Thursday, January 27, 2011

Modeling Class E/F RF Amplifiers, Part 2

[Part 1 can be found here] 

In Part 1 I evaluated a Class E/F amplifier using a center-tapped transformer to supply power to the drains of the two active devices. 

DC power can also be fed to the drains of the two MOSFETs via two inductors, thus simplifying the transformer because it no longer needs to be center-tapped. I was curious how such a circuit compared (in performance) to the circuit described in Part 1. 

Here's the SPICE model (using Linear Technology's free SPICE program: LTSpice):
   
(Click on image to enlarge)

Regarding the model: 

1. C4, at 1 Farad, provides a "stiff" AC ground for simulation purposes so that there's no voltage fluctuation at the R4, C4, L1, L2 node. 

2. L1 and L2 provide a high-impedance feed for DC power to the "Drains" of the MOSFETs. I chose these to be (initially) 25 uH. 

3. The MOSFETs are modeled with voltage-controlled switches. Their Ron is set via external resistors to be 0.15 Ω (plus 0.01 Ω within the Switch model). 

4. 500 pf capacitors mimic the MOSFETs' Coss (per the IRF 530 datasheet). Note that these capacitors are not voltage dependent. 

5. Refer to Part 1 for more detail on selecting the Transformer inductance of 400 nH. 

6. C3 was arrived at by trial-and-error (after initially setting its value to 3.655 nF per Part 1) by adjusting its value so that the voltage waveform at node "Va" is zero when the current waveform (through R1) is non-zero.
    
(Click on image to enlarge)

When adjust the value of C3 (or the transformer's inductance), the current and voltage waveforms maintain the same relationship to C3's value (or to the transformer's inductance) as was described in Part 1. Refer to the image below.
   
(Click on image to enlarge)
Lowering RMS current through the switches: 

I thought I'd see if I could lower power dissipation in the switching devices by lowering their RMS current (see Part 1 for more discussion on this technique). By lowering the value of inductance of L1 and L2, I could replicate current waveforms through the switching devices that look as though the amplifier is now a a "Class E/F2,odd" amplifier, but, when I made my measurements using the tools in LTSpice, there was really only minimal effect (if any) in power dissipation. 

Here's how the waveforms look with L1 and L2 reduced to 800 nH. As the table following this image shows, although the current-waveform changes significantly, there isn't much of a difference in switching-device power-dissipation (as measured across R1 and R2).
   
(Click on image to enlarge)

I used the LTSpice measurement functions to measure power and RMS current at different points in the circuit (for various values of L1, L2), and these measurements have been tabulated below.
   
(Click on image to enlarge)

Some observations from this table:
  • I really cannot see much change in R1's power dissipation as L1, L2 are varied. That is, the differences are small, and perhaps are within the tolerance of simulation errors.
  • Also, there's not much change in either Output Power (Pload) or efficiency for the range of values of L1 and L2 shown in the table. Therefore, a L1, L2 values in the range of, say, 25 uH - 100 uH, might be the simplest approach for my application.
  • As L1, L2 are decreased, C3 must be increased to compensate for their effect on the time-relationship of the voltage and current waveforms. However, an increase in the value of C3 increases the current through C3 (because voltage across the transformer is essentially constant, and we decrease C3's reactive impedance when we increase its value). Therefore, due to its internal ESR, C3 will dissipate more power when L1,L2 are reduced.
  • As the values of L1 and L2 is increased, it takes longer for the simulation waveforms to ramp up to their final values.
From the minimal change in R1 dissipation, and from the gross changes in C3 current (and thus, potentially, its power dissipation), it really doesn't make much sense (in my application) to attempt to "tune" L1 and L2 to minimize power dissipation in the switching devices by minimizing their RMS currents. 

It's possible, though, that in applications in which higher currents are present, some benefit might be achieved by "tuning" L1 and L2. However, the simulations don't bear this out in my application, and there seems to be little benefit. 

Overall, the power-out from this circuit topology is comparable to that of the center-tapped transformer topology described in Part 1. So, for my application, either circuit should do the job, and it'll just be a matter of determining which one is easiest to build. 

Modeling with the IRF530 MOSFET 

Actual MOSFETs can be simulated via LTSpice's "nmos" model, within which one can retrieve the SPICE parameters for a number of different MOSFETs, including the IRF530. (Note: these devices are not alphabetized in the "Pick New MOSFET" table.) 

Below is a circuit with IRF530 MOSFETs substituting for the original voltage-controlled switches. The final value of C3, as well as the AC source's amplitude and offset voltage (V2), were arrived at by trial-and-error. Also, two AC sources are used (the second with an offset of 180 degrees), because it seemed like an easy way to provide two sine-wave sources that were 180 degrees out of phase but with the same DC offset.
   
(Click on image to enlarge)

There are several differences between this circuit model and the original one with the voltage-controlled switches:
  1. The Sine Wave sources have an amplitude of 10 Vpeak and a DC offset of 5V (to bias the IRF530's near their turn-on point). This gives peak gate voltages of +15/-5 Volts (note that the max rating of the IRF530's gate voltage (VGS) is +/- 20V).
  2. C3 is now 3.69 nF.
Here is the Drain Voltage versus Source Current for one of the IRF530 MOSFETs:
   
(Click on image to enlarge)

Note: Source current is shown, rather than Drain current, because the Drain pin has quite a bit of current flowing through it even when the MOSFET is OFF. This current at the Drain pin is not dissipative current, though (that is, it isn't being dissipated as I2 * R heat through RDS(on). Instead, it's current passing through the Gate-Drain capacitance and out the Gate pin (LTSpice plots verify that I(drain) - I(gate) = I(source)). 

Here are some of the currents and powers (and overall efficiency) measured via LTSpice.
  • RF Power Out: 48.2 watts
  • DC Power In: 52.0 watts
  • Efficiency: η = 48.2/52.0 = 93%
  • IL1 = 1.14 ARMS
  • IC3 = 4.5 ARMS
  • IL3 = 5.2 ARMS
Note the high currents in C3 and L3. Care should be taken to use low-ESR devices! 

