Saturday, December 14, 2024

Do Common-mode Currents Split Equally across Circuit Conductors?

This post discusses common-mode currents in circuits, examining the commonly-held belief that a cable's common-mode currents are divided equally among the cable's conductors.  It is a companion piece to my blog post on transmission-line common-mode currents:  https://k6jca.blogspot.com/2024/12/do-common-mode-currents-split-equally.html

Introduction:

When designing electronic circuitry for products, one requirement is to ensure that any unintentionally radiated EMI (Electromagnetic Interference) is below limits set by EMC (Electromagnetic Compatibility) specifications.

Although EMI emissions can sometimes be generated by a voltage (e.g. high voltage nodes in off-line switching power supplies), in my experience EMI emissions are more often created by unintentional currents within the electronic system. These EMI-creating currents are commonly known as "common-mode currents," and they can have current-loop paths (the complete path that a current take, outbound from the source and then back again) whose encompassed-area is large, resulting in significant magnetic-field induced EMI.

On the other hand, "differential-mode" currents are currents of intentionally created signals and, if their routes are well designed, both the signal's path to its load and its return path will be in close proximity.  The loop area enclosed by the entire differential-mode current path should be small and thus magnetic-field EMI radiation from this signal minimized.

Given a two wire cable, it is commonly accepted that the differential-mode currents on such a cable will be of equal magnitude and flow in opposite directions, while common-mode currents, if they exist, will be of equal magnitude but flow in the same direction, as shown in the diagram, below (adapted from Introduction to Electromagnetic Compatibility, C. R. Paul, Wiley Interscience, NY, 1992):


But must common-mode currents across multiple conductors always be equal?


Unequal Common-mode Currents in Electronic Systems:

Below is an image (from a Murata Manufacturing Company, Ltd. document) illustrating how differential-mode and common-mode currents might exist within an electronic system.

Note that the differential-mode currents are equal in magnitude and travel in opposite directions on well-defined circuit paths.  They are driven by the "Signal Source," which is a differential-mode voltage.

In the circuit above, the "Noise Source" is a common-mode voltage.  The combination of this common-mode voltage (Vcm) and the differential-mode voltage (Vdm) would be modeled per the illustration below, such that Vcm attaches to the midpoint of Vdm, making Vdm balanced with respect to Vcm:

The circuit's common-mode currents, in this case generated by an unwanted noise voltage, travel in the same direction on the two conductors along a path that includes parasitic capacitance coupling of the load to ground.  But are these two common-mode currents equal?

In the diagram below I've added parasitic capacitances to ground at the load-end of each wire.  Note that the common-mode currents on the two wires will only be equal if the capacitances are equal -- that is, if the circuit load is balanced with respect to ground.  


Therefore, if the impedance-to-ground seen by Vcm looking into one wire is different from the impedance Vcm sees looking into the other wire, the common-mode currents will not be equal.


Analysis of the Circuit's Currents:

The following images provide a more in-depth analysis of the circuit:










So, we have shown that the equations: I1 = Ic + Id and I2 = Ic - Id (and therefore the equations Ic = (I1 + I2)/2 and Id = (I1 - I2)/2) are not always true, but instead their applicability is a function of circuit component values and the balance-about-ground of the differential-voltage driving the circuit.


Another Interesting Observation:

Recall that R3 is the load for our differential-mode signal, Vdm.  Let's examine the equation for the current through R3:


We do not want Vcm (i.e. a noise source) to corrupt the Vdm signal across R3!  To prevent this from occurring in the circuit above, R2*R4 should equal R1*R5.


SPICE Simulations:

The following SPICE simulations can give us actual values of the currents in each component.  With these currents, we can test if and when Ic = (I1 +I2)/2 and Id = (I1-I2)/2 are true.  


Note that the following values for current calculated with LTSpice are the same values as would be calculated using the equations derived with MATLAB, shown earlier in this post.

LTSpice with Balanced Vdm and Balanced Load:

In the circuit below Vdm equals 1 volt (i.e. Vdm1 = Vdm2 = 0.5V), and it is balanced with respect to ground (e.g. when Vcm is set to 0 for Vdm analysis).  That is, its midpoint is referenced to ground.

Also, the "parasitic loads to ground," R4 and R5, are of equal value and balanced with respect to ground.

To analyze the currents when driven by Vcm alone, Vdm is set to 0 (i.e. the Principle of Superposition) and the circuit simulated.

Similarly, to analyze the currents when the circuit is driven by Vdm, Vcm is set to 0 and the circuit is simulated.

