Negative feedback - How much?

[Jana]

The signal coming out of the secondary of the OT is in phase with the signal coming into the PI from the preamp.

The reason it is negative feedback is because the feedback from the OT is fed into the *Other* input of the PI, which the opposite phase input from the other input. I used opposite phase because I think we are getting confused with which is inverting and which is non-inverting.

[tubeswell]

The signal coming out of the secondary of the OT is in phase with the signal coming into the PI from the preamp.

Huh? That would be positive feedback. The NFB signal that is injected into the signal path has to be opposed to the signal path's signal to have a cancelling/smoothing effect. I think the phase inverter inversion is confusing the issue. Consider for a moment a SE amp like a 5F2A. In a 5F2A, the cathode of the driver stage is the same phase as that stage's grid, and the plate of the driver stage is 180 deg inverted (from both the cathode and the grid). This 180 inverted signal then goes into the output stage's grid and is inverted a further 180 degrees from that grid (so the signal at the output stage's plate is in-phase with the signal at the drivers stage's grid/cathode). If the OT secondary signal was in phase with the output stage's plate, it would be providing positive feedback to the drivers stage's cathode. Ipso facto, the secondary winding's signal has to be opposite to the outputs stage's plate signal.

Now consider PP amp with a paraphase. The first inverting stage (we'll call it the "A" stage) is still acting the same way in relation to the output secondary as the driver stage's plate would in a 5F2A. The 2nd stage of the paraphase inverter (we'll call it the 'B' stage) is also inverting the phase of the signal (from the A stage's plate), and feeding this to the B-side's output tube grid, and the plate of B-side output stage is inverting the signal a further 180 degrees, and then it enters the OT primary in an opposite phase to what the other (A) side of the OT primary is. But the OT secondary is in phase with this B side (remember that we are talking PP here now). However, in relation to where the NFB signal is being plugged back into the signal path (the paraphase A stage's cathode), the OT secondary has to still be 180 degrees out of phase. Otherwise it wouldn't be negative feedback.

The same thing applies to other types of PIs, namely; the B-side's output tube grid will always be in phase with the PI A-side's cathode, meaning that the B-side's output plate will be opposite to the PI A-side's cathode, and the B-side's output tube plate has to be the same phase as the OT secondary's NFB tap. But this still has to be in opposite (180 degree) phase to the point at which you inject the NFB back into the signal path.

[diagrammatiks]

The signal coming out of the secondary of the OT is in phase with the signal coming into the PI from the preamp.

Huh? That would be positive feedback. The NFB signal that is injected into the signal path has to be opposed to the signal path's signal to have a cancelling/smoothing effect. In a 5F2A, the cathode of the driver stage is the same phase as that stage's grid, and the plate of the driver stage is 180 deg inverted (from both the cathode and the grid). This 180 inverted signal then goes into the output stage's grid and is inverted a further 180 degrees from that grid (so the signal at the output stage's plate is in-phase with the signal at the drivers stage's grid/cathode). If the OT secondary signal was in phase with the output stage's plate, it would be providing positive feedback to the drivers stage's cathode. Ipso facto, the secondary winding's signal has to be opposite to the outputs stage's plate signal. I rest my case.

Has anyone considered what happens when the output tubes start clipping? Would it be true that for the duration of the clipped signal, the feedback would be less than what it would be if the output stage was not clipping? If true, then the total signal applied to the output tubes would increase when the output tubes start to clip, making them clip even harder. Is this the case?

no the long tailed pair is a differential amplifier. Page 177 of Merlin's 1st book.

If the two signals were out of phase the difference between them would be maximum.

Has anyone considered what happens when the output tubes start clipping? Would it be true that for the duration of the clipped signal, the feedback would be less than what it would be if the output stage was not clipping? If true, then the total signal applied to the output tubes would increase when the output tubes start to clip, making them clip even harder. Is this the case?

negative feedback collapses as the power tubes go into clipping. There will be a certain point where the loop falls completely.

[tubeswell]

no the long tailed pair is a differential amplifier. Page 177 of Merlin's 1st book.

With due respect, this is not relevant to the point I am making. I'm talking about the phase between the NFB signal and the PI's cathode (being out-of-phase). (Not discussing the difference between the PI's plates - that's another matter).

If the two signals were out of phase the difference between them would be maximum..

?? Not sure where you are coming from. The NFB signal is not the same strength as the PI's cathode signal. The NFB's signal is supposed to be a fraction of the strength of PI cathode's swing, to make sure that it won't cancel out or do anything weird. But the two signals still have to be more-or-less 180 degrees out-of-phase with each other. This smooths out some of the bumps that would otherwise be there if the NFB weren't connected. If the NFB were in phase with the PI cathode, you would bet a positive feedback boost that would put the amp into perpetual oscillation.

[diagrammatiks]

that would only apply if you were attempting to insert negative feedback into the same point that the original signal passes through.