References: 

Spice:
  • Free Download of LTSpice here.
  • LTSpice "Getting Started Guide" here.
Class E Amplifiers: Class E/F Amplifiers:

Caveats: 

1. These simulations are for educational purposes only. 

2. I could have easily have made a mistake, so please view (and use) these simulations accordingly.  

3. And a final caveat regarding SPICE modeling. Modeling results should always be taken with a bit of caution, for the ability of SPICE to mimic actual circuit behavior depends in large part on how accurately a circuit and its components have been modeled. Factors such as parasitic components and circuit non-linearities can all cause modeled SPICE performance to diverge from actual performance. SPICE can provide valuable insights into circuit operation, but a bit of skepticism, too, should be applied when evaluating results.
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Wednesday, January 26, 2011

Modeling Class E/F RF Amplifiers, Part 1

[Part 2 can be found here.] 

I've recently been thinking of replacing the Johnson Ranger which drives my 813 AM transmitter with something more modern and which consumes less power -- a homebrew, solid-state unit which would replace the Ranger's RF driver and Speech Amplifier. 

The Ranger drives my 813 AM transmitter with 36-37 watts of RF Power(SWR is 1.27:1, by the way). I have a DDS VFO module that I'd like to use as the frequency-control of my Ranger-replacement, but its output is probably on the order of milliwatts (not yet measured), so I'll need to boost it with (probably) several stages of RF amplification. 

With that thought in mind, I started researching possible amplifier circuits that might provide the required 40 - 50 watts of RF power on 75 meters. 

I'd like to maximize efficiency of the PA, and of course Class-C Amplifier topologies immediately leapt to mind. But while researching designs, I came across the Class-E mode of amplification which promised improved efficiencies over Class C. This looked very worthwhile, and the design procedures for such amplifiers seem to be well established. 

However, as I searched the literature, I stumbled across an interesting QEX article ("A 200 W Power Amplifier", in the Jan/Feb 2004 issue) describing a Class E/F amplifier for 40 meters. The Class E/F topology (being push/pull) offers benefits over Class E, such as:
  • Lowers the required Vdc power supply voltage for the same power out.
  • Lowers Vds across the MOSFETs for the same power out.
  • Improves harmonic suppression.
Further investigation was needed! 

Unfortunately, design methodology for the Class E/F amplifiers doesn't seem to be at the same state as that of Class E amplifiers. Not letting that deter me, I figured SPICE modeling of these amplifiers would allow me to play around with circuit parameters and component values, and I would (hopefully) gain some insight into their operation and possible component values for my application. (By the way, there's a free SPICE program available via the Linear Technology website). 

Building a SPICE Model of a Class E/F amplifier: 

Below is my SPICE model for a Class E/Fodd amplifier. It's topology is based upon the amplifier circuit described in the 7 MHz, 1.1 KW demonstration amplifier by Kee, Aoki, and Rutledge). My goal was to design a PA for 75 meters (3.87 MHz, typical frequency) and output power (at the 50 Ω load resistor) on the order of 40-60 watts.
   
(Click on image to enlarge)
(Note: Comments are in BLUE, SPICE directives are in BLACK).

How did I arrive at the component values in the above circuit? Let's look at my procedure: 

First, my givens: Rload = 50 Ω. Desired power out: 50 watts. 

Next, I wanted to select a power-supply voltage (Vdc) that would:
  1. Allow me to use a 1:1 impedance transformation from load to the differential connection across the PA MOSFET drains.
  2. Not be too high of a voltage so that I could use 100V (VDSS) MOSFETs that are in my junk-box (e.g. IRF 530) .
  3. Yet be high enough to minimize current through the MOSFETs and the resultant "i2 R" power loss through their RDS(on) resistance.
Per my design goal of an amplifier that delivers 50 watts into a 50 Ω load, this means that the voltage across the 50 Ω load is 70.7 volts peak (141.4 Vpp or 50 volts RMS). 

And per my goal of not requiring impedance transformation (either from high-to-low or low-to-high impedance) from the load to the MOSFETs' drains, I'm using a 1:1 transformer to present the 50 Ω load resistor as 50 Ω to the PA MOSFETs' drains (this load appears differentially between the two drains). 

From the Class E/F literature, the peak voltage across the drain of each MOSFET is ≈ π*Vdc, where Vdc is the power-supply voltage (refer to equations 37 & 38 in "The Class E/F Family of ZVS Switching Amplifiers", for example). So, in my case of an untransformed (i.e. 1:1) 50 Ω load, then the peak value of the voltage at the drain of either MOSFET is the same as the peak value across the 50 Ω load resistor. In other words, Vds(peak) = Vload(peak), where the load is R3 in the circuit model above. Using this equality, we can derive the following equation:
   
Vdc ≈ [ √(Pload * Rload) ] / [ π * (√2) / 2]

Where Pload = power delivered to Rload (R3), and therefore Pload = [ (√2 / 2)*π*Vdc]2 / Rload, using our definition of Vpeak = π*Vdc.

In this case in which Power = 50 watts and Rload = 50 Ω, Vdc calculates to be 22.5 volts. (Note: I'm using 22.8 volts for my modeling, which is the lower-limit of the output voltage adjustment of some off-the-shelf 24 volt switching supplies), and Vds (peak) across either MOSFET is π*Vdc, or 71.6 volts peak. The IRF 530 MOSFET has a rated VDSS of 100 volts, so there's almost 30 volts of margin. 

OK, now it's time to calculate the inductance that we'll need between the drains of the two MOSFETs. This is the inductance of the primary of the center-tapped 1:1 transformer. Because the transformer is 1:1, the inductance of the primary is the same as the inductance of the secondary (same number of turns in each). To determine the inductance of the primary, let's first assume it has a loaded Q of 5. (I'm actually going to use 5.1). Why 5? In fact, some of the Class E/F amplifiers in the literature use a Q of around 3 or even less. I'd like to keep Q sort of high so that harmonic suppression isn't too compromised, but this may need adjusting if/when I actually build the circuit. 

OK -- the inductance of the transformer's primary is loaded by Rload (R3), reflected through the transformer from the secondary. Thus, because the transformer is 1:1, this 50 Ω load appears as 50 Ω across the primary, and it is this 50 Ω that "loads" the Q. Because L and R are in parallel, we know that:
 
Q = |R / XL|

In this case Q is actually Qloaded, and so we can rewrite this equation to give us L:
   
L = R/(2 * π * f * Qloaded),

which, in our example, gives us the following:
   
L = 50/(2*3.14*3.87e6*5.1) 
= 400 nH
   
Now let's use this value to model our transformer...

