Basic question re: gNFB around 3 stages

rongon · 21st May 2012, 02:13 AM

I have a question about applying global feedback to a 3-stage amplifier. I hope someone can help me out of my confused state -- at least for the moment!

OK...

I take negative feedback from the secondary of the output transformer to the cathode of the first stage.

Let's say I've applied 6dB of negative feedback, so the input sensitivity of the amplifier has been reduced by half.

Open loop, the amplifier reaches full power at 500mV (peak). With the 6dB of gNFB applied as above, 1V pk will be required to drive the complete amp to full power.

The output tube has a grid bias of -45V.

The driver tube (2nd stage) has a gain (measured) of 7x, and has a grid bias of -6.5V. It can drive the approx. 45V pk needed into the output stage.

The input tube has a gain of 13x, and a grid bias of -2V. So, open loop, with 0.5V pk at its input, it supplies the 6.5V pk output to drive the 2nd (driver) stage to output the 45V pk needed to drive the final stage to its full output.

OK, here's my question.

When I apply that 6dB of NFB, is it only the first stage sensitivity that is reduced, so therefore only the first stage needs to be driven with twice as much input voltage?

What is the impact on the 2nd (driver) stage? Is it still only supplying its 7x gain, and only needs to see the 6.5V pk at its input to drive the output stage with 45V pk signal?

??

I tried to make that as clear as I could. Thanks to anyone who can help me understand this.
--

Chris Hornbeck · 21st May 2012, 03:51 AM

The short answer is yes. The long answer is that the conventional way feedback is applied is to put it in series with signal. All stages within the loop act the same with or without feedback, which is imagined as being external to the loop, with a magic "adder" stuck up front, adding the signal and the feedback. Good question and well presented.

All good fortune,
Chris

DF96 · 21st May 2012, 10:24 AM

You could argue that the first stage sensitivity is not reduced at all, but it now sees a partially corrected input signal. The total system sensitivity is reduced.

Simon B · 21st May 2012, 12:14 PM

The forward gain of the amp, 'A', remains unchanged - the sensitivity of each stage stays the same, though 'G', the overall system gain, does change.

The feedback network has a transfer function usually denoted by 'B'. Note that part of the feedback network is the impedance of the input to the summing node. In your case the summing node is the first valve. This is the "magic adder" or possibly "magic subtractor", as feeding a signal to the cathode opposes the action of a signal fed to the grid. Whilst the grid has a conveniently high impedance, the cathode doesn't.

From the very outset of the development of this stuff, there were two opposite conventions on whether the summing node should be viewed as adding or subtracting, which meant that two different equations for the final system gain 'G' came into being:

G = A/(1+AB) and G = A/1-AB)

It's just a matter of convention, but if you do delve further into this stuff, it's as well to be forewarned about it. I learnt it simultaneously from two different lecturers, one from each faction.......

......

rongon · 21st May 2012, 03:29 PM

Ah, many thanks. I think some of this is beginning to sink in.

So I'm correct visualizing this as in the picture below?
-

DF96 · 21st May 2012, 03:39 PM

Yes that is how feedback to the first stage cathode works.

One complication, which fortunately you can generally ignore, is the common-mode response of the first stage. To a first approximation this is 1/mu down from the wanted differential-mode response, so starting with a low mu valve will cause a small adjustment in your calculated closed-loop gain. Provided signal levels are not too high and the first valve is properly biassed then you should not get too much common-mode distortion. The snag is that any CM distortion you do get is not removed by the feedback loop.

rongon · 21st May 2012, 03:48 PM

Would the common mode distortion be effectively cancelled if all 3 stages were push-pull (LTP -> PP driver -> PP output)?
--

DF96 · 21st May 2012, 03:52 PM

Possibly, but you might just move the problem from the CMRR of the first stage to the CMRR of later stages where the signal levels are higher.