[martin manning]

TW, see pp 187-8 in Designing Preamps for an explanation and diagrams of the FB phase relationship in the LTP.

In the case of the 5F2-A, the signal at the grid of the driver is inverted twice (once by the driver and once by the power tube), so the output is in phase with the input signal. The FB signal is applied to the cathode of the driver, and so it is inverted only once (by the power tube), making it negative FB.

[statorvane]

Huh? That would be positive feedback

It certainly can be if you have the OT wires swapped.

[tubeswell]

I am keen to discuss this in case I am proved wrong.

So what happens when the OT secondary voltage swings 'up'? - The voltage across the NFB voltage divider surely increases (because more current is present across the resistors in the divider, because more current is sourced from the OT secondary. V = I x R). The knee of the voltage divider goes up by a proportional amount. Everything referenced to the knee (of the divider) will go up by the same amount as the knee goes up.

In the case of an LTP, the bias and tail resistors are both sitting on top of the (NFB divider) knee. The grid load resistors are also sitting on top of the knee (in fact they are bootstrapped to the tail/bias resistor junction). So what part of the signal isn't being directly affected by the NFB?

Even where the grid load resistors of the driver stage are not sitting on the NFB knee (as in a 5F2A driver or a driver in a cathodyne pair), what happens to the cathode voltage when the NFB voltage divider voltage swings up? Surely the increase in current through the cathode resistor (which is also the leg of the NFB voltage divider) results in an increase in cathode voltage? If this happens when the grid of that stage happens to be swinging down (and the plate-to-cathode voltage is peaking upwards i.e.: NFB in opposite phase to the signal), then surely the plate-to-cathode voltage in the stage will be lower (because of the increased tube current sourced from the NFB loop) in comparison to what it would otherwise be if there was no NFB upswing present?

When the NFB voltage drops, there is (otherwise) a net decrease in current through the cathode resistor. If the decrease in current sourced from the NFB loop happens when the grid voltage happens to be swinging up, then the cathode voltage will still tend to rise (along with increasing tube current), but probably not by as much as it would have done if there had been no NFB present, because the net effect of the NFB is to artificially boost tube current all round. Therefore because plate-to-cathode voltage is not overall as potentially high as it might otherwise be if there was no NFB present, the net overall effect must be a decrease in the gain of the stage. In other words, the opposing NFB must be acting as a 'brake' on change in tube current, inhibiting the 'raw' performance of the tube.

Please point out the error in my logic.

[FunkyE9th]

no the long tailed pair is a differential amplifier. Page 177 of Merlin's 1st book.

With due respect, this is not relevant to the point I am making. I'm talking about the phase between the NFB signal and the PI's cathode (being out-of-phase). (Not discussing the difference between the PI's plates - that's another matter).

If the two signals were out of phase the difference between them would be maximum..

?? Not sure where you are coming from. The NFB signal is not the same strength as the PI's cathode signal. The NFB's signal is supposed to be a fraction of the strength of PI cathode's swing, to make sure that it won't cancel out or do anything weird. But the two signals still have to be more-or-less 180 degrees out-of-phase with each other. This smooths out some of the bumps that would otherwise be there if the NFB weren't connected. If the NFB were in phase with the PI cathode, you would bet a positive feedback boost that would put the amp into perpetual oscillation.

As mentioned above, the LTP is a differential amplifier. It takes the difference (subtraction) between the 2 inputs and amplifies it.

For example:
Cosine(0) = 1:
Cosine(180) = -1 This is 180degrees out phase with respect to cosine(0).

So Cosine(0) - Cosine(180) = 2. A gain increase and not gain reduction and the difference is at maximum. So if you put a signal that is 180 degrees out of phase at the other input of the LTP you get positive feedback not negative feedback.

[tubeswell]

As mentioned above, the LTP is a differential amplifier. It takes the difference (subtraction) between the 2 inputs and amplifies it.

For example:
Cosine(0) = 1:
Cosine(180) = -1 This is 180degrees out phase with respect to cosine(0).

Yes I already knew that. But my argument is about negative feedback needing to be 180 Degrees out of phase with the input voltage. (and not about the difference in phase between the A and B sides of a PI).

Why does Merlin's book say on P201 that " If the voltage is fed back is in phase with the input voltage it will serve to increase it... (and in the next para) If the voltage fed back is 180 degrees out of phase with the input voltage it will serve to decrease it..."? How can these statements be reconciled with feeding a 0-degree (in phase) signal from the OT secondary back to a pre-amp cathode (which otherwise moves in phase with the input signal at its grid) and calling it an NFB loop? These are my questions.

So Cosine(0) - Cosine(180) = 2. A gain increase and not gain reduction and the difference is at maximum. So if you put a signal that is 180 degrees out of phase at the other input of the LTP you get positive feedback not negative feedback.