To create a center-tapped transformer in LTSpice, we need to use three coupled-inductors. Two of these inductors represent the turns on either side of the primary center-tap (and thus each has half the number of turns of the secondary, given that the transformer is 1:1), and the third inductor is the secondary. 

We know that the primary should have an inductance of 400 nH. Therefore, the secondary, because it has the same number of turns as the primary, will also have an inductance of 400 nH. 

The two inductors representing the two halves of the primary each has half the number of turns as the secondary. Therefore each has 1/4 of the inductance of the secondary (inductance has a turns-squared relationship). So we specify each of these two inductors, in SPICE, to have 100 nH inductance. This gives us a 1:1 center-tapped transformer. (Refer to LTSpice "Help" for more information on Transformer modeling, and an example of a model for a 1:3 transformer). 

Let's give these inductors some series resistance so that they represent "real" rather than "ideal" components -- that is, they have some amount (albeit small) of series resistance. The two inductors in the primary, because they're coupled, represent 400 nH overall inductance. Let's assume an unloaded Q of 150 for this overall inductance. Because this is a series R-L configuration, we use a slightly different formula for Q:
   
Q = |XL / R|

Solving for R:
   
R = 2 * π * f * L / Q 
= 2*3.14*3.87e6*400e-9/150 
= 0.065 Ω

I've rounded this down to 0.06 Ω, and I'm splitting it into two 0.03 Ω resistors which I place in series with either side of the transformer's primary. 

(Note: in the past (and perhaps still) SPICE programs sometimes had problems converging upon a solution if one specified only ideal components. From habit I tend to throw in small values of resistance in series with an ideal component, such as the 0.06 Ω in series with the inductor(s) described above, with the hope that they'll nip such problems in the bud.) 

What should be the value of the resonating capacitor (C3)? Unfortunately, I couldn't find an equation for calculating this value. All I know is that the overall impedance of the parallel L-C circuit represented by the transformer primary and C3 (in the circuit above) should be slightly inductive (per Class E/F references). This means that, because the circuit is a parallel-resonant circuit, the resonant frequency of the L-C combo should be slightly above the circuit's operating frequency. 

But how far above should it be? I don't know, and I cannot find an equation to calculate it. [For further discussion of how one might calculate C3, please refer to Note 3 in the Notes section at the end of this posting]. 

So perhaps we should run some simulations and see if we can find it by trial and error... 

But before I get to that, let me first fill in the remaining components in my SPICE circuit model: 

I've set the dc-feed series inductor (L1) to an arbitrarily large value, in this case 10 uH (for no good reason, I chose 25 times 400 nH). 

C4 is set to the humongous value of 1 Farad to provide a "stiff" AC ground. (At 1 uF, the value I'd originally set it to) there was some voltage fluctuation at the C4, L1, R4 node, and so I thought it better, for simulation purposes, to make the AC ground here as solid as possible). 

I'm modeling the MOSFETs with (almost) ideal voltage-controlled switches (LTSpice's "SW" model). Because I'm not concerned (at the moment) with modeling the MOSFET inputs and drive circuit, I can define the inputs to be whatever is most convenient for me. Thus, I've set the switching thresholds of the SW model to 0 volts, and the switches are driven by a 1-volt amplitude sine-wave that has a 0 volt DC offset. Each switch is on for one-half of a full cycle, and neither switch is on at the same time. 

The default SW model in LTSpice has an Ron of 1 Ω, which is much too large for our purposes. The IRF 530 MOSFET that I'd like to use has an RDS(on) of 0.15 Ω (depending upon which manufacturer's datasheet you look at). Although I could specify Ron in the SW model to be 0.15 Ω, I personally prefer to specify it as a separate component in series with each switch, and thus I also changed the SW model's Ron value from 1 Ω to an arbitrarily small value (in this case 0.01 Ω). 

Similarly, I've placed the MOSFETs' Drain-Source Output Capacitance (Coss) across each series combination of 0.15 Ω and SW model. In the case of the IRF 530, Coss is 500 pf, which I've shown here in my circuit model. 

So now we're back to C3. How do I determine its value?  

Let's use SPICE to find it by trial-and-error... Let's start by assuming it resonates with 400 nH (the transformer's primary inductance) at our operating frequency of 3.87 MHz. This gives us a value for C3 of 4.22 nF. 

Plugging this value into the model and running a simulation, it's quite apparent from the large current spikes that the voltage waveform at the "drain" of S1 is misaligned with respect to the current flowing through S1):
   
(Click on image to enlarge)

Playing around with the value of C3 and iterating to a solution, I found that a value of 3.655 nF for C3 gives a good result for the voltage and current waveforms. (This also results in a resonant frequency of 4.16 MHz).
   
(Click on image to enlarge)
Experiments! 

Now that we've created our SPICE model of a Class E/F amplifier, let's do some experiments. 

First, let's try decreasing and increasing C3 (and thus raising and lowering, respectively, the resonant frequency from the original value of 4.16 MHz).

Lowering the resonant frequency to 4.08 MHz by increasing C3 to 3.8 nF gives us this relationship between the voltage and current waveforms:
   
(Click on image to enlarge)

So if we see waveforms that looks like the above, we know that we need to decrease the capacitance of C3 to shift the voltage waveform to the left, relative to the current waveform. Now let's raise the resonant frequency to 4.25 MHz by decreasing C3 to 3.5 nF:
(Click on image to enlarge)

And if we see waveforms that look like the above, we know we need to increase the capacitance of C3 to shift the voltage waveform to the right, relative to the current waveform. 

Notice in the above images how the voltage waveform shifts with respect to the current waveform as we move the L-C network's resonant frequency up and down. Our goal is to adjust C3 so that we minimize the power dissipated within the MOSFETs (Vds is 0 volts when current is non-zero). This leads us to the following rule-of-thumb, as expressed in the image below:
   
(Click on image to enlarge)
Lowering RMS current through the switches: 

Reduction of the RMS current through the switching devices can be accomplished by resonating L1 with the switch's Coss (C1 or C2) near the second harmonic of the operating frequency (refer to section "C" in the 7 MHz, 1.1KW Amp reference for more information). This type of configuration is known as a "Class E/F2,odd" amplifier. 

Here's the resulting waveforms when L1 is changed from 10 uH to 500 nH. 

