: none; margin: 0px 0px 0px 40px; padding: 0px; text-align: left;"><div><b>SCR-274-N Series, "A" Revision:</b></div></blockquote><div style="text-align: left;"><ul><ul><li><span style="text-decoration-line: underline;"><u>Audio Line at Front Connector (pin 4):</u> </span> Yes</li><li><u>AVC Circuitry:</u> No </li><li><u style="text-decoration-line: underline;">Dynamotor (Filament )Voltage:</u> 28 VDC</li><li><span style="text-decoration-line: underline;"><u>Headset Impedance</u>:</span> <span style="color: #cc0000;"><b>High Impedance (e.g. 8000 ohms)</b></span></li></ul></ul></div><p> </p><blockquote style="border: none; margin: 0px 0px 0px 40px; padding: 0px; text-align: left;"><div><div><b>SCR-274-N Series, "B" Revision:</b></div></div></blockquote><div style="text-align: left;"><ul><ul><li><u>Audio Line at Front Connector (pin 4):</u> Yes</li><li><u>AVC Circuitry:</u> No</li><li><u style="text-decoration-line: underline;">Dynamotor (Filament) Voltage:</u> 28 VDC</li><li><span style="text-decoration-line: underline;"><u>Headset Impedance</u>:</span> <b><span style="color: #cc0000;">Can be wired for High or Low Impedance</span> </b> </li></ul></ul></div><p> </p><blockquote style="border: none; margin: 0px 0px 0px 40px; padding: 0px; text-align: left;"><div><div><b>AN/ARC-5 Series:</b></div></div></blockquote><div><ul style="text-align: left;"><ul><li><u>Audio Line at Front Connector (pin 4):</u> <b><span style="color: #cc0000;">No</span></b></li><li><u>AVC Circuitry:</u> <b><span style="color: #cc0000;">Yes</span></b></li><li><span style="text-decoration-line: underline;"><u>Dynamotor (Filament) Voltage:</u></span> 28 VDC <span style="color: #cc0000;">(Note: R-148 wired for 14 VDC)</span></li><li><u>Headset Impedance</u><u>:</u> Low Impedance (e.g. 600 ohms<u>)</u></li></ul></ul></div><p><b><u><br /></u></b></p><p><b><u>Identifying a Receiver:</u></b></p><p>The receiver should have an attached metal tag stating its series, receiver identification, and sometimes its operating voltage (i.e. dynamotor voltage). The tag for the ARA and AN/ARC-5 series should be on the receiver's top cover, while the tag for the SCR-274-N series should be on the right side of the chassis, near the front panel.</p><div>Now, suppose you have purchased a receiver at a swap-meet. How do you know which receiver you actually have? Covers might have been swapped over the years, etc. Do not depend upon a cover tag being the correct one!</div><p>Here are the steps I would take:</p><blockquote style="border: none; margin: 0px 0px 0px 40px; padding: 0px; text-align: left;"><p>1. Ignore the receiver tag on the top of the radio (if it exists); after all, someone might have put the wrong cover on the radio. Instead, <b>look at the front dial and identify its frequency range.</b> This narrows the selection down to one of four receivers (ARA, SCR-274-N ("-A" revision), SCR-274-N ("-B" revision), and AN/ARC-5).</p><p>2. Is the chassis bare aluminum or painted with black-wrinkle paint? </p><p>2.1. If bare aluminum, the receiver is either an SCR-274-N ("-A" revision) or an SCR-274-N ("-B" revision) receiver.</p><p>2.1.1. If the SCR-274-N receiver has no side-mounted ID tag, we will need to determine the receiver's revision by examining the audio transformer under the chassis. Remove the bottom cover. If the Audio Output transformer (T1) has an "ES-691027" part number, it is the "-B" version of an SCR-274-N receiver (e.g. BC-454-B). Otherwise, if the P/N is "6308", the receiver is an "-A" version.</p><p>If the radio is the "-B" version, now would be a good time to check if the audio transformer, T1, is wired for 8000 ohm or 600 ohm impedance.</p></blockquote><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhIfcDAuxG771012ilsHKjVv_aqnRj25BwtmuirZ7yPZ9c7O7qEhN1LZ08hyLtFo0-zgJ2j0Kpx55fHV4NGrUzQ1WuAUXJyU8pUeLR-ROK5qYmRLAzbj7SmOAHACe0mgW9ts9RGSsthmxbe-CIJQ37_hdRvTfsrTpos2orC6CxULPwaBZQmbFAegn_JHuE/s1190/BC%20B-version%20audio%20transformer.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="763" data-original-width="1190" height="205" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhIfcDAuxG771012ilsHKjVv_aqnRj25BwtmuirZ7yPZ9c7O7qEhN1LZ08hyLtFo0-zgJ2j0Kpx55fHV4NGrUzQ1WuAUXJyU8pUeLR-ROK5qYmRLAzbj7SmOAHACe0mgW9ts9RGSsthmxbe-CIJQ37_hdRvTfsrTpos2orC6CxULPwaBZQmbFAegn_JHuE/s320/BC%20B-version%20audio%20transformer.png" width="320" /></a></div><blockquote style="border: none; margin: 0px 0px 0px 40px; padding: 0px; text-align: left;"><p>2.2. Otherwise, if the chassis is not aluminum but instead painted with black-wrinkle paint, the receiver is either an ARA receiver or an AN/ARC-5 receiver, or it could even be an SCR-274-N (although I personally have not come across a "BC" prefixed receiver in black wrinkle, I have seen some pictures).</p></blockquote><blockquote style="border: none; margin: 0px 0px 0px 40px; padding: 0px;"><p style="text-align: left;">2.2.1. Is the black-wrinkle receiver of the SCR-274-N series? Check the side of the chassis for an identification tag. If there is one (and it starts with "BC"), then the receiver should be of the SCR-274-N series. You can double-check this conclusion by removing the bottom cover and examining the Part Number of the Audio Transformer, T1. If this P/N is either 6309 or ES-691027, then the receiver is of the SCR-274-N series.</p></blockquote><blockquote style="border: none; margin: 0px 0px 0px 40px; padding: 0px;"><p style="text-align: left;">2.2.2. The receiver will be in the AN/ARC-5 series if the second IF tube is a 12SF7 (either pull the tube out or check its filament wiring -- a 12SF7's filament pins are pins 7 and 8, while a 12SK7's filament pins are 2 and 7). </p></blockquote><blockquote style="border: none; margin: 0px 0px 0px 40px; padding: 0px;"><p>Alternatively, you can remove the bottom cover and check C26. If there is no resistor across C26, the receiver is an AN/ARC-5 receiver (with AVC). </p></blockquote><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi4NXz7_IJtjcBKA4yGOlbsCSifABGeXywm8V9Z5jDeEwlo19PODVjBH98aybA288OyzZzKL77vgdxcCeSsxml4YgloGQuTlAm1oz_H5mhxjRVIAjehGrgYreITZ53KglolKWF-sVX8wRqqIPANxlhujckfcP7fX97eB7hhLqqcu50uBL0OtildeiVPva4/s803/C26%20no%20R.jpg" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="590" data-original-width="803" height="235" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi4NXz7_IJtjcBKA4yGOlbsCSifABGeXywm8V9Z5jDeEwlo19PODVjBH98aybA288OyzZzKL77vgdxcCeSsxml4YgloGQuTlAm1oz_H5mhxjRVIAjehGrgYreITZ53KglolKWF-sVX8wRqqIPANxlhujckfcP7fX97eB7hhLqqcu50uBL0OtildeiVPva4/s320/C26%20no%20R.jpg" width="320" /></a></div><div><br /></div><blockquote style="border: none; margin: 0px 0px 0px 40px; padding: 0px;"><p style="text-align: left;">2.2.3. If the black-wrinkle receiver isn't an SCR-274-N receiver nor an ARC-5 receiver, then it is an ARA receiver. </p></blockquote><blockquote style="border: none; margin: 0px 0px 0px 40px; padding: 0px;"><p style="text-align: left;"><u>A caveat regarding these checks:</u> modifications made by later owners might have substituted parts from other series, changed circuitry, or...? There are any number of reason why a receiver that appears to be of one series was actually born of another.</p></blockquote><u>And finally, an important note regarding receiver identification!</u><br /><p style="text-align: left;">The three IF transformer housings should each have a color dot in one corner on their topside. These three dots should all be the same color, and they should match the front-panel's tuning dial per the following table:</p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi1Ip7UiJQl46NQrR7VZknAFNeFPRyr3EL-pYhOxj5gnL5Dy8hy-KHiPnNKeQP1duHcv8PRCLccXPcbhC4F94P0GRDqqp_iy_qHguzZ5q92uQ5ONbSkDyd35Oqg9Mn6s-EM3O6O8R_kpplkHo0sI-EG6n2yF0lK0u35b65OcCmSJg7vCjcFheuiqh-N8I4/s1609/Transformer%20dot%20color.jpg" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="1289" data-original-width="1609" height="256" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi1Ip7UiJQl46NQrR7VZknAFNeFPRyr3EL-pYhOxj5gnL5Dy8hy-KHiPnNKeQP1duHcv8PRCLccXPcbhC4F94P0GRDqqp_iy_qHguzZ5q92uQ5ONbSkDyd35Oqg9Mn6s-EM3O6O8R_kpplkHo0sI-EG6n2yF0lK0u35b65OcCmSJg7vCjcFheuiqh-N8I4/s320/Transformer%20dot%20color.jpg" width="320" /></a></div><div><br /></div>At least one published modification for an ARC-5 receiver changed its frequency range by swapping the three IF assemblies from one receiver for those of another. I would recommend this quick check to verify that the IF cans are indeed the correct ones for your receiver!<p><br /></p><div><b><u>Powering the Receiver:</u></b></div><div><br /></div><div><div>The diagram below shows the various ways by which power can be applied to an ARC-5 receiver.</div><div><br /></div><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh6uOFBIpCYnqRA8lX_ESzMqrHz1kUTdigiWit9nq4KLW_yI85KoikYPgWWH42rqcsqfe6igISlkVZpGNt0py3wWdtohpEWIRatxzrWBTEM1LrWC_GB1bOpww69BXzgYXialxfQJYgt3ck2xXrTokKkmCX3rD5erAtB4EkzZLudimAFyhyphenhyphen4lG0b1P4UhuY/s1317/Powering%20the%20receiver.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="851" data-original-width="1317" height="207" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh6uOFBIpCYnqRA8lX_ESzMqrHz1kUTdigiWit9nq4KLW_yI85KoikYPgWWH42rqcsqfe6igISlkVZpGNt0py3wWdtohpEWIRatxzrWBTEM1LrWC_GB1bOpww69BXzgYXialxfQJYgt3ck2xXrTokKkmCX3rD5erAtB4EkzZLudimAFyhyphenhyphen4lG0b1P4UhuY/s320/Powering%20the%20receiver.png" width="320" /></a></div><br /><div>Per the diagram's notes:</div><div><ul><li>If a dynamotor is used, its Low Voltage (LV) power (e.g. +28V) must be applied via the Rear Panel connector (J3).</li></ul><ul><li>If an external power supply is used (supplying both LV and HV), this power can be applied either via the Rear Panel connector (J3) or via the Dynamotor connector (J2). </li></ul><ul><li>If the Rear Panel connector is used for power <i>without the dynamotor installed</i>, <span style="color: #cc0000;">then the HV and LV pins on J2, the Dynamotor connector, should be covered (heat-shrink tubing?) to prevent accidentally shorts or<b> <u>shocks</u>!</b></span></li></ul><ul><li>Also, if the Rear Panel connector is used for power (with or without the dynamotor), pins 6 and 7 on the Front Panel connector (J1) <i>must be connected together</i> to route +LV to the tube filaments.</li></ul></div><div><br /></div><div>The image below shows an FT-230-A adapter (for the SCR274-N receiver series) which will short pins 6 & 7 of the Front Panel connector. It can be used if no signals need to be accessed via the front panel, that is, if the radio's controls are connected to the Rear Panel connector, instead.</div><div><br /></div><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj69ijtsuqf-t6RB4eZ-44qIoyJzgtI-fvITKTgPVbsXgBIFLRnfMLkKjSiR7V0t9OWl7OVcOokD6mM3OSlyvB_P2tGoynhZS8i0Yp7_QtTiX5fpQJITTuFPmLNeJVQ4W7QeC5dvGxBjEL4eUc2vceA551FQJWeAqLo4HhxfgucxCVA6LqzfP8xD2bRxRg/s804/shorting%20box.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="421" data-original-width="804" height="168" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj69ijtsuqf-t6RB4eZ-44qIoyJzgtI-fvITKTgPVbsXgBIFLRnfMLkKjSiR7V0t9OWl7OVcOokD6mM3OSlyvB_P2tGoynhZS8i0Yp7_QtTiX5fpQJITTuFPmLNeJVQ4W7QeC5dvGxBjEL4eUc2vceA551FQJWeAqLo4HhxfgucxCVA6LqzfP8xD2bRxRg/s320/shorting%20box.png" width="320" /></a></div><br /><div>Note: The FT-230-A adapter is interchangeable with the MX-2/ARC-5 adapter (AN/ARC-5 receiver series) and the CBY-49107 adapter (ARA series).</div></div><div><br /></div><div><b><u><br /></u></b></div><div><b><u>Dynamotor Connector:</u></b></div><div style="font-weight: bold; text-decoration-line: underline;"><b><u><br /></u></b></div><div>A dynamotor connects to the receiver via a 3-pin male connector on the receiver's rear "apron." Its pin assignments are:</div><div><br /></div><div>Pin 1: Ground</div><div>Pin 2: +LV Input (Low Voltage, e.g. 28 VDC)</div><div>Pin 3: +HV Output (High Voltage, e.g. 250 VDC)</div><div><br /></div><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEilIQT5jJhHsDWaIG23CorekPT9q6bKsGziugE_y_Fh8YsAnNGjQTIjMTfKuSt0MqYiY1U3H0-KLiWCCJpsG1hM6v1lQPeqUuEv-EA3_LysMkIZ9F38mON3WTsoGf9tNYK3UBri_8ooDmhHPkQ31k6yr82KoEZvfnNQlxAQSYl_Z5EaZkM5f9rNlB0EMac/s390/dynamotor%20conn%20pinout.jpg" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="320" data-original-width="390" height="263" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEilIQT5jJhHsDWaIG23CorekPT9q6bKsGziugE_y_Fh8YsAnNGjQTIjMTfKuSt0MqYiY1U3H0-KLiWCCJpsG1hM6v1lQPeqUuEv-EA3_LysMkIZ9F38mON3WTsoGf9tNYK3UBri_8ooDmhHPkQ31k6yr82KoEZvfnNQlxAQSYl_Z5EaZkM5f9rNlB0EMac/s320/dynamotor%20conn%20pinout.jpg" width="320" /></a></div><br /><div><br /></div><div><b style="text-decoration-line: underline;">Front Panel Connector:</b><div><br /></div><div>The Front Panel's connector (8 pins, male) has the following signal assignments:</div><div><p></p><ul><li>Pin 1: Gain Control Line</li><li>Pin 2: Ground </li><li>Pin 3: (not used)</li><li>Pin 4: Headset Audio Output <span style="color: #cc0000;">(Note: the AN/ARC-5 receivers do <b><i>not</i></b> have this signal)</span>.</li><li>Pin 5: CW Osc. (i.e. BFO) Shut-Off Line</li><li>Pin 6: +L.V. to receiver circuitry</li><li>Pin 7: +L.V. from rear-panel connector</li><li>Pin 8: (not used)</li></ul><p></p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgHsyypzUORPVoKlXq4UyNZF9NRJC0RphlT3za6tVs0mvazAc8vzDiRkHmHDysr7bBaM9XWqaXnHgyfq4AMNMv9GxMaKz4urBxzomve3mtWM9nyC1OYYn2XPBlrNQUTiLb6i4TV1IE1Bnba9G8MtQCGFD8Ts1d2m4Yczw-9XpMJYOmMQlN6QWQ4NSIpwRE/s520/front%20panel%20pins.jpg" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="304" data-original-width="520" height="187" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgHsyypzUORPVoKlXq4UyNZF9NRJC0RphlT3za6tVs0mvazAc8vzDiRkHmHDysr7bBaM9XWqaXnHgyfq4AMNMv9GxMaKz4urBxzomve3mtWM9nyC1OYYn2XPBlrNQUTiLb6i4TV1IE1Bnba9G8MtQCGFD8Ts1d2m4Yczw-9XpMJYOmMQlN6QWQ4NSIpwRE/s320/front%20panel%20pins.jpg" width="320" /></a></div><p>The following schematic shows how controls can be wired to this connector:</p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEib0mrfh8Xjo-PxfsNVmB1e2WV64U5NV84bDiTzYvIdyUKvST12ISaibV9q6v6Fbqz7el2ioCpKRoj6jCicOi8BarJcET09uNmpkQ260vfUouvrWFzjHbIjK05gjcrUglDPYyWdOZ1Xa_1rNzqQ5ghPqDx-Bq9jc3StkSsb8JIbVZZAMgqjzaBhZH0A2YM/s798/Front%20Panel%20Controls.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="737" data-original-width="798" height="296" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEib0mrfh8Xjo-PxfsNVmB1e2WV64U5NV84bDiTzYvIdyUKvST12ISaibV9q6v6Fbqz7el2ioCpKRoj6jCicOi8BarJcET09uNmpkQ260vfUouvrWFzjHbIjK05gjcrUglDPYyWdOZ1Xa_1rNzqQ5ghPqDx-Bq9jc3StkSsb8JIbVZZAMgqjzaBhZH0A2YM/s320/Front%20Panel%20Controls.png" width="320" /></a></div><p><u>Front Panel Connector Notes:</u></p><p></p><ul><li>ARA series and SCR274-N series of receivers provide audio to pin 4. Note that the load on the audio signal is expected to be either 8000 ohms or 600 ohms, depending upon the receiver.</li></ul><ul><li>AN/ARC-5 receivers <b>have no internal connection to pin 4</b>, and so audio cannot be accessed via the front of an AN/ARC-5 receiver.</li></ul><ul><li>If +LV (low voltage power, e.g. 28 V) is applied via J3 (the Rear Panel Connector), then pins 6 & 7 of the Front Panel connector <b>must be connected together</b> (either via a wire or a switch), to route +LV to the tube filaments (and Dynamotor +LV input).</li></ul><ul><li>Gain control is provided via a potentiometer (20K or 50K ohms), and should be wired so that, in its position of maximum gain, resistance is at its minimum.</li></ul><ul><li>The CW Oscillator (BFO) can be turned ON or OFF via pin 5. Note that the BFO is OFF when pin 5 is shorted to ground. </li></ul><p></p><p><br /></p><p><u>Connecting to the Front Panel Connector's pins:</u></p></div></div><div>I use 1/8" O.D. copper tubing to create homemade jacks to slip over the receiver's connector pins, a great idea from John Stanley's (K4ERO) Command Set article in the January 2016 issue of <i>QST</i>. I purchased a package containing a variety of tube diameters from Amazon (<a href="https://www.amazon.com/dp/B07VPWSJRC?psc=1&ref=ppx_yo2ov_dt_b_product_details">https://www.amazon.com/dp/B07VPWSJRC?psc=1&ref=ppx_yo2ov_dt_b_product_details</a>)</div><div><br /></div><div><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhn7MMXzAshoe46kidRdd-3aaT0BpC_SIMOl6I8S2q2lSixBFH_YjQHRmR9MKLnvu4ZipuJJswHJUtwMQVCHAvofBucKwWHSrKqjbY6KtRq0grD_U-z_tGi8gBWwikuP7xHnINsmmKHWs01kdoBgJ-VAx_z4MN2B200AyyD5ohmSTyv7na0s0ANbnEapN8/s1040/copper%20tubing.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="633" data-original-width="1040" height="195" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhn7MMXzAshoe46kidRdd-3aaT0BpC_SIMOl6I8S2q2lSixBFH_YjQHRmR9MKLnvu4ZipuJJswHJUtwMQVCHAvofBucKwWHSrKqjbY6KtRq0grD_U-z_tGi8gBWwikuP7xHnINsmmKHWs01kdoBgJ-VAx_z4MN2B200AyyD5ohmSTyv7na0s0ANbnEapN8/s320/copper%20tubing.png" width="320" /></a></div><p>Use a jeweler's saw (see below, available inexpensively from Amazon) to cut the 1/8" O.D. tubing to length!</p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjdjz5TDjc_6U2J7P-IQdFqsjI1xKWPAYnZGJx4R8g7waMfS4QJkzs5u8aVJ2yyQkTgjh1R8AKUxycj5e5mZ3bI0YOwQICNbG-7gNMhhqedIuo0oR9zcLTyb4_dHhQrti16ymE3AhNfHr-eGYUzCUvI3fMUrNrKqM4OGLMbkUcqxhLPvXKPaTuOleiCseg/s596/jewelers%20saw.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="502" data-original-width="596" height="270" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjdjz5TDjc_6U2J7P-IQdFqsjI1xKWPAYnZGJx4R8g7waMfS4QJkzs5u8aVJ2yyQkTgjh1R8AKUxycj5e5mZ3bI0YOwQICNbG-7gNMhhqedIuo0oR9zcLTyb4_dHhQrti16ymE3AhNfHr-eGYUzCUvI3fMUrNrKqM4OGLMbkUcqxhLPvXKPaTuOleiCseg/s320/jewelers%20saw.png" width="320" /></a></div><p>Here is my control panel for my BC-454-B receiver:</p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj68LwtvdmubYNw2uPYMj2A9IYDaSv2jtLfA2YiXKcJJmBAHtc8GF00PjmXmbWekxabuwn5VjvbSCGb9uhxtU_o9zYi36slH9qLyNLqUwneMrgCdInzbOkQc3SZLFc1SdrZ9mEufsyTM9zGL_C0wmFCVqYA7BfRpdvesHZ0DglmNJZs6cApDruFfx44dTI/s640/IMG_9889.jpg" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="480" data-original-width="640" height="240" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj68LwtvdmubYNw2uPYMj2A9IYDaSv2jtLfA2YiXKcJJmBAHtc8GF00PjmXmbWekxabuwn5VjvbSCGb9uhxtU_o9zYi36slH9qLyNLqUwneMrgCdInzbOkQc3SZLFc1SdrZ9mEufsyTM9zGL_C0wmFCVqYA7BfRpdvesHZ0DglmNJZs6cApDruFfx44dTI/s320/IMG_9889.jpg" width="320" /></a></div><div><br /></div><div>The image below shows the short lengths of 1/8" O.D. copper tubing, at the end of each wire, that slip over the front panel connector's male pins.</div><div><br /></div><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgD63cl8YnH8fkCWfwinzezgIDQ-sqK_i4ptYYMgV11rK0yoOS9hZkhuKIVGGn9MREaqSJAsdNY0HbWSrjkY_Sb6NED8GYVnxC0GNwp2cOOdrh0HtmU65M1G3La3Y2OMVbgafXktVZEMaNUVvAw7wgVD55XYyMr1KzHq6enTKmycGjC7XAVQQQ0naKjFq4/s1326/IMG_9897.jpg" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="1098" data-original-width="1326" height="265" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgD63cl8YnH8fkCWfwinzezgIDQ-sqK_i4ptYYMgV11rK0yoOS9hZkhuKIVGGn9MREaqSJAsdNY0HbWSrjkY_Sb6NED8GYVnxC0GNwp2cOOdrh0HtmU65M1G3La3Y2OMVbgafXktVZEMaNUVvAw7wgVD55XYyMr1KzHq6enTKmycGjC7XAVQQQ0naKjFq4/s320/IMG_9897.jpg" width="320" /></a></div><div><br /></div><div>Note that the Low-Voltage Power switch (that will connect to pins 6 & 7) is on the back of the potentiometer. If a dynamotor is used, the gauge of these two wires should be sized to handle an amp of current.</div><div><br /></div><div>And to minimize confusion when installing the Control Panel, the wires are color-coded to identify to which pin they should attach (e.g. brown to pin 1, red to pin 2, etc.).</div><div><div><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEilBZfpk6UhFz0XcPRTTvpMbBny11_qTpdKKK1PtqCh9-OI6EGj6RFmnnqO8EfbIYkano_nYpiAOBbg8I0UDVakSdi8QVRtLhM4shcMM03LIByfIp9ZtEcWdGTLETAkM7AtWsFRJ0powMZAu4rLvntdrlmJ5lavpWc4Hs1rXRLsMzgK0dKiKWSlrsGcqeA/s797/IMG_9753.jpg" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="660" data-original-width="797" height="265" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEilBZfpk6UhFz0XcPRTTvpMbBny11_qTpdKKK1PtqCh9-OI6EGj6RFmnnqO8EfbIYkano_nYpiAOBbg8I0UDVakSdi8QVRtLhM4shcMM03LIByfIp9ZtEcWdGTLETAkM7AtWsFRJ0powMZAu4rLvntdrlmJ5lavpWc4Hs1rXRLsMzgK0dKiKWSlrsGcqeA/s320/IMG_9753.jpg" width="320" /></a></div><br /><p><b><u>Rear Panel Connector:</u></b></p></div></div></div><div>There is a 7 pin female connector on the rear panel that has the following signal assignments:</div><div><p></p><ul><li>Pin 1: Ground</li><li>Pin 2: Headset Audio Output</li><li>Pin 3: Gain Control Line</li><li>Pin 4: CW Osc. (BFO) Shut-Off Line</li><li>Pin 5: + Screen Grid Voltage <span style="color: #cc0000;">(not used)</span></li><li>Pin 6: +L.V. (i.e. Low Voltage, e.g. 28 VDC or VAC)</li><li>Pin 7: +H.V. (i.e. High Voltage, e.g. 250 VDC)</li></ul><div><br /></div><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj3rHFza4vTCyYkzkfrigOgMPcNfHyd6bE7wXGE9enRxlGKMwPEmgk1djElah1AnAHIENSDAgIDAnIlsNjvv0beJbDktbFKNydnsBPAluA2AmPxSG8m0jnUu6kelOmp5nEV-sSMsV5_gakLPAh1f_E22_vTpBB0zJivpb2t6MKEvUv-LZG7N7tNrU42QHk/s468/rear%20panel%20pins.jpg" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="316" data-original-width="468" height="216" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj3rHFza4vTCyYkzkfrigOgMPcNfHyd6bE7wXGE9enRxlGKMwPEmgk1djElah1AnAHIENSDAgIDAnIlsNjvv0beJbDktbFKNydnsBPAluA2AmPxSG8m0jnUu6kelOmp5nEV-sSMsV5_gakLPAh1f_E22_vTpBB0zJivpb2t6MKEvUv-LZG7N7tNrU42QHk/s320/rear%20panel%20pins.jpg" width="320" /></a></div><br /><div>Gain Control, BFO On/Off, and Audio Out signals can all be accessed via the Rear Panel connector. In this situation, their front-panel pins should not be used.</div><div><br /></div><div><div><u>Rear Panel Connector Notes:</u></div><div><ul><li>Pin 2 is the Audio Out signal. Note that the load on the audio signal is expected to be either 8000 ohms or 600 ohms, depending upon the receiver.</li></ul><ul><li>Pin 3, Receiver Gain can be controlled via an external potentiometer (20K or 50K ohms), and it should be wired so that, in its position of maximum gain, resistance is at its minimum.</li></ul><ul><li>Pin 4, the CW Oscillator (BFO) can be turned ON or OFF via pin 4. Note that the BFO is ON when pin 4 is OPEN. </li></ul><ul><li>Pin 6 is the +LV pin (low voltage power, e.g. 28 V). Note that if +LV is applied via this pin, then pins 6 & 7 of the Front Panel connector must be connected together (either via a wire or a switch), to route +LV to the tube filaments (and Dynamotor +LV input). The image below shows an FT-230-A Adapter, which can provide this function.</li></ul><div><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjah5Rm0OypWYlM_7IPy0oQUw8dsc77jZMPITqEHSFrPpG9tPvj-kH-CvtonSNI8xAfimsdKWfxmNY2ono8Q83mVyXHwp7iYDCXhM6OgCCs384GlwGfLgh1S8q_vOlDWwnYJOy_7m1UKViL8GK4U0V89MhL65WDvFscVfVv5VrZm5wpuaNTunM6yTawRR4/s804/shorting%20box.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="421" data-original-width="804" height="168" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjah5Rm0OypWYlM_7IPy0oQUw8dsc77jZMPITqEHSFrPpG9tPvj-kH-CvtonSNI8xAfimsdKWfxmNY2ono8Q83mVyXHwp7iYDCXhM6OgCCs384GlwGfLgh1S8q_vOlDWwnYJOy_7m1UKViL8GK4U0V89MhL65WDvFscVfVv5VrZm5wpuaNTunM6yTawRR4/s320/shorting%20box.png" width="320" /></a></div></div><ul><li>Pin 7 is the +HV pin. No connection from an external high-voltage source should be made to this pin if a Dynamotor is installed. If there is no Dynamotor, then +HV (e.g. 250 VDC) can be applied to the radio via this pin.</li></ul></div><div><br /></div></div><div><div><u>Connecting to the Rear Panel Connector:</u></div><div><br /></div><div>Similar to an idea I saw mentioned in John Stanley's (K4ERO) Command Set article in the January 2016 issue of <i>QST,</i> I use 2.5 mm Banana Plugs (available through Amazon) to attach to the Rear Connector's female sockets (<a href="https://www.amazon.com/dp/B09S3VDSVK?psc=1&ref=ppx_yo2ov_dt_b_product_details">https://www.amazon.com/dp/B09S3VDSVK?psc=1&ref=ppx_yo2ov_dt_b_product_details</a>). I chose these after first discovering that 2.0 mm plugs were too narrow and 3.0 mm plugs were too wide.</div><p></p><p></p><p></p><div style="-webkit-text-stroke-width: 0px; color: black; font-family: "Times New Roman"; font-size: medium; font-style: normal; font-variant-caps: normal; font-variant-ligatures: normal; letter-spacing: normal; orphans: 2; text-align: left; text-decoration-color: initial; text-decoration-style: initial; text-decoration-thickness: initial; text-indent: 0px; text-transform: none; white-space: normal; widows: 2; word-spacing: 0px;"><div style="font-weight: 400;"><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgSEWh6ZCSZhSOd-YPkQULSDRZr74DTJbQ2_eUwmcCPSCxafDfABOwrc8f6RQEpHRlurqWVtP6HhPeaSfFtqpIWuUkR_NxUP2hiat04iysu6Dj60O59yY7MZTCOoOXBDYZSKfYQY5PqMiGRvnHRzITKnmyLj1RcoOrX7M_zA_iDzaSNnbpOrCAZ7mLhmDY/s1106/banana%20plugs.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="620" data-original-width="1106" height="179" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgSEWh6ZCSZhSOd-YPkQULSDRZr74DTJbQ2_eUwmcCPSCxafDfABOwrc8f6RQEpHRlurqWVtP6HhPeaSfFtqpIWuUkR_NxUP2hiat04iysu6Dj60O59yY7MZTCOoOXBDYZSKfYQY5PqMiGRvnHRzITKnmyLj1RcoOrX7M_zA_iDzaSNnbpOrCAZ7mLhmDY/s320/banana%20plugs.png" style="cursor: move;" width="320" /></a></div><div><br /></div></div><div style="font-weight: 400;"><br /></div><div><b><u>Attaching an 8-Ohm Speaker:</u></b></div></div></div><div><div><p>I prefer to use an 8-ohm speaker instead of a headset (even though I do have a 600-ohm headset) and I find that the receiver, connected to the 8-ohm speaker via an impedance transformer designed for the appropriate impedance ratio, can drive my 8-ohm speaker to quite loud levels.</p><p>With a 600:8 ohm Impedance Transformer between radio and speaker, and measuring voltage with an oscilloscope at the speaker, the speaker's signal starts clipping at at about 3 Vpp (i.e. about 1.07 Vrms if the signal were a sine wave), which equates to a speaker drive power of about 0.14 watts. With my speaker, this is VERY LOUD -- I need to wear ear plugs! A 1 to 1.5 Vpp signal is more comfortable.</p><p>The following Audio transformers, available from Mouser (if in stock), should be able to transform an 8 ohm speaker to 600 ohms (or 500 ohms, which should be close enough):</p><blockquote style="border: none; margin: 0px 0px 0px 40px; padding: 0px;"><p><a href="https://www.mouser.com/c/passive-components/audio-transformers-signal-transformers/?m=Xicon&series=42TL">Xicon 42TL series</a> (75 mW max output drive):</p></blockquote><p></p><ul><ul><li>42TL001-RC: 500 ohm primary, 8 ohm secondary.</li><li>42TL003-RC: 1.2K ohm primary (for 600 ohms, connect between one end of the primary and its center-tap), 8 ohm secondary.</li></ul></ul><p></p><blockquote style="border: none; margin: 0px 0px 0px 40px; padding: 0px;"><p><a href="https://www.mouser.com/c/?q=xicon%2042TM">Xicon 42TM series</a> (200 mW max output drive):</p></blockquote><p></p><p></p><p></p><p></p><ul><ul><li>42TM001-RC: 500 ohm primary, 8 ohm secondary.</li><li>42TM003-RC: 1.2K ohm primary (for 600 ohms, connect between one end of the primary and its center-tap), 8 ohm secondary.</li></ul></ul><blockquote style="border: none; margin: 0px 0px 0px 40px; padding: 0px;"><a href="https://www.mouser.com/c/?q=xicon%2042TU">Xicon 42TU series</a> (460 mW max output drive):</blockquote><div><ul><li>42TU400-RC: 500 ohm primary, 8 ohm secondary.</li></ul></div><p>The image below shows the relative sizes of these three transformer families:</p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEju-NK6bz8mW39xRtH4TMQbfg3k2mDGU6nQPL-8xtf3jUAC7MxDfVcumNv23Hgj1WynavP5ibsa_18Ey94dEtVGoDP9M6DGuyPRrBwGFv6ZKRz-L_8aGsbcxxhyphenhyphensDXm68LIqrfNKOqDolHNQZOYH5X0q0sS79amczieYrOYwWuSuL2ojw7dSONmO8mFFXc/s327/xfrmr%20sizes.jpg" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="325" data-original-width="327" height="318" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEju-NK6bz8mW39xRtH4TMQbfg3k2mDGU6nQPL-8xtf3jUAC7MxDfVcumNv23Hgj1WynavP5ibsa_18Ey94dEtVGoDP9M6DGuyPRrBwGFv6ZKRz-L_8aGsbcxxhyphenhyphensDXm68LIqrfNKOqDolHNQZOYH5X0q0sS79amczieYrOYwWuSuL2ojw7dSONmO8mFFXc/s320/xfrmr%20sizes.jpg" width="320" /></a></div><p>I chose the <a href="https://www.mouser.com/ProductDetail/Xicon/42TM003-RC?qs=NizYiyh06IFQCKndLl2QhQ%3D%3D">Xicon 42TM003-RC</a> because physical size wasn't an issue <i>and</i> it could handle the maximum output from the receiver, if I accidentally left the receiver unattended, set to maximum volume, and tuned to a carrier.</p><p>Below is a schematic showing a possible wiring diagram. Note that the receiver is connected to the primary's <i>center-tap </i>(i.e. 600 ohms), not across the entire primary winding (i.e. 1200 ohms).</p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjk2gGBLOhudD_eo4LFP3yBES1LjF7aabJq36QC3LvjNFWDBIk9cCawOxuqC5KH8eFEXGxmnPE3uSp0kIB-cznSZzm03V-1TuzFLmRcTe087oQ1ZKeYTbQu5meYbjl6g0AOj7ycVcV4oiYQztjVU3wbdW3ING5TsmsVUHPfXVmSTna1VAIa-SWZysQlqQg/s1186/impedance%20transformer.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="712" data-original-width="1186" height="192" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjk2gGBLOhudD_eo4LFP3yBES1LjF7aabJq36QC3LvjNFWDBIk9cCawOxuqC5KH8eFEXGxmnPE3uSp0kIB-cznSZzm03V-1TuzFLmRcTe087oQ1ZKeYTbQu5meYbjl6g0AOj7ycVcV4oiYQztjVU3wbdW3ING5TsmsVUHPfXVmSTna1VAIa-SWZysQlqQg/s320/impedance%20transformer.png" width="320" /></a></div><p>If using a 4-ohm speaker in lieu of 8 ohms, connect it to the secondary's center tap, rather than across the entire secondary.</p><p><a href="http://www.ad5x.com/images/Articles/ARC5convert.pdf">Adapting an ARC-5 Receiver for Ham Use</a> demonstrates how a small audio transformer (Xicon 42TL series) can be installed behind the front "adapter panel" and held in place with hot-melt glue. </p></div><div><b><u><br /></u></b></div><div><b><u>Steps to Bring Up a Receiver:</u></b></div></div><p><u>1. Verify Filament Wiring:</u></p><p>As I mentioned earlier, always first verify receiver filament wiring before applying power to a receiver. A common modification was to rewire 28 VDC filament circuitry for 14 VDC (in practice, 12.6 VAC) by connecting all six filaments in parallel. You do not want to apply 28 VDC to a receiver that has had its filaments rewired!</p><p>Also check if the original 12V tubes were replaced with 6V tubes.</p><p>For 28 VDC operation the following pairs of tubes were connected in series:</p><p></p><ul style="text-align: left;"><li>12SK7 RF amp & 12K8 Mixer</li><li>12SK7 1st IF Amp & 12SK7 (or 12SF7) 2nd IF Amp</li><li>12SR7 Det./CW Osc. & 12A6 Audio Amp</li></ul><p></p><p>The quickest way to verify if a receiver has all six filaments wired in parallel is to check the filament wiring of the following tubes:</p><p></p><ul style="text-align: left;"><li>12K8 Mixer, pins 2 and 7.</li><li>12A6 Audio Amp, pins 2 and 7.</li><li>12SK7 First IF Amp, pins 2 and 7.</li></ul><div><b><span style="color: #cc0000;">If any of these pins are <u>connected to ground</u>, then chances are the receiver has been wired for 14 VDC operation</span></b> (or 6 VDC operation, if the original 12 volt tubes have been swapped out for 6 volt filament tubes!).</div><div><br /></div><div>Note: Of my six receivers, only one has retained its original 28 V wiring (the BC-454-B). The other five all have their filaments wired in parallel, and two of these even had 6V tubes substituted for the original 12V tubes.</div><div><br /></div><div><p><u>2. Check Capacitors:</u></p><p>Over time capacitors can degrade and fail. All capacitors should be checked with a good capacitance meter (that measures loss, or dissipation factor, as well as capacitance -- I use a GenRad 1657 RLC Digibridge).</p><p>The most likely failures will be the electrolytic capacitors (C5, C30, and C32), so I would start with those. For the caps in the "tubs" screwed to the chassis, I simply unsolder the wires from their terminals (or terminal) and measure capacitance (and dissipation factor) from each terminal to chassis (ground).</p><p>Also look for capacitor "tubs" that might be leaking (these should be fairly obvious). Check these, too.</p><p>Some authors recommend replacing all of the capacitors. I prefer to just replace those that measure poorly. But replacing them all would be a suitable tactic if you don't have a good way to check the capacitors.</p><p>If you find faulty caps, replace them. For faulty capacitors contained in the metal tubs (for example, C6A, C6B, and C6C)), I would simply remove the tub and add a terminal strip with new capacitors wired to it (see: <a href="http://www.ad5x.com/images/Articles/ARC5convert.pdf">Adapting an ARC-5 Receiver for Ham Use</a>). I'll show an example later in this post, when I describe my efforts with my BC-454-B.</p><p>Some restorers prefer to stuff the original capacitor metal "tubs" with new capacitors to retain the look of the original receiver (see: <a href="https://www.antiqueradios.com/forums/viewtopic.php?f=5&t=288697&start=0">https://www.antiqueradios.com/forums/viewtopic.php?f=5&t=288697&start=0</a>). I am not such a purist, preferring an easier route, myself. </p><p><u>3. Check the Receiver's Audio Impedance Requirement and add an Audio Transformer, if necessary.</u></p><p>The load required by the receiver's audio amplifier will be either High Impedance (e.g. 8000 ohms) or Low Impedance (e.g. 600 ohms). Refer to the discussion earlier in this post regarding how to determine your receiver's requirement. Note that a radio might have been modified by a previous owner. It's good to verify!</p><p>If your headset or speaker is not the correct impedance (if it's close to the correct impedance, it should be OK), you should use an impedance transformer to transform its impedance to the impedance that the receiver wants to see. For example, see discussion, above, regarding transformers that will transform an 8 ohm speaker to 600 ohms.</p><p><u>4. Should I Attach Control & Audio Signals to the Front or to the Rear Connector?</u></p></div><div><div><p>Connect the Control functions (Gain Control, BFO On/Off) to either the Front Panel connector or the Rear Panel connector, per your preference. Refer to the appropriate sections, above.</p><p><u>5. How will the Receiver be Powered?</u></p><p>Determine how the receiver will be powered (refer to the "Powering the Receiver" section, above). Here are three choices:</p><p></p><ol><li>Dynamotor (installed on the receiver), with +LV applied from an external supply to the Rear Panel connector.</li><li>External power supply providing both +HV and +LV, attached to the Rear panel connector.</li><li>External power supply providing both +HV and +LV, attached to the 3-pin Dynamotor connector.</li></ol><p></p><p>Refer to the "Powering the Receiver" section earlier in this post.</p><p><b>Only after I've gone through all these steps do I then apply power.</b></p></div><div><br /><p><b><u>Example, BC-454-B Revivification:</u></b></p><p>Here are the steps I took to bring my BC-454-B back to life.</p><p><u>1. Verify Filament Wiring</u></p><p>The receiver is one of the SCR-274-N series of receivers and therefore should have its tube filaments wired for 28 volts. In other words, the filaments of each of three pairs of tubes should be wired in series.</p><p>I removed the bottom cover and visually checked to see if any tubes had <i>no</i> filament pins tied directly to ground. At least one tube fit this criteria, confirming that the radio was wired for 28 VDC.</p><p><u>2. Check capacitors</u></p><p>With the bottom cover off, I first did a visual inspection of all of the "tub" capacitors (note: anti-fungal coating can possibly mask leakage). The C7 capacitor tub (containing three caps: C7 A, B, C) looked like it had leaked a bit since the radio was first manufactured (in March, 1945), but unsoldering the wires to the three capacitor terminals and measuring their capacitance to ground revealed no problem. They all measured in the range of 60 nF (spec'd to be 50 nF (i.e. 0.05 MF). Close enough!</p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhXHYFvDNoqhe47yNtQofrKDFludbT_RYoyU1CgA4Xn3v8iRXFcwfM77Xs9JvKL5y3TwNtdL3G43rLYxnYkaopk44T29NxnWwThioEg-f1UxGrTl2Nfy0-itxdtQG9tsJcl_hdICD7EWD8bOXP2DfIyiB4gHua0tkVXs5nv62G2JsmXRZ0f-bjZWxi32sM/s389/leaky%20caps.jpg" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="303" data-original-width="389" height="249" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhXHYFvDNoqhe47yNtQofrKDFludbT_RYoyU1CgA4Xn3v8iRXFcwfM77Xs9JvKL5y3TwNtdL3G43rLYxnYkaopk44T29NxnWwThioEg-f1UxGrTl2Nfy0-itxdtQG9tsJcl_hdICD7EWD8bOXP2DfIyiB4gHua0tkVXs5nv62G2JsmXRZ0f-bjZWxi32sM/s320/leaky%20caps.jpg" width="320" /></a></div><p>Without any other tubs looking like they had leaked, I decided to focus next on the electrolytic caps, assuming that the other non-electrolytic caps were probably good. One electrolytic cap, C32 (spec'd 5 uF) measured 6.7 nF (off by a factor of 1000 !!!), with a Dissipation Factor of 1.78. </p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiFAL7o_P8BasNsIByOr4dgwx1wItNrWCHFz8ByS7LgrPxvdijCqOlEauvgvj0z7vhWMJsmJhe6ybiFXNe28V1lQMfsMVIq3nmET4ccVObbPrYbYGvpVSXmVMy2Qr5nVlEieBcjJQH-6SGeHwa1SHZ6lLn9wgQSdABTqoqE4L-ysC2qwdue92nY3g_jgJs/s446/bad%20cap.jpg" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="421" data-original-width="446" height="302" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiFAL7o_P8BasNsIByOr4dgwx1wItNrWCHFz8ByS7LgrPxvdijCqOlEauvgvj0z7vhWMJsmJhe6ybiFXNe28V1lQMfsMVIq3nmET4ccVObbPrYbYGvpVSXmVMy2Qr5nVlEieBcjJQH-6SGeHwa1SHZ6lLn9wgQSdABTqoqE4L-ysC2qwdue92nY3g_jgJs/s320/bad%20cap.jpg" width="320" /></a></div><p>The values of all other electrolytic caps measured OK, with DFs of sub 0.02.</p><p>Not willing to go through the hassle of trying to fit a new cap into the original tub, I removed C32 and replaced it with a 2-lug terminal strip. The cap filters the +HV line, so larger is better for this application. I dug up a 50 uF, 360 WVDC cap (from my junkbox), which I soldered in to replace the original 5 uF cap. </p><p>Note: the receiver manuals do not specify capacitor working-voltage, so I simply use parts that are rated (by a comfortable margin) above the receiver's +HV rating of 250 VDC. </p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEip6wjg59rSl4E1PUqlydI191d3BKV6nohHN2CqBs4BfqxPnN08oDh8eKpAXD_Dw4u6LPCaMssrLYEtb6nRCBseNmrw24bcNg53u7KMm3SC46vYRYktxmcAFLFnNY3-r4G3Km4zXHYbqIae-sz4REeg_I87BWVjCpJtofv05nO5JsT_-7rx88NLdKrVQHo/s640/IMG_9728.jpg" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="480" data-original-width="640" height="240" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEip6wjg59rSl4E1PUqlydI191d3BKV6nohHN2CqBs4BfqxPnN08oDh8eKpAXD_Dw4u6LPCaMssrLYEtb6nRCBseNmrw24bcNg53u7KMm3SC46vYRYktxmcAFLFnNY3-r4G3Km4zXHYbqIae-sz4REeg_I87BWVjCpJtofv05nO5JsT_-7rx88NLdKrVQHo/s320/IMG_9728.jpg" width="320" /></a></div><p><u>3. Check Audio Transformer T1 for Impedance Requirement.</u></p><p>A quick verification of T1's Part Number (P/N ES-691027) and a check of its wiring, per the schematic notes in the SCR-274-N manual (see illustration earlier in this post) confirmed that the audio transformer's wiring had not been modified and that it was setup for a low impedance (e.g. 600 ohm) load. </p><p>Thus confirming the decal on the side of the chassis!</p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjiAsq3KFZzrDH4FZw-5fEW02XUafDrjRyzlj_Ka5QAyfEgnxTCvge5oMZ47ZdnOK-l1hoPHbPsAM80KmfjDSozCU6l7l8_noLn3FE8yHA17EMe0IsqQLW1yY7B80nhPhoTpwoUuy0cKE_0gBA5guqRz2YSXFRfbcjVgz3Oa_E0YKiK9FdI80AmIq3UW6M/s640/decal.jpg" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="480" data-original-width="640" height="240" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjiAsq3KFZzrDH4FZw-5fEW02XUafDrjRyzlj_Ka5QAyfEgnxTCvge5oMZ47ZdnOK-l1hoPHbPsAM80KmfjDSozCU6l7l8_noLn3FE8yHA17EMe0IsqQLW1yY7B80nhPhoTpwoUuy0cKE_0gBA5guqRz2YSXFRfbcjVgz3Oa_E0YKiK9FdI80AmIq3UW6M/s320/decal.jpg" width="320" /></a></div><p><u>4. Fabricate and Connect the Front Control Panel</u></p><p>One of my six receivers had a filthy control panel with an old pot, BFO switch, and audio jack attached.</p><p>I cleaned up the panel, installed a new pot from the junkbox (with power switch on its back), new BFO switch, and new audio jack. Then I attached wires and home-made "pin-sockets" (from 1/8" O.D. copper tubing) to slip over the Front Panel connector's pins.</p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjJtb3e0bOtP-PioCblWwwDRPjEa8yVmmJkj4AVu_Lt38oSPHFP7FWrBlUDVZa_TZhEqBfOz4fNSV9r5cvhw3Ip4xWp3XdVz9OSa3Tz0J_mtXgD9P_agAtyROo6mi-ftgXeBRnUjaJMihB80F0gCTwMbnM7XuhH2Cug7AruiUlya-VkfB4xx2luz2aaT3A/s1326/IMG_9897.jpg" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="1098" data-original-width="1326" height="265" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjJtb3e0bOtP-PioCblWwwDRPjEa8yVmmJkj4AVu_Lt38oSPHFP7FWrBlUDVZa_TZhEqBfOz4fNSV9r5cvhw3Ip4xWp3XdVz9OSa3Tz0J_mtXgD9P_agAtyROo6mi-ftgXeBRnUjaJMihB80F0gCTwMbnM7XuhH2Cug7AruiUlya-VkfB4xx2luz2aaT3A/s320/IMG_9897.jpg" width="320" /></a></div><p>5. <u>Rear Panel Connections</u></p><p>To use an 8-ohm speaker, I chose a <a href="https://www.mouser.com/ProductDetail/Xicon/42TM003-RC?qs=NizYiyh06IFQCKndLl2QhQ%3D%3D">Xicon 42TM003-RC</a> transformer to transform an 8-ohm speaker to 600 ohms. The center-tap of the transformer's primary connects to pin 2 (Audio Out) of the Rear Panel connector. Pick one or the other of the remaining two primary pins (it doesn't matter which one) and attach it to pin 1 (ground) of the Rear Panel connector.</p><p>The '+' terminal of the 28 VDC supply connected to pin 6 (+LV) of the Rear Panel connector, and the supply's '-' terminal connected to pin 1 (ground).</p><p>6. I decided to first test the receiver my high-voltage "lab" supply (a Heathkit IP-32, set to 250 VDC), and so I connected its =HV to to pin 3 of the Dynamotor connector, with supply's ground attached to pin 1 of the same connector.</p><p>7. I attached my antenna to the Receiver's antenna connector.</p><p>8. And then the big moment -- I turned the Gain control on the front panel clockwise, turning on the power switch. +LV current (filament current) initially rose, but settled to under 0.5 amps (it might have been significantly under, I don't recall), and +HV current was on the order of 40 to 50 mA.</p><p>And as the tubes warmed up, I started hearing signals on the 80 meter band. It worked!</p><p>9. <u>Check with Dynamotor</u></p><p>Next, after powering down the receiver ,I disconnected the Heathkit High Voltage supply and attached the Dynamotor to the back of the receiver. It spun up as soon as I turned on power and again, as the tubes warmed up, signals began appearing.</p><p><u>Final Configuration:</u></p><p>The diagram, below, shows my BC-454-B setup, including power supply (but not the antenna).</p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjCqoWAB_FHMyWjePGQF7xzqYVUoZmzBHgRoImhy4OHYisdi5QWA_geDVw3NL-wv7TIYz3uYe4eZtu7Wlu4ONlx_GwaWWuYP-KX5F2ISRK7qkXpoA-NPfiR4Z488xYBTpJ-uwSRi0k-kyztQ_F04zgAo7iPJpRZcyyWM6dwvLABK3sNudCkeYlRsKfdcxU/s1462/bc454b%20wiring%20diagram.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="754" data-original-width="1462" height="165" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjCqoWAB_FHMyWjePGQF7xzqYVUoZmzBHgRoImhy4OHYisdi5QWA_geDVw3NL-wv7TIYz3uYe4eZtu7Wlu4ONlx_GwaWWuYP-KX5F2ISRK7qkXpoA-NPfiR4Z488xYBTpJ-uwSRi0k-kyztQ_F04zgAo7iPJpRZcyyWM6dwvLABK3sNudCkeYlRsKfdcxU/s320/bc454b%20wiring%20diagram.png" width="320" /></a></div><p><b><u><br /></u></b></p><p><b><u>My BC-454-B, On the Air (YouTube Video):</u></b></p><p>Below is a video showing my BC-454-B in operation, powered by its dynamotor. Both AM and LSB reception are displayed. </p><p>Note that this receiver has no AVC, so some "riding" of the Gain control is necessary. </p><p>Also, note that the tiny impedance transformer (600 to the speaker's 8 ohms) is temporarily connected between the radio and the speaker using clip leads.</p><div class="separator" style="clear: both; text-align: center;"><iframe allowfullscreen="" class="BLOG_video_class" height="266" src="https://www.youtube.com/embed/_Q3RApbayJo" width="320" youtube-src-id="_Q3RApbayJo"></iframe></div><div><br /></div>In summary, only one cap needed to be replace. Pretty amazing for a radio manufactured in March of 1945!</div><div><br /><p><b><u>Thoughts on Modifications:</u></b></p><p>Personally, as an engineer, my feeling is that if modifications can improve the performance of a radio, why not make them? But I'm also sympathetic to those who want to keep their equipment in original condition -- <i> <span face="Roboto, arial, sans-serif" style="background-color: white; color: #4d5156; font-size: 14px;">à chacun son goût.</span></i></p><p><u>1. Bringing Output Audio to the Front Connector:</u></p><p>I haven't tried this yet, but if you have an AN/ARC-5 version of the receiver without the audio-out connection to pin 4 of the Front Panel connector J1, it should be possible to run a wire from the Rear Panel connector pin 2 (J3) to pin 4 of J1.</p><p>This will require removing the RF Coil Unit (Z5A, Z5B, and Z5C). Hopefully this is straightforward. I plan to find out when I start working on my R-27 receiver.</p><p><u>2. Adding AVC:</u></p><p>Lack of good AVC in the BC-454-B is certainly noticeable, and it might be more annoying if this were my only receiver. </p><p>There is a simple modification for adding AVC to receivers without it.</p><p>From: '<a href="https://www.radionerds.com/images/0/03/Surplus_Radio_Conversion_Manual_Volume_3.pdf">Automatic Volume Control for Your Command Set,' Page 15, Orr, William I., <i>Surplus Radio Conversion Manual, Volume III</i>, Editors and Engineers, Summerland, CA, 1960</a></p></div></div></div><blockquote style="border: none; margin: 0px 0px 0px 40px; padding: 0px;"><div><div><div><p style="text-align: left;">All the essential a.v.c. components are incorporated in all receivers, but there is no connection between them. The purpose of this conversion is to provide a.v.c. action to the r.f. and i.f. amplifier stages by completing the a.v.c. circuit. It required two additional resistors and a capacitor. Refer to figures 11A and 11B (not shown in this post, k6jca). First, unground pin 5 of the 12SR7 (VT-133). Connect the 100 uuf (i.e. 100 pF, k6jca) capacitor across pins 4 and 5 of the tube socket. Connect the 470K resistor from pin 5 to an adjacent ground lug. Connect the second 470K resistor between pin 5 and the junction of C-15A and R-11. Remove R-11 from the circuit to increase the effect of the a.v.c. action.</p></div></div></div></blockquote><div><p>Note: I haven't tried this modification. I imagine it would be satisfactory for AM (where there is a carrier present), but perhaps not for CW or SSB signals. For those modes, you'll probably still need to keep one hand on the gain control.</p><p><br /></p><p><b><u>Resources:</u></b></p><p>A Google search for various ARC-5 terms will return a plethora of hits. Below are some of the sites I found useful.</p><p><a href="https://armyradio.com/Command_Sets.html">Command Set Summary.</a> A good summary (from Australia) of the various ARC-5 receivers.</p><p><a href="https://hangarthirteen.org/wp-content/uploads/2020/08/16-40SCR274-5-Maintenance-Instructions-for-SCR-274-N.pdf">T.O. 12R2-3SCR274-2 (formerly 16-40SCR274-5). Handbook, Maintenance Instructions, Radio Set SCR-274-N (revised 25 July 1956)</a>. Note that this handbook only applies to the SCR-274-N equipment (e.g. receivers BC-454, BC-455, etc.). As such, for example, all receiver schematics show the second I.F. tube as being a 12SK7, not a 12SF7, as seems to have been used in the ARC-5 (not SCR-274) series of receivers. 228 pages.</p><p><a href="http://www.vmarsmanuals.co.uk/archive/1181_ARC5_Maintenance.pdf">AN16-30ARC5-2. Handbook, Maintenance Instructions, AN/ARC-5 Aircraft Radio Equipment, LF, MF, HF Components</a>. Note that this handbook only applies to the AN/ARC-5 equipment (e.g. receivers R-25, R-26, etc.). As such, for example, all receiver schematics show the second I.F. tube as being a 12SF7, not a 12SK7, as seems to have been used in the BC-series of receivers. 354 pages.</p><p>The Instruction Book (including schematics and wiring diagrams) for the ARA/ATA series can be downloaded as five PDFs via the following links. Note that this is the manual for the R.A.A.F. (which I assume is "Royal Australian Air Force").</p><p></p><ul style="text-align: left;"><li><a href="https://www.tuberadio.com/robinson/Manuals/ATA_ARA_part1.pdf">ARA/ATA Manual, Part 1</a></li><li><a href="https://www.tuberadio.com/robinson/Manuals/ATA_ARA_part2.pdf">ARA/ATA Manual, Part 2</a></li><li><a href="https://www.tuberadio.com/robinson/Manuals/ATA_ARA_part3.pdf">ARA/ATA Manual, Part 3</a></li><li><a href="https://www.tuberadio.com/robinson/Manuals/ATA_ARA_part4.pdf">ARA/ATA Manual, Part 4</a></li><li><a href="https://www.tuberadio.com/robinson/Manuals/ATA_ARA_part5.pdf">ARA/ATA Manual, Part 5</a></li></ul><p></p><p><a href="http://radiomanual.info/schemi/Surplus_NATO/AN-ARC-5_alignment_procedure_NAVWEPS16-30ARC5-501_1949.pdf">NAVWEPS 16-30ARC5-501. Handbook, Bench Test and Alignment Procedure, Radio Equipment AN/ARC-5</a>. Specifically for the AN/ARC-5 series, but can probably be used for the SCR-274 series (e.g. receiver I.F. schematic shows 12SF7 as the second I.F. tube). 56 pages.</p><p><a href="https://armyradio.com/command_sets_list_of_differences.html">List of component differences between the various receivers.</a> Lists the C, L, and R component values for the different receivers in the ARA, SCR-274-N, and AN/ARC-5 series. (However, I don't believe this list identifies the differences in T1 between the series).</p><p><a href="http://www.ad5x.com/images/Articles/ARC5convert.pdf">Adapting an ARC-5 Receiver for Ham Use</a>. AD5X. Among other things, describes using 1/8" O.D. copper tubing for connectors and also a nice home-brew tuning knob.</p><p><a href="https://www.antiqueradios.com/forums/viewtopic.php?f=5&t=288697&start=0">https://www.antiqueradios.com/forums/viewtopic.php?f=5&t=288697&start=0</a> A thread showing an example of recapping a receiver using the original capacitor housing.<br /></p><p><a href="http://jlandrigan.com/files/BC%20453/The%20ARC-5'er%20Power%20Supply.pdf">http://jlandrigan.com/files/BC%20453/The%20ARC-5'er%20Power%20Supply.pdf</a> 1966 article of a power-supply design. Also discusses impedance transformation if using a speaker.<br /></p><p><a href="file:///C:/Users/jeff/Documents/aa%20-%20Projects/Radios/Military/Mil%20-%20Command%20Set/5702_RSGB_Command_Receivers.pdf">Stevens (G2BVN),</a> <a href="file:///C:/Users/jeff/Documents/aa%20-%20Projects/Radios/Military/Mil%20-%20Command%20Set/5702_RSGB_Command_Receivers.pdf">The Command Set Receivers, <i>R.S.G.B. Bulletin</i>, February, 1957.</a> Discusses the command set receivers (does not mention the variant with the 12SF7 as the second I.F.) and modifications thereof.</p><p>Stanley, J., K4ERO, "Using Unmodified Command Set ARC-5 Radios on the Ham Bands," <i>QST</i>, January 2016</p><p><a href="https://www.radionerds.com/index.php/Surplus_Conversion_Manuals">Surplus Radio Conversion Manual, Volumes 1, 2, and 3.</a> Originally published as paperbacks by Editors and Engineers. Note that Volume 3 contains a simple AVC mod on page 15 (for the SCR-274N receivers, such as the BC-454 and BC-455, that lack AVC).</p><p><a href="https://www.scribd.com/doc/225935431/Compilation-of-CQ-Articles-on-ARC-5-Command-Sets">Command Sets (<i>CQ </i>Magazine Article Reprints)</a>. Mostly transmitter mods, but some receiver information, too.</p><p><br /></p><p><b><u>Standard Caveat:</u></b></p><p>As always, I might have made a mistake in my equations, assumptions, drawings, or interpretations. If you see anything you believe to be in error or if anything is confusing, please feel free to contact me or comment below.<br /><br />And so I should add -- this information is distributed in the hope that it will be useful, but WITHOUT ANY WARRANTY; without even the implied warranty of MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.</p></div>Jeffhttp://www.blogger.com/profile/04853314106806116765noreply@blogger.com0tag:blogger.com,1999:blog-2257458838727315711.post-46855472882109115652023-11-02T10:20:00.017-07:002023-11-02T16:37:56.533-07:00Lubricating the Bearings of an ARC-5 Receiver's Dynamotor<p>Back when I was a freshman in high school (a <i>very</i> long time ago), a family friend gave me a BC-454 ARC-5 receiver with an AC power supply he had built for it.</p><p>It provided much listening pleasure for me, but in the fog of the intervening years it eventually disappeared to who knows where.</p><p>Over the past decade or so I've picked up a few ARC-5 receivers at various swap-meets, and now that I'm retired I thought I'd try to get them (or some of them) back on the air. With that goal in mind, I thought I'd try running them from dynamotors (as they were designed to be operated), rather than spending time to build an AC power supply.</p><p>The images below show one of my dynamotors (a little worse for wear!).</p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjWQrgXkOc_fu4ZOWge6R0weiNzHrYy5DSVLGfL3Jc-dJqhNYNk9V5ba_myyOMD8KzT-7h9HhUUAQbFuKJoQLx-ukouyasaUiF5ggbFY-jNWFl_JoA702yR5sDIbPu8Ml7K1Z5SInXau59m0RX4br5ayd6A9HJ0r72JTbqpi0ez3l57DF64wXiHnyd7ksw/s577/IMG_9730%20annotated.jpg" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="409" data-original-width="577" height="227" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjWQrgXkOc_fu4ZOWge6R0weiNzHrYy5DSVLGfL3Jc-dJqhNYNk9V5ba_myyOMD8KzT-7h9HhUUAQbFuKJoQLx-ukouyasaUiF5ggbFY-jNWFl_JoA702yR5sDIbPu8Ml7K1Z5SInXau59m0RX4br5ayd6A9HJ0r72JTbqpi0ez3l57DF64wXiHnyd7ksw/s320/IMG_9730%20annotated.jpg" width="320" /></a></div><p>Top view:</p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi8DdZTuA8pdl2I_M2rJuUZgZu0UgaK9StGYbGfQilIWaaQiKQ7WY5W19n2xumbI7fDKQ6lLjzYbcBlMd1tIZkJGtHV-Ne15MLtC_9kc0MOyStudHnB3ZWD0U9tpfuYovMIkEoLMMZxQDlXP6a8VycRWdzBolgVlqhwGRe8CMyvH43muurDeeuFFoenNHs/s560/IMG_9731%20annotated.jpg" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="405" data-original-width="560" height="231" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi8DdZTuA8pdl2I_M2rJuUZgZu0UgaK9StGYbGfQilIWaaQiKQ7WY5W19n2xumbI7fDKQ6lLjzYbcBlMd1tIZkJGtHV-Ne15MLtC_9kc0MOyStudHnB3ZWD0U9tpfuYovMIkEoLMMZxQDlXP6a8VycRWdzBolgVlqhwGRe8CMyvH43muurDeeuFFoenNHs/s320/IMG_9731%20annotated.jpg" width="320" /></a></div><p>Bottom view. Note that the dynamotor is labeled "DY-2/ARR-2", identifying it as a dynamotor for the AN/ARC-5 series of receivers (e.g. R-25, R-26, etc.). </p><p>(Note: The SCR-274 series of receivers (e.g. BC-454, BC-455, etc.) used the DM-32-A dynamotor (see table here: <a href="https://en.wikipedia.org/wiki/AN/ARC-5">Wikipedia</a>) -- the two dynamotors are essentially identical.)</p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj1KlN_Zn2Mw3abGQffTqNxmWX-E0cIWueAOmyo0zwacRV03jJoujUm9vzAitQTQXPu79vEwF8ADNUdc7n5MFNV3okUTz47wQZbDMMqmGWC5u8KaEVgXaLIX8LofM0wVvoaws_s6UsKCaQsh7obEb_AlPoj7md-dEQhZUr-wPy_6nIxZwM5fi-6yVY8bvY/s501/IMG_9732%20annotated.jpg" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="384" data-original-width="501" height="245" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj1KlN_Zn2Mw3abGQffTqNxmWX-E0cIWueAOmyo0zwacRV03jJoujUm9vzAitQTQXPu79vEwF8ADNUdc7n5MFNV3okUTz47wQZbDMMqmGWC5u8KaEVgXaLIX8LofM0wVvoaws_s6UsKCaQsh7obEb_AlPoj7md-dEQhZUr-wPy_6nIxZwM5fi-6yVY8bvY/s320/IMG_9732%20annotated.jpg" width="320" /></a></div><p>The figure, below, shows how the dynamotor's jacks are defined for High Voltage (H.V.), Low Voltage (L.V.) and Ground:</p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiybq8wq87_ZEiOaFYMUjrv90vGHT83NC5RXEwkPBZNIUEFuTPKlGUe6RkehwuIwfAMbltLrlEBBZzgib2TBZrGzGaaxPWccb3564OfyFD8YdfoDKpSfkEoob7rpq3GEGPQDTvAcVAzh-TaFPPCja2i0XXgj8cocT3gVBI-If8llfNxxHQ5lZoLxKELmPU/s501/annotated%20bottom%20plate.jpg" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="384" data-original-width="501" height="245" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiybq8wq87_ZEiOaFYMUjrv90vGHT83NC5RXEwkPBZNIUEFuTPKlGUe6RkehwuIwfAMbltLrlEBBZzgib2TBZrGzGaaxPWccb3564OfyFD8YdfoDKpSfkEoob7rpq3GEGPQDTvAcVAzh-TaFPPCja2i0XXgj8cocT3gVBI-If8llfNxxHQ5lZoLxKELmPU/s320/annotated%20bottom%20plate.jpg" width="320" /></a></div><p>The two dynamotors I had would not rotate when 28 volts was applied, implying that their bearing grease had solidified over the years. I would need to disassemble them and relubricate the dynamotor's bearings.</p><p><b><u><br /></u></b></p><p><b><u>Disassembling the Dynamotor and Cleaning its Bearings:</u></b></p><p>To access the bearings the dynamotor's two end covers must first be removed.</p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi3yuytfrrmiXo5-f-panByBEqaqQpgVW-vl0CstcTNvMNrNGxN_awYoTQ1nmYCUu2FI86-Ck6t01YOu8wYzbUVNk16-00R4a6LrOSHXgBj3FyhyphenhyphenWMnS7av31SF29DHN9NWz1KKPg25ouFv-9NVUMNABFQkB88hKNShGhMwvK-NiQmPGRpdUeRn8qDvpRM/s558/1.%20%20First%20remove%20the%20two%20end%20covers.jpg" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="425" data-original-width="558" height="244" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi3yuytfrrmiXo5-f-panByBEqaqQpgVW-vl0CstcTNvMNrNGxN_awYoTQ1nmYCUu2FI86-Ck6t01YOu8wYzbUVNk16-00R4a6LrOSHXgBj3FyhyphenhyphenWMnS7av31SF29DHN9NWz1KKPg25ouFv-9NVUMNABFQkB88hKNShGhMwvK-NiQmPGRpdUeRn8qDvpRM/s320/1.%20%20First%20remove%20the%20two%20end%20covers.jpg" width="320" /></a></div><div><br /></div><div>Note that the screws holding these two covers to the dynamotor body might be held in place with a "safety wire" (to, I assume, keep the screws from vibrating loose). Cut this wire and remove it, and then unscrew the two screws holding a cover to the body. Remove the two screws and their associated washers and lift off the cover off the dynamotor. Repeat for the second end cover.</div><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg9HbT-zxBJwycAnfi_q8gzURJfnboh8MESqrWKAHqSTcttlx7UxSwWbgmkRtzzwRLo-JDqArg-Cw0l-ZsaBSVgAqRi9PxMUi87lbMDUejuMVGQVWQzydzzUMuAi667TSO8d10rWnHrNXAs2v9nRwfsWU_PpmdPNJbb75MJ6slpTXWOI5UFN5UZeZx2scM/s532/2%20-%20cut%20the%20wire%20and%20remove%20cover.jpg" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="430" data-original-width="532" height="259" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg9HbT-zxBJwycAnfi_q8gzURJfnboh8MESqrWKAHqSTcttlx7UxSwWbgmkRtzzwRLo-JDqArg-Cw0l-ZsaBSVgAqRi9PxMUi87lbMDUejuMVGQVWQzydzzUMuAi667TSO8d10rWnHrNXAs2v9nRwfsWU_PpmdPNJbb75MJ6slpTXWOI5UFN5UZeZx2scM/s320/2%20-%20cut%20the%20wire%20and%20remove%20cover.jpg" width="320" /></a></div><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh5p75qIvGNYP1CuE9r8d9rHgJ6TZ95x7UqKOserCe5UNUzQE5IysmYUKims2daWpM74mOFrjCLhr_cJBxWJcZLqE1IpxgImSnAshyphenhyphencT7CBu9PEo7biolys8Aayof7Su2L8kW1K_vL3wAcX0CHJY_8kZx2xyW8Y3OWdi-N806GUOgEAAwBoTFzJtmfAeGU/s515/3-%20end%20cover%20off.jpg" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="451" data-original-width="515" height="280" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh5p75qIvGNYP1CuE9r8d9rHgJ6TZ95x7UqKOserCe5UNUzQE5IysmYUKims2daWpM74mOFrjCLhr_cJBxWJcZLqE1IpxgImSnAshyphenhyphencT7CBu9PEo7biolys8Aayof7Su2L8kW1K_vL3wAcX0CHJY_8kZx2xyW8Y3OWdi-N806GUOgEAAwBoTFzJtmfAeGU/s320/3-%20end%20cover%20off.jpg" width="320" /></a></div><p>Below is a picture of the dynamotor with its end covers removed:</p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgyzLMkZuqstkznpi12RfZoUpOOYUjewGNbyrjarG3k94OzCOs8DnXV8NydM86s87_5wG_BmzGHEEJGzYZb9mFg7kDE25RhovinjbviU9xIY2mkyovpplPemCZ85b0UJr9-Yt7vSYGCI3O6Nv-4tYw-tjIzF0eLsP-4nY1WQeJ1WzWmEFDCVpjerw9mS8Q/s640/IMG_9740.jpg" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="480" data-original-width="640" height="240" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgyzLMkZuqstkznpi12RfZoUpOOYUjewGNbyrjarG3k94OzCOs8DnXV8NydM86s87_5wG_BmzGHEEJGzYZb9mFg7kDE25RhovinjbviU9xIY2mkyovpplPemCZ85b0UJr9-Yt7vSYGCI3O6Nv-4tYw-tjIzF0eLsP-4nY1WQeJ1WzWmEFDCVpjerw9mS8Q/s320/IMG_9740.jpg" width="320" /></a></div><p>Because I will be cleaning and relubricating the bearings without removing the bearings from the dynamotor assembly, I first removed the four brushes (two on the L.V. side and two on the H.V. side) so that I would not get degreaser or grease/oil on them. When you remove your dynamotor's brushes, keep track of which brush goes into which brush holder, and note the orientation of each. Upon reassembly of the dynamotor you will need to return the brushes to their original brush-holders and in the proper orientation!</p><p>As an aid, note that the end brackets have embossed upon them "HV" or "LV" to identify which end of the dynamotor it is, along with "+" or "-" near each of the brush caps, as shown in the four images, below:<br /></p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg1_Ftpx2vIt0Cd421ujhHEkL1cGsQFpiIu7k1MdQ59rEfxrsfaO7MW-cf-k8uyKK3zUVUHNJTAQStriyYhmsZIUMxuIzrK0Wey0CoSh9YJ-ic6H9VfqrzokAlB_zrBqRCsNOe8HV3uclbj06i9jVH67cPAeKqSDdHHbuQBTCo5sc7Ga9dHdWLeIRfhpaE/s231/embossed%20LV%20plus.jpg" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="167" data-original-width="231" height="167" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg1_Ftpx2vIt0Cd421ujhHEkL1cGsQFpiIu7k1MdQ59rEfxrsfaO7MW-cf-k8uyKK3zUVUHNJTAQStriyYhmsZIUMxuIzrK0Wey0CoSh9YJ-ic6H9VfqrzokAlB_zrBqRCsNOe8HV3uclbj06i9jVH67cPAeKqSDdHHbuQBTCo5sc7Ga9dHdWLeIRfhpaE/s1600/embossed%20LV%20plus.jpg" width="231" /></a></div><p><br /></p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiN09PGz6URBvP0pS9nv8aMi-dMpcDS0OGQQEV9X07gaq_4VT_xgk827EF8TFkf6VW67uV8WxQ8Z7xEDLt0Qk-xBC18iDlmDO-Wd9QftZWosoLOCa1lcfANNrNJ_JlGQfenjT9vopV5Ye44x-oNxI0VH508rP3Ayl4ec-fDT1KWqtVPKby4QhpsxsMACSk/s225/embossed%20LV%20minus.jpg" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="225" data-original-width="205" height="225" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiN09PGz6URBvP0pS9nv8aMi-dMpcDS0OGQQEV9X07gaq_4VT_xgk827EF8TFkf6VW67uV8WxQ8Z7xEDLt0Qk-xBC18iDlmDO-Wd9QftZWosoLOCa1lcfANNrNJ_JlGQfenjT9vopV5Ye44x-oNxI0VH508rP3Ayl4ec-fDT1KWqtVPKby4QhpsxsMACSk/s1600/embossed%20LV%20minus.jpg" width="205" /></a></div><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhIXagnCb2oM7ToC9inG62C0k5-JzCfLXegLjL2TmPhk9DlN-wOiKfY1kyR6n-11lQ-AbzNJzMh5FVu3zvhTDD-PinUwUOlPsOFMbsOZmNC8h6FKu6nZjcit35ppRH5RJub2jdyk8dAv1nREmWPHe4VjaCKEvFRyL_ewIo1gSa_S94FR6Y5Qdj7hcNGO8s/s222/embossed%20HV%20minus.jpg" style="margin-left: 1em; margin-right: 1em;"><br /><img border="0" data-original-height="185" data-original-width="222" height="185" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhIXagnCb2oM7ToC9inG62C0k5-JzCfLXegLjL2TmPhk9DlN-wOiKfY1kyR6n-11lQ-AbzNJzMh5FVu3zvhTDD-PinUwUOlPsOFMbsOZmNC8h6FKu6nZjcit35ppRH5RJub2jdyk8dAv1nREmWPHe4VjaCKEvFRyL_ewIo1gSa_S94FR6Y5Qdj7hcNGO8s/s1600/embossed%20HV%20minus.jpg" width="222" /></a></div><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg2fMjW8_-9e32V3DQV7QXlDWK1MC0-C09ZZ6XLfDv6R2wNsxehaLHd829Nr7T9IhE4SOtpZ4TK4Ap7xuYBxRL8nfROOqwRW9pjBURCTwWWR4jmzk6efjsQeAvrH4Zs5kHds2c1d_xcPPqHWsayg7u6yGiPWW9c2TD_7Ofkh00lKSH3tzvd5RCPWxwGXvs/s214/embossed%20HV%20plus.jpg" style="margin-left: 1em; margin-right: 1em;"><br /><img border="0" data-original-height="199" data-original-width="214" height="199" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg2fMjW8_-9e32V3DQV7QXlDWK1MC0-C09ZZ6XLfDv6R2wNsxehaLHd829Nr7T9IhE4SOtpZ4TK4Ap7xuYBxRL8nfROOqwRW9pjBURCTwWWR4jmzk6efjsQeAvrH4Zs5kHds2c1d_xcPPqHWsayg7u6yGiPWW9c2TD_7Ofkh00lKSH3tzvd5RCPWxwGXvs/s1600/embossed%20HV%20plus.jpg" width="214" /></a></div><div><br /></div><div>Remove the brushes. Note which one goes into which brush holder, and the orientation of each brush. (In my dynamotors the brushes each have either a "+" or a "-" engraved on one side of each brush, and they have always been oriented in their respective holders with the engraving facing up, not down. So all I really need to remember is which two brushes go into the L.V. side and which two go into the H.V. side.).</div><div><br /></div><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhdJl7ZWlTXrVl8EKYgmS5czp_DGlaUotcgE4z8YaNBet9rihAB1vB4fYOAWQyxJ6hbbSO3q0_3VFkFfZrsHRwVyQiOZkDytfOnov_OJlT4Yx-hOhUGSV4qGxOqXpoMo6VxUM9v9RqwKc2YjfUrw7GmCvlq1lhk67LUJ5j83lxj3fZv_njEmFMpKFvz19g/s640/3p5%20removing%20the%20brushes.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="480" data-original-width="640" height="240" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhdJl7ZWlTXrVl8EKYgmS5czp_DGlaUotcgE4z8YaNBet9rihAB1vB4fYOAWQyxJ6hbbSO3q0_3VFkFfZrsHRwVyQiOZkDytfOnov_OJlT4Yx-hOhUGSV4qGxOqXpoMo6VxUM9v9RqwKc2YjfUrw7GmCvlq1lhk67LUJ5j83lxj3fZv_njEmFMpKFvz19g/s320/3p5%20removing%20the%20brushes.png" width="320" /></a></div><p>Next, remove the two screws of the end shield that covers the bearing, as shown in the image, below. Then remove the end shield itself and any washers (also known as shims) that were between it and the top of the bearing. (Note that on the L.V. side of my dynamotor there were four washers between the end-cap and the top of the bearing).</p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjSHv4diyi07BOS30Yx_S_CbzKim_hMqcHERWRMpaZbIufN017-JIuRP3SHVoaKnwoQRGBWYIkcEEpsNNkF2a-wnxKAoSt_U1A4l5XM8C_Q2YUbofONSOfMqY8RwM6yMoWyFVZDdrC0LYjT0jqp1mV9XN09eH9SMRfh2HMvxuRMHnrojdS_XLOHTZHOYHU/s914/5%20-%20removing%20end%20cap.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="794" data-original-width="914" height="278" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjSHv4diyi07BOS30Yx_S_CbzKim_hMqcHERWRMpaZbIufN017-JIuRP3SHVoaKnwoQRGBWYIkcEEpsNNkF2a-wnxKAoSt_U1A4l5XM8C_Q2YUbofONSOfMqY8RwM6yMoWyFVZDdrC0LYjT0jqp1mV9XN09eH9SMRfh2HMvxuRMHnrojdS_XLOHTZHOYHU/s320/5%20-%20removing%20end%20cap.png" width="320" /></a></div><div><br /></div><div>While work continues, let the end-cap and washers soak in a bit of the degreaser fluid.</div><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhLjBDLLUjtWHAMLY_SWXG8Vn3Txfk3r9gPmzAUGhwI5BdnJWuU5hKvgRLEihChkmRGWIT3d5v_HNLg9SI9IxDnQ5y0Td73E9sOlJa8HbEapcMe7wHpVNUDWHPLJFFdeaLzHS-uw3KWvOIqvabYpoBhvuZ7qFK0BoK26d0GEHuscYFy3yijeFfks-NGvPQ/s481/6%20-%20soak%20washers%20in%20degreaser.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="324" data-original-width="481" height="216" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhLjBDLLUjtWHAMLY_SWXG8Vn3Txfk3r9gPmzAUGhwI5BdnJWuU5hKvgRLEihChkmRGWIT3d5v_HNLg9SI9IxDnQ5y0Td73E9sOlJa8HbEapcMe7wHpVNUDWHPLJFFdeaLzHS-uw3KWvOIqvabYpoBhvuZ7qFK0BoK26d0GEHuscYFy3yijeFfks-NGvPQ/s320/6%20-%20soak%20washers%20in%20degreaser.png" width="320" /></a></div><div><br /></div><div>Next, take a look at the bearing grease covering the bearing. The bearing-grease in my dynamotor was, hard, dried-out. I used a toothpick to remove the grease from the top of the bearing. (Note that this will NOT remove all of the dried grease!)</div><div><br /></div><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhkguYBMGI-c4IWrT7Wxr6ldEMlwnA9-A2fBoNoPk8T4fSVWQVFDxfgytgHwDzFcterHutCSETPZX9u3hSwx4ZfP-io-I9N23CA1cZTrbz0svjRsCTJ-wPuWnaLdjCNWTAMwZ4Qu7-5caja12GiLfJESyx9os7iYU_3qwYuFtJo71uL4aR1kDrPkWyL__M/s393/7-%20pick%20out%20soldified%20grease.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="330" data-original-width="393" height="269" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhkguYBMGI-c4IWrT7Wxr6ldEMlwnA9-A2fBoNoPk8T4fSVWQVFDxfgytgHwDzFcterHutCSETPZX9u3hSwx4ZfP-io-I9N23CA1cZTrbz0svjRsCTJ-wPuWnaLdjCNWTAMwZ4Qu7-5caja12GiLfJESyx9os7iYU_3qwYuFtJo71uL4aR1kDrPkWyL__M/s320/7-%20pick%20out%20soldified%20grease.png" width="320" /></a></div><div><br /></div>After I had removed what grease I could with a toothpick, I used a toothbrush and degreaser fluid (in this case paint thinner) to scrub away the remaining hardened grease. This process will take some time. Continually turn the the armature while you scrub -- much of the bearing (and its grease) is not easily accessible.<div><br /></div><div>Note that while scrubbing I keep the bearing always pointed down to prevent the degreasing fluid from flowing onto the armature and possibly affecting the armature's lacquer, etc.</div><div><br /></div><div><b>Also, I recommend doing this degreasing outside (to, hopefully, minimize inhaling of fumes), and wear protective gear (e.g. gloves).</b><br /><div><br /></div><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhEVhRU_ywVmrU6DHlC-Rhmkd174neqrCrWMBrBr-269SHCH2CnuLhvXl0XrpXQM4eMCrn9mOqyPxuz8E82BWFQvkfFKh2Yg3uXMSTtB4u5TuzHMnFVcf-AtyBJfyPx60X_kPvQwIECcTkawZ7-Lmrkcmd543Gk3MWkHvi7RCASpYxam3USxKCL57zpcY4/s640/8%20-%20%20scrubbing%20bearings.jpg" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="480" data-original-width="640" height="240" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhEVhRU_ywVmrU6DHlC-Rhmkd174neqrCrWMBrBr-269SHCH2CnuLhvXl0XrpXQM4eMCrn9mOqyPxuz8E82BWFQvkfFKh2Yg3uXMSTtB4u5TuzHMnFVcf-AtyBJfyPx60X_kPvQwIECcTkawZ7-Lmrkcmd543Gk3MWkHvi7RCASpYxam3USxKCL57zpcY4/s320/8%20-%20%20scrubbing%20bearings.jpg" width="320" /></a></div><br /><div>Periodically use a clean cloth to examine at the degreasing fluid dripping from the bearing (I will set the bearing on top of a cloth to capture the dripping fluid, as shown below). The goal is fluid that is clear and not be dirty.</div><div><br /></div><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg4ddD0taMi16WOe8QUsHoxvGk5q3St70qQD9Z_p1ibkcWVfKQTBGT6qksgwTWSAOus_tl1yIyoCIN5UqClQvwAK24QoNYDdA32Xhqm_sgm9JG90GVp9PFo40Hz-BFrqscp_z9MrTDlolsMirN_G4LcB1crUZT_TD9KG8tSv0xyN_b1rmVFxG_VH1Cc_uE/s640/9%20-%20drain%20bearings.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="640" data-original-width="480" height="320" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg4ddD0taMi16WOe8QUsHoxvGk5q3St70qQD9Z_p1ibkcWVfKQTBGT6qksgwTWSAOus_tl1yIyoCIN5UqClQvwAK24QoNYDdA32Xhqm_sgm9JG90GVp9PFo40Hz-BFrqscp_z9MrTDlolsMirN_G4LcB1crUZT_TD9KG8tSv0xyN_b1rmVFxG_VH1Cc_uE/s320/9%20-%20drain%20bearings.png" width="240" /></a></div><br /><div>Below, after I finished the L.V. side, I started scrubbing away the grease on the H.V. side bearing. The piece of cloth helps prevent the degreaser from splattering on me, and the cloth is on top of a clay flower-pot base (to keep the degreaser-soaked cloth off of the table-top).</div><div><br /></div><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgzfIOpjyj6PBg3FF8W-_RAhZG-PxR13SrN2xyvr8wv-CRq-iwBJ5BkourZFo0moe4ycSKpmaYJynRwZSXUgLDlNJDut4HxBopBiNOIv9dPBiJqiMKacluJEK1LeI1wkL4O12Y4U2xhoD31eDOIpv4if8uEmXmnRVvLxs701qV1S0BRDD8B351sZmBErG0/s567/10%20-%20cleaning%20the%20bearings%20on%20the%20HV%20side.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="382" data-original-width="567" height="216" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgzfIOpjyj6PBg3FF8W-_RAhZG-PxR13SrN2xyvr8wv-CRq-iwBJ5BkourZFo0moe4ycSKpmaYJynRwZSXUgLDlNJDut4HxBopBiNOIv9dPBiJqiMKacluJEK1LeI1wkL4O12Y4U2xhoD31eDOIpv4if8uEmXmnRVvLxs701qV1S0BRDD8B351sZmBErG0/s320/10%20-%20cleaning%20the%20bearings%20on%20the%20HV%20side.png" width="320" /></a></div><div><br /></div>By the way, the image below shows the end-shield and single spring washer that covered the bearing on the high-voltage side of the dynamotor.<br /><div><br /></div><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh2ItWyFKHwAwTBumYRO8AgwKxZl1fKoLdMijhFpHVc9aNJpjjRWZ1-XB2S5MVO4Lp3q6lVgV0NA5-eh1pHNsuRy2LQgBmsRMUe8VDUdlzo3yQ3A4pklVJrp_C9tqcUKmwanQGvdFXMcZdQMeQ4govMMBDJwni97TnyiZwB0-yrrjaC73-CVqKHNyFHsHA/s530/11%20-%20HV%20spring%20washer.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="449" data-original-width="530" height="271" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh2ItWyFKHwAwTBumYRO8AgwKxZl1fKoLdMijhFpHVc9aNJpjjRWZ1-XB2S5MVO4Lp3q6lVgV0NA5-eh1pHNsuRy2LQgBmsRMUe8VDUdlzo3yQ3A4pklVJrp_C9tqcUKmwanQGvdFXMcZdQMeQ4govMMBDJwni97TnyiZwB0-yrrjaC73-CVqKHNyFHsHA/s320/11%20-%20HV%20spring%20washer.png" width="320" /></a></div><div><br /></div><div><br /></div><div><b><u>An Oops Moment!!!</u></b></div><div><br /></div><div>During the cleaning process I accidentally dropped the dynamotor onto the ground, denting one of the pot metal end brackets and bending one of the clamp screws (that is, one of the two long screws that hold the two end-brackets together). To repair the damage, I further disassembled the dynamotor in an attempt to un-dent and unbend the parts I had mucked up...</div><div><br /></div><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgxPo_Pb8lmz4nzrhZkg2W8ulHYUiaPxovUWNG4gmw2kJDv_4aN4Y06z961fa6MXO7775bnPIyXAU-miygl8svFuElZ6gCXVGzCW6mX3s8D-DI2CFF4PDoeIS1KA3dXKGZizNV7GCEHGXm97J2JWXUPei5GBTQ_HxqsFlcSJPVCIQhNEwsUvpHrrozDci8/s640/12%20-%20oops%20disassembly.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="480" data-original-width="640" height="240" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgxPo_Pb8lmz4nzrhZkg2W8ulHYUiaPxovUWNG4gmw2kJDv_4aN4Y06z961fa6MXO7775bnPIyXAU-miygl8svFuElZ6gCXVGzCW6mX3s8D-DI2CFF4PDoeIS1KA3dXKGZizNV7GCEHGXm97J2JWXUPei5GBTQ_HxqsFlcSJPVCIQhNEwsUvpHrrozDci8/s320/12%20-%20oops%20disassembly.png" width="320" /></a></div><br /><div>With the dynamotor disassembled, here's a look at the dynamotor's armature:</div><div><br /></div><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjRgL4ZqOxhKOSUMDiC5GqbTu-24rpcax7nV44q3uB_0fW_jKRi2eSKdWTMCRl_AAebV6dvpVrd350cpM1ZDEfdFqogBarLZvg1EhLBgd9ZOeU9d18xzvTCWziIaAMyS5AY83586LyIkcLa2I5m63XiO1JOS0_NCj0_pU4jEsjMIRpbdeQYBUV3q0QpiMo/s504/14%20-%20dm32%20armature.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="314" data-original-width="504" height="199" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjRgL4ZqOxhKOSUMDiC5GqbTu-24rpcax7nV44q3uB_0fW_jKRi2eSKdWTMCRl_AAebV6dvpVrd350cpM1ZDEfdFqogBarLZvg1EhLBgd9ZOeU9d18xzvTCWziIaAMyS5AY83586LyIkcLa2I5m63XiO1JOS0_NCj0_pU4jEsjMIRpbdeQYBUV3q0QpiMo/s320/14%20-%20dm32%20armature.png" width="320" /></a></div><br /><div><br /></div><div><b><u>Relubricating the Bearings:</u></b></div><div><br /></div><div>To relubricate the bearings, the <a href="https://hangarthirteen.org/wp-content/uploads/2020/08/16-40SCR274-5-Maintenance-Instructions-for-SCR-274-N.pdf">Maintenance Handbook</a> recommends, "Apply three or four drops of a light machine oil to the balls and repack the outer side of the bearing with a small amount of AN-G-15 grease. Add only enough grease to cover the bearing. Do <b>not</b> pack the bearing full."</div><div><br /></div><div>I used <i>3-in-One</i> oil (that I had on hand) for the recommended "light machine oil. I did not have any AN-G-15 grease, so instead I searched Amazon and found a polyurea grease, manufactured by Gennel (China?), which claims to be applicable to high-speed fan bearings (which turn at high speeds like dynamotors, right?). Unfortunately, there is no technical info on this grease, so I'm not sure how well it will work. Time will tell if my choices for oil and grease were good!</div><div><br /></div><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiPmZKNpeGH3YVxOv-F3FJuOKvuqRsYEIWbc0MdCoQcLgXaeOui5JiPNbGY5J-hllQB7vhuHGvkBmFN9VLPRJztCwhCYl6zIe7FbKNX7QVd1MYYyX0bb7Aqax0tRHwx0o71J3-_OEmOct84IFAh1s24GlV6CJwbRVMddCZe3rY0ULDVthsOBU4yVeI-_80/s548/15%20-%20relubing.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="548" data-original-width="479" height="320" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiPmZKNpeGH3YVxOv-F3FJuOKvuqRsYEIWbc0MdCoQcLgXaeOui5JiPNbGY5J-hllQB7vhuHGvkBmFN9VLPRJztCwhCYl6zIe7FbKNX7QVd1MYYyX0bb7Aqax0tRHwx0o71J3-_OEmOct84IFAh1s24GlV6CJwbRVMddCZe3rY0ULDVthsOBU4yVeI-_80/s320/15%20-%20relubing.png" width="280" /></a></div><div><br /></div>After the bearings are relubricated with new oil and grease, the washers and end-caps are re-installed on both the H.V. and L.V. ends of the dynamotor<div><div><br /></div><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhlubLKE9FM1tdKncM8JxH65ePnb8PkdsRTIKJ2KKb9dvfRomGa_c-Oz5Ennuxi_2G-SfMj1yOHCIcRGKyX3QxcgzTEjvqR8x8323Et4BlP9eAJwG6b7w0FaRwJM_4YbcxMGn05dqswPHroqHFvx56_ReA2d1CBrjnIWFNfq5rb5nJmOSrolsUP2AIiZA0/s475/16%20-%20reassembly.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="463" data-original-width="475" height="312" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhlubLKE9FM1tdKncM8JxH65ePnb8PkdsRTIKJ2KKb9dvfRomGa_c-Oz5Ennuxi_2G-SfMj1yOHCIcRGKyX3QxcgzTEjvqR8x8323Et4BlP9eAJwG6b7w0FaRwJM_4YbcxMGn05dqswPHroqHFvx56_ReA2d1CBrjnIWFNfq5rb5nJmOSrolsUP2AIiZA0/s320/16%20-%20reassembly.png" width="320" /></a></div><br /><div>Finally, the end-cap screws are locked down with some nail polish.</div><div><br /></div><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiTgiM6_2OoonRy12urzFewnSmZqCFPI6Q9BGnYv1RnXBfQIRiFdlsPwYyropqhC7Nn6IMS5CRvFPK_2R73BI6jAv0hQZ8OrIvo8xxQAPze0jPEtit1G9GG0N1A5qdt5B-N7bmcq0U8Q9V2jX72cCJA8fo1MK2UuUpdHMz2MghR5LT5g-MBNdwtg11BCfI/s500/17%20-%20lock%20in%20screws.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="417" data-original-width="500" height="267" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiTgiM6_2OoonRy12urzFewnSmZqCFPI6Q9BGnYv1RnXBfQIRiFdlsPwYyropqhC7Nn6IMS5CRvFPK_2R73BI6jAv0hQZ8OrIvo8xxQAPze0jPEtit1G9GG0N1A5qdt5B-N7bmcq0U8Q9V2jX72cCJA8fo1MK2UuUpdHMz2MghR5LT5g-MBNdwtg11BCfI/s320/17%20-%20lock%20in%20screws.png" width="320" /></a></div><br /><div>(Note: If you look closely at the picture, above, you will see that the two end-cap screws are Philips-head screws, not slotted screws. I somehow lost the original two screws (probably when I was banging on the pot-metal end bracket trying to straighten it), and I had to dig through my junk box to find two other screws that would fit).</div><div><br /></div><div><br /></div><div><b><u>Other Notes:</u></b></div><div><br /></div><div><b>1. SCR-274 vs. AN/ARC-5:</b> Per <a href="https://en.wikipedia.org/wiki/AN/ARC-5">Wikipedia</a>, there were three separate series of Command-Set gear: 1) the ARA-ATA series, 2) the SCR-274-N series, and 3) the AN/ARC-5 series. These three series have now become known, informally, under the umbrella term: ARC-5.</div><div><br /></div><div>Note that equipment designs can differ between series. For example, AN/ARC-5 receivers seem to use a 12SF7 tube for the second I.F. stage (this tube has an internal diode that is used to derive an AVC voltage), while the SCR-274-N series receivers seem to use a 12SK7 for the second I.F. (i.e. no diode, and thus no AVC).</div><div><b><br /></b></div><div><b>2.</b> <b>14V vs 28V dynamotor</b>: At least one version of the AN/ARC-5 receivers was set up to run from 14 volts in lieu of 28 volts. If your receiver came with a dynamotor, don't assume that it matches your receiver's voltage. Someone along the way might have rewired your receivers filaments, or replaced 12V tubes with 6V tubes, or added an unknown dynamotor.</div><div><br /></div><div>To check filament wiring, remove the receiver's bottom plate and verify if the the 12 Volt tube filaments are all wired in parallel (e.g. to be used with a 14V dynamotor), or if pairs of tube filaments are wired in series (i.e. to be used with a 28 V dynamotor). </div><div><br /></div><div>Similarly, if your dynamotor does not have any identifying information on it (as was the case for one of my three dynamotors), remove the L.V. end-cap and take a look at the voltage specification printed on the end of the armature:</div><div><br /></div><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi6FLmQqFC1J_eqeMRTAHqUy1Vg4mYzGoesgOiFIpzd4AKYszUc6Du4nqo59WCRbckeovXLcVKkTF2jlOtJIcI1tcyNfJv4135Gr6pr98QjhuvLyTWFWzl62kmmJp9I9tNiDEOoRtoFWP1wRtjDKYoIVs6XRLsFKNHD1Dvr-w944ZsxgG4orARUN7vtBSg/s499/armature%20voltage.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="442" data-original-width="499" height="283" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi6FLmQqFC1J_eqeMRTAHqUy1Vg4mYzGoesgOiFIpzd4AKYszUc6Du4nqo59WCRbckeovXLcVKkTF2jlOtJIcI1tcyNfJv4135Gr6pr98QjhuvLyTWFWzl62kmmJp9I9tNiDEOoRtoFWP1wRtjDKYoIVs6XRLsFKNHD1Dvr-w944ZsxgG4orARUN7vtBSg/s320/armature%20voltage.png" width="320" /></a></div><br /><div><br /></div><p><b><u>Resources:</u></b></p><p>A Google search for various ARC-5 terms will return a plethora of hits. Below are some of the sites I found useful.</p><p><a href="https://hangarthirteen.org/wp-content/uploads/2020/08/16-40SCR274-5-Maintenance-Instructions-for-SCR-274-N.pdf">T.O. 12R2-3SCR274-2 (formerly 16-40SCR274-5). Handbook, Maintenance Instructions, Radio Set SCR-274-N (revised 25 July 1956)</a>. Note that this handbook only applies to the SCR-274-N equipment (e.g. receivers BC-454, BC-455, etc.). As such, for example, all receiver schematics show the second I.F. tube as being a 12SK7, not a 12SF7, as seems to have been used in the ARC-5 (not SCR-274) series of receivers. 228 pages.</p><p><a href="http://www.vmarsmanuals.co.uk/archive/1181_ARC5_Maintenance.pdf">AN16-30ARC5-2. Handbook, Maintenance Instructions, AN/ARC-5 Aircraft Radio Equipment, LF, MF, HF Components</a>. Note that this handbook only applies to the AN/ARC-5 equipment (e.g. receivers R-25, R-26, etc.). As such, for example, all receiver schematics show the second I.F. tube as being a 12SF7, not a 12SK7, as seems to have been used in the BC-series of receivers. 354 pages.</p><p><a href="http://radiomanual.info/schemi/Surplus_NATO/AN-ARC-5_alignment_procedure_NAVWEPS16-30ARC5-501_1949.pdf">NAVWEPS 16-30ARC5-501. Handbook, Bench Test and Alignment Procedure, Radio Equipment AN/ARC-5</a>. Specifically for the AN/ARC-5 series, but can probably be used for the SCR-274 series (e.g. receiver I.F. schematic shows 12SF7 as the second I.F. tube). 56 pages.</p><p><a href="https://www.youtube.com/watch?v=iNJVMzGYUW4">Video, ARC-5 Receiver Dynamotor Repair and Demonstration</a>. Companion blog-post is here: <a href="https://hackaday.com/2019/12/12/wwii-aircraft-radio-roars-to-life-what-it-takes-to-restore-a-piece-of-history/">https://hackaday.com/2019/12/12/wwii-aircraft-radio-roars-to-life-what-it-takes-to-restore-a-piece-of-history/</a></p><p><a href="https://www.youtube.com/watch?v=crUNcNsiZEA">Video, Basic Dynamotor Maintenance</a>. (He describes the dynamotor for a BC-603, which looks similar to an ARC-5 Dynamotor).</p><p><a href="https://aafradio.org/docs/Dynamotors.html">Dynamotor Refurbishment</a>. Lots of good information on refurbishing old dynamotors.</p><p><a href="http://www.ad5x.com/images/Articles/ARC5convert.pdf">Adapting an ARC-5 Receiver for Ham use</a>. AD5X.</p><p><a href="file:///C:/Users/jeff/Documents/aa%20-%20Projects/Radios/Military/Mil%20-%20Command%20Set/5702_RSGB_Command_Receivers.pdf">Stevens (G2BVN),</a> <a href="file:///C:/Users/jeff/Documents/aa%20-%20Projects/Radios/Military/Mil%20-%20Command%20Set/5702_RSGB_Command_Receivers.pdf">The Command Set Receivers, <i>R.S.G.B. Bulletin</i>, February, 1957.</a> Discusses the command set receivers (does not mention the variant with the 12SF7 as the second I.F.) and modifications thereof.</p><p>Stanley, J., K4ERO, "Using Unmodified Command Set ARC-5 Radios on the Ham Bands," <i>QST</i>, January 2016</p><p><br /></p><p><b><u>Standard Caveat:</u></b></p><p>As always, I might have made a mistake in my equations, assumptions, drawings, or interpretations. If you see anything you believe to be in error or if anything is confusing, please feel free to contact me or comment below.<br /><br />And so I should add -- this information is distributed in the hope that it will be useful, but WITHOUT ANY WARRANTY; without even the implied warranty of MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.</p></div></div>Jeffhttp://www.blogger.com/profile/04853314106806116765noreply@blogger.com0tag:blogger.com,1999:blog-2257458838727315711.post-31556241409904266702023-10-17T15:29:00.009-07:002023-10-25T10:23:11.974-07:00Correcting QST's LPi Antenna's Unun Schematic (QST, November, 2023)<p>The November, 2023 issue of QST has an interesting antenna article titled, "The LPi Antenna," which describes a broadband antenna covering 21-29.7 MHz with an SWR of less than 3:1.</p><p>Given its asymmetrical shape and its high feed-point impedance, the LPi antenna is fed via both a common-mode choke (to reduce common-mode currents that might be introduced due to antenna asymmetry) and an unun configured to be a 4:1 impedance transformer (to transform a high load impedance to a lower impedance, e.g. a 200 ohm load impedance to 50 ohms). </p><p>A schematic representing these two transformers is shown in Figure 2 of the article:</p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi19NWWIKXJSXRVvJWLvJB4KpCI_cr9OYNFlB07RbqUWTpFMiHnC_ly1Oqx4Wk9cJY1c9wVUdikTVEPKA7vjzclYRZB8l-hKkIctoHKHVnABUNStG4R2rF-aMY1UYwjTqwzsiNCxeOy5cIF6DvDiIsADjOLz7b3R831NJMqkhx7RPJ8pfYz41WwPpgJxU0/s1314/Schematic,%20Balun%20plus%20Unun.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="792" data-original-width="1314" height="193" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi19NWWIKXJSXRVvJWLvJB4KpCI_cr9OYNFlB07RbqUWTpFMiHnC_ly1Oqx4Wk9cJY1c9wVUdikTVEPKA7vjzclYRZB8l-hKkIctoHKHVnABUNStG4R2rF-aMY1UYwjTqwzsiNCxeOy5cIF6DvDiIsADjOLz7b3R831NJMqkhx7RPJ8pfYz41WwPpgJxU0/s320/Schematic,%20Balun%20plus%20Unun.png" width="320" /></a></div><div class="separator" style="clear: both; text-align: center;"><br /></div>As drawn in the article's Figure 2, this circuit cannot work because both of the 'dotted' terminals of the 4:1 unun are driven by the same signal. <div><br /></div><div><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj_KcGFJehLU7SqAnWqJ7o7L99MUpxq_3J8WdSrS5okUYvReKfvKGxLNUKyZUTE82lR3k_IuXJaj53l11MYURDdFEj1v9FPgDbUKp-eawj1uCC99eYmIC6fr7Q2-2gzVOYs984ADcWlc7kOVpBLh8lGb76-J0DjcKTz5tjOTu9Jjw5DGa_VNR1dv_Pfr7s/s1314/Schematic,%20circled%20dotteed%20terminals%20Balun%20plus%20Unun.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="792" data-original-width="1314" height="193" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj_KcGFJehLU7SqAnWqJ7o7L99MUpxq_3J8WdSrS5okUYvReKfvKGxLNUKyZUTE82lR3k_IuXJaj53l11MYURDdFEj1v9FPgDbUKp-eawj1uCC99eYmIC6fr7Q2-2gzVOYs984ADcWlc7kOVpBLh8lGb76-J0DjcKTz5tjOTu9Jjw5DGa_VNR1dv_Pfr7s/s320/Schematic,%20circled%20dotteed%20terminals%20Balun%20plus%20Unun.png" width="320" /></a></div><br /><div>Clearly this effect was not the intent of the author and thus the schematic was incorrectly drawn. (Note: although no current will flow in an ideal unun if its two dotted terminals are connected together, in fact the unun is not an ideal transformer, so some current will flow through it even if both dotted terminals are tied together, but in this case the device will appear to be a series impedance and <u>not</u> an impedance transformer.)</div><div><br /></div><div>What should be the correct schematic? </div><div><br /></div><div>First, let's look at how the transformer dot convention defines the directions of a transformer's current. Per the dot convention, and given a two-winding transformer, if current is flowing 'into' the dotted terminal of Port 1, then it will flow 'out of' the dotted terminal of Port 2. </div><div><br /></div><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEidIRvbHjUZo787_D_j6d6JyC9gDKNTL3GpUNJ5N-BELlFRycG6IMhJRxmz1tdsX6IZR2bXhaY0UCOowqEZCrZfIh-uSaCpdSaGQYyZSqz79wK6-zIkJW6VBDpTUd9JA5uVUFOTUsbS9bNe51UQBqerJek91yEKwRrJtQbD8ytbXiB8PVueMQ61qfzT6d4/s779/Current%20flow%20per%20Dot%20Convention.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="521" data-original-width="779" height="214" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEidIRvbHjUZo787_D_j6d6JyC9gDKNTL3GpUNJ5N-BELlFRycG6IMhJRxmz1tdsX6IZR2bXhaY0UCOowqEZCrZfIh-uSaCpdSaGQYyZSqz79wK6-zIkJW6VBDpTUd9JA5uVUFOTUsbS9bNe51UQBqerJek91yEKwRrJtQbD8ytbXiB8PVueMQ61qfzT6d4/s320/Current%20flow%20per%20Dot%20Convention.png" width="320" /></a></div><br /><div>Given this convention, if both dotted terminals were tied together and driven in common, no current would flow through the transformer windings. We can show this result algebraically, in the figure, below:</div><div><br /></div><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEha1Ud7noBN_wN0-f__8IGQDiPmHDnfO1i8Uz9B3I-OCS8ElpcHJ76ucR_s2EktfufJsU8WoZ378YxSyE9IdWE-bRO9eL8F-pu8DRL3NRt0VC0FL-3RHnFUQhWMz04hhS20zvx3K01E4t5l-tet2174hm1ptACXjN6pKBH5b5fmHpJmZuRK-eaHujo7_GI/s1461/Schematic,%20Current%20Flow,%20Balun%20plus%20Unun.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="971" data-original-width="1461" height="213" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEha1Ud7noBN_wN0-f__8IGQDiPmHDnfO1i8Uz9B3I-OCS8ElpcHJ76ucR_s2EktfufJsU8WoZ378YxSyE9IdWE-bRO9eL8F-pu8DRL3NRt0VC0FL-3RHnFUQhWMz04hhS20zvx3K01E4t5l-tet2174hm1ptACXjN6pKBH5b5fmHpJmZuRK-eaHujo7_GI/s320/Schematic,%20Current%20Flow,%20Balun%20plus%20Unun.png" width="320" /></a></div><div><br /></div><div>What is the correct schematic for the choke/unun circuit? The dotted terminals for the two windings should be at opposite ends of the unun transformer, as shown in the diagram, below:</div><div><br /></div><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjrpkR5TGu1r0W1NsbcwLXPbo59-F6IY9EkKzhqOPmq_CSLVCcwsx0xZEf7vZg9arXKeytqfGLGkQ2lcVhmm7RxIEfdGhSNLDg4F0CnQH_kLUwWu6m6MrDwgginaixHrhv0Vbvo8K6wXlWQ21rVpte_YYR7vE56CF1sIP3-EIR_Z1zKHveNoyT7nq2ok20/s1447/Schematic,%20Corrected%20Current%20Flow,%20Balun%20plus%20Unun.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="967" data-original-width="1447" height="214" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjrpkR5TGu1r0W1NsbcwLXPbo59-F6IY9EkKzhqOPmq_CSLVCcwsx0xZEf7vZg9arXKeytqfGLGkQ2lcVhmm7RxIEfdGhSNLDg4F0CnQH_kLUwWu6m6MrDwgginaixHrhv0Vbvo8K6wXlWQ21rVpte_YYR7vE56CF1sIP3-EIR_Z1zKHveNoyT7nq2ok20/s320/Schematic,%20Corrected%20Current%20Flow,%20Balun%20plus%20Unun.png" width="320" /></a></div><div><br /></div>To verify that this is schematic is correct, we can compare it to the actual wiring of the choke/unun pair as shown in Figure 3 of the QST article. </div><div><br /></div><div><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjfZg7yQy8E0rEhsxMeGIjyEOeh9ewvAAAL6QTXYECGJqYv379Qk0SPB5xc3niNvVXA6ScXVcFZAG8bK7YysMlNDouledmQ4M6R_VP35wV6Q4XX9ztT1Ic9dhqwXd1YMInPWl-k14aF_kHbLpiMyG_yM3rD_Qfza3erX-b8MifUNy2JUKONnkhZvU3KwHA/s937/Wiring,%20Balun%20plus%20Unun.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="937" data-original-width="911" height="320" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjfZg7yQy8E0rEhsxMeGIjyEOeh9ewvAAAL6QTXYECGJqYv379Qk0SPB5xc3niNvVXA6ScXVcFZAG8bK7YysMlNDouledmQ4M6R_VP35wV6Q4XX9ztT1Ic9dhqwXd1YMInPWl-k14aF_kHbLpiMyG_yM3rD_Qfza3erX-b8MifUNy2JUKONnkhZvU3KwHA/s320/Wiring,%20Balun%20plus%20Unun.png" width="311" /></a></div><br /><div>To identify each transformer's "dotted" terminals, note that because both the common-mode choke and the unun are wound with coax, both the shield and the center-conductor at <u>one end</u> of the length of coax used for a transformer's windings can be considered to be its two dotted terminals. In the image, below, I've added two dots to one end of the common-mode choke's coax winding (one dot next to the center-conductor and one dot next to the shield) and also two dots to one end of the unun's coax winding. </div><div><br /></div><div><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhck-p01lOSzGyBXNg6mvkdnHQK_Nj6Au9FKpDJWtxbta98lOPRTczD1vTM-llFwSDzFc3PmvmfF51m4IdnVFw1_Ho3epv69qOiL2zebfS6LTRbgiwQEEeUo8a6YY9cKnkoZTylMyXtrUFM66n1-g0FVMipFsQGIXIQcY2E9VfqDUSMtT1fo5SWdCmAv50/s937/Wiring,%20dotted%20terminals,%20Balun%20plus%20Unun.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="937" data-original-width="911" height="320" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhck-p01lOSzGyBXNg6mvkdnHQK_Nj6Au9FKpDJWtxbta98lOPRTczD1vTM-llFwSDzFc3PmvmfF51m4IdnVFw1_Ho3epv69qOiL2zebfS6LTRbgiwQEEeUo8a6YY9cKnkoZTylMyXtrUFM66n1-g0FVMipFsQGIXIQcY2E9VfqDUSMtT1fo5SWdCmAv50/s320/Wiring,%20dotted%20terminals,%20Balun%20plus%20Unun.png" width="311" /></a></div><br />If you examine the connections of the 'dotted' shields and center-conductors of these two coax cables in the image, above, you will see that they match the wiring in my 'corrected' schematic of the choke plus unun circuit.</div><div><br /></div><div><br /></div><div><b><u>Alternative Representations of the Corrected Schematic:</u></b></div><div><br /></div><div>Per the comment from KE5FX (see comments section), I thought it would be useful to add a couple of alternative ways to draw the same corrected schematic. Note that I'm using the "dotted" transformer terminals as they were shown in the original article; I'm just changing how the unun is wired. </div><div><br /></div><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiYciAhqPjaKsexdTOHa9hP1R4NDhvD_ajCQGJU4g1BtopIQhWiKJ3o1fKGrEjkylvhLxUI9aWvy1XI4Sr9FQp2H2O1oaO_qwnvSMGgIrAhNFq2myghW3XeqqlXCmR4UpJ2xKPwuof_Us8kgS0rsJ2qNr083pB8FRTged-FBaYoi18u6glBN_k8sPmWXdY/s1314/Schematic,%20Balun%20plus%20Unun%20Alt1.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="792" data-original-width="1314" height="193" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiYciAhqPjaKsexdTOHa9hP1R4NDhvD_ajCQGJU4g1BtopIQhWiKJ3o1fKGrEjkylvhLxUI9aWvy1XI4Sr9FQp2H2O1oaO_qwnvSMGgIrAhNFq2myghW3XeqqlXCmR4UpJ2xKPwuof_Us8kgS0rsJ2qNr083pB8FRTged-FBaYoi18u6glBN_k8sPmWXdY/s320/Schematic,%20Balun%20plus%20Unun%20Alt1.png" width="320" /></a></div><div><br /></div>If we redraw the circuit above, we can see that the unun is actually a center-tapped auto-transformer, as shown, below:</div><div><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEixMLgFQnZ0VFC8gjzRt_MKvpfEy4Q6hn4Mo8mmN47EGz1sdtW_2Db-YK1VaaMcaIA-ydCorFetswfcIikWDwmW1w4rdtRtnv5r2FHBeU7GGxIGDz5ZNe9aGiO2n3X5nqW1BStRr8g43-ceTKVUrNh8LYeA2SOFwCMOn0vZwaL4zXl_ULbG3suHkQM7X7g/s1314/Schematic,%20Balun%20plus%20Unun%20Alt%202.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="792" data-original-width="1314" height="193" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEixMLgFQnZ0VFC8gjzRt_MKvpfEy4Q6hn4Mo8mmN47EGz1sdtW_2Db-YK1VaaMcaIA-ydCorFetswfcIikWDwmW1w4rdtRtnv5r2FHBeU7GGxIGDz5ZNe9aGiO2n3X5nqW1BStRr8g43-ceTKVUrNh8LYeA2SOFwCMOn0vZwaL4zXl_ULbG3suHkQM7X7g/s320/Schematic,%20Balun%20plus%20Unun%20Alt%202.png" width="320" /></a></div><br /><div>These two representations are both functionally equivalent to my corrected schematic earlier in this blog post.</div><div><br /></div><div><p><b><u>Standard Caveat:</u></b></p><p>As always, I might have made a mistake in my equations, assumptions, drawings, or interpretations. If you see anything you believe to be in error or if anything is confusing, please feel free to contact me or comment below.<br /><br />And so I should add -- this information is distributed in the hope that it will be useful, but WITHOUT ANY WARRANTY; without even the implied warranty of MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.</p></div></div>Jeffhttp://www.blogger.com/profile/04853314106806116765noreply@blogger.com1tag:blogger.com,1999:blog-2257458838727315711.post-37967662689220114802023-08-12T09:41:00.000-07:002023-08-12T09:41:13.598-07:00L-Network Equations for Any Impedance Transformation<p>
The equations presented in this blog post can be used to design L-Networks that will transform
any complex load impedance to be any other complex impedance.</p>
<p>
The only assumptions are:</p><p>1. The L-Network is lossless (equations for lossy
L-Networks are <i>much</i> more complex, see: <a href="https://k6jca.blogspot.com/search/label/Networks%3A%20L-Network%20Design%20Equations%20for%20Lossy%20Networks">Lossy L-Network Equations</a>).</p><p>2. The 'real' component of the impedances cannot be negative.</p>
<p>
These equations represent a continuation of the equations I had derived for
transforming a complex load to a real (with no imaginary component) impedance
(see either <a href="https://k6jca.blogspot.com/2022/06/l-network-design-equations.html">L-Network Equations</a> or my article, "Correcting a Common L-Network Misconception," in the
