feedback resistor math check please?

gadget73

junk junkie
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I'm changing the transformers on a little console amp from the 2.5 ohm originals to a set with 8 ohm output impedance. Unfortunately I don't have the originals any longer and I never thought to measure the feedback level so I'm working half blind here. This is the schematic I drew up a while ago for the stock amplifier with the 2.5 ohm outputs:

chan A resized.jpg

The feedback resistor is 47K, confirmed with measurement. Pretty sure the driver is a 12AX7, not a 12AU7 so ignore that for now. I was drawing this based on the tubes in the amp.

Anyway, if my math is right here, I think that 47K becomes an 83.7K resistor theoretically to maintain the same feedback level?

sqrt 8 = 2.82
sqrt 2.5 = 1.58
2.82 / 1.58 = 1.78
1.78 * 47000 = 83660

I think thats right. Anyone want to give me a sanity check?

Its very likely this will end up with some values changed but I just want a starting point. The transformers I'm using are from a Hammond AO-35.
 
I thought so, just wanted to make sure I hadn't gone off the rails. 83.7K seems like an awful high value to me, but 47K seems high. I'll probably throw and 82K in there since its the closest normal value and see what I get. The wax caps have to go before I even bother, I know not to waste my time with them.
 
The full calculation is: (sqrt(8/2.5)*(47K + 2.2K)) - 2.2K = 85.87K, so your estimate is off by a mere 2.6%, which is probably well within the tolerance of the parts used anyway. But it's good to note that so that you know you are within tolerance. And probably getting that calculation within 10% would have been good enough anyway. It's not like these are moon shot orbital calculations or anything.

That full calculation is based on the fact that if the feedback level is to be the same, then the voltage produced at the junction of the FB resistor and the cathode resistor must be the same. The rest (a few steps of algebra) falls out from there.
 
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I think you've pretty much described how those calcs are done if it is not stated what the feedback level is. In other words you can estimate with fairly good accuracy the FB level applied by measuring (or calculating) the open loop gain of the portion of the circuit encompassed within the FB loop (including the output stage and step down through the output transformer). Then knowing the voltage ratio applied through the FB voltage divider, you can determine FB level applied. You can manipulate that procedure to hone in on resistor sizes needed before you ordered the parts.

In an actual build I rarely do this though, because its so easy to just hook it up on the bench and measure it directly, but it is good to be aware of the theory behind what's going on so you have some intuition about how the circuit will change with change in FB components or change in FB tap point used.

There are several write ups on the details of these calculations. One I believe is on the valve wizard's web site. The other that I am more familiar with is in Morgan Jones' book Valve Amplifiers. It's described in the chapter on the Bevois Valley amp.
 
There are several write ups on the details of these calculations. One I believe is on the valve wizard's web site. The other that I am more familiar with is in Morgan Jones' book Valve Amplifiers. It's described in the chapter on the Bevois Valley amp.

Aiken Amps has a good writeup about it: http://www.aikenamps.com/index.php/designing-for-global-negative-feedback

...and the Amp Books web site has an analysis of the Fender 5E1 feedback circuit that provides a simple, concrete example: https://www.ampbooks.com/mobile/classic-circuits/fender-champ-5e1-negative-feedback/
 
If I still had the original transformers around I'd have simply measured it. Frankly I hate calculating things that a simple measurement will tell me. I'm less likely to screw up when reading the meter. Fortunately or unfortunately, someone here on AK had the same amp with a bad transformer and I sent them off without even considering making some measurements first.

This is something of a parts bin thing. The amp came to me with a bad power transformer. I replaced the power trafo with one Knockbill graciously gave me and then I figured out the output transformers were near to useless. I robbed the tubes for other purposes, then the transformers were sent off, leaving me with a compact little chassis that sat around for a year or so. I'm tired of looking at it, so I found a set of Hammond tone cab trafos that physically and electrically fit, so now I'm putting it back together so it works. One thing that bothers me, the output transformers are physically different in size despite the same mfg and EIA numbers, and being built within a few weeks of each other. I hope its just the outer can that differs in size. The bobbin itself is not visible.
 
