I'd like to try this so-called 'Schade' parallel feedback in a push-pull pentode amp. I was reading the posts here and some old articles about implementing this type of feedback. It seems the general consensus is that you want to apply 10% negative feedback to the amplifier to create triode-like behavior from the output stage. But I don't understand what exactly "10% negative feedback" means.
Does that mean 10x gain reduction (-10x gain)? That would be 20dB negative feedback.
Or do I have that wrong?
Thanks for any clarifications you can send my way.
This is what the circuit looks like so far. At this point it has 20dB gain reduction compared to open loop. The OPT is a Dynaco A470 (from a Stereo 70 amplifier).
Does that mean 10x gain reduction (-10x gain)? That would be 20dB negative feedback.
Or do I have that wrong?
Thanks for any clarifications you can send my way.
This is what the circuit looks like so far. At this point it has 20dB gain reduction compared to open loop. The OPT is a Dynaco A470 (from a Stereo 70 amplifier).
Traditionally the percentage indicates the fraction of the total winding that is used for feedback. 10% would mean that the winding is tapped at 10%.
But lots of would-be designers make up their own definition so it's a toss-up.
How many dB feedback that is depends on the open loop gain of the amp and the gain after the feedback loop is closed. It does not have a direct 1:1 relation on the tap percentage. If the gain of the amp without feedback is say 46dB, and with feedback it is 33 dB, the feedback is 13dB. Etc.
Jan
But lots of would-be designers make up their own definition so it's a toss-up.
How many dB feedback that is depends on the open loop gain of the amp and the gain after the feedback loop is closed. It does not have a direct 1:1 relation on the tap percentage. If the gain of the amp without feedback is say 46dB, and with feedback it is 33 dB, the feedback is 13dB. Etc.
Jan
Were you thinking of the negative feedback introduced through the screen tap/'ultralinear' tap on the OPT? If so, I'm sorry for confusing the issue with my schematic. Please ignore the OPT and the screen taps on its primary.
The question I'm asking is about the plate-to-grid local negative feedback from R15 and R16 in the schematic.
I chose 100k for those resistors so that gain of the whole amplifier was reduced by 20dB, or its output was reduced by 10x.
Describing the amount of negative feedback applied as a percentage rather than in dB was used by Schade in his original paper on the 6L6 from the late 1930s, on pages 360-61:
The penultimate paragraph mentions oscillograms taken "for 10, 20, and 30 per cent inverse feedback".
I'm confused by what Schade means by "30 per cent inverse feedback".
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The question I'm asking is about the plate-to-grid local negative feedback from R15 and R16 in the schematic.
I chose 100k for those resistors so that gain of the whole amplifier was reduced by 20dB, or its output was reduced by 10x.
Describing the amount of negative feedback applied as a percentage rather than in dB was used by Schade in his original paper on the 6L6 from the late 1930s, on pages 360-61:
The penultimate paragraph mentions oscillograms taken "for 10, 20, and 30 per cent inverse feedback".
I'm confused by what Schade means by "30 per cent inverse feedback".
- Does he mean a gain reduction of -30x?
- Or does he mean the closed loop gain of the amplifier is reduced by 30% compared to open loop?
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Ahh, I see.
Point c in fig 33 in your post seems to define it.
For 10%, it feeds back 10% of the output signal to the grid via a 90k-10k ratio resistive divider.
Or 45k - 5k.
Jan
Point c in fig 33 in your post seems to define it.
For 10%, it feeds back 10% of the output signal to the grid via a 90k-10k ratio resistive divider.
Or 45k - 5k.
Jan
Thanks Jan. Okay, I think I'm catching on. Schade is not quantifying the negative feedback in a circuit by the amount of gain reduction the applied NFB produces, but rather by the ratio of resistance values in the NFB voltage divider. In other words, a 10:1 ratio of impedances (resistances) means "10% inverse feedback".
So the next question is, if the general advice for applying parallel "Schade" NFB in a two-stage amplifier is to "aim for 10% negative feedback", does that mean the parallel feedback resistor value should be about 10% of the anode resistance of the driver stage?
If the driver stage uses a pentode (like the 6CB6 in the schematic I posted), does that define the output resistance of the driver stage?
Let's say I have a 6AU6 driver pentode with a 100k ohm anode load resister. The Zout of that 6AU6 driver stage is the 6AU6 internal anode resistance (which is very high) in parallel with the value of Rload (100k) and the grid leak of the output stage (let's say that's 330k ohms). That will end up being about 1M//100k//330k, or something like 72k ohms.
Do I select the value of the resistor going from anode-to-grid of the output tube based on that 72k ohms? What the heck does 10% of that mean? That would be 7200 ohms, which is not possible.