Total currents are calculated with both Vdm and Vcm set to 1 volt .  Note that the total currents should equal the sum of the Vcm-driven currents plud the Vdm-driven currents.


Recall our equations Ic = (I1 + I2)/2 and Id = (I1 - I2)/2.  I1 is equal to the total current through R1, which is 0.0131 Amps.

Similarly, I2 is equal to the total current through R2, which in the circuit above is -0.0087 Amps (negative because I2 is defined to travel from left-to-right, while the current through R2 is actually traveling right-to-left).

Per the equations for Ic and Id:

     Ic = (I1 + I2)/2 = (0.0131-0.0087)/2 = 0.0022 A

     Id = (I1 - I2)/2 = (0.0131+0.0087)/2 = 0.0109 A

Note that the calculated value for Ic exactly matches the simulated values of the Vcm-driven currents through R1 and R2.

Similarly, the calculated value for Id exactly matches the simulated values of the Vdm-driven currents through R1 and R2.

Therefore, for the circuit above, the equations Ic = (I1+I2)/2 and Id = (I1-I2)/2 are TRUE.


LTSpice with Unbalanced Vdm and Balanced Load:

Now let's examine the same circuit, but with the Vdm source unbalanced (i.e. single-ended drive).  Note that the Vdm drive remains unchanged at 1V.


Calculating Ic and Id, recall that I1 = IR1 and I2 = IR2.  Using the "Total Current" values from the LTSpice simulation:

     Ic = (I1 + I2)/2 = (0.0142-0.0075)/2 = 0.0034 A

     Id = (I1 - I2)/2 = (0.0142+0.0075)/2 = 0.0109 A

If we compare these values to the values generated with LTSpice, we can easily see that the common-mode currents through R1 and R2 ae each 0.0022 A, not the calculated value of Ic = 0.0034 A.

Similarly, the circuit's Vdm-driven currents through R1 and R2 do not equal Id.  In fact, a portion of the differential-mode current travels out through the parasitic resistors R4 and R5 and takes the long way back to Vdm (i.e. via Vcm), rather than returning via R2.

Therefore, for the circuit above, the equations Ic = (I1+I2)/2 and Id = (I1-I2)/2 are FALSE.


LTSpice with Balanced Vdm and Unbalanced Load:

Finally, suppose that, with respect to ground, Vdm is balanced but the load is unbalanced -- in this case, R4 = 500 ohms and R5 = 400 ohms.


Calculating Ic and Id, recall that I1 = IR1 and I2 = IR2.  Using the "Total Current" values from the LTSpice simulation:

     Ic = (I1 + I2)/2 = (0.0128-0.0085)/2 = 2.15 mA

     Id = (I1 - I2)/2 = (0.0128+0.0085)/2 = 10.65 mA

If we compare these values to the Vcm-driven values of IR1 and IR2 generated with LTSpice, we can easily see that the common-mode currents through R1 and R2 are 2.0 and 2.5 mA, respectively, not each 2.15 mA.

Similarly, the Vdm-driven currents through R1 and R2, are also unequal.  In fact, a portion of the differential-mode current travels out through Vcm and takes the long way back to Vdm2 via the parasitic resistor R5.

Also, in this circuit we can see that a small portion of the Vcm-driven current goes through R3.  If R3 is our load for the differential-drive signal Vdm, then, if Vcm is an unwanted noise source, we have corrupted Vdm's signal across R3 with this noise.

Therefore, for the circuit above, the equations Ic = (I1+I2)/2 and Id = (I1-I2)/2 are FALSE.


Conclusions:

Although there is often an assumption that common-mode current in a cable must be equally distributed across the cable's conductors, in fact it is possible for the conductors to have different amounts of common-mode current.  Ditto for the differential-mode current.

In addition, if a circuit is not perfectly "balanced", it is possibly for a common-mode noise voltage to add unwanted noise to a differential-signal's load.


Standard Caveat:

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.

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.

Thursday, December 5, 2024

Do Common-mode Currents Split Equally across Transmission-Line Conductors?

(This blog post is a companion piece to my post on circuit cable common-mode currents, found here: https://k6jca.blogspot.com/2024/12/do-common-mode-currents-split-equally_14.html)

Are common-mode currents equal on a transmission line's conductors?  Let's look at two cases: a coaxial cable transmission line and a two-wire (e.g. twin-lead) transmission line.

Coaxial Cable Transmission Line

If the transmission line is a coaxial cable, the answer is no.