2023 March-April issue of <i>QEX</i> magazine).
</p><p><br /></p>
<p>
<b><u>A Quick Review: The Eight L-Network Configurations:</u></b>
</p>
<p>
Note that there are 8 possible L-Network configurations, divided into 4
Series-Parallel L-Networks and 4 Parallel-Series L-Networks:
</p>
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiKrFpQCcJ8WQ5OCluor7R4OIgvZpRuyxwdxkHWOnpezKN0zLcnyU4tGjJoWNuqS616r4cS6h2-2g9ojsskBIcnWt0scnMPzm6cEt5HsAFxKElHqmukt02BgoSstqEM4Ybs3gVDdnWB9zq6c8shBypWAfd4RGVsILPAXa6L6r1qejh-wN8l_H1ucHRH/s657/fig%201,%20The%20four%20series-parallel%20l-networks.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="588" data-original-width="657" height="286" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiKrFpQCcJ8WQ5OCluor7R4OIgvZpRuyxwdxkHWOnpezKN0zLcnyU4tGjJoWNuqS616r4cS6h2-2g9ojsskBIcnWt0scnMPzm6cEt5HsAFxKElHqmukt02BgoSstqEM4Ybs3gVDdnWB9zq6c8shBypWAfd4RGVsILPAXa6L6r1qejh-wN8l_H1ucHRH/s320/fig%201,%20The%20four%20series-parallel%20l-networks.png" width="320" /></a>
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhoOd66ulu6PEbJ2vdNHRhx93Hish1hC-zv5FSYeE23GR8ATAlPzTrH1cR15LRFXZvd9W0ceUrySqSbcpTMLfH3oGQxZy9GoK-qXuHmjzWK1DQ2YCaL94PXIk1kyeVxKeJLgqfpb_pxENtyrT1o3zo3nWPjK1Vj5n8CHRM0LNU0XXDWvr46GKdZJWJC/s653/fig%202,%20The%20four%20parallel-series%20l-networks.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="590" data-original-width="653" height="289" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhoOd66ulu6PEbJ2vdNHRhx93Hish1hC-zv5FSYeE23GR8ATAlPzTrH1cR15LRFXZvd9W0ceUrySqSbcpTMLfH3oGQxZy9GoK-qXuHmjzWK1DQ2YCaL94PXIk1kyeVxKeJLgqfpb_pxENtyrT1o3zo3nWPjK1Vj5n8CHRM0LNU0XXDWvr46GKdZJWJC/s320/fig%202,%20The%20four%20parallel-series%20l-networks.png" width="320" /></a>
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<div class="separator" style="clear: both; text-align: center;"><br /></div>
<div>It is important to note that some impedance transformations can be accomplished with either Series-Parallel or Parallel-Series L-Networks -- the applicability of one network topology for a particular impedance transformation does <i>not</i> exclude the other.</div><div><br /></div><div>Let's start by defining the design equations for Series-Parallel L-Networks...</div><div><br /></div>
<p>
<b><u>Equations for Series-Parallel L-Network Solutions:</u></b>
</p>
<div>
First, we will define a Series-Parallel L-Network in terms of the <i>impedance</i> (X)
of its series-component and the <i>susceptance</i> (B) of its shunt component:
</div>
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</div>
<p>
Depending upon the particular desired impedance transformation, there will be
either two different Series-Parallel L-Network solutions for that impedance
transformation, a single Series-Parallel solution, or no Series-Parallel
solution. We can determine which of these results will be true by using the following criteria:</p>
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi5sAe1qhjuVCieGxXi10_iVSpuETsEYGu-BFL8zaxyllvZfVrfqWg88d9XIPagPXu7ZqgTFy6wJmDS_mviE6uVJCwjrExyMEAPgu0q3YbjObLeeVQBxcJm__xWG-35mhLgvr6vl5QPaVoF2apwV3WJsRwwWsmEuVOlomgRslh05C1AdrGnABqUvPEk/s1085/Series-Parallel%20Criteria.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="488" data-original-width="1085" height="144" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi5sAe1qhjuVCieGxXi10_iVSpuETsEYGu-BFL8zaxyllvZfVrfqWg88d9XIPagPXu7ZqgTFy6wJmDS_mviE6uVJCwjrExyMEAPgu0q3YbjObLeeVQBxcJm__xWG-35mhLgvr6vl5QPaVoF2apwV3WJsRwwWsmEuVOlomgRslh05C1AdrGnABqUvPEk/s320/Series-Parallel%20Criteria.png" width="320" /></a>
</div>
<p>
The equations for calculating a Series-Parallel L-Network's X and B values are
listed, below. Note that there are two sets of X and B equations,
meaning that there is the possibility of two different Series-Parallel
L-Network solutions for a particular impedance transformation.
</p>
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi3IqOm-2YlSNnJfVae7kUuXW7inBF7heKo5qKxA5ioKGVNFKzXLtPrx-__x5XeqM4vJ-F4oQ-YbuYJRn2JoJOoaOosmy4f7i7OsB0wn9vmzQ_WEfzUr9uiPwMg73hMQfllPO9WSpFE3Ri27q8HSYeTaG3HwGsg-HOl_qc8DANE2d2uO75AOQz05R3e/s1203/Series-Parallel%20Equations.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="814" data-original-width="1203" height="217" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi3IqOm-2YlSNnJfVae7kUuXW7inBF7heKo5qKxA5ioKGVNFKzXLtPrx-__x5XeqM4vJ-F4oQ-YbuYJRn2JoJOoaOosmy4f7i7OsB0wn9vmzQ_WEfzUr9uiPwMg73hMQfllPO9WSpFE3Ri27q8HSYeTaG3HwGsg-HOl_qc8DANE2d2uO75AOQz05R3e/s320/Series-Parallel%20Equations.png" width="320" /></a>
</div>
<br />
<p>
The signs of the calculated values of X and B determine the type of
Series-Parallel L-Network represented by each X, B pair:
</p>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgSTnMQO27eyaKX8ku6nstVVESQMyqX7j_COh_eaaXXi8Q5Z82888f7gJAXeZHp-L-HhZvvfvLy2oVx7YDKr7MyohbwluHJeSMOigure-yS9a8aVuoEgnNP2MSOvIuCKvQZVhTUBYDjMrepCYpURTCK05wnpOvKJBKzCOlydjTrbdmml1BGsIgD5sn9/s687/Four%20Series-Parallel%20networks%20vs%20X%20and%20B.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="641" data-original-width="687" height="299" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgSTnMQO27eyaKX8ku6nstVVESQMyqX7j_COh_eaaXXi8Q5Z82888f7gJAXeZHp-L-HhZvvfvLy2oVx7YDKr7MyohbwluHJeSMOigure-yS9a8aVuoEgnNP2MSOvIuCKvQZVhTUBYDjMrepCYpURTCK05wnpOvKJBKzCOlydjTrbdmml1BGsIgD5sn9/s320/Four%20Series-Parallel%20networks%20vs%20X%20and%20B.png" width="320" /></a>
</div>
<div><br /></div>
The component values can be calculated from the values of X and B:
<div><br /></div>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh17bCt_jo8erpW_B0k2-NLynK7QC-j-7YuDHk0dH0m32r99rT6Aj2DLDTKmvQtR1WB3fT1JLl1pIScAD7UZSxCOrSECNyUsMSWcqtrvIKEf2nnTDAIAjciM6OaRxupP6qnCGQjDellnTQ6sCkk2-Kl4epTHzPVUugZ1oOp2lcyzSfzTc0kDZCN1yBb/s895/Calculating%20the%20Values%20of%20L%20and%20C.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="492" data-original-width="895" height="176" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh17bCt_jo8erpW_B0k2-NLynK7QC-j-7YuDHk0dH0m32r99rT6Aj2DLDTKmvQtR1WB3fT1JLl1pIScAD7UZSxCOrSECNyUsMSWcqtrvIKEf2nnTDAIAjciM6OaRxupP6qnCGQjDellnTQ6sCkk2-Kl4epTHzPVUugZ1oOp2lcyzSfzTc0kDZCN1yBb/s320/Calculating%20the%20Values%20of%20L%20and%20C.png" width="320" /></a>
</div>
<div><br /></div>Note that because the L-Networks above are assumed to be lossless, the actual component values that provide the match will differ slightly from those calculated with the above equations, due to component loss. However, the higher the Q of the components, the closer the actual "real-world" match will be to the ideal "lossless" match.<div><br />
<div>
<p>
<b><u>Equations for Parallel-Series L-Network Solutions:</u></b>
</p>
<div>
Next, let's define the design equations for Parallel-Series L-Networks.</div><div><br /></div><div>First we will define a Parallel-Series L-Network in terms of
the <i>susceptance</i> (B) of its shunt component and
the <i>impedance</i> (X) of its series-component:
</div>
<div><br /></div>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEieOHCByaXEnSdaqXDzaXl615LYvkQ_FL7_KqgxNVgRm53IZ5CTyXlx95SVLWqAPWMCT7odTeW4J2h71pdTmCVvCv6xELDYwxmFBwgKEd_hCNeQRDVXr5pkTIjAnMUW33rZb5wpWSLyDw1syWSpZ1OZY9rK_GKTEc1BYCo3g6Osj70-09XoBjkAo6er/s721/Parallel-Series%20Network%20and%20Zin%20Equation.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="494" data-original-width="721" height="219" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEieOHCByaXEnSdaqXDzaXl615LYvkQ_FL7_KqgxNVgRm53IZ5CTyXlx95SVLWqAPWMCT7odTeW4J2h71pdTmCVvCv6xELDYwxmFBwgKEd_hCNeQRDVXr5pkTIjAnMUW33rZb5wpWSLyDw1syWSpZ1OZY9rK_GKTEc1BYCo3g6Osj70-09XoBjkAo6er/s320/Parallel-Series%20Network%20and%20Zin%20Equation.png" width="320" /></a>
</div>
<p>
Depending upon the particular desired impedance transformation, there will
be either two different Parallel-Series L-Network solutions for that impedance
transformation, a single Parallel-Series solution, or no Parallel-Series
solution. We can determine which of these results will be true by using the following criteria:
</p>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjrRiFl-v0SMlRyMD1ofGLRbSGW3AwDpOoQj3Cc_50o2plhk0--qev-JmmUwtcjyJjk6bEHJ-x03NCqCX-PYOmYmAE2dOKSRzjuJyrf9IAXyqN4zJpg9s9aFCmatrg3yaKWrp9a3m-_wG97jFslWk2FG2dXkLXjQr__eB_u2vLmY9wI_wMmzWdVXRm3/s1228/Parallel-Series%20Equations.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="846" data-original-width="1228" height="220" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjrRiFl-v0SMlRyMD1ofGLRbSGW3AwDpOoQj3Cc_50o2plhk0--qev-JmmUwtcjyJjk6bEHJ-x03NCqCX-PYOmYmAE2dOKSRzjuJyrf9IAXyqN4zJpg9s9aFCmatrg3yaKWrp9a3m-_wG97jFslWk2FG2dXkLXjQr__eB_u2vLmY9wI_wMmzWdVXRm3/s320/Parallel-Series%20Equations.png" width="320" /></a>
</div>
<p>
The equations for calculating a Parallel-Series L-Network's B and X values
are listed, below. Note that there are two sets of B and X equations,
meaning that there is the possibility of two different Parallel-Series
L-Network solutions for a particular impedance transformation.
</p>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgPi5CIVyCwe3I0EsZI_owOdpUo1wYFlKRENUd_HNAWgpsmyqgnHcuQlDBv1sXQod5WOiivHm1KmZmxBArqWYO6_cQrr-gIuXoZue2CSC3lAQ6ut8otoNnzqqYamuDVEw3rZZbB_cmr2ZFX38d6j4u2w_ltE5UhWNUFGRu7prfb9MxDRoXI4bWLtY_G/s1085/Parallel-Series%20Criteria.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="488" data-original-width="1085" height="144" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgPi5CIVyCwe3I0EsZI_owOdpUo1wYFlKRENUd_HNAWgpsmyqgnHcuQlDBv1sXQod5WOiivHm1KmZmxBArqWYO6_cQrr-gIuXoZue2CSC3lAQ6ut8otoNnzqqYamuDVEw3rZZbB_cmr2ZFX38d6j4u2w_ltE5UhWNUFGRu7prfb9MxDRoXI4bWLtY_G/s320/Parallel-Series%20Criteria.png" width="320" /></a>
</div>
<p>
The signs of the calculated values of B and X determine the type of
Parallel-Series L-Network represented by each B, X pair:
</p>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhdXlfkFN9OStZVRhMn_zg5trqwc2pmcBDCb4wp_oOLrNsIawssKlMs0IrXswWbXzajDCUqnhbcblSgOqzk7EIP3kRuEsiIDwPsArzQANcb6QNLK7RO8icpf-pGxb4gJHjHC3Xj_RwP1mdolW-NXuylEjmJwOJTa_wxAQUN-howitupLV3Dd4t_k1yZ/s677/Four%20Parallel-Series%20networks%20vs%20X%20and%20B.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="646" data-original-width="677" height="305" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhdXlfkFN9OStZVRhMn_zg5trqwc2pmcBDCb4wp_oOLrNsIawssKlMs0IrXswWbXzajDCUqnhbcblSgOqzk7EIP3kRuEsiIDwPsArzQANcb6QNLK7RO8icpf-pGxb4gJHjHC3Xj_RwP1mdolW-NXuylEjmJwOJTa_wxAQUN-howitupLV3Dd4t_k1yZ/s320/Four%20Parallel-Series%20networks%20vs%20X%20and%20B.png" width="320" /></a>
</div>
<div><br /></div>
The component values can be calculated from the values of B and X:
</div>
<div>
<br />
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEid4C2xO1XaivT3KZzHfLxAKjTKhueunI2s7fhKwR_rqLQ6kViqHMfTm78JtzLDHmaGfPnYOUNVTdOt-ITdfgix_6lWlCcLzVqspZELw3iGQroit4TMbUAFUtcsiOSzGAqSJZ5LkapXtE49-Ks4WsHAL1gQRS3GXEJktnnTOxJUrxggbqtDZIGzsT8T/s895/Calculating%20the%20Values%20of%20L%20and%20C.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="492" data-original-width="895" height="176" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEid4C2xO1XaivT3KZzHfLxAKjTKhueunI2s7fhKwR_rqLQ6kViqHMfTm78JtzLDHmaGfPnYOUNVTdOt-ITdfgix_6lWlCcLzVqspZELw3iGQroit4TMbUAFUtcsiOSzGAqSJZ5LkapXtE49-Ks4WsHAL1gQRS3GXEJktnnTOxJUrxggbqtDZIGzsT8T/s320/Calculating%20the%20Values%20of%20L%20and%20C.png" width="320" /></a>
</div>
<div><br /></div><div>Note that because the L-Networks above are assumed to be lossless, the actual component values that provide the match will differ slightly from those calculated with the above equations, due to component loss. However, the higher the Q of the components, the closer the actual "real-world" match will be to the ideal "lossless" match.</div><div><br /></div>
<div><br /></div>
<div>
<b><u>Example:</u></b>
</div>
<div><br /></div>
<div>
Let's suppose I would like to transform a load equal to 93+j25 ohms to
be 25-j74 ohms when looking into the input of an L-Network
terminated with the load impedance (and at a frequency of 10 MHz).</div><div><br /></div><div>There are actually four different L-Networks that can accomplish this transformation. Two of these L-Networks are Series-Parallel, and two are Parallel-Series.</div>
<div><br /></div>
<div>
Dick Benson, W1QG created a nice MATLAB GUI, using the equations above, to find the L-Networks that can accomplish this impedance transformation.</div><div><br /></div><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjVh0-VsrItR9MFNxWJoBZsId5Aqa-vU6iP7Cnqcc0t9KhoABDm7AFhiCqodZSZTr2-6fkUmXvSMYyFEVzz-nt6ead2H63AmNDEznX3uGRoTCNMFrwpal2OzhNg0i11Ac42T4Z_zF0ykhpFXFpB0Sw2d0rk22NED8NDcWnwJpylNX5RSwLN4LdOB3yO/s1035/230305%20w1qg%2093+j25%20to%2025-j74.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="888" data-original-width="1035" height="275" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjVh0-VsrItR9MFNxWJoBZsId5Aqa-vU6iP7Cnqcc0t9KhoABDm7AFhiCqodZSZTr2-6fkUmXvSMYyFEVzz-nt6ead2H63AmNDEznX3uGRoTCNMFrwpal2OzhNg0i11Ac42T4Z_zF0ykhpFXFpB0Sw2d0rk22NED8NDcWnwJpylNX5RSwLN4LdOB3yO/s320/230305%20w1qg%2093+j25%20to%2025-j74.png" width="320" /></a></div><br /><div>You can find this MATLAB application at the following MATLAB Central file exchange site:
<a href="https://www.mathworks.com/matlabcentral/fileexchange/115440-rf-impedance-matching-application?s_tid=srchtitle">W1QG MATLAB L-Network</a>. Note that you might need a MATLAB account to download it.</div></div>
<div><br /></div>
<div>
For those who don't have MATLAB, I have created an EXCEL Spreadsheet to perform the
same calculations. Below, it shows the results for this same impedance transformation:<br />
<div><br /></div><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj8AyNqchk3GAkPCWasPW72RBU1qXK2ef-YTqMF7hVtvOPZXz6arJ2ldFvVmmoFXWJqc9YabvbF_wtveGk0tK0heE--pOGl9B3dMi841nwpuZbFwJtUJDJxIL2ZmKAFNKJpFxXeTwF-8mNm1HUyOjTFXfE3FVnGdM0YigDm4tHZWq4QidQzvEKDvlG7/s915/230305%20Spreadshet%2093+j25%20to%2025-j74.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="643" data-original-width="915" height="225" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj8AyNqchk3GAkPCWasPW72RBU1qXK2ef-YTqMF7hVtvOPZXz6arJ2ldFvVmmoFXWJqc9YabvbF_wtveGk0tK0heE--pOGl9B3dMi841nwpuZbFwJtUJDJxIL2ZmKAFNKJpFxXeTwF-8mNm1HUyOjTFXfE3FVnGdM0YigDm4tHZWq4QidQzvEKDvlG7/s320/230305%20Spreadshet%2093+j25%20to%2025-j74.png" width="320" /></a></div><br /><div>You can download this spreadsheet from my GitHub site:
<a href="https://github.com/k6jca/L-Network-Design">K6JCA L-Network</a>.</div></div>
<div><br /></div>
<div>
The image, below, shows the four Smith-Chart plots for the four L-Network
solutions shown above (plotted using
<a href="http://www.ae6ty.com/smith_charts.html">SimSmith</a>).<br />
<div><br /></div><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjGctwotpKO7OiAvckxfYsfhe8-BOk534b_XbafGVe9DRHXaDN8__M3MUp--t-QYB7ahi0AzMgJix1zmiJQHnXKduziUcj_QKq6uqB720VhqlScitlJFNu4aCikq4_BNVllJCR51vr_mKEbLnStoMLB7n5N0wtRMfflmoJ1h9wYFsOC_vn1hbursUID/s1970/230305%20Smith%20Charts%2093+j25%20to%2025-j74.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="1970" data-original-width="1745" height="320" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjGctwotpKO7OiAvckxfYsfhe8-BOk534b_XbafGVe9DRHXaDN8__M3MUp--t-QYB7ahi0AzMgJix1zmiJQHnXKduziUcj_QKq6uqB720VhqlScitlJFNu4aCikq4_BNVllJCR51vr_mKEbLnStoMLB7n5N0wtRMfflmoJ1h9wYFsOC_vn1hbursUID/s320/230305%20Smith%20Charts%2093+j25%20to%2025-j74.png" width="283" /></a></div><br /><div><br /></div><div><b><u>Deriving the above equations:</u></b></div>
<p>
I had no success in my attempt to derive the above equations by hand.
However, MATLAB and its <i>Symbolic Math Toolbox</i> can go where this mere human cannot. Dick Benson, W1QG, a good friend and MATLAB expert, showed me how MATLAB and its <i>Symbolic Math Toolbox</i> can derive these equations.</p>
<p>Below is a MATLAB script that will derive these equations.</p>
<pre class="codeinput" style="background: rgb(247, 247, 247); border: 1px solid rgb(211, 211, 211); font-size: 12px; margin-bottom: 20px; margin-top: 0px; outline: 0px; padding: 10px; vertical-align: baseline;"><span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">% *****</span>
<span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">% L_Network_Equation_Derivation_k6jca_V1</span>
<span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">%</span>
<span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">% Symbolic Math derivation of L network equations when</span>
<span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">% Zin (looking into the L-Network) is defined to be a</span>
<span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">% complex impedance rather than an impedance that is</span>
<span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">% resistive only, with no reactance.</span>
<span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">%</span>
<span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">% Inspired by the MATLAB solution by Dick Benson, W1QG.</span>
<span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">% See "RF Impedance Matching Application" at:</span>
<span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">% https://www.mathworks.com/matlabcentral/profile/authors/6564033</span>
<span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">%</span>
<span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">% k6jca</span>
<span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">% *****</span>
clear; close; clc;
<span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">% Define the symbols in the equations. Note:</span>
<span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">% B is the susceptance of the L-Network shunt component.</span>
<span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">% X is the reactance of the L-Network series component.</span>
<span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">% Rld and Xld are the resistance and reactance of the load.</span>
syms <span class="string" style="background: transparent; border: 0px; color: #a020f0; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">B</span> <span class="string" style="background: transparent; border: 0px; color: #a020f0; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">X</span> <span class="string" style="background: transparent; border: 0px; color: #a020f0; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">Rin</span> <span class="string" style="background: transparent; border: 0px; color: #a020f0; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">Xin</span> <span class="string" style="background: transparent; border: 0px; color: #a020f0; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">Rld</span> <span class="string" style="background: transparent; border: 0px; color: #a020f0; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">Xld</span> <span class="string" style="background: transparent; border: 0px; color: #a020f0; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">real</span>
<span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">% *****</span>
<span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">% Derive the Series-Parallel L-Network Equations</span>
<span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">%</span>
<span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">% Start by defining the equation for Zin given the</span>
<span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">% Series-Parallel L-Network terminated with Zld = Rld+jXld:</span>
fprintf(<span class="string" style="background: transparent; border: 0px; color: #a020f0; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">'\nSeries-Parallel L-Network:\n'</span>)
Zin = 1i*X + 1/(1i*B + 1/(Rld + 1i*Xld))
<span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">% Next, solve the equation for B and X, given Zin = Rin+jXin</span>
S_SP = solve(Zin==Rin+1i*Xin,[B,X],<span class="string" style="background: transparent; border: 0px; color: #a020f0; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">'ReturnConditions'</span>,true);
<span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">% Then, use 'simplify()' to simplify the resulting equations</span>
series_par_B = simplify(S_SP.B,<span class="string" style="background: transparent; border: 0px; color: #a020f0; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">'Steps'</span>,50);
series_par_X = simplify(S_SP.X,<span class="string" style="background: transparent; border: 0px; color: #a020f0; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">'Steps'</span>,50);
<span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">% Finally, print the Series-Parallel X, B equation-pairs</span>
fprintf(<span class="string" style="background: transparent; border: 0px; color: #a020f0; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">'\nFirst Series-Parallel X, B pair:\n'</span>)
series_par_X1 = series_par_X(1,1)
series_par_B1 = series_par_B(1,1)
fprintf(<span class="string" style="background: transparent; border: 0px; color: #a020f0; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">'\nSecond Series-Parallel X, B pair:\n'</span>)
series_par_X2 = series_par_X(2,1)
series_par_B2 = series_par_B(2,1)
<span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">% *****</span>
<span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">% Derive the Parallel-Series L-Network Equations</span>
<span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">%</span>
<span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">% Start by defining the equation for Zin given the</span>
<span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">% Parallel-Series L-Network terminated with Zld = Rld+jXld:</span>
fprintf(<span class="string" style="background: transparent; border: 0px; color: #a020f0; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">'\n\n\nParallel-Series L-Network:\n'</span>)
Zin = 1/(1i*B + 1/(Rld + 1i*Xld + 1i*X))
<span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">% Next, solve the equation for B and X, given Zin = Rin+jXin</span>
S_PS = solve(Zin==Rin+1i*Xin,[B,X],<span class="string" style="background: transparent; border: 0px; color: #a020f0; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">'ReturnConditions'</span>,true);
<span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">% Then, use 'simplify()' to simplify the resulting equations</span>
par_series_B = simplify(S_PS.B,<span class="string" style="background: transparent; border: 0px; color: #a020f0; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">'Steps'</span>,50);
par_series_X = simplify(S_PS.X,<span class="string" style="background: transparent; border: 0px; color: #a020f0; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">'Steps'</span>,50);
<span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">% Finally, print the Parallel-Series B, X equation-pairs</span>
fprintf(<span class="string" style="background: transparent; border: 0px; color: #a020f0; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">'\nFirst Parallel-Series B, X pair:\n'</span>)
par_series_B1 = par_series_B(1,1)
par_series_X1 = par_series_X(1,1)
fprintf(<span class="string" style="background: transparent; border: 0px; color: #a020f0; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">'\nSecond Parallel-Series B, X pair:\n'</span>)
par_series_B2 = par_series_B(2,1)
par_series_X2 = par_series_X(2,1)
<span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">% END OF SCRIPT</span></pre>
<p><br /></p>
<p>
<b><u>Thanks!</u></b>
</p>
<p>
A huge thanks to Dick Benson, W1QG, who took the equations I had derived for
my article in the 2023 March-April issue of <i>QEX</i> and extended that concept to
L-Network equations that can transform any complex impedance to any other
complex impedance (not just to a real-only impedance).
</p>
<p><br /></p>
<p>
<b><u>Standard Caveat:</u></b>
</p>
<p>
As always, I might have made a mistake in my equations, assumptions,
drawings, or interpretations. If you see anything you believe to be in
error or if anything is confusing, please feel free to contact me or comment
below.<br /><br />And so I should add -- this information is distributed in
the hope that it will be useful, but WITHOUT ANY WARRANTY; without even the
implied warranty of MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.
</p>
</div>
</div>Jeffhttp://www.blogger.com/profile/04853314106806116765noreply@blogger.com0tag:blogger.com,1999:blog-2257458838727315711.post-25743523963200856472023-05-23T16:27:00.003-07:002023-05-23T16:28:37.911-07:00Transmission Lines: Measuring Zo, the Characteristic Impedance, Method 3: Circle of Constant SWR<p>
This third technique (the other two are
<a href="https://k6jca.blogspot.com/2023/05/transmission-lines-measuring-zo.html">here</a> and
<a href="https://k6jca.blogspot.com/2023/05/transmission-lines-measuring-zo_23.html">here</a>) for finding a transmission line's Zo is based upon the concept of "Circles
of Constant SWR"</p>
<p>
I've used this technique in the past to find the Zo of twisted-pair I had made
from 14 AWG Stranded THHN wire (see this 2018 <a href="https://k6jca.blogspot.com/2018/05/adventures-with-80-meter-loop-antenna_30.html">blogpost</a>), per the image, below:
</p>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEimztGrAzj2eek4NFi8AgPrt_Hbi0w4jp0_gRpmpnffR_EvMnS5NZ8nzVY6Uo2xLZu4nUTQ7_YMW7lCoStKadp4pqEZp82tjK0oDtvQRjX94GhvlbcWpSw_Xcf-d_wvt578fIFn7GdJC3SDEGMpNHXssbMmRdnDcqQSo72qUkURcbhw-Q7YBoKwfTFG/s1067/180529%20Twisted-Pair%20S11.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="850" data-original-width="1067" height="255" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEimztGrAzj2eek4NFi8AgPrt_Hbi0w4jp0_gRpmpnffR_EvMnS5NZ8nzVY6Uo2xLZu4nUTQ7_YMW7lCoStKadp4pqEZp82tjK0oDtvQRjX94GhvlbcWpSw_Xcf-d_wvt578fIFn7GdJC3SDEGMpNHXssbMmRdnDcqQSo72qUkURcbhw-Q7YBoKwfTFG/s320/180529%20Twisted-Pair%20S11.png" width="320" /></a>
</div>
<br />
<p>
<b><u>The Concept of Circles of Constant-SWR:</u></b>
</p>
<p>
If a transmission line is terminated in a load not equal to the line's
Characteristic Impedance Zo, there will be a voltage (and a current)
Standing-Wave on the transmission line, and the impedance seen at the
input end of the transmission line will be a function of the frequency, the line's length,
its characteristics (e.g. Zo, loss, and phase delay), and the load impedance.
</p>
<p>
As frequency changes, so does the impedance seen at the input end of the
transmission line, because the signal's round-trip phase shift on the
transmission line changes (i.e. as frequency changes so does the apparent length of the line, if length is defined in terms of wavelengths).
</p>
<p>
If we were plot on a Smith Chart, as frequency changes, the Reflection Coefficient (Γ) representing Zin of the transmission line, we would see a locus of points representing different Zin impedances.
</p>
<p>
If the transmission line is lossless, then as the frequency changes, the Reflection Coefficients representing these impedances form a
circle -- a Circle of Constant SWR, centered on Zo of the transmission line.
</p>
<p>
You might be familiar with circles of constant SWR circling around a Smith Chart's center-point (typically the 50 + j0 ohm point). </p>
<p>
An example of a circle of constant SWR centered on a 50-ohm Smith Chart is shown below, simulated with SimSmith. Note that the coax is an ideal
(lossless) 50 ohms and the load is 25 ohms.
</p>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgC8mnFo0pygAEOBnqAVFkL59jQoVuPH1eupo1-guwR1gBOUs4LKlX6HamSYRG9ujzvV2LGbD1NzAqsTOgXEBiVC7K8MEG62smjqKoKVOz0p5n635ztOz3hbk-QxsGt4W2mmb88xOOztATyt66WDBr5Qmq-blYsVyvIpqMNMlbDExUjRR2A7-0R6OVF/s1070/230523%20Circle%20of%20constant%20SWR%20Ideal%20Coax%20Zo%2050%20Zld%2025.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="818" data-original-width="1070" height="245" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgC8mnFo0pygAEOBnqAVFkL59jQoVuPH1eupo1-guwR1gBOUs4LKlX6HamSYRG9ujzvV2LGbD1NzAqsTOgXEBiVC7K8MEG62smjqKoKVOz0p5n635ztOz3hbk-QxsGt4W2mmb88xOOztATyt66WDBr5Qmq-blYsVyvIpqMNMlbDExUjRR2A7-0R6OVF/s320/230523%20Circle%20of%20constant%20SWR%20Ideal%20Coax%20Zo%2050%20Zld%2025.png" width="320" /></a>
</div>
<p>
At 0 Hz, (i.e. 9 o'clock on the chart) the measured S11 value represents a Zin
impedance of 25 + j0 ohms. And halfway around the circle, at the 3 o'clock position, the measured
S11 value represents a Zin impedance of 100 ohms. Note that all of the
impedances on the circle have the same SWR, referenced to a Zo of 50 ohms, and
these are all impedances we would see at the input of the lossless
transmission line as the frequency is changed.
</p>
<p>
To calculate Zo from the circle, it is simplest to chose the two points
opposite each other on the "real-only" axis of the Smith Chart, where the
reactance of each Zin is 0. Zo is then just the <i>geometric mean</i> of
the two Zin resistance values (because Zin has no reactance if on the
"real-only" axis).
</p>
<p>
In the example above, the two resistance values are 25 and 100 ohms. The
Zo calculation is:
</p>
<p style="text-align: center;">Zo = SQRT(25*100) = 50 ohms</p>
<p>This concept SWR circles can be extended to transmission lines of other impedances, too, even when plotted on a Smith Chart referenced to 50 ohms, but the circles will now be centered around Zo of the coax being measured, rather than the Smith Chart's reference Zo.</p><p><br /></p>
<p>
<b><u>Circles of Constant SWR when Zo of a Lossless Line is NOT 50 ohms:</u></b>
</p>
<p>
The concept of Circles of Constant SWR still applies even if Zo of the
Transmission Line is not 50 ohms. If the line is lossless, SWR depends
solely upon the line's Zo and the Load impedance.
</p>
<p>
If we were to measure the impedance looking into the input (Zin) side of the
Transmission Line (using a VNA, for example), as frequency changes we would
see Zin (calculated from measured S11) trace a circle. But the circle now would be
centered on the Zo of our transmission line.
</p>
<p>
And this is the third technique we can use to find Zo of a unknown
transmission line. Terminate the line with a known impedance
(<i>not</i> equal to Zo -- we need <i>some </i>SWR, otherwise there is no
circle!) and next plot S11 over frequency on the Smith Chart.</p>
<p>
If the line is lossless, the resulting plot will be a circle centered on the
new Zo, and Zo can be found by taking the square-root of the impedances calculated from the S11 measurements at these two points opposite
each other (180 degrees apart) on the circle. If the circle's center
lies on the Smith Chart's "real only" axis, then it is easiest just to use
the resistances at the 3 and 9 o'clock positions on the circle, where the reactance of either Zin value is 0.</p>
<p>
The example, shown below, shows circles of constant SWR for a transmission
line whose Zo is 75 ohms and which, in one case, is terminated with 50 ohms,
and in the other case terminated with 300 ohms. You can see that the
geometric means of impedances opposite each other on either unit circle
calculate to be 75 ohms, the Zo of the transmission line, independent of which SWR circle is used.
</p>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi86OnaJ4AlUqYrojqbceYmFoVK8ZcIQHdb4LFXXz3_b0_9ePv4EtnUxa4o80t-gTXNjr4nlSA33mE0qHtgvi7CxeCS3dCjpNUSEa0sqSkPRgbDntodSe9IhgrOWrGvF_2yyWBd7DDbTEYsDzG0MXxgbf-fwFnX2HzfTWdrv7UoEsNpu_I6tMjqnC8P/s1070/230523%20Circle%20of%20constant%20SWR%20Ideal%20Coax%20Zo%2075,%2050%20and%20300%20ohm%20terminations.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="818" data-original-width="1070" height="245" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi86OnaJ4AlUqYrojqbceYmFoVK8ZcIQHdb4LFXXz3_b0_9ePv4EtnUxa4o80t-gTXNjr4nlSA33mE0qHtgvi7CxeCS3dCjpNUSEa0sqSkPRgbDntodSe9IhgrOWrGvF_2yyWBd7DDbTEYsDzG0MXxgbf-fwFnX2HzfTWdrv7UoEsNpu_I6tMjqnC8P/s320/230523%20Circle%20of%20constant%20SWR%20Ideal%20Coax%20Zo%2075,%2050%20and%20300%20ohm%20terminations.png" width="320" /></a>
</div>
<br />
<p><b><u>Circles of Constant SWR for Lossy Lines:</u></b></p>
<p>
If the line is lossy the plotted S11 values will no longer trace a perfect circle on
the Smith Chart as frequency is varied. Instead, they will form a spiral which spirals, as frequency increases, inwards towards the Zo of the lossy
transmission line.</p>
<p>
Thus, we no longer have perfect circles of constant SWR (we have a spiral,
instead), and choosing two points opposite on the spiral (e.g. both on the real-only axis)
will give us a calculated value close to the transmission line's actual Zo, but it will not be perfect.
</p>
<p>The less the line loss, however, the more accurate the calculation will be.
</p>
<p>
Below is a plot showing a simulated example of Belden 8241 75 ohm coax
terminated with 50 ohms. Included is also the constant SWR circle (in yellow) for
the same load, but assuming ideal, lossless, coax.
</p>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjaF17dj-Py9OZP9k5UuODW5YFS0FCK9ccAf1bbUvfLyOH_7BFiAF0mKolpLEq7X13PaZvKxDCy_UFuXAYmqThM8aPD3GJki8Ssha3_SKch62EUC1YOEQGodT3uh9mOppyb1gKAi01cFxeGRZ9CFG3nipf-ZpmqCgnKmxBtjDpuGLpZf8BX8sZSrYSo/s1070/230522%2040ft%208241%2022%20MHz%2050ohm%20termination.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="818" data-original-width="1070" height="245" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjaF17dj-Py9OZP9k5UuODW5YFS0FCK9ccAf1bbUvfLyOH_7BFiAF0mKolpLEq7X13PaZvKxDCy_UFuXAYmqThM8aPD3GJki8Ssha3_SKch62EUC1YOEQGodT3uh9mOppyb1gKAi01cFxeGRZ9CFG3nipf-ZpmqCgnKmxBtjDpuGLpZf8BX8sZSrYSo/s320/230522%2040ft%208241%2022%20MHz%2050ohm%20termination.png" width="320" /></a>
</div>
<br />
<p>
If the load is changed from 50 ohms to 300 ohms, the lossy coax's spiral is
larger and thus more obvious.
</p>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhfdJrbr2WVSZ_r1it8_tG3MjSZUvZ97oe5v4xDnAXl602oSZGO8bUKoRh3_nP-bkk47Oe09AtkT4rxEcJm27FefIvyyuVOu8dthLqmqM7bTiD5sea12XEP-hm5w3JlzYGYD90WppvmLInFn41TbGdx1yMoZ-pEfsYL2K3d_hTkGVMPJqgw_LATdtRG/s1070/230522%20%2040ft%208241%2022MHz%20300ohm%20termination.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="818" data-original-width="1070" height="245" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhfdJrbr2WVSZ_r1it8_tG3MjSZUvZ97oe5v4xDnAXl602oSZGO8bUKoRh3_nP-bkk47Oe09AtkT4rxEcJm27FefIvyyuVOu8dthLqmqM7bTiD5sea12XEP-hm5w3JlzYGYD90WppvmLInFn41TbGdx1yMoZ-pEfsYL2K3d_hTkGVMPJqgw_LATdtRG/s320/230522%20%2040ft%208241%2022MHz%20300ohm%20termination.png" width="320" /></a>
</div>
<div><br /></div>
<div>But the calculation is very close to 75 ohms. Not too bad!</div>
<div><br /></div>
<div><br /></div>
<div>
<b><u>Mathematics:</u></b>
</div>
<div><br /></div>
<div>
I won't go into the mathematical under-pinings too deeply. Just
recognize that the fundamental equation for analyzing this behavior is the
equation for Zin for a terminated transmission line (whether the line be lossy or it be lossless).
</div>
<div><br /></div>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgYVseywMasUZjRD-UJkPOvhDNQkGsRdRtD0Zwfjs7rkxt4Z2mWTsEVMVVhStitwcC7ychoElux51pANJQjeitrRAQXEantmhqSAS61mC2W5tPBsIV8u5U4g9NWi_k27J3P6Rp-v9OmWKXiwqBQTzRLS-d8C9YFVO79tYtiO1VZrzRhQJao9DvhNEKX/s960/Slide1.PNG" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="720" data-original-width="960" height="240" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgYVseywMasUZjRD-UJkPOvhDNQkGsRdRtD0Zwfjs7rkxt4Z2mWTsEVMVVhStitwcC7ychoElux51pANJQjeitrRAQXEantmhqSAS61mC2W5tPBsIV8u5U4g9NWi_k27J3P6Rp-v9OmWKXiwqBQTzRLS-d8C9YFVO79tYtiO1VZrzRhQJao9DvhNEKX/s320/Slide1.PNG" width="320" /></a>
</div>
If the line is lossless, α equals 0 and Γ (of the load) is only modified by a
phase shift (its magnitude remains constant -- thus the "circle" of constant SWR).<div><br /></div><div>For more on the math assuming lossless coax, see <a href="https://onlinelibrary.wiley.com/doi/pdf/10.1002/9780470423875.app3">here</a>.<br /><div><br /></div><div>If the line is lossy, well, then things get more complicated. We have some of the same potential issues that we had with the 1/8th-Lambda technique (e.g. Zo being complex, not real-only). So any calculation of Zo using this method should be considered an estimation (albeit perhaps a very good estimation), but there will be some amount of uncertainty, too.</div><div><br /></div><div>If you have any question about your calculation, I would recommend recording several of the spiral loops and calculate Zo using multiple pairs of opposite-points to, hopefully, provide a more accurate Zo result.</div><div><br /></div><div><div>But, despite its potential drawbacks, I have found that I can quickly make the measurement on a VNA and get a reasonable result for Zo.</div><div><br /></div>
<p>
<b><u>Resources:</u></b>
</p>
<p>
More on the underlying "lossless line" math can be found, here: <u style="font-weight: bold;">Munk, Ben A., </u>Metamaterials: Critique and Alternatives, Wiley Online Library, <a href="https://onlinelibrary.wiley.com/doi/pdf/10.1002/9780470423875.app3">Appendix C, How to Measure the Characteristic Impedance and Attenuation of
a Cable</a>
</p>
<p><b><u><br /></u></b></p><p><b><u>Standard Caveat:</u></b></p>
<p>
As always, I might have made a mistake in my equations, assumptions, drawings,
or interpretations. If you see anything you believe to be in error or if
anything is confusing, please feel free to contact me or comment below.<br /><br />And
so I should add -- this information is distributed in the hope that it will be
useful, but WITHOUT ANY WARRANTY; without even the implied warranty of
MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.
<a href="https://onlinelibrary.wiley.com/doi/pdf/10.1002/9780470423875.app3">here</a></p>
</div></div>Jeffhttp://www.blogger.com/profile/04853314106806116765noreply@blogger.com0tag:blogger.com,1999:blog-2257458838727315711.post-10646085107367395592023-05-23T10:43:00.007-07:002023-06-05T05:42:06.351-07:00Transmission Lines: Measuring Zo, the Characteristic Impedance, Method 2: 1/8th Lambda<p>
In my previous post (<a href="https://k6jca.blogspot.com/2023/05/transmission-lines-measuring-zo.html">here</a>) I described the method of determining a transmission line's Characteristic
Impedance Zo from measurements of its input impedance when terminated with a
short and when terminated with an open.
</p>
<p>
In this new post I will describe a second method that is sometimes used
determine a transmission line's Zo (such as in some versions of the
NanoVNA's firmware). I will call this second method the "1/8th-Lambda
Method" (where "Lambda" is the wavelength of the signal on the transmission
line). I first learned of this method
<a href="https://groups.io/g/nanovna-users/message/32676">here</a>.
</p>
<p>
The 1/8th-Lambda method uses S11 measurements of a transmission line, and
these can be made with the line terminated either with an OPEN or with a
SHORT.
</p>
<p>
Assuming the transmission line is terminated with an OPEN, the principle
behind this technique is that the magnitude of the reactive-component 'Xin' of
Zin, the terminated transmission line's input impedance (as calculated from
S11), will equal Zo at a point 1/8th of a wavelength from the point at S11's
phase passes through 180 degrees (or 0 degrees, if the transmission line is
terminated with a SHORT).
</p>
<p>
To find this 1/8th-Lambda point, assuming the line is terminated with an OPEN
and the S11 frequency sweep is starting near 0 Hz:
</p>
<p>
1. Find the frequency at which the phase of S11 is -180 degrees (on the
left-hand side of the Smith Chart and on the x-axis). This is the
quarter-wavelength frequency of an OPEN-terminated transmission line.
</p>
<p>
2. Divide this frequency by 2. This new frequency is the
eighth-wavelength frequency of the transmission line. Note S11 at this
point.
</p>
<p>
3. Convert this S11 value to Zin (where Zin = Rin + jXin). The
value of Xin will be the negative of Zo, and therefore Zo = -Xin.
</p>
<p>
This same method can be used, with a few small changes, if the transmission
line is terminated with a SHORT:
</p>
<p>
1. Find the frequency at which the phase of S11 is 0 degrees on the
right-hand side of the Smith Chart. This is the quarter-wavelength
frequency for this length of transmission line (assuming the S11 scan starts
near 0 Hz).
</p>
<p>
2. Divide this frequency by 2. This new frequency is the frequency
for which the transmission line is an eighth of a wavelength long. Note
S11 at this point.
</p>
<p>
3. Convert this S11 value to Zin (Zin = Rin + jXin). The value of
Xin will be the negative of Zo, and therefore Zo = -Xin.
</p>
<div>
<u> DISCLAIMER:</u> This technique works best when the transmission line
is essentially lossless. Calculations made using measurements of lossy
lines will have some amount of error. I'll discuss this issue later in
this post.
</div>
<p><br /></p>
<p>
<b><u>The 1/8-Lambda Method and Lossless Transmission Lines:</u></b>
</p>
<p>
Let's look at the principle behind this technique assuming a
<i>lossless</i> transmission line...
</p>
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</div>
<b><u><div>
<b><u><br /></u></b>
</div>
<div>
<b><div class="separator" style="clear: both; text-align: center;">
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</div>
<div>
<b><u><br /></u></b>
</div>
Example: Lossless Transmission Line Terminated with an OPEN:</u></b><br />
<p>
I've used SimSmith to model 40 feet of ideal (lossless) 75-ohm coax terminated
with an OPEN. The simulation parameters are in the image, below.
</p>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj3oHEGqRmyI4DRuSjD4YHql6tWqo79N2rEdNMMzHo4be4SV6shslNUHgKrahRe8Q4_9QELsYoVZ9HUlmqtX39fTH1nRCNV5mFqKtu1n5aMcL-D3p15AWvgLwnO4YoWHMdR-IzFOvkT7hoz45Y3oylsSp1WW6QcrAVFe8HTIiQBOAFAiHuNi9pDrk5h/s1177/230522%20Simulink%20model%20open%20ideal%20lossless.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="721" data-original-width="1177" height="196" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj3oHEGqRmyI4DRuSjD4YHql6tWqo79N2rEdNMMzHo4be4SV6shslNUHgKrahRe8Q4_9QELsYoVZ9HUlmqtX39fTH1nRCNV5mFqKtu1n5aMcL-D3p15AWvgLwnO4YoWHMdR-IzFOvkT7hoz45Y3oylsSp1WW6QcrAVFe8HTIiQBOAFAiHuNi9pDrk5h/s320/230522%20Simulink%20model%20open%20ideal%20lossless.png" width="320" /></a>
</div>
<p>
With this simulation I created an S11 file (frequency spanning from 1 KHz to
6.001 MHz.). The S11 plot is shown below.
</p>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj6jb2TohwfWq9PuDyYZ2yywO5q_02_zrRWp_tral8-mO65H_acDqo8X0EMyOG9trbTVEkfKVZV9RR7azzlAj8S8J4RsFGIlvOR4T-HC0t9CYPtAmtqma15otpAdKLLK89SIkfa1PB2BBvcF_XzkyToYt6BtH2Byd5dSKWXYxJ4lfSNYojAVvlJHzw9/s1070/230522%20ideal%20open%206MHz.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="818" data-original-width="1070" height="245" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj6jb2TohwfWq9PuDyYZ2yywO5q_02_zrRWp_tral8-mO65H_acDqo8X0EMyOG9trbTVEkfKVZV9RR7azzlAj8S8J4RsFGIlvOR4T-HC0t9CYPtAmtqma15otpAdKLLK89SIkfa1PB2BBvcF_XzkyToYt6BtH2Byd5dSKWXYxJ4lfSNYojAVvlJHzw9/s320/230522%20ideal%20open%206MHz.png" width="320" /></a>
</div>
<p>
Note that S11, the Reflection-Coefficient, curve passes through -180 degrees
at about 4.057 MHz. At this frequency the transmission-line is a
quarter-wavelength long.
</p>
<p>
To find the eighth-wavelength point, we divide 4.057 MHz by 2, which results
in a frequency of 2.0285 MHz.
</p>
<p>
Convert the measured S11 data for this frequency to Zin: assuming the
VNA is calibrated for 50 ohms, Zin = 50*(1+S11)/(1-S11).
</p>
<p>
At this point on the Smith Chart, above, (actually, close to it -- the S11
capture skips 2.0285 MHz), Zin = -4e-8 - j75.04 ohms
</p>
<p>
From the derivation above for an OPEN termination, Zin = -jZo, therefore Zo =
jZin.
</p>
<p>Substituting in the value of Zin and solving...</p>
<p style="text-align: center;">
Zo = j*(-4e-8-j75.04) = 75.04 - j(-4e-8).
</p>
<p>The imaginary term is negligible. Therefore:</p>
<p style="text-align: center;">Zo = 75.04 ohms. </p>
<p>
<b><u><br /></u></b>
</p>
<p>
<b><u>Example: Lossless Transmission Line Terminated with a SHORT:</u></b><br />
</p>
<p>
We can do a similar calculation if the transmission line is terminated with a
SHORT. Below is the Smith Chart showing the Reflection Coefficient path.
</p>
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgbKJcIpmQ-4eEOIPULXJz4_NqPVIUptZT-W5dFnqF-b8V0Yv_MPJOqGBJfps6h-NVCLIkCAIrixzKCwnbXe5zgas9eX_sxGvUM0d1vHwCPr7SiN_j6j2IEWDBNYODhmyA8d_tM9yWRlj3hHQof0WlJe10pu8WRz9hSedD7Xoekyecb2mf5Glnx0H1f/s1070/230522%20ideal%20short%206MHz.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="818" data-original-width="1070" height="245" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgbKJcIpmQ-4eEOIPULXJz4_NqPVIUptZT-W5dFnqF-b8V0Yv_MPJOqGBJfps6h-NVCLIkCAIrixzKCwnbXe5zgas9eX_sxGvUM0d1vHwCPr7SiN_j6j2IEWDBNYODhmyA8d_tM9yWRlj3hHQof0WlJe10pu8WRz9hSedD7Xoekyecb2mf5Glnx0H1f/s320/230522%20ideal%20short%206MHz.png" width="320" /></a>
</div>
<div><br /></div>
<div>
The 1/8th Lambda frequency is again 2.0285 MHz, and again we will use 2.028
MHz on the above Smith Chart.
</div>
<div><br /></div>
<div>
Zin (calculated from the Reflection Coefficient) is -4e-8 + j74.96 ohms.
</div>
<div>
<p>
From the derivation above for a SHORT termination, Zin = jZo, therefore Zo =
-jZin.
</p>
<p>Substituting in the value of Zin and solving...</p>
<p>Zo = -j*(-4e-8 + j74.96) = 74.96 + j(4e-8). </p>
<p>The imaginary term is again negligible, and thus:</p>
<p style="text-align: center;">Zo = 74.96 ohms. </p>
<p>
<b><u>Conclusion:</u></b> For a lossless transmission line (or one whose loss is
insignificant) the 1/8-Lambda method works quite well at finding Zo of the
transmission line.
</p>
</div>
<div><br /></div>
<div>
<b><u>The 1/8-Lambda Method and Lossy Transmission Lines:</u></b>
</div>
<p>How well does this technique perform if the line is lossy?</p>
<p>
For a lossless transmission line, Zo is a real value and equal to the
SQRT(L/C). But for a lossy transmission line Zo is a complex value whose
real and imaginary parts are <i>both </i>frequency dependent, as shown in the
following two slides:
</p>
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg7F2WTUVqz7cgtG66eW1T9NGrQiU5LMCBtrMSAN3FbrIGXIjtqCtxk2K4vbWSKYcEUOa5thHKmgOk_9nmxHEbCi-ojXx93osRDnSqw4Zgfv9WT9GJJQThJII8FK0GWUJIE48Ep1TIQFIEeUPIPpMn2O9k7J_hkh1FNV4lAeSq3xK_bOyO_A37RoDQN/s812/230522%20eq%208%20complex%20Zo%201.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="362" data-original-width="812" height="143" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg7F2WTUVqz7cgtG66eW1T9NGrQiU5LMCBtrMSAN3FbrIGXIjtqCtxk2K4vbWSKYcEUOa5thHKmgOk_9nmxHEbCi-ojXx93osRDnSqw4Zgfv9WT9GJJQThJII8FK0GWUJIE48Ep1TIQFIEeUPIPpMn2O9k7J_hkh1FNV4lAeSq3xK_bOyO_A37RoDQN/s320/230522%20eq%208%20complex%20Zo%201.png" width="320" /></a>
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</div>
<div><br /></div>
<div>
We can see the frequency-dependency of Zo's real and imaginary components in
the following plot of Zo for Belden 8241 75 ohm coax. The plotted Zo is
calculated using Zo = SQRT(Zsc*Zoc), per the derivation in
<a href="https://k6jca.blogspot.com/2023/05/transmission-lines-measuring-zo.html">my previous blog post</a>, using S11 files generated with
<a href="http://www.ae6ty.com/smith_charts.html">SimSmith</a>.
</div>
<div><br /></div>
<div>
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</div>
<p>
Below is the plot of S11 for 40 feet of Belden 8241 coax terminated with an
OPEN. Note that as frequency increases the Reflection Coefficient is
spiraling in from the Smith Chart's unit-circle boundary.
</p>
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgGHLkfvqgubOiZbHU2L1WeZEsYN_7g03RUbzFyzo12yu2nT9Ofr_ew7Un43-zAxucl6ozvk7Rtc_PJPwst4wqOwbwIR4Up4nY8KMf7VvGkAhkhM1o-cNkonHmx2qyjl7ISwYa7Sbk4GZhA4rNIiMbTT_xtYcL-9K8q1a5O6m0osymBl2JOxondCiz0/s1177/230522%20Simulink%20model%20open%208241.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="721" data-original-width="1177" height="196" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgGHLkfvqgubOiZbHU2L1WeZEsYN_7g03RUbzFyzo12yu2nT9Ofr_ew7Un43-zAxucl6ozvk7Rtc_PJPwst4wqOwbwIR4Up4nY8KMf7VvGkAhkhM1o-cNkonHmx2qyjl7ISwYa7Sbk4GZhA4rNIiMbTT_xtYcL-9K8q1a5O6m0osymBl2JOxondCiz0/s320/230522%20Simulink%20model%20open%208241.png" width="320" /></a>
</div>
<p>
In the plot below I have applied the 1/8-Lambda technique to two different
S11 simulated-measurements. In one measurement the 40 feet of Belden
8241 coax was terminated with an OPEN, and in the other it was terminated
with a SHORT.
</p>
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjYsA0Wd5pyc2R9ZqYdsxogA8zJwyLFJHiyeyrb5yjx_El4Tq1c23QyPPYM3qrWPdIix62WNyGLt1RmpcN-th-V1qd1lsiEoc6e__GwMyX-gAhb-0rrkMkzODkrWOQHVGE4gOPq4zfs6lJy-SQgd-aBSWnKJLv_KcelZeOXa-VDMibotPMAd4r68fOj/s1070/230522%208241%20coax%20Zsc%20Zoc%201M999Hz.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="818" data-original-width="1070" height="245" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjYsA0Wd5pyc2R9ZqYdsxogA8zJwyLFJHiyeyrb5yjx_El4Tq1c23QyPPYM3qrWPdIix62WNyGLt1RmpcN-th-V1qd1lsiEoc6e__GwMyX-gAhb-0rrkMkzODkrWOQHVGE4gOPq4zfs6lJy-SQgd-aBSWnKJLv_KcelZeOXa-VDMibotPMAd4r68fOj/s320/230522%208241%20coax%20Zsc%20Zoc%201M999Hz.png" width="320" /></a>
</div>
<p>
Using the same procedure that I used for the lossless line, the 1/8th-Lambda
points are at 1.999 MHz (although there is some question as to this
frequency -- more on this in a bit...).
</p>
<p>
Converting the S11 values at 1.999MHz to Zin, let's use the equations
derived earlier (for a lossless line) to find Zo:
</p>
<div>
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiY1itYNpYLyp6BrsLAm3pOjbZHRS5reOGrplme9vvHso4CeYOUErUAaUTMbQyjptj_8j1-3QQrw2ZViMW_dPO2H0JzyTkZs-VotLOq9cglfx_CjzbPWXhDwLSwpRBjHxzZigpSW_2IdYVZQB7T7MgsLchyMshYu8KoS0a1P47G15ur4RJKbmQ9--yK/s815/230522%20eq%2010,%20Zo%20for%20OPEN.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="289" data-original-width="815" height="113" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiY1itYNpYLyp6BrsLAm3pOjbZHRS5reOGrplme9vvHso4CeYOUErUAaUTMbQyjptj_8j1-3QQrw2ZViMW_dPO2H0JzyTkZs-VotLOq9cglfx_CjzbPWXhDwLSwpRBjHxzZigpSW_2IdYVZQB7T7MgsLchyMshYu8KoS0a1P47G15ur4RJKbmQ9--yK/s320/230522%20eq%2010,%20Zo%20for%20OPEN.png" width="320" /></a>
</div>
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEind_8u-TeEd8ipjJjlhhSjcyHDo6fIu19hswtnm3eyq07qk7B_H9x1oZp4AVHXco6JWWmcmw-wv8-mYmmsZP_WfsuriDdiJYs-jUNfPqc56NNwUwBE65Y2NxnxWEewcp4tPyB-5G5qxeUwUQIslWr0gIjEnOZL_3ddmfBr1UAuONuZmOiqObrXyOLb/s815/230522%20eq%2011,%20Zo%20for%20SHORT.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="289" data-original-width="815" height="113" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEind_8u-TeEd8ipjJjlhhSjcyHDo6fIu19hswtnm3eyq07qk7B_H9x1oZp4AVHXco6JWWmcmw-wv8-mYmmsZP_WfsuriDdiJYs-jUNfPqc56NNwUwBE65Y2NxnxWEewcp4tPyB-5G5qxeUwUQIslWr0gIjEnOZL_3ddmfBr1UAuONuZmOiqObrXyOLb/s320/230522%20eq%2011,%20Zo%20for%20SHORT.png" width="320" /></a>
</div>
<p>
From the Smith Chart above, Zin at the 1/8th-Lambda points for the SHORT
and the OPEN are:.
</p>
<p></p>
<ul>
<li>
For the OPEN termination, Zin @ (1.999 MHz) = 1.721 - j75.98 ohms
</li>
<li>
For the SHORT termination, Zin @ (1.999 MHz) = 7.595 + j75.56 ohms.
</li>
</ul>
<p></p>
<div>
Using the above equations to transform Zin to Zo , the results are:
</div>
</div>
<div><br /></div>
<div style="text-align: center;">
Zo (OPEN @ 1/8 Lambda) = 75.98 + j1.721 ohms
</div>
<div style="text-align: center;">
Zo (SHORT @ 1/8 Lambda) = 75.56 - j7.595 ohms
</div>
<div style="text-align: center;"><br /></div>
<div>
I would expect Zo calculated with this method to be approximately equal to
the values found using the equation Zo = SQRT(Zsc*Zoc), which, per the plot
below (at the 1/8th-Lambda frequency) are:
</div>
</div>
<div><br /></div>
<div>
<div style="text-align: center;">Zo (@ 1.998 MHz) = 76.41 - j2.92 ohms</div>
<div style="text-align: center;"><br /></div>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgYQOoRni5QV5VvjgE88qarky97tNUjUL9yjLkkmj6dIfyUV4GE3KsFVgxzMIUL1UMB5-btKex8tOX6H5JqcbV31To1JiXJSgXRUS-gYNibCgI1WF91KP_lRv7YFvrolrELloKrETpTCPX29ChAztyh_NIKLUJ5o79pVSx10VUM9xfMPXQHV4bH4DZB/s789/230522%20Ro%20Xo%206MHz.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="602" data-original-width="789" height="244" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgYQOoRni5QV5VvjgE88qarky97tNUjUL9yjLkkmj6dIfyUV4GE3KsFVgxzMIUL1UMB5-btKex8tOX6H5JqcbV31To1JiXJSgXRUS-gYNibCgI1WF91KP_lRv7YFvrolrELloKrETpTCPX29ChAztyh_NIKLUJ5o79pVSx10VUM9xfMPXQHV4bH4DZB/s320/230522%20Ro%20Xo%206MHz.png" width="320" /></a>
</div>
<div><br /></div>
<div>
And the 'real' part of Zo is quite close (within about an ohm). But
the imaginary parts are significantly different -- the OPEN's Xo is
positive, rather than negative, and the SHORT's Xo is more than twice the
value calculated using Zo = SQRT(Zsc*Zoc).
</div>
<div><br /></div>
<div>
So, although the results are certainly acceptable, I wonder what could be
causing the discrepancies. There are a number of assumptions implied
with this method. Are these assumptions valid?
</div>
<div><br /></div>
<div>
<b><u>1/8th-Lambda Lossy Transmission Line Questions and Concerns:</u></b>
</div>
<p>
1. The 1/8th-Lambda method assumes that the quarter-wavelength
frequency is when the angle of S11 is -180 degrees (for an OPEN
termination). But for a lossy line, the quarter-wavelength point is
<u>not</u> where S11's phase is -180 degrees. Zo is a complex number
for lossy lines, and thus Zin at 1/4 wavelength would also have some amount
of reactance, meaning that S11, at 1/4 Lambda, will have an angle other than
180 or 0 degrees. How much of a difference would this make? I
don't know.
</p>
<p>
2. The 1/8th-Lambda method assumes that the frequency for the 1/8th
wavelength is half the frequency of the quarter-wavelength. But the
velocity factor of lossy transmission lines is not a constant, but varies
with frequency (see
<a href="file:///C:/Users/jeff/Downloads/28%20Lossy%20Transmittion%20Lines,%20Dispersion,%20and%20Group%20Velocity.pdf">here</a>
and
<a href="https://owenduffy.net/transmissionline/concept/mvf/index.htm">here</a>).
</p>
<p>
Therefore, a quarter-wavelength length of transmission line at one frequency
will <u>not</u> be an eighth-wavelength at half that frequency. How
much difference does this make? I don't know.
</p>
<p>
3. Note that the imaginary component Xo (the reactance of the
calculated Zo) is <i>negative</i> if the SHORT termination is used and
<i>positive</i> if the OPEN termination is used. This sign convention
will always be the case, irrespective of Zo, as long as the line is lossy,
because Rin is always positive, but it is multiplied by +j to calculate Zo
for an OPEN termination and by -j to calculate Zo for a SHORT termination
(see the Zin to Zo conversion equations for 1/8th-Lambda earlier in this
post).
</p>
<p>
Thus, the sign of Xo (the reactance of Zo) depends upon the load being an
open or a short, which makes this value questionable.
</p>
<p>
4. If we increase frequency to a point where loss is small, do the
results from the 1/8th-Lambda technique improve?
</p>
<p>
Below is an S11 plot of 40 feet of Belden 8241 75-ohm coax out to 100 MHz.
</p>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEihr5O0E3SylocQvROZDHmB7Mv_aCDSpc2anElnptop4iey16aQAeOlfRBmgoqaJfBboKRUBKCWRErlJKnYIUqWIh1_sKQYApknQ93qFkfmsNIB8YMJDyhQSDCgi-ZcmPmRbiwCz-MzkOdoHoAU0dyueVY2WUj4-3wphmtEobTKqHs7Q07CuIJ-AqNB/s1070/230522%208241%20coax%2090M926.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="818" data-original-width="1070" height="245" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEihr5O0E3SylocQvROZDHmB7Mv_aCDSpc2anElnptop4iey16aQAeOlfRBmgoqaJfBboKRUBKCWRErlJKnYIUqWIh1_sKQYApknQ93qFkfmsNIB8YMJDyhQSDCgi-ZcmPmRbiwCz-MzkOdoHoAU0dyueVY2WUj4-3wphmtEobTKqHs7Q07CuIJ-AqNB/s320/230522%208241%20coax%2090M926.png" width="320" /></a>
</div>
<br />
<p>
Note that the frequency where S11 crosses 0 degrees is about 88.901 MHz, and
the frequency where S11 crosses 180 degrees is about 92.951 degrees.
The frequency half-way between these two frequencies is the 1/8th-lambda
frequency, which, in this case, is 90.926 Mhz.
</p>
<p>
At this point S11 represents an input impedance, Zin, of 21.51-j72.29 ohms.
</p>
<p>
Given that the line is terminated with an OPEN, we use the following formula
to convert Zin to Zo at the 1/8th-Lambda frequency:
</p>
<p>Zo = -Xin + jRin = 72.29 + j21.51 ohms.</p>
<p>
However, using the equation Zo = SQRT(Zsc*Zoc) at this frequency, Zo, per
the plots below, calculates to be:
</p>
<p style="text-align: center;">Zo = 75.3 - j0.29 ohms</p>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhPdANTlKtOXEHpbjUHrLbeiGIISARL005IMXcoQGHTJ0yOuF64qIs3Pg8rDOFgWtEj3qZqJjRTcKCs0r_pEPQfPX_LVNPQ5tGscP_KaRrJ46uyV11GH43IOhq81PCjkbmofiSvbLWbvsYhfAvAlqTVn64VA9J4Z4rL5inTuc9t9bmRoN8Q-nPAICP9/s792/230522%20Ro%20Xo%20100MHz.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="602" data-original-width="792" height="243" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhPdANTlKtOXEHpbjUHrLbeiGIISARL005IMXcoQGHTJ0yOuF64qIs3Pg8rDOFgWtEj3qZqJjRTcKCs0r_pEPQfPX_LVNPQ5tGscP_KaRrJ46uyV11GH43IOhq81PCjkbmofiSvbLWbvsYhfAvAlqTVn64VA9J4Z4rL5inTuc9t9bmRoN8Q-nPAICP9/s320/230522%20Ro%20Xo%20100MHz.png" width="320" /></a>
</div>
<br />
<p>So there seems to an error at high frequencies, too.</p>
<p><br /></p>
<p>
<b><u>Lossy-Line Equations:</u></b>
</p>
<p>
In addition to the uncertainty in the quarter-wave frequency and in the
Velocity Factor, part of the problem is undoubtedly due to the equations
derived for a lossless transmission line are a simplification of the
equations for a lossy line.