In an actual build I rarely do this though, because its so easy to just hook it up on the bench and measure it directly, but it is good to be aware of the theory behind what's going on so you have some intuition about how the circuit will change with change in FB components or change in FB tap point used.
My crackpot inventions are normally constructed on breadboards first, because I'm usually expecting to joust with numerous demons. I discovered that it's helpful to include a switch in the feedback loop for instant closed-loop vs open-loop comparisons. I've been using miniature side-lever slide switches, attached with hot glue so the lever points straight up. Very handy.
 
6DZ7--as an example, let's use Gadget's schematic he posted in post #1, and see if we can determine how much feedback is applied given the 47K feedback resistor and 2.2K grid resistor as the return point for that feedback. We'll have to make some assumptions about stage gain (which are probably wrong). Ideally you'd either calculate this accurately using load lines, or you'd just measure it with a scope and signal generator. But then again if you're measuring, why not just also measure the feedback level and be done with it?

But anyway, here it goes...

Gadget's output stage uses 7189's. Typically that kind of stage can produce about 15 watts output. You'd need about 12V peak drive voltage, and I'll call this Vin, delivered at the grids of the 7189s to deliver 15 watts output. 12V peak drive would be Vin of 8.47V RMS drive voltage delivered to the grids of the output stage to deliver 15 watts output.

15 watts output across a 2.5 ohm loaded secondary would mean the RMS voltage seen at the speaker terminals would be Vout^2/2.5 = 15, or Vout = 6.12V.

Thus Vin of 8.47V delivers Vout of 6.12V at the speaker terminals (under 2.5 ohm load). Thus the gain of the output stage as a whole (including output transformer is Vout/Vin = 6.12/8.47 = 0.72.

The frontend of this amp utilizes a 12AX7 coupled to a split load inverter. We know the inverter in this configuration offers unity gain, so it can be ignored. The open loop gain of the 12AX7 stage is (guessing, given a 270K plate resistor and unbypassed 2.2K cathode resistor) about 50.

So the total open loop gain, I'll call it Aopen, of the amp is 50 * 0.72 = 36.

The feedback factor (I'll call it B for Beta) in the original schematic is 47K into 2.2K, or a voltage ratio of 2.2K/(47K+2.2K) = 0.045.

The closed loop gain, which I will call Aclosed, is Aopen/(1 + B*Aopen) = 36/(1+0.045*36) = 13.74

So the gain reduction factor is Aopen/Aclosed = 36/13.74 = 2.62

Converting to deciBells, 20*Log(2.62) = 8.34 dB.

So there is about 8 dB feedback applied--that is, assuming my open loop gain assumptions were anywhere accurate. You can work that process backwards given a level of feedback you want to apply, to determine what size of feedback resistor you would need.

===

While we're at it, we might as well calculate the sensitivity of this amp (how much input voltage delivered at the grids of the 12AX7 will produce full power output). I like to short cut this calculation because it turns out to be very accurate to look at the drive voltage needed at the output stage and work back from there.

We already know from above that the drive voltage needed at the output stage to deliver full power output is 8.47V (RMS). We also know the gain of the inverter is unity (can be ignored), and the gain of the 12AX7 is 50 under open loop conditions. We also know the gain reduction factor is 2.62, which is applied at the 12AX7 stage, so the closed loop gain of the 12AX7 stage is 50/2.62 = 19.1.

Thus the sensitivity of the amp is 8.47/19.1 = 0.44V RMS.

This all assumes my original estimations of stage gains were even close...but regardless, it shows the procedure.
 
This is fascinating stuff, and an area I've never really studied much in. Hopefully after dinner I can actually get it to pass signal and we can see how close the measurements are to the calculations. I'll be using an 82K resistor, so not precisely what is calculated but it still should be close enough to get an idea of whats going on.
 
Gadget, if you wanted to, you could split up the cathode resistor on the 12AX7 stage, and use a 2K upper resistor with 200 ohm lower resistor. Then, your feedback resistor would need to be 4.27K if pulling feedback from the 2.5 ohm tap (keeping FB levels the same). If you then pulled feedback from the 8 ohm tap instead, you'd take that 4.27K resistor value and multiply it up as per previous discussion. Might be something to try anyway if you don't mind messing with the cathode circuit in that 12AX7 stage. There will be a little error in 12AX7 stage bias when doing this, as compared to the original amp, since total cathode resistance as seen from the cathode looking to ground would be slightly higher because of the paralleling of the FB resistor with the lower cathode resistor, but the error would be small, about 5% I think.