Although a coax cable consists of two conductors, the center conductor and the shield, electrically it behaves as though there are three conductors:  the center conductor, the inner-surface of the shield, and the outer-surface of the shield.

The RF signal is applied between the center conductor and the shield.  Due to physics, the signal travels on the center conductor and the inner-surface of the shield (assuming the shield has adequate thickness).  For the purpose of transporting energy from the transmitter to the antenna, the outer-surface of the coax plays no role, and the EM fields of the forward and returning currents are completely contained within the coax -- these two currents do not radiate beyond the coax.

However, the outer-surface of the coax has an undesired effect where it attaches to the antenna -- it behaves as an extra wire that is attached between one side of the antenna and the distant transmitter, running the length of the coax transmission line.  As such, it becomes a radiating part of the antenna.

The current flowing on this outer surface (and radiating RF energy) is also commonly called a "common-mode current".  Note that this common-mode current is only on the outside of the coax cable, and only it is the cause of coax cable RF emissions.


So, on a coax cable, common-mode current is only carried on the shield (its outer surface) but not on the center conductor.  The common-mode current is not distributed equally across the coax cable's conductors.


Two-wire Transmission Line

Next, consider a two-wire (e.g. twin-lead) transmission line.

A two-wire transmission line does not have the same self-shielding properties as a coax cable, and there isn't a "third" conductive element (i.e. the coax shield's outer surface) that acts as an unintentional RF radiator.

But an antenna & transmitter system can have unequal common-mode currents on a two-wire transmission line if the system is unbalanced.

Unbalance can occur if any of the following conditions exist:

  1. The antenna elements are unmatched, either in length with respect to each other, or in position with respect to ground.
  2. The transmitter is not balanced with respect to ground.      
  3. The two wires of the transmission line are unequal in length, or one wire is closer to ground or conductive objects than the other wire.

Let's examine antenna imbalance and transmitter imbalance (skipping transmission line imbalance).  I'll use EZNEC for the antenna simulations.

The first simulation will be of a dipole that is balanced with respect to the length of its elements and their position with respect to ground.  The voltage source driving the antenna is also balanced with respect to ground in the sense that the voltage source's "midpoint" is tied to ground, so that, if one side of the transmission line is positive with respect to ground (e.g +0.5V), the other side is negative (e.g. -0.5V), and vice-versa.

This balanced voltage source is implemented with two identical in-phase voltage sources so that the total voltage swing is twice the amplitude of a single source.


In the EZNEC models, below, I'm going to accentuate any common-mode coupling paths and also minimize the effects of SWR by making the transmission line only 5 feet long.  At 3.5 MHz, this length is a very small fraction of a wavelength, nevertheless, it is still a transmission line.

Balanced Dipole Antenna and Transmitter:

In the antenna diagram below, you can see that the balanced voltage source consists of two series-connected 0.5 V sources (resulting in 1V of drive) with ground connected between them.  The dipole consists of two 50 foot elements, so it is 100 feet long.  But its height is only 5 feet above ground.

The transmission line is simply two #12 wires run in parallel with spacing of 1.2 inches (0.1 feet) between the two wires.  No thought was given to the line's characteristic impedance.  It is what it is.

The EZNEC simulation shows that for this balanced-antenna and balanced-source configuration, there are no common-mode currents -- the currents are differential only.  There is no current on the wire attached to ground.


EZNEC currents for this antenna:


Balanced Dipole, Unbalanced Transmitter:

Let's unbalance the voltage source driving the balanced antenna by removing the 0.5V source in wire 4 (as shown in the previous antenna diagram) and change the value of the remaining source from 0.5 volts to 1 volt, so that, overall, the magnitude of the driving voltage remains the same.

The dipole, itself, remains unchanged, that is, it remains balanced.

Below are the EZNEC results.  Clearly the currents in the transmission line are unbalanced, and the majority of current follows a path through wires 1, 3, 6, and 5 (the wire to ground).  This is common-mode current, and it is only on one of the transmission line's two wires (wire 3), not both.


EZNEC currents for this antenna:


By the way, the higher this antenna is raised, the smaller and smaller will be the current imbalance between the currents on the dipoles two radiating elements.  But these two currents will not be equal.

Unbalanced Dipole, Balanced Transmitter:

Let's return to a balanced voltage source driving the transmission line and change the balanced dipole to be unbalanced.  The simplest way to do this is to change the length of one of the dipole's elements, so let's change the length of the right-hand element to zero (that is, remove it).  The dipole is now a unipole.

Again, the EZNEC results show that the majority of the current flows through wires 1, 3, 6, and 5.  So again, there is significant common-mode current in this antenna system, but only on one wire (wire 3) of the two-wire transmission line.