</p>
<p>Let's derive those equations...</p>
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgb8ku2Y5ASI4VR1f9kJjYm8IV0VPS1GZ8PeD-LHasIcWc6lsy9ZNvbvJwtgVuHn66psMpRQH4QM2CotgRowtrwnVgZRMtO5Qt1hXZ3KakVdpSfysx_1XE9b8ZFilUFigvK7VpN_e3xV4m_2CJpIYQRtof0A_DYNguIrS7l4LQykpw_6JRPiof-6T6U/s815/230522%20eq%2013,%20start%20lossy%20derivation.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="323" data-original-width="815" height="127" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgb8ku2Y5ASI4VR1f9kJjYm8IV0VPS1GZ8PeD-LHasIcWc6lsy9ZNvbvJwtgVuHn66psMpRQH4QM2CotgRowtrwnVgZRMtO5Qt1hXZ3KakVdpSfysx_1XE9b8ZFilUFigvK7VpN_e3xV4m_2CJpIYQRtof0A_DYNguIrS7l4LQykpw_6JRPiof-6T6U/s320/230522%20eq%2013,%20start%20lossy%20derivation.png" width="320" /></a>
</div>
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiY8cStDLQ8_VTjj_27p9KYzmUhz06AgBQf1KQN3QEMnU9_o3W4lBVn3mvmeuxkxdOXpMz4uonxrSupdo8K2sJPn6DSj_Fgq27_vMyOvyywO3Pjh5coNBiMruOeIGm3frxAejTK9fOHGvWdgogu7_5qpL9TUPmzxWk9ibfUZzGt0QHtyFZFHzNMiKBZ/s815/230522%20eq%2014,%20continuiing%20lossy%20derivation.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="323" data-original-width="815" height="127" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiY8cStDLQ8_VTjj_27p9KYzmUhz06AgBQf1KQN3QEMnU9_o3W4lBVn3mvmeuxkxdOXpMz4uonxrSupdo8K2sJPn6DSj_Fgq27_vMyOvyywO3Pjh5coNBiMruOeIGm3frxAejTK9fOHGvWdgogu7_5qpL9TUPmzxWk9ibfUZzGt0QHtyFZFHzNMiKBZ/s320/230522%20eq%2014,%20continuiing%20lossy%20derivation.png" width="320" /></a>
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiOLf4Vkxbrb_kTI__CWtl4XtZ9O7_uHk5hWjI7H43tkDbxY_9I_SXKQPnZ_4FczUN71sCM2NAYuDzHJBI8gu8YYfALMgVaj26hcvbDOqwB049TFkp90GoLGP8RKbQ_X55pHVZgeSfv6kYPdLU10h7JBzP_SaIkjTyQFXyvH_-FFfpcHsmJvpkEaFgx/s815/230522%20eq%2015,%20equations%20with%20open%20load.png" style="margin-left: 1em; margin-right: 1em;"><br /><img border="0" data-original-height="361" data-original-width="815" height="142" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiOLf4Vkxbrb_kTI__CWtl4XtZ9O7_uHk5hWjI7H43tkDbxY_9I_SXKQPnZ_4FczUN71sCM2NAYuDzHJBI8gu8YYfALMgVaj26hcvbDOqwB049TFkp90GoLGP8RKbQ_X55pHVZgeSfv6kYPdLU10h7JBzP_SaIkjTyQFXyvH_-FFfpcHsmJvpkEaFgx/s320/230522%20eq%2015,%20equations%20with%20open%20load.png" width="320" /></a>
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhR20Y4Q64b3PO41Ih---1Gv6FZURjLQlNNky8DLbKhcwFLYhf0j_rYQJK916eQtQqa6Bk1PhHug9XTolIdKkDqKp5Zwi2gSxeg830fs__yBymCclODnmmWQvE-3VrdUm9AhvuDxYNV03m8lh9PK6AyStbjSxDVsciXrJaUCXjXvE0DaqguOY7Db96j/s815/230522%20eq%2016,%20quarter%20wave%20annotated.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="361" data-original-width="815" height="142" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhR20Y4Q64b3PO41Ih---1Gv6FZURjLQlNNky8DLbKhcwFLYhf0j_rYQJK916eQtQqa6Bk1PhHug9XTolIdKkDqKp5Zwi2gSxeg830fs__yBymCclODnmmWQvE-3VrdUm9AhvuDxYNV03m8lh9PK6AyStbjSxDVsciXrJaUCXjXvE0DaqguOY7Db96j/s320/230522%20eq%2016,%20quarter%20wave%20annotated.png" width="320" /></a>
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhsGH3umrZWXkyuTh2OGuRuqlm7K-DIoNhxAOSsHlCsrjNPAqBqCRQnJfLFvcuOaekC07zjpDrHQs3AVd7QMNVin7PK2QpocEqRQkh4rMCS1312X5MdpBr3_8a7tSxFEC-ab5TKmHALwchw0eMj8uARbkpnjxaQuLRCFMP5TnZI-yyUwH8bVttcLZfL/s815/230522%20eq%2017,%20eighth%20wave%20annotated.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="361" data-original-width="815" height="142" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhsGH3umrZWXkyuTh2OGuRuqlm7K-DIoNhxAOSsHlCsrjNPAqBqCRQnJfLFvcuOaekC07zjpDrHQs3AVd7QMNVin7PK2QpocEqRQkh4rMCS1312X5MdpBr3_8a7tSxFEC-ab5TKmHALwchw0eMj8uARbkpnjxaQuLRCFMP5TnZI-yyUwH8bVttcLZfL/s320/230522%20eq%2017,%20eighth%20wave%20annotated.png" width="320" /></a></div>
<p>
Given the last equation, above, if α is known, then the equation for Zo
can be modified. But α for the lossy line would first need to be
found. Plus, there are still the unresolved issues of Zo being complex
(and its angle at 1/4 Lambda, hence uncertainty in the quarter-wave
frequency) as well as frequency-dependent Velocity Factor.
</p>
<p>
In other words, the one-eighth-Lambda technique might get you close to the
actual Zo, but there will be some amount of error in the calculated value.
</p>
<p><br /></p>
<p>
<b><u>Conclusions:</u></b>
</p>
<p>
1. If the transmission line is lossy, expect some amount of error in
your Zo calculation. But you will probably be reasonably close to the
transmission line's actual Zo.
</p>
<p><br /></p><p><b><u>Other Notes:</u></b></p><p>Owen Duffy has an interesting post on an alternate method for finding the 1/8th-Lambda frequency and its associated Zo. See <a href="https://owenduffy.net/blog/?p=28718">here</a>. </p><p>Essentially, he measures the impedances of both a shorted line and an open line, and then, at the frequency where the |X| values of the two measurements intersect, he identifies this frequency as the 1/8th-Lambda point and defines Zo to be equal to |X| at this frequency.</p><p><br /></p>
<p>
<b><u>Resources:</u></b>
</p>
<p>
Transmission Line stuff:
<a href="https://122.physics.ucdavis.edu/sites/default/files/files/Electronics/TransmissionLinesPart_II.pdf">https://122.physics.ucdavis.edu/sites/default/files/files/Electronics/TransmissionLinesPart_II.pdf</a>
</p>
<p>
On frequency-dependent Velocity Factor:
<a href="https://scholar.valpo.edu/cgi/viewcontent.cgi?filename=28&article=1000&context=engineering_oer&type=additional">https://scholar.valpo.edu/cgi/viewcontent.cgi?filename=28&article=1000&context=engineering_oer&type=additional</a>
</p>
<p>
and here: <a href="https://owenduffy.net/transmissionline/concept/mvf/index.htm">https://owenduffy.net/transmissionline/concept/mvf/index.htm</a>
</p>
<p><br /></p>
<p>
<b><u>Standard Caveat:</u></b>
</p>
<p>
As always, I might have made a mistake in my equations, assumptions,
drawings, or interpretations. If you see anything you believe to be in
error or if anything is confusing, please feel free to contact me or comment
below.<br /><br />And so I should add -- this information is distributed in
the hope that it will be useful, but WITHOUT ANY WARRANTY; without even the
implied warranty of MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.
</p>
</div>
Jeffhttp://www.blogger.com/profile/04853314106806116765noreply@blogger.com0tag:blogger.com,1999:blog-2257458838727315711.post-86374887404558971822023-05-21T07:57:00.006-07:002023-05-29T08:07:54.979-07:00Transmission Lines: Measuring Zo, the Characteristic Impedance, Method 1: Zo=SQRT(Zsc*Zoc)<p>
Recently there was an interesting thread on measuring the
<a href="https://groups.io/g/nanovna-users/topic/measure_the_characteristic/98802426?p=,,,20,0,0,0::recentpostdate/sticky,,,20,2,0,98802426,previd%3D1684599114640891181,nextid%3D1683336731863963718&previd=1684599114640891181&nextid=1683336731863963718">characteristic impedance of a transmission line (Zo)</a>, on the
<a href="https://groups.io/g/nanovna-users">nanovna-users</a> groups.io site
</p>
<p>
There are several ways Zo can be calculated using S11 measurements. Owen
Duffy discusses one method:
<a href="https://owenduffy.net/blog/?p=28623">https://owenduffy.net/blog/?p=28623</a>. This method uses two S11 captures to calculate Zo -- one capturing the
transmission line's input impedance with the line terminated with a SHORT
(i.e. Zin = Zsc), and the other capture with the line terminated with an OPEN
(i.e. Zin = Zoc). From these two measurements Zo can be calculated using
the following formula:
</p>
<p style="text-align: center;">Zo = SQRT(Zsc*Zoc)</p>
<p>
Duffy's derivation uses tanh() and coth() functions. I've always
preferred using complex-exponential functions, rather than hyperbolic
trigonometry, for transmission line calculations. Either method will
give the same results; it is just that I find using complex-exponentials more
intuitive. But each to their own taste!
</p>
<p>
So in this blog post I will derive <span style="text-align: center;">Zo = SQRT(Zsc*Zoc) using</span> the
<a href="https://k6jca.blogspot.com/2023/05/transmission-lines-input-impedance-of.html">complex-exponential form</a>
of the equation for Zin, the input impedance of a terminated transmission
line. Please refer to the following images.
</p>
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi7xSxtBJ9ppwAkfWOt-pZSi2k1s4kz7vk1_agHR06lgJlAfXEhz2b6VFG2q8cKS1lofdFlYrTLltRlJDAiwNVIxOPoMb0aQx-3_XZ9P-xbHsP7O5ttxh1mME_R8gW03gd4Z6kpcm_yNNSd1kHU0RvwVT4SQBcISdy9XCjsINZBd7fI92CtBtSkrs2f/s960/Slide1.PNG" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="720" data-original-width="960" height="240" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi7xSxtBJ9ppwAkfWOt-pZSi2k1s4kz7vk1_agHR06lgJlAfXEhz2b6VFG2q8cKS1lofdFlYrTLltRlJDAiwNVIxOPoMb0aQx-3_XZ9P-xbHsP7O5ttxh1mME_R8gW03gd4Z6kpcm_yNNSd1kHU0RvwVT4SQBcISdy9XCjsINZBd7fI92CtBtSkrs2f/s320/Slide1.PNG" width="320" /></a>
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhDU4ojM0xOwfCwLiei7KkSWTjHgAGDpjweoraqgxiXXr33hLzuSfobX-pOGgpPpK-t2YF3DnygRGCBPC_dMg0JfQGp203ly0UO7OPhheQJnNQS8aR0lsYVCQ14zlZbI1k-iPowo2EsN7l8CacbpgLsx6Lv6X8_KYaN-Ic3BDdErq30m8hx7YlD2txa/s960/Slide2.PNG" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="720" data-original-width="960" height="240" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhDU4ojM0xOwfCwLiei7KkSWTjHgAGDpjweoraqgxiXXr33hLzuSfobX-pOGgpPpK-t2YF3DnygRGCBPC_dMg0JfQGp203ly0UO7OPhheQJnNQS8aR0lsYVCQ14zlZbI1k-iPowo2EsN7l8CacbpgLsx6Lv6X8_KYaN-Ic3BDdErq30m8hx7YlD2txa/s320/Slide2.PNG" width="320" /></a>
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEibm7H9Ti3KGlsx2CWg7dVhe1a4d_A5QOXAbFr-pzebqpmYvUSMy16waAR49DGeinwmqZUt1kYpluW7I6ejWC6tNdq9quGrFy2-PddtZSFesiFCmQlAoQEvwpTLg_Y_-hZrz7ohdNrZfJJC7HTDDK7LWI_5EViwl2VXOA9ERvND6T3rryDI619-Nxuv/s960/Slide3.PNG" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="720" data-original-width="960" height="240" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEibm7H9Ti3KGlsx2CWg7dVhe1a4d_A5QOXAbFr-pzebqpmYvUSMy16waAR49DGeinwmqZUt1kYpluW7I6ejWC6tNdq9quGrFy2-PddtZSFesiFCmQlAoQEvwpTLg_Y_-hZrz7ohdNrZfJJC7HTDDK7LWI_5EViwl2VXOA9ERvND6T3rryDI619-Nxuv/s320/Slide3.PNG" width="320" /></a>
</div>
<br />
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhc_elMmhWAGzOMSZdgq80xDmau70Q7TNVnNOggI5cbWk8tFpbUZ1fJkbT3ai3IwV8C16KhtF8GRN7xQUDqq_8Jd7hcdCfCqAsYLzPmOrYYH7uhUz2bjHit-mH8r6hKa5eftuGCDvbnVc2MpcL3LT_QPRYWQOf7rL7rlbP1FkzzFiZdnwaXc0SX1uax/s960/Slide4.PNG" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="720" data-original-width="960" height="240" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhc_elMmhWAGzOMSZdgq80xDmau70Q7TNVnNOggI5cbWk8tFpbUZ1fJkbT3ai3IwV8C16KhtF8GRN7xQUDqq_8Jd7hcdCfCqAsYLzPmOrYYH7uhUz2bjHit-mH8r6hKa5eftuGCDvbnVc2MpcL3LT_QPRYWQOf7rL7rlbP1FkzzFiZdnwaXc0SX1uax/s320/Slide4.PNG" width="320" /></a>
</div>
<p>
Note that the exponential terms in the equations for Zsc and Zoc cancel if the
line-length for all measurements are identical. Thus, if the
transmission line lengths (including the short and open) are identical, this
equation is <i>independent</i> of any non-ideal characteristic of the
transmission line (e.g. loss).
</p>
<p>
And because of this requirement that the length of the transmission line
be the same for both sets of S11 measurements. I would
<i>not</i> recommend just leaving the coax unterminated for the open
measurement. Ideally, you should use OPEN and SHORT calibration
standards that have the same length (i.e. the same Offset Delay) from the
connectors' Reference Plane to the actual short or open implemented within the
standard.
</p>
<p><br /></p>
<p>
<b><u>Example, Finding Zo of Belden 75-ohm Coax:</u></b>
</p>
<p>
In this example I'll use SimSmith to simulate two versions of 75 ohm coax --
one version of coax is perfect, without loss, and the other version is
Belden 9659 (which does have some loss) to demonstrate the calculation of Zo
and to show the effect of loss on Zo.
</p>
<p>
Both lengths of simulated-coax are 40 feet long, and I will use SimSmith to
create S11 files of the lines terminated with a short and with an open.
</p>
<p>
First, below is an S11 plot of the perfect 75 ohm coax terminated with a
SHORT. (The OPEN termination looks the same, except that it starts at "3
o'clock" on the Smith Chart). Note how the path (in green) follows the
boundary of the Smith Chart's unit circle. In this case, SWR is always
infinite.
</p>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgcKjZMzPsStUKT3sdBU3ZXJMmgfs9ZX-iPCk-SeGzlF80iO8UBMqMDkdflPPqv6jlihrx50c5LX5u3n6SeOzG0Ri4qelYGbbo14IIXTaxlI_6vRk1kVBW_KrtQR-JjBq9qHOg5-wwj5u2X_UX7_IT4s8U6VCcInS9hJV6DBM8j4LXBBbJJcAXYpGSo/s1177/230520%20SimSmith%20Perfect%2075%20SHORT.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="721" data-original-width="1177" height="196" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgcKjZMzPsStUKT3sdBU3ZXJMmgfs9ZX-iPCk-SeGzlF80iO8UBMqMDkdflPPqv6jlihrx50c5LX5u3n6SeOzG0Ri4qelYGbbo14IIXTaxlI_6vRk1kVBW_KrtQR-JjBq9qHOg5-wwj5u2X_UX7_IT4s8U6VCcInS9hJV6DBM8j4LXBBbJJcAXYpGSo/s320/230520%20SimSmith%20Perfect%2075%20SHORT.png" width="320" /></a>
</div>
<div class="separator" style="clear: both; text-align: center;"><br /></div>
Next is an S11 plot of the Belden 9659 75-ohm coax, terminated with a
short. Note that the path no longer follows the boundary of the Smith
Chart's unit circle. Instead, it spirals in (i.e. SWR improving as it
spirals in). This spiraling is due to coax-cable loss.
<div>
<br />
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi8XmsZc2_LzlyPJs_9XQmo7T8FZswc3M0TV4lKyHWTg7WEEX5xb16dGNmU-wAH08ig2XhGpZ6LnqDTq_-pK1jkL-R9yw57VH0XF4XYVdvg5boeVuZ1pmbI_mptCJ0UWMYeHqhgxBAEYgvfiHwOGDybrNOiJVT8UirxtbQNCUVV7_nDFvBwCxEvL9jR/s1177/230520%20SimSmith%20Belden%209659%2075%20SHORT.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="721" data-original-width="1177" height="196" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi8XmsZc2_LzlyPJs_9XQmo7T8FZswc3M0TV4lKyHWTg7WEEX5xb16dGNmU-wAH08ig2XhGpZ6LnqDTq_-pK1jkL-R9yw57VH0XF4XYVdvg5boeVuZ1pmbI_mptCJ0UWMYeHqhgxBAEYgvfiHwOGDybrNOiJVT8UirxtbQNCUVV7_nDFvBwCxEvL9jR/s320/230520%20SimSmith%20Belden%209659%2075%20SHORT.png" width="320" /></a>
</div>
<div class="separator" style="clear: both; text-align: center;"><br /></div>
And here is the S11 plot of the same Belden 9659 75-ohm coax terminated with
an OPEN. Again, note the spiraling-in due to loss.
</div>
<div>
<br />
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgajSm3fQ6utxDwWTyRKW66atXfi9iRGdc4Ihe1Lo9Bkfw0-DVQ9seVcYFqnulcwjns5fFiqV6xFGkhrUokdp_7jJw1K9sd3RfDTXJArkQG6Lak7g7S0RPiL-WqdVmSYczy5BFJWDo7kUB-uz8G01ROrquxco4Ao1RhUdrnbdJhO-rlogTKOx_q_yaj/s1177/230520%20SimSmith%20Belden%209659%2075%20OPEN.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="721" data-original-width="1177" height="196" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgajSm3fQ6utxDwWTyRKW66atXfi9iRGdc4Ihe1Lo9Bkfw0-DVQ9seVcYFqnulcwjns5fFiqV6xFGkhrUokdp_7jJw1K9sd3RfDTXJArkQG6Lak7g7S0RPiL-WqdVmSYczy5BFJWDo7kUB-uz8G01ROrquxco4Ao1RhUdrnbdJhO-rlogTKOx_q_yaj/s320/230520%20SimSmith%20Belden%209659%2075%20OPEN.png" width="320" /></a>
</div>
<div class="separator" style="clear: both; text-align: center;"><br /></div>
The following two plots show the calculation of Zo, from S11 data, using the
SQRT(Zsc*Zoc) method. Zo for both the perfect (lossless) transmission
line and for the Belden 9659 75-ohm transmission line are shown.
</div>
<div><br /></div>
<div>The first plot is to 40 MHz.</div>
<div>
<br />
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjmADac9l4lfIqEYKK3DkbW0DJTmQ7Jan5Xp-evP3Wte1VcV7AgEzTuqTbAsX7SfZSvvCUxVFr_pXy_9fPz7-LUCEPNh0Z8jGFIZipBGL4mT6J3_bAG5wez02-fig31_18At-E72r79VEjo9YJupa2h2lSOIgyB_0CR-ElK1B0_9LmmN9LDynqXfyyD/s562/230520%20plot%2040ft%2075ohm%20Zo%20to%2040%20MHz.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="506" data-original-width="562" height="288" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjmADac9l4lfIqEYKK3DkbW0DJTmQ7Jan5Xp-evP3Wte1VcV7AgEzTuqTbAsX7SfZSvvCUxVFr_pXy_9fPz7-LUCEPNh0Z8jGFIZipBGL4mT6J3_bAG5wez02-fig31_18At-E72r79VEjo9YJupa2h2lSOIgyB_0CR-ElK1B0_9LmmN9LDynqXfyyD/s320/230520%20plot%2040ft%2075ohm%20Zo%20to%2040%20MHz.png" width="320" /></a>
</div>
<div><br /></div>
<div>
And the second plot (to better show the Belden cable's Zo at low frequency)
is to 4 MHz.
</div>
<div><br /></div>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgM94Hnj5l1IoZQMnqjQ719LIGDlXobqOVN0m-cAAOHTGFrfFqU0gAFt5xH0qWnIbK8CculFZs1EiIcPZdjbSS_b11UGsvzaDFCfz5C7BHXaM-4dskTJCN2J2bKceX5ENU35Mo7lxCfDEy5Y7mL7q3HGbk9dZyi2A_qG6PyONwvyOU0MHn9YSxpxy0g/s562/230520%20plot%2040ft%2075ohm%20Zo%20to%204%20MHz.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="506" data-original-width="562" height="288" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgM94Hnj5l1IoZQMnqjQ719LIGDlXobqOVN0m-cAAOHTGFrfFqU0gAFt5xH0qWnIbK8CculFZs1EiIcPZdjbSS_b11UGsvzaDFCfz5C7BHXaM-4dskTJCN2J2bKceX5ENU35Mo7lxCfDEy5Y7mL7q3HGbk9dZyi2A_qG6PyONwvyOU0MHn9YSxpxy0g/s320/230520%20plot%2040ft%2075ohm%20Zo%20to%204%20MHz.png" width="320" /></a>
</div>
<br />
<div>
Note that Zo of the perfect coax is 75 + j0 ohms irrespective of
frequency. However, the Belden coax always has a reactive (imaginary)
component (whose value decreases as frequency increases), and at lower
frequencies its Zo deviates appreciably from the ideal 75 ohms.
</div>
</div>
<div><br /></div>
<div>
What causes this deviation from 75 ohms, especially at low frequencies?
</div>
<div><br /></div>
<div>
A transmission line's Zo is due to four physical characteristics of the
transmission line: resistance, conductance, inductance, and capacitance,
per the equation in the image, below.
</div>
<div>
<br />
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgvRGg0SEwETGBcl6EQdICWfT0E9n8qkXgYkkY41yng_7kkwu7PsKrfPPlJ72UJj8DXh6-j9MgHBwksG-UiCWW-DYfuGN01--PJMuFhwJXoNuUeZkNaOoGTM5eDxc_ZJrYTkjFqsU_qGiYHsol3YZsnZ1V9O04RKjKAOt1Y_EO1nc1Q8KGoNWG2zq9n/s811/Zo%20equations.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="545" data-original-width="811" height="215" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgvRGg0SEwETGBcl6EQdICWfT0E9n8qkXgYkkY41yng_7kkwu7PsKrfPPlJ72UJj8DXh6-j9MgHBwksG-UiCWW-DYfuGN01--PJMuFhwJXoNuUeZkNaOoGTM5eDxc_ZJrYTkjFqsU_qGiYHsol3YZsnZ1V9O04RKjKAOt1Y_EO1nc1Q8KGoNWG2zq9n/s320/Zo%20equations.png" width="320" /></a>
</div>
<div class="separator" style="clear: both; text-align: center;"><br /></div>
If the numerator and denominator under the top equation's square root are
expanded, they will form a 'real' term and an 'imaginary' term -- that is, a
complex number. A square-root of a complex number is a complex number,
and thus a transmission line's characteristic impedance is <i>always</i> a
complex value.
</div>
<div><br /></div>
<div>
At lower frequencies the greater will be the effect of the loss factors R and
G on the real and imaginary components of Zo. This effect can be seen
in the plots, above.
</div>
<div><br /></div>
<div>
However, if the frequency is high enough, or if R and G are negligible, we can
approximate Zo as a 'real' value, without any imaginary component.<br />
<p><br /></p>
<p>
<b><u>Real-World Issues:</u></b><br />
</p>
<p>
There was recently an interesting discussion about this method on the
nanovan-users group,
<a href="https://groups.io/g/nanovna-users/topic/measure_the_characteristic/98802426?p=,,,20,0,0,0::recentpostdate/sticky,,,20,2,0,98802426,previd%3D1685349043971057685,nextid%3D1683873957178087002&previd=1685349043971057685&nextid=1683873957178087002">here</a>.
</p>
<p>
François, F1AMM, posted the following image showing his calculation of the real part of Zo, where Zo was calculated using his measurements of Zsc and Zoc. It does not look anything like my SimSmith-simulated curves, above.</p>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjhqf1NMOzoMKv2dd2wARnk_sKfzEr9rk3bt5H9Mmi89XtX0nIPfEWTPKbxl27Jih_FIKqv80TjUXEeYfgsHAitl4GPGW3h7uAknhq4EJvBcdZqPXOmZ2jyrRv6W__XKeiHkwYGF4rtzPOgc4nw5fphnt_ANqkNKzCorUoK4MA5gmXMi7Z9XiKRTyQF/s638/F1AMM%20Image.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="396" data-original-width="638" height="199" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjhqf1NMOzoMKv2dd2wARnk_sKfzEr9rk3bt5H9Mmi89XtX0nIPfEWTPKbxl27Jih_FIKqv80TjUXEeYfgsHAitl4GPGW3h7uAknhq4EJvBcdZqPXOmZ2jyrRv6W__XKeiHkwYGF4rtzPOgc4nw5fphnt_ANqkNKzCorUoK4MA5gmXMi7Z9XiKRTyQF/s320/F1AMM%20Image.png" width="320" /></a>
</div>
<p>Why the difference?</p><p>I took my SimSmith-simulated Zsc and Zoc files for RG-142B and, using MATLAB, applied a frequency-dependent phase shift to S11 of one of the files (simulating an additional delay on that line). The result is plotted below:</p>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEib7swR7OrVaTQ1uVx0nbYrNIoWqW4pZDi-kS9Gab-6yN68FmUS7sZXwnvaBFYtq57oCwisRofMY_7QSoadi2fbSz8nDu8SEKVulSyxCCce0uCnMVCHE31iErnAmdpq0tSnKxMySCXZgHT4stju_QatGtuiSpXE27NdCH7dMaomEsmMAFU0oai0rHTw/s1000/230525%20increasing%20phase%20error.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="661" data-original-width="1000" height="212" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEib7swR7OrVaTQ1uVx0nbYrNIoWqW4pZDi-kS9Gab-6yN68FmUS7sZXwnvaBFYtq57oCwisRofMY_7QSoadi2fbSz8nDu8SEKVulSyxCCce0uCnMVCHE31iErnAmdpq0tSnKxMySCXZgHT4stju_QatGtuiSpXE27NdCH7dMaomEsmMAFU0oai0rHTw/s320/230525%20increasing%20phase%20error.png" width="320" /></a>
</div>
This was revealing, and I suspect that François' measurements were the result of his SHORT and OPEN standards being different lengths.</div><div><br /></div><div>I was curious if my homebrew SMA standards would perform better, and so I measured Zsc and Zoc of a 10 foot length of RG-142B using both my NanoVNA and my HP 8753C. The Zo-calculation results are shown below.</div><div><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj_-Btvg9qeEHNJGSCnZ5aGu1ujPmkmiSfxqm5nreWdc9wfoKbiNsIf5c-jTGOL0JxNOsqZdeuhbAiRZzAMoyTbWgb_cmOEKislv3R4QAp5UnAayKqH9SZdW4zQiv-TRSHRtCKEqAhOQF4ajKl9Q06Kdzd1OXceeWlrGzcR_fibncRIVB-Jm9WXvQgO/s939/230524%205ft1in%20RG-142%20SMA%20Ro%20Xo.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="730" data-original-width="939" height="249" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj_-Btvg9qeEHNJGSCnZ5aGu1ujPmkmiSfxqm5nreWdc9wfoKbiNsIf5c-jTGOL0JxNOsqZdeuhbAiRZzAMoyTbWgb_cmOEKislv3R4QAp5UnAayKqH9SZdW4zQiv-TRSHRtCKEqAhOQF4ajKl9Q06Kdzd1OXceeWlrGzcR_fibncRIVB-Jm9WXvQgO/s320/230524%205ft1in%20RG-142%20SMA%20Ro%20Xo.png" width="320" /></a></div><p>The errors aren't as dramatic as those of François, but they are greater than I had expected. What could be their cause?</p><p>To explore further, I wrote a MATLAB routine that simulated a transmission line and a "virtual" VNA, to allow me to simulate a 75-ohm transmission line with the following test options:</p><p></p><ol style="text-align: left;"><li>Calibrate the "virtual" VNA with Actual or Ideal SOL standards</li><li>Set the SOL Characteristics (within a "virtual VNA") to be their actual characteristics or their ideal characteristics.</li><li>Terminate the simulated coax with a Short or an Open whose characteristics are either actual or ideal.</li><li>Play around with various "actual" SOL parameters (e.g. capacitance of the Open, inductance of the Short, resistance of the Load, "Offset Delays" of each (see <a href="https://k6jca.blogspot.com/2019/12/vna-notes-on-12-term-error-model-and.html">here</a> for explanation).</li></ol><p></p><p>The MATLAB simulation can be downloaded from:</p><p><a href="https://github.com/k6jca/Simulation_of_Zo_sqrt-ZscZoc-_calculation">https://github.com/k6jca/Simulation_of_Zo_sqrt-ZscZoc-_calculation</a></p>
<p>The following plots are the results of some of these simulations.</p><p>First, verifying the simulation by calculating Zo assuming the SOL standards and the line-terminations are perfect. Not surprisingly, the plot looks as it should.</p>
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgUQ-DycVhaO0bZtLiDyj00e0lKHid9yJU-Q-4HyeXNd3yMiVuKC0a4l9ChqyeWWdiS2v7KdAxbT8vu9R4_9ufCkqPnPeSA65QNJgkYNJex-augPPvwGXwzUkwO7q-5fCPRqOZNI1mIlZexLEtEKVfAgDHHSkdAAiwhLSwZYDAY1TghPPbtZB64WFDS/s1062/fig%201%20everything%20perfect.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="786" data-original-width="1062" height="237" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgUQ-DycVhaO0bZtLiDyj00e0lKHid9yJU-Q-4HyeXNd3yMiVuKC0a4l9ChqyeWWdiS2v7KdAxbT8vu9R4_9ufCkqPnPeSA65QNJgkYNJex-augPPvwGXwzUkwO7q-5fCPRqOZNI1mIlZexLEtEKVfAgDHHSkdAAiwhLSwZYDAY1TghPPbtZB64WFDS/s320/fig%201%20everything%20perfect.png" width="320" /></a>
</div>
<div><br /></div>Next, I changed the LOAD standard from being a perfect 50 ohms to 51 ohms (i.e. it has 40 dB Return Loss), but keeping the VNA's characterization of the load at a perfect 50 ohms. </div><div><br /></div><div>The results surprised me. After all, how many of us run our VNAs this way -- assume that the LOAD, even though it might not be exactly 50 ohms, is close enough?</div><div><br /><div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEixRTyqCH0bvzS9IHXIyO5HxxeIyT2W5cWvBcaFRUY2XIAa-G9BWldNXemxBSiQ0601dCYL9KhwwmSjl2CiQmLIslecrUfQJ7te9eTVS6frAKtd_SZKqkD9iCXpAEVU3--tcCauktmdVYANQ7gwXeenhprOafQVrSNoDmRYccoRRzF1ZQWDaR3qB67p/s1058/fig%202%20everything%20perfect,%20but%20cal%20with%20actual%20LD.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="796" data-original-width="1058" height="241" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEixRTyqCH0bvzS9IHXIyO5HxxeIyT2W5cWvBcaFRUY2XIAa-G9BWldNXemxBSiQ0601dCYL9KhwwmSjl2CiQmLIslecrUfQJ7te9eTVS6frAKtd_SZKqkD9iCXpAEVU3--tcCauktmdVYANQ7gwXeenhprOafQVrSNoDmRYccoRRzF1ZQWDaR3qB67p/s320/fig%202%20everything%20perfect,%20but%20cal%20with%20actual%20LD.png" width="320" /></a>
</div>
<p>Next, I simulated the "virtual" VNA being calibrated with perfect standards, but the Zsc and Zoc measurements made with imperfect terminations (the OPEN had 50 femtoFarads of capacitance plus an offset-delay of 17.5 picoseconds, while the SHORT, although have no inductance, had 17.8 picoseconds of offset-delay).</p>
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhDM_joh1hOkkS7oA5eAE9CAt8WY__P3tHD-9fUYmnQdnkJSBXKRJGc8sX0sx0JO28HEy1WlcjO4FeA_89YgX5_O0p7wusYN2JW9VEV_E5dP0qvvwFbsSq8GEPJ6HcaiXW0bx5xpOfUYXZQFu6msPYiPdID6_YvGEAI4sH1S8Ud4uM--YFFqeNVbmPe/s1058/fig%203%20everything%20perfect,%20but%20test%20with%20actual%20SO.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="796" data-original-width="1058" height="241" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhDM_joh1hOkkS7oA5eAE9CAt8WY__P3tHD-9fUYmnQdnkJSBXKRJGc8sX0sx0JO28HEy1WlcjO4FeA_89YgX5_O0p7wusYN2JW9VEV_E5dP0qvvwFbsSq8GEPJ6HcaiXW0bx5xpOfUYXZQFu6msPYiPdID6_YvGEAI4sH1S8Ud4uM--YFFqeNVbmPe/s320/fig%203%20everything%20perfect,%20but%20test%20with%20actual%20SO.png" width="320" /></a>
</div>
<div><br /></div>I wondered if we could remove the plotted-error by using the imperfect SHORT's and OPEN's actual characteristics in the "virtual" VNA for calibration. Note that the LOAD was assumed to a perfect 50 ohms.</div><div><br /><div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEildsVG__AJFCO86Ncu0hqUtyxZEteVKX4lwCEDrxAJk8Do8Fr3Xg-nkRTeDB_imuVIHdBnNsSZtWH7sQf5XtXwg5iBInPxWOdddm2K2c8HImRU8B8jmD-JKfoFkG208b1YyZ5J9rGo22OLmBPmG4mTQHuzHrRnRa5WrTr7NrKdMHmzzjqjMsde4Yug/s1058/fig%204%20cal%20with%20actual%20SO,%20(ld%20perfect)%20test%20with%20actual%20SO.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="796" data-original-width="1058" height="241" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEildsVG__AJFCO86Ncu0hqUtyxZEteVKX4lwCEDrxAJk8Do8Fr3Xg-nkRTeDB_imuVIHdBnNsSZtWH7sQf5XtXwg5iBInPxWOdddm2K2c8HImRU8B8jmD-JKfoFkG208b1YyZ5J9rGo22OLmBPmG4mTQHuzHrRnRa5WrTr7NrKdMHmzzjqjMsde4Yug/s320/fig%204%20cal%20with%20actual%20SO,%20(ld%20perfect)%20test%20with%20actual%20SO.png" width="320" /></a>
</div>
<div><br /></div>The result still had errors. </div><div><br /></div><div>Finally, I ran a simulation in which the SOL standards were assumed by the "virtual" VNA to be perfect (i.e. their Characterizations), but the imperfect standards were used for calibration and for the Zsc and Zoc measurements, to simulate how I suspect many users (such as me) use their NanoVNAs).</div><div><br /><div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgK5xF6Swxrz7w5k3VIY3lpxkS3B7NlphegIZozxXbtxtdDJ5eFMhqY2esi8Zayx5jwEf1e3IctQ8MwnWNNC7mRU6AHRiBp5q5vYCMVdjEAw3NUG6jE0DEJ4JGGT0bCcA63k8xLLBoqM-lfgn3Sk8M-Jpf0Cd-9NuEsP7Fy15oitnZCMVPFY3CGiD46/s1058/fig%205%20SOL%20Char%20perfect,%20SOL%20cal%20Actual,%20Term%20Actual.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="796" data-original-width="1058" height="241" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgK5xF6Swxrz7w5k3VIY3lpxkS3B7NlphegIZozxXbtxtdDJ5eFMhqY2esi8Zayx5jwEf1e3IctQ8MwnWNNC7mRU6AHRiBp5q5vYCMVdjEAw3NUG6jE0DEJ4JGGT0bCcA63k8xLLBoqM-lfgn3Sk8M-Jpf0Cd-9NuEsP7Fy15oitnZCMVPFY3CGiD46/s320/fig%205%20SOL%20Char%20perfect,%20SOL%20cal%20Actual,%20Term%20Actual.png" width="320" /></a>
</div>
<p>Not surprisingly, there are still errors.</p><p><br /></p>
<p>
<b><u>Conclusion:</u></b>
</p>
<p>
The "SQRT(Zsc*Zoc)" method of calculating Zo will theoretically calculate accurate real and imaginary components of a transmission line's
Zo.
</p>
<p>However, from my plots above, in reality there always will be some amount of error in the Zo calculation, because it is impossible for the SHORT and OPEN standards to be perfect, and these imperfections <i>cannot</i> be calibrated away for the purposes of this method of Zo calculation.</p>
<p><br /></p>
<p>
<b><u>Some Useful Links:</u></b>
</p>
<p>
<a href="Determination of transmission line characteristic impedance from impedance measurements">Determination of transmission line characteristic impedance from
impedance measurements.</a> Owen Duffy blog post discussing this method of finding Zo.
</p>
<p>
<a href="https://athena.ecs.csus.edu/~milica/EEE161/lecturenotes/TransmissionLines.pdf">Waves on Transmission Lines.</a> (Useful on-line course notes from Sacramento State).
</p>
<p>
<a href="https://eng.libretexts.org/Bookshelves/Electrical_Engineering/Electronics/Microwave_and_RF_Design_II_-_Transmission_Lines_(Steer)/02%3A_Transmission_Lines/2.03%3A_The_Terminated_Lossless_Line">Section 2.3, The Terminated Lossless Line.</a> From on-line libretext. (Note error in the derivation of
equation 2.3.18 -- cos() is shown in the imaginary term. This should
be sin()).
</p>
<p>
<a href="https://eng.libretexts.org/Bookshelves/Electrical_Engineering/Electronics/Microwave_and_RF_Design_II_-_Transmission_Lines_(Steer)/02%3A_Transmission_Lines/2.05%3A_The_Lossy_Terminated_Line">Section 2.5, The Lossy Terminated Line.</a> From on-line libretext. (Note error in the derivation of
equation 2.5.5 -- cosh() is shown in the imaginary term. This should
be sinh()).
</p>
<p><br /></p>
<p>
<b><u>Standard Caveat:</u></b>
</p>
<p>
As always, I might have made a mistake in my equations, assumptions,
drawings, or interpretations. If you see anything you believe to be in
error or if anything is confusing, please feel free to contact me or comment
below.<br /><br />And so I should add -- this information is distributed in
the hope that it will be useful, but WITHOUT ANY WARRANTY; without even the
implied warranty of MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.
</p>
</div>
Jeffhttp://www.blogger.com/profile/04853314106806116765noreply@blogger.com0tag:blogger.com,1999:blog-2257458838727315711.post-80749175482693638632023-05-19T15:45:00.011-07:002023-05-21T15:29:56.488-07:00Transmission Lines: Input Impedance of a Terminated Line<div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiqVOXgj8-Xo9AKXG8QvW_xCF7m9IGWyV9_95evErb3TETYkkzcTMn3iAUensBEMu5zrRM-1_2ciphidO8KilwrdTo_uMuShWi5901ThfQtdSCrcffaOZeglcp5heL1A8naCH4zBJOHk92XCX5p9BrQcWJd_rAx0QJGXERo3wYDL1Q5mxiC2wTPstab/s786/blog%20header%20-%20%20terminated%20transmission%20line%202.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="427" data-original-width="786" height="174" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiqVOXgj8-Xo9AKXG8QvW_xCF7m9IGWyV9_95evErb3TETYkkzcTMn3iAUensBEMu5zrRM-1_2ciphidO8KilwrdTo_uMuShWi5901ThfQtdSCrcffaOZeglcp5heL1A8naCH4zBJOHk92XCX5p9BrQcWJd_rAx0QJGXERo3wYDL1Q5mxiC2wTPstab/s320/blog%20header%20-%20%20terminated%20transmission%20line%202.png" width="320" /></a></div><p>A common equation for calculating Zin of a terminated transmission line
uses complex exponentials. But now and again I come across an equivalent equation that instead uses the hyperbolic tangent function, tanh().</p>
<p>
I have <i>never</i> used hyperbolic trig functions as an engineer -- in fact, the only time I've come across them is with
respect to transmission lines, and personally, I find the complex-exponential form of these equations much
more intuitive, given how common complex exponentials are used in electrical
engineering.
</p>
<p>This blog post shows how the tanh() form of the equation for Zin, the input impedance of a terminated <i>lossy</i> transmission line, can be derived from the complex-exponential equation.</p>
<p>
Also, I will show how, for a lossless transmission line, the tanh() equation reduces to one using tangent (tan()) functions.</p>
<p>I'll start out this post with the lossy-line's complex-exponential equation for Zin, which is:</p>
<p style="text-align: center;">Zin = Zo*(1 + Γ*e^(-2*γ*<i>l</i>)) / (1 - Γ*e^(-2*γ*<i>l</i>)),</p><p>where Γ = (Zload - Zo) / (Zload + Zo), γ = α + jβ, and <i>l</i> = length. Note that α represents loss and jβ represents phase-shift as a signal travels along a transmission line.</p><p>And later in this post I'll do the same for the lossless line, whose complex-exponential equation is:</p><p style="text-align: center;">Zin = Zo*(1 + Γ*e^(-j2*<span style="text-align: left;">β</span>*<i>l</i>)) / (1 - Γ*e^(-j2*<span style="text-align: left;">β</span>*<i>l</i>)),</p><div>which is the same equation is the first one, above, but with α (which represents loss) set to 0.</div><div><br /></div><div>The rest will be explained in the following slides...</div><div><br /></div>
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<p>That's it!</p><p><br /></p><p><b><u>Standard Caveat:</u></b></p>
<p>
As always, I might have made a mistake in my equations, assumptions, drawings,
or interpretations. If you see anything you believe to be in error or if
anything is confusing, please feel free to contact me or comment below.<br /><br />And
so I should add -- this information is distributed in the hope that it will be
useful, but WITHOUT ANY WARRANTY; without even the implied warranty of
MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.</p>Jeffhttp://www.blogger.com/profile/04853314106806116765noreply@blogger.com0tag:blogger.com,1999:blog-2257458838727315711.post-83563811038955057472022-12-24T07:13:00.007-08:002022-12-25T19:07:08.663-08:00An FPGA SDR Transceiver (A PowerPoint Presentation)<p>These are slides from a PowerPoint presentation on my Homebrew Direct-Sampling FPGA Transceiver that I gave at the <a href="https://www.fars.k6ya.org/">Foothill Amateur Radio Society</a> (FARS) in November, 2022.</p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgWZZkam8Y_X3XsT42NqAe432Oj6tme3XdsuJapK5cRGwGIBn8L6Lt40k8G_r6_D5u6xJuRKi3WT-3m4YHmMK9RxT03bUsU381ohtx6v0r5aHgEgRm_RtI6L2KG-jtvWj1IYK2y_whC9en1sJk9ajnh71KZuA-LIXudi07W7vdXhLbCDNCoaWMRWDhg/s960/Slide1.PNG" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="720" data-original-width="960" height="240" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgWZZkam8Y_X3XsT42NqAe432Oj6tme3XdsuJapK5cRGwGIBn8L6Lt40k8G_r6_D5u6xJuRKi3WT-3m4YHmMK9RxT03bUsU381ohtx6v0r5aHgEgRm_RtI6L2KG-jtvWj1IYK2y_whC9en1sJk9ajnh71KZuA-LIXudi07W7vdXhLbCDNCoaWMRWDhg/s320/Slide1.PNG" width="320" /></a></div><br /><div class="separator" style="clear: both; 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text-align: center;"><br /></div>That's it! For more information on the FPGA transceiver please refer to the series of blog posts that delve more deeply into the design, starting with: <a href="https://k6jca.blogspot.com/2017/02/an-fpga-sdr-hf-transceiver-part-1.html">https://k6jca.blogspot.com/2017/02/an-fpga-sdr-hf-transceiver-part-1.html</a><div><br /></div><div>73,</div><div><br /></div><div>- Jeff, k6jca</div>Jeffhttp://www.blogger.com/profile/04853314106806116765noreply@blogger.com0tag:blogger.com,1999:blog-2257458838727315711.post-33349617882364305382022-06-30T05:02:00.027-07:002023-09-03T16:17:50.121-07:00L-Network Design Equations for Transforming a Complex Load to a Real Zo<p>This blog post summarizes L-Network equations I derived for transforming any complex load impedance (as long as its real part is not negative), to a real impedance (i.e. no imaginary component) when looking into the input port of the L-Network.</p><p>The derivation of these equations is described in a much longer blog post back in 2015 that can be found here:
</p>
<p>
<a href="https://k6jca.blogspot.com/2015/04/revisiting-l-network-equations-and.html">https://k6jca.blogspot.com/2015/04/revisiting-l-network-equations-and.html</a><br />
</p>
<p>These derivations can also be found in my article, "Correcting a Common L-Network Misconception," published in the March/April issue of <i>QEX</i>.</p><p>In addition, I also present an alternate expression of these equations in terms of Q.</p><p>Note that the equations described in this post are for
<i>lossless</i> L-Networks. Design equations for L-Networks with
<i>lossy</i> components are <i>significantly</i> more complex, and are
described in the following blog post <a href="https://k6jca.blogspot.com/2018/09/l-networks-new-equations-for-better.html">https://k6jca.blogspot.com/2018/09/l-networks-new-equations-for-better.html</a></p>
<p><u style="font-weight: bold;">And a final note:</u> I have recently posted L-Network equations for transforming any impedance to any other impedance, not just to a non-complex (i.e. real-only) impedance (as described in this blog post). You can find these equations here: <a href="https://k6jca.blogspot.com/2023/08/l-network-equations-for-any-impedance.html">https://k6jca.blogspot.com/2023/08/l-network-equations-for-any-impedance.html</a></p>
<p>
<b><u>Introduction:</u></b>
</p>
<p>
There are a total of eight possible lossless L-Network configurations,
consisting of four Parallel-Series L-Network configurations and four
Series-Parallel L-Networks configurations, as shown in the figure, below:
</p>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjRM_-hRNa5-6vGan2i61qglbP6N4eTLM5jRIUCS_t9eCHh2X5HDfK4hpdCZX-ruk0Hyiupj6gjl3BxHe3KHLQ2b9DeAxfW4HpxGt2RnFo7o1uuDOoeqVSF-WP3g2e7bkj0PieAb5gyQbzCTcngMs7ba_q26Dk9gc68Hm2YSaIxA1jD_vjMc9eJNaco/s1295/220630%20the%208%20possible%20L-networks.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="635" data-original-width="1295" height="157" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjRM_-hRNa5-6vGan2i61qglbP6N4eTLM5jRIUCS_t9eCHh2X5HDfK4hpdCZX-ruk0Hyiupj6gjl3BxHe3KHLQ2b9DeAxfW4HpxGt2RnFo7o1uuDOoeqVSF-WP3g2e7bkj0PieAb5gyQbzCTcngMs7ba_q26Dk9gc68Hm2YSaIxA1jD_vjMc9eJNaco/s320/220630%20the%208%20possible%20L-networks.png" width="320" /></a>
</div>
<p>
The range of impedances that each L-Network can match (to Zo) are shown as the
gray regions on the Smith Charts, below:
</p>
<p></p>
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjMfPOCDU3QwZ3jC7QOo-ajN_YgeqLIr-VwqbFpQxILuRTtPYpbUrLdsgvPFCPGde9DsClJAaBWCCaVpnK0SvTxVBTTLpNKqUCxKxEIqbiXn5TpkPQv3ZAAk15yyNmajKQ6ikaZOWWMumIjvzGH27VbuJjcgg91mx2wfX-94iCsJ0t6yWkAknZoh9lk/s1869/220706%20Parallel-Series%20four%20Impedance%20Regions.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="1869" data-original-width="1686" height="320" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjMfPOCDU3QwZ3jC7QOo-ajN_YgeqLIr-VwqbFpQxILuRTtPYpbUrLdsgvPFCPGde9DsClJAaBWCCaVpnK0SvTxVBTTLpNKqUCxKxEIqbiXn5TpkPQv3ZAAk15yyNmajKQ6ikaZOWWMumIjvzGH27VbuJjcgg91mx2wfX-94iCsJ0t6yWkAknZoh9lk/s320/220706%20Parallel-Series%20four%20Impedance%20Regions.png" width="289" /></a>
</div>
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiEm9xZ9oEUyHPaOZnk3L48Mtu1xgsWp49dEwhyTAXCyoK0wOaxoFVtWGMpfrHmvkMBEk8jqH3kyI-oJEPl21kv4hr7PKOU4uR8mGkZ0criW6CVGyxvL3FGVIF_ubK0bJdLYvP3xVDwXHNXQqmpe7VbrO20dQmp_IHDBmnSwZVMRDLlkDc4Uz7QZymF/s1869/220706%20Series-Parallel%20four%20Impedance%20Regions.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="1869" data-original-width="1695" height="320" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiEm9xZ9oEUyHPaOZnk3L48Mtu1xgsWp49dEwhyTAXCyoK0wOaxoFVtWGMpfrHmvkMBEk8jqH3kyI-oJEPl21kv4hr7PKOU4uR8mGkZ0criW6CVGyxvL3FGVIF_ubK0bJdLYvP3xVDwXHNXQqmpe7VbrO20dQmp_IHDBmnSwZVMRDLlkDc4Uz7QZymF/s320/220706%20Series-Parallel%20four%20Impedance%20Regions.png" width="290" /></a>
</div>
<br />Let us first look at the design equations for a lossless Parallel-Series
L-Network...
<p></p>
<p><u><b><br /></b></u></p><p>
<u><b>Design Equations for a Lossless Parallel-Series L-Network:</b></u>
</p>
<p>
A Parallel-Series L-network can transform Zload = Rload + jXload to a
target-resistance Zo if the following condition is satisfied:
</p>
<p style="text-align: center;">Rload ≤ Zo</p>
<p>
Assuming this condition is satisfied, the design equations for a
Parallel-Series L-network result in two sets of B, X pairs (B is the
susceptance of the "parallel" (i.e. shunt) component, and X is the reactance
of the series component).
</p>
<p>
Each B, X pair represents a unique Parallel-Series L-network
implementation. Thus, the impedance transformation can be implemented
with either of the two Parallel-Series L-Networks represented by these two B,
X pairs.
</p>
<p>
The figure below describes the equations for calculating the two B, X pairs
for a Parallel-Series L-Network: </p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEigeGvzsCLT7oD5MeCDB9eTFqr9wHxfVbGauVIrK57TA3u7C6SHaVWuB468kfRSXfOHEhIT1KyFXfc3TNr9T01uYL-Xw4HWUGwJs29sdzJHvoVqda5Re0KyE12k2qLWSkNfwEyWv3DCEIvYjTVF36_be-IMd7V2Zbdoe3zY0loCJhaz0H-y_rMHI7_l/s1111/220930%20Parallel-Series%20Equations.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="740" data-original-width="1111" height="213" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEigeGvzsCLT7oD5MeCDB9eTFqr9wHxfVbGauVIrK57TA3u7C6SHaVWuB468kfRSXfOHEhIT1KyFXfc3TNr9T01uYL-Xw4HWUGwJs29sdzJHvoVqda5Re0KyE12k2qLWSkNfwEyWv3DCEIvYjTVF36_be-IMd7V2Zbdoe3zY0loCJhaz0H-y_rMHI7_l/s320/220930%20Parallel-Series%20Equations.png" width="320" /></a></div><p>These same equations can also be expressed in terms of Q. However, unlike the usual L-Network equations using Q that only transform a non-complex (e.g. resistive) load to a real, non-complex impedance, e.g. Zo (see <a href="https://eng.libretexts.org/Bookshelves/Electrical_Engineering/Electronics/Microwave_and_RF_Design_III_-_Networks_(Steer)/06%3A_Chapter_6/6.4%3A_The_L_Matching_Network">here</a>, <a href="https://web.ece.ucsb.edu/~long/ece145a/Notes5_Matching_networks.pdf">here</a>, or <a href="https://www.electronicdesign.com/technologies/communications/article/21800910/back-to-basics-impedance-matching-part-2">here</a>), these new equations can be used to transform <i>any</i> complex impedance to Zo.</p><p>In order to restate the above equations in terms of Q, first recognize that for a Parallel-Series L-Network we can derive Q to be:</p><p style="text-align: center;">Q = (Zo/Rload - 1)^0.5 = ((Zo - Rload)/Rload)^0.5</p><p>We can therefore substitute Q into the original Parallel-Series susceptance and reactance equations, above, wherever the quantity “((Zo - Rload)/Rload)^0.5” occurs in each equation. This quantity will need to be teased out of the reactance equations, but it is there. Just recognize that (Rload*(Zo-Rload))^0.5 is the same as Road*((Zo-Rload)/Rload)^0.5. </p><p>The resulting equations are shown, below:</p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiSs3HLQwDPeTTnYZi5T6RqcCMNl9s5Ju7PhlEwPpDhqaiGNRpIoiIBtYfNVL30SBI7ouiEDWnMNHp_7sSUMFv-VZ0HzkI3P2a7zKDa_NfAyG25cWvB7xKyP9rF1Qj8OvRGKlu1B1RVmWaP4v0GOjIyuApzmMqhqA5RGpd1C5CGMfNffPyYrDhKtYGI/s1032/230314%20Parallel-Series%20equations%20with%20Q.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="917" data-original-width="1032" height="284" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiSs3HLQwDPeTTnYZi5T6RqcCMNl9s5Ju7PhlEwPpDhqaiGNRpIoiIBtYfNVL30SBI7ouiEDWnMNHp_7sSUMFv-VZ0HzkI3P2a7zKDa_NfAyG25cWvB7xKyP9rF1Qj8OvRGKlu1B1RVmWaP4v0GOjIyuApzmMqhqA5RGpd1C5CGMfNffPyYrDhKtYGI/s320/230314%20Parallel-Series%20equations%20with%20Q.png" width="320" /></a></div><p><b><u><br /></u></b></p><p><b><u>Design equations for a Lossless Series-Parallel L-Network:</u></b></p>
<p>
A Series-Parallel L-network can transform Zload = Rload + jXload to a
target-resistance Zo if the following condition is satisfied:
</p>
<p style="text-align: center;">Gload ≤ 1/Zo</p>
<div>
where Gload (and Bload) can be calculated from Rload and Xload per the final
equations in the following derivation:
</div>
<div><br /></div>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiby9jBoz-g3MMMi8NlLSZ0kYgDnIMt2UDLamG4hD0qsKweFJHmR6RBKwjCqzZDtjTN0MQhuXGM48sT9sBtjSL-MsrDzZauS6ZRPnakqBalg6VsXaw97lklzpikhUJFKx499yxbohzXtxIYq4vpy-5pf9-KchecW2hww2d_N_dD2inNvR26rGPk6kVH/s741/220629%20Equations%20for%20Gload%20and%20Bload.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="563" data-original-width="741" height="243" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiby9jBoz-g3MMMi8NlLSZ0kYgDnIMt2UDLamG4hD0qsKweFJHmR6RBKwjCqzZDtjTN0MQhuXGM48sT9sBtjSL-MsrDzZauS6ZRPnakqBalg6VsXaw97lklzpikhUJFKx499yxbohzXtxIYq4vpy-5pf9-KchecW2hww2d_N_dD2inNvR26rGPk6kVH/s320/220629%20Equations%20for%20Gload%20and%20Bload.png" width="320" /></a>
</div>
<div><br /></div>
<div>
Assuming the condition that <span style="text-align: center;">Gload ≤ 1/Zo </span>is satisfied, the design equations for a Series-Parallel L-network result in
two sets of B, X pairs (B is the susceptance of the "parallel" (i.e. shunt)
component, and X is the reactance of the series component).
</div>
<p>
Each B, X pair represents a unique Series-Parallel L-network
implementation. Thus, the impedance transformation can be implemented
with either of the two Series-Parallel L-Networks represented by these two B,
X pairs.
</p>
<p>
The figure below describes the equations for calculating the two B, X pairs
for a Series-Parallel L-Network:</p><p><b></b></p><div class="separator" style="clear: both; text-align: center;"><b><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiK5Sk6UnwwaWG6FjxB8hyNIbHG9dau5YYvfg11PmSt4YzpreGPLD7Zce6qU5Z5d7x1CmKAQgZMJn5H3aD0DUd_RCCFyObJmfKKysSCFtLeI9dCpIs16Jhs_OwHhXAgeur3O_kNbdNW-StDmpBn05xXkvLlW4VSom9H_9gZT8RxIM_3ntB4Nv1jFYHK/s1107/220930%20Series-Parallel%20Equations.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="750" data-original-width="1107" height="217" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiK5Sk6UnwwaWG6FjxB8hyNIbHG9dau5YYvfg11PmSt4YzpreGPLD7Zce6qU5Z5d7x1CmKAQgZMJn5H3aD0DUd_RCCFyObJmfKKysSCFtLeI9dCpIs16Jhs_OwHhXAgeur3O_kNbdNW-StDmpBn05xXkvLlW4VSom9H_9gZT8RxIM_3ntB4Nv1jFYHK/s320/220930%20Series-Parallel%20Equations.png" width="320" /></a></b></div><div style="font-weight: bold; text-decoration-line: underline; text-decoration: underline;"><b><u><br /></u></b></div><div><p>Like the Parallel-Series equations earlier, these same equations can also be expressed in terms of Q. </p><p>In order to restate the above equations in terms of Q, it is useful to derive Q for a Series-Parallel L-Network in terms of the load conductance (Gload) and Zo expressed as conductance (i.e. 1/Zo, as Zo is a real, not complex, impedance). The resulting Q equation is:</p><p style="text-align: center;">Q = ((1/Zo)/Gload - 1)^0.5 = ((1/Zo - Gload)/Gload)^0.5</p><p>We can therefore substitute Q into the original Series-Parallel susceptance and reactance equations, above, wherever the quantity “((1/Zo - Gload)/Gload)^0.5” occurs in each equation (recognize that for the susceptance equations (Gload*(1/Zo-Gload))^0.5 is the same as Goad*((1/Zo-Gload)/Gload)^0.5 ). </p><p>The resulting equations are shown, below:</p></div><div style="font-weight: bold; text-decoration-line: underline; text-decoration: underline;"><b><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh1r7OM3CyP3Iim05iMhexVWNsgYq9jyLsAF4f1dDZhvc8uZc6Ci5lpGzV1UEnMJ9oUW87ASRuIxVth-r-aYUsROA6Gjv9_4a_Ld5W_NyXc_kciqqtf5IOotUPTCe5Z7WzGpSN42tE9ul5Fqje4lpVr1FMGyz22Gcu4iHUji9gMNo9XR1GioxGgw7me/s1064/230314%20Series-Parallel%20equations%20with%20Q.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="922" data-original-width="1064" height="277" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh1r7OM3CyP3Iim05iMhexVWNsgYq9jyLsAF4f1dDZhvc8uZc6Ci5lpGzV1UEnMJ9oUW87ASRuIxVth-r-aYUsROA6Gjv9_4a_Ld5W_NyXc_kciqqtf5IOotUPTCe5Z7WzGpSN42tE9ul5Fqje4lpVr1FMGyz22Gcu4iHUji9gMNo9XR1GioxGgw7me/s320/230314%20Series-Parallel%20equations%20with%20Q.png" width="320" /></a></div><br /><u><br /></u></b></div><u style="font-weight: bold;">Identifying L-Network Type from the Signs of B and X:</u><p></p>
<p>
For a Parallel-Series L-Network solution, the signs of B and X of each pair
will determine <i>which</i> of the four possible Parallel-Series L-Networks
each B, X pair represents.
</p>
<p>
Similarly, given the two pairs of B, X solutions for a Series-Parallel
L-Network, the signs of B and X for each pair will determine which of the four
possible Series-Parallel L-Networks each B, X pair represents.
</p>
<p>
The table, below, summarizes network configuration versus the signs of B and
X:
</p>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjcmY7stmyh8GGDRLEEjZSaSvIoCdFO2SGWXR-_FEkvIa6OB6-gDCfz5blzkNOulRXLj-4q9xEKNIn8Qb12pHodsf0KiPmzzcWa132LCG_d_E2ANg7T0ZT3VKVVLJqf3MzsQkaqG7QwwyvBp5CbS1oLi7JxTxbSBYdmaEAQLxv12qgbVqJ9e_E74y3Q/s783/220629%20Type%20of%20network%20from%20B%20X%20signs.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="648" data-original-width="783" height="265" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjcmY7stmyh8GGDRLEEjZSaSvIoCdFO2SGWXR-_FEkvIa6OB6-gDCfz5blzkNOulRXLj-4q9xEKNIn8Qb12pHodsf0KiPmzzcWa132LCG_d_E2ANg7T0ZT3VKVVLJqf3MzsQkaqG7QwwyvBp5CbS1oLi7JxTxbSBYdmaEAQLxv12qgbVqJ9e_E74y3Q/s320/220629%20Type%20of%20network%20from%20B%20X%20signs.png" width="320" /></a>
</div>
<div>
<b><u><br /></u></b>
</div>
<div>
<b><u>Converting B and X to Inductance and Capacitance Values:</u></b>
</div>
<div>
<p>
Capacitor and Inductor component values are calculated from B and X using
the formulas in the table, below:
</p>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgA6CJEdub2H9UMutjbzhKl0X9b6Moqv7SdL9VjGAaeXzD9AtcZMZ3zcKGSh-T_GawvHpJDiBY4Zzjrm2-JN--7oWzGAu3NlNOc0PND0wlyJ-fd-QYmaXKptwrDUeDe__JNs37z8f5rXndF5gt0mkGLNVEQlmtUKnMqV2zaGAECcuYN-GIWfbAhNBVy/s918/220629%20Calculating%20Component%20Vaues%20from%20B%20and%20X.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="587" data-original-width="918" height="205" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgA6CJEdub2H9UMutjbzhKl0X9b6Moqv7SdL9VjGAaeXzD9AtcZMZ3zcKGSh-T_GawvHpJDiBY4Zzjrm2-JN--7oWzGAu3NlNOc0PND0wlyJ-fd-QYmaXKptwrDUeDe__JNs37z8f5rXndF5gt0mkGLNVEQlmtUKnMqV2zaGAECcuYN-GIWfbAhNBVy/s320/220629%20Calculating%20Component%20Vaues%20from%20B%20and%20X.png" width="320" /></a>
</div>
<div><br /></div>
<p>
<b><u>An Example: Transforming Zload = 20 + j40 ohms to 50 ohms:</u></b>
</p>
<p>
Let us find the L-Networks that will transform a load impedance Zload = 20 +
j40 ohms (at 10 MHz) to be a resistive value Zo, where Zo = 50 ohms:
</p>
<p>In other words:</p>
<blockquote style="border: none; margin: 0px 0px 0px 40px; padding: 0px; text-align: left;">
<p>Target Impedance: Zo = 50 ohms</p>
<p>Zload = 20 + j40 ohms</p>
<p>Frequency = 10 MHz</p>
</blockquote>
<p>By inspection of Zload:</p>
<blockquote style="border: none; margin: 0px 0px 0px 40px; padding: 0px; text-align: left;">
<p>Rload = 20 ohms</p>
<p>Xload = 40 ohms</p>
</blockquote>
<p>and we can calculate:</p>
<blockquote style="border: none; margin: 0px 0px 0px 40px; padding: 0px;">
<p style="text-align: left;">1/Zo = 0.02 mhos</p>
</blockquote>
<blockquote style="border: none; margin: 0px 0px 0px 40px; padding: 0px; text-align: left;">
<p>Gload = 0.01 mhos</p>
<p>Bload = -0.02 mhos</p>
</blockquote>
<p>
Next, we must verify which L-network topologies (Parallel-Series or
Series-Parallel) can transform Zload to Zo:
</p>
<blockquote style="border: none; margin: 0px 0px 0px 40px; padding: 0px;">
<p style="text-align: left;">
1. <u>Is Rload </u><span style="text-align: center;"><u>≤</u></span><u> Zo?</u> <b>Yes</b>, 20 is less than 50,
and so Zload can be transformed to Zo using a Parallel-Series L-network.
</p>
</blockquote>
<blockquote style="border: none; margin: 0px 0px 0px 40px; padding: 0px;">
<p style="text-align: left;">
2. <u>And is Gload </u><span style="text-align: center;"><u>≤</u></span><u> 1/Zo?</u> <b>Yes</b>, 0.01 is less
than 0.02, and so Zload can be transformed to Zo using a Series-Parallel
L-network.
</p>
</blockquote>
<p>
Because both selection criteria are satisfied, Zload = 20 + j40 ohms can be
transformed to Zo using any one of four L-networks (i.e. either one of the
two Parallel-Series networks or one of the two Series-Parallel L-networks).
</p>
<p>
The EXCEL spreadsheet, below, shows the B, X pairs and associated
configuration for these four L-networks and calculates the inductor and
capacitor component values:
</p>
<p></p>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiMYvtvgdUz6WYZP-vWPVMXAbwioHB3aHi8R9mPbsLUwSkDVE_MZnkvjnOlXyQypZEhlx_gT-Vb6IMgysTfn5E9HCiZmYMY99DIPnpMWjPj7GSeosZTlLHFfAS4KPh-P8xO3DsOXGf0YwXIbklLZBewZvGWS6eBWUaWdD6-Fq2RdpVDqdJW1BysY5eC/s1045/220629%20Component%20Values.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="688" data-original-width="1045" height="211" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiMYvtvgdUz6WYZP-vWPVMXAbwioHB3aHi8R9mPbsLUwSkDVE_MZnkvjnOlXyQypZEhlx_gT-Vb6IMgysTfn5E9HCiZmYMY99DIPnpMWjPj7GSeosZTlLHFfAS4KPh-P8xO3DsOXGf0YwXIbklLZBewZvGWS6eBWUaWdD6-Fq2RdpVDqdJW1BysY5eC/s320/220629%20Component%20Values.png" width="320" /></a>
</div>
<br />Below are these four L-Networks, shown in schematic form:
<p></p>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjIsDDwKb69PHoLJY_v9lJu5p62DcIifxHAUTmmHtF5UaFRr3wI7SQMGeWINUnGkcQWEtVM9sMFVplc6l7n3SYhG7asz9R4ZGu4CWUV3K1LWXl2hhUn5XdFcw3e1JgZmSoderxToKkPB5EGtbedCEQVO1BZFwiL_aDwzJiiY5OkRWc-vzODfE0wZdnM/s779/220629%20The%20four%20L%20networks.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="779" data-original-width="774" height="320" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjIsDDwKb69PHoLJY_v9lJu5p62DcIifxHAUTmmHtF5UaFRr3wI7SQMGeWINUnGkcQWEtVM9sMFVplc6l7n3SYhG7asz9R4ZGu4CWUV3K1LWXl2hhUn5XdFcw3e1JgZmSoderxToKkPB5EGtbedCEQVO1BZFwiL_aDwzJiiY5OkRWc-vzODfE0wZdnM/s320/220629%20The%20four%20L%20networks.png" width="318" /></a>
</div>
<div><br /></div>
<p>
Note that only two of the four possible Parallel-Series L-Networks and two
of the four possible Series-Parallel L-Networks can perform this particular
impedance transformation.
</p>
<p>
Therefore, there remain four L-Networks that can <b><i>not</i></b> transform
Zload = 20 + j40 ohms to 50 ohms. These networks are:
</p>
<p style="text-align: center;">LpLs, CpLs, LsLp, and CsLp</p>
<p>
<b><u><br /></u></b>
</p>
<p>
<b><u>Final Comments:</u></b>
</p>
<p>
If you examine the Series-Parallel L-Network equations for X and B and the
quantities under their square root signs, you will see that a matching
solution exists not only when G < 1/Zo, but also when G = Zo.
</p>
<p>
However, when G = Zo, there is no difference between the two sets of
Series-Parallel equations for X and B, and X becomes zero, leaving single
shunt susceptance, B, as the matching network, rather than a two-element
L-Network.
</p>
<p>
A similar situation exists for the Parallel-Series L-Network
equations. If Rload = Zo, the two sets of Parallel-Series equations
become identical, with B equal to zero and only a single series reactance,
X, as the matching network, rather than a two-element L-Network.
</p>
<p><br /></p>
<p>
<b><u>Standard Caveat:</u></b>
</p>
<p>
As always, I might have made a mistake in my equations, assumptions,
drawings, or interpretations. If you see anything you believe to be in
error or if anything is confusing, please feel free to contact me or comment
below.<br /><br />And so I should add -- this information is distributed in
the hope that it will be useful, but WITHOUT ANY WARRANTY; without even the
implied warranty of MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.
</p>
<p><br /></p>
</div>
Jeffhttp://www.blogger.com/profile/04853314106806116765noreply@blogger.com0tag:blogger.com,1999:blog-2257458838727315711.post-89143496138556545722022-05-09T19:41:00.046-07:002022-05-12T09:05:46.563-07:00IMD Uncertainty due to Distortion on Input Test Signal<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhZBRO4CwFOQ2Z78bM565Hbm7kiV3SX03JlONdFA2fz8SwLqmgCMyOJj9Ko3MYunuYCosr9tOKy0cmMVQXSrG78Q3F_GLvDD12n0RZ2IHOShSudkPWtYi0B2t4MDMCQ9I4bchcQy6x4IqXZTM7u1_h_Twabp-PseZXNOLAcPcDpA5yNR1K_HvjndA_e/s816/220507%20PA%20In,%20Out,%20Worst%20IMD3,%2080%20meters.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="542" data-original-width="816" height="213" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhZBRO4CwFOQ2Z78bM565Hbm7kiV3SX03JlONdFA2fz8SwLqmgCMyOJj9Ko3MYunuYCosr9tOKy0cmMVQXSrG78Q3F_GLvDD12n0RZ2IHOShSudkPWtYi0B2t4MDMCQ9I4bchcQy6x4IqXZTM7u1_h_Twabp-PseZXNOLAcPcDpA5yNR1K_HvjndA_e/s320/220507%20PA%20In,%20Out,%20Worst%20IMD3,%2080%20meters.png" width="320" /></a>
</div>
<p></p>
<p>
Several years ago, while I was building my
<a href="https://k6jca.blogspot.com/2019/12/a-500-watt-hf-pa-part-2-pa-and-bias.html">500 watt solid-state HF Power Amplifier</a>, I created an IMD test system to let me quickly perform two-tone IMD
measurements as a function of swept input power (and thus a function of swept output power).
</p><p>This was after I had discovered that at low to medium transmit levels the PA could have significant IMD (as shown in the image, below), and I needed a way to quickly perform distortion and gain measurements over a broad range of powers while I experimented with improving the PA's performance.</p><p></p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjHwMk9V6P67kM0sybQTtH-UuRIsZXjht_2LvJnJBy-VoodwaemIlddjdBGQcH46YTT0eUaG7ylzbjzKiDVVh7918Afr7jSnKwvp0RzrbSy8EfrPuBLa66r_yHqHZGZRLdH8x9937GDauactzDff3dJc6Ty9AzqyStGAv2Un7YLEAFZDyTX4UbPYGjn/s816/190807%2080%20Meter%20IMD3%20vs%20Bias%20Current.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="539" data-original-width="816" height="211" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjHwMk9V6P67kM0sybQTtH-UuRIsZXjht_2LvJnJBy-VoodwaemIlddjdBGQcH46YTT0eUaG7ylzbjzKiDVVh7918Afr7jSnKwvp0RzrbSy8EfrPuBLa66r_yHqHZGZRLdH8x9937GDauactzDff3dJc6Ty9AzqyStGAv2Un7YLEAFZDyTX4UbPYGjn/s320/190807%2080%20Meter%20IMD3%20vs%20Bias%20Current.png" width="320" /></a></div><p>The test system consisted of two HP 3335
generators whose outputs were summed together. The resulting two-tone signal was then amplified to higher power (sufficiently high enough to drive the PA to full output) using an ENI 525LA Linear
amplifier that is rated up to 25 watts output with 50 dB of gain. </p><p> Distortion was measured with an Agilent E4406A Spectrum Analyzer, and the
system was controlled via GPIB using a MATLAB program to automate the
power-sweeping and distortion measurement (see the link to my HF PA, above, for more info on the test system).</p><p></p>
<p>
At higher powers the ENI amplifier does introduce its own IMD distortion into
its output, but I already had this piece of equipment in my lab, and I decided
that for my "one-off" PA design, it made sense to use equipment I already had, rather than try to
purchase or build something that might be better performing but more
expensive and that I might never use again.
</p>
<p>
But there is an issue using a test source that introduces its own IMD products. If the test signal to the PA's input has distortion, these
distortion products will interact with the two-tone IMD products created <i>within </i>the PA, with the result that the <i>measured</i> IMD at the PA's
output might not represent the PA's <i>actual</i> IMD performance.
</p>
<p>
The block diagram, below, shows my IMD test system and conceptually describes the problem
when IMD is present on the two-tone test signal at the PA's input:</p><p></p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEicJVGwb9YNikJdv-hAGzY7wDZrPa-C4Gx2G-W58H_QdbW_jjElFHeO1c8XBAF5GnJWiieWhoAsz7kU11AXSAe_Nd-7EFOVYyzsG9CJGo20YZGS1igs7I5WEaDvabgtlJpMMUFR-oOS5ytdVC8Wrez7raFRPiexse1TvuJdQ5s8JVNDvOHP1fG966Qq/s1597/Visio-HF%20PA%20Test%20System%20showing%20input%20distortion%20path.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="789" data-original-width="1597" height="158" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEicJVGwb9YNikJdv-hAGzY7wDZrPa-C4Gx2G-W58H_QdbW_jjElFHeO1c8XBAF5GnJWiieWhoAsz7kU11AXSAe_Nd-7EFOVYyzsG9CJGo20YZGS1igs7I5WEaDvabgtlJpMMUFR-oOS5ytdVC8Wrez7raFRPiexse1TvuJdQ5s8JVNDvOHP1fG966Qq/s320/Visio-HF%20PA%20Test%20System%20showing%20input%20distortion%20path.png" width="320" /></a></div><br />At the time, I did not investigate any further the issue of the effect of source distortion on output distortion, but recently I came upon a
<a href="https://www.eham.net/community/smf/index.php/topic,136136.0.html">thread on the eHam.net</a> Amplifier forum that discussed IMD uncertainty and it prompted me to
revisit the issue with respect to my own PA design. (And more discussion can be found on this eHam thread (started by me): <a href="https://www.eham.net/forum/view?id=index.php/topic,136313">Calculating IMD Uncertainty</a>).<p></p><div>
<p>
<b><u>The Math of IMD Uncertainty:</u></b></p>
<p>As I mentioned, above, the two-tone signal (with distortion) from my ENI 525LA amplifier drives the PA. Thus, the two tones <i>and</i> IMD products from the preceding ENI amplifier are amplified by
the PA. During the amplification process the PA's own non-linearities also create their own distortion products from the two-tone signal being amplified, and these
products will sum (in some fashion) with the incoming IMD products from the ENI amp.
</p>
<p>
(This is a simplified description of what actually is occurring. In addition
to the PA creating distortion from the two-tone signal, there are also
distortion products created by the mixing of the incoming distortion
products from the ENI amplifier. I am going to assume that these
additional mixing products are at a low enough level to not significantly contribute further to the IMD at the amplifier's output.)
</p>
<p>
So, as an approximation of how the PA's internally generated IMD products
interact with the incoming IMD products from my ENI amplifier, we can
say that the PA's <i>measured</i> Output IMD (as measured on my Spectrum Analyzer) is the <i>sum</i> of the PA's <i>actual</i> IMD (that is generated solely by the two tone test signal) plus the <i>incoming</i> Source IMD (i.e. the ENI amp's IMD ).
</p>
<p>Both the measured IMD and the source IMD should be computed in terms of dBc, that is, with respect to the level of either one of the two tones of equal amplitude (if Source IMD and Measured IMD are both referenced to dBc, no additional scaling is required of the source IMD components with respect to PA gain).</p><p>If we consider the source and measured IMD components to be voltages, the Spectrum Analyzer is measuring the sum of the two:</p>
<div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj0JYQ3s1XGycz8DCLmutKg-d5ZhUM5T8dKPonksc62o376kYnTITWTMVQyakI_IqehCIcA73rwWPnNKUsMZAvLGrlnWT51JS1RH6clBq5KUv7cV6JJKJkjrtU0p8XU-gHUCOuU3O9otAvtWPKH8oQ_PHbnMYOPy6-WBb3eyI0fJo9roUly7ATcdurf/s857/220509%20Equation%201%20Vmeasured.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="442" data-original-width="857" height="165" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj0JYQ3s1XGycz8DCLmutKg-d5ZhUM5T8dKPonksc62o376kYnTITWTMVQyakI_IqehCIcA73rwWPnNKUsMZAvLGrlnWT51JS1RH6clBq5KUv7cV6JJKJkjrtU0p8XU-gHUCOuU3O9otAvtWPKH8oQ_PHbnMYOPy6-WBb3eyI0fJo9roUly7ATcdurf/s320/220509%20Equation%201%20Vmeasured.png" width="320" /></a></div><div><br /></div>The level of V<sub>IMDMeasured</sub> will be a function of the amplitudes and phases of V<sub>IMDActual</sub> and of V<sub>IMDSource</sub>.</div><div><br /></div><div>The maximum and minimum values of V<sub>IMDMeasured</sub> will occur when the phase difference between V<sub>IMDActual</sub> and V<sub>IMDSource</sub> is either 0 or 180 degrees.</div><div><br /></div><div><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiJ-ZeDQQlbKHZKZaPmqDLL2X8CJwDVvUd5jUQCKHNXcM82Mi64lAlPzxIiMSPFxTjdFKpwu7d4SBpWabbmDMBFiU4onMoWE-FMkEFtLeECupO1Ji0lVSOZC3v5hJEY_mrV6EUcK20tPcu_8XvoKPJzhPVM5P3_vk6ysOptzHCWMmD7cmZfOyHoSmJ1/s969/220509%20Equation%202%20relative%20phases.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="631" data-original-width="969" height="208" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiJ-ZeDQQlbKHZKZaPmqDLL2X8CJwDVvUd5jUQCKHNXcM82Mi64lAlPzxIiMSPFxTjdFKpwu7d4SBpWabbmDMBFiU4onMoWE-FMkEFtLeECupO1Ji0lVSOZC3v5hJEY_mrV6EUcK20tPcu_8XvoKPJzhPVM5P3_vk6ysOptzHCWMmD7cmZfOyHoSmJ1/s320/220509%20Equation%202%20relative%20phases.png" width="320" /></a></div><p>The images below depict a signal increasing in value due to Constructive Interference...</p>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhrrAZj3ZaV5RccFgbKGf7cgngugr6PSwwWaUAnCesWbeUgMK1NSA-1NTX_ZkxKv55w7N41fWXiP_XQyC-gPC5MasNdSGV4PV_-qAwAgG97CmZcvACzVmLh1EUaQmNG67B8mOzadm5lTNhzqJzwGAKkIgROyPrMpkJZJa13QmNetEq-uMENXPruM6pz/s438/220509%20Constructive%20Interference.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="371" data-original-width="438" height="271" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhrrAZj3ZaV5RccFgbKGf7cgngugr6PSwwWaUAnCesWbeUgMK1NSA-1NTX_ZkxKv55w7N41fWXiP_XQyC-gPC5MasNdSGV4PV_-qAwAgG97CmZcvACzVmLh1EUaQmNG67B8mOzadm5lTNhzqJzwGAKkIgROyPrMpkJZJa13QmNetEq-uMENXPruM6pz/s320/220509%20Constructive%20Interference.png" width="320" /></a>
</div>
<p>...and of a signal decreasing in value due to Destructive Interference:</p>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgpyrgXLtL40wFMKQmT4-0TTNKivikaaJC6Kvbe6YVI19znymZEGa6QcBEWt5Mm8Jg79AB93Dw6jyamP0GlbmDjwQHsXOjcjYyYhNy83sYg3E93JH77W0R4BnrXGbc4Rkz7NZMJrifZLiYqgv_JNctc09A-A46JcFEdh-kTxuHU2wEJa7BNXHQko19A/s438/220509%20Destructive%20Interference.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="371" data-original-width="438" height="271" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgpyrgXLtL40wFMKQmT4-0TTNKivikaaJC6Kvbe6YVI19znymZEGa6QcBEWt5Mm8Jg79AB93Dw6jyamP0GlbmDjwQHsXOjcjYyYhNy83sYg3E93JH77W0R4BnrXGbc4Rkz7NZMJrifZLiYqgv_JNctc09A-A46JcFEdh-kTxuHU2wEJa7BNXHQko19A/s320/220509%20Destructive%20Interference.png" width="320" /></a>
</div>
<p>Because the amplitude of the resulting signal depends upon the relative phase of the two signals being summed, there are a range of values that V<sub>IMDMeasured</sub> could be. And so V<sub>IMDMeasured</sub> no longer represents the PA's actual distortion (V<sub>IMDActual</sub>).</p><p>But it is V<sub>IMDActual</sub> that we are really interested in -- the <i>actual</i> distortion of the PA, not the now-uncertain <i>measured</i> distortion of the PA.</p><p>Rearranging the above two equations, we can create two new equations representing V<sub>IMDActual</sub> and its worst-case and best-case values as a function the measured IMD and the source IMD. These two values, representing the best and worst case possibilities of the <i>actual</i> IMD value, bound the region of possible values within which value of the PA's actual IMD could be.</p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgcKujWnanGhDelHEZUK3VYGcvhEWP9JqtdqWmNcAwUfl4-dyyoQXui2AD16rhFrP9AkThBypEiooKm9O2ogkAVuG16x7j5eCWDPgmOyCQKaaTKo0f4j4jS4vHxtDf7KTMHlCGdHZ2ZBgrSILZcX5Qrl4k7c39df5ZJhbaqU8Em7xOcJAKPrEacx22k/s988/220509%20Equation%203%20best%20and%20worst%20case.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="511" data-original-width="988" height="166" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgcKujWnanGhDelHEZUK3VYGcvhEWP9JqtdqWmNcAwUfl4-dyyoQXui2AD16rhFrP9AkThBypEiooKm9O2ogkAVuG16x7j5eCWDPgmOyCQKaaTKo0f4j4jS4vHxtDf7KTMHlCGdHZ2ZBgrSILZcX5Qrl4k7c39df5ZJhbaqU8Em7xOcJAKPrEacx22k/s320/220509%20Equation%203%20best%20and%20worst%20case.png" width="320" /></a></div><div><br /></div><div>Note two things about the best-case and worst-case equations, above, for V<sub>IMDActual</sub>.</div><div><br /></div><div>First, V<sub>IMDSource</sub> is assumed to be <i>less than</i> V<sub>IMDMeasured</sub>. (Later in this post I will discuss how the 'best case' equation for V<sub>IMDActual</sub> changes for the case when V<sub>IMDSource</sub> is <i>greater than</i> V<sub>IMDMeasured</sub>.)</div><div><br /></div><div>Second, note that the above equations are in terms of voltage. Because my IMD measurements are in dB, it is necessary to convert the results back to dB for comparison purposes. The following derivation shows how this is done:</div><div><br /></div><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjKLq8NhWZvWtL7u3UqugcT_lHiWcF32qUGAaGBlx4eJ0igi45Uikq1V42jKCA1MQ1tOoRmeUlRLXU0pI5oqDzqcLtpmUA1581B68MBNVX7uO5oV7edU-HWR3WkH4-xJwUHL-0ALtDKnMkUku68oYF6amDTAxMM-UjxNTg6a_ZRdoHx1IjoLFBX8gaz/s917/220509%20Equation%204%20converting%20to%20dB.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="620" data-original-width="917" height="216" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjKLq8NhWZvWtL7u3UqugcT_lHiWcF32qUGAaGBlx4eJ0igi45Uikq1V42jKCA1MQ1tOoRmeUlRLXU0pI5oqDzqcLtpmUA1581B68MBNVX7uO5oV7edU-HWR3WkH4-xJwUHL-0ALtDKnMkUku68oYF6amDTAxMM-UjxNTg6a_ZRdoHx1IjoLFBX8gaz/s320/220509%20Equation%204%20converting%20to%20dB.png" width="320" /></a></div></div><div><p>Continuing with the derivation:</p>
<div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgLZEcKQoNNo1CW0vWhXBJe7PW3Ng-ezpsjYhhPQM2dmAtOPbjo-eYL0lt4D_84ASGTcNd9tNBdBuzoGxcsTdTC3K2KasaZMYFqVVIMkRbwrB5I3ZNBeKzF984E44WMy6q8IYYPZieYpcmSxPV_aTpgGkK2ZWw4FZjRv631NMgRlmojoVOsz9j6aAGn/s1032/220509%20Equation%205%20final%20dB%20equation.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="696" data-original-width="1032" height="216" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgLZEcKQoNNo1CW0vWhXBJe7PW3Ng-ezpsjYhhPQM2dmAtOPbjo-eYL0lt4D_84ASGTcNd9tNBdBuzoGxcsTdTC3K2KasaZMYFqVVIMkRbwrB5I3ZNBeKzF984E44WMy6q8IYYPZieYpcmSxPV_aTpgGkK2ZWw4FZjRv631NMgRlmojoVOsz9j6aAGn/s320/220509%20Equation%205%20final%20dB%20equation.png" width="320" /></a></div><br /><div>This last equation is solely in terms of IMD measurements made in dB. Note that the left hand quantity represents the value that should be added to the <i>measured</i> IMD (in dB) to determine the <i>actual</i> IMD, in dB.</div><div><br /></div><div>(There is MATLAB code later in this post that describes all of the mathematical steps).</div><div><p>We can also create a plot of the worst-case and best-case values as a function of the difference in signal levels (as either voltage or power) between <i>measured</i> IMD and <i>source</i> IMD.</p></div><p>The plot below graphs the two possible outputs from the equation, above. Note that the x-axis is the value of "dB<sub>delta</sub>", i.e. the difference between the Source IMD and the Measured IMD. </p>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhQo5V2N2p5-nQVa_6039iiAYA2b0jGiXTdkw1dERiyrnEgmvY3zoGEcbCBsVok0dKSPVdb0RBdXfBb4CghUseEjpUPsnfMgtEYGwQVMLlNu8UExefT64FKOa3H9fnDMgPBwMqq974GLB0iIVLc7-NV-7ywKDXsE1b_uYgCAi7FO-4fCsvZHklIx6XD/s576/k6jca%20Plot%20of%20Uncertainy%20Curves.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="512" data-original-width="576" height="284" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhQo5V2N2p5-nQVa_6039iiAYA2b0jGiXTdkw1dERiyrnEgmvY3zoGEcbCBsVok0dKSPVdb0RBdXfBb4CghUseEjpUPsnfMgtEYGwQVMLlNu8UExefT64FKOa3H9fnDMgPBwMqq974GLB0iIVLc7-NV-7ywKDXsE1b_uYgCAi7FO-4fCsvZHklIx6XD/s320/k6jca%20Plot%20of%20Uncertainy%20Curves.png" width="320" /></a>
</div>
<br />
<div>The above equations assume the typical case of the Source IMD being less than the Measured IMD (and thus dB<sub>delta</sub> is negative). But this might not be the case. If Source IMD is greater than Measured IMD, then the Source and Measured terms must be swapped in the 'best-case' equation to prevent the argument of the log() function from becoming negative. (The 'worst-case 'equation remains unchanged)</div><div><br /></div><div>Therefore, if Source IMD is greater than Measured IMD, the 'best-case' equation becomes:<br /></div><div><br /></div><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgx0P5OT0M-InYvkuv-L9HQTejF-wvNMT2qHQYunCA1dWIymIu92hNssNJq7qBuag_6M2daWk0nWUrZ526e6iZ1awDq31M7d7Me2EkNektU7gC9nR54X2YR5aziB6X5trJ7HS8jAFNVG6N7h8nPDKHjvhQ5Y4-rk2c5cUL8RQfMTCm446xRoTU_P5mx/s836/220509%20Equation%206%20vsource%20greater%20than%20vmeasured.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="759" data-original-width="836" height="291" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgx0P5OT0M-InYvkuv-L9HQTejF-wvNMT2qHQYunCA1dWIymIu92hNssNJq7qBuag_6M2daWk0nWUrZ526e6iZ1awDq31M7d7Me2EkNektU7gC9nR54X2YR5aziB6X5trJ7HS8jAFNVG6N7h8nPDKHjvhQ5Y4-rk2c5cUL8RQfMTCm446xRoTU_P5mx/s320/220509%20Equation%206%20vsource%20greater%20than%20vmeasured.png" width="320" /></a></div><br /><div><br /></div>
<p>
<b><u>An Example:</u></b>
</p>
<p>Let us use as an example 80 Meter IMD measurements I made of my HF PA back in 2019:</p>
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiHVpPX9dVGFLDMcif3xP5VZxPiq84Wt6ltDJBFOoA9wRgYq-0KDscWWFtDWVb5I1Dc89gCiaI_EiJMkR6OnoIAuShVuSbhXUtadZD1EY0KKDNf7MOos7SQvR_SEfIqnkNlYQKS05ytSibpZSAu10RdVpWuZIDisVcPBHyDuV1ISRzSGktpETfAv3t5/s816/220507%20Measured%20IMD,%2080%20meters,%20annotated.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="542" data-original-width="816" height="213" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiHVpPX9dVGFLDMcif3xP5VZxPiq84Wt6ltDJBFOoA9wRgYq-0KDscWWFtDWVb5I1Dc89gCiaI_EiJMkR6OnoIAuShVuSbhXUtadZD1EY0KKDNf7MOos7SQvR_SEfIqnkNlYQKS05ytSibpZSAu10RdVpWuZIDisVcPBHyDuV1ISRzSGktpETfAv3t5/s320/220507%20Measured%20IMD,%2080%20meters,%20annotated.png" width="320" /></a>
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<p>At the same time, I also captured the IMD characteristics of the two-tone test signal from the ENI 525LA amplifier that was driving the PA's input. </p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhlMAoU3U11dw66Kke3bE4aCHd6HXmY0BB56uI_OKsyoo7wZV_1QR0dj_kNQOsaIqWJKEwUy-VEtaOENJie_XEoH4l6uDuqfT680PXFbdOis9odYdGcxgU-hJZU5WyCbDNFrOk6yZLpVttmJZ4qK3_3mMExzC5KoLW2IgRnZTI14XE5EOH1InbRrYYZ/s816/220507%20PA%20Input%20IMD%2080%20meters,%20annotated.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="542" data-original-width="816" height="213" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhlMAoU3U11dw66Kke3bE4aCHd6HXmY0BB56uI_OKsyoo7wZV_1QR0dj_kNQOsaIqWJKEwUy-VEtaOENJie_XEoH4l6uDuqfT680PXFbdOis9odYdGcxgU-hJZU5WyCbDNFrOk6yZLpVttmJZ4qK3_3mMExzC5KoLW2IgRnZTI14XE5EOH1InbRrYYZ/s320/220507%20PA%20Input%20IMD%2080%20meters,%20annotated.png" width="320" /></a>
</div>
<p>Note that the power-steps on the x-axis of the two plots, above, are exactly the same -- I generated the Source IMD plot by simply moving the Spectrum Analyzer's input from the PA's output (via a large attenuator) to a directional coupler attached to the PA's input and then I ran exactly the same test sequence (this second path is shown in the test system the block diagram earlier in this post). </p><p>(Using the directional coupler to measure ENI 525LA distortion allowed me to capture any distortion artifacts from the ENI 525LA that might have been caused by an impedance mismatch between the ENI 525LA's output and the PA's input, and therefore not present if I had simply terminated the ENI 525LA's output in 50 ohms.)</p><div>With the captured PA input data (i.e. 525LA output) and PA output data I have all the information I need to calculate PA IMD uncertainty.</div><div><br /></div><div>The plot below shows:</div><div><ol style="text-align: left;"><li>Measured PA Output IMD3.</li><li>Measured PA Source IMD3 (i.e. Input IMD3 to the PA's input from the EN 525LA's Output).</li><li>Worst-case <i>Actual</i> PA Output IMD.</li><li>Best-case <i>Actual</i> PA Output IMD.</li></ol><div><br /></div></div><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg01RRSymxxbXJjUF_pxXo7wMqjJP-iaUI84sfr1Ez42ReJwynOvSQ1PQm7RjqVzKAcaWBqb6iwPZuj72PZCji8xyHdRzCa9f7WXNehcbxeRABsUDBoV9khBHaS8_YgBsDvb1RGCw-y4Xjncf4GD4i8lLN7e6A42ZPTcCX100V9sfst9NkEYy0JqAHe/s816/220507%20PA%20In,%20Out,%20Worst%20IMD3,%2080%20meters.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="542" data-original-width="816" height="213" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg01RRSymxxbXJjUF_pxXo7wMqjJP-iaUI84sfr1Ez42ReJwynOvSQ1PQm7RjqVzKAcaWBqb6iwPZuj72PZCji8xyHdRzCa9f7WXNehcbxeRABsUDBoV9khBHaS8_YgBsDvb1RGCw-y4Xjncf4GD4i8lLN7e6A42ZPTcCX100V9sfst9NkEYy0JqAHe/s320/220507%20PA%20In,%20Out,%20Worst%20IMD3,%2080%20meters.png" width="320" /></a>
</div>
<div>The plot below is the same as the plot above, but I've added some data points on the four curves that correspond to a PA output of +57 dBm (i.e. 500 Watts PEP).</div></div><div>
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjJ5Uf9yAcmCZHpStK_MhIZDmn_O711STlrz28xO64ljx-0C9euZzi7CVrqZE-i2GF2r7PPn3EY92KkWvnW5rSV_tFvN5pIFVssm6f1whAQVpRawyjlB56LkL6929_PyGKDiGf1g_2l255RR3_E4Eyz9mfwEjgTxfcjXtiOip48vReIjC_LDfOhSUa9/s816/220507%20PA%20In,%20Out,%20Worst%20IMD3,%2080%20meters,%20with%20data%20tips.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="542" data-original-width="816" height="213" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjJ5Uf9yAcmCZHpStK_MhIZDmn_O711STlrz28xO64ljx-0C9euZzi7CVrqZE-i2GF2r7PPn3EY92KkWvnW5rSV_tFvN5pIFVssm6f1whAQVpRawyjlB56LkL6929_PyGKDiGf1g_2l255RR3_E4Eyz9mfwEjgTxfcjXtiOip48vReIjC_LDfOhSUa9/s320/220507%20PA%20In,%20Out,%20Worst%20IMD3,%2080%20meters,%20with%20data%20tips.png" width="320" /></a>
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<p>You can see that at 500 watts out (i.e. 57 dBm), there is almost exactly a 10 dB delta between the measured IMD of the PA and the Source (Input) IMD from the ENI 525LA amplifier. </p><p>And the worst-case value that the <i>actual</i> PA IMD could be is 2.4 dB above the <i>measured</i> value, and the best-case value is 3.3 dB below the measured value (i.e. -3.3 dB) .</p><p>(Note that +2.4 dB and -3.3 dB are the values found for a -10 dB difference in signal level on the Uncertainty plot of best and worst case amplitudes versus relative signal level, earlier in this post .)</p><div>You can see the delta between measured IMD and worst-case 'actual' IMD increases as power approaches (and exceeds) 500 watts, due to the worsening Source (Input) IMD (yellow line).</div><div><br /></div><div>Also, you can see that at low levels (i.e. below about 37 dBm output power) the best-case curve looks a bit wonky, and the worst-case curve begins to slope back up. And you see the best-case curve going <i>above</i> the measured curve. </div><div><br /></div><div>Although surprising, this is to be expected, because in this region the measured Source IMD actually goes <i>higher</i> than the measured Output IMD. But note that in this region we are no longer actually reading valid Source IMD numbers -- the Source IMD products have run into the noise floor of the Spectrum Analyzer, and it is the noise floor we are instead measuring. </div><div><br /></div><div>You can verify this by looking at the noise floor values for a given low-power point on the x axis (e.g. at 25 dBm) -- the noise floor of the Input is higher than the noise floor of the Output. For this reason, I would recommend discounting best-case and worst-case curves at powers below which Source IMD is in the noise floor.</div><div><br /></div><div>Below are the Best and Worst case IMD plots for my PA's 80 meter IMD3, IMD5, IMD7, and IMD9 distortion products:</div>
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiQjNc1lz_-5ZmdQk3RmhcsCSRGGnGn9BSNLQWpmGPhxfwJUOBjoEUlhFQ_5IYU0sDwDrByhPHNl_-pFtcO5rmpMXjBWv723abvtYdn-9PS8HSnbBlsSUy3kl6LWLuQjey7jpERHjd0nwE-HmQMrUAlIg6T3z18HGW-nKuPSBjo20olTQUbkmP8IdPf/s816/220507%2080%20meter%20Best%20and%20Worst%20IMDs.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="542" data-original-width="816" height="213" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiQjNc1lz_-5ZmdQk3RmhcsCSRGGnGn9BSNLQWpmGPhxfwJUOBjoEUlhFQ_5IYU0sDwDrByhPHNl_-pFtcO5rmpMXjBWv723abvtYdn-9PS8HSnbBlsSUy3kl6LWLuQjey7jpERHjd0nwE-HmQMrUAlIg6T3z18HGW-nKuPSBjo20olTQUbkmP8IdPf/s320/220507%2080%20meter%20Best%20and%20Worst%20IMDs.png" width="320" /></a>
</div>
<p>Personally, I find the above plot too busy, and it can be difficult to understand what is going on. So, recognizing that the 'best case' values are not of much use (compared to the worst case values), the plot below shows just the measured PA IMD and the worst-case values that the PA IMD could <i>actually</i> be.</p>
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgpSMrcWN2tqX_EVE2iez2FFTKLBUHuDMPik1KPrCd5F1XClXm3qwh99nIIV5aEvxyGWMxOZi6AV6mMo0jJU-a5Qq6iEgmQIyGwVrz_-9mMHZ5ETwi-M46TMwOOhIGrgATo9QsbVhZNrtVdk82LTIEgu01ycCbg3k3ra7o3f3AIYgZS4aIrTIVKygVy/s816/220507%2080%20meter%20Worst%20IMDs.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="542" data-original-width="816" height="213" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgpSMrcWN2tqX_EVE2iez2FFTKLBUHuDMPik1KPrCd5F1XClXm3qwh99nIIV5aEvxyGWMxOZi6AV6mMo0jJU-a5Qq6iEgmQIyGwVrz_-9mMHZ5ETwi-M46TMwOOhIGrgATo9QsbVhZNrtVdk82LTIEgu01ycCbg3k3ra7o3f3AIYgZS4aIrTIVKygVy/s320/220507%2080%20meter%20Worst%20IMDs.png" width="320" /></a>
</div>
<div><br /></div>(Note, again, that the worsening IMD at lower powers occurs because the <i>measured</i> IMD product or the <i>source</i> IMD product has disappeared into the noise-floor of the Spectrum Analyzer).<br />
<p><br /></p>
<p>
<b><u>Matlab Code:</u></b>
</p>
<p>I used MATLAB to generate the above 'uncertainty' plots from data captured with my E4406A Spectrum Analyzer. Data was captured in terms of dB.</p><p>Below are two different methods for calculating the best and worst case IMD values. Both give the same results.</p><p>The first method calculates the best and worst case IMD values using power (i.e. dB).</p><p>The second method calculates best and worst case IMD measurements using voltage. This second method might be more intuitively obvious, but it does involve converting the measurements that were made in dB to volts, performing the appropriate addition or subtraction, and then, when the calculations are finished, converting from volts back to dB.</p><p>How the conversion is made from dB to volts (and then back again) is arbitrary with respect to what is chosen as reference power and load resistance. The only rule is that the same reference power and load resistance that were used to convert from power to voltage must also be used to convert back from voltage to power.</p><p>For convenience, I chose dBm as my reference and 50 ohms as my load, simply because there are on-line calculators that allow me to quickly convert from dBm to Vrms and back again, so I could check for errors in my MATLAB calculations by compoaring with calculations I made separately using the on-line calculators.</p><p>And I set 0 dBc (i.e. the level of either of the two tones) to be equal to 0 dBm. Therefore, at the end when I converted voltage back to dB (actually, back to dBm), the final result was identical to power in dBc.</p><p><br /></p><div><pre class="codeinput" style="background: rgb(247, 247, 247); border: 1px solid rgb(211, 211, 211); font-size: 12px; margin-bottom: 20px; margin-top: 0px; outline: 0px; padding: 10px; vertical-align: baseline;"><span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">% % *************************************************************</span>
<span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">% % Below are two different methods to calculate IMD Uncertainty,</span>
<span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">% % given measurable IMD distortion on a two-tone test signal</span>
<span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">% % driving a PA-under-test.</span>
<span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">% % Both produce the same results.</span>
<span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">% % *************************************************************</span>
<span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">% Uncertainty Calculation, Method 1 (in terms of dB)</span>
<span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">% --------------------------------------------------</span>
<span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">% Note that the IMD measurements are in dB.</span>
<span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">%</span>
<span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">% Calculate Best-case and Worst-case IMD possibilities given an amplifier's</span>
<span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">% measured output IMD and the measured IMD of the source driving it.</span>
<span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">%</span>
<span class="keyword" style="background: transparent; border: 0px; color: blue; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">function</span> [BestCaseIMDdB,WorstCaseIMDdB] = IMDuncertainty(IMDsourcedB,IMDoutputdB)
IMD_delta = IMDsourcedB-IMDoutputdB; <span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">% neg. if source IMD < Output IMD</span>
expTerm = 10^(IMD_delta/20); <span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">% Vsource/Voutput</span>
<span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">% Next, dB of sum of Source and PA-Output IMD voltages,</span>
<span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">% referenced to PA-Output's IMD voltage.</span>
dB_of_sum = 20*log10(expTerm + 1);
WorstCaseIMDdB = IMDoutputdB + dB_of_sum;
<span class="keyword" style="background: transparent; border: 0px; color: blue; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">if</span> IMD_delta > 0 <span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">% Source IMD level > Measured Output IMD level</span>
dB_of_difference = 20*log10(expTerm - 1); <span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">% delta to Output IMD</span>
BestCaseIMDdB = IMDoutputdB + dB_of_difference;
<span class="keyword" style="background: transparent; border: 0px; color: blue; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">else</span>
<span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">% Source IMD level <= Measured Output IMD level</span>
dB_of_difference = 20*log10(1 - expTerm); <span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">% delta to Output IMD</span>
BestCaseIMDdB = IMDoutputdB + dB_of_difference;
<span class="keyword" style="background: transparent; border: 0px; color: blue; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">end</span>
<span class="keyword" style="background: transparent; border: 0px; color: blue; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">end</span>
<span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">% Uncertainty Calculation, Method 2 (in terms of voltages)</span>
<span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">% --------------------------------------------------</span>
<span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">% Note that the IMD measurements are in dB.</span>
<span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">%</span>
<span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">% Calculate Best-case and Worst-case IMD possibilities given an amplifier's</span>
<span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">% measured output IMD and the measured IMD of the source driving it.</span>
<span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">%</span>
<span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">% In order to calculate voltage from power, the IMD measurements are</span>
<span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">% assumed to be in dBm (referenced to 50 ohms), but this definition is</span>
<span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">% arbitrary! The final results would be the same irrespective of</span>
<span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">% reference power (e.g. 1 mW), or assumed load resistance (e.g. 50 ohms).</span>
<span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">%</span>
<span class="keyword" style="background: transparent; border: 0px; color: blue; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">function</span> [BestCaseIMDdB,WorstCaseIMDdB] = IMDuncertainty(IMDsourcedB,IMDoutputdB)
<span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">% First assume 0 dBc equals 0 dBm as a convenient, yet arbitrary</span>
<span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">% reference. Convert powers (measured in terms of dBc, now dBm)</span>
<span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">% to voltages:</span>
Vsource = (10^(IMDsourcedB/10)*50*0.001)^(1/2); <span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">% convert dBm to V</span>
Voutput = (10^(IMDoutputdB/10)*50*0.001)^(1/2); <span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">% convert dBm to V</span>
Vactual_WorstCase = Vsource + Voutput;
<span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">% convert V to dBm (actually dBc)</span>
WorstCaseIMDdB = 10*log10((Vactual_WorstCase^2)/(50*0.001));
<span class="keyword" style="background: transparent; border: 0px; color: blue; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">if</span> Vsource > Voutput <span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">% Source IMD level > Measured Output IMD level</span>
<span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">% subtract Vouput from Vsource, to keep the log argument positive</span>
Vactual_BestCase = Vsource - Voutput;
<span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">% convert V to dBm (actually dBc)</span>
BestCaseIMDdB = 10*log10((Vactual_BestCase^2)/(50*0.001));
<span class="keyword" style="background: transparent; border: 0px; color: blue; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">else</span>
<span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">% Source IMD level <= Measured Output IMD level</span>
<span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">% subtract Vsource from Voutput, to keep the log argument positive</span>
Vactual_BestCase = Voutput - Vsource;
<span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">% convert V to dBm (actually dBc):</span>
BestCaseIMDdB = 10*log10((Vactual_BestCase^2)/(50*0.001));
<span class="keyword" style="background: transparent; border: 0px; color: blue; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">end</span>
<span class="keyword" style="background: transparent; border: 0px; color: blue; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">end</span>
</pre><div>Note: If you are viewing the code on a small screen, you might not be able to see the full length of the lines of code. In this case, cut-and-paste the code into an application with a larger viewing pane (e.g. Notepad, Word, etc.).</div><div><br /></div>
<p>
<b><u>Resources:</u></b>
</p>
<p>
These two application notes discuss IMD Uncertainty (as it relates to
Spectrum Analyzer IMD measurements and IMD generated within the Spectrum Analyzer, itself):
</p>
<p>
<a href="https://cdn.rohde-schwarz.com/pws/dl_downloads/dl_application/application_notes/1ma219/1MA219_2e_IM_Distortion.pdf">Interaction of Intermodulation Products between DUT and Spectrum
Analyzer, Rohde & Schwarz</a>
</p>
<p>
<a href="https://www.keysight.com/us/en/assets/7018-06794/application-notes/5980-3079.pdf">Dynamic Range Optimization for Distortion Measurements, Keysight
Technologies</a>
</p>
<p><br /></p>
<p>
<b><u>Standard Caveat:</u></b>
</p>
<p>
As always, I might have made a mistake in my equations, assumptions,
drawings, or interpretations. If you see anything you believe to be
in error or if anything is confusing, please feel free to contact me or
comment below.<br /><br />And so I should add -- this information is
distributed in the hope that it will be useful, but WITHOUT ANY WARRANTY;
without even the implied warranty of MERCHANTABILITY or FITNESS FOR A
PARTICULAR PURPOSE.
