I'm reading Morgan Jones "Valve Amplifiers" chapter on cathode decoupling. I understand the mathematics but not the negative feedback concept itself.
We're talking about a cathode biased triode here and the benefits of using a bypass capacitor (around the cathode resistor) to decouple the A/C signal to reduce negative feedback and avoid the associated reduction in gain. He states that the feedback fraction (beta) = Rl/Rk. To me this implies a voltage divider at work involving the load resistor and the cathode resistor. But I can't make sense of that.
What I don't follow is the A/C signal path at work here. I would assume that the capacitor grounds out any A/C signal on the cathode; I get that. But just how does the negative feedback occur in the first place (without the cap)? Is it internal to the triode? I can't see it passing back to the signal at the grid through the 0V (negative) line. So how does it work?
We're talking about a cathode biased triode here and the benefits of using a bypass capacitor (around the cathode resistor) to decouple the A/C signal to reduce negative feedback and avoid the associated reduction in gain. He states that the feedback fraction (beta) = Rl/Rk. To me this implies a voltage divider at work involving the load resistor and the cathode resistor. But I can't make sense of that.
What I don't follow is the A/C signal path at work here. I would assume that the capacitor grounds out any A/C signal on the cathode; I get that. But just how does the negative feedback occur in the first place (without the cap)? Is it internal to the triode? I can't see it passing back to the signal at the grid through the 0V (negative) line. So how does it work?
. But just how does the negative feedback occur in the first place (without the cap)? Is it internal to the triode? I can't see it passing back to the signal at the grid through the 0V (negative) line. So how does it work?
Assume a standard triode gain stage with not caps, just a plate and cathode resistors. Place a sine wive in the grid. Now measure the cathode voltage. You should expect an AC signal on the cathode This is do to AC current through the tube and the voltage drop on the cathode resistor. What you have here is the cathode bias changing based on the signal at the grid --- feedback.
Sy, I think I get your point (I think you mean to say proportional to the ratio). Beta (the feedback fraction) is the ratio of R(k) to R(plate resistor). The higher the ratio the greater the feedback.
The question that's on my mind now is why the use of decoupling caps is so widespread in guitar amps and much less so in Hi-Fi.
For triodes not drawing grid current, i.e., 99% of the tube voltage amplifiers in the world.
What about them? It becomes clear once you realize that vacuum tube voltages must be referenced to the cathode, not ground. If the screen of a pentode draws a constant current with no AC component, that current is simply imposed on the cathode current. It will have no effect one way or the other as the cathode voltage varies due to varying current at the plate. If there is an AC component at the screen, then you have an additional source of degenerative feedback. This is often termed "ultralinear" operation, and it will cost you gain, as your pentode isn't working like a pentode anymore. May be that's a good thing and may be it isn't. That depends on what your design is trying to accomplish.
HelloI'm reading Morgan Jones "Valve Amplifiers" chapter on cathode decoupling. I understand the mathematics but not the negative feedback concept itself.
We're talking about a cathode biased triode here and the benefits of using a bypass capacitor (around the cathode resistor) to decouple the A/C signal to reduce negative feedback and avoid the associated reduction in gain. He states that the feedback fraction (beta) = Rl/Rk. To me this implies a voltage divider at work involving the load resistor and the cathode resistor. But I can't make sense of that.
What I don't follow is the A/C signal path at work here. I would assume that the capacitor grounds out any A/C signal on the cathode; I get that. But just how does the negative feedback occur in the first place (without the cap)? Is it internal to the triode? I can't see it passing back to the signal at the grid through the 0V (negative) line. So how does it work?
Understanding this one is like patting you’re your head with one hand and rubbing your stomach with the other. There are potentially two things going on at the same time.
With a cathode bypass capacitor in place the voltage of the cathode tends to stay constant as the grid voltage modulates. The operation of the tube with varying signal voltage tends to dance around a constant cathode to ground voltage.
Remove the cathode bypass capacitor and you have a different creature. With no cathode bypass capacitor in place the cathode to ground voltage behaves differently. As the grid voltage becomes more negative the current through the valve decreases the voltage drop across the cathode resistor decreases and the cathode to ground voltage decreases. As the grid voltage becomes more positive the current through the valve increases the voltage drop across the cathode resistor increases and the cathode to ground voltage increases.
The varying voltage on the cathode is the feedback mechanism for the non-bypassed cathode resistor grounded cathode amplifier.
DT
All just for fun!
Mile,Sy:
I was thinking of the case of the power pentode, assuming a perfect waveform driving the gate/cathode - would I still get a good transfer function via the anode voltage or would the screen grid current distort this grid/cathode current (as Ik = Ig + Ig2 + Ia) ?
And what are the implications of running a 10R balancing resistor on the cathode of a power pentode when using a fixed Vg2 and just getting the output from the anode?
Maybe this is irrelevant - but I just worried where any g2 current was 'going'!
I was thinking of the case of the power pentode, assuming a perfect waveform driving the gate/cathode - would I still get a good transfer function via the anode voltage or would the screen grid current distort this grid/cathode current (as Ik = Ig + Ig2 + Ia) ?
And what are the implications of running a 10R balancing resistor on the cathode of a power pentode when using a fixed Vg2 and just getting the output from the anode?
Maybe this is irrelevant - but I just worried where any g2 current was 'going'!
No, in the case of the pentode, you indeed have to think about Ip plus Ig2. You also have to think about the reference for g2- to ground or to the cathode? Unless you're in AB2, Ig1 will be zero.
As a practical matter, the 10 ohm cathode resistor will have very little effect. Let's take an EL34/6L6/6550 sort of tube as an example. A typical signal current swing will be 100mA at full throttle. That's a 1V swing across that resistor. The grid will be driven by roughly 40V, so the cathode signal voltage will only be 6% of the grid signal voltage. That represents (roughly) 0.2dB of feedback.
As a practical matter, the 10 ohm cathode resistor will have very little effect. Let's take an EL34/6L6/6550 sort of tube as an example. A typical signal current swing will be 100mA at full throttle. That's a 1V swing across that resistor. The grid will be driven by roughly 40V, so the cathode signal voltage will only be 6% of the grid signal voltage. That represents (roughly) 0.2dB of feedback.
For much of the signal cycle Ig2 is roughly proportional to Ia, but this changes when the anode dips below Vg2 - Ig2 then increases more quickly than Ia. So in a Class A small signal stage you can use cathode degeneration on a pentode without any problems*, but it gets more complicated for an output stage.
A 10R resistor is likely to be sufficiently small (much smaller than 1/gm) that it won't do any harm. Actually, too small to do any good?
* Actually is one problem: cathode degeneration on a triode can reduce noise as well as signal, but on a pentode it drops the signal much more than the noise - this is because there is no partition noise in the cathode current so the feedback doesn't see it.
A 10R resistor is likely to be sufficiently small (much smaller than 1/gm) that it won't do any harm. Actually, too small to do any good?
* Actually is one problem: cathode degeneration on a triode can reduce noise as well as signal, but on a pentode it drops the signal much more than the noise - this is because there is no partition noise in the cathode current so the feedback doesn't see it.
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ETF presentation: Just Excelent article for basic understanding inside of Tube Amp`s circuit.
Thanks
Written up extensively in Jan Didden's new publication. Interesting and useful, but I think it works differently than Frank thinks it does.
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