Is not the NFB signal, in being presented to both grids at the same time (in that case) being cancelled out? (leaving the NFB at the cathode to carry out its function of mitigating the input signal to the LTP's A stage?)

[FunkyE9th]

As mentioned above, the LTP is a differential amplifier. It takes the difference (subtraction) between the 2 inputs and amplifies it.

For example:
Cosine(0) = 1:
Cosine(180) = -1 This is 180degrees out phase with respect to cosine(0).

Yes I already knew that. But my argument is about negative feedback needing to be 180 Degrees out of phase with the input voltage. (and not about the difference in phase between the A and B sides of a PI).

Why does Merlin's book say on P201 that " If the voltage is fed back is in phase with the input voltage it will serve to increase it... (and in the next para) If the voltage fed back is 180 degrees out of phase with the input voltage it will serve to decrease it..."? How can these statements be reconciled with feeding a 0-degree (in phase) signal from the OT secondary back to a pre-amp cathode (which otherwise moves in phase with the input signal at its grid) and calling it an NFB loop? These are my questions.

So Cosine(0) - Cosine(180) = 2. A gain increase and not gain reduction and the difference is at maximum. So if you put a signal that is 180 degrees out of phase at the other input of the LTP you get positive feedback not negative feedback.

Is not the NFB signal, in being presented to both grids at the same time (in that case) being cancelled out? (leaving the NFB at the cathode to carry out its function of mitigating the input signal to the LTP's A stage?)

From page 187 of Merlins book, last paragraph...

"Since the long tail pair is a differential amplifier, the feedback must be in phase with the input signal at the 1st grid."

Page 201 is not referring to a differential amplifier.

The Dumble local feedback is not a using a differential amplifier. That one requires the feedback to be 180 degrees out of phase.

[martin manning]

Why does Merlin's book say on P201 that " If the voltage is fed back is in phase with the input voltage it will serve to increase it... (and in the next para) If the voltage fed back is 180 degrees out of phase with the input voltage it will serve to decrease it..."? How can these statements be reconciled with feeding a 0-degree (in phase) signal from the OT secondary back to a pre-amp cathode (which otherwise moves in phase with the input signal at its grid) and calling it an NFB loop?

For the 5F2-A, a positive swing in the input signal at the grid decreases the Vg-k difference. If that signal comes back around from the OT secondary with the same phase and is applied to the cathode, it will increase the Vg-k difference and reduce the effect of the original signal. Voila, negative feedback. It’s not the absolute phase of the FB signal but its effect that matters.

Is not the NFB signal, in being presented to both grids at the same time (in that case) being cancelled out? (leaving the NFB at the cathode to carry out its function of mitigating the input signal to the LTP's A stage?)

But what about the FB signal going directly to the grid of the non-inverting side through the capacitor connected to its grid? The roles of the two PI triodes with respect to inverting/not inverting the pre-amp signal are now reversed; with the net result being a reduction in the amplitude of both outputs.

[tubeswell]

As mentioned above, the LTP is a differential amplifier. It takes the difference (subtraction) between the 2 inputs and amplifies it.

For example:
Cosine(0) = 1:
Cosine(180) = -1 This is 180degrees out phase with respect to cosine(0).

Yes I already knew that. But my argument is about negative feedback needing to be 180 Degrees out of phase with the input voltage. (and not about the difference in phase between the A and B sides of a PI).

Why does Merlin's book say on P201 that " If the voltage is fed back is in phase with the input voltage it will serve to increase it... (and in the next para) If the voltage fed back is 180 degrees out of phase with the input voltage it will serve to decrease it..."? How can these statements be reconciled with feeding a 0-degree (in phase) signal from the OT secondary back to a pre-amp cathode (which otherwise moves in phase with the input signal at its grid) and calling it an NFB loop? These are my questions.

So Cosine(0) - Cosine(180) = 2. A gain increase and not gain reduction and the difference is at maximum. So if you put a signal that is 180 degrees out of phase at the other input of the LTP you get positive feedback not negative feedback.

Is not the NFB signal, in being presented to both grids at the same time (in that case) being cancelled out? (leaving the NFB at the cathode to carry out its function of mitigating the input signal to the LTP's A stage?)

From page 187 of Merlins book, last paragraph...

"Since the long tail pair is a differential amplifier, the feedback must be in phase with the input signal at the 1st grid."

Page 201 is not referring to a differential amplifier.

The Dumble local feedback is not a using a differential amplifier. That one requires the feedback to be 180 degrees out of phase.

Ahh that's it. Thankyou

[martin manning]

So, are you straightened out on the 5F2-A FB now too?

[tubeswell]

Yes thanks. I was forgetting about the cathode being positive w.r.t the grid in a cathode biased stage (All I was thinking about was the unbypassed unfed-back cathode following the grid swing, but not thinking about how increasing the cathode swing in sympathy would reduce the plate to cathode voltage further. Duhh)