(Note: C3 had to be changed from 3.655 nF to 3.798 nF to bring the voltage and current waveforms back into their proper time-relationship.)
   
(Click on image to enlarge)

A couple of points: 1. I played around with different values of L1 (modifying C3 each time to maintain the proper time relationship for the Voltage and Current waveforms) and measured, via LTSpice measurement functions, RMS current and power at different points in the circuit. Here's a table of my results:
   
(Click on image to enlarge)

Some observations from this table:
  • Current (RMS) through a switch (i.e. I(R1)) is minimized when L1 = 500 nH. BUT, the delta in power dissipation (from, say, the case in which L1 = 100 uH) is only about 0.02 watts.
  • Also, current through other devices increases as the value of L1 is decreased. There is no "sweet spot" for these other currents when L1 = 500 nH, and the other devices will dissipate more power (because of their higher currents) as the value of L1 is decreased to 500 nH.
  • Overall, there's not much change in either Output Power (Pload) or efficiency for the range of values of L1 shown. Therefore, using an L1 value in the range of, say, 10 uH - 100 uH might be the best approach for my application.
  • I can't explain why there's as much difference as there is when L1 = 240 nH.
  • As C3 is increased to compensate for a decrease in L1, C3's current increases. Therefore, due to its ESR, it will dissipate more power. (However, the delta in current when going from an inductance of 100uH to 500nH isn't too bad.)
  • As the value of L1 is increased, it takes longer for the simulation waveforms to ramp up to their final values.
2. L1, at 500 nH, resonates with the 500 pF value of C1 at about 10 MHz, which, although near to the second harmonic of 3.87 MHz, is actually a bit closer to the third harmonic. I don't know if there's a formula one can use to exactly calculate what L1 should be -- the literature seems to only state that the resonant frequency should be "near" the second harmonic. Iteration through trial-and error might be the only approach (at this time) for selecting an inductance for L1 to minimize switching-device power dissipation. 

3. From these results, it's doubtful that, for my application, tuning L1 for minimum IRMS, through the switches would be worth the effort. However, this approach might prove more worthwhile in applications in which significantly more current flows through the switching devices. 

Other notes: 

1. An interesting observation I made while experimenting with the SPICE model is that, once you've set the phase (i.e. time) relationship of the switches' "drain" voltage and current waveforms, this relationship is independent of the DC supply voltage. That is, as you raise or lower Vdc, you will not change where the drain voltage is, in time, with respect to the drain current. In other words, you shouldn't need to "retune" the resonant circuit as you raise or lower Vdc. This also implies that you can change the power out simply by raising or lowering Vdc. 

(This observed independence on Power Supply level probably depends, at least in part, on how independent the value of Coss is from the voltage across the Drain-to-Source of the MOSFET, which is something I don't know. Results might differ when using actual "real-world" components.) 

2. Regarding the calculation of C3. After I finished my simulations I went back and reviewed some of the literature. There is a hint of how to determine C3 in the Cal Tech Thesis of Scott Kee (available as a PDF here). Refer to equation 7.18 and the discussion at the top of page 130. 

The Thesis states, "it is apparent that the required fundamental frequency differential load is a resistance in parallel with an inductance having the same impedance at the fundamental frequency as the capacitance Cs does." Note that in our application Cs is the same as either C1 or C2 (Coss). 

To me this statement means that we can consider the transformer inductance to consist of two "virtual" parallel inductors, one inductor resonating with C3 at the operating frequency, and the other inductor resonating with Coss, also at the operating frequency. In other words, the two inductors, as well as C3 and Coss, can all be considered a parallel circuit consisting of these four components (all in parallel), resonant at the operating frequency. 

The 400 nH inductance is actually the combined inductance of the our two "virtual" parallel inductors. Let's look at the circuit configuration during the half of the cycle when S2 is closed: C2 is shorted and both one end of the transformer's inductance (400 nH) and C3 are grounded. This means that the 400 nH inductance, C1 (500 pf), and C3 (3.655 nF) are all in parallel. The equivalent capacitance is therefore 4.155 nF, which, in conjunction with 400 nH, gives us a resonant frequency of 3.90 MHz. 

3.90 MHz is pretty close to (although not exactly at) my operating frequency of 3.87 MHz. Coincidence? Should C3, when it's paralleled with Coss, be calculated to resonate with the transformer's inductance at the operating frequency? 

That is, should we use the equation:
   
F = 1/( 2 * π * √[ L(xfrmr) * (C3 + Coss) ] )

as a first-order approximation when deriving C3? 

References: 

 Spice:
  • Free Download of LTSpice here.
  • LTSpice "Getting Started Guide" here.
Class E Amplifiers: Class E/F Amplifiers:

Caveats: \

1. These simulations are for educational purposes only. 

2. I could have easily have made a mistake, so please view (and use) these simulations accordingly.  

3. And a final caveat regarding SPICE modeling. Modeling results should always be taken with a bit of caution, for the ability of SPICE to mimic actual circuit behavior depends in large part on how accurately a circuit and its components have been modeled. Factors such as parasitic components and circuit non-linearities can all cause modeled SPICE performance to diverge from actual performance. SPICE can provide valuable insights into circuit operation, but a bit of skepticism, too, should be applied when evaluating results.
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Tuesday, November 2, 2010

Improving the Heathkit HR-10B Receiver

UPDATE:  24 April 2012.  Just updated the schematic, which had an error (the source pin of the MPF102 oscillator should connect directly to T5 pin 2).  The new schematic is Rev. 2 (the old schematic was Rev. 1).

The Heathkit HR-10B is a 5-band, 7-tube amateur radio receiver manufactured from 1967 to 1975 and the companion to Heathkit's DX-60B transmitter. Essentially, the HR-10B design is the same as its predecessor, the HR-10 -- the only change seems to be that the top cover was painted with a wrinkly finish rather than the smooth finish of the original HR-10. It requires an external speaker or headphones.



I've always like the way the Heathkit HR-10 series receivers looked with their functional control layout and slide-rule dial. I picked this one up for a reasonable price, and I thought I'd give it a try. Powering it up, I immediately noticed a number of problems:
  • Slide Rule Dial tracking diverged greatly on 80 meters.
  • Receive frequency changed significantly as "RF Gain" was varied.
  • Couldn't use AGC (AVC) for SSB/CW modes.
  • Broad Selectivity.
  • Deaf on 15 and 10 meters.
From my internet research, these problems seemed to be common. Not a very good receiver, and it got me wondering...was there anything I could do to improve it?