</p><p><br /></p>
</div>
</div>
Jeffhttp://www.blogger.com/profile/04853314106806116765noreply@blogger.com1tag:blogger.com,1999:blog-2257458838727315711.post-3608420358118928482022-05-05T06:51:00.011-07:002022-05-10T09:17:23.890-07:00Schematic, Amazon Relay Module: 1 Channel, Optocoupler Isolation Hi/Low Trigger<p>A ham radio friend (Dan, KA6RCZ) recently purchased an inexpensive single channel relay module (made by HiLetgo, in this instance) from Amazon, and he wanted to know if it would work in an application he had planned. </p><p>Because of local high noise at his location, Dan uses a remote webSDR site (<a href="http://websdr1.kfsdr.com:8901/">KFS</a>) through his PC for the receive side of his communications, and he wanted some way to turn off the PC's audio (to an external speaker) whenever he was transmitting.</p><p>Unfortunately, the module came with no documentation, and so to verify if it would work in my friend's application, I decided to draw its schematic.</p><p>This schematic was created from tracing out the circuitry of the 12 V version of the module, but it is probably applicable to the other voltage versions of the same module (e.g. the 5V relay module), assuming that the only change between different modules is the voltage rating of the relay.</p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiKiskyCZRKPRvMtCvr1B3kV0G-ZxGbE2XpiowsZSYZKi5Z3ydsYjuV2cAQQXsyjUMJ3k8jD5EP-xZnxCqTkoTPtlLzTMAm49haNpsZJ07ALqHBYHZ4J4b1S8oLgChiiVaHsgEu4E4wURBGsOYZbNw_sjHpKQtQo_KyyJsOZZ5FoLzrE4TXO5_6EhVZ/s1650/HiLetgo%201%20Ch%20Relay%20Module%20Schematic.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="1275" data-original-width="1650" height="247" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiKiskyCZRKPRvMtCvr1B3kV0G-ZxGbE2XpiowsZSYZKi5Z3ydsYjuV2cAQQXsyjUMJ3k8jD5EP-xZnxCqTkoTPtlLzTMAm49haNpsZJ07ALqHBYHZ4J4b1S8oLgChiiVaHsgEu4E4wURBGsOYZbNw_sjHpKQtQo_KyyJsOZZ5FoLzrE4TXO5_6EhVZ/s320/HiLetgo%201%20Ch%20Relay%20Module%20Schematic.png" width="320" /></a></div><br /><p>Note that when the jumper is in the 'H' position, the relay turns ON when the IN voltage is raised to about 1.5 V above ground (i.e. 1.5 V above the voltage at the DC- connector). This threshold should be independent of the spec'd relay voltage, assuming that the only component that changes between different voltage-rated modules is the relay. </p><p>So, for example, a 5V relay module should turn on when the IN voltage is above about 1.5 V (note: IN voltage level referenced to the DC- pin). </p><p>When the jumper is in the 'L' position, the relay turns ON when the IN voltage is less than about 1.5 V below DCV (where 'DCV' is the DC voltage applied to the module between the DC+ and DC- inputs). </p><p>So, for a 5V relay module with 5 VDC applied between the DC+ and DC- connectors, I'd expect the relay to turn ON when the IN voltage is less than about 3.5 V (note: IN voltage level referenced to the DC- pin).</p><p><br /></p><p><u>A Circuit for Dan's Application:</u></p><p>I drew the connection diagram, below, for Dan's webSDR application. </p><p></p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEir4AjeV26B66zyRRKWsSvtW3rto-a1xR490kiXsKBwp0lN9K-50_aATVtFspggpAPNMqxM3dUqsBwUaZOBiL398yiRhQsJloqWuNEI2vndeT6avMkDXDn90YeDzi2Z5Ev01s8UMFtv9gidaGPq-uf6tFD6yKLv6Om2gLduqAWfUr3jvKJqic-wgI2O/s1977/IC-7300%20Audio%20Control%20with%20Amazon%20Relay%20Module,%20Rev%202.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="1201" data-original-width="1977" height="194" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEir4AjeV26B66zyRRKWsSvtW3rto-a1xR490kiXsKBwp0lN9K-50_aATVtFspggpAPNMqxM3dUqsBwUaZOBiL398yiRhQsJloqWuNEI2vndeT6avMkDXDn90YeDzi2Z5Ev01s8UMFtv9gidaGPq-uf6tFD6yKLv6Om2gLduqAWfUr3jvKJqic-wgI2O/s320/IC-7300%20Audio%20Control%20with%20Amazon%20Relay%20Module,%20Rev%202.png" width="320" /></a></div><br />Please note the following:<p></p><p></p><ol><li>An Icom 7300 controls the relay module's on/off state using the radio's back-panel "SEND" jack.</li><li>The module's JUMPER should be in the 'L' (not 'H') position.</li><li>DC power to the module is assumed to be 13.8 VDC (i.e. the same voltage as the radio's power).</li><li>An optional series Capacitor and Resistor can be added between the module's IN and DC- terminals to delay when the module turns <i>off </i>at the end of TX. This delay is to prevent you from hearing the end of your transmission when you transition back to RX (there is typically a delay in the webSDR audio). The value of the capacitor sets the "release" delay (47 uF gives roughly 300 msec of delay), and the resistor limits the current into the SEND jack at the start of TX, when the capacitor discharges (the SEND jack is rated at 0.5A, max).</li></ol><p></p><p><br /></p><p><b><u>Standard Caveat:</u></b></p><p>As always, I might have made a mistake in my equations, assumptions, drawings, or interpretations. If you see anything you believe to be in error or if anything is confusing, please feel free to contact me or comment below.<br /><br />And so I should add -- this information is distributed in the hope that it will be useful, but WITHOUT ANY WARRANTY; without even the implied warranty of MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.</p>Jeffhttp://www.blogger.com/profile/04853314106806116765noreply@blogger.com2tag:blogger.com,1999:blog-2257458838727315711.post-43932820549007266232022-04-12T19:40:00.034-07:002022-04-21T08:17:15.669-07:00Repair Log: HP 3314A, Part 2<p>
This post is part 2 of my HP 3314A Function Generator
troubleshooting posts, in which I continue troubleshooting my swap-meet find. </p><p>Part 1 is
<a href="https://k6jca.blogspot.com/2022/03/blog-post.html">here</a>.</p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEizg2s0ZWqiS3WpaM3o3uY9mqbBDKbIlKbQnqZg-TwfAAmWLoQn-T8Lc4SFLFDA_A33JSFmcOet5W-zEmoxBPIT-pH1lQ07MA--f_stHN7CYJpJNkhu6T1in9l0vXGwmiXgadAnWaAOnq05Mc-03qZ7lpZFagBzWhBCb9I1WjYcf59ZL3IvDvipxeCM/s640/working%20front%20view.jpg" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="480" data-original-width="640" height="240" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEizg2s0ZWqiS3WpaM3o3uY9mqbBDKbIlKbQnqZg-TwfAAmWLoQn-T8Lc4SFLFDA_A33JSFmcOet5W-zEmoxBPIT-pH1lQ07MA--f_stHN7CYJpJNkhu6T1in9l0vXGwmiXgadAnWaAOnq05Mc-03qZ7lpZFagBzWhBCb9I1WjYcf59ZL3IvDvipxeCM/s320/working%20front%20view.jpg" width="320" /></a></div><br /><p><b><u>Battery Replacement:</u></b></p>
<p>
The HP 3314A contains a Lithium battery to keep power applied to the CMOS RAMs
(U211 and U212) and to the Reset Circuitry (U203, an MC14584B Hex Schmitt
Trigger IC) when the 3314A is powered off.
</p>
<p>
The image below shows the dead battery that was in the 3314A when I
purchased it. Although not easily seen in the picture below, there are
two wires attached to the PCB that are pushed against the ends of the battery,
but they are not soldered to it. So electrical contact with the
battery's terminals depends upon the physical pressure of the two wires
against the battery's ends.
</p>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhFXxJWeU__NU27Lhsdo7ctq-oR75klt_WPdDKxnZP-NW6EiKXQbLSuGsEUEMzvUzle_1uqaczyslz7gVup63_j806Ck6hvHhx00pcrrB_8fdSJhzroUZeiCSJavSWVNz9Tj6tEodaHX2jLN4sWETW0uBgMF8aeYS0b73G465PdsIXLsSkYWH1pDwVR/s640/220328%20Original%20Battery.jpg" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="480" data-original-width="640" height="240" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhFXxJWeU__NU27Lhsdo7ctq-oR75klt_WPdDKxnZP-NW6EiKXQbLSuGsEUEMzvUzle_1uqaczyslz7gVup63_j806Ck6hvHhx00pcrrB_8fdSJhzroUZeiCSJavSWVNz9Tj6tEodaHX2jLN4sWETW0uBgMF8aeYS0b73G465PdsIXLsSkYWH1pDwVR/s320/220328%20Original%20Battery.jpg" width="320" /></a>
</div>
<p>
My guess is that the two wires attached to the PCB wereleads that were part of
an older, originally installed by HP, battery, and a previous owner simply
clipped these two leads from the ends of that original battery and then
pressed them against the terminals of a modern replacement battery that did
not have wires attached to it, hoping that this physical contact would provide
a reliable electrical contact.
</p>
<p>
I purchased a CR123A battery holder and new 123-style Lithium 3 volt battery
via Amazon and, after removing the original two wires on the board, I used two
new wires to connect the holder to the PCB. This assembly is held in
place with a tie-wrap.
</p>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEht3TAuxluZGoVhVoAlHDDVzJ94JV4dEDOz_938HQil7dT4M-qD__qIy0xvTV6AgFZpFS7TWQx0rkA7vLdUarEoq1G6CA_5Vtl_MdF9qCkUJzwV3Axq05r9P0gduYVkO8nHOTLUaHxY5D2bHRHP9gfHPXGOb-DxQLI4Kw4g6qlgfewoHRgrYQ3508dy/s284/New%20Battery.jpg" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="211" data-original-width="284" height="211" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEht3TAuxluZGoVhVoAlHDDVzJ94JV4dEDOz_938HQil7dT4M-qD__qIy0xvTV6AgFZpFS7TWQx0rkA7vLdUarEoq1G6CA_5Vtl_MdF9qCkUJzwV3Axq05r9P0gduYVkO8nHOTLUaHxY5D2bHRHP9gfHPXGOb-DxQLI4Kw4g6qlgfewoHRgrYQ3508dy/s1600/New%20Battery.jpg" width="284" /></a>
</div>
<br />
<p>
<b><u>ROM Failure:</u></b>
</p>
<div>
Even with the bad CMOS RAM temporarily replaced (see
<a href="https://k6jca.blogspot.com/2022/03/blog-post.html">Part 1</a>), the
unit would still end up with a blank seven-segment display after the power-up
count-down sequence. Why was the microprocessor failing to continue?
</div>
<div><br /></div>
<div>Perhaps there's another memory failure.</div>
<div><br /></div>
<div>
Fortunately, HP included a number of self-tests in the HP 3314A. One of
these is a memory test, which verifies the contents of the ROM memory (of
which there are six ROMs in my early version of the 3314A (U207-U210, U236,
U238)), as well as the RAM (U233, U234) and CMOS RAM (U211, U212; these last two being
the battery backed-up RAM).
</div>
<div><br /></div>
<div>
This memory test can be invoked by holding down the ARB button while turning
on the 3314A. It can take 20 to 30 seconds to complete, and, when
finished, it will display which memory devices test GOOD by turning ON a
front-panel LED for each good device. If the associated LED is OFF, then
the memory device failed the test.
</div>
<div><br /></div>
<div>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhkyEKcWyv3LuEF_SnQL_52-IKrPm0xKdFToIdCjaJThure2aJmbYTzer74hIpNr9VLmO_gAukCuoascW3hRSYVnf_Qx1MWdmieYL8G1ONLeWxHHYCHRi60Bn5Ld2_W_-1KsFQT6beOsa49hMxNCd20s9OkUG32zd94dziOL5gqmEq1wLtzCBvjbcjl/s724/Memory%20Tests.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="724" data-original-width="632" height="320" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhkyEKcWyv3LuEF_SnQL_52-IKrPm0xKdFToIdCjaJThure2aJmbYTzer74hIpNr9VLmO_gAukCuoascW3hRSYVnf_Qx1MWdmieYL8G1ONLeWxHHYCHRi60Bn5Ld2_W_-1KsFQT6beOsa49hMxNCd20s9OkUG32zd94dziOL5gqmEq1wLtzCBvjbcjl/s320/Memory%20Tests.png" width="279" /></a>
</div>
<br />When I invoked this test, the AMPLITUDE LED remained OFF, signifying
that U236 was bad.
</div>
<div><br /></div>
<div>
(Note -- when I first tried to invoke this test, prior to replacing U212 (the
bad CMOS RAM IC), all of the LEDs would turn OFF and the test would never
finish.)
</div>
<div><br /></div>
<div>
Pinout-wise, the six on-board ROMs seem equivalent to 2364 ROMs. These
are 8Kx8 bit memories. I have 2764 EPROMs in my junk box (from
designs I did 40 years ago) that are also 8Kx8 devices. Could I program one of my EPROMs to replace the bad on-board ROM?
</div>
<div><br /></div>
<div>
But first, were the 3314A ROM binary
files even available?
</div>
<div><br /></div>
<div>
A query to the HP/Agilent/Keysight groups.io site (<a href="https://groups.io/g/HP-Agilent-Keysight-equipment">https://groups.io/g/HP-Agilent-Keysight-equipment</a>) quickly got a reply to take a look at the
<a href="http://www.ko4bb.com/getsimple/index.php?id=manuals&dir=01_ROM_Images_and_Drivers/HP_3314A_Eprom">Ko4BB web-site</a>, which has files for both the early version of the 3314A (like mine) which
had six 8Kx8 ROMs, and the later revision which had a single 64Kx8 ROM.
</div>
<div><br /></div>
<div>
So binary files were available, but one more issue remained -- the
original ROMs were 24 pin devices, but the 2764 EPROM is a 28 pin device.
</div>
<div><br /></div>
<div>
I could wire up an adapter board, myself, but fortunately, one can purchase
adapter boards for this purpose, and eBay had just the thing.
(Apparently this was a common modification for the Commodore computer --
search for "2364 EPROM Adapter").
</div>
<div><br /></div>
<div>
Here's a photo of the adapter kit I purchased from eBay (this photo is from
the eBay listing):
</div>
<div><br /></div>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgyEpu4OOLqDbz80YDooo-1vdvtljCUjaHeCpfuRAeqksl1fm-HTuOMt5yvhlVLU-X2WBqgrs2CXuFgK80a_hZFx900Fn1Qb7ps-_Sq9X8g5JugPv6l7hrksMI28dNjQPfShc85IEPXeLGcd91YgA8YTcG1db_wj13LXiAHMtL4JGLH421Jb3r9eyFr/s497/Prom%20to%20eprom%20adapter.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="338" data-original-width="497" height="218" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgyEpu4OOLqDbz80YDooo-1vdvtljCUjaHeCpfuRAeqksl1fm-HTuOMt5yvhlVLU-X2WBqgrs2CXuFgK80a_hZFx900Fn1Qb7ps-_Sq9X8g5JugPv6l7hrksMI28dNjQPfShc85IEPXeLGcd91YgA8YTcG1db_wj13LXiAHMtL4JGLH421Jb3r9eyFr/s320/Prom%20to%20eprom%20adapter.png" width="320" /></a>
</div>
<br />
<div>
Note that pin 27 of the 2764 EPROM (its /PGM pin) should be tied to VCC for
normal READ operations. This pin is not normally tied high on the adapter board, but the board does include pads to accomplish
this -- I simply applied a solder-short from the "A14" pad to the "VCC" pad
(these pads are visible in the center of the board in the image, above).
</div>
<div><br /></div>
<div>
Pin 26 of the 2764 EPROM should remain floating (No Connection). Note
that this pin corresponds to the "A13" pad on the adapter board, and it should remain unconnected.
</div>
<div><br /></div>
<div>
With the adapter built and an EPROM programmed (using an inexpensive TL866A
PROM Programmer, available, for example, via Amazon), I soldered the adapter
to the PCB in place of the original ROM and inserted the programmed PROM into
its socket.
</div>
<div><br /></div>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgt1zzjWSnsP6DwXy3zVjOkSWAJ8TFXNzpA40LqSYi4_96FVSi3GROYkBjyCweIDqpbQH4YXhyHoKyHj7OJSe0co0fFaBwmSTexVwATD3_dF6W54X0yPgo7lEHc6V2P07l0W4xdvsXXCVSMM_Y29L83gZP-4eQiIrSB2VreuuF6KfbGARwCx5qbDxKp/s324/New%20PROM.jpg" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="242" data-original-width="324" height="239" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgt1zzjWSnsP6DwXy3zVjOkSWAJ8TFXNzpA40LqSYi4_96FVSi3GROYkBjyCweIDqpbQH4YXhyHoKyHj7OJSe0co0fFaBwmSTexVwATD3_dF6W54X0yPgo7lEHc6V2P07l0W4xdvsXXCVSMM_Y29L83gZP-4eQiIrSB2VreuuF6KfbGARwCx5qbDxKp/s320/New%20PROM.jpg" width="320" /></a>
</div>
<br />
<div>
I powered up and...hurray! The seven-segment display did <i>not</i> blank out after the initial count-down following power up! </div><div><br /></div><div>And when I invoked the Memory Test at
power-up, all memories passed.
</div>
<div><br /></div>
<div>
But there were still problems -- now the unit displayed the following errors
during its power-on calibration procedure:
</div>
<div><br /></div>
<blockquote style="border: none; margin: 0px 0px 0px 40px; padding: 0px; text-align: left;">
<div>E30 (No Frequency Detected)</div>
<div><br /></div>
<div>E34 (Signal Amplitude Outside Measurement Range)</div>
</blockquote>
<div><br /></div>
<div>So on to the next problem...</div>
<div><br /></div>
<div><br /></div>
<div>
<b><u>Error E30: No Frequency Detected:</u></b>
</div>
<div><br /></div>
<div>
The Service Manual has a flowchart for troubleshooting Frequency Calibration
errors. Following this chart led me to the box labeled "Common Mode
Rejection Circuitry":
</div>
<div><br /></div>
<div>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgXGlXhyZmOd1ConnOTYqRW175usu9x0oDr4yRVm8dP5HsoR94dlQMFcaEO__EVlSesOgf9NZXktEJ_LBRgS4uMFosLn9lOQRlUnKZedrRnJw1lFVv1T72Y1Y9M9v-DswmBkEtWtdxpeC-m8YhDKauHy4MgyA4MeqFFT6qkR79-5tKvXqEbZhXw605v/s1453/Frequency%20Error%20Troubleshooting.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="825" data-original-width="1453" height="182" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgXGlXhyZmOd1ConnOTYqRW175usu9x0oDr4yRVm8dP5HsoR94dlQMFcaEO__EVlSesOgf9NZXktEJ_LBRgS4uMFosLn9lOQRlUnKZedrRnJw1lFVv1T72Y1Y9M9v-DswmBkEtWtdxpeC-m8YhDKauHy4MgyA4MeqFFT6qkR79-5tKvXqEbZhXw605v/s320/Frequency%20Error%20Troubleshooting.png" width="320" /></a>
</div>
<br />The Common Mode Rejection Circuitry is part of the "Triangle Integrator"
schematic, below. The purpose of this circuitry is to keep the DC level
at the Source Pins of Q220 (i.e. pins 1 and 5) at about -5 Vdc.
</div>
<div><br /></div>
<div>
On my board I measured the "Common Mode Sense" voltage (at the base of Q211)
to be -3.4 volts, not the specified -5 volts. Clearly there was a
problem. But was the problem actually with the common-mode circuitry, or
was it somewhere else, with the symptoms exhibiting themselves as a
common-mode problem?
</div>
<div><br /></div><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEihDcHDXVs9qjL3Qjzx4rjwX_1pZ3zS8OHuuTlniBf8RvegBKD38Kt-cyu2elw0kvD2hjdnZkjoTXhar9fgqAVmxGzVxc55GCcIzmxBgLmEW56_XE8czN7cPVsLaTEtBubI4sn5KVrB2hF46jMDM4mQW9v1lksH3zfe8sKVRkub6qSE9uAJGos7RY7C/s914/HP%20Schematic,%20Common-Mode%20Circuit.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="717" data-original-width="914" height="251" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEihDcHDXVs9qjL3Qjzx4rjwX_1pZ3zS8OHuuTlniBf8RvegBKD38Kt-cyu2elw0kvD2hjdnZkjoTXhar9fgqAVmxGzVxc55GCcIzmxBgLmEW56_XE8czN7cPVsLaTEtBubI4sn5KVrB2hF46jMDM4mQW9v1lksH3zfe8sKVRkub6qSE9uAJGos7RY7C/s320/HP%20Schematic,%20Common-Mode%20Circuit.png" width="320" /></a></div><br /><div>(Important Sidebar Note: Although the Service Manual shows voltages at
Q220 pins 1 and 5 having a DC offset of -5 Vdc (Waveform 1, Figure 8-15,
although good luck discerning this level from the manual download from the
Kesight site), these two voltages can actually be a bit lower than -5 Vdc
(e.g. -5.4 Vdc), due to Q211 base current, even if the "Common Mode Sense"
signal at the base of Q211 is -5 Vdc)).</div>
<div>
<div><br /></div>
<div>
But back to the problem -- why was the Common Mode Sense level so far off?
</div>
<div><br /></div>
<div>
First, I needed to understand the circuit, starting with the Triangle
Integrator circuitry.
</div>
<div><br /></div>
<div>
The functional operation of the Triangle-wave generator can be understood
using the block diagram, below.
</div>
</div>
<div><br /></div>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjqfHqlL4FS5yPdRG-1HYrSJ36ZBUo14V3t4Q-ihDU4-X07zoKI5Su7osW15z0LB5Q5dd2pPtpvVhGXxTeLVTowG-qsu7aPVpN85cyF8E3QYgMVpSy26x7_G9ftIQniAvaa0Hvbvo8SNZNpO8R1-0RLZA5qg0bg2dCMDdeGyt4ArKvCwhV2mWyheJbC/s760/HP%20Integrator%20Block%20Diagram.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="760" data-original-width="752" height="320" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjqfHqlL4FS5yPdRG-1HYrSJ36ZBUo14V3t4Q-ihDU4-X07zoKI5Su7osW15z0LB5Q5dd2pPtpvVhGXxTeLVTowG-qsu7aPVpN85cyF8E3QYgMVpSy26x7_G9ftIQniAvaa0Hvbvo8SNZNpO8R1-0RLZA5qg0bg2dCMDdeGyt4ArKvCwhV2mWyheJbC/s320/HP%20Integrator%20Block%20Diagram.png" width="317" /></a>
</div>
<br />
<div>
<div>First, note that in normal operation (e.g. the generator's state after power-up), the path through the Q210 differential pair is not used. Instead, Q208 controls the generation of the triangle wave.</div><div><br /></div><div>Basically, a constant current is passed through the capacitor C (connected
across the collectors of the two differential pairs Q208 and Q210), first charging the
capacitor in one direction (Iup and Idn both passing through, for example, the left-hand transistor of the differential pair, Q208a, with Iup charging the capacitor).</div><div><br /></div><div>A downstream comparator (not shown in the block diagram, above), monitors the voltage across the capacitor and, when this voltage reaches a threshold, it flips the states of the Sup and Sdn control signals, forcing the capacitor to charge in the opposite direction (i.e. via Idn, with both Iup and Idn now passing through Q208b).
</div>
<div><br /></div>
<div>
If everything is working properly, this flipping back and forth of the charging path should result in a 1 volt
peak-to-peak triangle waveform at the output of the Triangle Integrator
circuitry (i.e. 1 Vpp at the emitter of Q217 and 1 Vpp at the emitter of Q218).
And because Q217 and Q218 are emitter followers) the same peak-to-peak
levels should be seen at each of Q220's two Source pins (pins 1 and
5).
</div>
<div><br /></div>
<div>
But when I measured the signals at Q220's Source pins, there wasn't a
1 Vpp signal present on either pin. Instead, one Source pin (pin 1) was stuck at 0.7 Vdc, while the other Source pin (pin 5) was sitting at -7.5 volts.
</div>
<div><br /></div>
<div>
<div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjNE7NrxY3VKueo2oqOQSGn5DCSsary863OCIKgC6GuCcSVfdUn-D0pg8hmo4ArbadYWVbcfD7wp6KAfcky4h7Q4TH0kfIz9lCB_mZIyHE6D2qKdzBWXPSm0HbC-poo3Kw-p8rDHpppqzyxlPliITTIweQTHxEALfWNF0j8g5jrjSDwjIRL_ig2Hg_B/s769/HP%20Schematic,%20Integrator%20Voltages.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="404" data-original-width="769" height="168" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjNE7NrxY3VKueo2oqOQSGn5DCSsary863OCIKgC6GuCcSVfdUn-D0pg8hmo4ArbadYWVbcfD7wp6KAfcky4h7Q4TH0kfIz9lCB_mZIyHE6D2qKdzBWXPSm0HbC-poo3Kw-p8rDHpppqzyxlPliITTIweQTHxEALfWNF0j8g5jrjSDwjIRL_ig2Hg_B/s320/HP%20Schematic,%20Integrator%20Voltages.png" width="320" /></a></div><br />The voltage delta between these two pins was 8.2 Vdc (much greater than the
1 volt delta it should have been). And it was because of this excessive voltage delta that the Common Mode Sense voltage measured to be -3.4 Vdc
and not the desired -5 Vdc (the Common-Mode Sense voltage, assuming that Q211's base
current is insignificant, is a function of the voltage divider consisting of R233, R234, R237, and R23 connected in series between the two Source pins of Q220, and thus this voltage should equal the average of the voltage between these two pins (i.e. (0.7 + (-7.5))/2 = -3.4 V).</div><div><br /></div>
<div>
So why was the voltage delta between the two source pins so large? And
why weren't the transistors in the Q208 differential pair flipping back and
forth to create the triangle wave? Was there a problem with the triangle generator, or maybe the comparator?</div><div><br /></div><div>This was looking a lot less like an issue with the Common Mode Rejection circuitry and instead an issue somewhere else.</div>
<div><br /></div>
<div>
Probing the circuit's transistor voltages gave me some very strange results when I looked at transistor voltages in aggregate, but after digging a bit further I found that my confusion was, in fact,
caused by a schematic error in which the transistor reference designators
for Q208a and Q208b were swapped, compared to the labels on the PCB
silkscreen. The correct reference designators are shown, below:</div>
</div>
<div><br /></div>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhwKkOBuD-SK38gtugWugyhD6efj7dCLIchJkfCwe_htqRaQfNy71H4By-JBMx-nCTgpgO2K5N1uJ-qepSdZSLj7FejxXHotZY381-aHok4oAyqbPuZ8lDES1TBwMePdd66rjgrPAZzUDov0w77c22YdV6ZSveyXZDzG1wAiQv0nkalIj2Lpijj6H_y/s1002/HP%20Schematic%20Error,%20Triangle%20Generator%201.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="548" data-original-width="1002" height="175" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhwKkOBuD-SK38gtugWugyhD6efj7dCLIchJkfCwe_htqRaQfNy71H4By-JBMx-nCTgpgO2K5N1uJ-qepSdZSLj7FejxXHotZY381-aHok4oAyqbPuZ8lDES1TBwMePdd66rjgrPAZzUDov0w77c22YdV6ZSveyXZDzG1wAiQv0nkalIj2Lpijj6H_y/s320/HP%20Schematic%20Error,%20Triangle%20Generator%201.png" width="320" /></a>
</div>
<br />
<div>
With that schematic issue resolved, the measured Integrator's voltages now made
sense, and the problem would seem to be further downstream, at the
comparator. (It is the comparator that flips the Sup and Sdn signals
back and forth, which in turn control the charging direction of the
integrating capacitor via Q208a or Q208b).<br />
</div>
<div><br /></div>
<div>
While troubleshooting the comparator's voltages I ran into another documentation error
-- this time in the Comparator Block Diagram:
</div>
<div><br /></div>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEif4ZrZpqrCMZ2xUzbqR_w48YjQSCd02ibCOEF2RWGDO9yOPkJErFIry-nkdVsbpmNu5mg20jMJt2qq0tK73YKXReCx3FKUMmt72TPLNA7ji_Y_SKLkOI-I--uChyFT9ei-ovCBh1FJJ-ZhE64wDidU1q1xIGa7r1Z4k4W-yfRez_wIwtqD7bgPcQg1/s773/HP%20Schematic%20Error,%20Comparator%20Block%20Diagram.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="726" data-original-width="773" height="301" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEif4ZrZpqrCMZ2xUzbqR_w48YjQSCd02ibCOEF2RWGDO9yOPkJErFIry-nkdVsbpmNu5mg20jMJt2qq0tK73YKXReCx3FKUMmt72TPLNA7ji_Y_SKLkOI-I--uChyFT9ei-ovCBh1FJJ-ZhE64wDidU1q1xIGa7r1Z4k4W-yfRez_wIwtqD7bgPcQg1/s320/HP%20Schematic%20Error,%20Comparator%20Block%20Diagram.png" width="320" /></a>
</div>
<br />
<div>
(And although it is not shown in this blog post, Figure 8-19 has the same pin
errors.)
</div>
<div><br /></div>
<div>
As I probed the DC voltages of U303 (a CA3102 dual differential amplifier IC)
it quickly became apparent that something was really messed up. For
example, the device's Substrate pin (pin 5) was sitting at -4 Vdc instead of
-15 Vdc. The substrate pins voltage should be the lowest voltage on the chip, and it clearly was not (see the voltages in the image, below).</div><div><br /></div><div><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh0dZr6ySzUv-S4JcYpE_wQGbQlRbh0ePXkCJ9Z9l92TQo-LopqtuTxsN7IwGBfC2QFx3PpHgglDBcCTfWd5rqQ5QHnLyI6YpCgG4nMdo5vas9StN7WdtzMLbOCy1ueb02E2V20GSLuiNLXVtnD0m8xv9WS-uWBmR4p5Ax0hoAg2OwQ9Mu2tR40lK5H/s1449/schematic,%20comparator%20voltages.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="857" data-original-width="1449" height="189" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh0dZr6ySzUv-S4JcYpE_wQGbQlRbh0ePXkCJ9Z9l92TQo-LopqtuTxsN7IwGBfC2QFx3PpHgglDBcCTfWd5rqQ5QHnLyI6YpCgG4nMdo5vas9StN7WdtzMLbOCy1ueb02E2V20GSLuiNLXVtnD0m8xv9WS-uWBmR4p5Ax0hoAg2OwQ9Mu2tR40lK5H/s320/schematic,%20comparator%20voltages.png" width="320" /></a></div><br />Also, the current into pin 11 (transistor base current) was a whopping 33 mA!</div><div><div><br /></div>It looked like U303 was bad, but I didn't have any replacements in my
junkbox. Fortunately, I found some on eBay and ordered a couple.
But in the meantime I decided to make a simple replacement for the CA3102
using six discrete 2N3904 transistors to replicate the CA3102's two
differential amplifiers. It wouldn't be perfect -- the 2N3904
transistors weren't matched and their fT is much lower than that of the
CA3102's transistors, but hopefully I could continue with my troubleshooting
while awaiting the arrival of the CA3102 devices.</div><div><br /></div><div>(Sidebar: Interestingly, the CA3102 was already mounted in a socket. I don't know if HP did this during manufacturing, or if someone later modified the board and added the socket. If the former, maybe the CA3102 devices needed to be hand-selected to get the required performance.)</div>
<div><br /></div>
<div>
Below is an image of my "Temporary CA3102" replacement, using six 2N3904
transistors and built onto a 14-pin DIP header that, in turn, is plugged into
a 14-pin socket at U303's location on the PCB:
</div>
<div><br /></div>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjzgsJBhhK0Ea2J3_Lnc9haHmguyccDcOfVwwRXRtKlH8aToNRjSTqXD1YwzrlBL8tnOz6Hhkstw_NX0sQ_Up4cRODVu-el-TvU1y5_BiiHYJngq5vRaDM1J5Xey8yPKTg-pwt9HalwM9PvMcRSujO5LCLCODMH3hil2hgJcVmxJCtsq31VpP0yCRZH/s322/U303%20Kludge.jpg" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="234" data-original-width="322" height="233" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjzgsJBhhK0Ea2J3_Lnc9haHmguyccDcOfVwwRXRtKlH8aToNRjSTqXD1YwzrlBL8tnOz6Hhkstw_NX0sQ_Up4cRODVu-el-TvU1y5_BiiHYJngq5vRaDM1J5Xey8yPKTg-pwt9HalwM9PvMcRSujO5LCLCODMH3hil2hgJcVmxJCtsq31VpP0yCRZH/s320/U303%20Kludge.jpg" width="320" /></a>
</div>
<div><br /></div>
<div>
With this temporary replacement installed on the PCB, there were no longer any
"E30" errors at power up, and there was now a sine wave present at the 3314A's
output!
</div>
<div><br /></div>
<div>
But, even though there was now a sine wave output, there were still errors
displayed at power up. These errors consisted of one E31 error
(frequency error exceeds correction capability) and four E34 errors (signal
amplitude outside measurement range) that were displayed during the
calibration sequence.
</div>
<div><br /></div>
<div>So on to the next problem...</div>
<div><br /></div>
<div><br /></div>
<div>
<b><u>Error E34: Signal Amplitude Outside Measurement Range:</u></b>
</div>
<div><br /></div>
<div>
Error E34 identifies the signal amplitude as being "outside measurement
range." But what is "outside measurement range"?
</div>
<div><br /></div>
<div>
Unfortunately, the Service Manual seems to be mute on this error condition,
but, if I measured the 3314A's output (across a 50 ohm load) with the 3314A's
amplitude set to 1.00 volts (peak-to-peak) and 0 volts DC offset, the output
level and offset were clearly not what they should have been, as the image
below shows:
</div>
<div><br /></div>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEje39A3vv1umzuZYXH3xzGzm51sFG99F33Hua3XiwUrxSpM-zgc9eRCK8QPP7CVlEsZJgrv_J5iTzza6gZez4k-z2Y9O_wrCl9DByPGld8htahLWHgOsxRCnXjTMqVJfnpoOT1yFHvhQOpA_DUX5BctYbVEeLVpoge2rchezrJS5QUGZLp9ChtwCgLn/s640/Bad%20Output.jpg" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="480" data-original-width="640" height="240" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEje39A3vv1umzuZYXH3xzGzm51sFG99F33Hua3XiwUrxSpM-zgc9eRCK8QPP7CVlEsZJgrv_J5iTzza6gZez4k-z2Y9O_wrCl9DByPGld8htahLWHgOsxRCnXjTMqVJfnpoOT1yFHvhQOpA_DUX5BctYbVEeLVpoge2rchezrJS5QUGZLp9ChtwCgLn/s320/Bad%20Output.jpg" width="320" /></a>
</div>
<br />
<div>
With the output amplitude set to 1.00 volts and no DC offset, the voltage
across a 50 ohm load measured to be 0.88 Vpp with an offset of -0.24
Vdc, not the desired 1.0 Vpp with 0 Vdc offset.
</div>
<div><br /></div>
<div>Clearly there was a problem.</div>
<div><br /></div>
<div>
Because E34 is an "Amplitude Calibration" error, I decided to follow the
Service Manual's "Amplitude Calibration Troubleshooting Flowchart" (figure
8-11B).
</div>
<div><br /></div>
<div>
Following its flow and making the recommended measurements at the appropriate
steps, I arrived at its conclusion that, "Maybe U502 or associated
cells/circuits" were at fault.
</div>
<div><br /></div>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEidwRoF-rlQWlaak9IqbbY8Yb6oa7GDI9eyW4r7Y2xw0WM6t_eJ7anYa9rAnDubvP7IoBfBiNOjrgt08LJfuYLcWIiod0tP-Bh8HEWGCJ_mQg-kuBhRDB0VInrmdE2H19NcjQdzumAXuFI88wHIspEO-darK_AGpL-8AEch8BEJM3STvv-YgvAQxwI3/s1393/Amplitude%20Error%20Troubleshooting.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="869" data-original-width="1393" height="200" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEidwRoF-rlQWlaak9IqbbY8Yb6oa7GDI9eyW4r7Y2xw0WM6t_eJ7anYa9rAnDubvP7IoBfBiNOjrgt08LJfuYLcWIiod0tP-Bh8HEWGCJ_mQg-kuBhRDB0VInrmdE2H19NcjQdzumAXuFI88wHIspEO-darK_AGpL-8AEch8BEJM3STvv-YgvAQxwI3/s320/Amplitude%20Error%20Troubleshooting.png" width="320" /></a>
</div>
<br />
<div>Well, that's pretty vague.</div><div><br /></div><div>And what is U502? It is HP's custom "Sine Shaper" IC that converts the
triangle wave into a reasonable facsimile of a sine wave. If this part
were bad, then I might as well abandon the project, because it was sure to be
"unobtanium."
</div>
<div><br /></div>
<div>
But it was too early to abandon hope, so I started troubleshooting by first
examining U502's signals.
</div>
<div><br /></div>
<div>
Well, U502's signals "seemed" good, in the sense that nothing was obviously
out of whack, but there wasn't much information in the Service Manual as to
what U502's signals should actually look like.
</div>
<div><br /></div>
<div>
Assuming (hopefully) that U502 wasn't the issue, could the problem be in an
earlier stage? For example, I had noticed that the DC levels at the
Integrator circuit's Q220 Source pins were about 0.4 volts lower than the
image for the waveforms shown in the Service Manual (i.e. they measured at
-5.4 volts instead of -5 volts), as shown, below:
</div>
<div>
<div><br /></div>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi4C2qUH0Xpsy38MXw59wwCe2-7xQCOPzxh2TwpKe5IqV-XX9e1Hlsxbg8RB8vNt0V_9CZnggVvUd5dBBiefoq_FkPI8CFDtAy_uK8JpnAt3YKSS75uFjHsaC5Mya-GlbFO1bRL7w2TvLd7inZBlvg7QR-e-86Dp5E5j2pGQST876iFnVPnnf6Q7msJ/s907/220409%20HP%20Schematic%20Triangle%20Generator%20--%20zoom%20into%20Common-mode.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="750" data-original-width="907" height="265" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi4C2qUH0Xpsy38MXw59wwCe2-7xQCOPzxh2TwpKe5IqV-XX9e1Hlsxbg8RB8vNt0V_9CZnggVvUd5dBBiefoq_FkPI8CFDtAy_uK8JpnAt3YKSS75uFjHsaC5Mya-GlbFO1bRL7w2TvLd7inZBlvg7QR-e-86Dp5E5j2pGQST876iFnVPnnf6Q7msJ/s320/220409%20HP%20Schematic%20Triangle%20Generator%20--%20zoom%20into%20Common-mode.png" width="320" /></a>
</div>
<div><br /></div>
</div>
<div>
Could this offset be the problem? A quick test (raising the Source
voltages of Q220 from -5.4 to -5 volts by paralleling R219 (in the Q211, Q212 circuit)
with a variable resistor) showed that changing this DC offset had absolutely
no effect on the 3314A's output voltage level or offset.
</div>
<div><br /></div>
<div>
Hmmm, could the problem be in the next stage, the Transconductance Amplifier?
</div>
<div><br /></div>
<div>
Did not seem so. DC voltages seemed reasonable, and signal levels at the
bases of Q203 were correct.
</div>
<div><br /></div>
<div>
So the signals going to the Sine Shaper seemed to be reasonable, at least as
far as I could determine (which was not very far).
</div>
<div><br /></div>
<div>
Again, hoping that the Sine Shaper IC was not the issue, perhaps the problem
was in a stage following it.
</div>
<div><br /></div>
<div>
The Sine Shaper drives the "Preamplifier" Stage (Figure 8-23 in the Service
Manual), which in turn drives the "Output Amplifier and Step Attenuator"
stage, whose output, in turn, goes to the output connector on the front panel.
</div>
<div><br /></div>
<div>
As I measured the DC levels of the Preamplifier from input to output, they
looked OK...but wait! The node common to R611, R612, R606, R607, and
R608 measured at 0 Vdc (which makes sense, as this should be the DC level of a
balanced amplifier when there is no DC offset), but the DC voltage at the
<i>other</i> side of R607 (i.e. at connector J1 to the Output Amplifier's
input) was at -0.24 volts.
</div>
<div><br /></div>
<div>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhOnyxIXLsYAeyJywuxWtJDbeDLS4ZL55owV7qt2-Nfb0pRyQF8x4mcfAN9VOqf6i3gbAeIk-6OVv2WSKCLN2ORsXnn9HM4MHz1Je64_ZhrRpsCfCxtuo_L3fYIHXwkK-wnhlXYng0BH21nMbWzLlQWvN94dCco_vP_aDippdylDwY0vZWynqDK1Pqx/s839/Schematic,%20Preamplifier.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="694" data-original-width="839" height="265" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhOnyxIXLsYAeyJywuxWtJDbeDLS4ZL55owV7qt2-Nfb0pRyQF8x4mcfAN9VOqf6i3gbAeIk-6OVv2WSKCLN2ORsXnn9HM4MHz1Je64_ZhrRpsCfCxtuo_L3fYIHXwkK-wnhlXYng0BH21nMbWzLlQWvN94dCco_vP_aDippdylDwY0vZWynqDK1Pqx/s320/Schematic,%20Preamplifier.png" width="320" /></a>
</div>
<br />What happens to the DC level at the Preamplifier's J1 connector if I
disconnect the cable between J1 and the input to the Output Amplifier?
The DC voltage at the output side of R607 goes to 0 Vdc, and the sine wave
amplitude increases such that, when this point is terminated with 50 ohms, its
amplitude is correct.
</div>
<div><br /></div>
<div>
That is, with the Preamplifier's output disconnected from the input of the
"Output Amplifier" stage, and with the 3314A's amplitude set to 1 Vpp and 0
Vdc offset, the sine wave at the Preamplifier Output (terminated with 50 ohms)
correctly measures to be 1 Vpp with 0 Vdc offset, instead of the 0.88 Vpp with
-0.24 Vdc offset that would have been measured at the <i>output</i> of the
Output Amplifier.
</div>
<div><br /></div>
<div>
So the Output Amplifier would seem to the be culprit! Here is an image
of that board:
</div>
<div><br /></div>
<div>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgsSDarLFLnYVp3FO16SVRVngHDhswetImDZZ-_L2gu856Gwy8jNO-3uZ2OnioVc_mW06kiAfB-DJ_0UCsZTTP4YqD6lRsuXxydwV-WRUx7p5WOdyfaVu-atWtlB95GMHiJATQc9whv8Cka7Dfjd07ij5LAPzEhxvEuq_8AdpJGu-2GOCBTUNvPiG7G/s619/A8%20Output%20Board.jpg" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="404" data-original-width="619" height="209" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgsSDarLFLnYVp3FO16SVRVngHDhswetImDZZ-_L2gu856Gwy8jNO-3uZ2OnioVc_mW06kiAfB-DJ_0UCsZTTP4YqD6lRsuXxydwV-WRUx7p5WOdyfaVu-atWtlB95GMHiJATQc9whv8Cka7Dfjd07ij5LAPzEhxvEuq_8AdpJGu-2GOCBTUNvPiG7G/s320/A8%20Output%20Board.jpg" width="320" /></a>
</div>
<br />While probing the Output Amplifier's DC voltages, it quickly became
apparent that Q10 was bad (an MPSH10 NPN transistor -- its Vbe measured to be
3.7 V).
</div>
<div><br /></div>
<div>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjFCsK5pSJr75OBZSH3zBywePQCYxrmT1-ys0Nn4OFAA-9E8dubn89XDpgu7JxVAyXxGR5ySwbtyKrErdK3vdb6m-iDepUm6MCITR1OML0ZjwVTNLn8mpsOEjShCQc5Ac_gaPsDMc4PYa3fpAX--ZqMgwWI9UolkhL4FGl0gxJUbTjllWREjyUSkkIz/s875/Schematic,%20Output%20Amplifier%20Failure.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="807" data-original-width="875" height="295" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjFCsK5pSJr75OBZSH3zBywePQCYxrmT1-ys0Nn4OFAA-9E8dubn89XDpgu7JxVAyXxGR5ySwbtyKrErdK3vdb6m-iDepUm6MCITR1OML0ZjwVTNLn8mpsOEjShCQc5Ac_gaPsDMc4PYa3fpAX--ZqMgwWI9UolkhL4FGl0gxJUbTjllWREjyUSkkIz/s320/Schematic,%20Output%20Amplifier%20Failure.png" width="320" /></a>
</div>
<br />I replaced Q10 and powered up the unit. Success! There were
no longer any E34 errors during power-on calibration.
</div>
<div><br /></div>
<div><br /></div>
<div>
<u>Important Note:</u> The scanned copy of the schematic for the "Output
Amplifier and Step Attenuator" board in the downloadable Service Manual on the
Keysight site (see Resource list, below) is missing some circuitry, which I've
shown in the image, below (copied from a later manual revision):
</div>
<div><br /></div>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj7wZyBFPe7_msz4RF1NEkcfB86bFiqpdWkT01F68ymN7wQtzP8f6yIC-kPVuVVb0WvjCtlbeSBL7_ABa5Nvjb1BcKOnx5URZbMf4lI6vbzn_YjgJUTZZV2LihXriW1QV64Wek_dumeo2YKWvzZK0RpFv8T0_fvDKnZsUzzY0VVaNagEf67pg64wCNB/s1483/220411%20Missing%20Circuitry,%20Output%20Amplifier.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="1483" data-original-width="1441" height="320" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj7wZyBFPe7_msz4RF1NEkcfB86bFiqpdWkT01F68ymN7wQtzP8f6yIC-kPVuVVb0WvjCtlbeSBL7_ABa5Nvjb1BcKOnx5URZbMf4lI6vbzn_YjgJUTZZV2LihXriW1QV64Wek_dumeo2YKWvzZK0RpFv8T0_fvDKnZsUzzY0VVaNagEf67pg64wCNB/s320/220411%20Missing%20Circuitry,%20Output%20Amplifier.png" width="311" /></a>
</div>
<div><br /></div>
<b><u>Conclusion:</u></b>
<div><br /></div>
<div>
With all fixes installed, the 3314A's Calibration Sequence (initiated at
power-up) successfully completes with no displayed errors. And it plays the Hallelujah Chorus! (See <a href="https://www.youtube.com/watch?v=H90d6sPUF6A">here</a> for an example).</div><div><br /></div><div>I have not done any of the Performance Tests listed in the Service Manual. I might do these at some future date, at which point, if there are any issues that need repair, there might be a Part 3 added to this 3314A troubleshooting series.</div>
<div><br /></div>
<div>Finally, note that all troubleshooting was done without the benefit of a Signature
Analyzer. However, if there had been problems with the digital logic, a
Signature Analyzer (e.g. HP 5004A) would have been invaluable.
</div>
<div>
<br />
<div><br /></div>
<div>
<b><u>Other Notes and Comments:</u></b>
</div>
<p><u>1. E31 Error:</u></p>
<p>
The E31 error (that began appearing during the Power-up Cal sequence after I
had replaced the bad U303 (CA3102) with my temporary "six 2N3904"
replacement), disappeared when I replaced the U303 temporary fix with an
actual CA3102 device.
</p>
<p><u>2. Replacing the CMOS RAM with higher-current devices:</u></p>
<p>
If uPD444 CMOS RAMs are not available for replacing U211 and/or U212, other
1Kx4 devices can be used, if they are pin compatible. Note, though,
that if the new RAMs require higher Vcc current, then to prevent battery drain their power pins should not be attached to their respective pin 18 PCB pads (which are powered by the battery when AC power is OFF). </p>
<p>
Instead, lift the power pins (pin 18) of the two devices and connect these
two lifted pins to +5 Vdc at the "front-panel" side of C17, as shown in the image, below
(the example, below, uses two Intel D2148H RAMs in lieu of the two original NEC
uPD444 RAMs):
</p>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi4PtIyASJ1IDQSbW92MjXEdkICkpH00E0POoptw-kmZdAj6FBW5JQqulnNi9i6AxU6nEOoQmFxQnd2Q3yOa9WLkgG8gWAZTINyaG9RFUmpPBpPRSjeBw5pkf82MIoSgdrGMsOiUi862IOstGs58gHa1W3sWIKoI8Inkj-yCEl4gd7RnSy-7YjlT3DR/s358/Intel%20RAM.jpg" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="275" data-original-width="358" height="246" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi4PtIyASJ1IDQSbW92MjXEdkICkpH00E0POoptw-kmZdAj6FBW5JQqulnNi9i6AxU6nEOoQmFxQnd2Q3yOa9WLkgG8gWAZTINyaG9RFUmpPBpPRSjeBw5pkf82MIoSgdrGMsOiUi862IOstGs58gHa1W3sWIKoI8Inkj-yCEl4gd7RnSy-7YjlT3DR/s320/Intel%20RAM.jpg" width="320" /></a>
</div>
<p>With this modification, the CMOS RAM will no longer retain stored information when the 3314A is turned off. And thus there will be an "E09" error (non-volatile
memory lost; battery down), displayed at the end of the Power-up calibration
sequence, every time the 3314A is powered up.
</p>
<p><br /></p>
<p>
<b><u>Resources:</u></b>
</p>
<p>
HP 3314A Manuals: <a href="https://www.keysight.com/us/en/support/3314A/programmable-function-generator.html">https://www.keysight.com/us/en/support/3314A/programmable-function-generator.html</a>
</p>
<p>
HP 3314A Repair in 3 parts: <a href="https://diysquared.blogspot.com/2021/04/fixing-hp-3314a-function-generator-part.html">https://diysquared.blogspot.com/2021/04/fixing-hp-3314a-function-generator-part.html</a>
</p>
<p>
HP 3314A Repair on YouTube: <a href="https://www.youtube.com/watch?v=wNxgBubSfH8">https://www.youtube.com/watch?v=wNxgBubSfH8</a>
</p>
<p>
HP 3314A Repair (on Antique Radios forum): <a href="https://antiqueradios.com/forums/viewtopic.php?f=8&t=360491">https://antiqueradios.com/forums/viewtopic.php?f=8&t=360491</a>
</p>
<p>
HP 3314A Teardown (EEVblog): <a href="https://www.eevblog.com/forum/testgear/hp-3314a-function-generator-teardown-explanation/">https://www.eevblog.com/forum/testgear/hp-3314a-function-generator-teardown-explanation/</a>
</p>
<p>
HP 3314A Playing the Hallelujah Chorus: <a href="https://www.youtube.com/watch?v=H90d6sPUF6A">https://www.youtube.com/watch?v=H90d6sPUF6A</a><br />(Note: Hold down FUNCTION (the blue key), SW/TR INTVL, and
START FREQ while powering up the unit).
</p>
<p>
HP 3314A ROM Images: <a href="http://www.ko4bb.com/getsimple/index.php?id=manuals&dir=01_ROM_Images_and_Drivers/HP_3314A_Eprom">http://www.ko4bb.com/getsimple/index.php?id=manuals&dir=01_ROM_Images_and_Drivers/HP_3314A_Eprom</a>
</p>
<p><br /></p>
<p>
<b><u>Standard Caveat:</u></b>
</p>
<p>
As always, I might have made a mistake in my equations, assumptions,
drawings, or interpretations. If you see anything you believe to be in
error or if anything is confusing, please feel free to contact me or comment
below.<br /><br />And so I should add -- this information is distributed in
the hope that it will be useful, but WITHOUT ANY WARRANTY; without even the
implied warranty of MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.
</p>
</div>
Jeffhttp://www.blogger.com/profile/04853314106806116765noreply@blogger.com0tag:blogger.com,1999:blog-2257458838727315711.post-34887932080520860042022-03-29T08:26:00.781-07:002022-04-14T12:42:42.511-07:00Repair Log: HP 3314A, Part 1<p></p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhkOWdK0hC30uQOziOi_xhV4G19-oDlcei62FPK17BkwYeEQLTyOJycWSBAlJdLTCuuCoBD6xNrB92Q0NpSmTYbHUsGWlI1FCTD7csAlRw0qpq37qFz_4RoB4gecvscGpYYI1K3wFMV3OBDnPUcZOYuF-18MPEpKelnqBfSvZRrrImJ2V-mEwgHftej/s375/IMG_7611.PNG" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="239" data-original-width="375" height="204" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhkOWdK0hC30uQOziOi_xhV4G19-oDlcei62FPK17BkwYeEQLTyOJycWSBAlJdLTCuuCoBD6xNrB92Q0NpSmTYbHUsGWlI1FCTD7csAlRw0qpq37qFz_4RoB4gecvscGpYYI1K3wFMV3OBDnPUcZOYuF-18MPEpKelnqBfSvZRrrImJ2V-mEwgHftej/s320/IMG_7611.PNG" width="320" /></a></div><br />At a recent swap-meet I came across an HP 3314A Programmable Function
Generator that was described as non-operational but was physically in good
shape.
<p></p>
<p>
In my opinion, HP cases make great project boxes (see
<a href="https://k6jca.blogspot.com/2015/06/antenna-auto-tuner-design-part-1.html">here</a>,
<a href="https://k6jca.blogspot.com/2017/02/an-fpga-sdr-hf-transceiver-part-1.html">here</a>, and
<a href="https://k6jca.blogspot.com/2019/12/a-500-watt-hf-pa-part-1-overview.html">here</a>), and this case looked like it would be an excellent case for a future,
yet-to-be-defined project.
</p>
<p>
And so I made an offer based solely on the idea of scrapping the 3314A's
electronics but keeping its case and internal chassis. The offer was
accepted, and the 3314A came home with me.
</p>
<p>
But after I had returned home, I had second thoughts about my plan to scrap
the unit. Why not try to fix it, if that were possible?
</p>
<p>
And thus this blog post, which describes my efforts to bring this HP 3314A
back to life.
</p>
<p>
<b><u><br /></u></b>
</p>
<p>
<b><u>Initial Power-On Error:</u></b>
</p>
<p>
The first problem I came across was at power-up -- the unit would cycle
through its initial count-down following power-up, and then it would
hang. The count numbers displayed during the Power-up cycle would often also
be screwed up. And at the end of this cycle, sometimes the 3314A would hang
with the LED numerical display being blank, and sometimes the display would
show Error Codes E13 or E43.
</p>
<p>
Looking in the manual (section 3-1), Error Code E13 does not exist, and,
although Error Code E43 exists, it is an HP-IB Error Code meaning "Invalid
Data." Given that the instrument was not in Remote mode (i.e. HP-IB
mode), the display of this last error code could be meaningless and a result
of some other error.
</p>
<p>
Note that immediately following power-up, the unit's 3-digit numerical display
should show the following sequential count-down:
</p>
<div style="text-align: center;">999</div>
<div style="text-align: center;">888</div>
<div style="text-align: center;">777</div>
<div style="text-align: center;">666</div>
<div style="text-align: center;">555</div>
<div style="text-align: center;">444</div>
<div style="text-align: center;">333</div>
<div style="text-align: center;">222</div>
<div style="text-align: center;">111</div>
<div style="text-align: center;">000</div>
<div><br /></div>
<p>The video, below, demonstrates the Count-down problem:</p>
<div class="separator" style="clear: both; text-align: center;">
<iframe allowfullscreen='allowfullscreen' webkitallowfullscreen='webkitallowfullscreen' mozallowfullscreen='mozallowfullscreen' width='320' height='266' src='https://www.blogger.com/video.g?token=AD6v5dzGk-NdnoRWZbNEklvIsR0r84MZ13yLskt55nnZK1ewv3qAcuTRQ38tZwrEsLaZSNz71U6XNJfTJnd1xQksfQ' class='b-hbp-video b-uploaded' frameborder='0'></iframe>
</div>
<p>You can see from the video that the displays are changing, so the 3314A is
attempting this count-down, but the digits of the first counts are screwed up.
</p>
<p>But before I go any further...</p>
<p>
<b><u>A quick note on my initial trouble-shooting:</u></b>
</p>
<p>
The HP Service Manual (downloadable from the Keysight website:
<a href="https://www.keysight.com/us/en/support/3314A/programmable-function-generator.html">HP 3314A Manuals</a>) recommends trouble-shooting using Signature Analysis. Unfortunately,
I was at my wife's place in Nevada City, CA, at the time of this initial
trouble-shooting, and my Signature Analyzer was back at my house in Silicon
Valley, CA.
</p>
<p>
(Plus, the SA unit has been sitting in a box, untested, for at least 20 years
(another swap-meet find), and I had no idea if it worked, or not, as I've
never had a need to use it!)
</p>
<p>
So, this initial phase of trouble-shooting was done with a 2-channel
oscilloscope...
</p>
<p>
<b><u><br /></u></b>
</p>
<p>
<b><u>Probing the Front Panel Assembly:</u></b>
</p>
<p>
Because segments of the LED seven-segment displays were either being
incorrectly kept off or, alternately, incorrectly illuminated, it made sense
to start my trouble-shooting at the Keyboard assembly where the seven-segment
displays reside.
</p>
<p>
But I immediately encountered several issues. The first issue is that
the HP schematic has a number of errors.
</p>
<p>
The image, below, shows my corrections, in red. You can see that U1's
(74LS273) inputs were incorrectly assigned. Also, the pin number of the
LED displays are incorrect.
</p>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjWOaKEQhnbdTgexBj0BGRHGm4n_e5aR6mPdoWDQqlT_7xyKyKx80Qa-OVUxGbHqY2LqmKXeiWG4Tkm0ar9gFgMXiWqvK9UABzgVK9UwtkjHcpLWLh2nkIY0sJ0ecJbJwdwMD2MDgUqDFOhjG2IkNBnBKVFKcOSQ4IChUWkOn8NRlWOLyUvDDcw-zlm/s1461/HP%20Schematic%20KBRD%20U1%20and%20Digits%20--%20ANNOTATED.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="833" data-original-width="1461" height="182" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjWOaKEQhnbdTgexBj0BGRHGm4n_e5aR6mPdoWDQqlT_7xyKyKx80Qa-OVUxGbHqY2LqmKXeiWG4Tkm0ar9gFgMXiWqvK9UABzgVK9UwtkjHcpLWLh2nkIY0sJ0ecJbJwdwMD2MDgUqDFOhjG2IkNBnBKVFKcOSQ4IChUWkOn8NRlWOLyUvDDcw-zlm/s320/HP%20Schematic%20KBRD%20U1%20and%20Digits%20--%20ANNOTATED.png" width="320" /></a>
</div>
<p>
Here's a closeup of U1. Its correct pin number assignments are shown in
red:
</p>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEixMh4eY4M2YHtyJXwJ-7I9vnXOIFFj-bZPltTi0P1LRAwfCNOxctDl4vrbr_DHtLyrN69ATac5mzs2_rqC_-uigxixAvPJh1mY-AyoFnwLpjH9e0S9Z9CNzeaLr0LYYouXCtVoViymUFTInPHW0Qb-9WZd0pf5_iv96U7R4boTc3ckAafnLgKqnf3m/s699/HP%20Schematic%20KBRD%20U1%20and%20Digits%20--%20Zoom%20on%20U1.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="483" data-original-width="699" height="221" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEixMh4eY4M2YHtyJXwJ-7I9vnXOIFFj-bZPltTi0P1LRAwfCNOxctDl4vrbr_DHtLyrN69ATac5mzs2_rqC_-uigxixAvPJh1mY-AyoFnwLpjH9e0S9Z9CNzeaLr0LYYouXCtVoViymUFTInPHW0Qb-9WZd0pf5_iv96U7R4boTc3ckAafnLgKqnf3m/s320/HP%20Schematic%20KBRD%20U1%20and%20Digits%20--%20Zoom%20on%20U1.png" width="320" /></a>
</div>
<p>
And here's a closeup of the correct seven-segment display pinout (in red):
</p>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgZcSx4YDpT9KbOi7c-K9IQkHOqcly88UngnfiHl7rdcWKF79T7dtA_uS8UakAaGkH0QuV7K1jvSdGN1jf1zwMS90-NXiXBSn2R0dvNVCCL6i_RLF6Cu1FO04oabuhI5an1vbM37T-umexCJTNijqA7ZRYefcmhtvgR4hqLRIWsJxNb3c-Hzear4Y6f/s696/HP%20Schematic%20KBRD%20U1%20and%20Digits%20--%20Zoom%20on%20digits.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="491" data-original-width="696" height="226" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgZcSx4YDpT9KbOi7c-K9IQkHOqcly88UngnfiHl7rdcWKF79T7dtA_uS8UakAaGkH0QuV7K1jvSdGN1jf1zwMS90-NXiXBSn2R0dvNVCCL6i_RLF6Cu1FO04oabuhI5an1vbM37T-umexCJTNijqA7ZRYefcmhtvgR4hqLRIWsJxNb3c-Hzear4Y6f/s320/HP%20Schematic%20KBRD%20U1%20and%20Digits%20--%20Zoom%20on%20digits.png" width="320" /></a>
</div>
<p>
(Note that I also corrected the names of the three signals driving the
common-cathode pins of these three displays to match the net names assigned to
their respective drivers.)
</p>
<p>
The image below shows the correct pin number (in red) and the associated data
bit from the 8-bit bus (in blue) for each segment of the LED.
</p>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEggXtNBFABJKuwyF2HWMfnUw_ITQLTq3PsuQ53mfVDHkT0JeBbO-SIjTXW4cILvquHnvAPuOolI16kHCVEUKdzd009EYMGwfeRTx772Xbv-X9Og8BpJgYf7R7AsWU-0DcHc8T_tZcgj6IpmiPhudAzh_49sIWFxUZmcbXFQN_M4DAWjRu-P7h7l-rOi/s471/HP%20Seven%20Segment%20Display%20Pin%20numbers%20and%20data%20bits.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="328" data-original-width="471" height="223" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEggXtNBFABJKuwyF2HWMfnUw_ITQLTq3PsuQ53mfVDHkT0JeBbO-SIjTXW4cILvquHnvAPuOolI16kHCVEUKdzd009EYMGwfeRTx772Xbv-X9Og8BpJgYf7R7AsWU-0DcHc8T_tZcgj6IpmiPhudAzh_49sIWFxUZmcbXFQN_M4DAWjRu-P7h7l-rOi/s320/HP%20Seven%20Segment%20Display%20Pin%20numbers%20and%20data%20bits.png" width="320" /></a>
</div>
<p>When count-down sequence first starts, the first count presented on each
seven-segment display is '9', and then this number counts down (8, 7, 6,
etc.). The numbers are the same on all three seven-segment displays, so
the displayed numbers should be, in sequence, 999, 888, 777, 666, 555, etc.
</p>
<p>
Thus, the segments of the seven-segment display associated with data-bits DL0,
DL1, DL2, and DL3 should be ON for the first count-down digits (9, 8, 7...), but, as
you can see in the video, above, they are OFF. These will later turn ON
for lower counts (e.g. 3, 2, 1, 0), but at the start of count-down they are OFF.
</p>
<p>The image below shows the problematic segments (highlighted in yellow):</p><p>
</p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg-0hYQWJLOXHjCpBT-yN4LRsv-kxwZfUABN_NBD8v-7b-VrUrNeBHFOAPUrSfM5sOowIQakDtOZfLvEy85lrh_N0iGoqw4LgxS4jvDSo-pszix81iCPUZrwx4ead1ggbcmt6Wh0iAaXj9d6LPnvPNyXDHyn9hkfjTAwJvbd5e-0-mr_IIRzrF9ILHB/s471/HP%20Seven%20Segment%20Display%20segments%20in%20error.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="328" data-original-width="471" height="223" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg-0hYQWJLOXHjCpBT-yN4LRsv-kxwZfUABN_NBD8v-7b-VrUrNeBHFOAPUrSfM5sOowIQakDtOZfLvEy85lrh_N0iGoqw4LgxS4jvDSo-pszix81iCPUZrwx4ead1ggbcmt6Wh0iAaXj9d6LPnvPNyXDHyn9hkfjTAwJvbd5e-0-mr_IIRzrF9ILHB/s320/HP%20Seven%20Segment%20Display%20segments%20in%20error.png" width="320" /></a></div><br />The Keyboard assembly is a multiplexed system with the segments of the
seven-segment displays sharing the same "row" signals (as in a row-column
matrix) as the other front-panel LEDs. Thus, probing a segment to see if
it is correctly turning on or off can be a bit of a challenge with a
two-channel scope, especially with the count-down's count quickly changing as
it counts down.
<p></p>
<p>
But by observing U1's outputs (the 'LS273) responsible for driving the
numerical segments immediately after power-on, it was clear that U1's outputs
were not being set to the correct state to drive these segments ON for the
first few counts (a U1 output bit must be set LOW to turn ON a segment).
</p>
<p>
Tracing this failure back...the data at U1's inputs, arriving at the Keyboard
from the Controller Board, was also in the incorrect state for these
counts. Thus, U1 was simply passing on incorrect data from the
Controller Board.<br />
</p>
<p>So time to move the probing to the Controller Board...</p><p><br /></p>
<p>
<b><u>Probing the Controller Board:</u></b>
</p>
<p>
At the controller board I started at the processor, looking at its Data Bus
while triggering the scope on the rising edge of the clock clocking data into
the U1 register on the Keyboard assembly. This probing showed that the
data from the microprocessor that was being written to the Keyboard assembly
was also incorrect.
</p>
<p>
So why was the processor writing incorrect data? Was it getting bad data
from the PROMs (of which there are six)? Or was RAM bad? Or was
something fighting the bus when the processor was attempting to read or write
this data?
</p>
<p>
Even though I did not have a Signature Analyzer, I decided to invoke one of
its tests (SA Test #1, in which switches 1, 2, 7, and 8 are turned ON),
because I assumed this test would sequentially count through all of the
microprocessor's addresses and I could look for errors either on the Address
Bus or on the Data Bus during this time.
</p>
<p>
Everything was looking good until I got to the CMOS SRAM, U211 and U212 (NEC
UPD444/6514 1Kx4 bit CMOS RAM). The four data I/O pins of U212 had flaky
looking levels for some of its read cycles. Plus, these flaky levels
would change with time.
</p>
<p>
The image below (lower scope trace) shows the levels one of these pins (pin
11, or I/O4). U212's outputs are enabled whenever the upper trace is
low.
</p>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhpJMmq_6A-3UGueHDVkspLpmb2qI5GY0STyCKvZMD8Nk-eq3rMsFXrtFR2jM1l1lgUHHXrLv-tTJB_73a0zHMicKiBuimhfKfJCUECKMksJRUTTFGgVishfVfePYxYLnxprqtqYT4rtXPj0rpVgezZL1NwKrB7vONHEVWNBuryaC5hAeEec1YYnCbx/s640/220328%20U212%20IO4%20nCS%20annotated.jpg" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="480" data-original-width="640" height="240" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhpJMmq_6A-3UGueHDVkspLpmb2qI5GY0STyCKvZMD8Nk-eq3rMsFXrtFR2jM1l1lgUHHXrLv-tTJB_73a0zHMicKiBuimhfKfJCUECKMksJRUTTFGgVishfVfePYxYLnxprqtqYT4rtXPj0rpVgezZL1NwKrB7vONHEVWNBuryaC5hAeEec1YYnCbx/s320/220328%20U212%20IO4%20nCS%20annotated.jpg" width="320" /></a>
</div>
<div><br /></div>
Compared to U212, the I/O pins of U211 looked fine -- it was only U212's four
data pins that had the issue. And these four data bits correspond to the
low-order four bits of the Data Bus, which were the problem bits driving the
Keyboard's seven-segment displays.
<div><br /></div>
<div>
Could a bad U212 be the culprit? I placed an order of a pair of new NEC
UPD444 RAMs on eBay, but while waiting for them to arrive I thought I'd continue testing...
</div>
<div><br /></div>
<div>
By the way -- the HP manual has some errors regarding SRAM power on the
Controller Board. My corrections are shown, below, in red:
</div>
<div>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgHH5dYMZGr2km9cOqWT3NzLn8bHtWqhi-6BejZmXfEJ_9oIaSOuWEh6v9RNHmIHWV56tTMEhqrrt5ummm8pzmYUxn6taIJdmrKvisRGXgDJqlSoCUPU6UZsVSOLGZ3XDrn4P5T2eCvDBSywQB2U-OS6gHsjRZ0NQWaV_yq4fRLo0sTrVy9kik3n_s5/s1439/HP%20Schematic,%20cmos%20ram%20corrections.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="787" data-original-width="1439" height="175" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgHH5dYMZGr2km9cOqWT3NzLn8bHtWqhi-6BejZmXfEJ_9oIaSOuWEh6v9RNHmIHWV56tTMEhqrrt5ummm8pzmYUxn6taIJdmrKvisRGXgDJqlSoCUPU6UZsVSOLGZ3XDrn4P5T2eCvDBSywQB2U-OS6gHsjRZ0NQWaV_yq4fRLo0sTrVy9kik3n_s5/s320/HP%20Schematic,%20cmos%20ram%20corrections.png" width="320" /></a>
</div>
<p><br /></p>
<p>
<b><u>Excessive Battery Current Drain:</u></b>
</p>
<p>
After ordering replacement UPD444 CMOS SRAM from eBay, I decided to look
around a bit more and see what other problems I could find while waiting for
the SRAM to arrive.
</p>
<p>
One issue I had noticed when I first opened up the unit was that the battery
backup was dead (the battery was sitting at about 0.45 volts), yet it was a
fairly new battery (with a 2023 date). Could it have been killed by
excessive current draw?
</p>
<p>Note that this battery serves two purposes:</p>
<p>
First, it keeps power applied to U203 of the RESET circuitry (CMOS
inverters), even when power is OFF, so that the RESET line can be properly
driven when power is first applied.
</p>
<p>
Second, the battery provides power to U211 and U212 (the CMOS SRAM), to
maintain the contents in these two SRAM chips even when the unit is powered
OFF.
</p>
<p>
Per section 5.2 of the manual, battery current drain when power was off
should be on the order of 1.35 uA (typical) to 18.5 uA (max), when measured
as an equivalent voltage drop across R13, a 1K ohm resistor -- e.g. 1.35 uA
of current would equal 1.35 mV across R13).
</p>
<p>
Replacing the dead battery with a bench supply set to 3.0 volts, I measured
0.495 volts across R13 -- in other words, current draw was 495 uA, not the
spec'd 18.5 uA max!
</p>
<p>No wonder the battery was dead!</p>
<p>
The battery-power circuit is fairly simple, which means there weren't many
suspects. And because U212 (one of the two CMOS SRAMS) was on my
"hit-list" of suspected bad parts, I decided to clip its VCC pin (pin 18)
and see what happened to battery current.
</p>
<p>
With the pin clipped (i.e. no VCC applied to U212), the voltage across R13
dropped way down (below the level that my Fluke DVM can measure) -- clearly
U212 was the culprit!
</p>
<p><br /></p>
<p>
<b><u>U212 Temporary Replacement Test:</u></b>
</p>
<div>
OK -- two different tests pointed to U212 being bad. Was there some
way that I could verify this while waiting for my eBay UPD444 CMOS RAM to
arrive?
</div>
<div><br /></div>
<div>
A friend gave me a National MM2114N-L 1Kx4 bit RAM from his junk box.
This device draws more supply current than the original NEC part, so it
would not be a satisfactory permanent replacement for the UPD444, but I
thought I would give it a try to test my theory that it was U212 that caused
the count-down errors.
</div>
<div><br /></div>
<div>
So I removed the original part, installed an 18-pin socket in its place, and
inserted the '2114 in its place (also, the battery had been removed, too (it
was dead) -- if the battery had been good, though, the MM2114N would have
presented a much larger than expected current drain, and I would have removed it before doing this test).
</div>
<div><br /></div>
<div>
I turned power on, and...nothing. The seven-segment displays remained
dark!
</div>
<div><br /></div>
<div>
Probing with the scope, I quickly discovered that with power ON, Vcc at the
'2114 was around 3 volts, not the expected 5 volts.
</div>
<div><br /></div>
<div>
Power to the CMOS RAM is a diode-OR'd combination of Vcc (via CR2, when the unit
is powered up) or the Battery voltage (via CR3, when the unit is powered down
and the CMOS RAM thus battery-powered).
</div>
<div><br /></div>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhAYH7Gjy2_hw6Pfvp9VndX1fBNdOVBIvnjkryhLbvhFQh0hO3wDhuYI9NCzoxtqyuCoG5DYVC6_d2ZuW61NO7wf45WDDwToYQ2v6zn0uFlv8ooBB2fsTL1tgQw3Yfy2x1DTMLGcwyjeNjKo6SRnEGyJCTjOZ-zYzb4efqzf0x2dLZApgSdo_-dNu04/s1153/HP%20R17%20and%20CR2.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="530" data-original-width="1153" height="147" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhAYH7Gjy2_hw6Pfvp9VndX1fBNdOVBIvnjkryhLbvhFQh0hO3wDhuYI9NCzoxtqyuCoG5DYVC6_d2ZuW61NO7wf45WDDwToYQ2v6zn0uFlv8ooBB2fsTL1tgQw3Yfy2x1DTMLGcwyjeNjKo6SRnEGyJCTjOZ-zYzb4efqzf0x2dLZApgSdo_-dNu04/s320/HP%20R17%20and%20CR2.png" width="320" /></a>
</div>
<p>There is a 49.9 ohm resistor (R17) in series with CR2. The Vcc current
of the '2114 RAM was enough to create a significant voltage drop across this
49.9 ohm resistor.</p>
<div>The obvious temporary solution was to short out the resistor for this test, which I
did.</div>
<div><br /></div>
<div>
<div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjru9st-rKANoNvBlzlD_TigTNXI2XaVLC7xNtVuOnVBxUzlRopsb3wewk6KhDzI7OIuNBhOChBlit6v0rrgIWVeLjBANJvAB-vAZfyG24fZkM619PQc839-l07CXlPEnDS3-VBwIpRJLJBWEJIxvLgXn4SwllrwvpslM36W1Ii60y88VyrhhH0N4n2/s640/HP%20Replacement%20U212%20and%20R17%20short.jpg" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="480" data-original-width="640" height="240" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjru9st-rKANoNvBlzlD_TigTNXI2XaVLC7xNtVuOnVBxUzlRopsb3wewk6KhDzI7OIuNBhOChBlit6v0rrgIWVeLjBANJvAB-vAZfyG24fZkM619PQc839-l07CXlPEnDS3-VBwIpRJLJBWEJIxvLgXn4SwllrwvpslM36W1Ii60y88VyrhhH0N4n2/s320/HP%20Replacement%20U212%20and%20R17%20short.jpg" width="320" /></a></div><br />Powered the unit back ON, and the count-down was now correct!
Therefore, the count-down problem was caused by a bad U212 CMOS RAM IC.
</div>
<div><br /></div><div>Here's a video of the correct power-up countdown, with the MM2114N-L as the temporary replacement for U212:</div><div><br /></div><div class="separator" style="clear: both; text-align: center;"><iframe allowfullscreen='allowfullscreen' webkitallowfullscreen='webkitallowfullscreen' mozallowfullscreen='mozallowfullscreen' width='320' height='266' src='https://www.blogger.com/video.g?token=AD6v5dyyWjUpMj7vVaMd5LC0cKqU2YwNWgo1T8VMlZDNv893LVATMOSsgEow7TmuhHxBUVA_mZVAU7xHdlKwwOXXBw' class='b-hbp-video b-uploaded' frameborder='0'></iframe></div><br /><div>Even thought the count-down is now correct, the unit still fails at the end of the count-down, either by blanking
the seven-segment displays (as shown in the video, above) or showing (sometimes briefly) the two error
codes mentioned above.</div><div><br /></div><div><br /></div>
<div>
(An interesting question (unanswered) is why did HP add R17 in
series with CR2? My guess is that it has to do with current-limiting
CR2's current at power-up, but why? Note, its P/N is 1902-0535. With 50 ohms in series and assuming a direct short from CR2's cathode to ground at power-up, CR2's initial current surge would be around 100 mA. Is the Schottky diode that sensitive to a temporary current surge?).
</div>
<div><br /></div>
<div><br /></div><div>More Troubleshooting continues in <b><a href="https://k6jca.blogspot.com/2022/04/repair-log-hp-3314a-part-2.html">Part 2</a></b>!</div>
<p><br /></p>
<p>
<b><u>Other Manual Issues:</u></b>
</p>
<p>
Another Keyboard schematic error: There are no net names assigned to
the inputs of the four U5 drivers ('LS05 devices), so it is impossible to
tell which output they actually connect to.
</p>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgA_FOg6V4v5n0RLJVvIx_Dqs_azYJQAN-0Qo8Vlkh5jbfiwAV9By4m95vMghI7JZKCREIfzisHFPyIenXar_Mi5cf8pNM0nLvymn55vR6zeKPX1aOHYRlIKfbjlN_KNj-kqanXCM-3dTzNIGMfHgX_gsy0Zz2mNCatcM1dQVUWG9KNw7-eYPlK5Lcl/s1235/HP%20Schematic%20Display%20Keyboard%20Drive,%20Annotated.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="857" data-original-width="1235" height="222" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgA_FOg6V4v5n0RLJVvIx_Dqs_azYJQAN-0Qo8Vlkh5jbfiwAV9By4m95vMghI7JZKCREIfzisHFPyIenXar_Mi5cf8pNM0nLvymn55vR6zeKPX1aOHYRlIKfbjlN_KNj-kqanXCM-3dTzNIGMfHgX_gsy0Zz2mNCatcM1dQVUWG9KNw7-eYPlK5Lcl/s320/HP%20Schematic%20Display%20Keyboard%20Drive,%20Annotated.png" width="320" /></a>
</div>
<p><br /></p>
<p><b><u>Resources:</u></b></p><p>HP 3314A Manuals: <a href="https://www.keysight.com/us/en/support/3314A/programmable-function-generator.html">https://www.keysight.com/us/en/support/3314A/programmable-function-generator.html</a></p><p>HP 3314A Repair in 3 parts: <a href="https://diysquared.blogspot.com/2021/04/fixing-hp-3314a-function-generator-part.html">https://diysquared.blogspot.com/2021/04/fixing-hp-3314a-function-generator-part.html</a></p><p>HP 3314A Repair on YouTube: <a href="https://www.youtube.com/watch?v=wNxgBubSfH8">https://www.youtube.com/watch?v=wNxgBubSfH8</a></p><p>HP 3314A Repair (on Antique Radios forum): <a href="https://antiqueradios.com/forums/viewtopic.php?f=8&t=360491">https://antiqueradios.com/forums/viewtopic.php?f=8&t=360491</a></p><p>HP 3314A Teardown (EEVblog): <a href="https://www.eevblog.com/forum/testgear/hp-3314a-function-generator-teardown-explanation/">https://www.eevblog.com/forum/testgear/hp-3314a-function-generator-teardown-explanation/</a></p><p>HP 3314A Playing the Hallelujah Chorus: <a href="https://www.youtube.com/watch?v=H90d6sPUF6A">https://www.youtube.com/watch?v=H90d6sPUF6A</a><br />(Note: Hold down FUNCTION (the blue key), SW/TR INTVL, and START FREQ while powering up the unit).</p><p>HP 3314A ROM Images: <a href="http://www.ko4bb.com/getsimple/index.php?id=manuals&dir=01_ROM_Images_and_Drivers/HP_3314A_Eprom">http://www.ko4bb.com/getsimple/index.php?id=manuals&dir=01_ROM_Images_and_Drivers/HP_3314A_Eprom</a></p><p><br /></p>
<p>
<b><u>Standard Caveat:</u></b>
</p>
<p>
As always, I might have made a mistake in my equations, assumptions,
drawings, or interpretations. If you see anything you believe to be in
error or if anything is confusing, please feel free to contact me or comment
below.<br /><br />And so I should add -- this information is distributed in
the hope that it will be useful, but WITHOUT ANY WARRANTY; without even the
implied warranty of MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.