EZNEC currents for this antenna:



Conclusions:

Although there is often an assumption that common-mode current in a cable must be equally distributed across the cable's conductors, in fact it is possible for the conductors to have different amounts of common-mode current.

Similarly, the two wires of a two-wire transmission line (e.g. twin-lead) can have mismatched amounts of common-mode current on them.


Standard Caveat:

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.

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.

Thursday, September 19, 2024

RF Directional Bridge: Operation versus Source and Detector Impedances

During recent correspondence with Owen Duffy, VK1OD, we discussed RF Directional Bridges and their design.

One question that arose was the effect of the values of the various bridge resistances on a bridge's operation, in this instance the effect of bridge Source and Detector impedances on directional bridge measurements.  

In the various bridge circuits I've seen, Source and Detector impedances (Rs and Rd) seem to ubiquitously be set to Zo (the bridge's reference impedance, typically 50 ohms).  But why?  I could see why this would be the case if one were using an external signal generator with a defined source impedance of 50 ohms as the signal source and measuring the resulting bridge voltage with an external detector having an input impedance of 50 ohms.

But suppose I were designing my own circuit incorporating source and detector with the bridge.  Was it a design requirement that Rs  = Rd = Zo?

As it turns out, no, this is not a design requirement.

Rs and Rd need not equal Zo, the bridge's reference impedance, but their values must satisfy the following equation, assuming all other bridge resistor values equal Zo:

Rs*Rd = Zo^2

This blog post will explain why.


RF Directional Bridge Circuit:

The classic RF Directional Bridge circuit, also known as a Return-Loss-Bridge (or RLB), is shown below:

This circuit is basically a Wheatstone Bridge.  As drawn, above, ZL is the unknown load impedance to be measured, located in the upper right-hand arm of the bridge.  (And although GND is shown at the bottom node of the circuit, it need not be assigned to that node, or to any node at all.)

The bridge is driven by a voltage source Vs with a source impedance Rs (I used Rs in lieu of Zs because it really is a resistive-only impedance).  The source impedance has been set to equal Zo, the bridge's Reference impedance.

The resistors in the bridge arms, R1, R2, and R3, all equal Zo, the bridge's Reference impedance.

Voltage measurement is made across the Detector's impedance, Rd (also equal to Zo).  This measurement is voltage 'Vdet', and its value is proportional to the Reflection Coefficient of ZL, referenced to Zo.

That is,

Vdet = A*(ZL - Zo)/(ZL + Zo)

where 'A' is a constant, and the quantity (ZL - Zo)/(ZL + Zo) represents ZL's Reflection Coefficient, also known as Gamma (Γ).


Directional Bridge Design Requirements:

There are two design requirements for Directional Bridges (that I am aware of):

1.  The "source impedance" looking back into the port to which ZL would be attached (but not currently attached-to) should equal Zo.

2.  The Vdet voltage if ZL = Zo should be 0.  And Vdet should be equal and opposite for ZL values of 0 and Infinity.

The bridge above satisfies these two requirements.  If we remove ZL and calculate (or simulate) the impedance looking into the port to which it had been attached, it equals Zo.  And, given Vs = 1 volt, Vdet is 0v for ZL = Zo, -0.5v for ZL = 0 ohms, and +0.5v for ZL = Open (infinite ohms).


An Equation for Vdet:

To understand how a Directional Bridge's component values affect its operation, we should derive an equation for Vdet that is a function of these components.  We can do this using basic circuit analysis of the circuit, shown below, which results in four equations (also shown below) that we can solve for Vdet:


Pushing pencil on paper to derive an equation for Vdet from these four equations, for me, quickly became prone to error, and so I turned to MATLAB's Symbolic Math Toolbox to ease the pain with the following MATLAB script:

The equation it returned was a complicated one:

vdet = -(Rd*Vs*(R1*R3 - R2*ZL))/(R1*R2*R3 + R1*R3*Rd + R2*R3*Rd + R1*R2*Rs + R1*R3*Rs + R1*Rd*Rs + R2*Rd*Rs + R3*Rd*Rs + R1*R2*ZL + R1*R3*ZL + R2*R3*ZL + R1*Rd*ZL + R2*Rd*ZL + R2*Rs*ZL + R3*Rs*ZL + Rd*Rs*ZL)

Slightly rearranging the equation:

