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Old 28th September 2010, 08:02 PM   #1
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Default Understanding Cathode Decoupling

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?
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Old 28th September 2010, 08:11 PM   #2
Yvesm is offline Yvesm  France
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Any tube reacts to the voltage applied between its grid and its cathode. Not ground !
And a DC + AC voltage appear at the cathode when connected to ground thru an unbypassed resistor and it subtracts from the grid to ground voltage.
That is a feedback.

Yves.
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Old 28th September 2010, 08:20 PM   #3
SY is offline SY  United States
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One more key hint- cathode current equals plate current. That means that the voltage swing at the cathode (wrt ground) is the ratio of the cathode resistor to the plate resistor.
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Old 28th September 2010, 08:47 PM   #4
ChrisA is offline ChrisA  United States
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Originally Posted by Captn Dave View Post
. 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.
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Old 28th September 2010, 08:53 PM   #5
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OK, I see that now. The signal voltage across the cathode resistor is in phase with the input signal and the triode responds to changes in the voltage between the gird and cathode.

And my earlier analysis of a voltage divider is not correct.

Thanks for the help!
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Old 28th September 2010, 09:14 PM   #6
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One more key hint- cathode current equals plate current. That means that the voltage swing at the cathode (wrt ground) is the ratio of the cathode resistor to the plate resistor.
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.
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Old 28th September 2010, 09:19 PM   #7
SY is offline SY  United States
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Higher gain. Linearity is less important in MI amps.
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Old 28th September 2010, 09:25 PM   #8
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One more key hint- cathode current equals plate current.
What about pentodes drawing screen(G2) current?
And in fact any tube where a +ve driven grid attracts current?
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Old 28th September 2010, 09:50 PM   #9
DF96 is offline DF96  England
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It is not quite true that the feedback fraction is Rk/R(anode), unless you include in R(anode) the load of the following stage. Cathode degeneration does not sample the output voltage but the output current.
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Old 28th September 2010, 10:04 PM   #10
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Originally Posted by Globulator View Post
What about pentodes drawing screen(G2) current?
And in fact any tube where a +ve driven grid attracts current?
For triodes not drawing grid current, i.e., 99% of the tube voltage amplifiers in the world.
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