</p>
</div>
Jeffhttp://www.blogger.com/profile/04853314106806116765noreply@blogger.com0tag:blogger.com,1999:blog-2257458838727315711.post-21030700297288437812022-03-21T10:25:00.008-07:002022-03-23T06:53:04.003-07:00Representing a Transmission Line as a 2-Port Network<p>
In this blog post I will derive equations representing a transmission line of characteristic impedance Zo and of length '<i>l</i>' as
a two-port network. As a result, there will be a total of four different pairs of equations, each pair representing the two-port network in a different way.</p><p>These pairs of equation will be:</p><p>1. V and I at Port 2, in terms of V and I at Port 1, for a lossy transmission line.</p><p>2. V and I at Port 1, in terms of V and I at Port 2, for a lossy transmission line.</p><p>3. V and I at Port 2, in terms of V and I at Port 1, for a <i>lossless</i> transmission line.</p><p>4. V and I at Port 1, in terms of V and I at Port 2, for a <i>lossless</i> transmission line.</p><p>(These last two sets of equations are special cases of the first two sets).</p><p>The image below shows these voltages and currents for both the transmission line and its two-port model.</p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/a/AVvXsEgwsGtcNobyinSEm0TlnT1m1uvCiQSG_rSskHDU6gEaz5XUFa4U-kbnWTEh0IKIBTHWtv02GGTU9HtFpcID7JkA1lRkilBBA5fRRT6GPNCadhsjQHC56UepMtyxiExiL7BOB2fyPGlKnZKuLs18rLTos0aFLO7anIB-BW0NzuhuNAb_695ynM3780yF=s461" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="333" data-original-width="461" height="231" src="https://blogger.googleusercontent.com/img/a/AVvXsEgwsGtcNobyinSEm0TlnT1m1uvCiQSG_rSskHDU6gEaz5XUFa4U-kbnWTEh0IKIBTHWtv02GGTU9HtFpcID7JkA1lRkilBBA5fRRT6GPNCadhsjQHC56UepMtyxiExiL7BOB2fyPGlKnZKuLs18rLTos0aFLO7anIB-BW0NzuhuNAb_695ynM3780yF=s320" width="320" /></a>
</div>
<div><br /></div>
<div>
The first two sets of equations that will be derived are summarized by the two matrix equations, below:
</div>
<div><br /></div>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/a/AVvXsEjbkIQc3Hlx1tfYJMpWVLotBY8ix2noVvpAXegrX22y-cZwzxI8aCarXf96YxbOEP7qMIt7gjWyQXysxD8HdQEYKpaXz5NXr0XmHwlXw_eI43-JCQw7qhPSQwuqFNkbdZxdiQE77lOnluEIuSHeGtgcLlFIiFM7qfXgRZkHn4ZNeioz0Xb_Hi5eW1z8=s604" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="432" data-original-width="604" height="229" src="https://blogger.googleusercontent.com/img/a/AVvXsEjbkIQc3Hlx1tfYJMpWVLotBY8ix2noVvpAXegrX22y-cZwzxI8aCarXf96YxbOEP7qMIt7gjWyQXysxD8HdQEYKpaXz5NXr0XmHwlXw_eI43-JCQw7qhPSQwuqFNkbdZxdiQE77lOnluEIuSHeGtgcLlFIiFM7qfXgRZkHn4ZNeioz0Xb_Hi5eW1z8=s320" width="320" /></a>
</div>
<br />
<div>To derive the two equations in the first set of these equations, let's start with some fundamental definitions...</div><div><br /></div>
<div> </div>
<div>
<b><u>Some fundamentals:</u></b>
</div>
<div>
<b><u><br /></u></b>
</div>
<div>
Assume that the transmission line has on it both a forward voltage wave (traveling
from left to right) and an independent reverse voltage wave (traveling from
right to left). At any point on the transmission line these forward and reverse voltage waves create forward and reverse current waves:
</div>
<div><br /></div>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/a/AVvXsEgH6B_hf3HrQKTMjyDFkmiui_Ua6VX2mAioOUJYp56mh9Tk-kdYK3_JOHT7JUJf5G6vo1okq7km8FiOvnIZoJ99q1dvM14IAdVHKCnwQl-WbV8ejwg6_M8yq88pxb3G14aEeGcraor6nx-N0f24omb0B0M5jWkTxBtQ0if8VBbxPhdzjBWf3OH-DH0E=s402" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="355" data-original-width="402" height="283" src="https://blogger.googleusercontent.com/img/a/AVvXsEgH6B_hf3HrQKTMjyDFkmiui_Ua6VX2mAioOUJYp56mh9Tk-kdYK3_JOHT7JUJf5G6vo1okq7km8FiOvnIZoJ99q1dvM14IAdVHKCnwQl-WbV8ejwg6_M8yq88pxb3G14aEeGcraor6nx-N0f24omb0B0M5jWkTxBtQ0if8VBbxPhdzjBWf3OH-DH0E=s320" width="320" /></a>
</div>
<div><br /></div>
<div>
If we were to measure the voltage at any point on the line, the voltage we
would measure at that point will be the <i>sum</i> of the forward voltage and the
reverse voltage at that point. </div><div><br /></div><div>Note that the forward voltage and the reverse voltage each has its own amplitude and phase at that point.</div>
<div><br /></div>
<div>
Similarly, the current at that point will be the <i>difference</i> between the
forward current and the reverse current (because the currents are flowing in opposite directions). </div>
<div><br /></div>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/a/AVvXsEiGyZ2IZLD8m4Xe1fqxzNbOLhrtGQy45a2Y3vS-_KDSkWRmyi4p9huzx6o1oThdNz2zfi_BX4PdDlWd31yH6dgRHXMltT-Z7RD9IMkDtR-scfAISJVZezItttsTTI3i3uoV_6mpJ8Upd12wpRxkf5i0uk2T1P3veD7cD148WS3NI63PCtATFZ50XC3o=s429" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="313" data-original-width="429" height="233" src="https://blogger.googleusercontent.com/img/a/AVvXsEiGyZ2IZLD8m4Xe1fqxzNbOLhrtGQy45a2Y3vS-_KDSkWRmyi4p9huzx6o1oThdNz2zfi_BX4PdDlWd31yH6dgRHXMltT-Z7RD9IMkDtR-scfAISJVZezItttsTTI3i3uoV_6mpJ8Upd12wpRxkf5i0uk2T1P3veD7cD148WS3NI63PCtATFZ50XC3o=s320" width="320" /></a>
</div>
<br />
<div>
Recognize that for either the forward wave or the reflected wave, their
voltage and currents follow ohms law, independently of the other wave (because of the Principle of Superposition).
E.g. I<sup>+</sup> = V<sup>+</sup>/Zo and I<sup>-</sup>=V<sup>-</sup>/Zo, where Zo is the transmission line's
characteristic impedance.
</div>
<div><br /></div>
<div>
So we can write equations for the Total Voltage and Current at each end of the
transmission line in terms of the forward and reverse voltages at those points:</div>
<br />
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/a/AVvXsEihfmHDKUDWGiun5Ua5RLi3g40Wkuu8sgPGH8HtW3-EtXIomKglIEEhjUnGVq_RDtRIujLYowTyWf1DRNsXZ79-qX67T_fhmn4CL5RKSBYzVGR2rR0Ptys6AmJU0tSWToltsfcSeEFmxBYRtga7Im-0LD73z73LRcpG2U5xTTGVyMyXozStrvjp5qd_=s539" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="353" data-original-width="539" height="210" src="https://blogger.googleusercontent.com/img/a/AVvXsEihfmHDKUDWGiun5Ua5RLi3g40Wkuu8sgPGH8HtW3-EtXIomKglIEEhjUnGVq_RDtRIujLYowTyWf1DRNsXZ79-qX67T_fhmn4CL5RKSBYzVGR2rR0Ptys6AmJU0tSWToltsfcSeEFmxBYRtga7Im-0LD73z73LRcpG2U5xTTGVyMyXozStrvjp5qd_=s320" width="320" /></a>
</div>
<p>We can also express the forward and reverse voltages at either end of the
transmission line as functions of the Total Voltage and Current at either end:
</p>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/a/AVvXsEhhqLDQ_JNSTQMnBh64LKa-9jAzxryLaS41A5s66hn9z9EBx4dPm3GJ9_FuPGBKQKLc07WpDYbeRF7wOkRB5_t9xPaXu3V-zSOZcD_SdEDluO_droYvNYZhv3J0V-dVtK0nVPr55RmCcCRb4oI0MEgu8il03oWIgJX6UNErM0oJc-UtgjARpfqlhWaJ=s578" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="405" data-original-width="578" height="224" src="https://blogger.googleusercontent.com/img/a/AVvXsEhhqLDQ_JNSTQMnBh64LKa-9jAzxryLaS41A5s66hn9z9EBx4dPm3GJ9_FuPGBKQKLc07WpDYbeRF7wOkRB5_t9xPaXu3V-zSOZcD_SdEDluO_droYvNYZhv3J0V-dVtK0nVPr55RmCcCRb4oI0MEgu8il03oWIgJX6UNErM0oJc-UtgjARpfqlhWaJ=s320" width="320" /></a>
</div>
<p>I will use equations 1, 2, 3, and 4 (in the figure, above) to derive the two-port network equations, but before I go any further in this derivation, let me first define a transmission
line's "Propagation Constant."
</p>
<p><br /></p>
<p>
<b><u>Sidebar on the exponential terms
e<sup>-γl </sup>and e<sup>γl</sup> :</u></b>
</p>
<div>
<p>
Assuming a sinusoidal signal on the transmission line, the exponential term
e<sup>-γl </sup> (or e<sup>γl</sup>) represents the effect on a sine wave's amplitude and phase, due to the transmission line's <a href="https://en.wikipedia.org/wiki/Propagation_constant">Propagation Constant</a>, γ, as that wave travels along the transmission line.</p><p>The other variable in the exponent is 'l', which represents the length that the wave has traveled (or will travel) along the transmission line.</p><p>The Propagation Constant is a complex number: γ = α + jβ, where α represents a "real" change in signal amplitude (either positive or negative, i.e. attenuation or gain), and jβ represents a phase shift of the sinusoidal signal, either a positive phase shift or a negative phase shift. </p><p>Both of these values are in terms of unit-length, and so when substituted into the exponential, multiplying α by the length of the line gives overall attenuation (or gain, depending upon reference point), and multiplying jβ by length gives phase-shift for the wave's journey along a defined length of line.</p>
<p>
If we were to take a snapshot in time and simultaneously compare the Forward Voltage V<sup>+</sup> at point 1 with the Forward Voltage V<sup>+</sup> at point 2, we would see that the voltage at point 2 has
a negative phase shift compared to the phase of V<sup>+</sup> at point 1, because the
voltage at point 2 represents an earlier version of V<sup>+</sup> that had been at point 1 at an
earlier point in time, compared to the V<sup>+</sup> that is currently at Point 1. </p><p>We would also see that the amplitude of V<sup>+</sup> at point 2 has been attenuated relative to its original amplitude at point 1 due to transmission line
loss as it traveled from left to right along the line. Thus, γ's "α" term (in the Propagation Constant's equation γ = α + jβ) is negative to represent this loss at point 2 compared to the amplitude at point 1.</p><p>The figure below demonstrates both the attenuation and the phase shift at point 2, compared to point 1, for the Forward moving voltage wave:</p>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/a/AVvXsEja0mVv34wUs5TPu3IL-0nDe50nmqUvyuKEZ5AaY1jHEpP8x6Mi3WYgn99PL0p6LoFbraJkNVcLUXNQBk6AitvjDieTHgwivFuAZ_lwhuMXpQPIDu7_mBKSpp2ODar1gFKZt9FAOV3QcK4g4xPzfKSpca-m1Sak3ns4lLRTSLOtYmbWP3QY416Yfcrt=s824" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="647" data-original-width="824" height="251" src="https://blogger.googleusercontent.com/img/a/AVvXsEja0mVv34wUs5TPu3IL-0nDe50nmqUvyuKEZ5AaY1jHEpP8x6Mi3WYgn99PL0p6LoFbraJkNVcLUXNQBk6AitvjDieTHgwivFuAZ_lwhuMXpQPIDu7_mBKSpp2ODar1gFKZt9FAOV3QcK4g4xPzfKSpca-m1Sak3ns4lLRTSLOtYmbWP3QY416Yfcrt=s320" width="320" /></a>
</div>
<div><br /></div><div>With the same "snapshot in time," we can also simultaneously compare the Reverse Voltage V<sup>-</sup> at point 1 with the Reverse Voltage V<sup>-</sup> at point 2. </div><div><br /></div><div>We would see that the voltage at point 2 has a <i>positive</i> phase shift compared to the phase of V<sup>-</sup> at point 1, because the voltage at point 2 represents a <i>later</i> version of V<sup>-</sup> that has yet to arrive at point 1, compared to the phase of V<sup>-</sup> that is currently at Point 1. </div><div><br /></div><div>We would also see that the amplitude of V<sup>-</sup> at point 2 has not yet been attenuated compared to the amplitude of V<sup>-</sup> at point 1, because it has not yet traveled from right-to-left along the line, and thus it has not yet been attenuated by line loss. Therefore, γ's "α" term (in the Propagation Constant equation γ = α + jβ) is positive for V<sup>-</sup> at point 2 , not negative, because the amplitude at point 2 is larger compared to the amplitude at point 1.</div><div><p>The figure below demonstrates both the amplitude and phase at point 2, compared to point 1, for the Reverse moving voltage wave.:</p></div>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/a/AVvXsEiVesiqd4RkQWUrxvZcSGOojz6CT10hgqkPc4WUEnV5iWkMkKLiZwZEoLj1Vj2Nt_yQOkf1_C1FZgcEyZ6bjK05aYDi1p4yBhjV7WhJfaqR-nnt-v0zLYXhfYMhrW6j4shOPCT48KRUsLOcTlYNPet4AXZoRVK4QRpsGs5CHGeZCt4_chmaXLFxBrZ5=s824" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="647" data-original-width="824" height="251" src="https://blogger.googleusercontent.com/img/a/AVvXsEiVesiqd4RkQWUrxvZcSGOojz6CT10hgqkPc4WUEnV5iWkMkKLiZwZEoLj1Vj2Nt_yQOkf1_C1FZgcEyZ6bjK05aYDi1p4yBhjV7WhJfaqR-nnt-v0zLYXhfYMhrW6j4shOPCT48KRUsLOcTlYNPet4AXZoRVK4QRpsGs5CHGeZCt4_chmaXLFxBrZ5=s320" width="320" /></a>
</div>
<p>With this information and the equations, above, we can derive network equations that represent a lossy transmission line of length 'l'.</p><p><br /></p>
<p><b><u>Lossy Transmission Line Equations for V<sub>2</sub> and I<sub>2</sub> in terms of V<sub>1</sub> and
I<sub>1</sub>:</u></b></p>
<p>From the previous two figures that describe phase-shift and attenuation of V<sup>+</sup> and V<sup>-</sup> at Point 2 compared to our reference at Point 1, we can create the identities represented by equations 5 and 6 in the figure, below, and, using these, represent V<sub>2</sub> as a function of Point 1's V<sup>+</sup> and V<sup>-</sup> :</p>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/a/AVvXsEhQWZDL6L9KmUtUmWw3ATcqLQ0f5juKlJxef4PSOfxotDHsGUMdhECPVxrmTv9mft4aCmxrBfBOAQxADYn8rO4Wogk0PfekOpStyULMXrp-sxs-XK53JRtQNc3xOAMhMQd5XmsRCJ6YhTg-mTTgHM6oRJi284VPFsWOhVs0XcAXGcEplEGQzwfjtk2V=s1044" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="559" data-original-width="1044" height="171" src="https://blogger.googleusercontent.com/img/a/AVvXsEhQWZDL6L9KmUtUmWw3ATcqLQ0f5juKlJxef4PSOfxotDHsGUMdhECPVxrmTv9mft4aCmxrBfBOAQxADYn8rO4Wogk0PfekOpStyULMXrp-sxs-XK53JRtQNc3xOAMhMQd5XmsRCJ6YhTg-mTTgHM6oRJi284VPFsWOhVs0XcAXGcEplEGQzwfjtk2V=s320" width="320" /></a>
</div><div><br /></div>
Similarly, we can represent the current I<sub>2 </sub>at Point 2 as function of Point 1's V<sup>+</sup> and V<sup>-</sup> :</div><div><br /><div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/a/AVvXsEhhh7v1OXQiYb94dBDPJ_y8dYoZMJ6JXpt6pnzGV51zJNJVJzGYYAh4SSPjV3pdjFm2hPG16BIPAXm5Sg-qinU6rRTPbCLA9iB_MMR-4Lg10tzPZY7SX4-TKwcIXnpJc0I-ToBSWhaCUM4fVaYx58NwykKaCdPwXWNS3cOzn9r6G-_j1Np6IGz2dIs9=s779" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="390" data-original-width="779" height="160" src="https://blogger.googleusercontent.com/img/a/AVvXsEhhh7v1OXQiYb94dBDPJ_y8dYoZMJ6JXpt6pnzGV51zJNJVJzGYYAh4SSPjV3pdjFm2hPG16BIPAXm5Sg-qinU6rRTPbCLA9iB_MMR-4Lg10tzPZY7SX4-TKwcIXnpJc0I-ToBSWhaCUM4fVaYx58NwykKaCdPwXWNS3cOzn9r6G-_j1Np6IGz2dIs9=s320" width="320" /></a>
</div>
<div><br /></div>Next we substitute equations 1 and 2 (introduced earlier, above) into the V<sup>+</sup> and V<sup>-</sup> terms of these last two equations, rearrange, and get the following two equations:</div><div><br /><div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/a/AVvXsEh7yW3YpWTaIZpXSeBnlrpI2_BW7zRKcUJRpoj15cdVgULZasvG53kKbpeB8mTh17i3JnBXfnXUNLN7IUqgV_C-FeTkBXJgwUolnLgKoKcYuNB0haBz5LULMgDU8Ak38VSo31EBzaV4FAnUE4T_QS1N6pZvTZ24XJETi8jatMg_cyPdQjZyYjfbKzIJ=s884" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="401" data-original-width="884" height="145" src="https://blogger.googleusercontent.com/img/a/AVvXsEh7yW3YpWTaIZpXSeBnlrpI2_BW7zRKcUJRpoj15cdVgULZasvG53kKbpeB8mTh17i3JnBXfnXUNLN7IUqgV_C-FeTkBXJgwUolnLgKoKcYuNB0haBz5LULMgDU8Ak38VSo31EBzaV4FAnUE4T_QS1N6pZvTZ24XJETi8jatMg_cyPdQjZyYjfbKzIJ=s320" width="320" /></a>
</div>
<div><br /></div>Equations 9 and 10 represent V<sub>2</sub> and I<sub>2</sub> in terms of V<sub>1</sub> and I<sub>1</sub>, and in themselves they satisfactorily represent the lossy transmission line as a two-port network (from the perspective of calculating V<sub>2</sub> and I<sub>2</sub> from V<sub>1</sub> and I<sub>1</sub>).</div><div><br /></div><div>But we can also represent these equations using Hyperbolic functions:</div><div><br /><div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/a/AVvXsEjRRc2PLJYX8z_HSzgTjeW1XbHM6WWnD-ohWFWePEdt86ZaE7gJHCQvqRdsdR1U5WoYHLfCjKMDCO2ZvTfTQahDqIOjDQ2XG6RhA6se0fJ_vz3HyvS3am2ktQgbZj87Knf8uhyW-7-JhzSpZSmzEgwzCBF8PQ8FsTGXbOGVZaofxfAyDKbhOkdx8G7h=s687" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="423" data-original-width="687" height="197" src="https://blogger.googleusercontent.com/img/a/AVvXsEjRRc2PLJYX8z_HSzgTjeW1XbHM6WWnD-ohWFWePEdt86ZaE7gJHCQvqRdsdR1U5WoYHLfCjKMDCO2ZvTfTQahDqIOjDQ2XG6RhA6se0fJ_vz3HyvS3am2ktQgbZj87Knf8uhyW-7-JhzSpZSmzEgwzCBF8PQ8FsTGXbOGVZaofxfAyDKbhOkdx8G7h=s320" width="320" /></a>
</div>
<p>The matrix form of these last two equations is:</p>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/a/AVvXsEjRvvmr_Aw8LaT12gTdf7PxSyGXffyQFglbHXhJ9v5EEX544POtlV8C6beSY2sXIYgjF_yQpen4fnJ01CGjK8SwmjBK_DzEt1ANc6MPC2fkj0YFPaRQgoryYJYA_La2L5KhbMiDO8oB6durSuHmr2Crmh6SeqAZw72N5n7lLsrVKHdhqNv0GgQnFWYt=s668" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="338" data-original-width="668" height="162" src="https://blogger.googleusercontent.com/img/a/AVvXsEjRvvmr_Aw8LaT12gTdf7PxSyGXffyQFglbHXhJ9v5EEX544POtlV8C6beSY2sXIYgjF_yQpen4fnJ01CGjK8SwmjBK_DzEt1ANc6MPC2fkj0YFPaRQgoryYJYA_La2L5KhbMiDO8oB6durSuHmr2Crmh6SeqAZw72N5n7lLsrVKHdhqNv0GgQnFWYt=s320" width="320" /></a>
</div>
<br />
<p>
<b><u>Lossy Transmission Line Equations for V<sub>1</sub> and I<sub>1</sub> in terms of V<sub>2</sub> and
I<sub>2</sub>:</u></b>
</p>
<div>
In the same way that I defined V<sub>2</sub> and I<sub>2</sub> in terms of V<sub>1</sub> and I<sub>1</sub>, I can define V<sub>1</sub> and I<sub>1</sub> in terms of V<sub>2</sub> and I<sub>2</sub>. (In fact, this is the typical form in which equations representing the transmission line as a two-port network appear).</div>
<p>This time, we begin by considering the Forward Voltage V<sup>+</sup> and the Reverse Voltage V<sup>-</sup> from Point 2 (rather than from Point 1, as we did for the first derivations).</p><p>This will give us V<sup>+</sup> and V<sup>-</sup> at Point 1 in terms of V<sup>+</sup> and V<sup>-</sup> at Point 2, as illustrated, below.</p><p>First, V<sup>+</sup> at Point 1 in terms of V<sup>+</sup> at Point 2: </p>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/a/AVvXsEjib1n8E1VpxzcmX6z020TFWvfTfZ3KoPkBRbsd4oQpsae7ezxefpqBWBZoyAwcU-QrvhsOl3CpzC1OrJT47utxhl41u_tO0zKlasfFiLFrhNxUBL_YONMBNBXA9_GeOwCNr9p0kp-LntBYpRdrExQwcthJmnA2g_2Vx9_OKB0asOpBdkHAw5DLhzG4=s824" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="647" data-original-width="824" height="251" src="https://blogger.googleusercontent.com/img/a/AVvXsEjib1n8E1VpxzcmX6z020TFWvfTfZ3KoPkBRbsd4oQpsae7ezxefpqBWBZoyAwcU-QrvhsOl3CpzC1OrJT47utxhl41u_tO0zKlasfFiLFrhNxUBL_YONMBNBXA9_GeOwCNr9p0kp-LntBYpRdrExQwcthJmnA2g_2Vx9_OKB0asOpBdkHAw5DLhzG4=s320" width="320" /></a>
</div>
<p>Next, V<sup>-</sup> at Point 1 in terms of V<sup>-</sup> at Point 2:</p>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/a/AVvXsEjV4799QN5138stEHNbE4saSqvy6EgqyY8VsYdfgIHc3n5F5Tw_EbTcnnhP3wxaOp3Bpx86qKkWgXJvYCKASKyLva5zcSK9sAg-gKJsMXGDOiEVHYPvg2Bh7L62s9NkB6KKWwHQBjcHiKBkLRM-f5zakMs8It8cNvX_pCLa-5qDkc254EM2SK8cxrKs=s824" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="647" data-original-width="824" height="251" src="https://blogger.googleusercontent.com/img/a/AVvXsEjV4799QN5138stEHNbE4saSqvy6EgqyY8VsYdfgIHc3n5F5Tw_EbTcnnhP3wxaOp3Bpx86qKkWgXJvYCKASKyLva5zcSK9sAg-gKJsMXGDOiEVHYPvg2Bh7L62s9NkB6KKWwHQBjcHiKBkLRM-f5zakMs8It8cNvX_pCLa-5qDkc254EM2SK8cxrKs=s320" width="320" /></a></div>
<div><br /></div><div>We can use these equations to represent the total Voltage at Port 1 in terms of V<sup>+</sup> and V<sup>-</sup> at Point 2:</div><br />
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/a/AVvXsEiBvx2SaeTB7juHtzU1eH6ckPO1I54xz_ssKEr5twZp6LzDO-zopRpn7H5l9xNM2hpbWQYpisZhkubySRsE7xY-7yH6YtBezX7xTSkzJL0fiEyMJ_7BmvyFbn7JOACu86ikBe0yWdKTW6kuQ1G75xxoKgTXLcSpm7ptiPmru7gFFRxrVBKSEHsjjCVk=s1027" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="511" data-original-width="1027" height="159" src="https://blogger.googleusercontent.com/img/a/AVvXsEiBvx2SaeTB7juHtzU1eH6ckPO1I54xz_ssKEr5twZp6LzDO-zopRpn7H5l9xNM2hpbWQYpisZhkubySRsE7xY-7yH6YtBezX7xTSkzJL0fiEyMJ_7BmvyFbn7JOACu86ikBe0yWdKTW6kuQ1G75xxoKgTXLcSpm7ptiPmru7gFFRxrVBKSEHsjjCVk=s320" width="320" /></a>
</div>
<p>Repeat for the total Current at Port 1:</p>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/a/AVvXsEheks7Et2pIHZrG04pnqhG2pSO_5KBinI38WdQ7uE2XNpGCvkFFIo73e7IRnHQJH_maPAXTQuCELzm0X2QRRjyldGUZJTC5zoCceddjvx1dcYm-Ov0Z1Z5vWlTwzhbfW51VF3aN7kKKE3H280a2ZReOusfu2hhMZ_dNULnAtxayn26t6lv8lSvx1pK6=s814" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="404" data-original-width="814" height="159" src="https://blogger.googleusercontent.com/img/a/AVvXsEheks7Et2pIHZrG04pnqhG2pSO_5KBinI38WdQ7uE2XNpGCvkFFIo73e7IRnHQJH_maPAXTQuCELzm0X2QRRjyldGUZJTC5zoCceddjvx1dcYm-Ov0Z1Z5vWlTwzhbfW51VF3aN7kKKE3H280a2ZReOusfu2hhMZ_dNULnAtxayn26t6lv8lSvx1pK6=s320" width="320" /></a>
</div>
<div><br /></div>Then, substituting into these two equations equations 3 and 4 (presented earlier) and rearrange. We arrive at two equations expressing V<sub>2</sub> and I<sub>2</sub> in terms of V<sub>1</sub> and I<sub>1</sub>:</div><div><br /><div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/a/AVvXsEiKJoKVi4rQ_7UQw6Ku5mWMEfTfXqenSzQP0QKF9Bv1CcxUeEo-8AmLw2BZgWr18UV4LgzpJNqESNvu6kWXQCYCiM7qUU2_NE-0u7AgjAzt-pPRdjUeK2VTMu-7ech59MjnZkOZ7eFJ7LP0ktQ_QAmGMtqT3G_jqxoxBF6h-RIuwfJP8fy2GbGu83jJ=s920" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="423" data-original-width="920" height="147" src="https://blogger.googleusercontent.com/img/a/AVvXsEiKJoKVi4rQ_7UQw6Ku5mWMEfTfXqenSzQP0QKF9Bv1CcxUeEo-8AmLw2BZgWr18UV4LgzpJNqESNvu6kWXQCYCiM7qUU2_NE-0u7AgjAzt-pPRdjUeK2VTMu-7ech59MjnZkOZ7eFJ7LP0ktQ_QAmGMtqT3G_jqxoxBF6h-RIuwfJP8fy2GbGu83jJ=s320" width="320" /></a>
</div>
<div><br /></div>The above form is a sufficient representation of these two equations, but often they will be represented in terms of Hyperbolic functions:</div><div><p></p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/a/AVvXsEg24OjUhmoeqR8NtWdH__fCK1k3M54xdCVT6VlGxyVpWiFPFXXK-vlVTi7rWOtfaKasU6XYQLwIhhaRLbGa0uBCbjGI2kCpX6KR3nxXUEroaTKbzJgZq9PcXcjBH0b8zpfkEKQ1uSR5ufktWCJUvlavCQ1naP5iBTUkKH4ecMCRIOySDkDgemrrGZ3M=s690" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="446" data-original-width="690" height="207" src="https://blogger.googleusercontent.com/img/a/AVvXsEg24OjUhmoeqR8NtWdH__fCK1k3M54xdCVT6VlGxyVpWiFPFXXK-vlVTi7rWOtfaKasU6XYQLwIhhaRLbGa0uBCbjGI2kCpX6KR3nxXUEroaTKbzJgZq9PcXcjBH0b8zpfkEKQ1uSR5ufktWCJUvlavCQ1naP5iBTUkKH4ecMCRIOySDkDgemrrGZ3M=s320" width="320" /></a></div><br />The matrix form of these last two equations is:<p></p>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/a/AVvXsEiq9HWRb6pXpDhXIum_p6o1okcG4GR1tRCpXbhS63HJgxMvSRdPIoLjn6qGVPk0zZsQMqzYlGej4rq-4xn75p7szVKklgX5s1J462FeEiVW4exJYMI8-qfmtWU4j55m_pAgJOxYtsTvAGJH2VD4Vr5JX5bdmesY8WkxVO7SHBi9oQhAnBHeqZdkIEts=s633" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="321" data-original-width="633" height="162" src="https://blogger.googleusercontent.com/img/a/AVvXsEiq9HWRb6pXpDhXIum_p6o1okcG4GR1tRCpXbhS63HJgxMvSRdPIoLjn6qGVPk0zZsQMqzYlGej4rq-4xn75p7szVKklgX5s1J462FeEiVW4exJYMI8-qfmtWU4j55m_pAgJOxYtsTvAGJH2VD4Vr5JX5bdmesY8WkxVO7SHBi9oQhAnBHeqZdkIEts=s320" width="320" /></a>
</div>
<p>
The above 2x2 matrix is in the form of an ABCD matrix (note that I2's
direction is defined as exiting the network, not entering it).</p>
<p><br /></p>
<p>
<b><u>Two-Port Network Equations for Lossless Lines:</u></b>
</p>
<p>
Sometimes it is convenient to assume a transmission line is lossless.
What do I mean by lossless?
</p>
<p>
Recall the equation for the Propagation Constant: γ = α + jβ. If the line is lossless, α = 0 and thus
<i>e<sup>-γl</sup></i> becomes <i>e<sup>-jβl</sup></i>.
</p>
</div>
<div>
Similarly, <i>e<sup>γl</sup></i> becomes <i>e<sup>jβl</sup></i>.
</div>
<div><br /></div>
<div>
In either case, because α = 0, there is no attenuation as a wave travels along
the transmission line, only phase shift. With jβ and -jβ in the exponents of equations 9, 10, 17, and 18, we can use Euler's Formula and substitute cos and sin for the exponential terms in these four equations.</div><div><br /></div><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/a/AVvXsEiQ7h_KHAown88wR4tG0MVOMY-NLwGvgaUIvfpGhVWwrrbDJwjQ38D6ZfFaT0Kplx37bg89wtag3FTaMJwiCAiVG405jwZ0Ph9YiBmudHwETfh-k_VZdpzhBDsrxDbSzt1NWEwlAQwNyCLUxaGK2-PbzTx_R9rDJqbSBJMf5vDT1UlBnC-4TVM3LTpk=s974" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="766" data-original-width="974" height="252" src="https://blogger.googleusercontent.com/img/a/AVvXsEiQ7h_KHAown88wR4tG0MVOMY-NLwGvgaUIvfpGhVWwrrbDJwjQ38D6ZfFaT0Kplx37bg89wtag3FTaMJwiCAiVG405jwZ0Ph9YiBmudHwETfh-k_VZdpzhBDsrxDbSzt1NWEwlAQwNyCLUxaGK2-PbzTx_R9rDJqbSBJMf5vDT1UlBnC-4TVM3LTpk=s320" width="320" /></a></div><div><br /></div>
<div>
You can see equations 23 and 24 used in C. L. Ruthroff's article, "Some
Broad-Band Transformers," in the August, 1959 issue of the "Proceedings of the
I.R.E."
</div>
<div><br /></div>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/a/AVvXsEhQO4pntVQmDN2S4H9JnHW_-Evj6yrkO1SbCIecIogOOgFEtsdi0V6wOJ4cACPLC0AfemKoLp4BgcH4ooUrx6ltTAKcFzkYHo_ek25YiqwJrz_0VAECy5y3afZVZW_UTlvKuLUz6VwMzhue37icgd9NQmvm_FQ_2U9emP70t1pe07SNOnzeHeciv6W3=s537" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="530" data-original-width="537" height="316" src="https://blogger.googleusercontent.com/img/a/AVvXsEhQO4pntVQmDN2S4H9JnHW_-Evj6yrkO1SbCIecIogOOgFEtsdi0V6wOJ4cACPLC0AfemKoLp4BgcH4ooUrx6ltTAKcFzkYHo_ek25YiqwJrz_0VAECy5y3afZVZW_UTlvKuLUz6VwMzhue37icgd9NQmvm_FQ_2U9emP70t1pe07SNOnzeHeciv6W3=s320" width="320" /></a>
</div>
<div><br /></div><div>In matrix form, these equations are:</div><div><br /></div><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/a/AVvXsEgsMZ6vYOxLDRBiYdrf4T0HXkdaaV6gwXstVZXqYxbSnzgLbXLLJtvSOkGYA7E-ZbsLp0R4yofhX8oYbIIltJO2uKRJsORUcLf-shey5FdKt7NY5Fu3b0sP-bV14WYhYoU7WQGpzWtdMg0IQfHtEKMUAkwhp_IPN-GdgYYc408wvSgXYj_nuH_6OQ3t=s620" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="545" data-original-width="620" height="281" src="https://blogger.googleusercontent.com/img/a/AVvXsEgsMZ6vYOxLDRBiYdrf4T0HXkdaaV6gwXstVZXqYxbSnzgLbXLLJtvSOkGYA7E-ZbsLp0R4yofhX8oYbIIltJO2uKRJsORUcLf-shey5FdKt7NY5Fu3b0sP-bV14WYhYoU7WQGpzWtdMg0IQfHtEKMUAkwhp_IPN-GdgYYc408wvSgXYj_nuH_6OQ3t=s320" width="320" /></a></div><div><br /></div><div>Note that these equations can also be expressed in terms of the Hyperbolic
functions cosh and sinh in lieu of cos and sin.</div>
<div><br /></div>
Given <i>γl</i> =
j<i>βl,</i> and using cos-cosh identities: cosh(<i>γl</i>) = cosh(j<i>βl</i>) = cos(j*j<i>βl</i>) =
cos(-<i>βl</i>) = cos(<i>βl</i>).<br /><br />The sinh term is similar:
sinh(<i>γl</i>) = sinh(j<i>βl</i>) = -j*sin(j*j<i>βl</i>) = -j*sin(-<i>βl</i>) =
j*sin(<i>βl</i>).<br />
<div><br /></div><p><b><u>Resources:</u></b><br /><br />For further reading...<br /><br /><a href="https://www.qsl.net/kp4md/ruthroff.pdf">"Some Broad-band Transformers", C.L. Ruthroff, Proceedings of the IRE,
August, 1959</a><br /><br /><a href="https://hamwaves.com/chokes/doc/guanella.1944.pdf">"New Method of Impedance Matching in Radio-Frequency Circuits", G.
Guanella, The Brown Boveri Review, September, 1944</a><br /><br /><a href="http://read.pudn.com/downloads145/ebook/631954/RFCD_07.pdf">"Transmission Line Transformers," Chapter Six of Radio Frequency Circuit
Design, W. Alan Davis, Krishna Agarwal, John Wiley & Sonse, Inc. 2001</a><br /><br /><a href="https://www.researchgate.net/publication/251964449_A_novel_topology_of_broad-band_coaxial_impedance_transformer">"A novel topology of Broad-band Coaxial impedance transformer", Centurelli,
Piatella, Tommasino, Trifiletti, Proceedings of the 40th European Microwave
Conference"</a><br /><br /><a href="http://www.ee.ic.ac.uk/hp/staff/dmb/courses/ccts1/01800_LinesB.pdf">Phasors and Transmission Lines</a><br /><br /><br />And viewing...<br /><br />Transmission Line YouTube Series,
Professor Gregory D. Durgin, Georgia Tech. Starting with this
video: <a href="https://www.youtube.com/watch?v=7Oz1sazpekM">https://www.youtube.com/watch?v=7Oz1sazpekM</a><br /><br /><br /><b><u>Standard Caveat:</u></b><br /><br /><span style="background-color: white; color: #333333; font-family: "georgia" , serif; font-size: 13px;">As always, I might have made a mistake in my equations, assumptions,
drawings, or interpretations. If you see anything you believe to be in
error or if anything is confusing, please feel free to contact me or comment
below.</span><br /><br style="background-color: white; color: #333333; font-family: georgia, serif; font-size: 13px;" /><span style="background-color: white; color: #333333; font-family: "georgia" , serif; font-size: 13px;">And so I should add -- this information is distributed in the hope that it
will be useful, but WITHOUT ANY WARRANTY; without even the implied warranty
of MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.</span>
</p>
Jeffhttp://www.blogger.com/profile/04853314106806116765noreply@blogger.com0tag:blogger.com,1999:blog-2257458838727315711.post-14484349509237788122022-02-11T09:21:00.030-08:002022-02-19T12:53:10.107-08:00Common-mode Chokes: Calculating the Inductance of a Ferrite Toroidal Inductor<p>
I was recently comparing measurements of a common-mode choke's impedance
(measured several different ways using a Vector Network Analyzer -- see
<a href="https://k6jca.blogspot.com/2022/01/common-mode-chokes-removing-vna-port.html">here</a>). Unfortunately, each method I used gave different results, such as
the two measurements shown in the image, below (one impedance derived from an
S11 measurement, and the other derived from S21):
</p>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/a/AVvXsEhdVzcTiaR6w-tioKK1QUwhaeI81qeH00_v-2nwiyBOjAxET7nfFZbv3j15cR9QL6u57XF4_XIcA2noY4Zy0sR-df0ZGZQQcsaPkjTpZ8ks8rjOKvpDVQWrq1LN-diEja3AvpfOdPOuPXbzwEdKwTq78KnagTCVR2W3b50ezcTr82EWsxs2csDdEoDD=s1083" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="759" data-original-width="1083" height="224" src="https://blogger.googleusercontent.com/img/a/AVvXsEhdVzcTiaR6w-tioKK1QUwhaeI81qeH00_v-2nwiyBOjAxET7nfFZbv3j15cR9QL6u57XF4_XIcA2noY4Zy0sR-df0ZGZQQcsaPkjTpZ8ks8rjOKvpDVQWrq1LN-diEja3AvpfOdPOuPXbzwEdKwTq78KnagTCVR2W3b50ezcTr82EWsxs2csDdEoDD=s320" width="320" /></a>
</div>
<p>
For comparison purposes, I thought it would be useful to calculate what the
impedance ought to be, given the toroid core geometry and the ferrite
mix. (In this example, the core is a Mix 31 FT-240 core.)
</p>
<p>But how to do this?</p>
<p>
Fortunately, the
<a href="https://www.fair-rite.com/">Fair-Rite website</a> has a number
of resources, such as a paper titled "<a href="https://www.fair-rite.com/wpapers_anotes/specifying-a-ferrite-for-emi-suppression/">Specifying a Ferrite for EMI Suppression,</a>" by Carole U. Parker. This paper has all of the information required
to express the equation for inductance given a ferrite toroidal inductor (e.g. the paper's equation 9 and the definition of L<sub>0</sub> at the back of the paper).
</p>
<p>
The site also has downloadable .CSV files containing the "relative
permeability" data, versus frequency, for various ferrite mixes.
</p>
<p>In other words, everything I needed was there.</p>
<p>
To tie it all together I wrote a MATLAB script to read the
relative-permeability .csv file for a particular mix and, given the ferrite
core's dimensions, calculate and then plot impedance as magnitude and phase as
well as resistance and reactance versus frequency.
</p>
<p>
<b><u><br /></u></b>
</p>
<p>
<b><u>But first, some theoretical background:</u></b>
</p>
<p>The basic formula for a toroidal inductor is:</p>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/a/AVvXsEiya3qyh8x4mGGkSXJvK9JSNOJ3-22dBHhXSxC8DmVDPRd3mHyz9LaK8jT6Ox-MEkJu0sLZzHDODbbn4Pmj6ZoaEgdu8CtBZTweALmN-Psp35J8aefVSkiRaBkNCJz7qwQjGue7rOFxi2fntA3MfX0iSHGnwdHNwVV5TnDOQyH4kU6OmbWQ2H8-3fHH=s544" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="302" data-original-width="544" height="178" src="https://blogger.googleusercontent.com/img/a/AVvXsEiya3qyh8x4mGGkSXJvK9JSNOJ3-22dBHhXSxC8DmVDPRd3mHyz9LaK8jT6Ox-MEkJu0sLZzHDODbbn4Pmj6ZoaEgdu8CtBZTweALmN-Psp35J8aefVSkiRaBkNCJz7qwQjGue7rOFxi2fntA3MfX0iSHGnwdHNwVV5TnDOQyH4kU6OmbWQ2H8-3fHH=s320" width="320" /></a>
</div>
<p>
The Fair-Rite paper uses log<sub>10</sub> rather than natural log in its
equation for inductance. I would like to use this convention so that I
can check my final derivation against Fair-Rite's formulas. So,
including this conversion, plus converting permeability to H/mm (because
the core dimensions are given by Fair-Rite in mm), I can express my final
equation for inductance, as shown, below.
</p>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/a/AVvXsEj7BD7a_7Ck_rA-VaFno-9WxoCTrLBJ8mrfDFMx4VayKwobqA8arD_6-QakqnYruDccqfN9G8TAa-ip36vwbmL-k6cxfMpjFEeyNHx0yTY1Dcgl46-2OYtz6us0Fk7AmaIeToWJEZGF2-L4uCWVEpB9mmHwyN422m_z5n2egpH0ro4FWguy9Mng0XNL=s577" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="396" data-original-width="577" height="220" src="https://blogger.googleusercontent.com/img/a/AVvXsEj7BD7a_7Ck_rA-VaFno-9WxoCTrLBJ8mrfDFMx4VayKwobqA8arD_6-QakqnYruDccqfN9G8TAa-ip36vwbmL-k6cxfMpjFEeyNHx0yTY1Dcgl46-2OYtz6us0Fk7AmaIeToWJEZGF2-L4uCWVEpB9mmHwyN422m_z5n2egpH0ro4FWguy9Mng0XNL=s320" width="320" /></a>
</div>
<div><br /></div>
<div>
Note that ferrite's relative permeability is a complex value consisting of
both real (μ') and imaginary (μ'') terms. Here is an example of these
values, versus frequency, for Mix 31, from Fair-Rite's
"31-Material-Fair-Rite.csv" file (downloadable on
<a href="https://www.fair-rite.com/31-material-data-sheet/">this page</a>):
</div>
<div><br /></div>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/a/AVvXsEgB8SItmxetu1bk2c_R2dVA4WwDUvu8rK5Zjyr9cgPnaMLqUtxw2FZmOiUrlHBY4-8bC7yLaY74jum70rcMjwFy85F3eEryaIRIsYrxH7J3K8f-mt3UFHND53lPCX94TJcxXmXwAjvBhum3YV-BCQ8r13PGwb1EwVxtR6YEmzI6dfZxXDIo20KXHH0C=s794" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="485" data-original-width="794" height="195" src="https://blogger.googleusercontent.com/img/a/AVvXsEgB8SItmxetu1bk2c_R2dVA4WwDUvu8rK5Zjyr9cgPnaMLqUtxw2FZmOiUrlHBY4-8bC7yLaY74jum70rcMjwFy85F3eEryaIRIsYrxH7J3K8f-mt3UFHND53lPCX94TJcxXmXwAjvBhum3YV-BCQ8r13PGwb1EwVxtR6YEmzI6dfZxXDIo20KXHH0C=s320" width="320" /></a>
</div>
<div><br /></div>
<br />
<div>
<b><u>Calculating Impedance:</u></b><br />
<p>
The inductance equation, above, represents a complex inductance consisting
of both real and imaginary parts.
</p>
<p>
The impedance of this inductor can be calculated by multiplying inductance
by jω. Note that ω (in radians/sec) can be replaced by 2π*frequency
(where frequency is in Hz).
</p>
<p>
The resulting impedance equation is shown, below. I've also expanded
the equation to show the series resistance and reactance components of this
impedance. </p><p></p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/a/AVvXsEjbuLm84yI_qai4YnET0dszO8PcGo-YJG2ex0e_H0lbLZcQbozQDwhTAlat0tTBy764uHxqpcwqmUzKdNt-VascvhjeN8-b9IeQR8PS1xuWrF8X3a_3qVsZuA1WDyf0YC2aCQZ5j1qBfcBMHoI1LdGyt-5PpbRAwNywSetY2Dc51_J88mGm3URhfh1V=s744" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="744" data-original-width="702" height="320" src="https://blogger.googleusercontent.com/img/a/AVvXsEjbuLm84yI_qai4YnET0dszO8PcGo-YJG2ex0e_H0lbLZcQbozQDwhTAlat0tTBy764uHxqpcwqmUzKdNt-VascvhjeN8-b9IeQR8PS1xuWrF8X3a_3qVsZuA1WDyf0YC2aCQZ5j1qBfcBMHoI1LdGyt-5PpbRAwNywSetY2Dc51_J88mGm3URhfh1V=s320" width="302" /></a></div><br />(Note that u'', the relative-permeability's <i>imaginary</i> component, determines the inductor's <i>resistive</i> losses, while u', the relative-permeability's <i>real</i> component, determines the inductor's <i>reactance.</i>)<p></p></div><div><br /></div><div><br /></div><div><b><u>Including Parasitic Capacitance:</u></b></div><div><br /></div><div>But we are not finished. I also need to include the effect of
parasitic capacitance (e.g. inter-winding capacitance) on the overall
impedance. This capacitance can be modeled as a shunt capacitor in
parallel with the inductor whose equation was defined, above, and its presence
affects the inductor's self-resonant frequency (SRF):</div><div><br /></div><div><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/a/AVvXsEhsnc2CrsMBuVxX8_ZUQK8gbmpqotlkEdL12sO0rMJu2-nUz-FYtBbBkDk4ONVEpID7yeIk5RK3sbzhHgTmDXKFccfHxy9NtaCF1DWcpMPFMO8gTbf2_jPbl0-lCTbiWPEIqds5ZnquisFiSquHRV7G25UGJntpxj6IOAc3VA8b4iP15W5NjYoIOM5-=s635" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="635" data-original-width="598" height="320" src="https://blogger.googleusercontent.com/img/a/AVvXsEhsnc2CrsMBuVxX8_ZUQK8gbmpqotlkEdL12sO0rMJu2-nUz-FYtBbBkDk4ONVEpID7yeIk5RK3sbzhHgTmDXKFccfHxy9NtaCF1DWcpMPFMO8gTbf2_jPbl0-lCTbiWPEIqds5ZnquisFiSquHRV7G25UGJntpxj6IOAc3VA8b4iP15W5NjYoIOM5-=s320" width="301" /></a></div><br />Taking into account this parasitic capacitance, the <i>actual</i> impedance of
the ferrite toroidal inductor is the calculated complex impedance of the
inductor in parallel with the shunt capacitor's impedance:</div><div><br /></div><div>
<div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/a/AVvXsEgX012OzITNYVzKPuJnMPwrCe0x4jYwAOs8NzcA8pkv2ii2mjevF5Q0nikZTOIvkHv291qP0HB4lCEPYi09DIPiLc3olnZz_0RfHPVpVPPH-dRqUWhPgyjWOQ3gT88ZWeNv89ZO3lA_UgrtSFXckHT5ig-SK3SmkqTxuCpjNJptG4JIMn66PYNYm0G0=s724" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="584" data-original-width="724" height="258" src="https://blogger.googleusercontent.com/img/a/AVvXsEgX012OzITNYVzKPuJnMPwrCe0x4jYwAOs8NzcA8pkv2ii2mjevF5Q0nikZTOIvkHv291qP0HB4lCEPYi09DIPiLc3olnZz_0RfHPVpVPPH-dRqUWhPgyjWOQ3gT88ZWeNv89ZO3lA_UgrtSFXckHT5ig-SK3SmkqTxuCpjNJptG4JIMn66PYNYm0G0=s320" width="320" /></a></div><br />If using MATLAB and its matrix-based math to calculate the actual
impedance versus frequency, it is convenient to first convert impedance to
admittance, sum these, and then convert the sum back to the actual impedance
(matrix) of the device:
</div>
<div>
<div><br /></div>
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<a href="https://blogger.googleusercontent.com/img/a/AVvXsEjDTgh7I6eVCMy6COp_ZEODjplezCl9HDcAsvil0aAd2pxM9N0bIhYz5Di_1I8Uk9SsDzy_wNk1tWJ0QbJDe8x-o8SmPPyKtN4OMaxN_yVTzouPDKttxDzLW65WLU86sN1HSVcPVj_sgyY2KZ_bxVxhmtKIn2ZYUazUS4SJ1KZ7EwxHmUuwu-dZUV3-=s507" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="497" data-original-width="507" height="314" src="https://blogger.googleusercontent.com/img/a/AVvXsEjDTgh7I6eVCMy6COp_ZEODjplezCl9HDcAsvil0aAd2pxM9N0bIhYz5Di_1I8Uk9SsDzy_wNk1tWJ0QbJDe8x-o8SmPPyKtN4OMaxN_yVTzouPDKttxDzLW65WLU86sN1HSVcPVj_sgyY2KZ_bxVxhmtKIn2ZYUazUS4SJ1KZ7EwxHmUuwu-dZUV3-=s320" width="320" /></a>
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<div>
The section below is my MATLAB code for calculating the combined impedance
of a ferrite toroidal inductor shunted with parasitic capacitance.
</div>
<div><br /></div>
<div>
The code first reads the Fair-rite .csv file containing the ferrite
relative-permeability data (versus frequency -- note that I modified the
original file to shorten the frequency range to just those frequencies I am
interested in).
</div>
</div>
<div><br /></div>
<div>
Inductance is then calculated for both the Ferrite inductor and for the shunt
parasitic capacitance. These are then converted to admittances, summed,
and then the sum inverted to give the actual impedance.
</div>
<div>
<br />
<pre class="codeinput" style="background: rgb(247, 247, 247); border: 1px solid rgb(211, 211, 211); font-size: 12px; margin-bottom: 20px; margin-top: 0px; outline: 0px; padding: 10px; vertical-align: baseline;"><span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">% Calculating Impedance of an Inductor wound on a Ferrite Toroid core.</span>
<span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">% Date: 220209</span>
<span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">% k6jca</span>
<span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">%</span>
<span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">% Inductance is calculated using a formula derived from equations in the</span>
<span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">% paper: "Specifying a Ferrite for EMI Suppression," by Carole U. Parker</span>
<span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">% of Fair-Rite Products. This paper appeared in the June, 2008 issue of</span>
<span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">% "Conformity", but should be found on the Fair-Rite site (Google title).</span>
<span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">%</span>
<span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">% Self-Resonant-Frequency (SRF) is simulated by specifying a shunt</span>
<span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">% capacitance value paralleled with the inductor.</span>
<span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">%</span>
<span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">% The ferrite mix's u' and u'' values (vs frequency) come from a .CSV file</span>
<span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">% downloaded from the Fair-Rite website.</span>
<span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">%</span>
<span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">% The inductor analyzed in this example is 12 turns wound on a Mix 31</span>
<span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">% FT-240 core.</span>
<span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">%</span>
<span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">% Run on MATLAB Version R2020a</span>
clear;
clc;
close <span class="string" style="background: transparent; border: 0px; color: #a020f0; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">all</span>;
comment1=<span class="string" style="background: transparent; border: 0px; color: #a020f0; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">'12 tight turns on FT-240 Mix 31 Core'</span>; <span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">% for Plot annotation</span>
<span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">% The ferrite mix u' and u'' data is in the following .CSV file. Note that</span>
<span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">% the data in Fair-rite's downloadable .CSV file spans the frequency range</span>
<span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">% of 10 KHz to 1 GHz (much more than I need), and so I trimmed it down</span>
<span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">% to cover only 1-60 MHz and renamed the file:</span>
ftoread = <span class="string" style="background: transparent; border: 0px; color: #a020f0; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">'31-Material-Fair-Rite_1MHz-60MHz.csv'</span>; <span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">% File with Mix data</span>
N = 12; <span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">% inductor's number of turns</span>
<span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">% FT240 dimensions</span>
OD = 61; <span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">% outer-diameter, in mm</span>
ID = 35.55; <span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">% inner-diameter, in mm</span>
HT = 12.7; <span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">% height, in mm</span>
<span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">% Define the inductor's shunt capacitance (which affect the inductor's</span>
<span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">% self-resonant-frequency (SRF).</span>
<span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">% (One can manually adjust so that calculated SRF is similar to</span>
<span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">% measured SRF).</span>
Cp = 0.65e-12; <span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">% in Farads</span>
Cp_text = num2str(Cp*1e12); <span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">% For plot annotation</span>
<span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">% Read Fair-Rite's "mix" data from the CSV file and store in matrices.</span>
<span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">% Note that the Excel file is in a directory parallel with the directory</span>
<span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">% holding this matlab script.</span>
A = readmatrix([<span class="string" style="background: transparent; border: 0px; color: #a020f0; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">'..\Excel\'</span>,ftoread]); <span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">% CSV file is in EXCEL directory</span>
f = A(:,1); <span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">% frequency</span>
u1 = A(:,2); <span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">% Ferrite's u' value</span>
u2 = A(:,3); <span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">% Ferrite's u'' value</span>
jw = 1i*2*pi*f; <span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">% convert frequency to radians</span>
<span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">% Calculate the inductor's impedance.</span>
<span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">% (Formula derived from equations in the paper: "Specifying a Ferrite</span>
<span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">% for EMI Suppression," by Carole U. Parker of Fair-Rite Products.)</span>
Zl = jw*4.6052e-10*(N^2).*(u1-1i*u2)*HT*log10(OD/ID);
<span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">% The shunt capacitance is in parallel with the inductor, and is</span>
<span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">% the source of the inductor's Self-Resonant-Frequency.</span>
<span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">%</span>
<span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">% Because it is in parallel, a simple way to calculate its effect</span>
<span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">% on impedance is to convert the inductor's and capacitor's impedances</span>
<span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">% to admittances (admittance is just the inverse of impedance), add them,</span>
<span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">% and then convert the sum back to impedance.</span>
<span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">% MATLAB's 'RF Toolbox' has two nice routines for doing the matrix</span>
<span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">% inversions: z2y() and y2z().</span>
<span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">%</span>
YZl = squeeze(z2y(Zl)); <span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">% inductor's admittance</span>
YCp = (jw*Cp); <span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">% capacitor's admittance</span>
Yactual = YZl + YCp; <span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">% sum admittances</span>
Zactual = squeeze(y2z(Yactual));<span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">% Final impedance (Z(inductor) paralleled</span>
<span class="comment" style="background: transparent; border: 0px; color: forestgreen; margin: 0px; outline: 0px; padding: 0px; vertical-align: baseline;">% with Z(cap)) is the inversion of Y</span></pre>
<p>
The MATLAB script can be downloaded from the following github directory:
</p>
<p>
<a href="https://github.com/k6jca/Calculating_Ferrite_Toroid_Inductor_Impedance">https://github.com/k6jca/Calculating_Ferrite_Toroid_Inductor_Impedance</a><br />
</p>
</div>
<div>
<p>
Note that the script in the directory contains the MATLAB code, above,
<i>plus</i> code (<i>not</i> shown above) to generate the plot, below.
</p><p>I ran my code using MATLAB revision R2020a, but you will need at least MATLAB revision R2019a (it's required for the readmatrix() function). Of course, you can always use an earlier version, but you'll need to replace readmatrix() with something else.</p><p>The same caution applies to the sgtitle() function. This function first appears in MATLAB revision R2018b.</p>
<p>
<b><u><br /></u></b>
</p>
<p>
<b><u>Impedance of 12 turns on a Mix 31 FT-240 Ferrite Core:</u></b>
</p>
<p>
Below are curves shown the calculated impedance of an inductor created by
winding 12 turns around a Mix 31 FT-240 core. Impedance is shown in
terms of Magnitude and Phase, and also in terms of series Resistance and
Reactance.
</p>
<p>
The shunt capacitance was set to a value of 0.65 pF. (The next section
explains why 0.65 pF was chosen.)
</p>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/a/AVvXsEgF8vX7FxBKx60Bxxyso5pwr4JHrAhXfTuPUS8DqdkN4sQqMTD6OQ44oQZXtEtKVbAbPw1ZSG9dglT4pb9dtqNQQMXg6w4tTti8Xw0DOUKLQaK5cJvoCWBWtRtsC6ufevZNZmzpUaKsvPbSXnxVB0FDlPPi2mFhNyS1WTdIViPHqXpM74SzX__lUox3=s1016" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="759" data-original-width="1016" height="239" src="https://blogger.googleusercontent.com/img/a/AVvXsEgF8vX7FxBKx60Bxxyso5pwr4JHrAhXfTuPUS8DqdkN4sQqMTD6OQ44oQZXtEtKVbAbPw1ZSG9dglT4pb9dtqNQQMXg6w4tTti8Xw0DOUKLQaK5cJvoCWBWtRtsC6ufevZNZmzpUaKsvPbSXnxVB0FDlPPi2mFhNyS1WTdIViPHqXpM74SzX__lUox3=s320" width="320" /></a>
</div>
<br />
<p>
<b><u>Selecting a Shunt Capacitance value:</u></b>
</p>
<p>Why choose 0.65 pF? Why use any capacitance?</p>
<p>If there were no shunt capacitor (i.e. its capacitance = 0 pF), then there is no self-resonant point for this inductor (at least out to 60 MHz) -- the inductor remains inductive over the frequency range of 1 to 60 MHz:</p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/a/AVvXsEj40t9EJ0c_Hcjje32U1BlTAxCZssItryX1hRWLSC8WgJW_J09FMhHVcHW59xL5exEQeA7-tuloHMOV1EOCOOyo7mkT_w_z9EcGUSfnKROfQKPTfj0AjC35EuCqnq7oyU45J8oYMEu8BPt5OTQlElHvI4V5XHGcLSKtAuZki8mRuDOBQgmKDEEEEzWL=s1016" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="759" data-original-width="1016" height="239" src="https://blogger.googleusercontent.com/img/a/AVvXsEj40t9EJ0c_Hcjje32U1BlTAxCZssItryX1hRWLSC8WgJW_J09FMhHVcHW59xL5exEQeA7-tuloHMOV1EOCOOyo7mkT_w_z9EcGUSfnKROfQKPTfj0AjC35EuCqnq7oyU45J8oYMEu8BPt5OTQlElHvI4V5XHGcLSKtAuZki8mRuDOBQgmKDEEEEzWL=s320" width="320" /></a></div><p>But measurements made with a VNA <i>exhibit a resonance</i>. </p><p>And so I chose a shunt capacitor value (in this case 0.65 pF) that makes the <u>resonant frequency</u> of the ferrite inductor's calculated impedance the same as the resonant frequency of the Y21-derived impedance (derived from the VNA measurements), as shown below (i.e. both the calculated phase and the impedance-from-Y21 phase cross 0 degrees at the same frequency).</p>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/a/AVvXsEjaJFt3HnEMn06RvRBw79Qa0XbJ8KPwocEXNNUEKsYrMwI7ns1kA5Wb9hDkw-gaw1Ibpo9tQ8exTSiWtY6L2QMbwyWU28QBYvxxuIfWYSDBULA__kvnQAbyid6B8GPMmLPOTiKZ4_eGSkUbTbsQ5wQaZtCLJBQ2V6BSF5nCgqfNqn5d2vyZtrwQMgK0=s1083" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="759" data-original-width="1083" height="224" src="https://blogger.googleusercontent.com/img/a/AVvXsEjaJFt3HnEMn06RvRBw79Qa0XbJ8KPwocEXNNUEKsYrMwI7ns1kA5Wb9hDkw-gaw1Ibpo9tQ8exTSiWtY6L2QMbwyWU28QBYvxxuIfWYSDBULA__kvnQAbyid6B8GPMmLPOTiKZ4_eGSkUbTbsQ5wQaZtCLJBQ2V6BSF5nCgqfNqn5d2vyZtrwQMgK0=s320" width="320" /></a>
</div>
<p>
As you can see, the calculated impedance, although not exactly the same as
the impedance found via the Y21 method, is in the ballpark. Differences
between calculated and measured impedances could be caused by any of a
number of factors -- dimensional variations in core size, mix variations,
actual winding style vs. ideal winding style, measurement error, etc.
</p>
<p>
(Note: also shown is the impedance measured using S11, which I consider to
be an inferior method of measuring common-mode choke impedance.)
</p>
<p><br /></p>
<p>
<b><u>Conclusions and Notes:</u></b>
</p>
<p>1. I consider the <a href="https://k6jca.blogspot.com/2020/07/the-y21-method-of-measuring-common-mode.html">Y21 method</a> of measuring a common-mode choke's impedance to be superior to the <a href="https://k6jca.blogspot.com/2018/06/transmit-common-mode-chokes-11-current.html">G3TXQ method</a> of using only a VNA's S21 measurement, but, in the range of 1 to 30 MHz, the difference is not that significant (in my experience).</p>
<p>2. The value of the parallel parasitic capacitance is a subjective choice. I chose a value to make the resonant frequency of the calculated impedance the same as the resonant frequency of the Y21-derived impedance. But is this the best choice? I do not know.</p><p>3. If a multi-pole RLC ladder-network model is synthesized from the measured impedance (i.e. the <a href="https://k6jca.blogspot.com/2020/07/the-y21-method-of-measuring-common-mode.html">Y21-derived</a> impedance), and the capacitance of this synthesized model compared to the shunt capacitance I added to the Fair-Rite calculated inductance, the values are quite close (i.e. 0.62 pF for the synthesized model versus 0.65 pF for the Fair-Rite calculated Z).</p><p>The synthesized RLC ladder-network, created by Dick Benson, W1QG, using a custom MATLAB tool that will synthesize a network from the S-parameter measurements of an inductor (e.g. my VNA measurements for the 12-turns on the Mix 31 FT-240 core), looks like this:</p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/a/AVvXsEiR16P67BVVPcyX7-xDiEdnypQxwPfUfh0PNumlprrpotELtnjEjITjmGlagZuuUD82_e2Y9fLGWmaOksKqWarmUN4N-cuslDlC3w511zMkEf4lif5EjlzKNKUXIYGxb-fGTamCLb7YJDdoPh3Z1T7IRxlN0qzi4RVtbLN6FMKSRoIHPp2uS-Hj-7rD=s566" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="353" data-original-width="566" height="200" src="https://blogger.googleusercontent.com/img/a/AVvXsEiR16P67BVVPcyX7-xDiEdnypQxwPfUfh0PNumlprrpotELtnjEjITjmGlagZuuUD82_e2Y9fLGWmaOksKqWarmUN4N-cuslDlC3w511zMkEf4lif5EjlzKNKUXIYGxb-fGTamCLb7YJDdoPh3Z1T7IRxlN0qzi4RVtbLN6FMKSRoIHPp2uS-Hj-7rD=s320" width="320" /></a></div><br />You can see that the derived capacitance is 0.62 pF. Quite close to the 0.65 pF I chose for the Fair-Rite derived model.</div><div><br /></div><div>How accurate is this multi-pole synthesized RLC model to the Y21-derived measured impedance? Comparing their plots below (yellow = Y21-derived, cyan = synthesized), you can see that they are quite close.</div><div><br /></div><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/a/AVvXsEhxpYrwcwroVByNfKp5al-7yJDX6_TlVw33ems1_cs7grlZwk7I0m5Ks9miRnGzbCUFQNo2vwdFOi36em-_QgMCGSFuBUSyRQ_M5CvYzjRjhtshf7jAFSGWbWbkXWeqhTTvgkmEj0bebNvvh1fWwak6LfQMA1NKaPq7Yy_UKu3XLxHpTu_37bVd-2_K=s1084" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="824" data-original-width="1084" height="243" src="https://blogger.googleusercontent.com/img/a/AVvXsEhxpYrwcwroVByNfKp5al-7yJDX6_TlVw33ems1_cs7grlZwk7I0m5Ks9miRnGzbCUFQNo2vwdFOi36em-_QgMCGSFuBUSyRQ_M5CvYzjRjhtshf7jAFSGWbWbkXWeqhTTvgkmEj0bebNvvh1fWwak6LfQMA1NKaPq7Yy_UKu3XLxHpTu_37bVd-2_K=s320" width="320" /></a></div><br /><div><br /><p><b style="font-family: new; font-size: 16px;"><u>My Balun (and 80-Meter Loop) posts:</u></b></p><div style="font-family: new; font-size: 16px;"><div style="font-family: "times new roman"; font-size: medium;"><div style="font-family: new; font-size: 16px;"><div style="font-family: "times new roman"; font-size: medium;"><div style="font-family: new; font-size: 16px;"><div style="font-family: "times new roman"; font-size: medium;"><div>
<div style="font-family: new; font-size: 16px;"><br /></div>
<div style="font-family: new; font-size: 16px;">
80 Meter Loop, Part 1: <a href="http://k6jca.blogspot.com/2018/05/adventures-with-80-meter-loop-antenna.html">http://k6jca.blogspot.com/2018/05/adventures-with-80-meter-loop-antenna.html</a>
</div>
<div style="font-family: new; font-size: 16px;"><br /></div>
<div style="font-family: new; font-size: 16px;">
80 Meter Loop, Part 2: <a href="http://k6jca.blogspot.com/2018/05/adventures-with-80-meter-loop-antenna_30.html">http://k6jca.blogspot.com/2018/05/adventures-with-80-meter-loop-antenna_30.html</a>
</div>
<div style="font-family: new; font-size: 16px;"><br /></div>
<div style="font-family: new; font-size: 16px;">
Balun Power Dissipation: <a href="http://k6jca.blogspot.com/2018/06/common-mode-chokes-baluns-power.html">http://k6jca.blogspot.com/2018/06/common-mode-chokes-baluns-power.html</a>
</div>
<div style="font-family: new; font-size: 16px;"><br /></div>
<div style="font-family: new; font-size: 16px;">
Notes on 1:1 Baluns: <a href="http://k6jca.blogspot.com/2018/06/transmit-common-mode-chokes-11-current.html">http://k6jca.blogspot.com/2018/06/transmit-common-mode-chokes-11-current.html</a>
</div>
<div style="font-family: new; font-size: 16px;"><br /></div>
<div style="font-family: new; font-size: 16px;">
Notes on Common-Mode Currents: <a href="http://k6jca.blogspot.com/2018/07/thoughts-and-notes-common-mode-current.html">http://k6jca.blogspot.com/2018/07/thoughts-and-notes-common-mode-current.html</a>
</div>
<div style="font-family: new; font-size: 16px;"><br /></div>
<div style="font-family: new; font-size: 16px;">
Y21 Method of Measuring Common-Mode Impedance: <a href="https://k6jca.blogspot.com/2020/07/the-y21-method-of-measuring-common-mode.html">https://k6jca.blogspot.com/2020/07/the-y21-method-of-measuring-common-mode.html</a><br /><br />A 3-Port Method for Characterizing Baluns:
<a href="http://k6jca.blogspot.com/2020/06/another-method-to-characterize-baluns.html">http://k6jca.blogspot.com/2020/06/another-method-to-characterize-baluns.html</a>
</div>
</div>
<div>
<span style="font-family: inherit;"><br /></span><span style="font-family: inherit;"><br /></span>
</div>
</div>
<div style="font-family: "times new roman"; font-size: medium;"></div>
<b><u>Standard Caveat:</u></b>
</div>
</div>
</div>
</div>
</div>
<p>
I might have made a mistake in my designs, equations, schematics, models,
etc. If anything looks confusing or wrong to you, please feel free to
comment below or send me an email.<br /><br />Also, I will note:<br /><br />This
design and any associated information is distributed in the hope that it
will be useful, but WITHOUT ANY WARRANTY; without even the implied warranty
of MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.
</p>
</div>
Jeffhttp://www.blogger.com/profile/04853314106806116765noreply@blogger.com0tag:blogger.com,1999:blog-2257458838727315711.post-91029048704258298222022-01-20T06:59:00.142-08:002022-02-14T05:33:29.773-08:00Common-Mode Chokes: Removing Capacitance Effects from S11 Impedance Measurements<p>
One way to verify the impedance of a common-mode choke (i.e. CM choke) is with
a Vector Network Analyzer (VNA), such as the Agilent 8753 series or the
popular nanoVNA.
</p>
<p>Using such a VNA, a common mistake is to try to determine a CM choke's impedance using an S11 measurement -- this method of determining CM choke impedance can lead to very inaccurate results unless the user is aware of the potential pitfalls and how to correct for them. (Determining CM choke impedance with an S21 measurement is usually a better method.)</p>
<p>
An example of an erroneous S11 CM choke measurement is shown in the figure, below.
The choke consists of 12 turns around a Mix 33 FT-240 core. (Note that
this isn't actually a common-mode choke -- it is simply a choke, but its
impedance would be the same as a common-mode choke wound the same way.)
</p>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/a/AVvXsEiNmT0vTkiQZov8gJkPtCUP5rCto5BVhrrnhswtm4Dxg8I61LP9KHPKQWTaZrHqhByQ_AY8HXexsnx2WNrFMyIg14QJDK9sdCOsUMCgrPWmkzMzHAC90tYkMHsFe_j4pkKcFBkL5wG0jlozEXarOrO3sHcNq2Mc3zPBw8xKSlDAYR7mcMULr8p3eSC-=s1083" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="758" data-original-width="1083" height="224" src="https://blogger.googleusercontent.com/img/a/AVvXsEiNmT0vTkiQZov8gJkPtCUP5rCto5BVhrrnhswtm4Dxg8I61LP9KHPKQWTaZrHqhByQ_AY8HXexsnx2WNrFMyIg14QJDK9sdCOsUMCgrPWmkzMzHAC90tYkMHsFe_j4pkKcFBkL5wG0jlozEXarOrO3sHcNq2Mc3zPBw8xKSlDAYR7mcMULr8p3eSC-=s320" width="320" /></a>
</div>
<p>
The plot shows a maximum impedance of about 5600 ohms at 7 MHz, after which impedance falls off with frequency. </p><p>But does
this plot really represent the impedance of the choke?</p>
<p>
If I use the G3TXQ S21 method of measuring a common-mode choke (this method is
described <a href="http://www.karinya.net/g3txq/chokes/s21.pdf">here</a>
and
<a href="http://k6jca.blogspot.com/2018/06/transmit-common-mode-chokes-11-current.html">here</a>), the |Z| plot looks completely different, with a peak of 7400 ohms at 25
MHz:
</p>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/a/AVvXsEji7t-CRxt7sj0VJNxyWr00CyfWr80dvN-4cBDBjyX298y60fEndU6Xudm9JGmaLN26nPYaJenKCn4GLM7HjeY_i6DAAkRuCnGzawXLRx2cpIByyfReEctSHo8okZvqtrhnoUMLegvoRSOuWfQg55mh-zPMtPa9B6UaWER0FMyl06clTvMRREjmAiBG=s1083" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="759" data-original-width="1083" height="224" src="https://blogger.googleusercontent.com/img/a/AVvXsEji7t-CRxt7sj0VJNxyWr00CyfWr80dvN-4cBDBjyX298y60fEndU6Xudm9JGmaLN26nPYaJenKCn4GLM7HjeY_i6DAAkRuCnGzawXLRx2cpIByyfReEctSHo8okZvqtrhnoUMLegvoRSOuWfQg55mh-zPMtPa9B6UaWER0FMyl06clTvMRREjmAiBG=s320" width="320" /></a>
</div>
<p>
Why the difference? And can I modify my S11 measurement to give the same (or close to the same) results as the S21 measurement?
</p>
<p><br /></p>
<p>
<b><u>The Effect of External Capacitance:</u></b>
</p>
<p>
A choke (common-mode or other type) is a two-terminal inductor that, when
including its own parasitic capacitance and series resistance, can be modeled as shown
in the figure, below:
</p>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/a/AVvXsEhcxpZnUOQdIXDax-FH2xoyo0MXwoV90ZA19-rSbIJiA4yixZeI-mcLy-iTS8gMhL9z4WDk9ozHRCgTpT5gb1Ebmt62CoSP0vwXnC0O66f8M-Rc2gUZTA7y_c9tQFz70ztDrBucUtIPqDIwWjLLFo6PSx0fpQZIU3YUJYhe885H9hS1OxAbhfbHDF3Y=s635" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="444" data-original-width="635" height="224" src="https://blogger.googleusercontent.com/img/a/AVvXsEhcxpZnUOQdIXDax-FH2xoyo0MXwoV90ZA19-rSbIJiA4yixZeI-mcLy-iTS8gMhL9z4WDk9ozHRCgTpT5gb1Ebmt62CoSP0vwXnC0O66f8M-Rc2gUZTA7y_c9tQFz70ztDrBucUtIPqDIwWjLLFo6PSx0fpQZIU3YUJYhe885H9hS1OxAbhfbHDF3Y=s320" width="320" /></a>
</div><div><br /></div>
To measure the device's impedance using a VNA S11 measurement, I connect it across Port 1 of the
VNA. But note that the test setup has different sources of parasitic capacitance. <div><br /></div><div>For example, there is the VNA port's intrinsic capacitance (i.e. internal port parasitic capacitance), as well as parasitic capacitance external to the VNA (caused by, for example, coupling between the DUT's leads, coupling of the DUT to the instrument (e.g. chassis ground), etc.).<div><div>
<br />
<div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/a/AVvXsEgN5SgBNTj1OgSGCl7XX-7uD6MzzKSEDD_s5S977QemXynqTZmmhA53i8nUXJCJ3u7my6UvOt5ZgF7CksH0w0mMYOY7Wc-0mVBy-x9GgxiMhP04WXfdxn2Y3HLu0kA0-NHJeXSBslR_v0y5Q86AUwmQgxGHLDGFk4LaHnFNjdFEHECo8ntnJUVIKKZ_=s781" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="469" data-original-width="781" height="192" src="https://blogger.googleusercontent.com/img/a/AVvXsEgN5SgBNTj1OgSGCl7XX-7uD6MzzKSEDD_s5S977QemXynqTZmmhA53i8nUXJCJ3u7my6UvOt5ZgF7CksH0w0mMYOY7Wc-0mVBy-x9GgxiMhP04WXfdxn2Y3HLu0kA0-NHJeXSBslR_v0y5Q86AUwmQgxGHLDGFk4LaHnFNjdFEHECo8ntnJUVIKKZ_=s320" width="320" /></a></div><div class="separator" style="clear: both; text-align: center;"><br /></div>
<div>The effect of the VNA's internal port capacitance upon the DUT measurement can be removed by performing the normal VNA Short-Open-Load calibration. Assuming that the Open calibration standard is close to a perfect Open (or that its imperfections have been programmed into the VNA, so that the VNA can compensate for them), then, following calibration, a 10 pF cap (for example) placed across the VNA's Port should measure to be 10 pF, despite the value of the VNA port's internal capacitances.</div><div><br /></div><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/a/AVvXsEiwApQb6UvKb8Re4UN1ykdADyFYObq3bIrmKTHLv-9hxh7LPtp4-Q7jkT_hQYIE2kNPxhP1EpO42EAfE17okfXKH9NkC2R4N8zgpBc4ux0s-c2eRx8MHRugQ6VXJSgaRaQx1Tke-zsWUbox-gOpY7guy0VI0HfvH4W0t5JnTIH9fVS8nlG8amufy7f2=s859" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="640" data-original-width="859" height="238" src="https://blogger.googleusercontent.com/img/a/AVvXsEiwApQb6UvKb8Re4UN1ykdADyFYObq3bIrmKTHLv-9hxh7LPtp4-Q7jkT_hQYIE2kNPxhP1EpO42EAfE17okfXKH9NkC2R4N8zgpBc4ux0s-c2eRx8MHRugQ6VXJSgaRaQx1Tke-zsWUbox-gOpY7guy0VI0HfvH4W0t5JnTIH9fVS8nlG8amufy7f2=s320" width="320" /></a></div><div class="separator" style="clear: both; text-align: center;"><br /></div><div>But the <i>external</i> parasitic capacitance is <i>not</i> removed by the VNA calibration process. And this capacitance will change the self-resonant frequency of the DUT. </div><div><br /></div><div>In the case of my 12-turn choke on the mix 31 core, if I
compare the S11 |Z| measurement to the S21 |Z| measurement, I see that the
external parasitic capacitance has lowered the self-resonant frequency of the
choke from 25 MHz to 7 MHz. A significant change! </div>
<div><br /></div>
<div><br /></div>
<b><u><div></div>
Removing the Effect of the External Parasitic Capacitance:</u></b><br />
<p>
We can remove the effect of the external parasitic capacitance from an S11 impedance
measurement <i>if we know (or can estimate) its capacitance value</i>.
</p>
<p>
Per the figure, above, the impedance that is actually measured via S11 is:
</p>
<p style="text-align: left;">
Z(S11) = Z(DUT) || Z(Cexternal) = 1/(1/Z(DUT) + 1/Z(Cexternal))
</p>
<p>We can express this measurement in terms of admittances:</p>
<p style="text-align: left;"> Y(S11) = Y(DUT) + Y(Cexternal),</p>
<p style="text-align: left;">where:</p>
<p style="text-align: left;">
Y(DUT) = 1/Z(DUT) and Y(Cexternal) = jωCexternal</p>
<p style="text-align: left;">Rearranging the admittance equations:</p>
<p style="text-align: left;"> Y(DUT) = Y(S11) - Y(Cexternal)</p>
<p style="text-align: left;">
and Z(DUT) is easily found by simply inverting Y(DUT): Z(DUT) =
1/Y(DUT)
</p>
<p>
If I make the assumption that the external capacitance is around
2 pF (I'm going to use 1.95 pF for the plot, below) and calculate Z(DUT) per
the following procedure, I get a "compensated" value of S11 impedance is now
much closer to the S21 impedance measurement, as shown by the third line in
the plot, below:
</p>
<div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/a/AVvXsEhx_zBmHbe2eQ2WnYQDO-G5gXHfwfg4X1bkaR-LP0aO5hZPz6sXSVagor5Rq7Sfbwsu6x2zQ_NBun5UvB5GK0gWhfecJjH0dKtNwfRKSkHILaOkL8GH0aBHb-BXOS33BQmeNxhBAiMKPNvP-vuBhIx0fky8DYYpb5XWHXy2yYk-KWAHJZs5w_eM3lp7=s1083" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="759" data-original-width="1083" height="224" src="https://blogger.googleusercontent.com/img/a/AVvXsEhx_zBmHbe2eQ2WnYQDO-G5gXHfwfg4X1bkaR-LP0aO5hZPz6sXSVagor5Rq7Sfbwsu6x2zQ_NBun5UvB5GK0gWhfecJjH0dKtNwfRKSkHILaOkL8GH0aBHb-BXOS33BQmeNxhBAiMKPNvP-vuBhIx0fky8DYYpb5XWHXy2yYk-KWAHJZs5w_eM3lp7=s320" width="320" /></a></div><div class="separator" style="clear: both; text-align: center;"><br /></div>
<p>Why 1.95 pf? I'll explain this in the next section, below.</p>
<p>
<b><u><br /></u></b>
</p>
<p>
<b><u>R and X versus |Z|:</u></b>
</p>
<p>
I'm really not interested in the <i>magnitude</i> of the impedance of a
common-mode choke. Although a high magnitude of impedance might seem
great, it really tells me nothing about how the choke will behave when
installed in an antenna system, because the choke's reactance (whose value
is not known if we only examine the <i>magnitude</i> of the impedance) might
actually <i>worsen</i> common-mode rejection. (More on a CM choke
worsening CM rejection <a href="http://www.karinya.net/g3txq/chokes/">here</a>
and
<a href="http://k6jca.blogspot.com/2018/06/transmit-common-mode-chokes-11-current.html">here</a>).
</p>
<p>
Unless you can characterize the impedance of common-mode paths of your
antenna system, then measurement of the magnitude of Z (i.e. |Z|) is not very useful.</p><p>Instead, calculate the common-mode choke's <i>resistance</i>, as it
is this component of the common-mode choke's impedance that is guaranteed to
reduce common-mode currents, even if the choke's reactance happens to
unfortunately series-resonate with, say, a coax-cable's common-mode path
impedance.</p>
<p>
For this reason, I always characterize a common-mode choke's impedance in
terms of R and X, and I ignore |Z|.<br />
</p>
<p>
The plot below shows the DUT's impedance converted into R and X for S11,
S21, and the compensated S11 measurements:
</p><p></p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/a/AVvXsEgPAJvGrPHPAvqYZ-mLZo2R1CRwsn0146DvaVS3Slvkscw00L-HpxMX_U6nH0sWrpJp5G1P4gbd0wMBr1vI_JGdqbUZ9zDRkEscthttuFLp7uoPZax37aqqrcn-ZmP_K6qtKQiwdBE5Mv1nG2vKvH5rMcAeaFHe9i7fE2iOinmMYCUSPTH8gIbGkFOw=s1083" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="759" data-original-width="1083" height="224" src="https://blogger.googleusercontent.com/img/a/AVvXsEgPAJvGrPHPAvqYZ-mLZo2R1CRwsn0146DvaVS3Slvkscw00L-HpxMX_U6nH0sWrpJp5G1P4gbd0wMBr1vI_JGdqbUZ9zDRkEscthttuFLp7uoPZax37aqqrcn-ZmP_K6qtKQiwdBE5Mv1nG2vKvH5rMcAeaFHe9i7fE2iOinmMYCUSPTH8gIbGkFOw=s320" width="320" /></a></div><br />Let's examine the plots, above, and get back to the question of why did I
chose 1.95 pF for my shunt capacitance value...<p></p></div>
<div>
Later in this post I will show that the shunt capacitance measures to be
around 2 pF. I don't know the exact value because the Y21 method I use
to calculate shunt capacitance is a lumped-element approximation of what is
actually a <i>distributed</i> circuit. And, because this model is a
circuit approximation, its values are also approximations.
</div>
<div><br /></div>
<div>
I chose 1.95 pF because it gives R and X values that are fairly close (to my
eye) to the R and X values found via the S21 method, as you can see in the
plot, above.
</div>
<div><br /></div>
<div>
If I make the capacitance value smaller, the peak of R will shift to the left,
aligning it better with the peak found via S21, but the higher frequency
values of R will fall further away from the values found via S21.
</div>
<div><br /></div>
<div>
And if I increase C, those higher frequency values will align better, but the
peaks move further apart.
</div>
<div><br /></div>
<div>
So I chose 1.95 pF as a reasonable compromise. The next section shows
better how R and X vary with choice of shunt capacitance value.
</div>
<div><br /></div>
<div>
<p>
<b><u>Sensitivity of Z, R, and X to External Parasitic Capacitance:</u></b>
</p>
<p>
How sensitive is the plot of the "compensated" S11 impedance measurement to
the value of the external parasitic capacitance?
</p>
<p>
In addition to the original S11 and S21 |Z| plots, the figure below has
three |Z| plots of the <i>compensated</i> S11 measurement using three
different values of capacitance. I've selected 2.0 pF as the "nominal"
capacitance, with the other two capacitances being the this nominal
capacitance varied by +/- 5% (i.e. 1.9 pF and 2.1 pF). You can see
that there is an appreciable change in |Z| at the higher frequencies. </p><p></p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/a/AVvXsEhMoYxfeuauvlGv618p3bXJqLE3ZS714NFEngEN7WSCAiUn9qAk98x8JqCPmNkERMccIlCtQ_dlTthJF5a-TzB17X0F5KYAqzocVQ_TKePNggQJ61mjkwSjjhIMLIyj6Sq8AIVB6u50mMH_0TNeP17DcaAuuScuS-mhV8iqLGqGOOC3WXQRgjp-ADIT=s1083" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="759" data-original-width="1083" height="224" src="https://blogger.googleusercontent.com/img/a/AVvXsEhMoYxfeuauvlGv618p3bXJqLE3ZS714NFEngEN7WSCAiUn9qAk98x8JqCPmNkERMccIlCtQ_dlTthJF5a-TzB17X0F5KYAqzocVQ_TKePNggQJ61mjkwSjjhIMLIyj6Sq8AIVB6u50mMH_0TNeP17DcaAuuScuS-mhV8iqLGqGOOC3WXQRgjp-ADIT=s320" width="320" /></a></div><br />The figure below shows the R and X components for the original S21 plot as
well as for the compensated S11 measurement using the two +/- 5% values of
external capacitance (i.e. 1.9 and 2.1 pf).<p></p><p></p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/a/AVvXsEixUyoJa6guXIZPKWTKk7nPxyqXpnMsmyJdshNSmJxmdSGhG4ev9l6e8pWng63gyh-bYRcFqo0gi2ZYnIxNdJKxhAj6lBcjX7TmW9yUkv0F56IwiNPJoc95BqUDjFHmlNUiCWoZsOAulaui29pjWn82Jrmn8oocCDL24WJSvJ8PfuDTSIuIyqOq-MZp=s1083" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="759" data-original-width="1083" height="224" src="https://blogger.googleusercontent.com/img/a/AVvXsEixUyoJa6guXIZPKWTKk7nPxyqXpnMsmyJdshNSmJxmdSGhG4ev9l6e8pWng63gyh-bYRcFqo0gi2ZYnIxNdJKxhAj6lBcjX7TmW9yUkv0F56IwiNPJoc95BqUDjFHmlNUiCWoZsOAulaui29pjWn82Jrmn8oocCDL24WJSvJ8PfuDTSIuIyqOq-MZp=s320" width="320" /></a></div><br />Note that in this example, a change in capacitance of 10 % results in
roughly a 20% change in the frequency of peak choke resistance!<p></p>
<p><br /></p>
<p>
<b><u>Measuring Shunt Capacitance:</u></b>
</p>
<p>
I can get an idea of the value of the parasitic shunt capacitance of my S21
Fixture by using the "<a href="http://k6jca.blogspot.com/2020/07/the-y21-method-of-measuring-common-mode.html">Y21 method</a>" to calculate three impedances of the fixture: the series impedance
between ports 1 and 2, and the shunt impedances (to ground) at either port.
</p>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/a/AVvXsEhr-EkxuAEhGybpbtME7rRb0SuadBvqUNIHx79yCmIss-1IWl_sZ7t5_DowLvEKpOBeQegtuLYG4svDtLajYDhdf2QKApCH1yLji7s-QI-ml4_MMPqX-db3tyoqdsCBg_mxw3z1ne5v9BJq0miIj53jTECXIarJRb07WbgF1ZeZ8zmY-BqipaQ2czkQ=s371" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="237" data-original-width="371" height="204" src="https://blogger.googleusercontent.com/img/a/AVvXsEhr-EkxuAEhGybpbtME7rRb0SuadBvqUNIHx79yCmIss-1IWl_sZ7t5_DowLvEKpOBeQegtuLYG4svDtLajYDhdf2QKApCH1yLji7s-QI-ml4_MMPqX-db3tyoqdsCBg_mxw3z1ne5v9BJq0miIj53jTECXIarJRb07WbgF1ZeZ8zmY-BqipaQ2czkQ=s320" width="320" /></a>
</div>
<p>
To perform the S-parameter measurements for the Y21 calculations, I first
perform a full 2-port VNA calibration in which the THRU standard is a short
BNC barrel:
</p>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/a/AVvXsEgyhSL_hfjWjtUIde2Oj_odJP2rCyd0cd-7Y6mKYnd0dCmJkB-VFpiPUrPPMRHd8xEwAQE7TDX7lR-9PVjrmGnoTKGOlZ0vvF5JBbWHogtWIcuEEmV7kGrCaQBpLcbA370eU0g3Xs0aCV3SHLiyueDWrfue_-qwSc2LPs6i4SyFBSGAZMw_bDIyEn9C=s1280" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="1280" data-original-width="960" height="320" src="https://blogger.googleusercontent.com/img/a/AVvXsEgyhSL_hfjWjtUIde2Oj_odJP2rCyd0cd-7Y6mKYnd0dCmJkB-VFpiPUrPPMRHd8xEwAQE7TDX7lR-9PVjrmGnoTKGOlZ0vvF5JBbWHogtWIcuEEmV7kGrCaQBpLcbA370eU0g3Xs0aCV3SHLiyueDWrfue_-qwSc2LPs6i4SyFBSGAZMw_bDIyEn9C=s320" width="240" /></a>
</div>
<br />
<p>
And then I replace this barrel with the "S21 Fixture" and measure the
fixture's S-parameters:
</p>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/a/AVvXsEjU2SffZnrcFjk-wCJbI240k1yjtxJswwi2aLXyAQJj29qsujNf9BX4NrnUfd3GYvXBqEdXAyqy0MIJcU4wH1-GBUcZshCJMPEqn5_5It3pRLScWXGFIcGjWD4bJofBV8dvW3IcS2yNPeLnshcJRKPDyk7kOuNxfCpqE5yrklEs4yV5w0OPMNMEvWn_=s1280" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="1280" data-original-width="960" height="320" src="https://blogger.googleusercontent.com/img/a/AVvXsEjU2SffZnrcFjk-wCJbI240k1yjtxJswwi2aLXyAQJj29qsujNf9BX4NrnUfd3GYvXBqEdXAyqy0MIJcU4wH1-GBUcZshCJMPEqn5_5It3pRLScWXGFIcGjWD4bJofBV8dvW3IcS2yNPeLnshcJRKPDyk7kOuNxfCpqE5yrklEs4yV5w0OPMNMEvWn_=s320" width="240" /></a>
</div>
<p>The fixture's S-parameters, with shorting wire attached, are:</p>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/a/AVvXsEiJUju6ghfRtHiFRPnBQwXBdjnkFT4D9Gil8zPGw2ONwM_BHMn9FTTPG_rRmyJ-83sVo-EI4bPWl5pa0FFf9qlXauKjLOOxckNV_DKZgWIhewmW92ltMfK7bqv7Sc6JWWP6GrKefcsS4cifm7jUM0lMz2NQ2ic12qNgY-3m4Ci5UDUJkC8uG3AX72Uo=s1075" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="813" data-original-width="1075" height="242" src="https://blogger.googleusercontent.com/img/a/AVvXsEiJUju6ghfRtHiFRPnBQwXBdjnkFT4D9Gil8zPGw2ONwM_BHMn9FTTPG_rRmyJ-83sVo-EI4bPWl5pa0FFf9qlXauKjLOOxckNV_DKZgWIhewmW92ltMfK7bqv7Sc6JWWP6GrKefcsS4cifm7jUM0lMz2NQ2ic12qNgY-3m4Ci5UDUJkC8uG3AX72Uo=s320" width="320" /></a>
</div>
<p>
The figure below shows the fixture's series-impedance (displayed as R and X)
as calculated using two methods: G3TXQ's
<a href="http://k6jca.blogspot.com/2018/06/transmit-common-mode-chokes-11-current.html">S21 method</a>
and the
<a href="http://k6jca.blogspot.com/2020/07/the-y21-method-of-measuring-common-mode.html">Y21 method</a>.