Vdet = (Rd*Vs*(R2*ZL - R1*R3)) /

        (ZL*R1*R2 + ZL*R1*R3 + ZL*R2*R3* + ZL*Rd*Rs +

         ZL*R1*Rd + ZL*R2*Rd + ZL*R2*Rs + ZL*R3*Rs +

         R1*R3*Rd + R2*R3*Rd + R1*R2*Rs + R1*R3*Rs +

         R1*Rd*Rs + R2*Rd*Rs + R3*Rd*Rs + R1*R2*R3)

Our goal is to take the above equation and from it produce a final equation in the form of:

Vdet = A*(ZL - Zo)/(ZL + Zo)

where 'A' is a constant, and the quantity (ZL - Zo)/(ZL + Zo) represents ZL's "Reflection Coefficient", also known as Gamma (Γ).

The numerator Rd*Vs*(R2*ZL - R1*R3) is already in the form of a Constant times (ZL - Zo), if we assume R1*R3 = R2*Zo, in which case the numerator becomes:  Vs*Rd*R2*(ZL - Zo).

Note that if R1 = R2 = R3 = Zo, then the requirement that R1*R3 = R2*Zo is satisfied.

So all we need to do is to whip the denominator into the shape of a second Constant times (ZL + Zo).  You can see that the 16 denominator terms have been grouped into 8 that contain ZL and 8 that don't contain ZL -- the latter will become, in some way, the 'Zo' part of the equation (ZL + Zo).


Arriving at Vdet's Relationship with Rs and Rd, the Source and Detector Impedances:

From LTSpice simulations it appeared that the values of Rs and Rd resulted in a circuit that met the two "Directional Bridge Design Requirements" mentioned, above, if the values of Rs and Rd satisfied the following equation:

Rs*Rd = Zo^2

and with R1 = R2 = R3 = Zo.

But a demonstration with LTSpice is not a mathematical proof.  This goal is easily achieved, though, by substituting Rs*Rd = Zo^2 and R1 = R2 = R3 = Zo into the above equation for Vdet.  Doing so, the Vdet equation reduces to:

Vdet = (Vs*Rd/(2*(2*Zo + Rs + Rd)))*((ZL - Zo)/(ZL + Zo))

We can see that this equation is now in the form of 

Vdet = A*(ZL - Zo)/(ZL + Zo) = A*Γ

where A = Vs*Rd/(2*(2*Zo + Rs + Rd)).


An LTSpice Simulation Example:

Let's let Zo = 50 ohms, Rs = 1 ohm and Rd = 2500 ohms (thus satisfying Rs*Rd = Zo^2), and Vs = 1 volt.  Here are the results of an LTSpice simulation of this circuit:


You can see that the port impedance measured looking into the port to which ZL would attach is 50 ohms, and that Vdet = 0 when ZL = Zo, and +/- 0.481 volts when ZL = Infinite or 0 ohms, respectively.


Taking it to the Limit:

So now we know that Rs and Rd need not always equal Zo.  What happens if we take these two values to their limits by reducing Rs to 0 and increasing Rd to infinity (i.e. we remove Rd)?

The equation for Vdet reduces to:

Vdet = (Vs/2)*(ZL - Zo)/(ZL + Zo)


We will see something interesting if I add a signal-port (and ground) in lieu of ZL and rearrange this circuit by flipping all components (except for the voltage source) upside down to become:


I've replaced ZL with a test port to which a load (ZL) would attach.  If a transmission line were attached to this port, the voltage at the R3/Port node would represent the sum of the forward voltage wave's amplitude (Vf) and the reflected voltage wave's amplitude (Vr) at this point.

That is, the voltage at the test port equals Vf + Vr.

Note that Vf travels out the ZL port onto the transmission line, and Vr, representing reflection of Vf from the far-end of the transmission line, comes into the port from the outside world.

Vr, arriving at the port from the other end of the transmission line, sees R3 as the transmission line's near-end termination.  And so its power absorbed, with no re-reflections, by R3, because the latter's value equals the transmission line's Zo.

The voltage at the R1/R2 node is simply a simulcra of Vf traveling out the ZL port.  Assuming the ZL port is attached to a transmission line of Zo and that there are no re-reflections of Vr from R3 (because R3 = Zo), Vf will always be Vs*R3/(Zo+R3), or Vs/2, given R3 = Zo, irrespective of the value of Vr.

Given that R1 and R2 both equal Zo, the voltage at their common node is also Vs/2, the same as Vf going out the port.  

Vdet is the difference between these two nodes, or 'Vf+Vr' - Vf.  The result is:  Vdet = Vr.

