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.
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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)
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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.