Slide Rule Dial not tracking on 80 Meters.

If I calibrated the HF oscillator at the 3.5 MHz mark on the dial and then tuned up in frequency , I found that as I tuned towards 4 MHz, I would hear a 4 MHz signal at about the 3.96 MHz dial mark. In other words, as I tuned through the band the oscillator diverged significantly from the scale markings.

The tuning capacitor, when the dial is at 3.5 MHz, is at maximum capacitance. The fact that I'm receiving a 4 MHz signal at the 3.96 MHz dial tick tells me that the capacitance of the tuning capacitor has decreased too much as I rotated the dial.

One way to fix this is to add additional parallel capacitance to the 80-meter oscillator tank circuit so that , as the variable capacitor is tuned, the the overall "delta" in capacitance is reduced. I found that, for the amount of divergence that I was experiencing, paralleling C30/C66 with a 6 pf Silver Mica capacitor brought the dial into close enough calibration for my purposes. (There is a bit of divergence at around 3.6 MHz, but there's nothing I can do about that).


Receive Frequency shifts with changes in RF gain.

This seems to be a common problem with the HR-10 series receiver, and dynamic variation of the "125" plate-voltage line (i.e. the junction of R44, R43, and C56 in the schematic) seems to be the source of the problem. This voltage is generated by dropping the DC from the cathode of the rectifier through a series 1500 ohm, 10 watt resistor. Thus, because any change in RF gain changes plate current, plate voltage will also change because the the voltage drop through the 1500 ohm resistor has changed.

Unfortunately, as plate voltage varies, so does the frequency of the oscillator(s).

One way to fix the frequency shifting is to stabilize (regulate) the plate voltage. I added a series-string of four 5 watt Zener Diodes from the R44, R43, and C56 junction to ground (there's also a series 10-ohm resistor so that I can measure current, and thus power-dissipation, through the zeners). These diodes consist of three 33V, 5 watt diodes and one 18V diode, for a total voltage of 117 volts (prior to adding the zeners this node measured 144 VDC instead of the spec'd 125 volts, so there's headroom). Power dissipation in the 33V diodes measured to be about 1.2 watts apiece, so there's plenty of margin, dissipation-wise.

These diodes are shown at the top of the schematic below.

(Click on schematic to enlarge)

Note: Reference Designators in the schematic reference the original Heathkit parts. Parts without reference designators are new parts.


AGC (AVC) for SSB and CW

The HR-10B suffers from the standard problem with receivers designed pre-SSB: the AGC is worthless for SSB. Instead the user is advised to turn off the AVC, set the AF Gain to 3 o'clock (i.e. HIGH!), and then adjust the RF gain for an appropriate signal level.

In other words, there is no AGC for SSB or CW!

Before getting further into modifications, let's first take a step back and try to understand why this is...

Vintage receivers (prior to the days of product-detectors) typically used their diode-detector to detect both AM as well as SSB/CW detection. In SSB/CW mode this detector is driven by the output of a basic heterodyne mixer in which this output contains the BFO signal that it is driven with, as well as the beat products. These beat products form an envelope on the output waveform which is detected by the diode-detector. There are several problems with this method of demodulation for SSB and CW.

First, because the SSB or CW signal is demodulated with an envelope detector, the BFO signal must be quite a bit larger than the IF signal if there is to be minimal distortion on either CW or SSB. You can get an idea of why this is so by looking at the image below and comparing the envelopes of the two waveforms (Es = Eo and Es = 0.5Eo).


(Click on image to enlarge)
(Terman, F. E. Radio Engineers' Handbook, First Ed., McGraw-Hill Book Co., 1943, Page 567)

Imagine that Es is the IF signal representing a CW signal and Eo is the oscillator. If the amplitude of Es is significantly less than Eo, then the envelope on the resultant mixed waveform looks close to a sine-wave (look at the envelope of the Es = 0.5 Eo signal). And because this envelope is detected with the diode-detector, it will sound fairly undistorted.

But as the amplitude of Es approaches that of Eo, the envelope becomes much more distorted (look at the envelope of the Es = Eo signal), and thus the resultant detected output will be full of harmonics and sound grossly distorted.

So the IF signal must always be appreciably less than the BFO signal. But...this introduces another problem. Because AVC is also derived from the signal at the output of this mixer (which contains the BFO signal if its on), if the BFO is on this BFO component at the output of the mixer will swamp the AVC circuit and thus severely attenuate the receiver.

For this reason the operator manuals for older receivers state that, when receiving CW (or SSB) signals, the AVC should be turned OFF, the Audio Gain turned UP, and the RF Gain manually adjusted to provide a comfortable signal level. Not very convenient nor friendly to your ears when a very strong signal suddenly pops up nearby, and an excellent reason for adding a product detector and upgrading the AVC circuitry.

So...I decided to update the AGC circuit and at the same time add a product detector in lieu of the original CW detection scheme.

First thing I did was to replace V5, a triple-diode tube (6BJ7) with three 1N4148 diodes. (I had some DC voltage on the AVC line even with no input signal that I attributed to "leakage" in the 6BJ7 tube. Replacing the tube with diodes fixed this problem, and, of course, also lowered power dissipation).

For the AGC circuit I added a 1N4148 diode to change the AGC voltage-doubler configuration from a "Villard" circuit to a "Greinacher" circuit, which has better ripple characteristics. I increased the AGC decay time by paralleling a new 0.22 uF cap with the existing 0.05 uF cap (C29) and moving the location of the 1M resistor (R26) to increase the decay resistance from the original 2.2M ohms to 3.2M ohms.

With the 1M ohm resistor that had controlled the charge-rate moved, I replaced its function with a much smaller 9.1K ohm resistor (this value doesn't need to be exact -- in fact, you can probably get away with just using a jumper in lieu of this resistor).

I used an NE602 for the product detector -- my original goal being to use its on-chip oscillator for the BFO. Unfortunately, when I tried this (using the "stock" HR-10B BFO components) I found that the BFO frequency would "pull" with incoming signal strength (e.g. as RF Gain or AVC varied). I couldn't discover why this was happening, so I worked around it by replacing the on-chip oscillator function with a simple external oscillator using an MPF102 FET, and it worked much better.