</p>
<p>
G3TXQ's method does not take into account the VNA port shunt
impedances, which can result in errors, such as the resistance component of the series-impedance going <i>negative</i>, as shown in the next plot. But if I use the Y21 method, I can model the measurement circuit as a three-element Pi network and calculate both the
series impedance and the two shunt impedances:</p><p></p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/a/AVvXsEhSMnxio8_e6mBBGnHUMEC4JjReQl7bID7wpiTBWyYZYUm4oXjl_ZVQXvhwgtuIRBjxF4HzqPGs50YgAWQuAUjaFVR_ozHm9O2kcDEUmu3j9u0SRFjqlSfManJb98nD6_sH81XrnkIV4-fW1EHc_suZdJv8g5rlRABbd6iL-vwKZmMF6r1SoV4dLcEH=s371" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="237" data-original-width="371" height="204" src="https://blogger.googleusercontent.com/img/a/AVvXsEhSMnxio8_e6mBBGnHUMEC4JjReQl7bID7wpiTBWyYZYUm4oXjl_ZVQXvhwgtuIRBjxF4HzqPGs50YgAWQuAUjaFVR_ozHm9O2kcDEUmu3j9u0SRFjqlSfManJb98nD6_sH81XrnkIV4-fW1EHc_suZdJv8g5rlRABbd6iL-vwKZmMF6r1SoV4dLcEH=s320" width="320" /></a></div><br />Using the Y21 method we see that the fixture, when shorted with a wire,
looks like a series impedance of 0.21 + jω200e-9 ohms (i.e. it is about 200
nH of inductance in series with about 0.2 ohms of resistance):<p></p>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/a/AVvXsEjE-TDSNCcUn_TuY-gycGhCkThAHSiquD208m1mvhvIl4YLqmnq5RhV-0e7rfvG8W6Kw67lBUEb7cfBHKRaorGSFZa-Uk5QvOJdhCXHNeXhK9AxVgugo-SJxr2viXvxTKbVMkI0EvFue9L_uHfMPe1QXVxsQENRnfKVtK_3-1DX8Zy1l9BezDc289I7=s1083" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="758" data-original-width="1083" height="224" src="https://blogger.googleusercontent.com/img/a/AVvXsEjE-TDSNCcUn_TuY-gycGhCkThAHSiquD208m1mvhvIl4YLqmnq5RhV-0e7rfvG8W6Kw67lBUEb7cfBHKRaorGSFZa-Uk5QvOJdhCXHNeXhK9AxVgugo-SJxr2viXvxTKbVMkI0EvFue9L_uHfMPe1QXVxsQENRnfKVtK_3-1DX8Zy1l9BezDc289I7=s320" width="320" /></a>
</div>
<p>And the shunt capacitances at either port are about 2.1 pF:</p>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/a/AVvXsEjB-sKf4SO9eajVlapFo-9NxsxwBQ70arw40ToaF7SS-QWh6OvP3VTRC7O0VFlLInPG0VtF70D1oRtA6W79Kd9Deb55m86SoOFN7eAsfQpirXugRVAzxxB6Prh2ksoBLdCsPUDSIirutDSB_mkk8sjmTUOx-SYyaPQVWSenuf4d4u8iddmpqEI5K-i3=s576" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="512" data-original-width="576" height="284" src="https://blogger.googleusercontent.com/img/a/AVvXsEjB-sKf4SO9eajVlapFo-9NxsxwBQ70arw40ToaF7SS-QWh6OvP3VTRC7O0VFlLInPG0VtF70D1oRtA6W79Kd9Deb55m86SoOFN7eAsfQpirXugRVAzxxB6Prh2ksoBLdCsPUDSIirutDSB_mkk8sjmTUOx-SYyaPQVWSenuf4d4u8iddmpqEI5K-i3=s320" width="320" /></a>
</div>
<p>
Note that these shunt capacitances are modeled as <i>lumped-elements,</i> approximating what is actually happening with the fixture's <i>distributed</i>
circuit.
</p>
<p><br /></p>
<p>
<b><u>Impedance Measurement using the Y21 Method:</u></b>
</p>
<p>
Let's use the Y21 method and calculate the series and shunt impedances when
my 12-turn choke is connected to the S21 fixture.
</p>
<p>
The figure below shows R and X for the choke as calculated from S11, S21,
S11 compensated for external port capacitance (1.95 pF), and the Y21 method. </p><p></p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/a/AVvXsEgAe1iveYEqQkbrehkn9RNQZ2S0-oDnTWQKznTO0ox0V7lyhYDPfWBnNKwdhd9FPxNRel6K5i7HBStT7DNfK0o6lz4oeGbDEld__WINV78HIrzWBYPZf5_pe79UzAQnCBlTGautHyolg9FJ3nXF0wQEcvEttiPLlym1ozpHTtYyp-WXqB-AYisUfjT6=s1083" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="759" data-original-width="1083" height="224" src="https://blogger.googleusercontent.com/img/a/AVvXsEgAe1iveYEqQkbrehkn9RNQZ2S0-oDnTWQKznTO0ox0V7lyhYDPfWBnNKwdhd9FPxNRel6K5i7HBStT7DNfK0o6lz4oeGbDEld__WINV78HIrzWBYPZf5_pe79UzAQnCBlTGautHyolg9FJ3nXF0wQEcvEttiPLlym1ozpHTtYyp-WXqB-AYisUfjT6=s320" width="320" /></a></div><br />The figure below shows the external shunt capacitance values calculated with the Y21
method. Note that these are now closer to 3 pF, rather than 2
pF. This is possibly due to distributed coupling from the physical
structure of the choke to ground, which then becomes part of the 3-component
lumped-element Y21 circuit approximation.
<p></p>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/a/AVvXsEgWXyg6Qg73ZeAQH8wHYdq9IgiQ4NP1YBAEe4TP_rryb8HDcaUpXB8VOr8G4jDZtDocdWRkrmK9JUgdmqhr0EqoP_l2I8j4QIys3O2Ymcd-BifEvpjX4B3Q9hH4jPOWHEsQTbNJ7aZokoqP6NFNEqlgCQhxHLmr4l4pnEGAaGW1YQ0RB-0oEnCNZmFK=s576" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="512" data-original-width="576" height="284" src="https://blogger.googleusercontent.com/img/a/AVvXsEgWXyg6Qg73ZeAQH8wHYdq9IgiQ4NP1YBAEe4TP_rryb8HDcaUpXB8VOr8G4jDZtDocdWRkrmK9JUgdmqhr0EqoP_l2I8j4QIys3O2Ymcd-BifEvpjX4B3Q9hH4jPOWHEsQTbNJ7aZokoqP6NFNEqlgCQhxHLmr4l4pnEGAaGW1YQ0RB-0oEnCNZmFK=s320" width="320" /></a>
</div>
<p><br /></p>
<p>
<b><u>Conclusions:</u></b>
</p>
<p>1. Stray shunt capacitance can greatly affect the measured impedance of a common-mode choke if measuring it using a VNA's S11 measurement. If you know the approximate value of your stray shunt capacitance, you can remove its effect from the measurement by converting impedances to admittances, and then subtracting from the measured admittance the admittance of the stray shunt capacitance.</p><p>But if you do not have a reasonable idea of what your stray capacitance value is, your results might not represent the choke's actual impedance.</p><p>And this is exactly the problem with the S11 method of measuring common-mode choke impedances -- what <i>is</i> the shunt capacitance?</p><p>2. G3TXQ's S21 measurement method provides a better measure of common-mode choke impedance than S11 measurements. But the Y21 method of measuring impedance (from an S21 measurement) is superior to G3TXQ's method.</p>
<p><br /></p>
<div style="font-family: new; font-size: 16px;">
<div style="font-family: "times new roman"; font-size: medium;">
<div>
<div>
<div>
<div style="font-family: new; font-size: 16px;">
<div style="font-family: "times new roman"; font-size: medium;">
<div>
<div>
<div>
<div style="font-family: new; font-size: 16px;">
<div style="font-family: "times new roman"; font-size: medium;">
<div>
<div>
<div>
<div style="font-family: new; font-size: 16px;">
<b><u>My Balun (and 80-Meter Loop) posts:</u></b>
</div>
<div style="font-family: new; font-size: 16px;">
<br />
</div>
<div style="font-family: new; font-size: 16px;">
80 Meter Loop, Part 1: <a href="http://k6jca.blogspot.com/2018/05/adventures-with-80-meter-loop-antenna.html">http://k6jca.blogspot.com/2018/05/adventures-with-80-meter-loop-antenna.html</a>
</div>
<div style="font-family: new; font-size: 16px;">
<br />
</div>
<div style="font-family: new; font-size: 16px;">
80 Meter Loop, Part 2: <a href="http://k6jca.blogspot.com/2018/05/adventures-with-80-meter-loop-antenna_30.html">http://k6jca.blogspot.com/2018/05/adventures-with-80-meter-loop-antenna_30.html</a>
</div>
<div style="font-family: new; font-size: 16px;">
<br />
</div>
<div style="font-family: new; font-size: 16px;">
Balun Power Dissipation: <a href="http://k6jca.blogspot.com/2018/06/common-mode-chokes-baluns-power.html">http://k6jca.blogspot.com/2018/06/common-mode-chokes-baluns-power.html</a>
</div>
<div style="font-family: new; font-size: 16px;">
<br />
</div>
<div style="font-family: new; font-size: 16px;">
Notes on 1:1 Baluns: <a href="http://k6jca.blogspot.com/2018/06/transmit-common-mode-chokes-11-current.html">http://k6jca.blogspot.com/2018/06/transmit-common-mode-chokes-11-current.html</a>
</div>
<div style="font-family: new; font-size: 16px;">
<br />
</div>
<div style="font-family: new; font-size: 16px;">
Notes on Common-Mode Currents: <a href="http://k6jca.blogspot.com/2018/07/thoughts-and-notes-common-mode-current.html">http://k6jca.blogspot.com/2018/07/thoughts-and-notes-common-mode-current.html</a>
</div>
<div style="font-family: new; font-size: 16px;">
<br />
</div>
<div style="font-family: new; font-size: 16px;">
Y21 Method of Measuring Common-Mode Impedance:
<a href="https://k6jca.blogspot.com/2020/07/the-y21-method-of-measuring-common-mode.html">https://k6jca.blogspot.com/2020/07/the-y21-method-of-measuring-common-mode.html</a><br /><br />A 3-Port Method for
Characterizing Baluns: <a href="http://k6jca.blogspot.com/2020/06/another-method-to-characterize-baluns.html">http://k6jca.blogspot.com/2020/06/another-method-to-characterize-baluns.html</a>
</div>
</div>
<div>
<span style="font-family: inherit;"><br /></span><span style="font-family: inherit;"><br /></span>
</div>
</div>
</div>
</div>
<div style="font-family: "times new roman"; font-size: medium;"></div>
<b><u>Standard Caveat:</u></b>
</div>
</div>
</div>
</div>
</div>
</div>
</div>
</div>
</div>
</div>
</div>
<p>I might have made a mistake in my designs, equations, schematics,
models, etc. If anything looks confusing or wrong to you, please feel
free to comment below or send me an email.<br /><br />Also, I will note:<br /><br />This
design and any associated information is distributed in the hope that it
will be useful, but WITHOUT ANY WARRANTY; without even the implied warranty
of MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.
</p>
</div>
</div></div>Jeffhttp://www.blogger.com/profile/04853314106806116765noreply@blogger.com0tag:blogger.com,1999:blog-2257458838727315711.post-702144693849102582021-10-03T20:33:01.615-07:002022-01-20T08:57:11.025-08:00Revisiting Maxwell's Tutorial on Wave Interference and Impedance Matching<div class="separator" style="clear: both; text-align: center;">
<div>(The Great Wave off Kanagawa,by Katsushika Hokusai)</div>
</div>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://1.bp.blogspot.com/-ygzddvjnPTQ/YVssSLuTk3I/AAAAAAAAMbk/LZGb5B1ywUQGTyJ9Nuhydzpg9ke8YkD6gCLcBGAsYHQ/s554/The%2BGreat%2BWave%2Boff%2BKanagawa%252C%2Bby%2BKatsushika%2BHokusai.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="429" data-original-width="554" height="248" src="https://1.bp.blogspot.com/-ygzddvjnPTQ/YVssSLuTk3I/AAAAAAAAMbk/LZGb5B1ywUQGTyJ9Nuhydzpg9ke8YkD6gCLcBGAsYHQ/w320-h248/The%2BGreat%2BWave%2Boff%2BKanagawa%252C%2Bby%2BKatsushika%2BHokusai.png" title="The Great Wave off Kanagawa, by Katsushika Hokusai" width="320" /></a>
</div>
<div style="text-align: left;"><br /></div>
<p>
After I had published my previous blog-post demonstrating that source
impedance does not affect SWR on a transmission line (<a href="http://k6jca.blogspot.com/2021/09/does-source-impedance-affect-swr.html">see here</a>), someone reminded me of the three-part series on Wave Mechanics, written by
Dr. Steven Best, VE9SRB, that appeared in <i>QEX</i> magazine in 2001 (the Jan/Feb,
July/Aug, and Nov/Dec 2001 issues), and Walter Maxwell's (W2DU) rebuttal to this
series that appeared in the July/August 2004 issue of <i>QEX</i>.
</p>
<p>
I recall reading Best's approach at the time and agreeing with it, but I did
not remember reading Maxwell's rebuttal. So why not reread
both?</p>
<p>
After finishing Best's three-part series I still agreed with his approach analyzing the wave mechanics as currently-generated waves interacting with waves consisting of an infinite series of past reflections --
it was an approach I've used in a number of my own blog posts (see the list at
the end of this post).
</p>
<p>
Unfortunately, by the third paragraph of Maxwell's rebuttal it was clear from
his statements that he had misunderstood Best's equations.
</p>
<div>
(Maxwell's article, "A Tutorial Dispelling Certain Misconceptions Concerning
Wave Interference in Impedance Matching," can be found
<a href="http://www.w1npp.org/ares/REFLEC~1/REFLEC~2/QEXJUL~1.PDF">here</a>).
</div>
<p>
Maxwell disputes a number of Best's statements, beginning with Best's
derivation of the equation for Vfwd (the total voltage traveling forward on a
transmission line, whose derivation is represented by equations 6, 7, and 8 in
Part 1 of Best's series), and ending with a refutation of Best's analysis of
the wave interaction that creates a steady-state effective Reflection
Coefficient of 0 (i.e. effective SWR = 1:1) at the input of a T-network tuner.
</p>
<p>Best's claim, which he backed up mathematically, is that it is a rearward travelling wave, passing through the tuner from output port to input port, that, when it reaches the tuner's input port, is exactly equal in amplitude but 180 degrees out of phase with the source-voltage's reflection at the T-network's input. These two rearward-traveling waves cancel, and it is this cancellation that causes "the effective
steady-state input impedance to be 50 ohms at the input to the T-network."</p><p>On the other hand, Maxwell countered that it is two <i>conjugately</i>-related rearward traveling waves (i.e. one wave is the conjugate of the other -- real components equal, imaginary components having opposite signs, rather than the two waves having equal amplitudes but opposite phases) that, through their interaction, creates a virtual open or short, reflecting all rearward traveling power back towards the load.</p>
<p>
Unfortunately, Best did not include any simulations for his T-network example
that would have verified his math and his assertions.
</p>
<p>
This lack is unfortunate because it is difficult to argue against a mathematical approach if simulations (or better yet, measurements) back it up.
</p>
<p>Later in this post, simulations will show that Best's analysis is correct (for example, that the two rearward waves traveling towards the source are equal in amplitude and opposite in phase). But first, I need to tweak Best's Tuner example a bit...</p>
<p>
For although simulating Best's T-network tuner system is straightforward using
Simulink, some of the values he chose (e.g. a transmission line
that is 1.2 wavelengths long) make it difficult to discern exactly what is
going on with the reflections and re-reflections on the transmission line.
</p>
<p>Let me demonstrate the problem...</p>
<p>First, the Simulink model of Best's T-network system:</p>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://1.bp.blogspot.com/-h95lcEBqn3k/YVxVnTyVA5I/AAAAAAAAMeo/QpaUlMvE5w0GIcfbQlF8IVwyuKbqVlLzwCLcBGAsYHQ/s1430/211003_Best_T-Network_Simulink_Model.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="952" data-original-width="1430" height="213" src="https://1.bp.blogspot.com/-h95lcEBqn3k/YVxVnTyVA5I/AAAAAAAAMeo/QpaUlMvE5w0GIcfbQlF8IVwyuKbqVlLzwCLcBGAsYHQ/s320/211003_Best_T-Network_Simulink_Model.png" width="320" /></a>
</div>
<br />
<div>
Below is a snapshot of the simulated directional coupler Vf and Vr (forward and reverse) voltage
waveforms at the T-network's input port.
</div>
<div><br /></div>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://1.bp.blogspot.com/-XQnFC5satas/YVmqYPPTAHI/AAAAAAAAMX8/jbAJsuZHwB0sr5iksNFKX_ZAQuCIiAIQACLcBGAsYHQ/s1071/211003_Best_T-Network_Simulink_Tuner_In_VR_VF.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="877" data-original-width="1071" height="262" src="https://1.bp.blogspot.com/-XQnFC5satas/YVmqYPPTAHI/AAAAAAAAMX8/jbAJsuZHwB0sr5iksNFKX_ZAQuCIiAIQACLcBGAsYHQ/s320/211003_Best_T-Network_Simulink_Tuner_In_VR_VF.png" width="320" /></a>
</div>
<div><br /></div>
<div>
Because the reflected voltage is decaying quickly (exponentially) over time,
it is difficult to compare the simulation voltages to calculated
voltages. Where do I place the display's cursors?
</div>
<div><br /></div>
<div>
But if I were to run a slightly different simulation that generates the
results shown below (where the voltage changes every 4 cycles, rather than
continuously), it is very easy to use the display cursors make voltage
measurements and compare these measurements to calculated results.
</div>
<div><br /></div>
<div>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://1.bp.blogspot.com/-nb56nnM2L3E/YVxYEcJqH_I/AAAAAAAAMew/qOVkMJ3r_yg6l_Yg75Ok_HfwI0Ev1hdBQCLcBGAsYHQ/s1071/211005%2BLC%2BNetwork%2Bwaveform%2Bexample.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="877" data-original-width="1071" height="262" src="https://1.bp.blogspot.com/-nb56nnM2L3E/YVxYEcJqH_I/AAAAAAAAMew/qOVkMJ3r_yg6l_Yg75Ok_HfwI0Ev1hdBQCLcBGAsYHQ/s320/211005%2BLC%2BNetwork%2Bwaveform%2Bexample.png" width="320" /></a>
</div>
</div>
<div><br /></div>
<div>
The display, above, was created by changing Best's T-network to an L-network
and lengthening the transmission line from 1.2 to 2 wavelengths, extending the
end-to-end travel time. The source's frequency was changed from 21.2 MHz to 10
MHz).
</div>
<div><br /></div>
<div>Although I've changed Best's example to allow better comparison with
computed results, <u>it is important to note</u> that the calculated voltage
amplitudes of this new model <b><i>are the same</i></b> as those calculated by
Best for his T-network.
</div>
<div><br /></div>
<div>But first -- if I am going to compare simulated results to my L-network's calculated values, I need to calculate those values, which means I need to define the system.</div>
<div><br /></div>
<div>And so on to the next step...</div><div><br /></div><div><br /></div>
<div>
<u><b>L-network Impedance Matching with Mismatched Load:</b></u>
</div>
<div>
<p>
Consider a lossless transmission line whose length is two wavelengths long,
with a characteristic impedance Zo equal to 50 ohms.
</p>
<p>
Let the load connected to the line's output be 150 ohms, resistive.
Therefore, in steady-state, the effective impedance seen at the input of the
transmission line is also 150 ohms, resistive, given the two-wavelength
length of the line.
</p>
<p>
And let there be a lossless, lumped-element L-network at
the input of the transmission line that will transform the 150 ohms
connected to its output port to 50 ohms, as measured looking into the L-network's input port, in steady-state.
</p><p>This L-network will be implemented as an LC network (series L, parallel C -- note that I will use "L-network" and "LC network" interchangeably in this post)).</p>
<p>
At the input and the output of this L-network let us insert directional
couplers that will measure, respectively, the total forward and reflected voltages
at the input of the L-network and the total forward and reflected voltages at the
output of the L-network.
</p>
<p>
The transmitter driving this network will be represented by a voltage source
of 141.4 VRMS with a source impedance of 50 ohms, resistive.
</p>
<p>
This transmitter will connect to the L- network's input via a very short
length of 50-ohm coax (in fact, we don't even need this coax, as the
directional coupler, being a lumped-element circuit itself, will still
display the correct Vf and Vr voltages as though there were a transmission
line attached).
</p>
<p>Here is the system:</p>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://1.bp.blogspot.com/-3Fpivy-67o4/YVnFC0TYyuI/AAAAAAAAMYU/FZAa36p0XzU87uF5HMg7XLknrWBM8kJGwCLcBGAsYHQ/s898/211003%2BLC%2BNetwork%2BGeneral%2Bto%2BSpecific.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="611" data-original-width="898" height="218" src="https://1.bp.blogspot.com/-3Fpivy-67o4/YVnFC0TYyuI/AAAAAAAAMYU/FZAa36p0XzU87uF5HMg7XLknrWBM8kJGwCLcBGAsYHQ/s320/211003%2BLC%2BNetwork%2BGeneral%2Bto%2BSpecific.png" width="320" /></a>
</div>
<div><br /></div>
The LC matching network consists of a series 1.1254 uH inductor and a parallel
150.05 pF capacitor which, at 10 MHz, will, in steady-state, transform the 150
ohm load connected to its output port to 50 ohms looking into its input
port.<br />
<p>
<b><u><br /></u></b>
</p>
<p>
<b><u>Calculating the System's Reflection and Transmission Coefficients:</u></b>
</p>
<p>
In a transmission line system, Transmission and Reflection Coefficients are
useful for calculating forward and reflected voltages. The figure
below shows the equations for these coefficients when two transmission lines
of different characteristic impedances are connected in series:
</p>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://1.bp.blogspot.com/-hgiXzQxLct0/YVnVWvOJ6LI/AAAAAAAAMYs/YXapt29GZe8wNoTfIDHwu771Lvf9Y5PrQCLcBGAsYHQ/s506/211003%2BGeneral%2BTrans%2Band%2BRefl%2BCoefficients.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="506" data-original-width="505" height="320" src="https://1.bp.blogspot.com/-hgiXzQxLct0/YVnVWvOJ6LI/AAAAAAAAMYs/YXapt29GZe8wNoTfIDHwu771Lvf9Y5PrQCLcBGAsYHQ/s320/211003%2BGeneral%2BTrans%2Band%2BRefl%2BCoefficients.png" width="319" /></a>
</div>
<p>
Note that these definitions of Transmission and Reflection Coefficients (see
<a href="https://en.wikipedia.org/wiki/Transmission_coefficient">here</a>,
for example) are <i>not</i> a function of the loads that might be
connected at the far ends of the two transmission lines. The
coefficients are solely a function of the impedances <i>immediately</i> at
either side of the impedance discontinuity.
</p>
<p>
In the LC-network system I have defined, above, there are a total of six
Reflection and Transmission Coefficients.
</p>
<p>
The load at the far end of the two-wavelength transmission line has one
Reflection and one Transmission Coefficient.
</p>
<p>
And the LC network, being a two-port lumped-element (i.e. essentially 0
delay) network with a transmission line attached to each of its two ports,
can be represented with two Reflection and two Transmission Coefficients.
</p>
<p>
These six Reflection and Transmission Coefficient are shown in the figure,
below:
</p>
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<a href="https://1.bp.blogspot.com/-m0AiqsMe4rY/YVnPAJkeuTI/AAAAAAAAMYc/TmIZoVLVpVELsJKNAyZGn6qBcKKHgs2BgCLcBGAsYHQ/s927/211003%2Bsystem%2Breflection%2Band%2Btransmission%2Bcoefficients.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="537" data-original-width="927" height="185" src="https://1.bp.blogspot.com/-m0AiqsMe4rY/YVnPAJkeuTI/AAAAAAAAMYc/TmIZoVLVpVELsJKNAyZGn6qBcKKHgs2BgCLcBGAsYHQ/s320/211003%2Bsystem%2Breflection%2Band%2Btransmission%2Bcoefficients.png" width="320" /></a>
</div>
<br />
<p><u>Calculating Load Reflection and Transmission Coefficients:</u></p>
<p>
The load's Reflection and Transmission Coefficients are easily calculated
using the following formulas:
</p>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://1.bp.blogspot.com/-qDZn0mtLD0I/YVnTLX-j5NI/AAAAAAAAMYk/MU6yEUCLC8QQKtcxhmVdPrXUDDao4KCcACLcBGAsYHQ/s403/211003%2Bload%2Bxmission%2Band%2Breflection%2Bcoefficients.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="403" data-original-width="392" height="320" src="https://1.bp.blogspot.com/-qDZn0mtLD0I/YVnTLX-j5NI/AAAAAAAAMYk/MU6yEUCLC8QQKtcxhmVdPrXUDDao4KCcACLcBGAsYHQ/s320/211003%2Bload%2Bxmission%2Band%2Breflection%2Bcoefficients.png" width="311" /></a>
</div>
<div><br /></div>
Given Zload = 150 + j0 ohms and Zo = 50 ohms:
</div>
<div><br /></div>
<blockquote style="border: none; margin: 0px 0px 0px 40px; padding: 0px; text-align: left;">
<div style="text-align: left;">Γ<sub>ld</sub> = 0.5 + j0</div>
<div style="text-align: left;">Τ<sub>ld</sub> = 1.5 + j0</div>
</blockquote>
<div style="text-align: center;"><br /></div>
<div>
<u>Calculating LC Network Reflection and Transmission Coefficients:</u>
</div>
<div><br /></div>
<div>
Equations to calculate Reflection and Transmission Coefficients for a lossless
two-port lumped-element network can be derived using the techniques described
in this blog post:
</div>
<div>
<a href="http://k6jca.blogspot.com/2021/05/lc-network-reflection-and-transmission.html">http://k6jca.blogspot.com/2021/05/lc-network-reflection-and-transmission.html</a><br />
<p>
The general method I use for calculating these Coefficients is described in
the figure, below:</p><p>
</p><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-1c0IRwT5MlE/YWM0NppebkI/AAAAAAAAMg4/Q_WswPhxy-wn8V1HoxYLpbmzLAOyRmzWwCLcBGAsYHQ/s728/General%2BNetwork%2BEquations.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="610" data-original-width="728" height="268" src="https://1.bp.blogspot.com/-1c0IRwT5MlE/YWM0NppebkI/AAAAAAAAMg4/Q_WswPhxy-wn8V1HoxYLpbmzLAOyRmzWwCLcBGAsYHQ/s320/General%2BNetwork%2BEquations.png" width="320" /></a></div><br />Note that, for the following calculations, the impedances that are connected
to either the input or the output port of the two-port network are the
impedances seen <i>immediately</i> at either port (e.g. the characteristic
impedances of the two transmission lines). These impedances
are <i>not</i> a function of the impedances connected to the far ends
of the two lines.
<p></p>
<p>
Given my example's LC network topology and component values, we can derive
<span style="text-align: center;">Γ</span><sub style="text-align: center;">1</sub><span style="text-align: center;"> </span><span style="text-align: center;">and </span><span style="text-align: center;">Τ</span><sub style="text-align: center;">21</sub><span style="text-align: center;"> (looking </span>into Port 1 of
the two-port LC network) <span style="text-align: center;">as described, below (note that </span><span style="text-align: center;">I've changed the name of </span><span style="text-align: center;">Γ</span><sub style="text-align: center;">11</sub><span style="text-align: center;">, used in the "general form" described in the previous figure,
to </span><span style="text-align: center;">Γ</span><sub style="text-align: center;">1</sub><span style="text-align: center;"> for this example</span><span style="text-align: center;">) </span><span style="text-align: center;">:</span>
</p>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://1.bp.blogspot.com/-aL2J5P0DIEA/YVnZM2CC5BI/AAAAAAAAMY0/OJ7LTxKgSRk9X4zxD3T7AyJ3UR_HhwZ3ACLcBGAsYHQ/s777/211003%2BLC%2BIn%2BTrans%2Band%2BRefl%2BCoeff.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="777" data-original-width="687" height="320" src="https://1.bp.blogspot.com/-aL2J5P0DIEA/YVnZM2CC5BI/AAAAAAAAMY0/OJ7LTxKgSRk9X4zxD3T7AyJ3UR_HhwZ3ACLcBGAsYHQ/s320/211003%2BLC%2BIn%2BTrans%2Band%2BRefl%2BCoeff.png" width="283" /></a>
</div>
<div><br /></div>
Similarly, looking into Port 2 of the two-port LC network, we can
derive <span style="text-align: center;">Γ</span><sub style="text-align: center;">2</sub><span style="text-align: center;"> and </span><span style="text-align: center;">Τ</span><sub style="text-align: center;">12</sub><span style="text-align: center;"> using a similar procedure:</span><br />
<p></p>
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<a href="https://1.bp.blogspot.com/-CrYiMwbiQjA/YVnZnazN4lI/AAAAAAAAMY8/dK-idaq-trUBsizlkCXe8wfLui62e3higCLcBGAsYHQ/s777/211003%2BLC%2BOut%2BTrans%2Band%2BRefl%2BCoeff.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="777" data-original-width="687" height="320" src="https://1.bp.blogspot.com/-CrYiMwbiQjA/YVnZnazN4lI/AAAAAAAAMY8/dK-idaq-trUBsizlkCXe8wfLui62e3higCLcBGAsYHQ/s320/211003%2BLC%2BOut%2BTrans%2Band%2BRefl%2BCoeff.png" width="283" /></a>
</div>
<div><br /></div>
Next, given the following values:
</div>
<div>
<ul style="text-align: left;">
<li>
Transmission lines' characteristic impedances Za and Zb both equal 50
ohms.
</li>
<li>Inductor impedance, Zl, at 10 MHz is + j70.7110 ohms.</li>
<li>Capacitor impedance, Zc, at 10 MHz is - j106.07 ohms.</li>
</ul>
</div>
<div>
We can calculate the LC network's two Reflection Coefficients and two
Transmission Coefficients to be:
</div>
<div><br /></div>
<blockquote style="border: none; margin: 0px 0px 0px 40px; padding: 0px; text-align: left;">
<div>
<div style="text-align: left;">
Γ<sub>1</sub> = 0.1667 + j0.4714
</div>
</div>
<div>
<div style="text-align: left;">Τ<sub>21</sub> = 0.5 - j0.7071</div>
</div>
<div><div style="text-align: left;">Γ<sub>2</sub> = 0.5 + j0</div></div>
<div>
<div style="text-align: center;">
<div style="text-align: left;">Τ<sub>12</sub> = 0.5 - j0.7071</div>
</div>
</div>
</blockquote>
<div>
<div style="text-align: center;">
<div><br /></div>
<div style="text-align: left;">
Although not necessary for the calculations later in this post, here are
the magnitudes of these four Reflection and Transmission Coefficients:
</div>
<div style="text-align: left;"><br /></div>
</div>
</div>
<blockquote style="border: none; margin: 0px 0px 0px 40px; padding: 0px; text-align: left;">
<div>
<div style="text-align: center;">
<div>
<div><div style="text-align: left;">|Γ<sub>1</sub>|= 0.5</div></div>
</div>
</div>
</div>
<div>
<div style="text-align: center;">
<div>
<div><div style="text-align: left;">|Γ<sub>2</sub>|= 0.5</div></div>
</div>
</div>
</div>
<div>
<div style="text-align: center;">
<div><div style="text-align: left;">|Τ<sub>21</sub>| = 0.866</div></div>
</div>
</div>
<div>
<div style="text-align: center;">
<div><div style="text-align: left;">|Τ<sub>12</sub>| = 0.866</div></div>
</div>
</div>
</blockquote>
<div>
<div style="text-align: center;">
<div><div></div></div>
</div>
</div>
<div>
<div style="text-align: center;">
<div><div></div></div>
</div>
</div>
<div>
<div style="text-align: center;">
<div><div></div></div>
</div>
</div>
<div>
<div style="text-align: center;">
<div><div></div></div>
<div><br /></div>
</div>
<p>
<b><u>Calculating Forward and Reflected Voltages:</u></b>
</p>
<p><span style="color: #2b00fe;">First, a note regarding voltage names...</span></p><p><span style="color: #2b00fe;">I derived my equations, below, independently from Best's articles, without referencing his voltage labels. Later in this post, at the point where I compare Best's calculated voltages to my voltages, there will be a table linking Best's voltage labels to my labels. </span></p><p>Continuing...</p><p>Using the three Reflection Coefficients and three Transmission Coefficients
calculated earlier, we can start calculating system forward and reflected voltages.
</p><p>The system is driven by a gated sine-wave source. To understand how an
infinite series of reflections can build up, let's examine what happens when
this source is first turned on, using a Lattice Diagram to show how a single
point on the Vs waveform moves through the system over time: </p>
<p></p>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://1.bp.blogspot.com/-VyuKgOy0fSs/YVzVjMV72pI/AAAAAAAAMf4/ocebi-UDK_UjH3HkEKhVY7V7XAugmkv7wCLcBGAsYHQ/s1137/211003%2Bstart%2Bof%2Bforward%2Band%2Breflected%2Bwaves%252C%2Bno%2Bexponential.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="787" data-original-width="1137" height="221" src="https://1.bp.blogspot.com/-VyuKgOy0fSs/YVzVjMV72pI/AAAAAAAAMf4/ocebi-UDK_UjH3HkEKhVY7V7XAugmkv7wCLcBGAsYHQ/s320/211003%2Bstart%2Bof%2Bforward%2Band%2Breflected%2Bwaves%252C%2Bno%2Bexponential.png" width="320" /></a>
</div>
<br />Thus, the voltage of each reflection, re-reflection, or transmitted
voltage can be found simply by multiplying the value of the voltage arriving
at the impedance discontinuity by the appropriate Reflection Coefficient or
Transmission Coefficient.
<p></p>
<p>
To calculate Vf1: note that the voltage leaving the source impedance
and entering the first transmission line, at the instant it enters this
line, sees the transmission line's characteristic impedance Zo as its
terminating impedance. Thus, the voltage entering the line, Vf1, can
be calculated as a simple voltage divider:
</p>
</div>
<div>
<p></p>
<p style="text-align: center;">Vf1 = Vs*Zo/(Zo + Zs)</p>
<p>
This is true for every point of the Vs waveform arriving at the
Zs-to-transmission line junction.
</p>
<p>Therefore, given a Vs of 141.4 volts, Vf1 will always be 70.7 volts.</p>
<p>
When Vf1 (the forward voltage on the transmission line traveling towards the
LC network's input) arrives at this input port, it sees an impedance
mismatch represented by Γ<sub>1</sub>, the LC network's input port
Reflection Coefficient.
</p>
<p></p>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://1.bp.blogspot.com/-mEwYQW9EFF4/YVyLXH7kSkI/AAAAAAAAMfA/k2VguDssr7kSKVWJxc1A4t32dzBTN2WmACLcBGAsYHQ/s918/211005%2BVr1%2BVf2.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="450" data-original-width="918" height="157" src="https://1.bp.blogspot.com/-mEwYQW9EFF4/YVyLXH7kSkI/AAAAAAAAMfA/k2VguDssr7kSKVWJxc1A4t32dzBTN2WmACLcBGAsYHQ/s320/211005%2BVr1%2BVf2.png" width="320" /></a>
</div>
<br />Part of Vf1 is transmitted back to the source. This reflected
voltage is Vr1, where Vr1 = Vf1* Γ<sub>1</sub>.
</div>
<div><br /></div>
<div>
And part of Vf1 continues forward through the LC network to the second
transmission line. This forward-traveling voltage is Vf2, where Vf2 =
Vf1*Τ<sub>21</sub>.
<p></p>
<p>
The part reflected back to the source, if the source impedance equals the
transmission line's characteristic impedance, stops there and is not
re-reflected back to the load.
</p>
<p>
A quick note -- the source voltage, itself, plays no role in the reflection
of a voltage wave arriving back at the source. This reflection is
solely a function of the source <i>impedance</i>, not the source
<i>voltage</i>, and it is an outcome of the <a href="https://en.wikipedia.org/wiki/Superposition_principle">Superposition Principle</a>:
</p>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://1.bp.blogspot.com/-Ez5g9gwZt6o/YVxlGe5TAiI/AAAAAAAAMe4/TM3CC3qg6JMwppHAbR7UnGeM8foPCsT2ACLcBGAsYHQ/s595/211005%2BReflected%2Bwave%2Bback%2Btowards%2Bsource%2Bsees.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="561" data-original-width="595" height="302" src="https://1.bp.blogspot.com/-Ez5g9gwZt6o/YVxlGe5TAiI/AAAAAAAAMe4/TM3CC3qg6JMwppHAbR7UnGeM8foPCsT2ACLcBGAsYHQ/s320/211005%2BReflected%2Bwave%2Bback%2Btowards%2Bsource%2Bsees.png" width="320" /></a>
</div>
<p>
Vf2, the part of Vf1 that has continued forward through the LC network and onto the second transmission line,
continues forward on the second transmission line until it reaches the load
mismatch, at which point part of it is reflected back to the source (that
is, towards the output port of the LC network) with a voltage of
Vf2*Γ<sub>ld</sub> , and part continues forward into the load (voltage
of Vf2*T<sub>ld</sub>).
</p>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://1.bp.blogspot.com/--pnMKgkuewo/YVzWUnSOOjI/AAAAAAAAMgA/5V9O3N-7bTkbnJuc62eCkJ1nZEthKpqbACLcBGAsYHQ/s978/211005%2Bstart%2Bof%2Bvld%2Band%2Bvrld%252C%2Bno%2Bexponentials.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="535" data-original-width="978" height="175" src="https://1.bp.blogspot.com/--pnMKgkuewo/YVzWUnSOOjI/AAAAAAAAMgA/5V9O3N-7bTkbnJuc62eCkJ1nZEthKpqbACLcBGAsYHQ/s320/211005%2Bstart%2Bof%2Bvld%2Band%2Bvrld%252C%2Bno%2Bexponentials.png" width="320" /></a>
</div>
<br />The reflection traveling back from the load, when it arrives at the
<i>output</i> port of the LC network, sees another impedance mismatch, and it will be partially re-reflected back
towards the load (i.e. it will move in the forward direction) with a voltage
of Γ<sub>2</sub>*(Vf2*Γ<sub>ld</sub>) and partially transmitted through the LC
network towards the source (voltage T<sub>12</sub>*(Vf2*Γ<sub>ld</sub>)) .
<p></p>
<p></p>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://1.bp.blogspot.com/-_nF1jQurPQM/YVzWejP88JI/AAAAAAAAMgE/00X4u1I8n_IBMq45lofn48OMs-SaNSpmQCLcBGAsYHQ/s963/211005%2Bstart%2Bof%2Bvfld%2Band%2Bvrs%252C%2Bno%2Bexponentials.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="626" data-original-width="963" height="208" src="https://1.bp.blogspot.com/-_nF1jQurPQM/YVzWejP88JI/AAAAAAAAMgE/00X4u1I8n_IBMq45lofn48OMs-SaNSpmQCLcBGAsYHQ/s320/211005%2Bstart%2Bof%2Bvfld%2Band%2Bvrs%252C%2Bno%2Bexponentials.png" width="320" /></a>
</div>
<br />The re-reflection back towards the load will, when it arrives at the
load mismatch, again be partially reflected back towards the LC network and
partially transmitted into the load.
<p></p>
<p>
This ping-ponging back and forth on the transmission line between the output
port of the LC network and the load continues <i>ad infinitum</i>. And
every point on the Vs waveform undergoes this ping-ponging.
</p>
<p>
Thus, in steady-state (after Vs has been on for awhile), if we were to take
a <i>snapshot in time </i>of the reflections and re-reflections at either
ends of the transmission line and the LC network, we would see both the
voltage Vf2 from the source entering the line as well as all of the past
reflections and re-reflections from previous values of Vf2:
</p>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://1.bp.blogspot.com/-LeQ77pSvRe8/YVoANSah-NI/AAAAAAAAMZ0/A-8dUMIDEMMYNjmd_ci0xsFY24MeqxP5ACLcBGAsYHQ/s961/211003%2BSnapshot%2Bof%2Breflections%2Band%2Brereflections.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="869" data-original-width="961" height="289" src="https://1.bp.blogspot.com/-LeQ77pSvRe8/YVoANSah-NI/AAAAAAAAMZ0/A-8dUMIDEMMYNjmd_ci0xsFY24MeqxP5ACLcBGAsYHQ/s320/211003%2BSnapshot%2Bof%2Breflections%2Band%2Brereflections.png" width="320" /></a>
</div>
<p>
There are four important infinite series in the diagram, representing the
past reflections that are still traveling back and forth on the transmission
line at the time of this "snapshot".
</p>
<p>
By the Principle of Superposition, we know, for example, that the
<i>total</i> voltage traveling forward towards the load on the second
transmission line is the vector sum of <i>all</i> existing voltages
(past re-reflections and present) on the line traveling in this
direction.
</p>
<p>
Therefore, to calculate these voltages that are a function of ongoing past
reflections, we will need to sum these series.
</p>
<p>
But before I get to these four infinite series, a quick note on the
exponential terms in the above diagram...
</p>
<div>
<p>
<b><u>Sidebar on the Propagation Constant e</u><sup>-γl </sup>:</b>
</p>
<p>
The exponential terms that appear in the equations (e.g.
e<sup>-γl </sup> or e<sup>-2γl</sup>) after a voltage has
traveled along the transmission line) represent the effect of the
transmission line's <a href="https://en.wikipedia.org/wiki/Propagation_constant">Propagation Constant</a> on the voltage's amplitude (i.e. attenuation) and phase.
</p>
<p>
If we were to take a snapshot in time and compare the voltage Vf at point
A with the voltage Vf at point B, we would see that the voltage at point B
has a negative phase shift compared to the phase of Vf at point A (because
the voltage at point B began earlier in time), as well as some attenuation
due to transmission line loss, per the figure, below:
</p>
<p></p>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://1.bp.blogspot.com/-1yZPjIhRbc0/YVn4RLv09FI/AAAAAAAAMZk/TvcOjpOMbXQadsSPmhnSOsgpEUmfva6dgCLcBGAsYHQ/s824/210918%2Bpropagation%2Bconstants.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="647" data-original-width="824" height="251" src="https://1.bp.blogspot.com/-1yZPjIhRbc0/YVn4RLv09FI/AAAAAAAAMZk/TvcOjpOMbXQadsSPmhnSOsgpEUmfva6dgCLcBGAsYHQ/s320/210918%2Bpropagation%2Bconstants.png" width="320" /></a>
</div>
<div><br /></div>
If the line is lossless, the attenuation factor is 1, and if the line is an
integral number of wavelengths long (as it is in this example), then the
phase shift is 0. In other words, for this example, the propagation
constant equals 1.
</div>
<div><br /></div>
<div><br /></div>
<div>
<b><u>The Four Infinite Series:</u></b>
</div>
<p>
Examining the earlier "snapshot-in-time" figure, there are four infinite
series. They are defined in the figure, below:
</p>
<div class="separator" style="clear: both; text-align: center;">
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</div>
<p>Let's write series-equations for these four infinite-series.</p>
<p>
First, Vld, the infinite series representing the voltage across the load:
</p>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://1.bp.blogspot.com/-NAdaXcRKRis/YVoH2Ry_afI/AAAAAAAAMZ8/h3ccgXx2plkEI4-S0e_aFj__vulX4LqgwCLcBGAsYHQ/s778/211003%2BVld%2Bseries.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="270" data-original-width="778" height="111" src="https://1.bp.blogspot.com/-NAdaXcRKRis/YVoH2Ry_afI/AAAAAAAAMZ8/h3ccgXx2plkEI4-S0e_aFj__vulX4LqgwCLcBGAsYHQ/s320/211003%2BVld%2Bseries.png" width="320" /></a>
</div>
<div><br /></div>
Next, Vrld, the infinite series representing the voltage reflected back
towards the LC network's output port from the load's mismatch:
</div>
<div><br /></div>
<div>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://1.bp.blogspot.com/-v2cGyvI3hUY/YVoIMbxbIGI/AAAAAAAAMaE/3aJe217aewc3iriUzHWiJ6v7zAiSoCyIACLcBGAsYHQ/s740/211003%2BVrld%2Bseries.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="276" data-original-width="740" height="119" src="https://1.bp.blogspot.com/-v2cGyvI3hUY/YVoIMbxbIGI/AAAAAAAAMaE/3aJe217aewc3iriUzHWiJ6v7zAiSoCyIACLcBGAsYHQ/s320/211003%2BVrld%2Bseries.png" width="320" /></a>
</div>
<div><br /></div>
Next, Vfld, the infinite series representing the load's reflected voltage
<i>re-reflected</i> back towards the load (i.e. now traveling forwards) by the
LC network's output impedance mismatch:
</div>
<div><br /></div>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://1.bp.blogspot.com/-0lGJ0R1qzFk/YVoIdg27QVI/AAAAAAAAMaM/wBISh12SvF4Wn-9lYf8eTmZZGun5PuvawCLcBGAsYHQ/s737/211003%2BVfld%2Bseries.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="326" data-original-width="737" height="142" src="https://1.bp.blogspot.com/-0lGJ0R1qzFk/YVoIdg27QVI/AAAAAAAAMaM/wBISh12SvF4Wn-9lYf8eTmZZGun5PuvawCLcBGAsYHQ/s320/211003%2BVfld%2Bseries.png" width="320" /></a>
</div>
<br />
<div>
And finally, Vrs, the infinite series representing the reflected voltage, from
the load, after it has passed backwards through the LC network (output port to
input port) towards the source:
</div>
<div><br /></div>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://1.bp.blogspot.com/-94EQB3v8luY/YVoI5pxFrkI/AAAAAAAAMaU/Cr-d43x1gv48k6gnADyjIsqIQl0QeEC6QCLcBGAsYHQ/s828/211003%2BVrs%2Bseries.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="381" data-original-width="828" height="147" src="https://1.bp.blogspot.com/-94EQB3v8luY/YVoI5pxFrkI/AAAAAAAAMaU/Cr-d43x1gv48k6gnADyjIsqIQl0QeEC6QCLcBGAsYHQ/s320/211003%2BVrs%2Bseries.png" width="320" /></a>
</div>
<br />
<div>
Note that both Vfld and Vrs can be expressed in terms of Vrld (per the last
line in each of the derivations, above). So let's find simpler
expressions for Vld and Vrld. We can then use these simpler expressions
to express Vfld and Vrs.
</div>
<div><br /></div>
<div>First, regarding infinite geometric series...</div>
<div><br /></div>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://1.bp.blogspot.com/-bN9sQNqSL60/YVoJzd2dSXI/AAAAAAAAMac/Etsm5uuEqFcR4i8bEexXTVnEo0oRmCIDgCLcBGAsYHQ/s611/211003%2Bconvergence%2Bof%2Ban%2Binfinite%2Bseries.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="351" data-original-width="611" height="184" src="https://1.bp.blogspot.com/-bN9sQNqSL60/YVoJzd2dSXI/AAAAAAAAMac/Etsm5uuEqFcR4i8bEexXTVnEo0oRmCIDgCLcBGAsYHQ/s320/211003%2Bconvergence%2Bof%2Ban%2Binfinite%2Bseries.png" width="320" /></a>
</div>
<div><br /></div>
Thus, subject to the restriction noted below, we can reduce Vld and Vrld from
values represented by infinite series to values easily calculated with the
following formulas:
<div><br /></div>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://1.bp.blogspot.com/-hLxWvrq2pbI/YVoKcjPHjWI/AAAAAAAAMak/0hGMmuyTKvYrt_cMG4oVsjZPiDdTIM0bwCLcBGAsYHQ/s355/211003%2Bconverged%2Bsolution%2Bfor%2BVld%2Band%2BVrld.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="346" data-original-width="355" height="312" src="https://1.bp.blogspot.com/-hLxWvrq2pbI/YVoKcjPHjWI/AAAAAAAAMak/0hGMmuyTKvYrt_cMG4oVsjZPiDdTIM0bwCLcBGAsYHQ/s320/211003%2Bconverged%2Bsolution%2Bfor%2BVld%2Band%2BVrld.png" width="320" /></a>
</div>
<div><br /></div>
Note that our other two infinite-series, Vfld and Vrs, are simply the Vrld
series multiplied by the appropriate Reflection or Transmission Coefficient:
<div><br /></div>
<blockquote style="border: none; margin: 0px 0px 0px 40px; padding: 0px; text-align: left;">
<div>
<div style="text-align: left;">Vfld = Γ<sub>2</sub> * Vrld</div>
</div>
<div>
<div style="text-align: left;">Vrs = Τ<sub>12</sub> * Vrld</div>
</div>
</blockquote>
<div>
<div><br /></div><div>The figure below summarizes the definitions of the steady-state traveling-wave voltages in the system:</div><div><br /></div><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-JeuNAuXlIt0/YWRiL_q3uQI/AAAAAAAAMhk/QAzfng5gPtceEmcefS_pZh370n2DCDGtwCLcBGAsYHQ/s1032/211011%2Bsteady%2Bstate%2Btraveling%2Bvoltage%2Bsummary.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="804" data-original-width="1032" height="249" src="https://1.bp.blogspot.com/-JeuNAuXlIt0/YWRiL_q3uQI/AAAAAAAAMhk/QAzfng5gPtceEmcefS_pZh370n2DCDGtwCLcBGAsYHQ/s320/211011%2Bsteady%2Bstate%2Btraveling%2Bvoltage%2Bsummary.png" width="320" /></a></div><div><br /></div>
<div><br /></div>
<div>
<b><u>Calculating Steady-state System Voltages:</u></b><br />
<div><br /></div>
<div>First, let's state the example's given values:</div>
<div><br /></div>
</div>
</div>
<blockquote style="border: none; margin: 0px 0px 0px 40px; padding: 0px; text-align: left;">
<div>
<div><div>Vs = 141.4 volts</div></div>
</div>
</blockquote>
<blockquote style="border: none; margin: 0px 0px 0px 40px; padding: 0px; text-align: left;">
<div><div style="text-align: left;">Zs = 50 ohms</div></div>
<div><div style="text-align: left;">Zo = 50 ohms</div></div>
<div>
<div style="text-align: left;">
Zl (i.e. inductor) = 0.0000 + j70.7110
</div>
</div>
<div>
<div style="text-align: left;">
Zc (i.e. capacitor) = 0.0000 - j106.07
</div>
</div>
<div><div style="text-align: left;">Zld (i.e. load) = 150 ohms</div></div>
<div>
<div style="text-align: left;">
e<sup>-γl </sup>= e<sup>-2γl</sup> = 1 (propagation
constants)
</div>
</div>
</blockquote>
<div>
<div><br /></div>
<div>
We can calculate the example's Reflection and Transmission Coefficients:
</div>
<div><br /></div>
</div>
<blockquote style="border: none; margin: 0px 0px 0px 40px; padding: 0px; text-align: left;">
<div><div>Γ<sub>1</sub> = 0.1667 + j0.4714 </div></div>
<div><div>Τ<sub>21</sub> = 0.5 - j0.7071</div></div>
<div><div>Γ<sub>2</sub> = 0.5 + j0</div></div>
<div>
<div>
<div style="text-align: left;">
<div style="text-align: center;">
<div style="text-align: left;">
Τ<sub>12</sub> = 0.5 - j0.7071
</div>
</div>
</div>
</div>
</div>
</blockquote>
<blockquote style="border: none; margin: 0px 0px 0px 40px; padding: 0px; text-align: left;">
<div>
<div style="text-align: left;">
<div style="text-align: center;">
<div style="text-align: left;">Γ<sub>ld</sub> = 0.5 + j0</div>
</div>
</div>
</div>
<div>
<div style="text-align: left;">
<div style="text-align: center;">
<div style="text-align: left;">
<div style="text-align: left;">Τ<sub>ld</sub> = 1.5 + j0</div>
</div>
</div>
</div>
</div>
</blockquote>
<div>
<div style="text-align: left;">
<div style="text-align: center;">
<div style="text-align: left;">
<div style="text-align: left;"><br /></div>
<div style="text-align: left;">
Next, using the equations developed, above, we can calculate the
following voltages:
</div>
<div style="text-align: left;"><br /></div>
</div>
</div>
</div>
</div>
<blockquote style="border: none; margin: 0px 0px 0px 40px; padding: 0px; text-align: left;">
<div><div>Vf1 = Vs*(Zo/(Zs+Zo) = 70.71 volts</div></div>
<div>
<div>
Vr1 = <span style="text-align: center;">Γ</span><sub style="text-align: center;">1</sub> * Vf1 = 11.7849 +
j33.3316
</div>
</div>
<div>
<div>
Vf2 = <span style="text-align: center;">Τ</span><sub style="text-align: center;">21</sub> * Vf1 = 35.3537 - j50
volts
</div>
</div>
<div><div>Vld = 70.7073 - j100 volts</div></div>
<div><div>Vrld = 23.5691 - j33.332 volts</div></div>
<div><div>Vfld = 11.7846 - j16.666 volts</div></div>
<div><div>Vrs = -11.7841 - j33.3314 volts</div></div>
</blockquote>
<div>
<div><br /></div>
<div>Calculating magnitudes of the complex voltages:</div>
<div><br /></div>
</div>
<blockquote style="border: none; margin: 0px 0px 0px 40px; padding: 0px; text-align: left;">
<div><div>|Vr1| = 35.35 volts</div></div>
<div><div>|Vf2| = 61.23 volts</div></div>
<div><div>|Vld| = 122.5 volts</div></div>
<div><div>|Vrld| = 40.82 volts</div></div>
<div><div>|Vfld| = 20.41 volts</div></div>
<div><div>|Vrs| = 35.35 volts</div></div>
</blockquote>
<div>
<div><br /></div>
<div><br /></div>
<div>
<b><u>Steady-state <i>Total</i> Forward and Reflected Voltages:</u></b>
</div>
<div><br /></div>
<div>
The simulations I will present later in this post display forward and
reflected voltages as sampled by two directional couplers (one at the LC
network input port, and the other at the LC network's output port).
</div>
<div><br /></div>
<div>
These voltages represent the <i>total</i> forward and reflected voltages on a
transmission line at the<i> point in time</i> that the the measurement
is made.
</div>
<div><br /></div>
<div>
The diagram, below, summarizes the <i>steady-state</i> forward and reflected
voltages on the transmission lines described earlier, and it shows how
voltages traveling in the same direction can be summed to create the
<i>total</i> forward and reflected voltages seen by the two directional
couplers.
</div>
<div><br /></div>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://1.bp.blogspot.com/-pZBmqC2iLhM/YVuUQbtZ0EI/AAAAAAAAMeg/qnyGoe6hMtUutK_1P6BrQm-srIeTBFaKACLcBGAsYHQ/s1051/211004%2BSystem%2BVoltages%2Bin%2BSteady-state.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="911" data-original-width="1051" height="277" src="https://1.bp.blogspot.com/-pZBmqC2iLhM/YVuUQbtZ0EI/AAAAAAAAMeg/qnyGoe6hMtUutK_1P6BrQm-srIeTBFaKACLcBGAsYHQ/s320/211004%2BSystem%2BVoltages%2Bin%2BSteady-state.png" width="320" /></a>
</div>
<br />
<div>
Note that summed voltages, above, are vector sums, not amplitude sums, and
thus their values must be calculated at the same point on the transmission
line.
</div>
<div><br /></div>
<div><u>Directional Coupler Forward and Reflected Voltages:</u></div>
<div><br /></div>
<div>
The two directional couplers each measure the total forward voltage (Vf) and the
total reflected voltage (Vr) passing through a point -- they cannot separate
out the different forward waves that might make up these total
voltages.
</div><div><br /></div><div>(Note, though -- although the directional couplers cannot separate out the individual voltages traveling in the same direction that might combine make up a total voltage in steady-state, we can measure these individual voltages during the transient phase when the source is turned on, or off. More on this later, in the simulation section.)</div>
<div><br /></div>
<div>During Steady-state, Vf and Vr are measured by the Directional Coupler at the LC Network's Input
Port, and equal:
</div>
<div>
<ul style="text-align: left;">
<li>Vf_LC_In = Vf1 = 70.71 volts</li>
<li>Vr_LC_In = Vr1 + Vrs = 0 volts</li>
</ul>
</div>
<div>Note that Vr_LC_In, the total reflected voltage from the network's input port, is the sum of two voltages.</div><div><br /></div><div>Similarly, steady-state Vf and Vr are measured by the Directional Coupler at the LC Network's Output
Port, and equal:
</div>
<div>
<ul style="text-align: left;">
<li>Vf_LC_Out = Vf2 + Vfld = 47.1382 - j66.663 volts</li>
<li>Vr_LC_Out = Vrld = 23.5691 - j33.332 volts</li>
</ul>
</div>
<div>Note that Vf_LC_Out, the total forward voltage moving towards the load from the output port of the network, is the sum of two voltages.</div><div><br /></div>
<div>Magnitudes of these voltages:</div>
<div><br /></div>
</div>
<blockquote style="border: none; margin: 0px 0px 0px 40px; padding: 0px; text-align: left;">
<div>
<div>
<div>
<div><div>|Vf_LC_In| = 70.71 volts</div></div>
</div>
</div>
</div>
</blockquote>
<blockquote style="border: none; margin: 0px 0px 0px 40px; padding: 0px; text-align: left;">
<div>
<div>
<div style="text-align: left;">|Vr_LC_In| = 0 volts</div>
</div>
</div>
<div>
<div><div style="text-align: left;">|Vf_LC_Out| = 81.65 volts</div></div>
</div>
<div>
<div><div style="text-align: left;">|Vr_LC_Out| = 40.82 volts</div></div>
</div>
</blockquote>
<div>
<div>
<div><br /></div><div><br /></div><div><b><u>Cancellation of the Reflected Wave to Source:</u></b></div>
</div>
<div><br /></div><div>As I mentioned in the introduction to this post, Best's conclusion is that, in steady state, the reflection of the source voltage off of the LC network's input impedance mismatch (i.e. Vr1) is cancelled by the rearward traveling wave from the load passing through the LC network from output port to input port towards the source (i.e. Vrs).</div><div><br /></div><div>For Vr1 to cancel Vrs, they need to be of equal amplitude but 180 degrees out of phase, per Best. Let's verify this:</div><div><br /></div></div><blockquote style="border: none; margin: 0px 0px 0px 40px; padding: 0px;"><div><div><div style="text-align: left;">|Vr1| = 35.35</div></div></div></blockquote><blockquote style="border: none; margin: 0px 0px 0px 40px; padding: 0px;"><div><div><div style="text-align: left;">Angle(Vr1) = 70.53 degrees</div></div></div></blockquote><div><div><div><br /></div></div></div><blockquote style="border: none; margin: 0px 0px 0px 40px; padding: 0px;"><div><div><div style="text-align: left;">|Vrs| = 35.35</div></div></div></blockquote><blockquote style="border: none; margin: 0px 0px 0px 40px; padding: 0px;"><div><div><div style="text-align: left;">Angle(Vrs) = -109.47 degrees</div></div></div></blockquote><div><div><div><br /></div><div>Magnitudes are the same. Let's check the difference in angle:</div><div><br /></div></div></div><blockquote style="border: none; margin: 0px 0px 0px 40px; padding: 0px;"><div><div><div style="text-align: left;">Delta_angle = Angle(Vr1) - Angle(Vrs) = 180 degrees.</div></div></div></blockquote><div><div><div><br /></div><div>Excellent. Amplitudes are identical, and angles are exactly 180 degrees out of phase, as predicted by the math.</div><div><br /></div><div><br /></div></div><div><br /></div>
<div>
<u><b>Steady-state Total Forward and Reflected Powers:</b></u>
</div>
<div><br /></div>
<div>
Calculating steady-state powers, given power delivery by the source of 100
watts:
</div>
<div><br /></div>
<div>
Total steady-state forward power towards the LC network's input port:
</div>
<div><br /></div>
</div>
<blockquote style="border: none; margin: 0px 0px 0px 40px; padding: 0px;">
<div>
<div>
<div>
<div style="text-align: left;">
P(Vf_LC_In) = (Vf1^2) / 50 = 100 watts
</div>
</div>
</div>
</div>
</blockquote>
<p>Total steady-state reflected power from the LC network's input port:</p>
<blockquote style="border: none; margin: 0px 0px 0px 40px; padding: 0px;">
<p style="text-align: left;">P(Vr_LC_In) = 0 watts</p>
</blockquote>
<div>
<div>
<div>
<div>Total steady-state reflected power towards the load:</div>
<div><br /></div>
</div>
</div>
</div>
<blockquote style="border: none; margin: 0px 0px 0px 40px; padding: 0px;">
<div>
<div>
<div>
<div style="text-align: left;">
P(Vf_LC_Out) = (abs(47.1382 - j66.663)^2)/50 = 133.3 watts
</div>
</div>
</div>
</div>
</blockquote>
<p> Total steady-state reflected power from the load:</p>
<blockquote style="border: none; margin: 0px 0px 0px 40px; padding: 0px;">
<div>
<div>
<div>
<div style="text-align: left;">
P(Vr_LC_Out) = (abs(23.5691 - j33.332)^2)/50 = 33.33 watts
</div>
</div>
</div>
</div>
</blockquote>
<div>
<div>
<div>
<div><br /></div>
<div>
<div>Total steady-state load power:</div>
<div><br /></div>
</div>
</div>
</div>
</div>
<blockquote style="border: none; margin: 0px 0px 0px 40px; padding: 0px;">
<div>
<div>
<div>
<div>
<div style="text-align: left;">
P(Vld) = (abs(Vld)^2)/Zld = 100 watts
</div>
</div>
</div>
</div>
</div>
</blockquote>
<div>
<div>
<div>
<div>
<div><br /></div>
<div>
As you can see, steady-state powers are exactly what is to be
expected.
</div>
<div><br /></div>
</div>
<div><br /></div>
<div>
<b><u>Comparison of Best's T-Network System Voltages with my LC-Network
System Voltages:</u></b>
</div>
<div><br /></div>
<div>
The Reflection Coefficients of Best's T-Network and my LC Network have
the same magnitude, but they do not have the same angle.
</div>
<div><br /></div>
<div>
Therefore, a direct comparison between Best's calculated <i>complex</i>
voltages values and mine cannot be made, because they will have
different phase angles.
</div>
<div><br /></div>
<div>
However, I can compare the magnitudes of these voltages, using MATLAB to
convert Best's complex numbers to magnitudes, similar to what I did
earlier in this post for my complex voltage values.
</div>
<div><br /></div>
<div>
The table below contains the magnitudes of the seven voltages named by
Best in his T-Network analysis. To their right I have added my
names for the equivalent seven voltages and the magnitudes that I have
calculated for them:
</div>
<div><br /></div>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://1.bp.blogspot.com/-mFTOOiowty8/YVyb5dDkSOI/AAAAAAAAMfo/OM_sRodmaQYUI22VMtYZybHb67GTGNnkgCLcBGAsYHQ/s957/Comparing%2Bve9srb%2Bto%2Bk6jca%2Bvoltages.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="620" data-original-width="957" height="207" src="https://1.bp.blogspot.com/-mFTOOiowty8/YVyb5dDkSOI/AAAAAAAAMfo/OM_sRodmaQYUI22VMtYZybHb67GTGNnkgCLcBGAsYHQ/s320/Comparing%2Bve9srb%2Bto%2Bk6jca%2Bvoltages.png" width="320" /></a>
</div>
<br />
<div>
When expressed as magnitudes, my calculations of system voltages
<i>are identical</i> to Best's. Therefore, any complaints that
Maxwell had regarding Best's analysis and results should also apply to
my analysis.
</div>
<div><br /></div>
<div>
So let's verify my results via simulation using
<a href="https://www.mathworks.com/products/simulink.html">Simulink</a>.
</div>
<div></div>
<div>
<u><br /></u>
</div>
<div>
<b><u><br /></u></b>
</div>
<div>
<b><u>Verification through Simulation:</u></b>
</div>
<div><br /></div>
<div><u>Simulink Model:</u></div>
<div><br /></div>
<div>
The Simulink Model, below, has directional couplers at the input and
output ports of the LC matching network. With these two
directional couplers I can examine forward and reflected voltages
simultaneously at both ports of the LC network.
</div>
<div><br /></div>
<div>
Note that the forward and reflected voltages at the LC network input
are:
</div>
<div><br /></div>
</div>
</div>
</div>
<blockquote style="border: none; margin: 0px 0px 0px 40px; padding: 0px; text-align: left;">
<div>
<div>
<div><div style="text-align: left;">Vf_LC_In</div></div>
</div>
</div>
<div>
<div>
<div><div style="text-align: left;">Vr_LC_In</div></div>
</div>
</div>
</blockquote>
<div>
<div>
<div>
<div style="text-align: center;"><br /></div>
<div>
Similarly, the forward and reflected voltages at the LC network output
are:
</div>
<div><br /></div>
</div>
</div>
</div>
<blockquote style="border: none; margin: 0px 0px 0px 40px; padding: 0px; text-align: left;">
<div>
<div>
<div><div style="text-align: left;">Vf_LC_Out</div></div>
</div>
</div>
<div>
<div>
<div><div style="text-align: left;">Vr_LC_Out</div></div>
</div>
</div>
</blockquote>
<div>
<div>
<div>
<div style="text-align: center;"><br /></div>
<div>
Drive is from a gated sine-wave. Gating allows display of signal
buildup and decay after the source has been turned on or off, which will
be used to find the values of Vr1, Vrs, Vf2, and Vfld.
</div>
<div><br /></div>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://1.bp.blogspot.com/-7kZQpLa2NbM/YVpbjIXlATI/AAAAAAAAMbM/ZL1gEq0kD6wK6WL3q-mai_60RcZk3mBlQCLcBGAsYHQ/s1430/211003%2BSimulink%2BModel.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="952" data-original-width="1430" height="213" src="https://1.bp.blogspot.com/-7kZQpLa2NbM/YVpbjIXlATI/AAAAAAAAMbM/ZL1gEq0kD6wK6WL3q-mai_60RcZk3mBlQCLcBGAsYHQ/s320/211003%2BSimulink%2BModel.png" width="320" /></a>
</div>
<br />
<div>
Note: the directional coupler model is described <a href="http://k6jca.blogspot.com/2021/09/does-source-impedance-affect-swr.html">here</a>.
</div>
<div><br /></div><div><br /></div><div><u>Simulation Goals:</u></div><div><br /></div><div>The goals of the simulations are to find the values of Vf1, Vr1, Vf2, Vfld, Vrld, Vrs, and Vload and verify that they match the calculated results.</div><div><br /></div><div>I will also examine a few of the voltages as they step up (or down) when the source voltage transitions from off to on or from on to off, and verify that the voltage amplitudes at each step are the predicted values.</div><div><br /></div><div>The Simulink model's Directional Couplers only measure <i>total</i> forward or reflected voltages on a transmission line. If two (or more) voltages are moving in the same direction on a line, the Directional Couplers cannot separate out the individual voltages.</div><div><br /></div><div>But, by gating the source on or off and having the second transmission line be long enough that it takes 4 cycles for a signal to travel round-trip from the output of the LC network, to the load, and then back again, the Directional Couplers can reveal single voltages during this transient state.</div><div><br /></div><div>For example, in steady-state, Vr of the Directional Coupler at the <i>input</i> of the LC network is the sum of Vr1 and Vrs. But for the four cycles after the source has just been turned on, Vr is equal to only Vr1, because the reflections that create Vrs have not yet arrived back from the load.</div><div><br /></div><div>And just after the source is turned off (after being ON for a while), Vr of the Directional Coupler at the LC input is now equal to only Vrs for the first four cycles (before it begins to decay), because Vr1 is now 0.</div><div><br /></div><div>The figure below demonstrates this:</div><div><br /></div><div><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-rOYrFyIETW4/YWQ5p03CtjI/AAAAAAAAMhc/Rl-ROt91HU0X8exOxkeg_WRv2OvwkPycACLcBGAsYHQ/s1071/211011%2BVr_LC_in.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="877" data-original-width="1071" height="262" src="https://1.bp.blogspot.com/-rOYrFyIETW4/YWQ5p03CtjI/AAAAAAAAMhc/Rl-ROt91HU0X8exOxkeg_WRv2OvwkPycACLcBGAsYHQ/s320/211011%2BVr_LC_in.png" width="320" /></a></div><br />Similarly, in steady-state, Vf of the Directional Coupler at the <i>output</i> of the LC network is the sum of Vf2 and Vfwd. And during the first four cycles after the source has been turned on, Vf is solely a function of Vf2. When the source is turned off, the first four cycles of Vf equals Vfld.</div><div><br /></div><div>The figure below demonstrates this:</div><div><br /></div><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-6OcB1aLEdIs/YWQ5d6wsuDI/AAAAAAAAMhU/Ckv5O7ELd-426FwMmSJ68l2Ntspe7_N8wCLcBGAsYHQ/s1071/211011%2BVf_LC_out.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="877" data-original-width="1071" height="262" src="https://1.bp.blogspot.com/-6OcB1aLEdIs/YWQ5d6wsuDI/AAAAAAAAMhU/Ckv5O7ELd-426FwMmSJ68l2Ntspe7_N8wCLcBGAsYHQ/s320/211011%2BVf_LC_out.png" width="320" /></a></div><br /><div><br /></div><div><u>Steady-state Voltages:</u></div>
<div><br /></div>
<div>Let's first look at the mathematically calculated values:</div>
<div><br /></div>
<div>
The voltages shown in the figure, below, represent the the system's
steady-state voltages (after the reflection build-ups have essentially
finished). These values were calculated using the equations
developed earlier in this post.
</div>
<div><br /></div>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://1.bp.blogspot.com/-YPZMdosZgBs/YVtpMyn_OlI/AAAAAAAAMcM/uSUVOwVrRdky2QR_F--me4lOLiJbOEOewCLcBGAsYHQ/s1051/211004%2BSystem%2BVoltages%2Bin%2BSteady-state.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="911" data-original-width="1051" height="277" src="https://1.bp.blogspot.com/-YPZMdosZgBs/YVtpMyn_OlI/AAAAAAAAMcM/uSUVOwVrRdky2QR_F--me4lOLiJbOEOewCLcBGAsYHQ/s320/211004%2BSystem%2BVoltages%2Bin%2BSteady-state.png" width="320" /></a>
</div>
<div><br /></div>
<div>As you can see, the calculated steady-state voltages are:</div>
<div><br /></div>
</div>
</div>
</div>
<blockquote style="border: none; margin: 0px 0px 0px 40px; padding: 0px; text-align: left;">
<div>
<div>
<div><div>Vf_LC_In = Vf1 = 70.71 V</div></div>
</div>
</div>
<div>
<div>
<div><div>Vr_LC_In = 0V</div></div>
</div>
</div>
<div>
<div>
<div><div>Vf_LC_Out = Vf2 + Vfld = 81.65 V</div></div>
</div>
</div>
<div>
<div>
<div><div>Vr_LC_Out = Vrld = 40.82 V</div></div>
</div>
</div>
<div>
<div>
<div><div>Vld = 122.5 V </div></div>
</div>
</div>
</blockquote>
<p>Therefore, we know that:</p>
<blockquote style="border: none; margin: 0px 0px 0px 40px; padding: 0px; text-align: left;">
<div>
<div><div>Vf1 = 70.71 volts</div></div>
</div>
<div>
<div><div>Vrld = 40.82 volts</div></div>
</div>
</blockquote>
<div>
<div>
<div><br /></div>
<div>
Let's compare these calculated voltages with the results of the
simulation.
</div>
<div><br /></div>
<div>
<span style="color: #2b00fe;">First, though, an important note regarding the voltages shown in the
simulation. The simulated voltages are peak amplitudes, not RMS
values. I chose to use peak amplitudes because these are easily
found using the cursors in the simulation display windows.</span>
</div>
<div>
<span style="color: #2b00fe;"><br /></span>
</div>
<div>
<span style="color: #2b00fe;">Thus, the simulation source voltage Vs is 141.4 Vpeak, rather than
141.4 VRMS, and the cursor readouts will show the same numerical value
as was calculated mathematically using RMS voltages, allowing quick
comparisons between calculated and simulated results.</span>
</div>
<div>
<span style="color: #2b00fe;"><br /></span>
</div>
<div>
<span style="color: #2b00fe;">I will compare voltage magnitudes, only.</span>
</div>
<div><br /></div>
<div>Simulated Steady-state results:</div>
<div><br /></div>
First: the steady-state forward and reflected voltages at the LC
network input. These voltages are:
</div>
<div><br /></div>
</div>
<blockquote style="border: none; margin: 0px 0px 0px 40px; padding: 0px; text-align: left;">
<div>
<div><div style="text-align: left;">Vf_LC_In</div></div>
</div>
<div>
<div><div style="text-align: left;">Vr_LC_In</div></div>
</div>
</blockquote>
<div>
<div>
<div>
<div style="text-align: left;"><br /></div>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://1.bp.blogspot.com/-zdXOyN9aKVw/YVtpgj2KDmI/AAAAAAAAMcY/J_zd_E5Ixi0zJvqlB9OKjXiXbIfJULlrQCLcBGAsYHQ/s1071/211004%2BSimulated%2BWaveforms%252C%2BSteady-State%252C%2BVf%2Band%2BVr%2BLC%2Bin.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="877" data-original-width="1071" height="262" src="https://1.bp.blogspot.com/-zdXOyN9aKVw/YVtpgj2KDmI/AAAAAAAAMcY/J_zd_E5Ixi0zJvqlB9OKjXiXbIfJULlrQCLcBGAsYHQ/s320/211004%2BSimulated%2BWaveforms%252C%2BSteady-State%252C%2BVf%2Band%2BVr%2BLC%2Bin.png" width="320" /></a>
</div>
<br />
<div>Vf_LC_In (i.e. Vf1) is 70.7 volts, and Vr_LC_In
is 0 volts -- the same amplitudes as were calculated mathematically.
</div>
<div><br /></div>
<div>
Next, let's look at the steady-state forward and reflected voltages at
the output of the LC matching network. These voltages are:
</div>
<div><br /></div>
</div>
</div>
</div>
<blockquote style="border: none; margin: 0px 0px 0px 40px; padding: 0px; text-align: left;">
<div>
<div>
<div><div style="text-align: left;">Vf_LC_Out</div></div>
</div>
</div>
<div>
<div>
<div><div style="text-align: left;">Vr_LC_Out</div></div>
</div>
</div>
</blockquote>
<div>
<div>
<div>
<div><br /></div>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://1.bp.blogspot.com/-_VhS54hdHNk/YVt7PTOkpcI/AAAAAAAAMd0/l94l059cidE4cYxhq6_yf3jvG4MEA-UXACLcBGAsYHQ/s1071/211004%2BSimulated%2BWaveforms%252C%2BSteady-State%252C%2BVf%2Band%2BVr%2BLC%2Bout.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="877" data-original-width="1071" height="262" src="https://1.bp.blogspot.com/-_VhS54hdHNk/YVt7PTOkpcI/AAAAAAAAMd0/l94l059cidE4cYxhq6_yf3jvG4MEA-UXACLcBGAsYHQ/s320/211004%2BSimulated%2BWaveforms%252C%2BSteady-State%252C%2BVf%2Band%2BVr%2BLC%2Bout.png" width="320" /></a>
</div>
<br />
<div>
<div>Vf_LC_Out is 81.6 volts, and Vr_LC_Out (i.e.
Vrld) is 40.8 volts -- again, the same amplitudes as were calculated
mathematically.
</div>
<div><br /></div>
</div>
<div>
And finally, let's look at the steady-state voltage Vload -- the voltage
across the 150 ohm load:
</div>
<div><br /></div>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://1.bp.blogspot.com/-IeUPUDnMIME/YVtp50IexeI/AAAAAAAAMcw/tXa4TrNYIqYhOfhy_SqVb5z9WXzHMJysACLcBGAsYHQ/s1074/211004%2BSimulated%2BWaveforms%252C%2BSteady-State%252C%2BVload.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="880" data-original-width="1074" height="262" src="https://1.bp.blogspot.com/-IeUPUDnMIME/YVtp50IexeI/AAAAAAAAMcw/tXa4TrNYIqYhOfhy_SqVb5z9WXzHMJysACLcBGAsYHQ/s320/211004%2BSimulated%2BWaveforms%252C%2BSteady-State%252C%2BVload.png" width="320" /></a>
</div>
<br />
<div>
In steady state the load voltage is 122 volts -- essentially the same as
the calculated Vld voltage of 122.5 volts.
</div>
<div><br /></div>
<div>
Therefore, for Vrld, Vf1, and the four directional-coupler voltages,
steady-state simulated voltages match the calculated voltages.
</div>
<div><br /></div>
<div><br /></div>
<div><u>Voltages at Startup, just after Vs has been gated ON:</u></div>
<div><br /></div>
<div>
<div>As stated earlier, to find the steady-state values of Vf2 and Vr1 we can examine the
voltages of the two directional couplers at startup, after Vs has been
gated ON, and just before reflections from the load have arrived back
at the LC network's output port.
</div>
<div><br /></div>
<div>
In other words, we can measure Vf2 and Vr1 during the first 4 cycles
of the source voltage Vs being on.
</div>
<div><br /></div>
<div>
The figure, below, summarizes the these voltages and their calculated
startup amplitudes.
</div>
</div>
<div><br /></div>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://1.bp.blogspot.com/-JF5KLOZAocw/YVtoZ1RTiXI/AAAAAAAAMbw/O4sNfI3yXS0QdPfswWGcQ2grDeRynaZRACLcBGAsYHQ/s1067/211004%2BSystem%2Bvoltages%2Bin%2Bstartup.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="911" data-original-width="1067" height="273" src="https://1.bp.blogspot.com/-JF5KLOZAocw/YVtoZ1RTiXI/AAAAAAAAMbw/O4sNfI3yXS0QdPfswWGcQ2grDeRynaZRACLcBGAsYHQ/s320/211004%2BSystem%2Bvoltages%2Bin%2Bstartup.png" width="320" /></a>
</div>
<div><br /></div>
<div>
The calculated startup voltages, at the first step of their buildup (or decay) are:
</div>
<div><br /></div>
<div>
<blockquote style="border: none; margin: 0px 0px 0px 40px; padding: 0px;">
<div>Vf_LC_In = Vf1 = 70.71 V</div>
<div>Vr_LC_In = Vr1 = 35.35 V</div>
<div>Vf_LC_Out = Vf2 = 61.23 V</div>
<div>Vr_LC_Out = 30.62 V</div>
<div>Vld = 91.85 V</div>
</blockquote>
<p>Thus, because during these 4 cycles Vr_LC_In = Vr1, and Vf_LC_Out = Vf2, we know that:</p>
</div>
</div>
</div>
</div>
<blockquote style="border: none; margin: 0px 0px 0px 40px; padding: 0px; text-align: left;">
<div>
<div>
<div><div>Vr1 = 35.35 V</div></div></div></div>
<div>
<div>
<div><div>Vf2 = 61.23 V</div></div>
</div>
</div>
</blockquote>
<div>
<div>
<div>
<div><br /></div>
<div><u>Simulated Startup Results:</u></div>
<div><br /></div>
<div>
First, let's examine the startup of the forward and reflected voltages
at the <i>input</i> of the LC network:
</div>
<div><br /></div>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://1.bp.blogspot.com/-_wH8CjMctl4/YVt_O_NoAHI/AAAAAAAAMeA/RvXjD0K8uUk3N7PyyG9X_zWiWY-g3tSkQCLcBGAsYHQ/s1071/211004%2BSimulated%2BWaveforms%252C%2BStart-up%252C%2BVf%2BVr%2BLC%2Bin.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="877" data-original-width="1071" height="262" src="https://1.bp.blogspot.com/-_wH8CjMctl4/YVt_O_NoAHI/AAAAAAAAMeA/RvXjD0K8uUk3N7PyyG9X_zWiWY-g3tSkQCLcBGAsYHQ/s320/211004%2BSimulated%2BWaveforms%252C%2BStart-up%252C%2BVf%2BVr%2BLC%2Bin.png" width="320" /></a>
</div>
<br />
<div>
Note that Vf_LC_In (i.e. Vf1) always equals 70.7 volts. In other
words, its initial voltage is the same as its steady-state voltage (it
neither builds nor decays).
</div>
<div><br /></div>
<div>
Vr1 equals the first four cycles of Vr_LC_In, and it measures to
be 35.32 volts.</div>
<div><br /></div>
<div>
Note that the measured voltage for Vr1 is the same as its calculated
amplitude.
</div>
<div><br /></div>
<div>
After these first four cycles Vr_LC_In drops to 8.83 volts due to
the reflection from the <i>load</i> mismatch (traveling backwards
towards the source) finally passing through the LC network from output
port to input port, and destructively summing with the Vr1 reflection
that is also traveling towards the source.
</div>
<div><br /></div>
<div>For every additional four cycles Vr_LC_In will continue to step down,
converging on its steady-state value of 0 volts.
</div>
<div><br /></div>
<div>
The mathematical calculation for this summation, over time (as more
reflections arrive back from the load) is shown as a running sum, below,
on the left-hand side of the figure:
</div>
<div><br /></div>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://1.bp.blogspot.com/-moeUSNToQ08/YVuCSR0pfJI/AAAAAAAAMeI/CgeBa-RoRa0p_otzf4ozVqXZay64zkDfQCLcBGAsYHQ/s1035/211004%2Bcalculated%2Brunning%2Bsum%2Bfor%2Bvr_LC_in.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="889" data-original-width="1035" height="275" src="https://1.bp.blogspot.com/-moeUSNToQ08/YVuCSR0pfJI/AAAAAAAAMeI/CgeBa-RoRa0p_otzf4ozVqXZay64zkDfQCLcBGAsYHQ/s320/211004%2Bcalculated%2Brunning%2Bsum%2Bfor%2Bvr_LC_in.png" width="320" /></a>
</div>
<br />
<div>
Next, let's compare the simulated startup results for the forward and
reflected voltages at the output of the LC network with their calculated
amplitudes.
</div>
<div><br /></div>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://1.bp.blogspot.com/-J8keloQjWV0/YVuC2vHDaWI/AAAAAAAAMeQ/47ZfLM5Kc2wrHhatA6AeB7KKB42U3pAUwCLcBGAsYHQ/s1071/211004%2BSimulated%2BWaveforms%252C%2BStart-up%252C%2BVf%2BVr%2BLC%2Bout.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="877" data-original-width="1071" height="262" src="https://1.bp.blogspot.com/-J8keloQjWV0/YVuC2vHDaWI/AAAAAAAAMeQ/47ZfLM5Kc2wrHhatA6AeB7KKB42U3pAUwCLcBGAsYHQ/s320/211004%2BSimulated%2BWaveforms%252C%2BStart-up%252C%2BVf%2BVr%2BLC%2Bout.png" width="320" /></a>
</div>
<div><br /></div>
Vf2 equals the first four cycles of Vf_LC_Out after Vs is gated
on, and its amplitude is 61.2 volts.</div>
<div><br /></div>
<div>
Note that the first four cycles of the other directional coupler voltage,
Vr_LC_Out, equals the first term of the Vrld infinite series as it arrives
at the output port of the LC network from the load. This amplitude
is 30.6 volts. Again, this amplitude is the same as the magnitude of the calculated value of, i.e. | 17.6768 - j25 |.
</div>
<div><br /></div>
<div>
Vrld's amplitude will continue to step up every 4 cycles, converging on
its steady-state value of 40.8 volts.
</div>
<div><br /></div>
<div>
Now let's check Vload at startup:<br />
<div><br /></div>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://1.bp.blogspot.com/-smKG6KX_dY4/YVtpB_cSZvI/AAAAAAAAMcA/E22afcSGiTs6aXV2b5PPDWruXCtmnaa1ACLcBGAsYHQ/s1071/211004%2BSimulated%2BWaveforms%252C%2BStart-up%252C%2BVload.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="877" data-original-width="1071" height="262" src="https://1.bp.blogspot.com/-smKG6KX_dY4/YVtpB_cSZvI/AAAAAAAAMcA/E22afcSGiTs6aXV2b5PPDWruXCtmnaa1ACLcBGAsYHQ/s320/211004%2BSimulated%2BWaveforms%252C%2BStart-up%252C%2BVload.png" width="320" /></a>
</div>
<div><br /></div>
The simulated Vload voltage (starting 2 cycles after Vs is gated on) is
91.8 V for four cycles. This matches the calculated voltage of 91.8
volts.
</div>
<div><br /></div>
<div>
The amplitude of Vload then steps up to 115 volts for the next four
cycles.
</div>
<div><br /></div>
<div>
It will continue stepping up every 4 cycles, converging to its
steady-state value of 122 volts.
</div>
<div><br /></div>
<div>
The calculated Vload amplitudes for the first 3 four-cycle voltage
amplitudes are shown below, on the right-hand side of the figure:
</div>
<div><br /></div>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://1.bp.blogspot.com/-MJ37RNqPWCI/YVuFaIjhF_I/AAAAAAAAMeY/cTNzCmkGVWMvudJAekmXeWwCwRjRxyEOACLcBGAsYHQ/s1035/211004%2Bcalculated%2Brunning%2Bsum%2Bfor%2Bvload.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="889" data-original-width="1035" height="275" src="https://1.bp.blogspot.com/-MJ37RNqPWCI/YVuFaIjhF_I/AAAAAAAAMeY/cTNzCmkGVWMvudJAekmXeWwCwRjRxyEOACLcBGAsYHQ/s320/211004%2Bcalculated%2Brunning%2Bsum%2Bfor%2Bvload.png" width="320" /></a>
</div>
<br />
<div>
You can see the the simulated amplitudes match the calculated amplitudes.
<div><br /></div>
<div>
<div>
Therefore, startup simulated voltages match the calculated voltages.
</div>
<div><br /></div>
</div>
<div><br /></div>
<div><u>Voltages at Shutdown, just after Vs has been gated OFF:</u></div>
<div><br /></div>
<div>
<div>
Finally, the steady-state values of Vrs and Vfld can be found by
examining the directional coupler voltages during the first four
cycles just after the voltage source Vs has been gated OFF (i.e. there
no longer is a drive voltage going to the input of the LC network and
thus there no longer is a Vf1, Vr1, or Vf2).
</div>
<div><br /></div>
<div>Calculated results will be compared with simulation results.</div>
<div><br /></div>
<div>
The figure, below, summarizes the these voltages and their calculated
startup amplitudes.
</div>
<div><br /></div>
</div>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://1.bp.blogspot.com/-d-fT8AYJOMs/YVtsSBgXsUI/AAAAAAAAMc4/MRtX8Xd0qwEA49QrmHVPMIVlSeV1xTJIACLcBGAsYHQ/s1067/211004%2BSystem%2Bvoltages%2Bin%2Bshutdown.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="917" data-original-width="1067" height="275" src="https://1.bp.blogspot.com/-d-fT8AYJOMs/YVtsSBgXsUI/AAAAAAAAMc4/MRtX8Xd0qwEA49QrmHVPMIVlSeV1xTJIACLcBGAsYHQ/s320/211004%2BSystem%2Bvoltages%2Bin%2Bshutdown.png" width="320" /></a>
</div>
<div><br /></div>
Note that Vfld and Vrld, the "continually echoing" forward and reflected
voltages on the transmission line, continue to ping-pong back and forth
between the two ends of the transmission line even with Vs off, but with
each bounce off of an impedance discontinuity their amplitudes diminish.
</div>
<div><br /></div>
<div>Let's look at the first step of Vload's decay.</div>
<div><br /></div>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://1.bp.blogspot.com/-5I3w3VSoh0U/YVttTgYpW1I/AAAAAAAAMdg/nZNEPDxOgsYhcDhCt8clfuTz8w9f_P6cQCLcBGAsYHQ/s1071/211004%2BSimulated%2BWaveforms%252C%2BShut-down%252C%2BVload.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="877" data-original-width="1071" height="262" src="https://1.bp.blogspot.com/-5I3w3VSoh0U/YVttTgYpW1I/AAAAAAAAMdg/nZNEPDxOgsYhcDhCt8clfuTz8w9f_P6cQCLcBGAsYHQ/s320/211004%2BSimulated%2BWaveforms%252C%2BShut-down%252C%2BVload.png" width="320" /></a>
</div>
<br />
<div>
To calculate the first decayed step of Vload, recognize that the
contribution to Vld from the source (i.e. Vf2*T<sub>ld</sub>) has just
disappeared with Vs being gated off.
</div>
<div><br /></div>
<div>
Therefore, simply subtract Vf2*T<sub>ld</sub> (the steady-state
complex value, not zero) from the steady-state complex value of Vld.
This first "decayed" amplitude of Vld calculates to be 30.6 V, which
matches the measurement, above.
</div>
<div><br /></div>
<div><br /></div>
<div>Next, let's measure Vrs.</div>
<div><br /></div>
<div>
<div>
To do this, let's examine reflected voltage at the LC network input,
Vr_LC_In, when Vs is gated off:
</div>
<div><br /></div>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://1.bp.blogspot.com/-Z85BAgphI7U/YVts7uZjJUI/AAAAAAAAMdU/I91aAilAjWUoAaF29zQ0PtM5r8AMHBCkgCLcBGAsYHQ/s1069/211004%2BSimulated%2BWaveforms%252C%2BShut-down%252C%2BVf%2BVr%2BLC%2Bin.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="875" data-original-width="1069" height="262" src="https://1.bp.blogspot.com/-Z85BAgphI7U/YVts7uZjJUI/AAAAAAAAMdU/I91aAilAjWUoAaF29zQ0PtM5r8AMHBCkgCLcBGAsYHQ/s320/211004%2BSimulated%2BWaveforms%252C%2BShut-down%252C%2BVf%2BVr%2BLC%2Bin.png" width="320" /></a>
</div>
<br />
<div>First, note that Vf_LC_In, and thus Vf1, is 0 immediately after Vs is gated
off.
</div>
<div><br /></div>
<div>
With Vf1 equal to 0, Vr1 is also 0, and thus the first 4 cycles of
Vr_LC_In are Vrs, which measures to be 35.3 V, which is the same as its
calculated value.</div>
<div><br /></div>
<div><br /></div>
<div>
<div>
Finally, let's measure Vfld using the Vf_LC_Out waveform, below:
</div>
<div><br /></div>
</div>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://1.bp.blogspot.com/-H11aONbSyeQ/YVtsjhyGXZI/AAAAAAAAMdE/elCguVoH0O4cb7hKBdXL_64y2rcLvy-qwCLcBGAsYHQ/s1071/211004%2BSimulated%2BWaveforms%252C%2BShut-down%252C%2BVf%2BLC%2Bout.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="877" data-original-width="1071" height="262" src="https://1.bp.blogspot.com/-H11aONbSyeQ/YVtsjhyGXZI/AAAAAAAAMdE/elCguVoH0O4cb7hKBdXL_64y2rcLvy-qwCLcBGAsYHQ/s320/211004%2BSimulated%2BWaveforms%252C%2BShut-down%252C%2BVf%2BLC%2Bout.png" width="320" /></a>
</div>
<div><br /></div>
</div>
<div>
<div>
With Vs now 0, Vf2 is also 0, and therefore Vf_LC_Out equals
Vfld, which measures to be 20.4 volts, equaling the
calculated value of Vfld.</div>
<div><br /></div>
<div>Finally, a quick look at Vr_LC_Out, just to show the decay of this
voltage when Vs is gated off.</div>
</div>
<div><br /></div>
<div>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://1.bp.blogspot.com/-nT0rUFDNlE8/YVttG5ZQl_I/AAAAAAAAMdY/qPPrd3KIO5I0sZGmuAQFuj8XJ-1dDIhlwCLcBGAsYHQ/s1071/211004%2BSimulated%2BWaveforms%252C%2BShut-down%252C%2BVr%2BLC%2Bout.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="877" data-original-width="1071" height="262" src="https://1.bp.blogspot.com/-nT0rUFDNlE8/YVttG5ZQl_I/AAAAAAAAMdY/qPPrd3KIO5I0sZGmuAQFuj8XJ-1dDIhlwCLcBGAsYHQ/s320/211004%2BSimulated%2BWaveforms%252C%2BShut-down%252C%2BVr%2BLC%2Bout.png" width="320" /></a>
</div>
<div><br /></div>
<div><br /></div>
<div>
<div>
Thus, the simulated shutdown voltages match the calculated voltages.