One of the definitions of Reflection Coefficient is  Γ = Vr/Vf.  So if we normalize our measured Vr by Vf (the voltage at the R1/R2 node), we will have measured Gamma, the Reflection Coefficient of the impedance ZL, measured at the test port.

We arrive at the same result if we used our previous equation, derived above:

Vdet = (Vs/2)*(ZL - Zo)/(ZL + Zo)

Substitute Vf for Vs/2 and Γ for (ZL - Zo)/(ZL + Zo).  The resulting equation becomes:

Vdet = Vf*Γ

And therefore:

Γ = Vdet/Vf

One more note regarding the circuit, above:  because R1 and R2 play no role in the impedance seen looking into the test port from the outside world, they can be any value, not just Zo.  The only rule is that their resistance values be the same so that the "simulated" Vf  at the R1/R2 node equals Vs/2.

And why, you might wonder, do R1 and R2 play no role in the port's impedance?  This is simply the result of the Superposition Principle, which states that, when analyzing how a circuit reacts to a voltage or current source, the effects of all other voltage and current sources connected in the circuit should first be removed.

So, to remove the effect of these other source, the other voltage sources are replaced with shorts, and the other current sources are replaced with opens (i.e. the current sources are simply removed).

So, to analyze the impedance that Vr sees as it enters the test port, first short Vs.  Clearly, the two series resistors R1 and R2 are now shorted to ground, and the impedance looking into the bridge's test port from the outside world consists of a single resistor, R3, shunting the port to ground and thus terminating it with an impedance Zo.


Standard Caveat:

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.

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.

Monday, August 12, 2024

Repair Log: Fluke 6060B (RF Output On/Off Switch)

 

I recently discovered my Fluke's 6060B's "RF Output On/Off" switch was not working:

This switch is an elastomeric push-button switch, in which a small conductive disk on the backside of the rubber pushbutton "shorts" two contacts on the switch PCB when the pushbutton is pressed.

Chances are that the PCB switch contacts needed to be cleaned, but to do this the Front Panel Assembly must be removed from the 6060B chassis and then disassembled to access the push-button's circuit-board pads.  Section 4B-4 of the Fluke 6060B manual describes a procedure for removing the Front Panel Assembly, but this procedure is incomplete.

Below are the steps required to remove the Front Panel Assembly from the chassis and then disassemble it so that the pushbutton switch contacts can be cleaned: 


Removing the Front Panel Assembly:








Disassembling the Front Panel Assembly:




If there is physical interference between the edge of the PCB and the metal bracket, as shown above, consider filing a small notch in the PCB to allow it to clear the bracket.


Cleaning the PCB's Switch Contacts:

The switch contacts on the PCB can be cleaned with either Isopropyl Alcohol (I used a 99% solution).  They can also be cleaned with distilled water.  In either case, wet the tip of a Q-tip with your preferred solution and gently wipe away any loose residue on the PCB's switch pads.  Use a clean Q-tip to dry the pads.

I would also recommend a gentle cleaning of the conductive disk that is attached to the rubber push-button (it is this disk that completes the switch contact when the pushbutton is pressed).

To test the resistance of the RF Output On/Off switch when it is depressed, place an ohm-meter across the two PCB vias shown in the image, below, and press the switch.  After cleaning, I measured about 1.5K Ohms when the switch was pressed).  Other switches measured in the range of 500 to 1200 ohms.  Note that this resistance seems to be the resistance across the small conductive "pad" on the back of the rubber elastomeric switch.


Reassembly:

To reassemble, perform steps 1 through 9 in reverse order!


A Tip:  

Keep the screws for each step in separate cups.  This will make reassembly much easier.


Standard Caveat:

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.

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.

Sunday, July 28, 2024

Repair Log: HP 5216A 7-Digit Counter

 


This blog post chronicles my efforts to repair an older HP 5216A Nixie tube counter.

The HP 5216A is a 7-digit counter specified to count frequencies between 3 Hz and 12.5 MHz.  In addition to frequency measurements, it can also perform period measurements, period-average measurements, ratio measurements, totalizing measurements, and time-interval measurements.

Documentation:

Although an Operating-Service Manual for the HP 5216A can be downloaded from the Keysight website, this version is of very poor quality with the schematics essentially unusable.  I would instead recommend purchasing a manual from Artek Manuals.  The one I purchased (and obtained via electronic download) is of excellent quality and at the time of this writing only $12.50.


