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Globulator 27th February 2012 09:58 PM

Primary current feedback - any good? Help!
This may be more suited to an SE design but I don't see why it should not scale to a PP half and it's associated driver.

Assume that each output pentode in the PP amp has it's own SRPP driver, and that the phase splitter is earlier in the signal chain and does not affect the driver/tube pairs.

So one way of getting feedback from the primary to the driver is to stick a biggish cap onto the anode and feed it back with a resistor into the (un-bypassed) SRPP cathode resistor at the bottom. Like a 22uF cap into a 27k resistor tied to the top of a 1k cathode resistor. For example.

This however needs a 22uF ish capacitor from the anode: i.e. get another capacitor in the signal path, and not even a film one.

So I was thinking that if I had a (for instance) 10ohm cathode resistor at the bottom of the output tube (usually used for fusing/bias measuring), I could just tap straight off that to give me a feedback signal. In this case it would however be a voltage signal of the current in the primary - rather than the voltage across it.

So the question is: Is this useful feedback into the driver stage, or will it try to change the OPT/amp into a current drive machine which would sound rubbish?

DF96 27th February 2012 10:15 PM

If you sample the output current, or a good proxy for it, then you will raise output impedance. In most cases this is not what you want.

Globulator 27th February 2012 10:20 PM


Originally Posted by DF96 (
If you sample the output current, or a good proxy for it, then you will raise output impedance. In most cases this is not what you want.

Yes, I want to lower output impedance so you are correct: this is not what I want at all.

If the sampling occurs through a 10ohm cathode resistor (at 50mA idle that would create a 0.5V swing for output tube cut-off) I would a reasonable signal, at 100ohms I think it may be noticeable.

At 500V and 50mA I have a tube with a resistance of 500/50e-3 = 10k, would a 10 or 100ohm really make a difference?

DF96 27th February 2012 10:34 PM

It is not the direct effect of the cathode resistance (although it does have a local degeneration effect) but the feedback to an earlier stage which makes the difference. The local effect scales with 1/gm, not Ra. The total effect depends on loop gain, like all feedback.

Globulator 27th February 2012 10:51 PM

1 Attachment(s)
So it's because it's current feedback, and R = V/I, so using negative current feedback tends to raise R, where negative voltage feedback lowers R.

That makes sense.

Found this link:

which has the diagram below which has negative voltage feedback and positive current feedback, to lower the output impedance to zero or make it negative.

I also note the 'snubbing' of the transformer to make it behave like a resistor.

smoking-amp 27th February 2012 10:55 PM

If this is planned for a P-P output stage, there may be some use for current sensing to reduce crossover distortion. See TubeCad Dec 20 and 30th, 2006:
2nd to last schematic:
European Triode Festival and Crossover Notch Distortion and New OTL Design

Broskie presents some somewhat obtuse methods to transcribe this to tube circuitry. But all thats needed is a differential driver stage to take the difference in the cathode current feedbacks. The differential feedback has higher gain when in class A overlapped mode, and lower gain in class B. So can be taken advantage of to smooth the output stage gain during a class AB crossover. Not much help for lowering the output Z in general though.

Yeah, positive current feedback should lower the output Z, as long as you don't use too much and turn it into an oscillator (negative output Z). If you use two of them (P-P and differential) in positive feedback mode though, it will do the opposite for class AB crossover distortion.

Usually when I see current feedback, they are trying to get critical damping of the speaker, which is at a higher Zo than the usual neg. V fdbk gives.

Dimitris AR 27th February 2012 11:42 PM

Positive current feedback will increase the gain of the earlier stage and make the amount of the global negative feedback higher( which is placed on that stage in post's #5 scheme ) and then lower the output impedance of the amp , so the problem is if that stage can stand this higher gain or not . IMO we have to design an gain stage biased like an high mu stage but with lower amplification then this stage can stand the effect of the positive current feedback ( I hope you understand what I mean because English is not my best ) .

jcx 28th February 2012 12:37 AM

just exactly the right negative Z to cancel xfmr pri resistance gives a large reduction in low frequency distortion from the xfmr

riccoryder 28th February 2012 05:06 AM

The effect is to cancel some of the internal feedback of the input triode, making it act more like a pentode. If one envisions the triode plate voltage influence on cathode current as plate to cathode transconductance, the a.c. plate voltage times gm(pk) is the feedback current reducing change at the cathode. We're just cancelling some of that current to raise gain.

The same thing is done between the phase splitter cathode and input cathode in the Dynaco ST-35.

artosalo 28th February 2012 07:13 AM


...will it try to change the OPT/amp into a current drive machine which would sound rubbish?
This is fact.. ?
Or would it be myth instead ?

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