If the BFO is on when in AM mode, you can hear it heterdyning with the carrier of the incoming signal, so it's necessary to turn the BFO off when receiving AM signals. To disable the oscillator the low-end of the oscillator tank circuit, T5, is removed from ground using a 2N7000 transistor. To turn the BFO on, this transistor must first be turned on to short pin 1 of T5 to ground.

While experimenting I ended up with quite a bit of attenuation at the input of the NE602 (the capacitive divider). I'm not sure if this much attenuation is needed; I added it because, during my testing I was experiencing some distortion issues and this seemed to help. However, I was making a number of changes around this time, and I could easily have over-compensated. Don't take these values as being the final word -- experiment!

The demodulated output from the NE602 drives two separate paths -- the audio path and the AGC path. I wanted to decouple the audio-path gain from the AGC-path gain (this is an audio-derived AGC circuit) just in case I needed different gains for the two paths. The op-amp inputs are fed via simple low-pass filters (to remove any residual RF from the output of the NE602). Gain of the audio path is about 37 dB, while gain of the AGC path is about 39 dB. This isn't much of a difference, and one could probably use the same op-amp to drive both paths.

The TL082 op-amp has a max power-supply rating of 35 volts (when powered with a single supply). I powered it with 30 volts to ensure that I'd have plenty of headroom when experimenting with gains -- the op-amps are biased at 15 volts, which give them about a +/- 12 volt swing (the TL082 output limits when within (roughly) 2-3 volts of either power-supply rail).

AGC gain is set to give me the same S-meter reading (roughly) when in either AM or SSB mode (BFO Off or On).

Audio gain is set to give the same audio ouput at the speaker (very roughly) when in either AM or SSB mode.

A relay is used to select between AM (no BFO) and SSB (BFO) modes. In AM mode, the HR-10B demodulation and AGC circuitry is the same as the "stock" receiver (with the exception of the changes to the voltage-doubler and RC time-constants described above). A 48V coil for the relay is used to minimize current drain (and thus power dissipation) -- it only draws 4 mA when on.

No changes were made to the Noise Limiter (ANL).

Here's a photo showing where and how I mounted the op-amps and the NE602. You can also see the string of zener diodes I added for oscillator stability near the top of the photo.


Note: Reference Designators in the schematic reference the original Heathkit parts. Parts without reference designators are new parts. And for many of these parts the value isn't critical -- I usually just pulled parts out of the junkbox that were in the ballpark of what I wanted.


Broad Selectivity

The HR-10B receiver has a two crystal crystal- lattice filter spec'd at 3 KHz down at 6 dB at an IF frequency of 1681 KHz.

Although 3 KHz might seem narrow, I've found that the skirts of the filter (on my receiver) are not very steep at all. This gentle roll-off of the filter skirts results in audio that is fairly broad, and, in fact, for AM reception I find that the receiver actually sounds pretty good.

I did find an article in Electric Radio regarding modification of the HR-10B crystal filter (as well as crystal filters in other receivers -- refer to the Electric Radio articles in the "Resource" section below). I decided not to attempt these mods at this time.


Deaf on 15 and 10 Meters

Lack of sensitivity on the high bands is a common complaint for this receiver, and mine is no different. I've poked around at this, and it looks like it's caused by a couple of things.

1. The RF Preamp (V1 and associated circuitry) has appreciably lower gain on 15 and 10 meters.

2. On 15 and 10 meters the HF Oscillator, rather than beating the incoming signal with the fundamental of the oscillator to get the IF frequency, instead beats the incoming signal with the second harmonic of the oscillator frequency. The amplitude of the second harmonic will always be less than that of the fundamental frequency, and, depending upon how the second harmonic is generated, the second harmonic might be significantly lower in amplitude.

Signal level at the output of the mixer is a function of the level of the input oscillator, so a lower-level oscillator signal will result in a lower-level output (and this can be exasperated if there's a square-law (or higher!) function in the mixing process.

I haven't yet looked into improving the performance on 15 and 10, given that there's sure to be a stability issue, too, given that we're using the second harmonic for the conversion. In other words, jitter or drift at the fundamental frequency means twice the jitter or drift at the second harmonic, and thus twice the degradation in stability!


Other Notes:

1. There's an optional Crystal Oscillator (HRA-10-1) which can be plugged in. This is a useful option!

2. Others have mentioned that alignment of the RF/Oscillator section can change when the bottom steel plate is reinstalled after completion of the alignment procedure. It has been recommend that holes be drilled in the bottom plate so that the receiver can be aligned with the plate in-place. I've not yet done this.


Resources:

Electric Radio Magazine Articles:
  • "Resurrection of a Heath HR-10B Receiver," Paschall, Issue 210, Nov. '06
  • "The Heathkit HR-10 Receiver," Hanlon, Issue 232, Sept. '08
  • "Heathkit HR-10 Receiver Update," Stock, Issue 234, Nov. '08
  • "Modifying Heathkit Crystal Filters," Stock, Issue 231, Aug. '08
Search Heath listserve archives here.

HR-10B Schematic here.

HR-10B Modifications here.


Standard Caveats

There might be mistakes. I cannot guarantee that everything is accurate. Use at your own risk!

The HR-10B has high voltages -- use caution whenever working on it!
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Tuesday, October 26, 2010

WRL Duo-Bander 84

I came upon a pair of these radios (plus an AC power supply) a few months ago while visiting a ham-radio store that specialized in older, used gear. It's a Duo-Bander 84, manufactured by WRL (World Radio Laboratories, Inc.) sometime in the late 60's.
   
(Click on image to enlarge)

The Duo-bander 84 is designed to be a sideband-only rig for 75 and 40 meters, and thus coverage on 75 meters is 3.8 to 4.0 MHz, and coverage on 40 is 7.1 to 7.3 MHz. 

Tuning-up the transceiver's Transmitter is a single knob operation (rather than the usual peak-the-grid, dip-the-plate, and adjust-loading). Simply insert carrier into the Transmit signal by un-nulling the carrier balance (using the "NULL" control), and then adjust the "TUNE" knob for maximum output on the meter. (The "NULL" control (carrier-balance) will then need to be re-nulled). 

The PA consists of a pair of 6HF5 sweep-tubes. With about 700 volts on the plates (the radio requires a separate power supply, by the way), I find that my peak power out is a bit more than 100 watts.

(A copy of the manual can be found via the link in the Resources section, below.)
   