Issues with my 5216A:

  • The counter, when it counts, only counts in "Check" mode (left-hand front panel rotary switch in the "Check" position), but it doesn't count when a signal is applied to the front-panel input BNC.
  • Sometimes the counter doesn't count at all, irrespective of mode, and the gate lamp never illuminates.  And sometimes the power doesn't seem to come on.
  • The most-significant (MS) digit does not illuminate reliably -- it will briefly flicker on, but remains off most of the time.

Repair Efforts:

1.  Power Supply Recapping:

The serial number of my 5216A has a three-digit prefix of 748, implying it was manufactured in 1967 (i.e. take the first two digits of a four-digit prefix (in this case "07") and add them to "1960").

Given that this counter was almost 60 years old, I decided to recap all of the electrolytic capacitors on the power supply board, using caps from my collection of miscellaneous electrolytic caps.

My replacement caps were often the wrong form factor, having radial leads rather than axial, but I was able to attach them without too much effort.  And if I did not have the exact capacitance or working-voltage specified in the manual for a capacitor, I would choose one with a higher capacitance and/or higher working-voltage rating.

With these caps replaced, the issues listed above still existed.


2.  Only counts in "Check" mode, but not when a signal is attached to the front panel input BNC.

The counter would only count when the left-hand rotary switch was placed in "Check" mode, letting it count the counter's internal frequency reference (or a divided-down product thereof).

To trouble shoot, I applied an external signal to the counter's front panel BNC input and followed it through the circuitry.  It appeared to not make it through IC11 on the Main Board (A4).  


As shown in the 'red' annotation on the schematic, above, the input signal appears at IC11 pin 4, but not at its output, pin 6.

Yet pin 5 was high (i.e. TTL input-floating level of around 2 V), which should have gated the signal on pin 4 through.  And also pin 3 was 0, so the other AND gate was cut off and should not interfere with the propagation of the signal from pin 4 to pin 6.

So most likely IC11 (HP part number 1820-0072) was bad.

P/N 1820-0072 is equivalent to an SN7450.  I was able to obtain one via eBay.  (Note, if you cannot locate an SN7450, you can use an SN7451 in a pinch.  It should support all counter functions except those requiring that the 'Time Interval' switch on the counter's back panel be set to "TIME INT" in lieu of "FREQ-PER").

Replacing IC11 fixed the problem of the counter not counting when a signal was applied to the counter's input.


3.  Intermittent Problems:

Sometimes the counter would not count at all in either "Check" mode or in Normal mode.  Sometimes it did not seem to even turn-on.

All of these problems were traced to wires that had broken at their solder joints.  I'm theorizing a previous owner  (one or more), when trying to repair the counter (there were several repairs that had been made previously to the main board, for example), overstressed the wiring harnesses while disassembling the unit, and these wires eventually broke from over flexing.

Here's an example:


Repairing the various broken wires (there were three or four) fixed the intermittent problems.


4.  Left-most Nixie digit not working.

The Nixies are driven by decoder-drivers ICs, HP part number 1820-0092.  I verified that the input to IC15, the decoder-driver for the MS (Most-Significant) digit was correct (a four-bit BCD-encoded low-true value), but the correct Nixie digit representing that BCD code was not being illuminated.

It looked to me as though the Nixie tube was soldered to the PCB (important note: IT IS NOT!), and so I decided that trying to replace the 1820-0092 would be a better path to take.

Unfortunately, a replacement 1820-0092 is almost impossible to find, so I designed an equivalent circuit using available parts, in this case an SN74141 decoder-driver (or its equivalent, a Russian K155ID1 IC, available on Amazon and eBay), and a 74LS04 hex-inverter to convert the 4-bit low-true BCD code input to high-true for the 74141 (or its equivalent) input.

This circuit is described in this blog post:  https://k6jca.blogspot.com/2024/07/hp-1820-0092-nixie-decoderdriver.html

I removed the suspect 1820-0092 and replaced it with an adapter board on which I'd built the new replacement circuit.  The Nixie still did not work.  Oh oh -- was the problem with my replacement design, or with the Nixie tube, itself?

(Here is an image of the Main PCB with my 1820-0092 replacement circuit installed).


Poking around a bit more, I discovered that the Nixie tubes are NOT soldered onto the PCB, but instead they are inserted into pin-sockets, and so they are easy to remove.  I swapped the bad tube with a working tube.  The problem followed the bad Nixie tube, and I was able to show that my 1820-0092 replacement circuit worked properly for decoding digits 0-9 and blanking the tube for non-valid input BCD codes.

These Nixie tubes (HP part number 1970-0025) display their digits "upside down" and are very difficult to find.  Luckily, I have an old HP 5221B counter that uses the same Nixie tubes, but it is only a 5 digit counter, not 7, and it is spec'd to 10 MHz instead of the 5216A's higher frequency of 12.5 MHz.