21st May 2012, 02:13 AM	#1
rongon diyAudio Member Join Date: Jul 2009 Location: Across the river from Rip's big old tree...	Basic question re: gNFB around 3 stages I have a question about applying global feedback to a 3-stage amplifier. I hope someone can help me out of my confused state -- at least for the moment! OK... I take negative feedback from the secondary of the output transformer to the cathode of the first stage. Let's say I've applied 6dB of negative feedback, so the input sensitivity of the amplifier has been reduced by half. Open loop, the amplifier reaches full power at 500mV (peak). With the 6dB of gNFB applied as above, 1V pk will be required to drive the complete amp to full power. The output tube has a grid bias of -45V. The driver tube (2nd stage) has a gain (measured) of 7x, and has a grid bias of -6.5V. It can drive the approx. 45V pk needed into the output stage. The input tube has a gain of 13x, and a grid bias of -2V. So, open loop, with 0.5V pk at its input, it supplies the 6.5V pk output to drive the 2nd (driver) stage to output the 45V pk needed to drive the final stage to its full output. OK, here's my question. When I apply that 6dB of NFB, is it only the first stage sensitivity that is reduced, so therefore only the first stage needs to be driven with twice as much input voltage? What is the impact on the 2nd (driver) stage? Is it still only supplying its 7x gain, and only needs to see the 6.5V pk at its input to drive the output stage with 45V pk signal? ?? I tried to make that as clear as I could. Thanks to anyone who can help me understand this. --

21st May 2012, 12:14 PM	#4
Simon B diyAudio Member Join Date: Dec 2011 Location: North-East England	The forward gain of the amp, 'A', remains unchanged - the sensitivity of each stage stays the same, though 'G', the overall system gain, does change. The feedback network has a transfer function usually denoted by 'B'. Note that part of the feedback network is the impedance of the input to the summing node. In your case the summing node is the first valve. This is the "magic adder" or possibly "magic subtractor", as feeding a signal to the cathode opposes the action of a signal fed to the grid. Whilst the grid has a conveniently high impedance, the cathode doesn't. From the very outset of the development of this stuff, there were two opposite conventions on whether the summing node should be viewed as adding or subtracting, which meant that two different equations for the final system gain 'G' came into being: G = A/(1+AB) and G = A/1-AB) It's just a matter of convention, but if you do delve further into this stuff, it's as well to be forewarned about it. I learnt it simultaneously from two different lecturers, one from each faction............. __________________ Engineer: One who can do for 10 shillings what any fool can do for 10 pounds.

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21st May 2012, 03:51 AM	#2
Chris Hornbeck diyAudio Member Join Date: May 2011 Location: Little Rock	The short answer is yes. The long answer is that the conventional way feedback is applied is to put it in series with signal. All stages within the loop act the same with or without feedback, which is imagined as being external to the loop, with a magic "adder" stuck up front, adding the signal and the feedback. Good question and well presented. All good fortune, Chris

21st May 2012, 10:24 AM	#3
DF96 diyAudio Member Join Date: May 2007	You could argue that the first stage sensitivity is not reduced at all, but it now sees a partially corrected input signal. The total system sensitivity is reduced.

21st May 2012, 03:39 PM	#6
DF96 diyAudio Member Join Date: May 2007	Yes that is how feedback to the first stage cathode works. One complication, which fortunately you can generally ignore, is the common-mode response of the first stage. To a first approximation this is 1/mu down from the wanted differential-mode response, so starting with a low mu valve will cause a small adjustment in your calculated closed-loop gain. Provided signal levels are not too high and the first valve is properly biassed then you should not get too much common-mode distortion. The snag is that any CM distortion you do get is not removed by the feedback loop.

21st May 2012, 03:48 PM	#7
rongon diyAudio Member Join Date: Jul 2009 Location: Across the river from Rip's big old tree...	Would the common mode distortion be effectively cancelled if all 3 stages were push-pull (LTP -> PP driver -> PP output)? --

21st May 2012, 03:52 PM	#8
DF96 diyAudio Member Join Date: May 2007	Possibly, but you might just move the problem from the CMRR of the first stage to the CMRR of later stages where the signal levels are higher.