===

6DZ7, here's the basic procedure if doing it in reverse.

1. Things you must know before you start:
a) Open loop gain of each stage that will be inside the feedback loop. Treat the output stage as one block consisting of the output tube(s) and the output transformer.
b) The max power the output stage will deliver. You would need to either calculate this using power tube load lines knowing the primary impedance and biasing conditions of the output stage (don't forget to factor in the power loss due to heat). I think you could also determine this knowing the turns ratio. Best would be just to measure it, or have that value already known from the published schematic.
c) The impedances of the secondary taps of the output transformer, and which tap you think you will use for the FB signal (usually either the 4 or 16 ohm tap).
d) The amount of feedback you want to apply. Typical amounts in tube amps are from 12 to 20 dB.

So, let's assume the following things:
  • The amp is a 6FQ7 gain stage, coupled to a 6FQ7 long tail pair inverter, driving EL84 outputs in PP AB1.
  • The cathode circuit where FB connects to the first stage consists of a 620 ohm resistor in series with a 47 ohm resistor. The FB will connect at the junction of those two resistors, so that the 47 ohm resistor will be the "lower" resistor in the FB voltage divider.
  • OL gain of first stage is 16.5.
  • OL gain of inverter (long tail pair this time) is 8.25 from each output to ground.
  • Output power is 16 watts per channel, single channel driven.
  • Output stage needs grid to cathode biasing of -13.8V.
  • You want 17 dB of global feedback applied around the entire amp.

1. First, I'd calculate input sensitivity with 17 dB feedback applied, to see if that's in the ballpark you can live with.

We know the output stage needs 13.8V peak drive to drive the output stage to full power (we know this because the stage is biased at -13.8V grid to cathode and a 13.8V peak drive signal will bring on the onset of grid current). 13.8V peak drive is 9.76V RMS.

17 dB global feedback will reduce the gain of the first stage by 17 dB, or in other words by a factor of: 17 dB = 20*Log(gain factor reduction). And solve for "gain factor reduction". It is 7.08.

Closed loop gain of first stage is then 16.5/7.08 = 2.33

Input sensitivity is then 9.76V/(2.33 * 8.25) = 0.51V (RMS). That will work well for driving the amp directly from a line source (so this amp is probably marketed as an "integrated" type, and has a volume control on it).

Now that we know that 17 dB feedback is appropriate (that is, assuming the amp is stable with that amount of feedback), we can move on to estimate the FB resistor size that will deliver that much feedback.

2. Calculate gain of the output stage, including output transformer (or preferably just measure it): This is where some error could occur because of not accounting for transformer loss. Best is to just measure this part if you can.

9.76V input to the grids of the output tubes delivers 16 watts output at the speaker terminal. We are going to tap feedback from the 4 ohm tap.

P = V^2/R

R = 4 ohms
P = 16 Watts

16 watts = V^2/4, solving for V, it equals 8V. Thus 9.76V input delivers 8V output to 4 ohm tap.

So gain of output stage is Vout/Vin = 8V/9.76V = 0.812

Now that we know the gain of the output stage for the FB tap selected, we can calculate the open and closed loop gains Aopen and Aclosed for the entire amp.

3. Gain = first stage gain * second stage gain * output stage gain

Aopen = 16.5 * 8.25 * 0.812 = 110.53
Aclosed = 2.33 * 8.25 * 0.812 = 15.61

Double checking that this gain reduction should equal 17 dB: 20*Log(110.53/15.61) = 17.0 dB. Yes.

Now we can calculate the FB factor Beta (B)

4. Aclosed = Aopen/(1 + B*Aopen)

15.61 = 110.53/(1 + B*110.53)

B = 0.055

5. Now knowing the FB factor B (Beta), we can calculate the FB resistor size.

R2/(R1+R2) = B

R2 is the lower cathode resistor: 47 ohms
R1 is the FB resistor we are solving for
B = 0.055

47/(R1 + 47) = 0.055

Solve for R1 (FB resistor). R1 = 808 ohms.