Top view (click on image to enlarge)

Bottom View (click on image to enlarge)


Bottom view of a later version.


Note the new holes in the chassis under the IF board, and the crystal filter is sans case. I don't know if these changes were done in the factory, or were made later in the field by an end-user. 

Notes: 

1. Adjusting PA bias. Per the manual, one should set the PA bias so that, when the carrier is properly nulled and there is no voice-excitation (and the meter switch on the back panel is set to "Bias"), the meter needle, when transmitting, is on the "bias" calibration mark on the meter scale. 

Unfortunately, one of my two radios had a broken meter. How should I then set its PA bias? 

Well, I wasn't too sure how accurate the meter was on the rig that was operating properly (and note, the meters are not calibrated in mA), so I didn't want to use it as a calibration reference. I did a bit more research and discovered that the Swan 350 also used a pair of 6HF5 tubes in its PA. Their manual is a bit more explicit, and they state that the PA bias should be adjusted to 50 mA idle current when in Transmit mode.  

I decided that if 50 mA is good enough for the Swan, it's probably good for the Duo-Bander, too! So I verified that the PA cathode resistance to ground was 2.5 ohms (there are four 10 ohm resistors in parallel), and adjusted the bias-pot on the back panel so that the PA cathode-to-ground voltage was around 0.125 volts. 

And with the PA Idle Bias adjusted for 50 mA, the meter needle sits at the "Bias" marking on the meter faceplate (when the meter is in BIAS mode)! 

2. No ALC circuit. The Duo-bander 84 does not have an ALC circuit to limit voice-peaks. To prevent excessive overdriving, I prefer, while monitor the output RF waveform with a 'scope, to adjust the mic's gain until the peaks are just at the peak-power out (this point will become evident as you adjust the mic-gain past this point -- the RF peaks will not get any higher, and you'll see more flat-topping). 

I've thought about adding an ALC circuit, but decided that it wasn't worth the effort. For those who are interested in experimenting, a good place to start would be to look at the schematics for the Swan 350, the Galaxy V Mark II, as well as other radios (Heathkit HW-12A, Galaxy GT-550, etc.) that use sweep tubes in their finals. You'll see an ALC circuit that's common to all of these radios and which consists of a pair of diodes used as a negative peak detector to generate a negative ALC voltage based upon the PA grid voltage. 

3. Distortion on transmit audio. 

Both of my radios exhibited significant distortion on their transmit audio when I was first testing them. On both, I traced the cause back to a bad C9 capacitor. This is a 2 uF, 50V cap that acts as an AC ground for the collector-load of Q4. The balanced-modulator (Q6 and Q7) requires that the audio-drive to it (from Q4) consist of two signals 180 degrees out-of-phase and of equal amplitude. Thus the phase-splitter's (Q4) emitter and collector loads should be identical. If C9 is not a good AC ground for R11, then the amplitudes will not be equal, and there can also be a phase difference between the two that is not equal to 180 degrees. 

On both radios I replaced their C9 caps with 4.7uF, 63V axial electrolytic caps that I had in my junk box, and the distortion problems greatly improved. (Note: it's OK to use a 4.7uF to replace the 2uF in this application. Larger value caps provide a "stiffer" AC ground for the audio signals, due to their lower impedance at audio frequencies). 

Here's C9 in the schematic:
   
(Click on image to enlarge)

4. Excessive Transmit Audio Low Frequency Roll-off. 

Per the alignment instructions, the filter passband can be "shifted" in frequency by adjusting the Carrier Crystal frequency with trimmer C39 (mounted next to the carrier crystal at the back of the radio). I found that, even with the crystal adjusted to shift the filter passband as close to the carrier as I could, I still had excessive roll-off in the low-frequency audio, so much so that it seemed as though the low-frequency cut-off was around 500 Hz. 

Poking around the audio path with a 'scope, I discovered that there was excessive low-frequency roll-off occurring just after the mic-jack coupling capacitor C7 (0.01 uF). I paralleled C7 (0.01 uF) with a 0.1 uF cap (or you could simply replace C7 with a 0.1 uF cap), and this removed the excessive low-frequency roll-off.
   
Bandwidth is now 300 - 3000 Hz. 

5. Carrier Crystal Oscillator stops oscillating in Transmit. 

While I was trying to adjust the Carrier Crystal oscillator frequency to shift the filter passband so that the low-frequency cut-off was around 300 Hz rather than 500 Hz, I discovered that, as I rotated trimmer C39 towards one of its limits, the oscillator would stop oscillating when I was transmitting (but there was no problem in Receive mode). 

As an experiment, I paralleled C15 (150 pf silver mica cap) with a 100 pf silver mica cap, and this seems to have cured the problem. My reasoning for adding this cap is: the oscillator would stop oscillating as its frequency was lowered, and so I assumed this meant that the trimmer cap was approaching its maximum capacitance. I decided to add capacitance across C15 (that is, to the fixed-cap side of the voltage divider formed by C15 and C39) in order to return the capacitance ratio between these two caps back to a value where the oscillator still oscillated. It seems to have worked, but I can't say that this is an optimal solution. Consider it a band-aid which fixed the problem for this particular transceiver.
   
(Note that you can easily add this capacitor by simply mounting it across the coax (from the Carrier oscillator) connected to the two pins at the back of the printed-circuit board, very close to the trimmer cap C39.) 

6. Receive Distortion due to AGC: 

While operating this radio I noticed that, for some signals, there was some subtle distortion on the receive audio. This problem seemed to manifest itself with stations who were using wide-band audio (e.g. lots of low frequencies). Fortunately, most signals I copied didn't seem to have this problem. 

But the distortion was noticeable enough on a couple of stations with whom I talk regularly, and so I decided to look into it. 

I noticed that, per the schematic, there are actually two AGC lines: an "AVC RF" line (to the grid of V6) and an "AVC IF" line (to the grids of V3 & V4). 

The AVC IF signal has a very fast decay time constant (essentially, the decay of C21, a 0.02 uF cap, is controlled by R22, a 33K ohm resistor, which is C21's discharge path into C22, a much larger 0.22 uF cap). 

 AVC RF, on the other hand, has a much slower discharge -- C22's decay is controlled by R23 in series with R24.
   