</div>
<div><br /></div><div><br /></div><div><b><u>Angle between Vr1 and Vrs:</u></b></div>
<div><br /></div><div>The mathematics predicts that, in steady-state, the amplitudes of the two voltage waves traveling back to the source, Vr1 and Vrs, will be equal, and they will be 180 degrees out of phase.</div><div><br /></div><div>The simulation below shows that this is true:</div><div><br /></div><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-r0ztqToTc8Y/YWOboDhYpAI/AAAAAAAAMhI/Vv5bnRBwfrEDX6vlpBo2jhYzWOYW51KMwCLcBGAsYHQ/s1851/211010%2Bsimulink%2Bmeasured%2Bangle%2Bbetween%2BVr1%2Band%2BVrs.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="871" data-original-width="1851" height="151" src="https://1.bp.blogspot.com/-r0ztqToTc8Y/YWOboDhYpAI/AAAAAAAAMhI/Vv5bnRBwfrEDX6vlpBo2jhYzWOYW51KMwCLcBGAsYHQ/s320/211010%2Bsimulink%2Bmeasured%2Bangle%2Bbetween%2BVr1%2Band%2BVrs.png" width="320" /></a></div><div><br /></div>Note that these two measurements are of the steady-state values of Vr1 and Vrs, even though they are taken just after V_source turns on or turns off.</div><div><br /><div><div>This method of cancellation is an important conclusion, because it validates Best's description of the steady-state wave mechanics at the input of the T network (and in my case, the input of the LC network). The interaction of Vr1 and Vrs, <i>over time</i>, creates a Reflection Coefficient of 0 (at the input port of the T or LC Network) in steady state, due to the cancellation of the steady-state values of Vr1 and Vrs.</div><div><br /></div><div>In other words, if there is no reflected voltage, as measured by a directional coupler, returning from the input of the T (or LC) network (i.e. the reflected voltage is 0 volts), then the port's Reflection Coefficient must be 0.</div><div><br /></div><div><br /></div></div>
<div>
<b><u>Simulation Summary:</u></b>
</div>
<div><br /></div>
<div>To summarize the simulation results: the simulated voltages match the calculated voltages.</div><div><br /></div><div>I've
included in the table, below, the simulation "capture" (that is, which of the three simulation states (startup, steady-state, or shutdown) and which directional coupler voltage) to identify where each simulation of a calculated voltage can be found:
</div>
<div><br /></div>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://1.bp.blogspot.com/--22afgDg-Xo/YVywzKTSF8I/AAAAAAAAMfw/YApyhS1riXk12v0CallPpVdbIIGLfuPdQCLcBGAsYHQ/s988/211005%2BComparing%2Bcalculated%2Band%2Bsimulated%2Bvoltages.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="397" data-original-width="988" height="129" src="https://1.bp.blogspot.com/--22afgDg-Xo/YVywzKTSF8I/AAAAAAAAMfw/YApyhS1riXk12v0CallPpVdbIIGLfuPdQCLcBGAsYHQ/s320/211005%2BComparing%2Bcalculated%2Band%2Bsimulated%2Bvoltages.png" width="320" /></a>
</div>
<div><br /></div>
(Note that this table does not include the simulated Load voltages nor
other voltages not described by Best in his analysis of his T-network,
although these voltages were calculated, verified via simulation, and
described in the simulation section, above).
</div>
<div>
<br />
<div><br /></div>
</div>
<div>
<b><u>Conclusions:</u></b>
</div>
<div><br /></div>
<div>
I have derived a set of mathematical equations, using Transmission and
Reflection Coefficients representing the LC network and the load,
to calculate forward and reflected voltages at the input and output
ports of an LC impedance matching network that transforms 150 ohms attached
to its output to be 50 ohms looking into its input port.
</div>
<div><br /></div>
<div>
These equations are used to calculate the values of these voltages from
startup, through the build-up transition, to steady state, and then,
when the source was gated off, the voltages changing from steady-state,
through the decay process, to zero-state.
</div>
<div><br /></div>
<div>
The amplitudes of these voltages match the amplitudes of the voltages
calculated by Dr. Best for his T-network example in part 3 of his
three-part series.
</div><div><br /></div><div>And these amplitudes are verified via Simulink simulations.</div>
<div><br /></div>
<div>Note that when the voltage source is first turned on, the source's voltage (after it has entered the first transmission line) is reflected by the impedance mismatch at the LC network's input port. This reflected voltage, Vr1, is not 0.</div><div><br /></div><div>But over time, the voltage traveling back to the source from the LC's input port becomes smaller. This decrease is due Vrld (representing the collection of past rearward traveling reflections from the load), after it passes through the LC network from output port to input port and becomes Vrs, interacting with the LC input port's reflection of the source voltage and cancelling, to some extent, Vr1.</div><div><br /></div><div>As Vrld increases over time, so does Vrs, increasing the amount of Vr1 being cancelled. And so, over time, the total voltage returning to the source from the input of the LC network converges to 0, the steady-state value.</div><div><br /></div><div>This growth in cancellation is exactly predicted by the description of the wave mechanics and the equations developed above.</div><div><br /></div><div>One important conclusion verified by the simulations is that Vr1 and Vrs, in steady-state, have equal amplitudes but are 180 degrees out of phase, so that when they are added, they cancel, and there no longer is a reflected wave traveling back from the LC network input. </div><div><br /></div><div>Thus, the L-network's input's effective Reflection Coefficient during steady-state is 0.</div><div><br /></div><div><br /></div>
<div>
In summary, the calculated voltages of Dr Best's
T-network system have been verified with Simulink simulations of my L-network system, validating Dr. Best's conclusions regarding the wave
mechanics of a system with an impedance-matching network.
</div>
<div><br /></div><div><br /></div><div><b><u>Other Comments:</u></b></div><div><br /></div><div><u>1. Match Independent from Source Impedance:</u></div><div><br /></div><div><div><div style="text-align: center;"><div style="text-align: left;">In steady-state, the impedance looking into the LC network's input port can be shown mathematically to be independent of source impedance. This conclusion is probably obvious, given that the input impedance of a two-port lumped-element matching network depends solely on the network's components and the load attached to its output port.</div><div style="text-align: left;"><br /></div><div style="text-align: left;">But the use of transmission and reflection coefficients to calculate transmission-line system voltages can obscure this fact, given that some of these transmission and reflection coefficients are functions of the source impedance (i.e. Za, as defined below).</div><div style="text-align: left;"><br /></div><div style="text-align: left;">Let's use the LC matching network, above, as an example, and let Zin equal the impedance looking into the LC network's input port.</div><div style="text-align: left;"><br /></div><div style="text-align: left;"><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-gIRcLi1L4lc/YWMx88hnPYI/AAAAAAAAMgw/LiyZvH08Q40OkT6eUYp5pYcYUIUrM6jrgCLcBGAsYHQ/s956/211010%2BZin.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="514" data-original-width="956" height="172" src="https://1.bp.blogspot.com/-gIRcLi1L4lc/YWMx88hnPYI/AAAAAAAAMgw/LiyZvH08Q40OkT6eUYp5pYcYUIUrM6jrgCLcBGAsYHQ/s320/211010%2BZin.png" width="320" /></a></div><br />Let Za equal the impedance looking back (towards the source) from the input of the two-port network. In this example, to meet this definition I've defined both the source impedance and the characteristic impedance of the transmission line connecting source to LC network input to be Za.</div><div style="text-align: left;"><br /></div><div style="text-align: left;">From Best (part 1, eq. 15): </div><div style="text-align: left;"><br /></div></div></div><blockquote style="border: none; margin: 0px 0px 0px 40px; padding: 0px;"><div style="text-align: center;"><div style="text-align: left;">Zin = V(x)/I(x) = Za * (Vfwd + Vref) / (Vfwd - Vref)</div></div></blockquote><div><div style="text-align: center;"><div style="text-align: left;"><br /></div><div style="text-align: left;">(Note that I have used Za in lieu of Best's Zo because I've defined (above) the characteristic impedance of the transmission line at the LC network's input to be Za, and location 'x' on the first transmission line is defined to be at the end of the line, next to the two-port network's input port).</div><div style="text-align: left;"><br /></div><div style="text-align: left;">From earlier in this blogpost I know:</div><div style="text-align: left;"><br /></div></div></div><blockquote style="border: none; margin: 0px 0px 0px 40px; padding: 0px;"><div style="text-align: center;"><div style="text-align: left;">Vfwd = Vf1</div></div></blockquote><blockquote style="border: none; margin: 0px 0px 0px 40px; padding: 0px;"><div style="text-align: center;"><div style="text-align: left;">Vref = Vr1 + Vrs</div></div></blockquote><div><div style="text-align: center;"><div style="text-align: left;"><br /></div><div style="text-align: left;">where:</div><div style="text-align: left;"><br /></div></div></div><blockquote style="border: none; margin: 0px 0px 0px 40px; padding: 0px;"><div style="text-align: center;"><div style="text-align: left;">Vr1 = Γ<sub>1</sub> * Vf1 </div></div></blockquote><div><div style="text-align: center;"><div style="text-align: left;"><br /></div><div style="text-align: left;">and </div><div style="text-align: left;"><br /></div></div></div><blockquote style="border: none; margin: 0px 0px 0px 40px; padding: 0px;"><div style="text-align: center;"><div style="text-align: left;">Vrs = Τ<sub>12</sub> * Vrld</div></div></blockquote><blockquote style="border: none; margin: 0px 0px 0px 40px; padding: 0px;"><div style="text-align: center;"><div style="text-align: left;">= Τ<sub>12 </sub>* Γ<sub>ld</sub> * Τ<sub>21 </sub>* Vf1 / (1-Γ<sub>2</sub> * Γ<sub>ld</sub> )</div></div></blockquote><div><div style="text-align: center;"><div style="text-align: left;"><br /></div><div style="text-align: left;">(assuming e<sup>-2γl</sup> = 1, which it does for my LC network example)</div><div><br style="text-align: left;" /></div><div style="text-align: left;">Substituting the equations for Vr1 and Vrs into the equation for Vref , and substituting Vf1 for Vfwd, then substituting these new equations for Vref and Vfwd into the equation, above, for Zin, and cancelling out Vf1, I get:</div><div style="text-align: left;"><br /></div></div></div><blockquote style="border: none; margin: 0px 0px 0px 40px; padding: 0px;"><div style="text-align: center;"><div style="text-align: left;"><div>Zin = Za*(1+Γ<sub>1</sub>-Γ<sub>2</sub>*Γ<sub>ld</sub>-Γ<sub>1</sub>*Γ<sub>2</sub>*Γ<sub>ld</sub>+Τ<sub>12</sub>*Τ<sub>21</sub>*Γ<sub>ld</sub>)/(1-Γ<sub>1</sub>-Γ<sub>2</sub>*Γ<sub>ld</sub>+Γ<sub>1</sub>*Γ<sub>2</sub>*Γ<sub>ld</sub>-Τ<sub>12</sub>*Τ<sub>21</sub>*Γ<sub>ld</sub>)</div></div></div></blockquote><div><div style="text-align: center;"><div style="text-align: left;"><div><br /></div><div>The goal is to reduce this large equation so that, hopefully, the Za term disappears (i.e. Zin becomes independent of source impedance). </div><div><br /></div><div>To do so, however, requires that I express the Transmission and Reflection Coefficients in terms of their underlying impedances. The equation becomes even more complicated!</div><div><br /></div><div>Fortunately, MATLAB's Symbolic Math Toolbox comes to the rescue. Using it, the equation for Zin (when the load is attached to the output port through a lossless transmission line two wavelengths long) is determined to be:</div></div><div style="text-align: left;"><div><br /></div></div></div></div><blockquote style="border: none; margin: 0px 0px 0px 40px; padding: 0px;"><div style="text-align: center;"><div style="text-align: left;">Zin = Zc + Zl + (Zc^2*(Γ<sub>ld</sub> - 1)) / (Zc - Γ<sub>ld</sub>*Zc + Zb*(Γ<sub>ld</sub> + 1))</div></div></blockquote><p>where Zc is the capacitor's impedance, Zl is the inductor's impedance, Zb is the characteristic impedance of the transmission line between the LC network's output and the load, and Γ<sub>ld</sub> is the reflection coefficient of the load. (For other transmission line lengths, first multiply Γ<sub>ld</sub> by the Propagation Constant, e<sup>-2γl</sup>, appropriate for that length before calculating Zin).</p><p>A more general form in terms of a series impedance (Zs) and a parallel impedance (Zp) with a load at the end of a transmission line of characteristic impedance Zb and length <i>l</i> (in wavelengths) is:</p><p style="text-align: center;">Zin_ZsZp = Zp + Zs + (Zp^2*(Γ<sub>ld</sub>e<sup>-2γl</sup> - 1))/(Zp - Γ<sub>ld</sub>e<sup>-2γl</sup>*Zp + Zb*(Γ<sub>ld</sub>e<sup>-2γl</sup> + 1))</p><div style="text-align: left;"><span style="color: #2b00fe;"><br /></span></div><div style="text-align: left;"><span style="color: #2b00fe;">Note that Zin is not dependent upon Za! </span></div><div><div style="text-align: center;"><div style="text-align: left;"><br /></div><div style="text-align: left;">A Simulink example demonstrates that source impedance does not affect the LC network's effective input impedance (i.e. looking into its input port) during steady-state. This effective impedance remains 50 ohms, resistive, and the reflected voltage at the LC network's input, as measured with a directional coupler <i>referenced to 50 ohms</i>, still goes to 0 during steady-state.</div></div></div></div><div><br /></div><div>Also, if the voltage source amplitude is adjusted so that 100 watts is dissipated by the load, the amplitudes of the total forward and total reflected voltages on the transmission line between the LC network and the load are the same as the matched source impedance example of this post.</div><div><br /></div><div>The simulation, below, demonstrates this independence from Source Impedance. Note that both the source impedance and the characteristic impedance of the transmission line connecting source to LC network input have been set to 25 ohms, and the source voltage adjusted so that the power dissipated at the load is 100 watts:</div><div><br /></div><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-e4yR3GmX320/YV8dss45zUI/AAAAAAAAMgY/UHtbp1Sl3eQH6pkoWr8jmE-9weuRWUuuACLcBGAsYHQ/s1430/211007%2B25%2Bohm%2Bsimulink.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="952" data-original-width="1430" height="213" src="https://1.bp.blogspot.com/-e4yR3GmX320/YV8dss45zUI/AAAAAAAAMgY/UHtbp1Sl3eQH6pkoWr8jmE-9weuRWUuuACLcBGAsYHQ/s320/211007%2B25%2Bohm%2Bsimulink.png" width="320" /></a></div><div><br /></div>Note that although there is now a mismatch between 25 ohm line and the input of the LC network (so that the actual SWR is no longer 1:1), the directional coupler, measuring at a single point and <i>referenced to 50 ohms</i>, in steady-state still sees a 50 ohm match at the input of the LC network.</div><div><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-NmWWx31yHmk/YV8dwuuvz0I/AAAAAAAAMgc/cYUoLRWlfkU4VJnwDNQcKFZb1C1tLT-DQCLcBGAsYHQ/s1071/211007%2B25%2Bohm%2Bsimulation%2Binput%2Bvf%2Band%2Bvr.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="877" data-original-width="1071" height="262" src="https://1.bp.blogspot.com/-NmWWx31yHmk/YV8dwuuvz0I/AAAAAAAAMgc/cYUoLRWlfkU4VJnwDNQcKFZb1C1tLT-DQCLcBGAsYHQ/s320/211007%2B25%2Bohm%2Bsimulation%2Binput%2Bvf%2Band%2Bvr.png" width="320" /></a></div><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-D5zlwb5JMlI/YV8d4GM6LGI/AAAAAAAAMgg/-CSQ2bR3Fj0UkL5R1S3FETCVjM4HRUnJgCLcBGAsYHQ/s1071/211007%2B25%2Bohm%2Bsimulation%2Boutput%2Bvf%2Band%2Bvr.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="877" data-original-width="1071" height="262" src="https://1.bp.blogspot.com/-D5zlwb5JMlI/YV8d4GM6LGI/AAAAAAAAMgg/-CSQ2bR3Fj0UkL5R1S3FETCVjM4HRUnJgCLcBGAsYHQ/s320/211007%2B25%2Bohm%2Bsimulation%2Boutput%2Bvf%2Band%2Bvr.png" width="320" /></a></div><br /><div>The Load Reflection and Transmission Coefficients have not changed:</div><div><div><br /></div><blockquote style="border: none; margin: 0px 0px 0px 40px; padding: 0px;"><div>Γ<sub>ld</sub> = 0.5 + j0</div><div>Τ<sub>ld</sub> = 1.5 + j0</div></blockquote><div style="text-align: center;"><br /></div></div><div style="text-align: left;">But the LC network's Reflection and Transmission Coefficients <i>have </i>changed:</div><div style="text-align: left;"><br /></div><div><blockquote style="border: none; margin: 0px 0px 0px 40px; padding: 0px;"><div><div>Γ<sub>1</sub> = 0.5285 + j0.3679</div></div><div><div>Τ<sub>21</sub> = 0.4878 - j0.9658</div></div><div><div>Γ<sub>2</sub> = 0.61 + j0.207</div></div><div><div style="text-align: center;"><div style="text-align: left;">Τ<sub>12</sub> = 0.2439 - j0.4829</div></div></div></blockquote><div><div style="text-align: center;"><div><br /></div><div style="text-align: left;">Here are the magnitudes of these six Reflection and Transmission Coefficients:</div><div style="text-align: left;"><br /></div></div></div><blockquote style="border: none; margin: 0px 0px 0px 40px; padding: 0px;"><div><div style="text-align: center;"><div style="text-align: left;"><div style="text-align: center;"><div style="text-align: left;">|Γ<sub>ld</sub>|= 0.5</div></div></div><div style="text-align: left;"><div style="text-align: center;"></div></div><div style="text-align: left;"><div style="text-align: center;"><div style="text-align: left;">|Γ<sub>1</sub>|= 0.644</div></div></div><div style="text-align: left;"><div style="text-align: center;"><div style="text-align: left;">|Γ<sub>2</sub>|= 0.644</div></div></div><div style="text-align: left;"><div style="text-align: center;"></div></div><div style="text-align: left;"><div style="text-align: center;"><div style="text-align: left;">|Τ<span style="font-size: 13.3333px;">ld</span>| = 1.5</div></div></div><div style="text-align: left;"><div style="text-align: center;"></div></div><div style="text-align: left;"><div style="text-align: center;"><div style="text-align: left;">|Τ<sub>21</sub>| = 1.08</div></div></div><div style="text-align: left;"><div style="text-align: center;"></div></div><div style="text-align: left;">|Τ<sub>12</sub>| = 0.54</div></div></div></blockquote><div><div style="text-align: center;"><div></div></div></div><div><div style="text-align: center;"><div></div></div></div><div><div style="text-align: center;"><div></div></div></div><div><div style="text-align: center;"><div><div></div></div><div style="text-align: left;"><br /></div><div style="text-align: left;"><br /></div></div></div></div></div></div></div><div><div><div>
<div><br /></div>
<div>
<b><u>Simulink File:</u></b>
</div>
<div><br /></div>
<div>
The Simulink file for my LC Network matching example can be found here:
</div>
<div>
<a href="https://github.com/k6jca/Maxwell-Revisit">https://github.com/k6jca/Maxwell-Revisit</a><br />
</div>
<div><br /></div>
<div>
<b><u><br /></u></b>
</div>
<div>
<b><u>References:</u></b>
</div>
<div><br /></div>
<div>
Best, S., Wave Mechanics of Transmission Lines, Part 1: Equivalence of
Wave Reflection Analysis and the Transmission-line
Equation, <i>QEX</i>, Jan/Feb, 2001
</div>
<div><br /></div>
<div></div>
<div>
Best, S., Wave Mechanics of Transmission Lines, Part 2: Where Does
Reflected Power Go?, <i>QEX</i>, July/Aug, 2001
</div>
<div><br /></div>
<div></div>
<div>
Best, S., Wave Mechanics of Transmission Lines, Part 3: Power Delivery
and Impedance Matching, <i>QEX</i>, Nov/Dec, 2001
</div>
<div><br /></div>
<div>
<div>
Maxwell, W., A Tutorial Dispelling Certain Misconceptions Concerning
Wave Interference in Impedance Matching, <i>QEX</i>, July/Aug,
2004
</div>
<div><br /></div>
<div></div>
</div>
<div><br /></div>
<div></div>
<div>
<b style="text-align: center;"><u>Other Transmission-Line Posts:</u></b>
</div>
<div>
<div>
<p>
<a href="http://k6jca.blogspot.com/2021/02/antenna-tuners-transient-and-steady.html" style="text-align: center;">http://k6jca.blogspot.com/2021/02/antenna-tuners-transient-and-steady.html</a>. This post analyzes the transient and steady-state response
of a simple impedance matching system consisting of a wide-band
transformer. I calculate the system's impulse response and
find the time-domain response by convolving this impulse-response
with a stimulus signal.
</p>
<p>
<span style="text-align: center;"><a href="http://k6jca.blogspot.com/2021/02/the-quarter-wave-transformer-transient.html">http://k6jca.blogspot.com/2021/02/the-quarter-wave-transformer-transient.html</a>. <span style="text-align: left;"> This post analyzes the transient and steady-state
response of a Quarter-Wave Transformer impedance matching
device. I calculate the system's impulse response and find
the time-domain response by convolving this impulse-response
with a stimulus signal.</span></span>
</p>
<div>
<a href="http://k6jca.blogspot.com/2021/03/useful-swr-voltage-and-power-equations.html" style="text-align: center;">http://k6jca.blogspot.com/2021/03/useful-swr-voltage-and-power-equations.html</a>. This post lists (in an easily accessible location that I
can find!) some equations that I find useful
</div>
<p>
<span style="text-align: center;"><a href="http://k6jca.blogspot.com/2021/05/antenna-tuners-lumped-element-tuner.html">http://k6jca.blogspot.com/2021/05/antenna-tuners-lumped-element-tuner.html</a>. This post analyzes the transient and steady-state
reflections of a lumped-element tuner (i.e. the common antenna
tuner). I describe a method for making these calculations,
and I note that the tuner's match is independent of the source
impedance.</span>
</p>
<p>
<span style="text-align: center;"><a href="http://k6jca.blogspot.com/2021/05/lc-network-reflection-and-transmission.html">http://k6jca.blogspot.com/2021/05/lc-network-reflection-and-transmission.html</a>. This post describes how to calculate the "Transmission
Coefficient" through a lumped-element network (and also its
Reflection Coefficient) if it were inserted into a transmission
line. <br /></span>
</p>
<p>
<a href="http://k6jca.blogspot.com/2021/09/does-source-impedance-affect-swr.html">http://k6jca.blogspot.com/2021/09/does-source-impedance-affect-swr.html</a>. This post shows mathematically that source impedance does
not affect a transmission line's SWR. This conclusion is then
demonstrated with Simulink simulations.<br />
</p>
<p><a href="https://k6jca.blogspot.com/2021/10/revisiting-maxwells-tutorial-concerning.html">https://k6jca.blogspot.com/2021/10/revisiting-maxwells-tutorial-concerning.html</a> This posts revisits Walt Maxwell's 2004 QEX rebuttal of Steven Best's 2001 3-part series on Transmission Line Wave Mechanics. In this post I show simulation results which support Best's conclusions.</p><p><br /></p>
<p>
<b><u>Standard Caveat:</u></b>
</p>
<div>
I might have made a mistake in my designs, equations, schematics,
models, etc. If anything looks confusing or wrong to you, please
feel free to comment below or send me an email.<br /><br />Also, I
will note:<br /><br />This design and any associated information is
distributed in the hope that it will be useful, but WITHOUT ANY
WARRANTY; without even the implied warranty of MERCHANTABILITY or
FITNESS FOR A PARTICULAR PURPOSE.
</div><div><br /></div>
</div>
</div>
</div>
</div>
</div>
Jeffhttp://www.blogger.com/profile/04853314106806116765noreply@blogger.com0tag:blogger.com,1999:blog-2257458838727315711.post-43340853755332496742021-09-16T16:15:00.061-07:002023-12-23T17:07:29.641-08:00Does Source Impedance affect SWR?<p>
<span style="color: #2b00fe;">[This post is a response to the article, "SWR Dependence on Output
Impedance," by Wright, M., W6PAP, in the September-October, 2021, issue of
<i>QEX</i> magazine. Although that article focused on source impedance affecting the <i>measurement </i>of SWR (a conclusion I do not agree with), I thought I should verify mathematically for myself that a transmission line's SWR is truly independent of source impedance.]</span>
</p>
<p>
Does a source-impedance mismatch at the input of a transmission line affect
the transmission line's SWR?
</p>
<p>
The answer is no, it does not. Let's examine this conclusion both
mathematically and with Simulink simulations.
</p>
<p>
<u><b>Mathematical Derivation of SWR with Mismatched Load and Source
Impedances:</b></u>
</p>
<p>
Consider a transmission line of characteristic impedance Zo and a length of
<i>l. </i>Let the source impedance Zs and the load impedance Zld be
mismatched from Zo, with reflection coefficients Γs and Γld,
respectively:
</p>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://1.bp.blogspot.com/-l9UXC9waRtI/YUUg7CWp40I/AAAAAAAAMVQ/Uv2DsWXAnbYzVj2ZhYYKmitRosYGSi5vQCLcBGAsYHQ/s705/210916%2Bbasic%2Bsystem.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="418" data-original-width="705" height="190" src="https://1.bp.blogspot.com/-l9UXC9waRtI/YUUg7CWp40I/AAAAAAAAMVQ/Uv2DsWXAnbYzVj2ZhYYKmitRosYGSi5vQCLcBGAsYHQ/s320/210916%2Bbasic%2Bsystem.png" width="320" /></a>
</div>
<p>The Reflection Coefficients are:</p>
<p style="text-align: left;"></p>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://1.bp.blogspot.com/-lyXYsBRCQ8o/YUOy28ME5qI/AAAAAAAAMRk/5jVOvNXOPIcjqa5DYA3z_DfXprkoZ673gCLcBGAsYHQ/s506/210916%2BReflection%2BCoefficient%2Bequations.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="302" data-original-width="506" height="191" src="https://1.bp.blogspot.com/-lyXYsBRCQ8o/YUOy28ME5qI/AAAAAAAAMRk/5jVOvNXOPIcjqa5DYA3z_DfXprkoZ673gCLcBGAsYHQ/s320/210916%2BReflection%2BCoefficient%2Bequations.png" width="320" /></a>
</div>
<br />
<p>
Voltage Vfs is the voltage entering the transmission line from the source
Vs. Imagine that Vs (a sine-wave source) has been on long enough to
allow any transient conditions to have settled. In other words, the
transmission line voltages are in steady-state.
</p>
<p>
Now let's consider a point on the Vfs waveform just as it enters the
transmission line. Even though there may be other voltages that
represent reflections or re-reflections on the transmission line, from the
point of view of this single point on Vfs, it only sees the transmission
line's impedance Zo (per the
<a href="https://en.wikipedia.org/wiki/Superposition_principle">Principle of Superposition</a>).
</p>
<p>
Therefore, one can calculate Vfs using the equation of a simple voltage
divider:
</p>
<p style="text-align: center;">Vfs = Vs*(Zo/(Zo+Zs))</p>
<div>
This point on Vfs will travel down the transmission line until it reaches the
mismatch at the load end (at which point it will be reflected back to the
source (with a changed value)).
</div>
<div><br /></div>
<div>
Before I go any further with reflections, though, let me quickly introduce
transmission-line Propagation Constants. These will be useful for
calculating the values of the reflections.
</div>
<div><br /></div>
<div><u>Propagation Constants γ, α, and β:</u></div>
<div><br /></div>
<div>
If one <i><b>simultaneously</b></i> compares (at, say, time t = t<sub>0</sub> ) the
phase of the steady-state signal at the source-end of the transmission line to
the phase of the signal further down the line, there will be a phase
difference between the two, due to the time it took the earlier Vfs to travel
along the transmission line.</div>
<div><br /></div>
<div>
Similarly, if the transmission line is lossy, the amplitude of the voltage
will decrease (due to loss) as it travels along the transmission line,
compared to its original value at the source end.
</div>
<div><br /></div>
<div>
We can represent this attenuation and phase-shift with the term
e<sup>-γl</sup>. This term is the one-way attenuation and phase-shift on
a transmission line of length <i>l</i>. (Round-trip attenuation and
phase-shift is simply e<sup>-2γl</sup>).
</div>
<div><br /></div>
<div>
Note that γ = α + j*β, where α represents loss in nepers per unit-length, and
β represents phase-shift in radians per unit-length.
</div>
<div><br /></div>
<div>
Therefore, e<sup>-γl</sup> = e<sup>-(α+jβl)</sup> = e<sup>-αl </sup>
* e<sup>-j*βl</sup>. If there is loss (α not equal to 0), then
e<sup>-αl </sup> will be a positive real number whose value is less
than 1. And e<sup>-j*βl</sup> is simply a phasor of magnitude 1 and
angle of -β<i>l</i>
radians.
</div><div><br /></div><div>Note that there is a negative sign in front of β<i>l</i> because the signal at the further point represents an <i>earlier</i> version (in time) of Vf, given it being measured simultaneously with Vf at the near end. That is, the phase of Vf at the further point would be negative with respect to the phase at the near end.</div>
<div><br /></div>
<div>The figure below demonstrates their meaning:</div>
<div><br /></div>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://1.bp.blogspot.com/-YP76Yr3ugRU/YUZwZtv2NqI/AAAAAAAAMWY/AsjMRUyDJhcZmtaMVVpX6HI27dad3zy8wCLcBGAsYHQ/s824/210918%2Bpropagation%2Bconstants.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="647" data-original-width="824" height="251" src="https://1.bp.blogspot.com/-YP76Yr3ugRU/YUZwZtv2NqI/AAAAAAAAMWY/AsjMRUyDJhcZmtaMVVpX6HI27dad3zy8wCLcBGAsYHQ/s320/210918%2Bpropagation%2Bconstants.png" width="320" /></a>
</div>
<br />
<div>
(You can find more on these propagation constants here:
<a href="https://www.microwaves101.com/encyclopedias/propagation-constant">https://www.microwaves101.com/encyclopedias/propagation-constant</a>)
</div>
<div><br /></div>
<div><br /></div>
<div><u>Reflections on the Transmission Line:</u></div>
<div><br /></div>Because of the mismatched source and load impedances, this point on the Vfs waveform, when it arrives at the load end of the transmission line, will be reflected back towards the source by the load mismatch. When it arrives at the source end of the line, it will be reflected back to towards the load by the source mismatch. <div><br /></div><div>The result -- this point will bounce back and forth
between the two mismatched ends
<i>ad infinitum.</i>
<div><br /></div>
<div>
The figure below shows the change in Vfs from the time when it enters the
transmission line to when arrives at one end of the line and is reflected back
towards the other end, etc. You can see that each new amplitude and
phase is a function of only three factors: the two reflection coefficients (Γs, Γld), and the
one-way transmission line loss (e<sup>-αl</sup>).
</div>
<div>
<div><br /></div>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://1.bp.blogspot.com/-DFoIuaaLXcc/YUiPLLCF7NI/AAAAAAAAMWw/HaK0uym38BkTgTqhEAqmJuUPi4DJaW0EwCLcBGAsYHQ/s993/210917%2Bbuilding%2Breflections2.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="528" data-original-width="993" height="170" src="https://1.bp.blogspot.com/-DFoIuaaLXcc/YUiPLLCF7NI/AAAAAAAAMWw/HaK0uym38BkTgTqhEAqmJuUPi4DJaW0EwCLcBGAsYHQ/s320/210917%2Bbuilding%2Breflections2.png" width="320" /></a>
</div>
<br />
<div>(Note that transmission line loss and delay is always referenced with respect to the Vfs that is <i>currently</i> entering the transmission line at the source).</div><div><br /></div><div>In steady-state there will be large (approaching infinite) number of past reflections on the
transmission line (because both ends are mismatched), and each reflection
represents a version of the Vfs waveform that had entered the transmission
line at some time earlier than the current Vfs entering the line.
</div>
<div><br /></div>
<div>
If we were to take a "snapshot" in time of the state of the current Vfs entering the line and all of
its reflections (from previous versions of itself) on the line, we would see
that each reflection has a phase offset associated with it -- the
phase offset represents the amount of time each reflection has "been on the
line", compared to the current Vfs entering the line). This phase
offset, -βl, is incorporated into the exponential function for line loss (as a phasor), and the
exponential function becomes a combination of line loss and line delay, represented as e<sup>-γl</sup> , where γ = α + j*β.
</div>
<div><br /></div>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://1.bp.blogspot.com/-pKzsdCWl2w8/YUiQJOeTN-I/AAAAAAAAMW4/hPxMerdtxZE1tyh2Lrpj93uVwvrrIwiAgCLcBGAsYHQ/s1003/210917%2Bsnapshot%2Bof%2Breflections.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="668" data-original-width="1003" height="213" src="https://1.bp.blogspot.com/-pKzsdCWl2w8/YUiQJOeTN-I/AAAAAAAAMW4/hPxMerdtxZE1tyh2Lrpj93uVwvrrIwiAgCLcBGAsYHQ/s320/210917%2Bsnapshot%2Bof%2Breflections.png" width="320" /></a>
</div><div><br /></div>
(Note that phase = βl (i.e. radians per unit-length times length), and phase also equals ωt<sub>d</sub>, where ω is the frequency (in
radians per second) of Vfs and t<sub>d</sub> is time-delay (in seconds). Therefore, because ωt<sub>d</sub> = βl, the
transmission line's time delay from end to end is: t<sub>d</sub> = βl/ω.)</div><div>
<div><br /></div>
<div>
The Lattice Diagram, below, shows a continuation of these reflections:
</div>
<div>
<p></p>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://1.bp.blogspot.com/-XILKHVngjPA/YUOvdI0Qc9I/AAAAAAAAMRE/XdOngQSLPr0rS1vZihbx_AG6cDPBPxe2QCLcBGAsYHQ/s1038/Reflections.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="1038" data-original-width="740" height="320" src="https://1.bp.blogspot.com/-XILKHVngjPA/YUOvdI0Qc9I/AAAAAAAAMRE/XdOngQSLPr0rS1vZihbx_AG6cDPBPxe2QCLcBGAsYHQ/s320/Reflections.png" width="228" /></a>
</div>
<div><br /></div>
<div>Here's a zoomed-in view of the reflections:</div>
<div>
<p style="text-align: left;"></p>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://1.bp.blogspot.com/-XCnIgPEAq1M/YUPhKNVV8kI/AAAAAAAAMUg/0S0HYF6HzeI1qb2iqwp84zbd05PvuVlZgCLcBGAsYHQ/s633/210916%2BReflections%2Bzoomed.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="633" data-original-width="581" height="320" src="https://1.bp.blogspot.com/-XCnIgPEAq1M/YUPhKNVV8kI/AAAAAAAAMUg/0S0HYF6HzeI1qb2iqwp84zbd05PvuVlZgCLcBGAsYHQ/s320/210916%2BReflections%2Bzoomed.png" width="294" /></a>
</div>
<br />(The voltage values of the reflections in the lattice diagram,
above, are assumed to be measured at the
<i><u>source</u> end of the transmission line</i> (i.e. at the SWR
meter's position in the earlier figure). Also, if there is
some loss in the system (either in Zs, Zld, or within the transmission
line, itself), these reflections will decay towards 0 over time.)
</div>
<div><br /></div>
<div>
So, in steady state, the total voltage traveling from the source-end of
the transmission line to the load is the voltage Vfs (at a point in time,
say, t = t<sub>0</sub> ) summed with the infinite series of all of
the re-reflections that have just been reflected back towards the load by
the source mismatch at that same point in time (i.e. at time t =
t<sub>0</sub> ) .
</div>
<div><br /></div>
<div>
The sum of these voltages is Vf, which represents the total voltage of the
wave traveling from source to load.
</div>
<div><br /></div>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://1.bp.blogspot.com/-lgXTfNxzITc/YUY2IFdaM_I/AAAAAAAAMWI/PYKOB65OhJ4-aYqVSSoMU7J3Ze8Z3nuDwCLcBGAsYHQ/s633/210918%2Bfirst%2Bseries%2Bterms.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="633" data-original-width="581" height="320" src="https://1.bp.blogspot.com/-lgXTfNxzITc/YUY2IFdaM_I/AAAAAAAAMWI/PYKOB65OhJ4-aYqVSSoMU7J3Ze8Z3nuDwCLcBGAsYHQ/s320/210918%2Bfirst%2Bseries%2Bterms.png" width="294" /></a>
</div>
<br />
<div>This sum can be expressed as the following infinite series:</div>
<div>
<p></p>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://1.bp.blogspot.com/-c8Bq4sxySiM/YUOx1ZqAa_I/AAAAAAAAMRM/-_xWvWDjMMYIP-G_OYQOgkgepVtpKZ8MQCLcBGAsYHQ/s807/210916%2BVf%2Bseries.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="283" data-original-width="807" height="112" src="https://1.bp.blogspot.com/-c8Bq4sxySiM/YUOx1ZqAa_I/AAAAAAAAMRM/-_xWvWDjMMYIP-G_OYQOgkgepVtpKZ8MQCLcBGAsYHQ/s320/210916%2BVf%2Bseries.png" width="320" /></a>
</div>
<p style="text-align: left;"></p>
Similarly, the steady-state total voltage flowing from the load back to
the source, Vr, can be calculated by summing all of the reflections from
the load at that same point in time (i.e. t = t<sub>0</sub> ):
</div>
<div>
<p></p>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://1.bp.blogspot.com/-h-_a8rdb_1k/YUOyF0z0o0I/AAAAAAAAMRU/eEFPn1ERfvwvHkkWoABgSMY4NQ2b24iQQCLcBGAsYHQ/s747/210916%2BVr%2Bseries.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="262" data-original-width="747" height="112" src="https://1.bp.blogspot.com/-h-_a8rdb_1k/YUOyF0z0o0I/AAAAAAAAMRU/eEFPn1ERfvwvHkkWoABgSMY4NQ2b24iQQCLcBGAsYHQ/s320/210916%2BVr%2Bseries.png" width="320" /></a>
</div>
<p style="text-align: left;">
We can further reduce the equations for Vf and Vr:
</p>
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<a href="https://1.bp.blogspot.com/-Exy76WAemWA/YUO2INDtLNI/AAAAAAAAMR0/n15tslaiq8Aj6PO_r398bnVsxzHKPCSBACLcBGAsYHQ/s666/210916%2Bfinal%2Bvf%2Band%2Bvr%2Bequations.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="666" data-original-width="593" height="320" src="https://1.bp.blogspot.com/-Exy76WAemWA/YUO2INDtLNI/AAAAAAAAMR0/n15tslaiq8Aj6PO_r398bnVsxzHKPCSBACLcBGAsYHQ/s320/210916%2Bfinal%2Bvf%2Band%2Bvr%2Bequations.png" width="285" /></a>
</div>
<p style="text-align: left;"><b><u>Important Note!</u></b> If we were to calculate Γ as being Vr/Vf, using the two equations above, we would see that <b>Γ is <i>independent</i> of Γs</b>. In other words, Gamma is independent of Source Impedance.</p><p style="text-align: left;">Let's check if SWR is also independent of Γs (it should be). Given the following equation for SWR (from Wikipedia):</p>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://1.bp.blogspot.com/-eOScqN9fx7Q/YUO1jQJdoeI/AAAAAAAAMRs/dABhvvltGXgSbxWg0-lYqDN8VYCzQjCAwCLcBGAsYHQ/s307/210916%2Bequation%2Bfor%2BSWR.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="124" data-original-width="307" height="124" src="https://1.bp.blogspot.com/-eOScqN9fx7Q/YUO1jQJdoeI/AAAAAAAAMRs/dABhvvltGXgSbxWg0-lYqDN8VYCzQjCAwCLcBGAsYHQ/s0/210916%2Bequation%2Bfor%2BSWR.png" width="307" /></a>
</div>
<p>
Let's substitute our equations for Vf and Vr into the above
equation. I'll use "D" to represent the denominator of Vf and
Vr:<br />
</p>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://1.bp.blogspot.com/-AdAbooMmXjU/YUPk4MV8SKI/AAAAAAAAMUo/6sJuYUbK9EAEEC-QbbqVbuB58yIwal6iQCLcBGAsYHQ/s390/210916%2Bequation%2Bfor%2BSWR%2Bwith%2Bequations%2Bsubstituted.png" style="margin-left: 1em; margin-right: 1em;"></a></div><p></p><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-0Lvd8WF8oQ8/YVhUkRZlKUI/AAAAAAAAMXU/EyKPc04DREcdc9Wm67bOB46q6YpRA1K_wCLcBGAsYHQ/s390/210916%2Bequation%2Bfor%2BSWR%2Bwith%2Bequations%2Bsubstituted.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="318" data-original-width="390" height="261" src="https://1.bp.blogspot.com/-0Lvd8WF8oQ8/YVhUkRZlKUI/AAAAAAAAMXU/EyKPc04DREcdc9Wm67bOB46q6YpRA1K_wCLcBGAsYHQ/s320/210916%2Bequation%2Bfor%2BSWR%2Bwith%2Bequations%2Bsubstituted.png" width="320" /></a></div><br />Let's start simplifying this equation:<p></p>
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<a href="https://1.bp.blogspot.com/-DzPJ_sEEKuw/YUPlo_mNyaI/AAAAAAAAMU4/wDHhVILXjjUGI0fhKbEz9qhsdVJuFkTZQCLcBGAsYHQ/s740/210916%2Bremoving%2BD%2Band%2BVfs.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="441" data-original-width="740" height="191" src="https://1.bp.blogspot.com/-DzPJ_sEEKuw/YUPlo_mNyaI/AAAAAAAAMU4/wDHhVILXjjUGI0fhKbEz9qhsdVJuFkTZQCLcBGAsYHQ/s320/210916%2Bremoving%2BD%2Band%2BVfs.png" width="320" /></a>
</div>
<p>Already we can see that SWR is not a function of Γs.</p><p>Next, let's look at the exponential part of the equation:</p>
<div>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://1.bp.blogspot.com/-F6PjZQfAnaA/YUO24tzZx8I/AAAAAAAAMSM/rDSanGnbJFMSHLVFRXVLpyfLUKYF4P76wCLcBGAsYHQ/s621/210916%2Bremoving%2Bexponential%2Bterms.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="273" data-original-width="621" height="141" src="https://1.bp.blogspot.com/-F6PjZQfAnaA/YUO24tzZx8I/AAAAAAAAMSM/rDSanGnbJFMSHLVFRXVLpyfLUKYF4P76wCLcBGAsYHQ/s320/210916%2Bremoving%2Bexponential%2Bterms.png" width="320" /></a>
</div>
<p>
Note the delay term. We can remove this from the equation
because its magnitude is 1 (and in the SWR equation we only care about
magnitudes):
</p>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://1.bp.blogspot.com/-Zvxs8b-KZwc/YUO3QhOpBdI/AAAAAAAAMSU/wyzKWdL8c_MrCc0aHSa0hlfaTfjyEKuDACLcBGAsYHQ/s715/210916%2Bremoving%2Bphasor.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="269" data-original-width="715" height="120" src="https://1.bp.blogspot.com/-Zvxs8b-KZwc/YUO3QhOpBdI/AAAAAAAAMSU/wyzKWdL8c_MrCc0aHSa0hlfaTfjyEKuDACLcBGAsYHQ/s320/210916%2Bremoving%2Bphasor.png" width="320" /></a>
</div>
<div><br /></div>
<u>SWR on a Lossy Transmission Line:</u>
</div>
<div><br /></div>
<div>
Therefore, SWR for a <i>lossy</i> transmission line, measured at the
source-end of the line, is:
</div>
<div><br /></div>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://1.bp.blogspot.com/-DPYcmNmaEFA/YUfGUn5fc0I/AAAAAAAAMWg/GkPpH3XtUo40TbGa5kpUQMwWt_lHJ3XlgCLcBGAsYHQ/s539/210916%2BSWR%2Bfor%2Blossy%2Blines.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="243" data-original-width="539" height="144" src="https://1.bp.blogspot.com/-DPYcmNmaEFA/YUfGUn5fc0I/AAAAAAAAMWg/GkPpH3XtUo40TbGa5kpUQMwWt_lHJ3XlgCLcBGAsYHQ/s320/210916%2BSWR%2Bfor%2Blossy%2Blines.png" width="320" /></a>
</div>
<div><br /></div>
where α is line loss in nepers/unit-length, and <i>l</i> is line length.
</div>
<div><br /></div>
<div>
The SWR of a lossy transmission line will vary along its length.
Thus, the general form of SWR for a lossy transmission line, as a function
of position along that line, is:
</div>
<div><br /></div>
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<a href="https://1.bp.blogspot.com/-EDmr3Q2xZFs/YUfG8TynIMI/AAAAAAAAMWo/uOGK8ANH1gMpgacg1t5DcCoANmZHzNQXACLcBGAsYHQ/s779/210916%2Bgeneral%2BSWR%2Bequation%2Bfor%2Blossy%2Blines.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="350" data-original-width="779" height="144" src="https://1.bp.blogspot.com/-EDmr3Q2xZFs/YUfG8TynIMI/AAAAAAAAMWo/uOGK8ANH1gMpgacg1t5DcCoANmZHzNQXACLcBGAsYHQ/s320/210916%2Bgeneral%2BSWR%2Bequation%2Bfor%2Blossy%2Blines.png" width="320" /></a>
</div>
<br />
<div><u>SWR on a Lossless Transmission Line:</u></div>
<div><br /></div>
<div>If the line is lossless:</div>
<div>
<div><br /></div>
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</div>
<div>
<div><br /></div>
<div>
Thus, if the line is lossless (or essentially lossless), we can remove
the exponential term representing line loss, resulting in the
following SWR equation for lossless lines. Note that this
equation is independent of position along the line:
</div>
<div><br /></div>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://1.bp.blogspot.com/--o__gi2r0Nc/YUO5pmtiNII/AAAAAAAAMS8/UWrqeWc7ck0NzIa_y0cZ_A80O1tAzarfQCLcBGAsYHQ/s356/210916%2BSWR%2Bfor%2Blossless%2Bline.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="157" data-original-width="356" height="141" src="https://1.bp.blogspot.com/--o__gi2r0Nc/YUO5pmtiNII/AAAAAAAAMS8/UWrqeWc7ck0NzIa_y0cZ_A80O1tAzarfQCLcBGAsYHQ/s320/210916%2BSWR%2Bfor%2Blossless%2Bline.png" width="320" /></a>
</div>
<br />
<div>
Most importantly, in <i>none</i> of these SWR equations does the SWR
depend upon the source's Reflection Coefficient. Therefore, SWR
is <i>independent</i> of source impedance.
</div>
<div><br /></div>
<div>
<p>
Regarding the infinite geometric series constraint earlier in this
post:
</p>
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</div>
<p>
This condition is met if the line is lossy (and the magnitude of
each reflection coefficient is less than or equal to 1), or, if the
line is lossless, at least one of the reflection coefficients has a
magnitude less than 1 (which is pretty much always, in real
life).
</p>
</div>
<div><br /></div>
<div>
<b><u>Simulating SWR Independence from Source Impedance:</u></b>
</div>
<div><br /></div>
<div>
To simulate SWR independence from source impedance, I'm going to tweak
a Simulink model found in this blog post of mine: <a href="http://k6jca.blogspot.com/2021/05/antenna-tuners-lumped-element-tuner.html">http://k6jca.blogspot.com/2021/05/antenna-tuners-lumped-element-tuner.html</a>
</div>
<div><br /></div>
<div>
This simulation model consists of a voltage source, a series source
impedance, a lumped-element directional coupler (tandem-match
topology) referenced to 50 ohms, a lossless, two-wavelength long 50
ohm transmission line, and a load impedance. Below is the
simulation model with the source impedance set to 50 ohms and a
mismatched load impedance:
</div>
<div><br /></div>
<div>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://1.bp.blogspot.com/-1JmJWfS1cPo/YUO7kD7z1-I/AAAAAAAAMTI/6dkOQIJrxDshr8MSRq6VJb3E2sZlzZrzgCLcBGAsYHQ/s1515/210910_QEX_Zs_50j0_Zld_75par3p05uH.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="952" data-original-width="1515" height="201" src="https://1.bp.blogspot.com/-1JmJWfS1cPo/YUO7kD7z1-I/AAAAAAAAMTI/6dkOQIJrxDshr8MSRq6VJb3E2sZlzZrzgCLcBGAsYHQ/s320/210910_QEX_Zs_50j0_Zld_75par3p05uH.png" width="320" /></a>
</div>
<br />This model performs a time-domain simulation in which the source
is a 10 MHz gated sine-wave whose amplitude is 1.09 volts (both the
gating and the voltage are hold-overs from my previous modeling).
</div>
<div><br /></div>
<div>
The directional coupler is modeled using a "tandem match" topology,
implemented with ideal transformers having a 1:N turns ratio (N in
this case being 50), with the outputs amplified by this same amount,
N, to compensate for the coupler's "coupling factor."
</div>
<div><br /></div>
<div>
Thus, the directional coupler's Vfwd and Vref outputs are scaled to be
a 1:1 representation of the transmission line's actual Vfwd and Vref
voltages.
</div>
<div><br /></div>
<div>Here is the model of my directional coupler:</div>
<div><br /></div>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://1.bp.blogspot.com/--rw7zAkkjko/YUO87kpMbHI/AAAAAAAAMTQ/ryxWPPsOay0FveRTvppHiZXw4jZ0WcJugCLcBGAsYHQ/s1409/Directional%2BCoupler%2BModel.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="952" data-original-width="1409" height="216" src="https://1.bp.blogspot.com/--rw7zAkkjko/YUO87kpMbHI/AAAAAAAAMTQ/ryxWPPsOay0FveRTvppHiZXw4jZ0WcJugCLcBGAsYHQ/s320/Directional%2BCoupler%2BModel.png" width="320" /></a>
</div>
<br />
<div><br /></div>
<div>
The Simulink model lets me plot the time-domain values of Vfwd and
Vref as "measured" by the directional coupler at the transmission
line's input.
</div>
<div><br /></div>
<div>
From the plots of these two waveforms I can manually find their peak
values (using cursors to measure the waveforms when they are in steady
state, after transient behavior has died down), and then calculate
SWR.
</div>
<div><br /></div>
<div>
To demonstrate, below are two simulations, each with a different
source impedance but the same mismatched load impedance. A third
simulation will verify the phase-shift predicted by the math.
</div>
<div><br /></div>
<div><u>Simulation 1:</u></div>
<div><br /></div>
<div>
First, let's match the source impedance to the transmission line's Zo
(i.e. 50 ohms). Note that the load consists of a 75 ohm resistor
in parallel with a 3.049 uH inductor, so it is mismatched from the
transmission line's Zo of 50 ohms (i.e. Zld = 65.033 + j25.46 ohms).
</div>
<div><br /></div>
<div>Using the model from above:</div>
<div><br /></div>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://1.bp.blogspot.com/-s31wLhVYiOo/YUO-KU90bRI/AAAAAAAAMTY/Ax9Nr87dnAQG-Kro-6ENWio9i_F_X-JHgCLcBGAsYHQ/s1515/210910_QEX_Zs_50j0_Zld_75par3p05uH.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="952" data-original-width="1515" height="201" src="https://1.bp.blogspot.com/-s31wLhVYiOo/YUO-KU90bRI/AAAAAAAAMTY/Ax9Nr87dnAQG-Kro-6ENWio9i_F_X-JHgCLcBGAsYHQ/s320/210910_QEX_Zs_50j0_Zld_75par3p05uH.png" width="320" /></a>
</div>
<br />
<div>Here are the simulation's time-domain waveforms:</div>
<div><br /></div>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://1.bp.blogspot.com/-zRpFcMJ4pKk/YUO-Xe6FQyI/AAAAAAAAMTc/GrQ9tV993agy7DO1_28qjjpf4hLXtz8GACLcBGAsYHQ/s1071/210910_VrVf_QEX_Zs_50j0_Zld_75par3p05uH.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="792" data-original-width="1071" height="237" src="https://1.bp.blogspot.com/-zRpFcMJ4pKk/YUO-Xe6FQyI/AAAAAAAAMTc/GrQ9tV993agy7DO1_28qjjpf4hLXtz8GACLcBGAsYHQ/s320/210910_VrVf_QEX_Zs_50j0_Zld_75par3p05uH.png" width="320" /></a>
</div>
<br />
<div>Zooming in and measuring peak voltage values:</div>
<div><br /></div>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://1.bp.blogspot.com/-3EK5fyWYhzs/YUO-gn5vKoI/AAAAAAAAMTk/2I-JmQhLtYAzP5PD_SPeMqeZLCfXpi2ywCLcBGAsYHQ/s1071/210910_ZoomVrVf_QEX_Zs_50j0_Zld_75par3p05uH.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="792" data-original-width="1071" height="237" src="https://1.bp.blogspot.com/-3EK5fyWYhzs/YUO-gn5vKoI/AAAAAAAAMTk/2I-JmQhLtYAzP5PD_SPeMqeZLCfXpi2ywCLcBGAsYHQ/s320/210910_ZoomVrVf_QEX_Zs_50j0_Zld_75par3p05uH.png" width="320" /></a>
</div>
<br />
<div>
Note that the SWR is 1.67 with the <i>matched</i> source impedance.
</div>
<div><br /></div>
<div>
As a check, let's solve the equations presented earlier for Vf, Vr,
and SWR, given Vs = 1.09, Zs = 50, and Zld = 65.033 + j25.46 ohms
(i.e. Γld of 0.1713 + j0.1834):
</div>
<div>
<ul style="text-align: left;">
<li>Vfs = Vs*(Zo/(Zo+Zs) = 0.545 volts.</li>
<li>Γs = (Zs-Zo)/(Zs+Zo) = 0.</li>
<li>Γld = (Zld-Zo)/(Zld+Zo) = 0.1713 + j0.1834.</li>
<li>
<b><div style="font-weight: 400;">
For the exponential factor e<sup>-2γl</sup>, let α = 0 (no
loss) and β<i>l</i> = 4pi radians (because the
transmission line is 2 wavelengths long, the one-way phase
delay is 2λ*2pi/λ = 4pi radians). So
e<sup>-2γl</sup> = e<sup>-j2β<i>l</i></sup> = e<sup>-j2*(4pi)</sup> = 1.
</div>
<ul style="font-weight: 400;"></ul></b>
</li>
<li>Vf = (Vfs)/(1-Γs*Γld*e<sup>-2γl</sup>) = 0.545 volts.</li>
<li><b>|Vf| = 0.545 volts (Check).</b></li>
<li>
<b><div style="font-weight: 400;">
Vr = (Vfs*Γld)/(1-Γs*Γld*e<sup>-2γl</sup>) = 0.0933
+ j0.1 volts.
</div></b>
</li>
<li><b>|Vr| = 0.1368 volts. (Check).</b></li>
<li><b>SWR = (0.545+0.1368)/(0.545-0.1368) = 1.67 (Check).</b></li>
<li>
<b>Time delay from Vf(peak) to Vr(peak) =
(angle(Vf)-angle(Vr))/(2pi)/10MHz = -13.04 ns (Check, given that
simulation's cursor resolution is 0.238 ns, and so the
measurement of -13.3 ns is within an increment of the cursor's
resolution).</b>
</li>
</ul>
<div>So, the simulated results match the mathematically-derived results!</div></div>
<div><br /></div><div><br /></div>
<div><u>Simulation 2:</u></div>
<div><br /></div>
<div>
Now let's change the source impedance so that it is 17.928 + j31.423
ohms (i.e. no longer matched to 50 ohms), while keeping the load the
same mismatched value that was used above.
</div>
<div><br /></div>
<div>Here's the model:</div>
<div><br /></div>
<div>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://1.bp.blogspot.com/-nC0ZMHFkl0w/YUO_2aF7I8I/AAAAAAAAMTw/2I6nK-KHeYUsSPHdmY0-ALcDNCL2VkP1wCLcBGAsYHQ/s1515/210910_QEX_Zs_17p928%252C0.5uH_Zld_75par3p05uH.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="952" data-original-width="1515" height="201" src="https://1.bp.blogspot.com/-nC0ZMHFkl0w/YUO_2aF7I8I/AAAAAAAAMTw/2I6nK-KHeYUsSPHdmY0-ALcDNCL2VkP1wCLcBGAsYHQ/s320/210910_QEX_Zs_17p928%252C0.5uH_Zld_75par3p05uH.png" width="320" /></a>
</div>
<br /> And here are the time-domain plots of Vfwd and Vref:
</div>
<div><br /></div>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://1.bp.blogspot.com/-8Y20TM7p3EE/YUO_-eELglI/AAAAAAAAMT0/15vwQeZ5MnUB1Ov7YWvPtFZGtTfMzkpDwCLcBGAsYHQ/s1071/210910_VrVf_QEX_Zs_17p928%252C0.5uH_Zld_75par3p05uH.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="792" data-original-width="1071" height="237" src="https://1.bp.blogspot.com/-8Y20TM7p3EE/YUO_-eELglI/AAAAAAAAMT0/15vwQeZ5MnUB1Ov7YWvPtFZGtTfMzkpDwCLcBGAsYHQ/s320/210910_VrVf_QEX_Zs_17p928%252C0.5uH_Zld_75par3p05uH.png" width="320" /></a>
</div>
<br />
<div>Zooming in...</div>
<div><br /></div>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://1.bp.blogspot.com/-g3LttDyec8Y/YUPALzmT7DI/AAAAAAAAMT8/0cChmyoKnqU8FdA5_AQHDcsTrrwdM2AsACLcBGAsYHQ/s1071/210910_ZoomVrVf_QEX_Zs_17p928%252C0.5uH_Zld_75par3p05uH.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="792" data-original-width="1071" height="237" src="https://1.bp.blogspot.com/-g3LttDyec8Y/YUPALzmT7DI/AAAAAAAAMT8/0cChmyoKnqU8FdA5_AQHDcsTrrwdM2AsACLcBGAsYHQ/s320/210910_ZoomVrVf_QEX_Zs_17p928%252C0.5uH_Zld_75par3p05uH.png" width="320" /></a>
</div>
<br />
<div>
Again, the SWR is unchanged at 1.67, even though the source impedance
is now mismatched from the transmission line's Zo.
</div>
<div><br /></div>
<div>
Again, let's calculate Vf, Vr, and SWR using the mathematical formulas
presented earlier in this post and compare them to the simulated
values, above:
</div>
<div>
<div>
<ul>
<li>
Vfs = Vs*(Zo/(Zo+Zs) = 1.09*(50/(50+17.928+j31.423) = 0.6609 -
j0.3057 volts.
</li>
<li>Γs = (Zs-Zo)/(Zs+Zo) = -0.2126 + j0.561.</li>
<li>Γld = (Zld-Zo)/(Zld+Zo) = 0.1713 + j0.1834.</li>
<li>
<b><div style="font-weight: 400;">
For the exponential factor e<sup>-2γl</sup>, let α = 0 (no
loss) and β<i>l</i> = 4pi radians (because the
transmission line is 2 wavelengths long, the one-way phase
delay is 2λ*2pi/λ = 4pi radians). So
e<sup>-2γl</sup> = e<sup>-j2β<i>l</i></sup> = e<sup>-j2*(4pi)</sup> = 1.
</div>
<ul style="font-weight: 400;"></ul></b>
</li>
<li>
<b><div style="font-weight: 400;">
Vf = (Vfs)/(1-Γs*Γld*e<sup>-2γl</sup>) = 0.5920 - j0.2387
volts.
</div></b>
</li>
<li><b>|Vf| = 0.6383 volts (Check).</b></li>
<li>
Vr = (Vfs*Γld)/(1-Γs*Γld*e<sup>-2γl</sup>) = 0.1452 + j0.0677
volts.
</li>
<li><b>|Vr| = 0.1602 volts. (Check).</b></li>
<li>
<b>SWR = (0.6383+0.1602)/(0.6383-0.1602) = 1.67 (Check).</b>
</li>
<li>
<b><b>Time delay from Vf(peak) to Vr(peak) =
(angle(Vf)-angle(Vr))/(2pi)/10MHz = </b> -13.04 ns (Check, given that simulation's cursor
resolution is 0.238 ns, and so the cursor measurement of -12.6
ns is within an increment or two of the cursor's
resolution).</b>
</li>
</ul>
</div>
<div>
Again, the simulated results match the mathematically-derived
results!
</div>
</div>
<div><br /></div>
<div><br /></div>
<div><u>Simulation 3:</u></div>
<div><br /></div>
<div>
In the two simulations, above, the transmission line was two
wavelengths long. This means that the one-way end-to-end phase
shift was 4pi, and therefore the quantity e<sup>-γl</sup>, given a
lossless line, equals
</div>
<div>e<sup>-j(4pi)</sup> which equals 1.</div>
<div><br /></div>
<div>
Let's change Simulation 2's transmission line length so that
e<sup>-γl</sup> does not equal 1 and verify that the mathematical
model still correctly predicts Vf, Vr, and SWR. Note that I am
not changing Simulation 2's source and load impedances.
</div>
<div><br /></div>
<div>
I'll make the length equal to 0.3λ. This will represent an
end-to-end phase shift of 0.6pi radians (i.e. 0.3λ*2pi radians/λ).
</div>
<div><br /></div>
<div>Here's the model:</div>
<div><br /></div>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://1.bp.blogspot.com/-TjIr-KvynHY/YUXXjOfx7dI/AAAAAAAAMVY/0VddkoPzaewvnLyX1kGOq8WvmWZ4kVRUgCLcBGAsYHQ/s1515/210917_QEX_Zs_0p3Lambda_17p928%252C0.5uH_Zld_75par3p05uH.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="952" data-original-width="1515" height="201" src="https://1.bp.blogspot.com/-TjIr-KvynHY/YUXXjOfx7dI/AAAAAAAAMVY/0VddkoPzaewvnLyX1kGOq8WvmWZ4kVRUgCLcBGAsYHQ/s320/210917_QEX_Zs_0p3Lambda_17p928%252C0.5uH_Zld_75par3p05uH.png" width="320" /></a>
</div>
<br />
<div>And here is the model's time-domain response:</div>
<div><br /></div>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://1.bp.blogspot.com/-381hi49v06g/YUXYFmX8qnI/AAAAAAAAMVg/6_nF0uxrKdYWV4CoYmWtjKjreCwgYDZtQCLcBGAsYHQ/s1071/210917_VrVf_0p3Lambda_QEX_Zs_17p928%252C0.5uH_Zld_75par3p05uH.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="792" data-original-width="1071" height="237" src="https://1.bp.blogspot.com/-381hi49v06g/YUXYFmX8qnI/AAAAAAAAMVg/6_nF0uxrKdYWV4CoYmWtjKjreCwgYDZtQCLcBGAsYHQ/s320/210917_VrVf_0p3Lambda_QEX_Zs_17p928%252C0.5uH_Zld_75par3p05uH.png" width="320" /></a>
</div>
<br />
<div>And the time-domain response, zoomed in:</div>
<div><br /></div>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://1.bp.blogspot.com/-l4gBce146Zc/YUXYMkOJtgI/AAAAAAAAMVk/ZURtOoNruY4mEs1WbvaemR4hjrdMpuMdACLcBGAsYHQ/s1071/210917_ZoomVrVf_0p3Lambda_QEX_Zs_17p928%252C0.5uH_Zld_75par3p05uH.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="792" data-original-width="1071" height="237" src="https://1.bp.blogspot.com/-l4gBce146Zc/YUXYMkOJtgI/AAAAAAAAMVk/ZURtOoNruY4mEs1WbvaemR4hjrdMpuMdACLcBGAsYHQ/s320/210917_ZoomVrVf_0p3Lambda_QEX_Zs_17p928%252C0.5uH_Zld_75par3p05uH.png" width="320" /></a>
</div>
<br />
<div>Note that the simulated SWR is still 1.67.</div>
<div><br /></div>
<div>
Let's now use the mathematical equations derived above to run some
calculations. I will check them by comparing the calculated values to
the values in the zoomed-in time-domain plots, above.
</div>
<div><br /></div>
<div>
First, recall that Zs = 17.928 + j31.423 ohms, Zld = 65.033 +
j25.46 ohms, Vs = 1.09 volts, Zo = 50 ohms, and and β<i>l</i> =
0.6pi.
</div>
<div>
<div>
<ul>
<li>
Vfs = Vs*(Zo/(Zo+Zs) = 1.09*(50/(50+17.928+j31.423) = 0.6609 -
j0.3057 volts.
</li>
<li>Γs = (Zs-Zo)/(Zs+Zo) = -0.2126 + j0.561.</li>
<li>Γld = (Zld-Zo)/(Zld+Zo) = 0.1713 + j0.1834.</li>
<li>
<b><div style="font-weight: 400;">
For the exponential factor e<sup>-2γl</sup>, let α = 0 (no
loss) and β<i>l</i> = 0.6pi radians (because the
transmission line is 0.3 wavelengths long, the one-way phase
delay is 0.3λ*2pi/λ = 0.6pi radians). So
e<sup>-2γl</sup> = e<sup>-j2β<i>l</i></sup> = e<sup>-j2*(0.6pi)</sup> = -0.809 + j0.5878.
</div>
<ul style="font-weight: 400;"></ul></b>
</li>
<li>
<b><div style="font-weight: 400;">
Vf = (Vfs)/(1-Γs*Γld*e<sup>-2γl</sup>) = 0.6588 - j0.4236
volts.
</div></b>
</li>
<li><b>|Vf| = 0.7832 volts (Check).</b></li>
<li>
Vr = (Vfs*Γld)/(1-Γs*Γld*e<sup>-2γl</sup>) = -0.1825 + j0.0729
volts.
</li>
<li><b>|Vr| = 0.1966 volts. (Check).</b></li>
<li>
<b>SWR = (0.7832+0.1966)/(0.7832-0.1966) = 1.67 (Check).</b>
</li>
<li>
<b style="font-weight: bold;">Time delay from Vf(peak) to Vr(peak) =
(angle(Vf)-angle(Vr))/(2pi)/10MHz = </b><b> 46.956 ns (Check).</b>
</li>
</ul>
</div>
</div>
<div>
Again, SWR is unchanged at 1.67, and the magnitudes of Vf and Vr, as
well as the time delay between their peaks, are confirmed by the
Simulink simulation.
</div>
<div><br /></div>
<div><br /></div>
<div>
<u><b>Conclusion:</b></u>
</div>
<div><br /></div>
<div>
I have mathematically derived an equation for SWR that shows that SWR
is independent of source impedance. And I have simulated three circuits, one with the source impedance matched to the transmission
line, one with the source impedance mismatched, and one with the source impedance still mismatched but with a different transmission line length. These three simulations give values for Vf, Vr, and SWR that match the values
calculated using the equations derived above.
</div>
<div><br /></div>
<div>And SWR is the same for all.</div>
<div><br /></div>
<div><br /></div>
<div>
<b style="text-align: center;"><u>Other Transmission-Line Posts:</u></b>
</div>
<div>
<p>
<a href="http://k6jca.blogspot.com/2021/02/antenna-tuners-transient-and-steady.html" style="text-align: center;">http://k6jca.blogspot.com/2021/02/antenna-tuners-transient-and-steady.html</a>. This post analyzes the transient and steady-state response
of a simple impedance matching system consisting of a wide-band
transformer. I calculate the system's impulse response and
find the time-domain response by convolving this impulse-response
with a stimulus signal.
</p>
<p>
<span style="text-align: center;"><a href="http://k6jca.blogspot.com/2021/02/the-quarter-wave-transformer-transient.html">http://k6jca.blogspot.com/2021/02/the-quarter-wave-transformer-transient.html</a>. <span style="text-align: left;"> This post analyzes the transient and steady-state
response of a Quarter-Wave Transformer impedance matching
device. I calculate the system's impulse response and find
the time-domain response by convolving this impulse-response
with a stimulus signal.</span></span>
</p>
<div>
<a href="http://k6jca.blogspot.com/2021/03/useful-swr-voltage-and-power-equations.html" style="text-align: center;">http://k6jca.blogspot.com/2021/03/useful-swr-voltage-and-power-equations.html</a>. This post lists (in an easily accessible location that I
can find!) some equations that I find useful
</div>
<p>
<span style="text-align: center;"><a href="http://k6jca.blogspot.com/2021/05/antenna-tuners-lumped-element-tuner.html">http://k6jca.blogspot.com/2021/05/antenna-tuners-lumped-element-tuner.html</a>. This post analyzes the transient and steady-state
reflections of a lumped-element tuner (i.e. the common antenna
tuner). I describe a method for making these calculations,
and I note that the tuner's match is independent of the source
impedance.</span>
</p>
<p>
<span style="text-align: center;"><a href="http://k6jca.blogspot.com/2021/05/lc-network-reflection-and-transmission.html">http://k6jca.blogspot.com/2021/05/lc-network-reflection-and-transmission.html</a>. This post describes how to calculate the "Transmission
Coefficient" through a lumped-element network (and also its
Reflection Coefficient) if it were inserted into a transmission
line. <br /></span>
</p>
<p>
<a href="http://k6jca.blogspot.com/2021/09/does-source-impedance-affect-swr.html">http://k6jca.blogspot.com/2021/09/does-source-impedance-affect-swr.html</a>. This post shows mathematically that source impedance does
not affect a transmission line's SWR. This conclusion is then
demonstrated with Simulink simulations.<br />
</p>
<p><a href="https://k6jca.blogspot.com/2021/10/revisiting-maxwells-tutorial-concerning.html">https://k6jca.blogspot.com/2021/10/revisiting-maxwells-tutorial-concerning.html</a> This posts revisits Walt Maxwell's 2004 QEX rebuttal of Steven Best's 2001 3-part series on Transmission Line Wave Mechanics. In this post I show simulation results which support Best's conclusions.</p><p><br /></p>
<p>
<b><u>Standard Caveat:</u></b>
</p>
<div>
I might have made a mistake in my designs, equations, schematics,
models, etc. If anything looks confusing or wrong to you, please
feel free to comment below or send me an email.<br /><br />Also, I
will note:<br /><br />This design and any associated information is
distributed in the hope that it will be useful, but WITHOUT ANY
WARRANTY; without even the implied warranty of MERCHANTABILITY or
FITNESS FOR A PARTICULAR PURPOSE.
</div>
</div>
</div>
</div>
</div>
</div>
</div>Jeffhttp://www.blogger.com/profile/04853314106806116765noreply@blogger.com1tag:blogger.com,1999:blog-2257458838727315711.post-53455122685874401762021-05-14T09:05:00.014-07:002021-10-12T06:30:00.958-07:00LC Network Reflection and Transmission Coefficients<p>
This blog post is inspired by a
<a href="https://k6jca.blogspot.com/2021/05/antenna-tuners-lumped-element-tuner.html">previous blog post</a>
in which I looked at the reflections and re-reflections on a transmission line
given an LC tuner circuit placed at the transmitter.
</p><p>In other words, the combined LC tuner and transmitter were considered to be a single "lumped-element" network.</p><p>I was wondering how these reflections would change if the LC tuner was moved
further away from the transmitter and connected to it with a second
transmission line (rather than being connected directly to the transmitter
<i>at</i> the transmitter's output). And I realized, although I knew how
to calculate the Reflection Coefficients at the LC network's two ports, I did
not know how to calculate the Transmission Coefficients <i>through</i> the
network (either from the input port to the output port, or from the output
port to the input port).</p>
<p>
A quick googling of the internet did not reveal anything, so I did some pencil
and paper noodling and came up with a way to calculate these coefficients for
the general case of a lossless, two-port network consisting of any
combination of lumped-element inductors, capacitors, and transformers.
</p><p><span style="color: #2b00fe;">(Note: this technique should be applicable to lossy two-port networks, too).</span></p>
<p>And thus this post.</p>
<p>
First, a review the Reflection and Transmission Coefficients when two
transmission lines of different characteristic impedances are connected in
series:
</p>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://1.bp.blogspot.com/-dRi-AkonU9g/YJ1Kz-vp5vI/AAAAAAAAMFw/696g2DUxhM8JBPgC7Gx1KW6ye43t8VXtgCLcBGAsYHQ/s514/TL%2Brefl%2B%2526%2Btrans.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="504" data-original-width="514" src="https://1.bp.blogspot.com/-dRi-AkonU9g/YJ1Kz-vp5vI/AAAAAAAAMFw/696g2DUxhM8JBPgC7Gx1KW6ye43t8VXtgCLcBGAsYHQ/s320/TL%2Brefl%2B%2526%2Btrans.png" width="320" /></a>
</div>
<p>
In the system above energy must be conserved. If the system is lossless
(ideal transmission lines with no loss), then the power of the incident wave
will equal the sum of the powers of the forward and the reflected waves:
</p>
<p style="text-align: center;">P(incident) = P(forward) + P(reflected)</p>
<p>Therefore, for the diagram, above:</p>
<p style="text-align: center;">
(|V(incident)|^2 ) / Za = (|V(forward)|^2) / Zb + (|V(reflected)|^2 ) / Za
</p>
<p style="text-align: left;"><br /></p>
<p style="text-align: left;">
But suppose I insert a lumped-element, lossless, two-port network between the
two transmission lines? How do I calculate the new reflection and
transmission coefficients if this network is now in-line?
</p>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://1.bp.blogspot.com/-JDV_bM6RO3s/YJ1KvYJ6c1I/AAAAAAAAMFs/S0jVH6GhZmwin9jyEvkHRoYHtwTZYuWVgCLcBGAsYHQ/s500/network%2Brefl%2B%2526%2Btrans%2Bquestion.png" style="margin-left: 1em; margin-right: 1em;"></a></div><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-h1Ml0l5stTw/YWM0nBbzEpI/AAAAAAAAMhA/W2SzdFqAqT8fkBkzZ_iDtmgYcpoOkXNiwCLcBGAsYHQ/s728/General%2BNetwork%2BEquations.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="610" data-original-width="728" height="268" src="https://1.bp.blogspot.com/-h1Ml0l5stTw/YWM0nBbzEpI/AAAAAAAAMhA/W2SzdFqAqT8fkBkzZ_iDtmgYcpoOkXNiwCLcBGAsYHQ/s320/General%2BNetwork%2BEquations.png" width="320" /></a></div><br />The Reflection Coefficient should be obvious -- it is just the normal equation
for a Reflection Coefficient but using the network's Zin (with the network
terminated with Zb) in place of the second transmission line's Zo.
<div><br /></div>
<div>
The Transmission Coefficient is not so obvious. But considering that the
incident wave sees the network's Zin at the network's input, I make the
assumption that the transmitted wave is first transformed using the standard
"Transmission Coefficient" equation using the network's Zin.
</div>
<div><br /></div>
<div>
But more is required than just this calculation, -- a quick check of the
results will show that energy (calculated as power) is <i>not</i> conserved.
</div>
<div><br /></div>
<div>
Another factor is needed, and that is the voltage attenuation (or gain) from
the network's input port to its output port when the output port is terminated
in the characteristic impedance of the second transmission line.
</div>
<div><br /></div>
<div>
Including this second factor will result in voltage values in which energy is
conserved.
</div>
<div><br /></div>
<div>The formulas for calculating the Reflection and Transmission Coefficients for
general 2-port, lossless, lumped-element networks are shown in the figure,
below:
</div>
<div><br /></div>
<div>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://1.bp.blogspot.com/-QXLLwIXuhHM/YJ1LAmmLtWI/AAAAAAAAMF8/WTZoVZ4Oms8KJMDv3u7XLDvnAVvriCgwQCLcBGAsYHQ/s728/General%2BNetwork%2BEquations.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="610" data-original-width="728" src="https://1.bp.blogspot.com/-QXLLwIXuhHM/YJ1LAmmLtWI/AAAAAAAAMF8/WTZoVZ4Oms8KJMDv3u7XLDvnAVvriCgwQCLcBGAsYHQ/s320/General%2BNetwork%2BEquations.png" width="320" /></a>
</div>
<br />
<p>Next, an example applying these equations...</p>
<p><br /></p>
<p>
<b><u>1. Example, LC Network:</u></b>
</p>
<p>
Let's insert a lossless LC network between the two transmission lines.
We can calculate the resultant Reflection and Transmission Coefficients as
shown in the figure, below:</p><p><b></b></p><div class="separator" style="clear: both; text-align: center;"><b><a href="https://1.bp.blogspot.com/-mBnUEbbl64A/YVmasrQf1AI/AAAAAAAAMXs/xXn9-Cr7QWofXHgQBxfwzEG9t1GEjYjhwCLcBGAsYHQ/s786/LC%2BNetwork%2Bequations.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="786" data-original-width="670" height="320" src="https://1.bp.blogspot.com/-mBnUEbbl64A/YVmasrQf1AI/AAAAAAAAMXs/xXn9-Cr7QWofXHgQBxfwzEG9t1GEjYjhwCLcBGAsYHQ/s320/LC%2BNetwork%2Bequations.png" width="273" /></a></b></div><b><u><p><b><u>2. Results:</u></b></p></u></b><p></p>
<p>
I will assign L = 1.378 uH and C = 137.8 pF. At 10 MHz their impedances
are:
</p>
<p style="text-align: center;"> Zl = + j86.58 ohms</p>
<p style="text-align: center;">Zc = -j115.6 ohms</p>
<p>And I will define Za = 50 ohms and Zb = 25 ohms.</p>
<p>
Using the equations in the LC network figure, above, let's do the following
calculations:
</p>
<p><br /></p>
<p><u>Calculating Γ<sub>11</sub>:</u></p>
<p>To calculate Γ<sub>11 </sub>(the Reflection Coefficient looking
into the network's input,) we first need to calculate Zin of the LC Network. Using the equation in the figure, above, the result is:
</p>
<p style="text-align: center;">
Zin = 23.88 + j81.41 ohms (using MATLAB).
</p>
<p>
Therefore, using the equation for Γ<sub>11</sub> and substituting in
the values for Zin and Za, the result is:
</p>
<p style="text-align: center;">
Γ<sub>11</sub> = 0.389 + j0.674.
</p>
<p>
<u><br /></u>
</p>
<p><u>Calculating T<sub>21</sub>:</u></p>
<div><br /></div>
<div>
To calculate T<sub>21</sub> I first calculate T<sub>1</sub> using
Zin and Za. The result is:
</div>
<div> </div>
<div style="text-align: center;">
T<sub>1</sub> = 1.389 + j0.674
</div>
<p>
Then I calculate the voltage gain, A<sub>V21</sub>, with the network
terminated with Zb. The result is::
</p>
<p style="text-align: center;">
A<sub>V21</sub> = 0.0207 - j0.2872
</p>
<p>
<u><br /></u>
</p>
<p>Next, let's calculate voltages...</p>
<p><u><br /></u></p><p><u>Set Vincident = 1 volt and Calculate Vf and Vr:</u></p>
<p style="text-align: center;">
Vr = Γ<sub>11</sub> * Vincident = 0.389 + j0.674 volts</p>
<p style="text-align: center;">
Vf = Γ<sub>11</sub> * A<sub>V21</sub> * Vincident =
0.222 - j0.385 volts</p>
<p>And note that the magnitudes of these voltages are:</p>
<p style="text-align: center;">|Vincident| = 1 volt</p>
<p style="text-align: center;">|Vr| = 0.778 volts </p>
<p style="text-align: center;">|Vf| = 0.445 volts</p>
<div><br /></div>
<div><u>Check if Energy is Conserved:</u></div>
<p>Let's verify that power, and thus energy, is conserved:</p>
<p style="text-align: center;">Pincident = (1^2) / 50 = 20 mW.</p>
<p style="text-align: center;">Pr = (|Vr|^2) / 50 = 12.1 mW</p>
<p style="text-align: center;">Pf = (|Vf|^2) / 25 = 7.9 mW</p>
<p style="text-align: center;">
Pf + Pr = 7.9 + 12.1 mW = 20 mW = Pincident.
</p>
<p style="text-align: left;">Energy is conserved!</p>
<p><br /></p>
<p><u>Verifying the results using Simulink:</u></p>
<span style="text-align: left;">To verify my results, I'll use the following Simulink model: </span><div><br /></div>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://1.bp.blogspot.com/-S0tNgGXe7Sk/YJ6Wmw0AWSI/AAAAAAAAMGY/51oRkTwUrq0vgMCayHzsrgmplAMaj-IugCLcBGAsYHQ/s1415/Simulink%2BModel.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="715" data-original-width="1415" src="https://1.bp.blogspot.com/-S0tNgGXe7Sk/YJ6Wmw0AWSI/AAAAAAAAMGY/51oRkTwUrq0vgMCayHzsrgmplAMaj-IugCLcBGAsYHQ/s320/Simulink%2BModel.png" width="320" /></a>
</div>
<br /><div><span>(For more information regarding its Directional Coupler models, please refer to this post: </span><a href="http://k6jca.blogspot.com/2021/05/antenna-tuners-lumped-element-tuner.html">http://k6jca.blogspot.com/2021/05/antenna-tuners-lumped-element-tuner.html</a><span>).</span></div><div><br /></div><div>The Simulation waveforms are below (note that the simulation voltages are
spec'd as peak values, not RMS):
</div>
<div><br /></div>
<div>
First, note that Vincident = 1 V (peak). The simulated Vf equals 0.444
volts (peak), which is essentially identical to the value calculated, above.
</div>
<div><br /></div>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://1.bp.blogspot.com/-WwnlfbiJ9fc/YJ6Xk6NtVmI/AAAAAAAAMGg/rjGa8tjK3GUmRY4t_3ZHF2a7wq4KK9MpwCLcBGAsYHQ/s1071/Simulink%2BVf.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="784" data-original-width="1071" src="https://1.bp.blogspot.com/-WwnlfbiJ9fc/YJ6Xk6NtVmI/AAAAAAAAMGg/rjGa8tjK3GUmRY4t_3ZHF2a7wq4KK9MpwCLcBGAsYHQ/s320/Simulink%2BVf.png" width="320" /></a>
</div>
<div><br /></div>
Below, Vr equals 0.777 volts peak, which is essentially the same as the result
calculated, above.<br />
<div><br /></div>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://1.bp.blogspot.com/-4wbe8Rm9e5Q/YJ6Xn55bpCI/AAAAAAAAMGk/MFeKscEEmeIDbGi-KPcPBor_koBWndlgQCLcBGAsYHQ/s1071/Simulink%2BVr.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="784" data-original-width="1071" src="https://1.bp.blogspot.com/-4wbe8Rm9e5Q/YJ6Xn55bpCI/AAAAAAAAMGk/MFeKscEEmeIDbGi-KPcPBor_koBWndlgQCLcBGAsYHQ/s320/Simulink%2BVr.png" width="320" /></a>
</div>
<br />
<div>So simulated results match calculated results!</div>
<p>
(Note that these are the steady-state values. You can see that there
is a short initial transient in both Vr and Vf when the sine-wave source is
first gated on. The length and amplitude of these transients are
related to the network type (e.g. L-network, PI, T, etc.), the component
values, and impedances seen by each network port.)
</p>
<p><br /></p>
<p>
<b><u>Other Notes:</u></b>
</p>
<p>
1. The technique I present, above, has not been rigorously proven
mathematically, so take it with a grain of salt. Never the less, in
the few cases where I have applied it, the calculated results match the
simulated results.
</p>
<p>
2. For Reflection and Transmission Coefficients from the network's Port 2 to its Port 1 (in other words, in the opposite direction), use the same technique. Note that the impedance (Zin)
looking into the network's <i>output</i> (Port 2) equals Zc || (Zl + Za),
where Za is the impedance of the transmission line connected to Port 1,
and <span style="text-align: center;">A</span><sub style="text-align: center;">V12</sub><span style="text-align: center;"> (the voltage gain from Port 2 to Port 1) = Za / (Za + Zl).</span>
</p>
<p>
<span style="text-align: center;">3. If Za = Zb = Zo, then T21 equals S21 and T12 equals S12.</span></p><p><span style="text-align: center;"><b><u>Other Transmission-Line Posts:</u></b></span></p><p><a href="http://k6jca.blogspot.com/2021/02/antenna-tuners-transient-and-steady.html" style="text-align: center;">http://k6jca.blogspot.com/2021/02/antenna-tuners-transient-and-steady.html</a>. This post analyzes the transient and steady-state response of a simple impedance matching system consisting of a wide-band transformer. I calculate the system's impulse response and find the time-domain response by convolving this impulse-response with a stimulus signal.</p><p><span style="text-align: center;"><a href="http://k6jca.blogspot.com/2021/02/the-quarter-wave-transformer-transient.html">http://k6jca.blogspot.com/2021/02/the-quarter-wave-transformer-transient.html</a>. <span style="text-align: left;"> This post analyzes the transient and steady-state response of a Quarter-Wave Transformer impedance matching device. I calculate the system's impulse response and find the time-domain response by convolving this impulse-response with a stimulus signal.</span></span></p><div><a href="http://k6jca.blogspot.com/2021/03/useful-swr-voltage-and-power-equations.html" style="text-align: center;">http://k6jca.blogspot.com/2021/03/useful-swr-voltage-and-power-equations.html</a>. This post lists (in an easily accessible location that I can find!) some equations that I find useful</div><p><span style="text-align: center;"><a href="http://k6jca.blogspot.com/2021/05/antenna-tuners-lumped-element-tuner.html">http://k6jca.blogspot.com/2021/05/antenna-tuners-lumped-element-tuner.html</a>. This post analyzes the transient and steady-state reflections of a lumped-element tuner (i.e. the common antenna tuner). I describe a method for making these calculations, and I note that the tuner's match is independent of the source impedance.</span></p><p><span style="text-align: center;"><a href="http://k6jca.blogspot.com/2021/05/lc-network-reflection-and-transmission.html">http://k6jca.blogspot.com/2021/05/lc-network-reflection-and-transmission.html</a>. This post describes how to calculate the "Transmission Coefficient" through a lumped-element network (and also its Reflection Coefficient) if it were inserted into a transmission line. <br /></span></p><p><a href="http://k6jca.blogspot.com/2021/09/does-source-impedance-affect-swr.html" style="color: #2b00fe;">http://k6jca.blogspot.com/2021/09/does-source-impedance-affect-swr.html</a><span style="color: #2b00fe;">. </span>This post shows mathematically that source impedance does not affect a transmission line's SWR. This conclusion is then demonstrated with Simulink simulations.<br /></p><p><a href="https://k6jca.blogspot.com/2021/10/revisiting-maxwells-tutorial-concerning.html">https://k6jca.blogspot.com/2021/10/revisiting-maxwells-tutorial-concerning.html</a> This posts revisits Walt Maxwell's 2004 QEX rebuttal of Steven Best's 2001 3-part series on Transmission Line Wave Mechanics. In this post I show simulation results which support Best's conclusions.</p><p><br /></p>
<p>
<b><u>Standard Caveat:</u></b>
</p>
<div>
I might have made a mistake in my designs, equations, schematics, models,
etc. If anything looks confusing or wrong to you, please feel free to
comment below or send me an email.<br /><br />Also, I will note:<br /><br />This
design and any associated information is distributed in the hope that it
will be useful, but WITHOUT ANY WARRANTY; without even the implied warranty
of MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.
</div>
</div>
Jeffhttp://www.blogger.com/profile/04853314106806116765noreply@blogger.com0tag:blogger.com,1999:blog-2257458838727315711.post-82904800182966271722021-05-10T16:49:00.020-07:002021-10-10T14:27:11.682-07:00Antenna Tuners: Lumped-Element Tuner & Transmission Line Wave Interactions<div class="separator" style="clear: both; text-align: left;"><span style="color: #2b00fe;"><u>Summary:</u> This blog post examines the operation of a typical "lumped-element" antenna tuner (for example, an L-network tuner) and its effect on forward and reflected wave interactions on a transmission line.</span></div><div class="separator" style="clear: both; text-align: left;"><span style="color: #2b00fe;"><br /></span></div><div class="separator" style="clear: both; text-align: left;"><span style="color: #2b00fe;">I also analyze at the effect of source impedance (i.e. a transmitter's impedance looking into its output port) on the transmission line waves and conclude that they are only affected by the source impedance during transient conditions, but not in steady-state. For an impedance matching network that has been "tuned" to provide a match, for example, to transform a load to 50 ohms, the source impedance has no effect on the matching-network's input impedance in steady-state.</span></div><div class="separator" style="clear: both; text-align: left;"><br /></div><div class="separator" style="clear: both; text-align: center;">
<a href="https://1.bp.blogspot.com/-TSlEuvr2Nkw/YJkeXEMWx3I/AAAAAAAAMCk/xA2WcTg6HjwLQSvkAKn8R0-kk6zOJKaDwCLcBGAsYHQ/s820/Circuit%2BModel.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="548" data-original-width="820" src="https://1.bp.blogspot.com/-TSlEuvr2Nkw/YJkeXEMWx3I/AAAAAAAAMCk/xA2WcTg6HjwLQSvkAKn8R0-kk6zOJKaDwCLcBGAsYHQ/s320/Circuit%2BModel.png" width="320" /></a>
</div>
<br />
<p>I usually don't care what the impedance of an antenna tuner is <i>looking backwards into its output port</i> (towards the transmitter). I adjust the tuner so that its input impedance looks like 50 ohms, and with that accomplished, I'm a happy camper.</p><p>But recently I was wondering if there was a way to calculate this impedance (assuming a load attached to the tuner's output port via a transmission line) and how this impedance would affect the forward and reflected voltages on the transmission line. And how the transmitter's output impedance (i.e. the system's source impedance) might affect the value of the tuner's output impedance.</p><p>And thus this post.</p><p>This post is a continuation of two earlier posts exploring transient and
steady-state responses in Antenna Tuners (i.e. impedance matching devices)
attached to transmission lines.
</p>
<p>
The
<a href="http://k6jca.blogspot.com/2021/02/antenna-tuners-transient-and-steady.html">first post</a>
calculates the transient and steady-state responses of an Antenna Tuner that,
for ease of analysis, consists of a single wide-band transformer.
</p>
<p>
The transformer, being a wide-band device, allows one to easily calculate the
system's impulse response, from which the transient and steady-state responses
are calculated using the convolution integral.
</p>
<p>
The
<a href="http://k6jca.blogspot.com/2021/02/the-quarter-wave-transformer-transient.html">second post</a>
is a continuation of the first post in which a wide-band matching device is
used. In this case, though, the wide-band matching device is a
quarter-wave transformer. The quarter-wave transformer, in this sense,
is a wide-band device in that its frequency response is wide band (there is no
attenuation if the transmission line is assumed to be ideal), although it only
matches impedances over specific narrow bands of frequency.
</p>
<p>
Because of its wide-band frequency response, the quarter-wave transformer's
impulse response (in a simple system) is easy to determine by hand and
transient and steady-state responses for a sinusoidal stimulus easily
calculated.