(now, I can't guarantee this whole thing is error free...someone feel free to review it and correct if you see anything wrong)

===

You may wonder why I picked this example with these apparently odd values. It's because this is an actual build of mine from last year. But in the actual build, the FB resistor size turned out to be 1K which delivered about 18 dB feedback. So there is some error in this procedure. Still, it gets you into the ball park. The errors are probably in two places: a) the estimate of output power at the speaker terminals is off, since output transformers are lossy (lose power to heat). b) This approach doesn't model the currents through the cathode circuit or back flowing through the FB loop at all. I think modeling those would produce a more accurate result. That's what Morgan Jones did. I will refer you to his book though, for the details there. (it's harder than this approach).

Bottom line, this is why I prefer to just do this on the bench. But as mentioned, it's good to go through this once or twice so you understand the theory behind it.
 
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well, it makes signal after fixing a transformer phasing problem. Very quick and dirty results, but here we go

full power out, 1kc, 8 ohm load, single channel driven
9 watts

input sensitivity
0.43 volts

feedback
12.7 db with an 83K feedback resistor

Pretty damn close all in all.

Plate volts:
287

screen volts
279

cathode volts
8.8v

phase inverter / voltage amp supply
275

12ax7 voltage amp plate
143v

12ax7 inverter plate
232

12ax7 inverter cathode
31


I'll see what the square waves look like tomorrow and see what else it may want. I think I've decided to eliminate the 130 ohm resistors between the output of the voltage doubler and the first filter cap. Numbers above are without them, with them everything drops about 10 volts at idle and some amount more under load. I see no point in this. Its not the original trafo so the voltages are probably a little bit different but not unreasonable for a 6bq5 cathode bias amp.
 
Input sensitivity
.43 volts

feedback
12.7 db with an 83K feedback resistor

Well figure that! Might just be dumb luck with my estimates. I was off by 3 dB on the feedback level. :(
===

6DZ7, Good observation! That's a slick way to determine open loop gain if you know feedback factor (B) and closed loop gain.

The full formula in abstract form would be:

Aopen = Aclosed/(1 - B*Aclosed)

Applying the formula to our example problem,

Aopen = 13.74/(1 - 0.045*13.74)
Aopen = 36
 
well, its nothing impressive but this is the best I could do and keep it stable. I can get lovely square waves but it goes completely ballistic with no or a cap-only load. Removing all tuning (phase cap and the HF cap off the voltage amp) keeps it stable with a cap but it goes bonkers with no load. Changing the 100pf cap off the voltage amp to 180pf helped both square wave presentation and made it absolutely stable. It holds its own no matter what I do to the output. No worries with LF, it settles immediately.

Upper trace is the resistor-only channel, lower is a 0.22uf cap only.


I think the next step is to see what EFB will do for me. I may also experiment with more feedback, just to see if it goes horribly unstable. Handily since this has the cathode resistors split from the factory, I can do a very easy side by side of stock vs modified performance. I have it hooked up right now, and honestly it sounds pretty decent.
 

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Same experience here. Awesome looking square waves = unstable amp. (With garden variety output traffos anyway).

Just curious what make/model of output transformers are currently installed in the amp?

It would be interesting to know if the stock amp was stable or where its quirks were just as a guage of how much you have improved it.

That's one thing I am going to start doing better--take a full battery of tests to "base line" the performance before I start tweaking.
 
They're from a Hammond AO-35 tone cabinet. Basically I needed something that physically fit and was electrically and wallet-compatible. These happened to do all of that. They're relatively small, 3 1/4" mounting centers in the same style as most console amps. Its a small chassis so anything bigger than the stock ones simply were not going to fit.

Basically my goal here was to make it useful so it can get out of my way. I may knock together an EFB regulator for one channel tonight to see what that does for me. I've got some speaker jacks in the mail. End goal is to just make it useful, its not going to win any performance or beauty contests but I want it working, reliable, and respectable sounding.
 
well, EFB makes no appreciable difference. Very slight improvement to distortion figures and about 1/4 watt more. Honestly, not enough to justify adding the pot, adjustments, or test points. The power supply in this thing really doesn't have much sag, I guess there simply isn't very much for it to do. I guess now I move on to figuring out what I'm going to do about rebuilding the power supply and waiting for the speaker terminals to show up. Might give trainbuffony's Scott can restuff instructional a try. If it goes horribly, frankly I don't care all that much.
 
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