(Click on image to enlarge)

Looking at the AGC (or AVC, if you prefer) signals with a 'scope, I noticed a potential problem with the AVC IF line. If you look at the top photo below, you'll notice there are a lot of "spikes" on its waveform (as the AGC goes more negative, there is more attenuation). These spikes are voice peaks at C21, and they quickly decay via R22. 

One often sees this sort of AGC response in older receivers. I believe this fast-decay AGC is to limit the gain of fast transients (e.g. static crashes?) yet not have them affect the longer-delay AGC. Unfortunately, in my experience, these sorts of spikes can "modulate" the received signal and produce perceptible receive audio distortion, and often the receive audio will sound better if one can eliminate this sort of AGC modulation. (Of course, recognize that, in doing so, there can be a trade-off with limiting the gain of fast-transient noise). 

For the WRL Duo-Bander 84, one way to clean up this distortion is to add more capacitance to C21 -- that is, make it larger. But rather than add another cap, one can simply short-out R22, the 33K ohm resistor between C22 and C21. This means that the AVC IF line's decay time constant becomes the same as the AVC RF line's time constant, and you can see the result on the AVC IF line in the lower photo, below.
   
(One issue with this sort of mod, though, in which the capacitance is greatly increased, is that the attack time will slow down (because you're charging more capacitance). The Duo-Bander's Receive AGC is audio-derived, and therefore, as with many other receivers with audio-derived AGC, you can get an audible attack "pop" at the start of strong signals. In theory, the mod I've made should exacerbate this type of popping, but I haven't noticed much difference, if any, in attack pops with or without this mod.) 

The mod can be made easily without removing the PCB: To short-out R22, jumper the top of C22 to the top of R37, as shown in the photo below.

 
Again, I want to stress -- the original distortion that I was hearing was very subtle and only manifested itself on a couple of signals that I listened to regularly. For the most part, all other receive signals sounded fine. So this modification is certainly not necessary, and, if you do try it, it's quite possible that you won't notice any difference on the majority, if not all, of the signals you copy. 

. Other problems... Here are some of the other problems I found in my two Duo-Bander radios:
  • Very little gain on receive. I traced this to R46 (47K, 1W plate-load for V7b) reading infinite ohms. Replaced with a 47K, 2W resistor from my junk-box.
  • Very low TX audio. I traced this to a faulty R9 (270K, 1/2 watt) in the Mic Preamp circuit (it read infinite ohms with an ohmmeter). Replaced with same value resistor.
8. Measuring resistance per the Manual's Resistance Chart. 

While trouble-shooting the rig I discovered that a number of resistances listed in the Resistance Chart were reading infinite ohms, despite their non-infinite values in the table. 

These were typically plate resistors (or other resistors connected to the +400V B+ supply), as well as, for example, cathode-resistors in the TX path (e.g. V7 cathode). The reason why they were reading infinite with my ohmmeter was that they had no path to ground. (Note that the manual specifies its resistance tables in the chart to be "resistance to ground"). 

To make an accurate comparison of all resistances, per the chart, I'd recommend that you actually measure across the specified resistors, or you could ground the appropriate relay pins (e.g. pins 2 and 6 for Receive and Transmit B+, and pin 7 for TX cathodes) and then make your resistance measurements (in either case, though, be sure that all power is first disconnected from the radio -- remove the connector from the power supply to the Transceiver's P1 Power Plug!). 

9. Power Supply: The manual states that the power supply requirements are:
  • HV: 800VDC @ 400 mA
  • Low B+: 325/375 VDC @ 200 mA
  • Neg: -100 VDC @ 30 mA
  • 12VDC @ 200 mA
  • 12VAC (or DC) @ 5 A
Per one of the links in the Resource section below, three power supplies were available for the Duo-Bander 84:
  • Deluxe 400 Watt AC supply AC384A $89.95;
  • 400 Watt DC Supply DC384A $99.95;
  • 250 Watt AC Power supply AC48A $49.95.
Dale, W4OP, adds this note:

"I thought it might be nice to have the WRL supply. What I ended up buying on eBay was a WRL Duo-Power 300. I only found out what it was after arrival. While it has a nice  metal WRL label on it, there is no mention of a model #. W9RAN came to the rescue, identifying the supply. It is a compact AC supply but has  two clasps that allow for  a vibrator add on supply for mobile operation.
Sure didn’t look that big in the photos. Recapped it and it runs as it should. You might add it to your list of power supplies for the Duo-Bander."

In addition, other Galaxy power supplies allegedly can be used, too (I've not personally verified this, though), per this post on the Electric Radio Forum:

 
12/05/02 08:47 PM | 0 Good Guy Alerts Vote Edit Reply | WRL Galaxy Duo-Bander 84 power supply info wtd:
n3ibx New Member 1 Post 0 Good Guy Alerts Washington Crossing N3IBX Ignore User Hello All, Would anyone know if the AC supply from a Galaxy 5 MkII will work with the Duo-Bander 84? Any assistance will be appreciated. Mod-U-Later, Joe Cro N3IBX
Joseph Cro

12/06/02 07:38 PM | 0 Good Guy Alerts Vote Edit Reply | Galaxy Power Supply
Stu New Member 1 Post 0 Good Guy Alerts Elmira NY K4BOV Ignore User Yes Joe,

Any Galaxy power supply except the PSA-300 (came with the Galaxy 300) will run the Duo-bander without modification. The PSA-300 will certainly run a Duo-bander; but, the 12 pin Jones plug must be rewired to conform with the later Galaxy pin configuration.

If you want to send me the serial number of your Duo-Bander, I may have some service bulletins for your particular series. Might want to give me the serial number of the power supply as well. They all work as I indicated; but, some are more capable than others.

Stu/K4BOV

10. Some other radios using 6HF5 tubes in their finals:
  • Swan 350 and Swan 400
  • Drake 2NT
  • Galaxy 300, Galaxy V, and Galaxy V Mark II
  • Galaxy 2000 (Amplifier)
  • Hallicrafters HT-46 and SR-400
Resources: 

1. Specifications and more information on the Duo-Bander 84 can be found here and here. (Admittedly these are both sketchy). 

2. An Instruction Manual (in PDF format, and including alignment instructions and schematic) for the Duo-Bander 84 can be found here. 


Standard Caveat: 

 Of course, I may have made a mistake, so use my suggestions at your own risk! Also, this radio uses high-voltages that can kill you. Always use caution when working on a radio of this type.
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