</p>
<p>
This new post examines the startup and steady-state wave interactions with the more
general case of a "Lumped-element" Antenna Tuning network. The tuner
might be an L-network, a T-network, a PI-network, or some other topology --
the approach described below should work for any two-port lumped-element impedance matching
topology.
</p><p>Lumped-element networks such as these are band-limited networks, and an accurate impulse response is significantly more painful to calculate (because an impulse is no longer an impulse after passing through such a network).</p><p>And because of this frequency-dependency, a true time-domain transient response (that includes, for example network "ringing") is difficult to calculate by hand. But a "limited" transient response that simply shows how reflections and re-reflections build over time on a transmission line is easier to calculate. This is the path this post will.</p>
<p>
The goal of these posts is to demonstrate the underlying wave
mechanics that occur on the transmission lines and at the impedance matching
devices (rather than a deep-dive into a rigorous mathematical derivation), and some assumptions are
made to keep the math manageable.
</p>
<p>These assumptions are:</p>
<p></p>
<ol style="text-align: left;">
<li>Transmission lines are assumed to be lossless.</li>
<li>
Transmission line delays (and thus their lengths) are assumed to delay a
sine-wave by multiples of 90 degrees or 180 degrees, depending upon the
application. E.g. the quarter-wave transformer has a length that adds
a phase-shift of 90 degrees when comparing the phase of the signal at its
input versus the phase of the signal at its output.
</li>
<li>
Antenna Tuning networks are assumed to consist of ideal components.
E.g. wide-band lossless transformers, or lossless inductors and capacitors. No resistors.
</li>
<li>
Transmitters are assumed to be ideal, consisting of a voltage source with a
source impedance Zs.
</li>
<li>Loads are assumed to be resistive, with no reactive component.</li>
<li>
Directional Couplers (i.e. SWR meters) are assumed to be lumped-element
circuits consisting of ideal components (e.g. ideal lossless transformers
and resistors without parasitics).
</li>
<li>
Unless otherwise specified, Directional Couplers are assumed to be
referenced to 50 ohms. That is, the voltage from their "Reflected
Voltage" port will be zero when the impedance connected to their Output port
is 50 + j0 ohms.
</li>
</ol>
<p></p>
<p>
Let's look at a basic station setup consisting of a Transmitter, an SWR Meter,
an Antenna Tuner, and a long Transmission Line connecting the Tuner's output
to a load (antenna).
</p>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://1.bp.blogspot.com/-MTLjHyxJLag/YJbqNThWHHI/AAAAAAAAMAs/2PxYqiIMBmgOEGTPSCpy8mwVJYVMirYogCLcBGAsYHQ/s826/Intro%2B--%2B1%2BTransmitter%2Bsystem.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="272" data-original-width="826" src="https://1.bp.blogspot.com/-MTLjHyxJLag/YJbqNThWHHI/AAAAAAAAMAs/2PxYqiIMBmgOEGTPSCpy8mwVJYVMirYogCLcBGAsYHQ/s320/Intro%2B--%2B1%2BTransmitter%2Bsystem.png" width="320" /></a>
</div>
<div><br /></div>
<div>
If the interconnects between the transmitter, SWR meter, and antenna tuner are
short (my rule-of thumb is that lengths are less than 1/20 of the wavelength of the highest
frequency), then these system elements and their interconnects can be treated
as a lumped-element system, and lumped-element circuit analysis techniques can be used
to calculate the voltages and currents at the various nodes of this part of
the system.
</div>
<div><br /></div>
<div>
On the other hand, the transmission line portion of the system is best
analyzed using the familiar concepts of reflection and transmission
coefficients of transmission-line systems.
</div>
<div>
<br />
<div class="separator" style="clear: both; text-align: center;">
<a href="https://1.bp.blogspot.com/-LqHNF8y9g9Y/YJbqVqIu1DI/AAAAAAAAMAw/rCZOOaoskuMkcKU81Q5rVyVSBO759Eu3wCLcBGAsYHQ/s828/Intro%2B--%2B2%2BLumped%2Bvs%2Bdistributed.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="308" data-original-width="828" src="https://1.bp.blogspot.com/-LqHNF8y9g9Y/YJbqVqIu1DI/AAAAAAAAMAw/rCZOOaoskuMkcKU81Q5rVyVSBO759Eu3wCLcBGAsYHQ/s320/Intro%2B--%2B2%2BLumped%2Bvs%2Bdistributed.png" width="320" /></a>
</div>
<p>
<b><u>Goals of this Post:</u></b>
</p>
<p>
<u>One goal</u> of this post is to show a method of calculating, and then
calculate, the following voltages:
</p>
<p></p>
<ol style="text-align: left;">
<li>
The Forward and Reflected voltages on the transmission line and how they
change from startup to steady-state.
</li>
<li>
The Forward and Reflected voltages that would be measured by the
lumped-element Directional Coupler (i.e. SWR meter) inserted between the
Transmitter and the Antenna Tuner, and how these voltages change from startup to
steady-state.
</li>
</ol>
<p></p>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://1.bp.blogspot.com/-YByAy0w3CaA/YJbri1dMasI/AAAAAAAAMBE/PToZ4d1In_8G7bgJKolsTT-jbHs3hF1BACLcBGAsYHQ/s828/Intro%2B--3%2Bvf%2Band%2Bvr.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="303" data-original-width="828" src="https://1.bp.blogspot.com/-YByAy0w3CaA/YJbri1dMasI/AAAAAAAAMBE/PToZ4d1In_8G7bgJKolsTT-jbHs3hF1BACLcBGAsYHQ/s320/Intro%2B--3%2Bvf%2Band%2Bvr.png" width="320" /></a>
</div>
<div><br /></div>
<div>
Note that in a lumped-element system there actually aren't forward and
reflected waves. But the SWR meter, itself being a lumped-element
circuit, does not know that they do not exist -- it is measuring the
voltages and currents at a single point (node) and from these deriving what the
forward and reflected voltages would be if it were connected in a 50 ohm
transmission line. (More on this topic, below).
</div>
<div><br /></div>
<div>
Vf and Vr (forward and reflected) voltages derived with the equations I will present will be verified with
simulated results generated with Simulink.
</div>
<div><br /></div>
<div>
<u>Another goal</u> is to determine, when the Antenna Tuner is tuned for a
1:1 SWR at its input (i.e. the impedance looking into the antenna tuner's input port is 50
ohms), the value of the impedance looking back into the OUT port of the antenna tuner
(towards the transmitter). This is the impedance that reflections from
the load will encounter as they travel backwards on the transmission line towards the tuner, and this is the impedance that then re-reflects
these reflections back to the load.
</div>
<div><br /></div>
<div>
In other words, when the system is in steady-state with the Antenna Tuner
tuned so that its input impedance is 50 + j0 ohms...
</div><div><br /></div>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://1.bp.blogspot.com/-XaW-DJfmUOg/YJbb8TnRQzI/AAAAAAAAMAk/1so1rkOz_j4nUPWBAtz9D1pTwGalxnfawCLcBGAsYHQ/s807/Tuner%2BOut%2BImpedance.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="807" data-original-width="742" height="320" src="https://1.bp.blogspot.com/-XaW-DJfmUOg/YJbb8TnRQzI/AAAAAAAAMAk/1so1rkOz_j4nUPWBAtz9D1pTwGalxnfawCLcBGAsYHQ/s320/Tuner%2BOut%2BImpedance.png" /></a>
</div>
<p>
<b><u>Foundation Concepts:</u></b>
</p>
<p>
The following concepts form the foundation for my equations and
calculations later in this post.
</p>
<p>
1. On a Transmission Line, the voltage at any point on that line is
the sum of the Forward Voltage and the Reflected Voltage:
</p>
<p style="text-align: center;">V(at point) = Vf(at point) + Vr(at point)</p>
<p>
2. On a Transmission Line, when a traveling wave arrives at an
impedance discontinuity, this "incident" wave splits into two parts.
One part is reflected back (Vr), and the other part continues forward (Vf)
through the discontinuity. The amplitudes of Vr and Vf with respect to
the incident wave's voltage are such that energy is conserved.
</p>
<p>
The amplitudes of the signals traveling in the two directions can be
determined using the calculated values of the Reflection Coefficient and the
Transmission Coefficient at the discontinuity
</p>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://1.bp.blogspot.com/-UKx9F5NcOco/YJlj_ahtg8I/AAAAAAAAMDs/PwqzpuDciAEXm1qxn9eNlH5oy_nG1pXMgCLcBGAsYHQ/s586/Gamma%2Band%2BT%2Bdefinitions.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="364" data-original-width="586" src="https://1.bp.blogspot.com/-UKx9F5NcOco/YJlj_ahtg8I/AAAAAAAAMDs/PwqzpuDciAEXm1qxn9eNlH5oy_nG1pXMgCLcBGAsYHQ/s320/Gamma%2Band%2BT%2Bdefinitions.png" width="320" /></a>
</div>
<div><br /></div><div>We can use the same V(transmitted) equation for a transmission line with a
resistive load. After all, a resistor simply looks like an infinitely long
transmission line whose characteristic impedance equals the resistor's
resistance:</div><div><br /></div>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://1.bp.blogspot.com/-JTMgX4lklGA/YJlkbkbnTZI/AAAAAAAAMD0/5BHvm4jbKWMi7QY1pfpRNIEu5zoeiZKcgCLcBGAsYHQ/s469/Reflection%2Band%2Btransmission%2Bcoeff%252C%2BR%2Bload.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="429" data-original-width="469" src="https://1.bp.blogspot.com/-JTMgX4lklGA/YJlkbkbnTZI/AAAAAAAAMD0/5BHvm4jbKWMi7QY1pfpRNIEu5zoeiZKcgCLcBGAsYHQ/s320/Reflection%2Band%2Btransmission%2Bcoeff%252C%2BR%2Bload.png" width="320" /></a>
</div>
<br />
<p>
3. On a Transmission Line, delay can be expressed as a complex number:
</p>
<p>
On an ideal lossless transmission line with a sinusoidal waveform, we can
mathematically treat delay as e<sup>-jφ</sup>, where φ is the phase shift
introduced by the transmission line's delay (note: e<sup>-jφ</sup> = cos(φ) - j*sin(φ)).
</p>
<p>
For example, if a transmission line is one wavelength long, a sine-wave signal's phase will
shift by 360 degrees (2 pi radians). And so in this case, the delay
factor would be e<sup>-j2π</sup> = 1 + j0.
</p>
<p>
If the transmission line were 1/4 λ long, the delay factor would be
e<sup>-jπ/2</sup> = 0 + j1.
</p>
<p><br /></p>
<p>4. The Principle of Superposition: </p>
<p>
The "Principle of Superposition" is a fundamental principle of circuit
analysis which states that, in a circuit with multiple voltage and current
sources, we can calculate the voltage and current at any node in the system by
first calculating the individual contributions of each voltage or current
source to the voltage (or current) at that node (by first turning all other sources off -- replacing voltage sources with shorts and current sources
with opens), and then summing the individual contributions together to get
the final value.
</p>
<p>
This same principle can be applied to a transmission line upon which
multiple reflections might exist. For example, on a mismatched
transmission line, the total reflected voltage on the line can be
represented by the sum of an infinite series of individual past reflected
voltages. (More on this in a bit).
</p>
<p><br /></p>
<p>
5. An ideal lumped-element Directional Coupler can be represented by the
following circuit:
</p>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://1.bp.blogspot.com/-FB7wEP5QjGE/YJfgCzbKBEI/AAAAAAAAMBU/G2dwow4BT1k7wBrErpPdTJOmiFIfHk9oQCLcBGAsYHQ/s1384/dir%2Bcplr%2Bckt.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="919" data-original-width="1384" src="https://1.bp.blogspot.com/-FB7wEP5QjGE/YJfgCzbKBEI/AAAAAAAAMBU/G2dwow4BT1k7wBrErpPdTJOmiFIfHk9oQCLcBGAsYHQ/s320/dir%2Bcplr%2Bckt.png" width="320" /></a>
</div>
<p>
The circuit above is the Directional Coupler I will use in my Simulink
simulations.
</p>
<p>
Lumped-element directional couplers measure current and voltage at a single
point. As such, they have no knowledge of the direction of wave
travel or even the existence of waves. But they will generate two voltages that would represent Vfwd
and Vref on a transmission line if the directional coupler were installed in
a transmission line whose characteristic impedance Zo equals the directional
coupler's "Rref".
</p>
<p>
Another way to look at this circuit's operation is to consider it to be a lumped-element bridge circuit, and the voltage Vref will equal zero when Zload (connected
to the directional coupler's Vout port) equals 50 + j0 ohms (assuming Rref =
50 ohms).
</p>
<p>
Therefore, if this directional coupler is attached to the input port of an
antenna tuner and the tuner adjusted until Vref equals 0, then the impedance
looking into the antenna tuner's input will be 50 + j0 ohms.
</p>
<p>
Note that for large transformer turns-ratios, the voltage drop across the current-sense transformer will essentially be 0, and Vin will essentially equal
Vout (which we want). We can derive the following equations for Vfwd and
Vref based upon the voltage at Vout and the load impedance attached to it. (Note that the turns ratio 'n' results in the attenuation of Vfwd and Vref and so my "ideal" directional coupler adds multiplication factors that remove this attenuation effect):
</p>
<p style="text-align: center;">Vfwd = Vout *(Zload + Rref)/(2*Zload)</p>
<p style="text-align: center;">Vref = Vout *(Zload - Rref)/(2*Zload)</p>
<p>(More on the derivation of these equations, here: <a href="http://k6jca.blogspot.com/2015/01/notes-on-directional-couplers-for-hf.html">http://k6jca.blogspot.com/2015/01/notes-on-directional-couplers-for-hf.html</a> )</p><p>If we know Vfwd and Vref, we can derive the Zload connected to the directional coupler's output port:</p>
<p style="text-align: center;">
Zload = Rref * (Vfwd + Vref) / (Vfwd - Vref)
</p>
<p>The drawing below illustrates these equations:</p>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://1.bp.blogspot.com/-4Qr9VqUXEEY/YJfm_LVgD5I/AAAAAAAAMBc/maysUiD82OI4u0wKHxjnSNL8nBm4f0x2QCLcBGAsYHQ/s1585/dir%2Bcplr%2Bckt%2Bwith%2Bload%2Band%2Bequations.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="919" data-original-width="1585" src="https://1.bp.blogspot.com/-4Qr9VqUXEEY/YJfm_LVgD5I/AAAAAAAAMBc/maysUiD82OI4u0wKHxjnSNL8nBm4f0x2QCLcBGAsYHQ/s320/dir%2Bcplr%2Bckt%2Bwith%2Bload%2Band%2Bequations.png" width="320" /></a>
</div>
<p><br /></p>
<p>
6. Any waveform can be represented by an infinite series of impulses, and we can examine the operation of a system by taking these impulses, one at a time, and investigating their effect on the system's voltages, and then summing their contributions:
</p>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://1.bp.blogspot.com/-4Anut49bazM/YJf3d_x8awI/AAAAAAAAMBk/zljMHFNXIV4sMhKiC9fVpgcustwxdbr6gCLcBGAsYHQ/s934/decomposing%2Bsne.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="709" data-original-width="934" src="https://1.bp.blogspot.com/-4Anut49bazM/YJf3d_x8awI/AAAAAAAAMBk/zljMHFNXIV4sMhKiC9fVpgcustwxdbr6gCLcBGAsYHQ/s320/decomposing%2Bsne.png" width="320" /></a>
</div>
<div><br /></div>
With these definitions out of the way, let's apply these principles to an
example...<br />
<p>
<b><u><br /></u></b>
</p>
<p>
<b><u>Example with a Source Impedance Equal to 50 Ohms and a 200
Ohm Load:</u></b>
</p>
<p>
Let's take an example with a 200 ohm load at the end of a 50-ohm
transmission line that is one wavelength long (taking its velocity factor
into account).
</p>
<p>
<u>Defining the Circuit Model:</u>
</p>
<p>
I will represent the transmitter as a voltage source with a series source
impedance Zs. In this example I will set Zs to 50 + j0 ohms.
</p>
<p>The transmission line's characteristic impedance will be defined to be 50 ohms, and the line lossless. Because the transmission line is one wavelength long, the load impedance
presented by the input of the line to the output port of the Antenna Tuner is also, conveniently, 200
ohms. </p>
<p>
The Transmitter, SWR Meter, and Antenna Tuner are assumed to be connected
together with very short interconnects, so I will combine these three
elements together and treat them as a single lumped-element network.
</p>
<p>
I will represent the antenna tuner with an L-network that will transform the 200
+ j0 ohms at its output port to 50 + j0 ohms looking into its input port.
</p>
<p>In the illustration, below, I've removed the directional coupler (in this example considered to be
ideal) because it has no effect on system voltages and currents (but I will present
equations that allow the calculation of the Vf and Vr it would generate, given the voltage at a node into which it would be inserted and the load impedance it
would see at that node.)
</p>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://1.bp.blogspot.com/-aT73sybFw78/YJgeHI-uWCI/AAAAAAAAMBs/ZftPJv9pipQNaqGK_wEAF6JK_DcL89LNwCLcBGAsYHQ/s820/Circuit%2BModel.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="548" data-original-width="820" src="https://1.bp.blogspot.com/-aT73sybFw78/YJgeHI-uWCI/AAAAAAAAMBs/ZftPJv9pipQNaqGK_wEAF6JK_DcL89LNwCLcBGAsYHQ/s320/Circuit%2BModel.png" width="320" /></a>
</div>
<p>Within the lumped-element network there are two nodes whose voltages I will
use for calculations. One is node 'A' (at the L-network's input), and
the other is node 'b' (at the L-network's output, which is also the
transmission line's input). A directional coupler could be inserted into the circuit at either of these two nodes to determine either the forward and reflected voltages at the input end of the transmission line (node B) or the "equivalent" forward and reflected voltages at node A.</p>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://1.bp.blogspot.com/-SVJ-4GpWjDc/YJgeMhMCTFI/AAAAAAAAMBw/qFFimEBUB1YurCQFKI-hgjaIzOHEMlT3gCLcBGAsYHQ/s818/Model%2527s%2BVa%2Band%2BVb.png" style="margin-left: 1em; margin-right: 1em;"></a></div><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-W6-ZgGXRWGY/YJrd6w90AcI/AAAAAAAAMFk/HfXm4sTw0387oCexM64zk0AR5ye3cRILACLcBGAsYHQ/s818/Model%2527s%2BVa%2Band%2BVb%252C%2BV2.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="373" data-original-width="818" src="https://1.bp.blogspot.com/-W6-ZgGXRWGY/YJrd6w90AcI/AAAAAAAAMFk/HfXm4sTw0387oCexM64zk0AR5ye3cRILACLcBGAsYHQ/s320/Model%2527s%2BVa%2Band%2BVb%252C%2BV2.png" width="320" /></a></div><div><br /></div><div>The Transmission Line's Reflection and Transmission Coefficients are shown,
below.</div>
<div><br /></div>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://1.bp.blogspot.com/-3gJELTDE7rw/YJgeRQ4Qr5I/AAAAAAAAMB0/KsDluefjrgoo_LTvf9ZnzWdP-yZbPdo6gCLcBGAsYHQ/s802/network%2Breflection%2Bcoefficients.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="248" data-original-width="802" src="https://1.bp.blogspot.com/-3gJELTDE7rw/YJgeRQ4Qr5I/AAAAAAAAMB0/KsDluefjrgoo_LTvf9ZnzWdP-yZbPdo6gCLcBGAsYHQ/s320/network%2Breflection%2Bcoefficients.png" width="320" /></a>
</div>
<div><br /></div>The illustration below shows how the Reflection Coefficient looking into the Output port of the
L-network was calculated:</div>
<div><br /></div>
<div>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://1.bp.blogspot.com/-wg5vDt3nCXc/YJlZZi7UxqI/AAAAAAAAMDY/tZi7YnqN2mkoOaQzseE7mnrKXTZfNr3GwCLcBGAsYHQ/s806/Calculating%2BZ_LCout.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="454" data-original-width="806" src="https://1.bp.blogspot.com/-wg5vDt3nCXc/YJlZZi7UxqI/AAAAAAAAMDY/tZi7YnqN2mkoOaQzseE7mnrKXTZfNr3GwCLcBGAsYHQ/s320/Calculating%2BZ_LCout.png" width="320" /></a>
</div>
<br />
<p>SimSmith can be used as a quick check of this value:</p>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://1.bp.blogspot.com/-AorxLdpeBWM/YJlZe9dOniI/AAAAAAAAMDc/EstdL0Knr4IUdtP3vvvWCx-UY_pYXJhKQCLcBGAsYHQ/s1353/SimSmith%2BGamma%2BLooking%2Binto%2BLC%2BOut%252C%2BZs%2B50.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="927" data-original-width="1353" src="https://1.bp.blogspot.com/-AorxLdpeBWM/YJlZe9dOniI/AAAAAAAAMDc/EstdL0Knr4IUdtP3vvvWCx-UY_pYXJhKQCLcBGAsYHQ/s320/SimSmith%2BGamma%2BLooking%2Binto%2BLC%2BOut%252C%2BZs%2B50.png" width="320" /></a>
</div>
<br />
<p>
<b><u>Calculating the Model's Voltages and Wave Reflections:</u></b>
</p>
<p><u>1. Startup:</u></p>
<p>
Let's say that at time t = 0 a sine-wave of amplitude 2 volts is started. From a lumped-element circuit analysis perspective, the
voltages in the lumped-element network can be calculated.
</p>
<p>
First, note that the output of the L-network is attached to the transmission
line. At startup the impedance that the L-network sees at this output (labeled below
as "Z_TLin") is Zo, because there are not yet any reflections on the
transmission line.
</p>
<p>
Given this load impedance, the voltages at the two nodes ('a' and 'b') can be calculated, and from these we can calculated the "equivalent" Vfwd
and Vref voltages that ideal "Tandem-match" lumped-element directional
couplers would generate if placed at either node 'a' or node 'b' (the latter
being the Transmission line's input). </p><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-7_U20QYnkhM/YJqBYKfAZ8I/AAAAAAAAMFc/jWTNQzXDhicTGba6IQ560pis3Sz7lGyTgCLcBGAsYHQ/s686/Lumped%2Belement%2BZ%2Band%2BV%2Bformulas%2BV2.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="508" data-original-width="686" src="https://1.bp.blogspot.com/-7_U20QYnkhM/YJqBYKfAZ8I/AAAAAAAAMFc/jWTNQzXDhicTGba6IQ560pis3Sz7lGyTgCLcBGAsYHQ/s320/Lumped%2Belement%2BZ%2Band%2BV%2Bformulas%2BV2.png" width="320" /></a></div><br /><p>(Note: after startup, as reflections on the transmission line arrive
back at the L-network's output, the impedance seen at the transmission
line's input by the L-network will change. The equation to calculate
this new impedance is shown at the bottom of the illustration, above).</p>
<p>
<u><br /></u>
</p>
<p><u>2. Reflections on the Transmission Line:</u></p>
<div class="separator" style="clear: both; text-align: left;">For a sine-wave entering a transmission line, we can take a point on that sine-wave and follow it as it travels down the line to the load and then reflects back towards the source.</div><div class="separator" style="clear: both; text-align: left;"><br /></div><div class="separator" style="clear: both; text-align: left;">If the impedance at the source end of the line is also mismatched from Zo, this point will re-reflect and head back towards the load.</div><div class="separator" style="clear: both; text-align: left;"><br /></div><div class="separator" style="clear: both; text-align: left;">These reflections and re-reflections of this point on the sine-wave will continue indefinitely over time, with each reflection becoming smaller (assuming a source or load impedance with some resistive element, because I've defined the transmission line to be lossless). And the amplitude of each new reflection can be calculated using a Lattice Diagram.</div><div class="separator" style="clear: both; text-align: left;"><br /></div><div class="separator" style="clear: both; text-align: left;">So, at any time 't', the forward voltage on a transmission line can be calculated by summing all of the contributions either from all of the past reflections traveling in the forward direction, plus the voltage currently being sent by the source. </div><div class="separator" style="clear: both; text-align: left;"><br /></div><div class="separator" style="clear: both; text-align: left;">Similarly, the reflected voltage on the transmission line can be calculated by summing all of the reflections traveling in the reverse direction (from the load).</div><div class="separator" style="clear: both; text-align: left;"><br /></div><div class="separator" style="clear: both; text-align: left;">At startup, given the initial Vfwd on the transmission line (calculated with
the lumped-element equations above), the resulting amplitude and phase
of the reflections and re-reflections can be calculated using the equations shown in the Lattice
Diagram, below.</div>
<div class="separator" style="clear: both; text-align: left;"><br /></div>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://1.bp.blogspot.com/--uVOYiezrdY/YJg-zRAPSZI/AAAAAAAAMCE/RTczCZAA97UOAfyCLQDcLsJSWSTtaXXMwCLcBGAsYHQ/s783/Zs_50%2BModel%2Band%2BLattice.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="600" data-original-width="783" src="https://1.bp.blogspot.com/--uVOYiezrdY/YJg-zRAPSZI/AAAAAAAAMCE/RTczCZAA97UOAfyCLQDcLsJSWSTtaXXMwCLcBGAsYHQ/s320/Zs_50%2BModel%2Band%2BLattice.png" width="320" /></a>
</div>
<p>Here's the Lattice Diagram by itself:</p>
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<a href="https://1.bp.blogspot.com/-9BCKbVNb_6M/YJg_BEAclkI/AAAAAAAAMCM/ZImvnd4hFncsp5gEsSr2MdkXE6WgXYHWwCLcBGAsYHQ/s500/General%2BLattice%2BDiagram.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="500" data-original-width="495" height="320" src="https://1.bp.blogspot.com/-9BCKbVNb_6M/YJg_BEAclkI/AAAAAAAAMCM/ZImvnd4hFncsp5gEsSr2MdkXE6WgXYHWwCLcBGAsYHQ/s320/General%2BLattice%2BDiagram.png" /></a>
</div>
<div class="separator" style="clear: both;"><br /></div>
<div class="separator" style="clear: both;">
<div class="separator" style="clear: both;">
Note that equations in the lattice diagram are easily determined and are
based solely upon the calculated Reflection coefficients at either end of
the Transmission line and also the Transmission coefficient at the load.
</div>
<div class="separator" style="clear: both;"><br /></div>
</div>
<div class="separator" style="clear: both;">
Note that the e<sup>-jφ</sup> factor is simply the phase-shift of the
signal due to the transmission line's delay. In this example in which the
transmission line is one wavelength long, the delay is 360 degrees, and so the delay factor e<sup>-j2π</sup> equals 1 + j0. In other words, it has no effect</div>
<p>
Note that the reflections described by the lattice diagram continue
forever. Thus, when summed, they are infinite series.
</p>
<p>
Never the less, these infinite series converge to values (see
<a href="http://k6jca.blogspot.com/2021/02/antenna-tuners-transient-and-steady.html">http://k6jca.blogspot.com/2021/02/antenna-tuners-transient-and-steady.html</a> for the math describing how to determine the equations to calculate the values
to which they converge).
</p>
<p>For the Lattice Diagram, above, the voltages converge to the following values:</p>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://1.bp.blogspot.com/-rKsfGsCSMik/YJnBPtUiBdI/AAAAAAAAMFM/TD2-PAR1tBkEpEkRB2Z1xYVmQNytjawLgCLcBGAsYHQ/s671/Steady-state%2Binfinite%2Bseries%2Bsolutions.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="541" data-original-width="671" src="https://1.bp.blogspot.com/-rKsfGsCSMik/YJnBPtUiBdI/AAAAAAAAMFM/TD2-PAR1tBkEpEkRB2Z1xYVmQNytjawLgCLcBGAsYHQ/s320/Steady-state%2Binfinite%2Bseries%2Bsolutions.png" width="320" /></a>
</div>
<div><br /></div>
Regarding Vf2 in the equation above, it represents the sum of <i>all
re-reflections</i> (reflected from the Output port of the impedance matching
network) traveling towards the load. Thus it excludes the very first
forward signal (called Vf1) which represents the current signal from the
generator, passing through the matching network, to the transmission line.<br />
<p><br /></p>
<p><u>3: Procedure to Perform Calculations</u></p>
<p>
I calculate the various voltages (nodal and transmission-line) in time
increments of Td, where Td is the period of the 10 MHz signal. The
Transmission Line in this example is ideal (thus the velocity factor equals
1), and so Td is also the amount of time it takes a "point" on the sine-wave
to travel from one end of the one-wavelength long transmission line to the
other end (and so a round-trip on the transmission line is 2*Td).
</p>
<p>
Below is the procedure I used to calculate the system's voltages over time.
</p>
<ol style="text-align: left;">
<li>
Calculate the reflection and transmission coefficients Γ<sub>1</sub>,
Γ<sub>2</sub>,, and T<sub>ld</sub>.
</li>
<li>
At startup (time 't' = 0) calculate impedances Za and Zb, and voltages Va
and Vb, using the lumped-element equations and impedances, with Z_TLin
equal to 50 ohms.
</li>
<li>
Given Va, Vb, and Z_TLin (= 50 ohms at startup), calculate the
"equivalent" forward and reflected voltages that lumped-element
Directional Couplers would generate: Vf_LCin, Vr_LCin, and Vf_TLin
using the lumped-element equations (Vr_TLin will be 0 as there are no reflections from the load yet).
</li>
<li>Set Vf1 equal to Vf_TLin. Vf1 represents the forward voltage on the transmission line generated at that moment by the source.</li>
<li>
Create three values for keeping track of running sums of Transmission Line
Forward and Reflected voltages and the Load voltage: Vf, Vr, and
Vload. For time t = 0, Set Vf(t) equal to Vf1 and set Vr(t) and Vload(t) to 0 (there are no reflections yet, and no signal has yet arrived at the load)</li>
<li>
Increment time t by Td (t = t + Td). Calculate Vld(t) using the
Lattice diagram equation for this time increment.
</li><li>Add Vld(t) to the Vload running sum: Vload(t) = Vload(t-1) + Vld(t).</li>
<li>
Increment time t by Td (t = t + Td). Calculate Vr_TLin(t) and Vf_TLin(t) using Lattice diagram equations for this time increment. Vr_TLin represents the reflection from the load when it arrives back at the transmission line input (i.e. the line's "source" end) and Vf_TLin represents the re-reflection of the Vr_TLin signal off of the impedance discontinuity at the transmission line's input as it starts its return journey on the line back towards the load.</li>
<li>
Add these values to the Vf, Vr, and Vload running sums: Vf(t) =
Vf (t-1) + Vf_TLin(t); Vr(t) = Vr(t-1) + Vr_TLin(t).
</li>
<li>
Calculate the new Z_TLin using Vf, Vr, and Rref (see lumped-element
equations).
</li>
<li>
Using this new Z_TLin, calculate impedances Za, Zb, and voltages Va, Vb,
Vf_LCin(t), and Vr_LCin(t) using lumped-element equations.
</li>
<li>Go to Step 6 and repeat.</li>
</ol>
<div>
(Note: tracking running sums is not required for the voltages Vf_LCin
and Vr_LCin because their values are based upon Z_TLin, which itself is
calculated from running sums of the Transmission line Vf and Vr voltages).
</div>
<p></p>
<p><u>4. Equation Results:</u></p>
<p>
Below are the results from the procedure described above (calculated using
MATLAB and tabulated in an Excel spreadsheet):
</p>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://1.bp.blogspot.com/-H--_XTJDysc/YJlB-L9pK5I/AAAAAAAAMDQ/OIP1e7aFc9ICkY_ICu1hXoFjpzQWIU_3ACLcBGAsYHQ/s1311/Calculated%2BResults%252C%2BZs%2B50%252C%2BZld%2B200.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="730" data-original-width="1311" src="https://1.bp.blogspot.com/-H--_XTJDysc/YJlB-L9pK5I/AAAAAAAAMDQ/OIP1e7aFc9ICkY_ICu1hXoFjpzQWIU_3ACLcBGAsYHQ/s320/Calculated%2BResults%252C%2BZs%2B50%252C%2BZld%2B200.png" width="320" /></a>
</div>
<div><br /></div>
As a check, let's take a look at the steady-state results predicted by the following Lattice Diagram equations:
</div>
<div><br /></div>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://1.bp.blogspot.com/-fkI72qorAPQ/YJnCImQlOdI/AAAAAAAAMFU/FfYikMurENQLYkE41gFwHZvwegCFonMvQCLcBGAsYHQ/s671/Steady-state%2Binfinite%2Bseries%2Bsolutions.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="541" data-original-width="671" src="https://1.bp.blogspot.com/-fkI72qorAPQ/YJnCImQlOdI/AAAAAAAAMFU/FfYikMurENQLYkE41gFwHZvwegCFonMvQCLcBGAsYHQ/s320/Steady-state%2Binfinite%2Bseries%2Bsolutions.png" width="320" /></a>
</div>
<div><br /></div>
The results of these equations are:<div>
<br />
<div style="text-align: center;">Vr = 0.8*0.6/(1 - 0.6*0.6) = 0.75</div>
<div style="text-align: center;"><br /></div>
<div style="text-align: center;">Vld = 0.8*1.6/(1 - 0.6*0.6) = 2.0</div>
<div style="text-align: center;"><br /></div>
<div style="text-align: center;">Vf2 = 0.8*0.6*0.6/(1 - 0.6*0.6) = 0.45</div>
<div style="text-align: center;"><br /></div>
<div style="text-align: center;">Vf = Vf1 + Vf2 = 0.8 + 0.45 = 1.25</div>
<div>
<div style="text-align: left;"><br /></div>
<div style="text-align: left;">
These results match the steady-state values shown in the Excel table, above.
</div>
<div style="text-align: left;"><br /></div>
<p><u>5. Simulink Results:</u></p>
<p>
The results derived with the above procedure can be verified with a Simulink
Simulation.
</p>
<p>Here is the Simulink model that I used for verification:</p>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://1.bp.blogspot.com/-rL-ME9hgnQc/YJk9tMLCLDI/AAAAAAAAMC0/a3cMVs5OcGgz9mwizqEwqtAY-erBUq9fwCLcBGAsYHQ/s1287/Simulink%2BModel%252C%2BZs%2B50%252C%2BZld%2B200.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="634" data-original-width="1287" src="https://1.bp.blogspot.com/-rL-ME9hgnQc/YJk9tMLCLDI/AAAAAAAAMC0/a3cMVs5OcGgz9mwizqEwqtAY-erBUq9fwCLcBGAsYHQ/s320/Simulink%2BModel%252C%2BZs%2B50%252C%2BZld%2B200.png" width="320" /></a>
</div>
<p>(Note that the model for the Directional Couplers has been shown earlier
in this post).
</p>
<p>The input is a gated sine-wave. This lets me determine Vf1 and Vf2
from the simulation results. Vf1 is the forward voltage on the
Transmission line at the start of the gated sine-wave. And Vf2 is
the forward voltage on the Transmission line <i>immediately after</i> the
gated-sine has been turned off (in other words, there is no longer a Vf1,
but the earlier re-reflections are still traveling from the tuner-end of the transmission line towards the load).</p>
<p>
You can see below that the amplitude of Vf1 is 0.8 volts and Vf2 is 0.45 volts. These
sum to the steady-state value Vf value of 1.25.</p>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://1.bp.blogspot.com/-3qDy1e68DGE/YJk94WYyM5I/AAAAAAAAMC8/6GGH3A4anuEPi7xUZelD2SRvnbyYmljrwCLcBGAsYHQ/s1076/Vf_TLin%252C%2BVr_TLin%252C%2BZs%2B50%252C%2BZld%2B200%252C%2Bannotation.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="792" data-original-width="1076" src="https://1.bp.blogspot.com/-3qDy1e68DGE/YJk94WYyM5I/AAAAAAAAMC8/6GGH3A4anuEPi7xUZelD2SRvnbyYmljrwCLcBGAsYHQ/s320/Vf_TLin%252C%2BVr_TLin%252C%2BZs%2B50%252C%2BZld%2B200%252C%2Bannotation.png" width="320" /></a>
</div>
<p>(<u>Note:</u> the magnitude of Vf, |Vf| only equals |Vf1| + |Vf2| if the reflection coefficients Γ<sub>1</sub> and Γ<sub>1</sub> at the two ends of the transmission line are real, without imaginary terms, and if the transmission line is a multiple of a half-wavelength in length. Otherwise, phase-shift is introduced between the two ends of the line and it is very likely that Vf2 will not be in phase with Vf1. In this case, vector addition still works (i.e. Vf = Vf1 + Vf2), but the sum of the <i>magnitudes</i> of Vf1 and Vf2 no longer equals the magnitude of Vf.)</p><p>And below we can see that the directional coupler's "equivalent Vref" at
the L-network's input decays to 0 in steady-state, indicating that the
L-network's input impedance looks like 50 ohms in steady-state.
</p>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://1.bp.blogspot.com/-YkRp_fWV41g/YJk9zbuguBI/AAAAAAAAMC4/U1GF2J1_SuUGP-31WP7fAUNJOExHXyi6QCLcBGAsYHQ/s1071/Vf_LCin%252C%2BVf_LCout%252C%2BZs%2B50%252C%2BZld%2B200%252C%2Bannotation.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="784" data-original-width="1071" src="https://1.bp.blogspot.com/-YkRp_fWV41g/YJk9zbuguBI/AAAAAAAAMC4/U1GF2J1_SuUGP-31WP7fAUNJOExHXyi6QCLcBGAsYHQ/s320/Vf_LCin%252C%2BVf_LCout%252C%2BZs%2B50%252C%2BZld%2B200%252C%2Bannotation.png" width="320" /></a>
</div>
<div><br /></div><p>The resulting Simulink wave-forms and their amplitudes match my calculated results.</p><div><br /></div>
<p><u>6. Summary of Forward and Reflected Voltages:</u></p>
<p>The diagram below illustrates the steady-state Forward and Reflected voltages and power when Zs equals 50 ohms:</p>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://1.bp.blogspot.com/-oL_UYjYn0Ms/YJldPkc46fI/AAAAAAAAMDk/xjSc2Yi78M8u_Ded-nZftD0t9aPAb7snACLcBGAsYHQ/s825/Transmission%2BLine%2BVf%252C%2BVf1%252C%2BVf2%252C%2BVr%2BGREEN.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="401" data-original-width="825" src="https://1.bp.blogspot.com/-oL_UYjYn0Ms/YJldPkc46fI/AAAAAAAAMDk/xjSc2Yi78M8u_Ded-nZftD0t9aPAb7snACLcBGAsYHQ/s320/Transmission%2BLine%2BVf%252C%2BVf1%252C%2BVf2%252C%2BVr%2BGREEN.png" width="320" /></a>
</div>
<div><br /></div>
<div><br /></div>
<p>
<b><u>Example with a Source Impedance <i>Not Equal</i> to 50 Ohms:</u></b>
</p>
<p>Let's set Zs to 5 ohms. Here's the new circuit:</p>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://1.bp.blogspot.com/-ub3BnYjFTKQ/YJmXoSd2gkI/AAAAAAAAMD8/WVAzmYmjSeA6NW_BwIuadiw4nWPgFCqQgCLcBGAsYHQ/s719/Circuit%252C%2BZs%2B%253D%2B5.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="248" data-original-width="719" src="https://1.bp.blogspot.com/-ub3BnYjFTKQ/YJmXoSd2gkI/AAAAAAAAMD8/WVAzmYmjSeA6NW_BwIuadiw4nWPgFCqQgCLcBGAsYHQ/s320/Circuit%252C%2BZs%2B%253D%2B5.png" width="320" /></a>
</div><div><br /></div>
The Reflection and Transmission Coefficients at the load remain
unchanged. But the Reflection Coefficient looking into the output of
the L-network (towards the source) has changed from 0.6 to the complex number 0.8995 +
j0.2621.
</div>
<div><br /></div>
<div>Note that this reflection coefficient can be found two ways:</div>
<div><br /></div>
<div>1. Circuit analysis method (in which Vs is replaced by a short):</div>
<div><br /></div>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://1.bp.blogspot.com/-AExz4q242q4/YJmYksnxKkI/AAAAAAAAMEE/0EvuNufdyZ0sQHCrzwWSKDUbH0oX-L4DACLcBGAsYHQ/s798/Calculating%2BGamma1%252C%2BZs%2B5.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="527" data-original-width="798" src="https://1.bp.blogspot.com/-AExz4q242q4/YJmYksnxKkI/AAAAAAAAMEE/0EvuNufdyZ0sQHCrzwWSKDUbH0oX-L4DACLcBGAsYHQ/s320/Calculating%2BGamma1%252C%2BZs%2B5.png" width="320" /></a>
</div>
<br />
<div>2. SimSmith method:</div>
<div><br /></div>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://1.bp.blogspot.com/-0LiS-O_Ht00/YJmYqMSGgvI/AAAAAAAAMEI/n203RCycCrIYi9H00mn8_ykvYAjNB5h3ACLcBGAsYHQ/s1353/SimSmith%2Bgamma%2B1%252C%2B5%2Bohms.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="927" data-original-width="1353" src="https://1.bp.blogspot.com/-0LiS-O_Ht00/YJmYqMSGgvI/AAAAAAAAMEI/n203RCycCrIYi9H00mn8_ykvYAjNB5h3ACLcBGAsYHQ/s320/SimSmith%2Bgamma%2B1%252C%2B5%2Bohms.png" width="320" /></a>
</div>
<br />
<div><u>Calculated Results:</u></div>
<div><br /></div>
<div>
The following results were calculated using the procedure described above
(and MATLAB).
</div>
<div><br /></div>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://1.bp.blogspot.com/-w5blBZEvUlQ/YJmmBgupz5I/AAAAAAAAMEk/l34dI2RcFBgIVtWqKjx7s2uDvhr6Pc24gCLcBGAsYHQ/s902/Excel%2BTable%252C%2BZs%2B5%2Bohms.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="705" data-original-width="902" src="https://1.bp.blogspot.com/-w5blBZEvUlQ/YJmmBgupz5I/AAAAAAAAMEk/l34dI2RcFBgIVtWqKjx7s2uDvhr6Pc24gCLcBGAsYHQ/s320/Excel%2BTable%252C%2BZs%2B5%2Bohms.png" width="320" /></a>
</div>
<br />
<div>Note the <i>steady-state</i> magnitudes of Vf, Vr, and Vload are the same
irrespective of Zs.</div>
<div><br /></div>
<div>
Examining the calculated values in their complex form, I can derive the
following diagram showing the steady-state voltages and power on the
transmission line.
</div>
<div><br /></div>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://1.bp.blogspot.com/-GGPtJo-SCkg/YJm0RtDFe1I/AAAAAAAAMFE/K4p8EEUHdc4ElJDSEVBrx2kXlvYeVjGPgCLcBGAsYHQ/s810/Steady-state%2BTransmission%2BLine%2BVoltages%2Band%2BPower%252C%2BZs%2B5.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="486" data-original-width="810" src="https://1.bp.blogspot.com/-GGPtJo-SCkg/YJm0RtDFe1I/AAAAAAAAMFE/K4p8EEUHdc4ElJDSEVBrx2kXlvYeVjGPgCLcBGAsYHQ/s320/Steady-state%2BTransmission%2BLine%2BVoltages%2Band%2BPower%252C%2BZs%2B5.png" width="320" /></a>
</div><div><br /></div>
Note that Vf1 is now a complex number (compared to the value calculated when
Zs was 50 ohms). Never the less, I can still calculate Vf2 and, with Vr,
calculate the Reflection Coefficient looking into the LC network's Output
port. Specifically, I made this calculation as follows:
<div>
<ul style="text-align: left;">
<li>Vf1 (calculated at time t = 0) = 0.1166 - j0.5911</li>
<li>
Vf (at time t = 28*Td, which is as far as I took my calculations) = 0.6199
- j1.0724.
</li>
<li>
Subtracting these two quantities, Vf2 = Vf - Vf1 = 0.5033 - j0.4814
</li>
<li>Vr (at time t = 28*Td) = 0.3721 - j0.6435</li>
</ul>
</div>
<div>
I can derive Γ<sub>1</sub> by dividing Vf2 by Vr: Γ<sub>1</sub> = Vf2/Vr = 0.8995
+ j0.2621, which is the same value calculated using circuit analysis
techniques.</div>
<div><br /></div>
<div>
(Note that because Γ<sub>1</sub> now has an imaginary part, the <i>magnitudes</i> of Vf1 and Vf2 no longer sum to be the magnitude of Vf, as it did in the Zs = 50 ohm example, above. But Vf still equals Vf1 + Vf2, but one needs to use vector arithmetic for the calculations.)</div><div>
<div>
<div><br /></div>
<div><br /></div>
<div><u>Simulated Results:</u></div>
<div><br /></div>
<div>The circuit was simulated with Simulink:</div>
<div><br /></div>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://1.bp.blogspot.com/-JHMMXFxGKc0/YJmbhC1lUZI/AAAAAAAAMEc/XdD1FLjvTkYq5xBnNxKjaW4fpQP19JdEQCLcBGAsYHQ/s1278/Simulink%2Bckt%252C%2BZs%2B5%2Bohms.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="619" data-original-width="1278" src="https://1.bp.blogspot.com/-JHMMXFxGKc0/YJmbhC1lUZI/AAAAAAAAMEc/XdD1FLjvTkYq5xBnNxKjaW4fpQP19JdEQCLcBGAsYHQ/s320/Simulink%2Bckt%252C%2BZs%2B5%2Bohms.png" width="320" /></a>
</div>
<br />
<div>I adjusted the source amplitude from 2 to 1.09 to set the steady-state Load
voltage very close to 2.0 volts peak -- this is the same Vload level for the 50 ohm Zs simulations).
</div>
<div><br /></div>
<div>
Note that I've increased the length of the transmission line from 1
lambda to 2 lambda. This gives the signal a bit more time (i.e. 4
cycles) to settle down before the next reflection arrives (it takes a
bit longer for the signal to settle when Zs is 5 ohms).<br />
<p>
The Simulation results are below. Note that an SWR meter at the
input to the LC network would still indicate an SWR of 1:1 after a
dozen or so cycles following startup:
</p>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://1.bp.blogspot.com/-NReW8cMv2Hs/YJmqxO6k2dI/AAAAAAAAMEs/4GvdB8GMdSYpnB0NKV_iULwDDA9EkL41ACLcBGAsYHQ/s1071/Va%2BVF%2Band%2BVr%252C%2BZs%2B5%2Bohms.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="784" data-original-width="1071" src="https://1.bp.blogspot.com/-NReW8cMv2Hs/YJmqxO6k2dI/AAAAAAAAMEs/4GvdB8GMdSYpnB0NKV_iULwDDA9EkL41ACLcBGAsYHQ/s320/Va%2BVF%2Band%2BVr%252C%2BZs%2B5%2Bohms.png" width="320" /></a>
</div>
<p>The Transmission Line voltages Vf and Vr, and the output voltage
Vload, have the same <i>steady-state</i> values as they had in the Zs = 50
ohm simulation. But the transient voltage levels at startup and shut-down differ from the
50 ohm simulation. However, they are consistent with the
calculations tabulated for Zs = 5 ohms, above. So we can say
that the simulation validates the calculations performed using the
lumped-element and transmission line models.
</p>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://1.bp.blogspot.com/-prTiP3InHDk/YJmtH3lYHpI/AAAAAAAAME8/63Mh_D5YOUAI23zlUNt8dBTnlwRE6TipwCLcBGAsYHQ/s1071/TL%2BVF%2Band%2BVr%252C%2BZs%2B5%2Bohms.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="792" data-original-width="1071" src="https://1.bp.blogspot.com/-prTiP3InHDk/YJmtH3lYHpI/AAAAAAAAME8/63Mh_D5YOUAI23zlUNt8dBTnlwRE6TipwCLcBGAsYHQ/s320/TL%2BVF%2Band%2BVr%252C%2BZs%2B5%2Bohms.png" width="320" /></a>
</div>
<div><br /></div>From these plots we can still get an idea of the <i>magnitude</i> of reflection coefficient
looking into the L-network's output port (we cannot get the angle because these plots don't show us the phase relationship between Vr and Vf2). Note
that |Vf2| is about 0.69 volts (this is measured on the "back porch" of
Vf_TLin, just after the source turns off (and waiting a cycle or two for
the level to stabilize), while |Vr| (in steady-state) is about 0.744
volts.</div>
<div><br /></div>
<div>
|Γ<sub>1</sub>| equals |Vf2|/|Vr| = 0.69/0.74 = 0.93, which is quite
close to the magnitude of the calculated Γ<sub>1</sub> (=
0.9369)<br /><br />
<p>
<b><u>Conclusions:</u></b>
</p>
<p>
1. The Reflection Coefficient looking <i>into the output</i> of
the Impedance Matching network (Antenna Tuner) is a function of the
source impedance and, for a source modeled as a voltage with a series source impedance, can be calculated using basic circuit analysis
techniques.
</p>
<p>
2. This Reflection Coefficient may or may not be the Complex
Conjugate of the load impedance connected to the Output port of the
Impedance Matching network. If the source impedance is 50 ohms
(and there is no other loss in the system), then it is the complex conjugate and thus there is a Conjugate
Match (this can be quickly demonstrated with SimSmith). But for
all other possible source impedances, there is no Conjugate Match.
</p>
<p>
3. Irrespective of presence or absence of a Conjugate Match, an
SWR meter at the input of the Impedance Matching network will show an
SWR of 1:1 when the Impedance Matching network has been tuned so that
the impedance <i>looking into</i> the Impedance Matching network's
input port is 50 + j0 ohms. The transmitter's source resistance has no effect on this tuning.</p>
<p>
<b><br /></b>
</p>
<p><b><u>Notes:</u></b></p>
<p>
If you'd like to play with the Simulink models, you can find them
here: <a href="https://github.com/k6jca/Antenna-Tuner-Simulink-Models">https://github.com/k6jca/Antenna-Tuner-Simulink-Models</a>
</p>
<p>They were created with Simulink R2020a (version 10.1), so if you have an earlier version of Simulink, you might not be able to run them.</p>
<p><br /></p><p><span style="text-align: center;"><b><u>Other Transmission-Line Posts:</u></b></span></p><p><a href="http://k6jca.blogspot.com/2021/02/antenna-tuners-transient-and-steady.html" style="text-align: center;">http://k6jca.blogspot.com/2021/02/antenna-tuners-transient-and-steady.html</a>. This post analyzes the transient and steady-state response of a simple impedance matching system consisting of a wide-band transformer. I calculate the system's impulse response and find the time-domain response by convolving this impulse-response with a stimulus signal.</p><p><span style="text-align: center;"><a href="http://k6jca.blogspot.com/2021/02/the-quarter-wave-transformer-transient.html">http://k6jca.blogspot.com/2021/02/the-quarter-wave-transformer-transient.html</a>. <span style="text-align: left;"> This post analyzes the transient and steady-state response of a Quarter-Wave Transformer impedance matching device. I calculate the system's impulse response and find the time-domain response by convolving this impulse-response with a stimulus signal.</span></span></p><div><a href="http://k6jca.blogspot.com/2021/03/useful-swr-voltage-and-power-equations.html" style="text-align: center;">http://k6jca.blogspot.com/2021/03/useful-swr-voltage-and-power-equations.html</a>. This post lists (in an easily accessible location that I can find!) some equations that I find useful</div><p><span style="text-align: center;"><a href="http://k6jca.blogspot.com/2021/05/antenna-tuners-lumped-element-tuner.html">http://k6jca.blogspot.com/2021/05/antenna-tuners-lumped-element-tuner.html</a>. This post analyzes the transient and steady-state reflections of a lumped-element tuner (i.e. the common antenna tuner). I describe a method for making these calculations, and I note that the tuner's match is independent of the source impedance.</span></p><p><span style="text-align: center;"><a href="http://k6jca.blogspot.com/2021/05/lc-network-reflection-and-transmission.html">http://k6jca.blogspot.com/2021/05/lc-network-reflection-and-transmission.html</a>. This post describes how to calculate the "Transmission Coefficient" through a lumped-element network (and also its Reflection Coefficient) if it were inserted into a transmission line. <br /></span></p><p><a href="http://k6jca.blogspot.com/2021/09/does-source-impedance-affect-swr.html" style="color: #2b00fe;">http://k6jca.blogspot.com/2021/09/does-source-impedance-affect-swr.html</a><span style="color: #2b00fe;">. </span>This post shows mathematically that source impedance does not affect a transmission line's SWR. This conclusion is then demonstrated with Simulink simulations.<br /></p><p><a href="https://k6jca.blogspot.com/2021/10/revisiting-maxwells-tutorial-concerning.html">https://k6jca.blogspot.com/2021/10/revisiting-maxwells-tutorial-concerning.html</a> This posts revisits Walt Maxwell's 2004 QEX rebuttal of Steven Best's 2001 3-part series on Transmission Line Wave Mechanics. In this post I show simulation results which support Best's conclusions.</p><p><br /></p>
<p><b>Standard Caveat:</b></p>
<p>
As always, I might have made a mistake in my equations, assumptions,
drawings, or interpretations. If you see anything you believe to
be in error or if anything is confusing, please feel free to contact
me or comment below.
<br /><br />
And so I should add -- this information is distributed in the hope
that it will be useful, but WITHOUT ANY WARRANTY; without even the
implied warranty of MERCHANTABILITY or FITNESS FOR A PARTICULAR
PURPOSE.
</p>
</div>
</div>
</div>
</div>
Jeffhttp://www.blogger.com/profile/04853314106806116765noreply@blogger.com0tag:blogger.com,1999:blog-2257458838727315711.post-79796425819462188582021-03-23T06:58:00.006-07:002021-03-26T06:05:36.708-07:00Notes on Antenna Tuners: The Elecraft T1 (by Kai Siwiak, KE4PT)<p>
<span style="color: #2b00fe;">In addition to his analysis of the </span><a href="http://k6jca.blogspot.com/2021/03/notes-on-antenna-tuners-icom-ah-4-by.html"><span style="color: black;">Icom AH-4 Tuner</span></a><span style="color: #2b00fe;">, Kai Siwiak has also analyzed the Elecraft T1 Automatic Antenna
Tuner. With his permission, his analysis appears, below:</span>
</p>
<p>
<span style="color: #2b00fe;"><br /></span>
</p>
<div style="text-align: left;">
<div style="text-align: center;">
<b>Elecraft T1 Automatic Tuner </b>
</div>
<div style="text-align: center;">Kai Siwiak, KE4PT </div>
<div style="text-align: center;"><br /></div>
</div>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://1.bp.blogspot.com/-Xlk-DRZyJMY/YFkMdfLFmuI/AAAAAAAAL-I/es91YfBMDd05Bsgat17DtAYKQHfa4oyAACLcBGAsYHQ/s757/Kai%252C%2BSchematic.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="335" data-original-width="757" src="https://1.bp.blogspot.com/-Xlk-DRZyJMY/YFkMdfLFmuI/AAAAAAAAL-I/es91YfBMDd05Bsgat17DtAYKQHfa4oyAACLcBGAsYHQ/s320/Kai%252C%2BSchematic.png" width="320" /></a>
</div>
<p>
<b><u>Tuner Description:</u></b>
</p>
<p>
The Elecraft T1 tuner is an L-network tuner whose variable inductance is in
series between the Radio-side and the Antenna-side ports of the tuner.
This inductance consists of 7 fixed inductors that allow the inductance to be
varied from 0 to 7.5 uH in 0.05 uH steps. The tuner's variable
capacitance consists of 7 fixed capacitors that allow the capacitance to be
varied from 0 to 1300 pF in 10 pF steps . This variable capacitance can
be switched to connect either between the tuner's input and ground, or between
the tuner's output and ground.
</p>
<p>
These seven fixed inductors and seven fixed capacitors, along with the ability
to connect the capacitance to either the input or the output of the tuner, are
switched with latching relays, and results in a total of 2^7 x 2^7 x 2 =
32,768 different tuning combinations.
</p>
<p>
Latching relays are used so the tuner draws no power from the battery except
when tuning.
</p>
<p>
A Stockton bridge circuit detects forward and reflected power, from which the
SWR is calculated. A modulated SSB transmission can be used for tuning, with
almost as good accuracy as a constant carrier.
</p>
<p>
The microprocessor tries a coarse tuning algorithm to roughly determine the
antenna impedance. This is followed by fine and very fine algorithms to seek
the best possible match (unlike many auto-ATUs which stop searching once they
have achieved an SWR below a certain level). The settings and band are stored,
allowing the T1 to return to this setting instantly the next time that band is
used.
</p>
<p>
Unlike many other auto-ATUs, the T1 is switched off once it has found a
match. Thus, it does not constantly monitor the SWR in order to
automatically re-tune if the SWR changes. The user must manually initiate a
re-tune if required.
</p>
<p>
The T1 can also be turned on using the external control interface, which can
provide information about the band selected on the transceiver, allowing the
T1 to automatically tune the antenna using the previously stored settings.
Currently, this is only possible with the Yaesu FT-817 transceiver and
optional adapter, but Elecraft has provided information about the serial data
protocol used, to allow interfaces for other low-power transceivers to be
designed. source: http://www.g4ilo.com/t1.html
</p>
<p><br /></p>
<p>
<b><u>Tuner Analysis:</u></b>
</p>
I terminated the “Radio side” of the tuner with a "perfect" 50 ohm resistor and
used MathCAD to calculate the impedances on the “Antenna side” while stepping
through the allowed component values (both L and C) for two cases: 1) the
0 – 1300 pF capacitor bank on the Antenna-side of the tuner, then 2), this same
capacitor bank connected to the Radio-side of the tuner.
<p></p>
<p>
I then plotted on a Smith chart the complex conjugate of the calculated
impedances on the “Antenna side”. This represents the impedances, when
connected to the Antenna-side of the tuner, that the T1 can match to an SWR of
1:1.
</p>
<p>Analysis covers the ham bands from 1.8 to 54 MHz.</p>
<p>Please note:</p>
<p>
Assuming that the frequency derivative of the antenna impedance is not so
large that the tuning algorithm can’t home in on it …
</p>
<p></p>
<ul style="text-align: left;">
<li>
The 32,768 blue and magenta points on the Smith charts show which impedances
at the tuner antenna port can be transformer to 50 ohms
</li>
<li>
16,384 Blue points are for the bank of Caps on the antennas side
</li>
<li>
16,384 Magenta points are for the bank of Caps on the transmitter side
</li>
</ul>
<p></p>
<p>If there is a length of coax between the tuner and the radiator … </p>
<p></p>
<ul style="text-align: left;">
<li>The coax losses will increase slightly </li>
<li>
The points indicated on the Smith chart would need to be rotated
<u><i>counter-clockwise</i></u> at a rate of one turn per electrical
effective half wavelength of coax.
</li>
</ul>
<div>
The range of input SWRs that the T-1 can match to an SWR of 1:1 on 160 and 80
meters is limited, as shown in the plots, below. The T1's designer,
Wayne Burdick, N6KR, says that a constructor whose interests lean towards LF
could double the value of each inductor to improve the matching range on 160m,
at the expense of 6m.
</div>
<div><br /></div>
<p></p>
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<br />
<p><br /></p>
<p>
<b><u><span style="color: #2b00fe;">Some Additional Notes (by Jeff, K6JCA)</span></u></b>
</p>
<div>
<span style="color: #2b00fe;">1. Actual component values (due to manufacturing tolerances, etc.) will
manifest as unequal spacing between some of the impedances mapped onto the
Smith Chart, compared to the simulated values. </span><span style="color: #2b00fe;">(For an example, compare the two plots below from: </span><a href="http://k6jca.blogspot.com/2015/05/notes-on-antenna-tuners-elecraft-kat500.html"><span style="color: black;">k6jca -- KAT500 notes</span></a><span style="color: #2b00fe;">).</span>
</div><div><span><span style="color: #2b00fe;"><br /></span></span></div><div><span><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-ShDgjB1DjDc/YFuR3PsYXNI/AAAAAAAAL-o/Fdmr0ic_2UwcdmWcMaVa55OvEh2YfEXQQCLcBGAsYHQ/s1488/kat500%2Bideal%2Bvs%2Bactual%2Bparts%2Bvalues.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="735" data-original-width="1488" src="https://1.bp.blogspot.com/-ShDgjB1DjDc/YFuR3PsYXNI/AAAAAAAAL-o/Fdmr0ic_2UwcdmWcMaVa55OvEh2YfEXQQCLcBGAsYHQ/s320/kat500%2Bideal%2Bvs%2Bactual%2Bparts%2Bvalues.png" width="320" /></a></div><span><span style="color: #2b00fe;"><div><span><span><span style="color: #2b00fe;"><br /></span></span></span></div>2. The L-network is assumed to be lossless. In reality, there
will be some loss, and this loss introduces errors into the
"complex-conjugate" technique that both Kai and I (in other posts) use to
calculate the impedances that the tuner will match. You can find
more on this topic in my Drake MN-4 post: </span><a href="http://k6jca.blogspot.com/search/label/Antenna%20Tuner%3A%20Drake%20MN-4"><span style="color: black;">k6jca -- Drake MN-4</span></a></span><span style="color: #2b00fe;">. Look for the subheading "Using MATLAB to Examine Loss Effects in
Match-space Plots", towards the end of the post.</span></span></div>
<div>
<span style="color: #2b00fe;"><br /></span>
</div><div><span style="color: #2b00fe;">3. On 6 meters the T1 might have a difficult timing finding an acceptable match, given the parts values listed in its Bill of Materials. The plot below shows the <i>best</i> SWR that loads with an SWR of 10:1 (or better) can be tuned to at 54 MHz. (The outer ring of the "donut" represents loads with a 10:1 SWR, while the inner ring represents loads with a 1.5:1 SWR.)</span></div><div><span style="color: #2b00fe;"><br /></span></div><div><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-QBcwsOCbXkM/YF3ZJ2yvzAI/AAAAAAAAL-0/kjGpKap0NPA2o84m_1sY8ppaRJMVDSWyQCLcBGAsYHQ/s787/210325%2BElecraft%2BT1%2BWorst-case%2BSWR%252C%2B54MHz.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="787" data-original-width="772" height="320" src="https://1.bp.blogspot.com/-QBcwsOCbXkM/YF3ZJ2yvzAI/AAAAAAAAL-0/kjGpKap0NPA2o84m_1sY8ppaRJMVDSWyQCLcBGAsYHQ/s320/210325%2BElecraft%2BT1%2BWorst-case%2BSWR%252C%2B54MHz.png" /></a></div><br /><span style="color: #2b00fe;">And below is the same plot for 30 MHz. All loads with an SWR of 10:1 or better now tune to SWRs better than 2:1 (per the scale on the right-hand side).</span></div><div><span style="color: #2b00fe;"><br /></span></div><div><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-bdUJLmh-j-k/YF3ZrbJFSPI/AAAAAAAAL-8/RRUoKApwb80KjRYaJ8PHqyjgcp_cOu1jwCLcBGAsYHQ/s787/210325%2BElecraft%2BT1%2BWorst-case%2BSWR%252C%2B30MHz.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="787" data-original-width="772" height="320" src="https://1.bp.blogspot.com/-bdUJLmh-j-k/YF3ZrbJFSPI/AAAAAAAAL-8/RRUoKApwb80KjRYaJ8PHqyjgcp_cOu1jwCLcBGAsYHQ/s320/210325%2BElecraft%2BT1%2BWorst-case%2BSWR%252C%2B30MHz.png" /></a></div><div><br /></div><span><span style="color: #2b00fe;">And a plot for 3.5 MHz:</span></span></div><div><span><span style="color: #2b00fe;"><br /></span></span></div><div><span><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-IwTIpOQcv0E/YF3bK1ZZIvI/AAAAAAAAL_E/3nWHKE78x9M9bu67zkmMAlZjPzgaKGptQCLcBGAsYHQ/s787/210325%2BElecraft%2BT1%2BWorst-case%2BSWR%252C%2B3M5Hz.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="787" data-original-width="772" height="320" src="https://1.bp.blogspot.com/-IwTIpOQcv0E/YF3bK1ZZIvI/AAAAAAAAL_E/3nWHKE78x9M9bu67zkmMAlZjPzgaKGptQCLcBGAsYHQ/s320/210325%2BElecraft%2BT1%2BWorst-case%2BSWR%252C%2B3M5Hz.png" /></a></div><span style="color: #2b00fe;"><div><span><span style="color: #2b00fe;"><br /></span></span></div>Plots made using the MATLAB code discussed here: </span><span style="color: black;"><a href="http://k6jca.blogspot.com/search/label/Smith%20Charts%3A%203-D%20Data%20Plots">k6jca - 3D Smith Charts</a>.</span></span></div><div><br /></div>
<div><span style="color: #2b00fe;">- Jeff, k6jca</span></div>
<div><br /></div>
<div>
<p>
<b><u>Antenna Tuner Blog Posts:</u></b><br /><br />A quick tutorial on Smith Chart basics:<br /><a href="http://k6jca.blogspot.com/2015/03/a-brief-tutorial-on-smith-charts.html">http://k6jca.blogspot.com/2015/03/a-brief-tutorial-on-smith-charts.html</a><br /><br />Plotting Smith Chart Data in 3-D:<br /><a href="http://k6jca.blogspot.com/2018/09/plotting-3-d-smith-charts-with-matlab.html">http://k6jca.blogspot.com/2018/09/plotting-3-d-smith-charts-with-matlab.html</a><br /><br />The L-network:<br /><a href="http://k6jca.blogspot.com/2015/03/notes-on-antenna-tuners-l-network-and.html">http://k6jca.blogspot.com/2015/03/notes-on-antenna-tuners-l-network-and.html</a><br /><br />A correction to the usual L-network design constraints:<br /><a href="http://k6jca.blogspot.com/2015/04/revisiting-l-network-equations-and.html">http://k6jca.blogspot.com/2015/04/revisiting-l-network-equations-and.html</a><br /><br />Calculating L-Network values when the components
are <i>lossy</i>:<br /><a href="http://k6jca.blogspot.com/2018/09/l-networks-new-equations-for-better.html">http://k6jca.blogspot.com/2018/09/l-networks-new-equations-for-better.html</a><br /><br />A look at highpass T-Networks:<br /><a href="http://k6jca.blogspot.com/2015/04/notes-on-antenna-tuners-t-network-part-1.html">http://k6jca.blogspot.com/2015/04/notes-on-antenna-tuners-t-network-part-1.html</a><br /><br />More on the W8ZR EZ-Tuner:<br /><a href="http://k6jca.blogspot.com/2015/05/notes-on-antenna-tuners-more-on-w8zr-ez.html">http://k6jca.blogspot.com/2015/05/notes-on-antenna-tuners-more-on-w8zr-ez.html</a> (Note that this tuner is also discussed in the highpass T-Network
post).<br /><br />The Elecraft KAT-500:<br /><a href="http://k6jca.blogspot.com/2015/05/notes-on-antenna-tuners-elecraft-kat500.html">http://k6jca.blogspot.com/2015/05/notes-on-antenna-tuners-elecraft-kat500.html</a><br /><br />The Nye Viking MB-V-A tuner and the Rohde Coupler:<br /><a href="http://k6jca.blogspot.com/2015/05/notes-on-antenna-tuners-nye-viking-mb-v.html">http://k6jca.blogspot.com/2015/05/notes-on-antenna-tuners-nye-viking-mb-v.html</a><br /><br />The Drake MN-4 Tuner:<br /><a href="http://k6jca.blogspot.com/2018/08/notes-on-antenna-tuners-drake-mn-4.html">http://k6jca.blogspot.com/2018/08/notes-on-antenna-tuners-drake-mn-4.html</a><br /><br />The Icom AH-4 Tuner (by Kai Siwiak, KE4PT):<br /><a href="http://k6jca.blogspot.com/2021/03/notes-on-antenna-tuners-icom-ah-4-by.html">http://k6jca.blogspot.com/2021/03/notes-on-antenna-tuners-icom-ah-4-by.html</a><br /><br />The Elecraft T1 Tuner (by Kai Siwiak, KE4PT):<br /><a href="https://k6jca.blogspot.com/2021/03/notes-on-antenna-tuners-elecraft-t1-by.html">https://k6jca.blogspot.com/2021/03/notes-on-antenna-tuners-elecraft-t1-by.html</a><br /><br />Measuring a Tuner's "Match-Space":<br /><a href="http://k6jca.blogspot.com/2018/08/notes-on-antenna-tuners-determining.html" style="font-family: "times new roman";">http://k6jca.blogspot.com/2018/08/notes-on-antenna-tuners-determining.html</a><br /><br />Measuring Tuner Power Loss:<br /><a href="http://k6jca.blogspot.com/2018/08/additional-notes-on-measuring-antenna.html">http://k6jca.blogspot.com/2018/08/additional-notes-on-measuring-antenna.html</a><br /><br />
</p>
<p>
<b><u>Standard Caveat:</u></b>
</p>
<div>
I might have made a mistake in my designs, equations, schematics, models,
etc. If anything looks confusing or wrong to you, please feel free to
comment below or send me an email.<br /><br />Also, I will note:<br /><br />This
design and any associated information is distributed in the hope that it
will be useful, but WITHOUT ANY WARRANTY; without even the implied warranty
of MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.
</div>
</div>
Jeffhttp://www.blogger.com/profile/04853314106806116765noreply@blogger.com0tag:blogger.com,1999:blog-2257458838727315711.post-33490541228122653712021-03-22T07:17:00.002-07:002021-03-22T14:47:28.855-07:00Notes on Antenna Tuners: The Icom AH-4 (by Kai Siwiak, KE4PT)<p>
<span style="color: #2b00fe;">Kai Siwiak, KE4PT, forwarded to me his analysis of the Icom AH-4 Remote
Tuner. With his permission I have added it to this blog.</span>
</p>
<p>
<span style="color: #2b00fe;"><br /></span>
</p>
<p></p>
<div style="text-align: left;">
<div style="text-align: center;">
<b>Icom AH-4 Tuner Analysis and Performance</b>
</div>
<div style="text-align: center;">Kai Siwiak, KE4PT</div>
<div style="text-align: center;"><br /></div>
</div>
<p>I analyzed the Icom AH-4 tuner using the circuit of Figure 1. </p>
<p></p>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://1.bp.blogspot.com/-_5poR4LOXhk/YFfk9xISH1I/AAAAAAAAL7o/rv40t7aS2yga-OrLoylF3QGU13BolrZiQCLcBGAsYHQ/s1129/Kai%2B-%2Bfigure%2B1.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="512" data-original-width="1129" src="https://1.bp.blogspot.com/-_5poR4LOXhk/YFfk9xISH1I/AAAAAAAAL7o/rv40t7aS2yga-OrLoylF3QGU13BolrZiQCLcBGAsYHQ/s320/Kai%2B-%2Bfigure%2B1.png" width="320" /></a>
</div>
<div class="separator" style="clear: both; text-align: center;"><br /></div>
<p>
<b><u>Tuner Analysis:</u></b>
</p>
I terminated the “Radio side” of the tuner with a "perfect" 50 ohm resistor and
used MathCAD to calculate the impedances on the “Antenna side” while stepping
through the allowed component values (both L and C) for two cases (0 – 2400 pF
capacitor bank on the output, then on the input), and across ham bands from 1.8
to 54 MHz.
<p></p>
<p>
I then plotted on a Smith chart the complex conjugate of the calculated
impedances on the “Antenna side”. This maps the impedances that the AH-4
can match, given the range of component values in Figure 1. Figure 2 shows the
range of impedances the AH-4 can match plotted on a Smith chart.
</p>
<p></p>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://1.bp.blogspot.com/-irXuIEuKyQU/YFfoXjFSmDI/AAAAAAAAL74/POoeo6w9Ytg9ETMrtxkzCwR1UgKgo2BTQCLcBGAsYHQ/s1045/Kai%2B-%2Bfigure%2B2.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="841" data-original-width="1045" src="https://1.bp.blogspot.com/-irXuIEuKyQU/YFfoXjFSmDI/AAAAAAAAL74/POoeo6w9Ytg9ETMrtxkzCwR1UgKgo2BTQCLcBGAsYHQ/s320/Kai%2B-%2Bfigure%2B2.png" width="320" /></a>
</div>
<br />Notice how the coverage on the Smith chart gets whittled away as the
frequency decreases. It indicates that the component values do not have
sufficient range to cover the whole chart at those frequencies!
<p></p>
<p>
Impedance coverage is not continuous, but depends on the step size of the
components, and that granularity is revealed here by the inductor and
capacitor steps in Figure 1.
</p>
<p>
What this implies is that the tuner range of matching can be completely
specified by:
</p>
<p>
(1) the tuner topology, here in Figure 1, a reversible L network with
additional switched capacitance on the antenna side;
</p>
<p>
(2) the range of component values and the component step size; usually the
minimum value of the inductor and minimum value of the capacitor in the
capacitor bank.
</p>
<p>
Also what is not revealed here is the tuning algorithm and tuning strategy.
These are typically proprietary features. Given estimates of the inductor Q it
would be possible to calculate the tuner losses across the Smith chart at
various frequencies. The AH-4 inductors are air-wound, so will be lower loss
than ferrite inductors.
</p>
<p>
On tuning strategy, the AH-4 initially sets the connected compatible radio RF
power to 10 W, and switches in a 10 dB attenuator between the radio and the
tuner. Thus no more than 1 W is every supplied to the antenna during tuning,
and the radio transmitter always “sees” more than 20 dB return loss (SWR <
1.2:1) during the tuning process, ensuring a valid tuning
solution.
</p>
<p>
I have similarly analyzed the Elecraft T1 miniature ATU, which also uses a
reversible L network with 0 to 1300 pF by 10 pF and 0 to 7.5 uH by 0.055 uH
(32,768 combinations). Its range of component values is smaller than that of
the AH-4, so it cover less of the Smith chart. Its inductors are
<i>not</i> air core, so losses will be higher than with the AH-4.
</p>
<p>Kindest regards, </p>
<p>Kai Siwiak, KE4PT</p>
<p><br /></p>
<p>
<b><u><span style="color: #2b00fe;">Some Additional Notes (by Jeff, K6JCA)</span></u></b>
</p>
<p>
<span style="color: #2b00fe;">To verify Kai's calculations, I analyzed the AH-4 using a MATLAB script I
had written for analyzing tuners (e.g. the KAT-500). If you compare my
plotted results, below, with Kai's Figure 2, you will see that they match
quite well.</span>
</p>
<p></p>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://1.bp.blogspot.com/-FAL4aHpbdU8/YFfzXUdXcKI/AAAAAAAAL8k/tFz_dBQlg9scMPzdJ93C941PYB2xqXD-wCLcBGAsYHQ/s576/k6jca%2Bah-4%2B1M8Hz.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="512" data-original-width="576" src="https://1.bp.blogspot.com/-FAL4aHpbdU8/YFfzXUdXcKI/AAAAAAAAL8k/tFz_dBQlg9scMPzdJ93C941PYB2xqXD-wCLcBGAsYHQ/s320/k6jca%2Bah-4%2B1M8Hz.png" width="320" /></a>
</div>
<div class="separator" style="clear: both; text-align: center;"><br /></div>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://1.bp.blogspot.com/-Oqags0_wAFA/YFfzrj2WHzI/AAAAAAAAL80/SCbznfUHfU89HrPaXgBbUj8LCuRhGie9QCLcBGAsYHQ/s576/k6jca%2Bah-4%2B4MHz.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="512" data-original-width="576" src="https://1.bp.blogspot.com/-Oqags0_wAFA/YFfzrj2WHzI/AAAAAAAAL80/SCbznfUHfU89HrPaXgBbUj8LCuRhGie9QCLcBGAsYHQ/s320/k6jca%2Bah-4%2B4MHz.png" width="320" /></a>
</div>
<div class="separator" style="clear: both; text-align: center;"><br /></div>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://1.bp.blogspot.com/-hs-rVc1QqKo/YFfzeMiQvdI/AAAAAAAAL8s/nAEzYS23hzULX-QLHo7hxgkpnE62jtEoQCLcBGAsYHQ/s576/k6jca%2Bah-4%2B7MHz.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="512" data-original-width="576" src="https://1.bp.blogspot.com/-hs-rVc1QqKo/YFfzeMiQvdI/AAAAAAAAL8s/nAEzYS23hzULX-QLHo7hxgkpnE62jtEoQCLcBGAsYHQ/s320/k6jca%2Bah-4%2B7MHz.png" width="320" /></a>
</div>
<p></p>
<div>
<span style="color: #2b00fe;">And here's a plot at 54 MHz. You can see the effect of component
step-size:</span>
</div>
<div>
<span style="color: #2b00fe;"><br /></span>
</div>
<div>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://1.bp.blogspot.com/-f0BNYUXAwoQ/YFfzTjAUgXI/AAAAAAAAL8g/aUwLeuM12jsZJNJdvXJWVpNJjZqsWk9JACLcBGAsYHQ/s576/k6jca%2Bah-4%2B54MHz.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="512" data-original-width="576" src="https://1.bp.blogspot.com/-f0BNYUXAwoQ/YFfzTjAUgXI/AAAAAAAAL8g/aUwLeuM12jsZJNJdvXJWVpNJjZqsWk9JACLcBGAsYHQ/s320/k6jca%2Bah-4%2B54MHz.png" width="320" /></a>
</div>
<br /><span style="color: #2b00fe;"><br /></span>
</div>
<div>
<span style="color: #2b00fe;">Some notes regarding these plots:</span>
</div>
<div>
<span style="color: #2b00fe;"><br /></span>
</div>
<div>
<span style="color: #2b00fe;">1. Step-size is assumed to be uniform. But in reality this is
not the case -- component values might have been selected for their ability
to be sourced, rather than their "ideal" value, and these component values
will also have tolerances. </span><span style="color: #2b00fe;">(See the plots towards the end of the KAT500 analysis for the effect of
"real" components upon a tuner's match-space: </span><a href="http://k6jca.blogspot.com/2015/05/notes-on-antenna-tuners-elecraft-kat500.html"><span style="color: black;">k6jca -- KAT500 notes</span></a><span style="color: #2b00fe;">).</span>
</div>
<div>
<span style="color: #2b00fe;"><br /></span>
</div>
<div>
<span style="color: #2b00fe;">2. I have </span><span style="color: #2b00fe;">ignored the 13 pF and the 100 pF capacitors that can be switched in and out
of the circuit at the Antenna connector side of the tuner. I suspect
these are used to compensate for parasitic impedances when the tuner is in
the CpLs mode, or possibly they bring the tuner's SWR close to 1:1 when the
tuner is in "bypass" mode).</span>
</div>
<div>
<span style="color: #2b00fe;"><br /></span>
</div>
<div>
<span><span style="color: #2b00fe;">3. The network is assumed to be lossless. In reality, there
will be some loss, and this loss introduces errors into the
"complex-conjugate" technique that both Kai and I use to calculate the
impedances that the tuner will match. You can find more on this topic
in my Drake MN-4 post: </span><a href="http://k6jca.blogspot.com/search/label/Antenna%20Tuner%3A%20Drake%20MN-4"><span style="color: black;">k6jca -- Drake MN-4</span></a></span><span style="color: #2b00fe;">. Look for the subheading "Using MATLAB to Examine Loss Effects in
Match-space Plots", towards the end of the post.</span>
</div>
<div>
<span style="color: #2b00fe;"><br /></span>
</div>
<div><span style="color: #2b00fe;">- Jeff, k6jca</span></div>
<div><br /></div>
<div>
<p>
<b><u>Antenna Tuner Blog Posts:</u></b><br /><br />A quick tutorial on Smith Chart basics:<br /><a href="http://k6jca.blogspot.com/2015/03/a-brief-tutorial-on-smith-charts.html">http://k6jca.blogspot.com/2015/03/a-brief-tutorial-on-smith-charts.html</a><br /><br />Plotting Smith Chart Data in 3-D:<br /><a href="http://k6jca.blogspot.com/2018/09/plotting-3-d-smith-charts-with-matlab.html">http://k6jca.blogspot.com/2018/09/plotting-3-d-smith-charts-with-matlab.html</a><br /><br />The L-network:<br /><a href="http://k6jca.blogspot.com/2015/03/notes-on-antenna-tuners-l-network-and.html">http://k6jca.blogspot.com/2015/03/notes-on-antenna-tuners-l-network-and.html</a><br /><br />A correction to the usual L-network design constraints:<br /><a href="http://k6jca.blogspot.com/2015/04/revisiting-l-network-equations-and.html">http://k6jca.blogspot.com/2015/04/revisiting-l-network-equations-and.html</a><br /><br />Calculating L-Network values when the components
are <i>lossy</i>:<br /><a href="http://k6jca.blogspot.com/2018/09/l-networks-new-equations-for-better.html">http://k6jca.blogspot.com/2018/09/l-networks-new-equations-for-better.html</a><br /><br />A look at highpass T-Networks:<br /><a href="http://k6jca.blogspot.com/2015/04/notes-on-antenna-tuners-t-network-part-1.html">http://k6jca.blogspot.com/2015/04/notes-on-antenna-tuners-t-network-part-1.html</a><br /><br />More on the W8ZR EZ-Tuner:<br /><a href="http://k6jca.blogspot.com/2015/05/notes-on-antenna-tuners-more-on-w8zr-ez.html">http://k6jca.blogspot.com/2015/05/notes-on-antenna-tuners-more-on-w8zr-ez.html</a> (Note that this tuner is also discussed in the highpass T-Network
post).<br /><br />The Elecraft KAT-500:<br /><a href="http://k6jca.blogspot.com/2015/05/notes-on-antenna-tuners-elecraft-kat500.html">http://k6jca.blogspot.com/2015/05/notes-on-antenna-tuners-elecraft-kat500.html</a><br /><br />The Nye Viking MB-V-A tuner and the Rohde Coupler:<br /><a href="http://k6jca.blogspot.com/2015/05/notes-on-antenna-tuners-nye-viking-mb-v.html">http://k6jca.blogspot.com/2015/05/notes-on-antenna-tuners-nye-viking-mb-v.html</a><br /><br />The Drake MN-4 Tuner:<br /><a href="http://k6jca.blogspot.com/2018/08/notes-on-antenna-tuners-drake-mn-4.html">http://k6jca.blogspot.com/2018/08/notes-on-antenna-tuners-drake-mn-4.html</a><br /><br />The Icom AH-4 Tuner:<br /><a href="http://k6jca.blogspot.com/2021/03/notes-on-antenna-tuners-icom-ah-4-by.html">http://k6jca.blogspot.com/2021/03/notes-on-antenna-tuners-icom-ah-4-by.html</a><br /><br />Measuring a Tuner's "Match-Space":<br /><a href="http://k6jca.blogspot.com/2018/08/notes-on-antenna-tuners-determining.html" style="font-family: "times new roman";">http://k6jca.blogspot.com/2018/08/notes-on-antenna-tuners-determining.html</a><br /><br />Measuring Tuner Power Loss:<br /><a href="http://k6jca.blogspot.com/2018/08/additional-notes-on-measuring-antenna.html">http://k6jca.blogspot.com/2018/08/additional-notes-on-measuring-antenna.html</a><br /><br />
</p>
<p>
<b><u>Standard Caveat:</u></b>
</p>
<div>
I might have made a mistake in my designs, equations, schematics, models,
etc. If anything looks confusing or wrong to you, please feel free to
comment below or send me an email.<br /><br />Also, I will note:<br /><br />This
design and any associated information is distributed in the hope that it
will be useful, but WITHOUT ANY WARRANTY; without even the implied warranty
of MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.
</div>
<p></p>
<p></p>
</div>
Jeffhttp://www.blogger.com/profile/04853314106806116765noreply@blogger.com0tag:blogger.com,1999:blog-2257458838727315711.post-79243680353399608422021-03-15T09:43:00.008-07:002022-11-16T16:24:43.269-08:00Useful SWR, Voltage, and Power Equations<p>Here are some equations for calculating maximum voltage and current on a transmission line when the load is mismatched. These can be used, for example, for calculating maximum flux densities in directional-coupler current and voltage sense transformer.</p>
<p>First, equations for the Reflection Coefficient (Gamma) and SWR:</p>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://1.bp.blogspot.com/-eU6Yr05gMo8/YE9_YXH07GI/AAAAAAAAL64/6jtr9E8GXF4mKbyS1Dk4zmCQmUhSbdCTACLcBGAsYHQ/s756/Gamma.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="613" data-original-width="756" src="https://1.bp.blogspot.com/-eU6Yr05gMo8/YE9_YXH07GI/AAAAAAAAL64/6jtr9E8GXF4mKbyS1Dk4zmCQmUhSbdCTACLcBGAsYHQ/s320/Gamma.png" width="320" /></a>
</div>
<p>Equations for Forward Power:</p>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://1.bp.blogspot.com/-Ub7784pQjDA/YE9_cWcBPcI/AAAAAAAAL68/pNkTEtbFFugOs1xYm6TKZ3WrG1SQ2Y67gCLcBGAsYHQ/s666/Pforward.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="512" data-original-width="666" src="https://1.bp.blogspot.com/-Ub7784pQjDA/YE9_cWcBPcI/AAAAAAAAL68/pNkTEtbFFugOs1xYm6TKZ3WrG1SQ2Y67gCLcBGAsYHQ/s320/Pforward.png" width="320" /></a>
</div>
<br />
<p>
Using the above equations, we can derive equations for Forward and Reflected Voltages on a transmission line, based upon Gamma and Power
delivered to Load:</p><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-u38jIleRENI/YE-NfGz5lPI/AAAAAAAAL7Y/uEnT4YsIMpI-1fEktLwz3EzszNdu3teIACLcBGAsYHQ/s715/Vf%2Band%2BVr.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="406" data-original-width="715" src="https://1.bp.blogspot.com/-u38jIleRENI/YE-NfGz5lPI/AAAAAAAAL7Y/uEnT4YsIMpI-1fEktLwz3EzszNdu3teIACLcBGAsYHQ/s320/Vf%2Band%2BVr.png" width="320" /></a></div><br /><p>And finally, equations to calculate Vmax, Vmin, Imax, and Imin:</p>
<div class="separator" style="clear: both; text-align: center;">
<a href="https://1.bp.blogspot.com/-bTA2g3RZgPg/YE9_mREk6WI/AAAAAAAAL7E/0AFN1TA8ONQ_Ds3UX3PgB8HsgXuH0NHdACLcBGAsYHQ/s628/Vmax%2Band%2BImax.png" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="423" data-original-width="628" src="https://1.bp.blogspot.com/-bTA2g3RZgPg/YE9_mREk6WI/AAAAAAAAL7E/0AFN1TA8ONQ_Ds3UX3PgB8HsgXuH0NHdACLcBGAsYHQ/s320/Vmax%2Band%2BImax.png" width="320" /></a>
</div>
<br />
<p>Let's take an example...</p>
<p>
I want to design a directional coupler so that the flux densities of its
current and voltage sense transformers don't cause the ferrite cores to
overheat (see this blogpost: <a href="http://k6jca.blogspot.com/search/label/Directional%20Couplers%3A%20Calculating%20Flux%20Density">Calculating Directional Coupler Flux Densities</a>) for a given maximum load power and a maximum SWR.
</p>
<p>
These flux densities are a function of |Vmax| for the voltage-sense transformer and |Imax| for the current-sense transformer. I want to
calculate these values using my known values of load power and SWR.
</p>
<p>
Let's say that the maximum power I'll deliver to my load is 100 watts to my load and that the load's SWR is 4:1.
</p>
<p>First, I'll calculate the magnitude of Gamma, the Reflection Coefficient:</p>
<p style="text-align: center;">|Γ| = (SWR - 1)/(SWR + 1) = 3/5 = 0.6</p>
<p>
Next, I'll calculate the magnitude of the forward voltage on the transmission
line, given the 100 watts being delivered to (and dissipated by) the load, a transmission line Zo of 50 ohms, and |Γ| equal to 0.6.
</p>
<p style="text-align: center;">Vf = ((Zo*Pload) / (1 - Γ^2))^0.5
</p>
<p style="text-align: center;">Vf = ((50*100) / (1-(0.6^2))^0.5</p>
<p style="text-align: center;">Vf = 88.39 V</p>
<p>(Note that Γ^2 and |Γ|^2 are equivalent).</p><p>Next, using Vf , |Γ|, and Zo, I can calculate |Vmax| and |Imax|:</p><p style="text-align: center;">|Vmax| = (1 + |Γ|)*Vf = (1+0.6)*88.39 = 141.4 V</p><p style="text-align: center;">|Imax| = |Vmax|/Zo = 141.4/50 = 2.83 A<br /></p>
<p>With these values I can then calculate my transformer maximum flux densities.</p>
<p><br /></p><p><b style="text-align: center;"><u>Other Transmission-Line Posts:</u></b></p><p><a href="http://k6jca.blogspot.com/2021/02/antenna-tuners-transient-and-steady.html" style="text-align: center;">http://k6jca.blogspot.com/2021/02/antenna-tuners-transient-and-steady.html</a>. This post analyzes the transient and steady-state response of a simple impedance matching system consisting of a wide-band transformer. I calculate the system's impulse response and find the time-domain response by convolving this impulse-response with a stimulus signal.</p><p><span style="text-align: center;"><a href="http://k6jca.blogspot.com/2021/02/the-quarter-wave-transformer-transient.html">http://k6jca.blogspot.com/2021/02/the-quarter-wave-transformer-transient.html</a>. <span style="text-align: left;"> This post analyzes the transient and steady-state response of a Quarter-Wave Transformer impedance matching device. I calculate the system's impulse response and find the time-domain response by convolving this impulse-response with a stimulus signal.</span></span></p><div><a href="http://k6jca.blogspot.com/2021/03/useful-swr-voltage-and-power-equations.html" style="text-align: center;">http://k6jca.blogspot.com/2021/03/useful-swr-voltage-and-power-equations.html</a>. This post lists (in an easily accessible location that I can find!) some equations that I find useful</div><p><span style="text-align: center;"><a href="http://k6jca.blogspot.com/2021/05/antenna-tuners-lumped-element-tuner.html">http://k6jca.blogspot.com/2021/05/antenna-tuners-lumped-element-tuner.html</a>. This post analyzes the transient and steady-state reflections of a lumped-element tuner (i.e. the common antenna tuner). I describe a method for making these calculations, and I note that the tuner's match is independent of the source impedance.</span></p><p><span style="text-align: center;"><a href="http://k6jca.blogspot.com/2021/05/lc-network-reflection-and-transmission.html">http://k6jca.blogspot.com/2021/05/lc-network-reflection-and-transmission.html</a>. This post describes how to calculate the "Transmission Coefficient" through a lumped-element network (and also its Reflection Coefficient) if it were inserted into a transmission line. <br /></span></p><p><a href="http://k6jca.blogspot.com/2021/09/does-source-impedance-affect-swr.html" style="color: #2b00fe;">http://k6jca.blogspot.com/2021/09/does-source-impedance-affect-swr.html</a><span style="color: #2b00fe;">. </span>This post shows mathematically that source impedance does not affect a transmission line's SWR. This conclusion is then demonstrated with Simulink simulations.<br /></p><p><a href="https://k6jca.blogspot.com/2021/10/revisiting-maxwells-tutorial-concerning.html">https://k6jca.blogspot.com/2021/10/revisiting-maxwells-tutorial-concerning.html</a> This posts revisits Walt Maxwell's 2004 QEX rebuttal of Steven Best's 2001 3-part series on Transmission Line Wave Mechanics. In this post I show simulation results which support Best's conclusions.</p><p><br /></p>
<p>
<b><u>Standard Caveat:</u></b>
</p>
<div>
I might have made a mistake in my designs, equations, schematics, models, etc.
If anything looks confusing or wrong to you, please feel free to comment below
or send me an email.<br /><br />Also, I will note:<br /><br />This design and
any associated information is distributed in the hope that it will be useful,
but WITHOUT ANY WARRANTY; without even the implied warranty of MERCHANTABILITY
or FITNESS FOR A PARTICULAR PURPOSE.
</div>
Jeffhttp://www.blogger.com/profile/04853314106806116765noreply@blogger.com2