So I decided it would be the sacrificial lamb, giving one of its Nixie tubes to replace the bad Nixie in the 5216A.  (Now I'm trying to find a replacement Nixie for the HP 5221B).


Other Notes:

1.  Creating a 1 MHz Reference Frequency for the 5216A from 10 MHz:

The HP 5216A's External Frequency Reference Input Specification is:
  • 1 MHz Sine Wave (note: square-wave should be adequate, too).
  • 1 V rms into 1000 Ohms (10V rms maximum).

I have a "house" 10 MHz frequency reference (based on GPS) that I use to maintain the frequency accuracy of various pieces of test gear in my lab.  But I would need to divide this down to 1 MHz for the 5216A counter.

Here's the design I came up with.


Schematic Notes:
  • C2 AC-couples the 10 MHz signal at the input BNC connector to the input of the inverter (NC7S14M5X).
  • Resistors R1 and R2 set  the inverter's input DC level to midway between VCC and GND.
  • Diodes D1 and D2 are clamping diodes to clamp overshoots and undershoots on the input signal, with R3 acting as a current limiter.  (The inverter also has integrated internal clamp diodes, but they are limited to 20 mA).  
  • U2 is a CMOS inverter with a Schmitt-Trigger Input (to prevent false counting of the counter U1 from noise on the input signal).  I used an NC7S14M5 device because I had it in my junkbox.  A 74AHC1G14, or equivalent, should work just fine.
  • U1 is a dual divide-by-10 counter, each counter consisting of two stages:  first a divide-by-5 stage that produces an asymmetrical 2 MHz intermediary signal, and then a divide-by-2 stage that also "squares up" the asymmetric signal; the result being a 1 MHz square wave.  Only one of the two divide-by-10 counters is used.
  • C1 AC couples U1's output to the output BNC connector.
  • R4 both limits output current and also provides a one-pole low-pass filter in conjunction with C4, removing ringing on the output signal, and its value is not large enough to significantly affect signal level, given the 5216A's input impedance of 1K ohms for the reference signal.  Note, too, that given the low-frequency of operation and the short lengths of coax that might be used to connect the 1 MHz output to a downstream input, R4's value does not need to equal the characteristic impedance of the coax.
Implementation:

I built the circuit on a small perf board that could be inserted into a three-BNC Pomona Box, as shown below.  Note that the NC7S14M5 inverter is in a SOT23-5 package and thus mounted on a small adapter board of its own to make signal connection easier.



The three-BNC Pomona Box housing the circuit is shown below.  Note that the 5.1 VDC power comes from a BNC on the front panel of the HP 5216A counter (a modification added to the counter's front panel by a previous owner), which connects to the counter's internal 5.1 VDC rail.


The 10 MHz input:


And the 1 MHz output:


Below is the counter using the 1 MHz External Reference.  The gray coax cable provides 5.1 VDC from the (added) BNC on the counter's front panel to the divide-by-10 circuitry attached to the External Frequency Reference Input BNC on the counter's rear panel.



2.  Removing the Main PCB (A4) for repair/troubleshooting.

Unless you have an Operating-Service Manual, it isn't clear how to remove Main PCB to get access to all of the IC pins and components for trouble-shooting or repair.  When mounted within the chassis, significant parts of the PCB are difficult to access with, for example, a scope probe.

But removal of this PCB is actually straightforward.  Per the manual's section 5-14:

First, remove both side panels and the top cover.

Then, remove the transparent colored plastic front-panel window by sliding it out either side of the unit.


Next, reach through the sides of the chassis and gently lift the sides of the main PCB.  Pull the board forward with your fingers.

And after the board is started, remove connector XA4 (the connector on the left-side of the rear of the PCB, when viewed from the front of the counter).


(This works better with two hands -- I was holding my camera with my other hand).

Push or pull the board out of the counter, being careful to keep the board moving in a straight line.

With the PCB now outside of the counter, you can reattach it to connector XA4 for trouble-shooting with power applied.  But there are two important caveats!

First, be sure to place something under the PCB (and behind connector XA4) to prevent either the board or the connector from shorting to the case.

Second, there is high voltage on the board, and it is easy to touch.  (Been there, done that!)  I now wear  gloves when making measurements on the board,


In the image above, several sheets of paper from a yellow legal-pad serve to temporarily isolate the board and connector XA4 from the chassis.


Standard Caveat:

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.

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.