Or should the value of that feedback resistor be 720k ohms, so that the Zout of the 6AU6 driver is 10% of the feedback resistor value?
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So the next question is, if the general advice for applying parallel "Schade" NFB in a two-stage amplifier is to "aim for 10% negative feedback", does that mean the parallel feedback resistor value should be about 10% of the anode resistance of the driver stage?
If the driver stage uses a pentode (like the 6CB6 in the schematic I posted), does that define the output resistance of the driver stage?
Let's say I have a 6AU6 driver pentode with a 100k ohm anode load resister. The Zout of that 6AU6 driver stage is the 6AU6 internal anode resistance (which is very high) in parallel with the value of Rload (100k) and the grid leak of the output stage (let's say that's 330k ohms). That will end up being about 1M//100k//330k, or something like 72k ohms.
Do I select the value of the resistor going from anode-to-grid of the output tube based on that 72k ohms? What the heck does 10% of that mean? That would be 7200 ohms, which is not possible.
Or should the value of that feedback resistor be 720k ohms, so that the Zout of the 6AU6 driver is 10% of the feedback resistor value?
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What's 'parallel Schade feeedback'? I'm not a tube guy.
I only know that 10% feedback means that 10% of the signal at the point where the feedback is taken off is returned.
Of course, the feedback resistors load down the signal at the take off point so as such has already a lowering effect on the gain, if that is what you mean.
Jan
I only know that 10% feedback means that 10% of the signal at the point where the feedback is taken off is returned.
Of course, the feedback resistors load down the signal at the take off point so as such has already a lowering effect on the gain, if that is what you mean.
Jan
"Parallel Schade feedback" is a confusing term people came up with because it was written about in the paper by Schade I showed in my earlier post. It's just parallel negative feedback applied from anode to grid of a pentode output stage. It's sometimes called 'anode to anode' NFB because it's applied from the anode of the output tube to the anode of the driver tube, as in the infamous RH84 design, but the feedback is really applied from the anode of the output tube back to the output tube's grid. The proper name would be parallel (or shunt) negative feedback, just like when you apply NFB around an inverting opamp, or in the (also confusingly named) 'anode follower', which is simply a tube as an inverting amplifier with feedback applied from its anode (output) back to its grid (input).
Well put. That's the answer right there. Sorry I've been slow to catch on, but thanks, that clarifies it nicely.
To directly answer the OPs original question, in Schade's terms it means a particual feedback network where the input resistance is 10% of the value of the feedback resistance. It can be confusing because in Schade's circuit, the input resistance is the often output resistance of the preceding stage.
Cheers
Ian
Cheers
Ian
Thanks Ian. That's quite clear.
So, if I have a 6CB6 input stage with 33k ohm anode load resistor, I can be pretty sure the output resistance of that input stage is about 30k ohms, correct?
Does that then mean the anode-to-grid feedback resistor should be 10X that, or about 300k?
Does the value of the feedback resistor define the feedback resistance, or is it more complicated than that?
In simulation, using a 100k ohm anode-grid feedback resistor yields very good results, without too much high frequency resonance. On the other hand, using a 300k feedback resistor yields a lot more gain, but not great performance (higher THD).
In the amplifier shown in the schematic, the open loop gain is 30X ( 29.5dB).
Which one of those is what you'd want to see for a circuit like the one in post #1?
So, if I have a 6CB6 input stage with 33k ohm anode load resistor, I can be pretty sure the output resistance of that input stage is about 30k ohms, correct?
Does that then mean the anode-to-grid feedback resistor should be 10X that, or about 300k?
Does the value of the feedback resistor define the feedback resistance, or is it more complicated than that?
In simulation, using a 100k ohm anode-grid feedback resistor yields very good results, without too much high frequency resonance. On the other hand, using a 300k feedback resistor yields a lot more gain, but not great performance (higher THD).
In the amplifier shown in the schematic, the open loop gain is 30X ( 29.5dB).
- with a 100k ohm feedback resistor, closed loop gain is 8X (18dB), so there is 11.5dB of gain reduction from NFB.
- with a 300k ohm feedback resistor, or 10X the output resistance of the driver stage, closed loop gain is 16X (24dB) which means only about 6dB of gain reduction from application of NFB.
- With a 47k ohm feedback resistor, closed loop gain is 4X (12dB), which means there's 17.5dB of gain reduction from NFB.
Which one of those is what you'd want to see for a circuit like the one in post #1?
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I believe you'll use (c), but scale R1 (Rg res. rather than R2 (plate feedback res.) according to n feedback factor. If R1(Rg) is so scaled and it's to low (say 10k), you can add a series resistor before R1 to make up correct load for driver stage to avoid overloading the driver. Some insertion loss is expected.
Hi, if you have 30k, the feedback resistor must be 270k, so nine times, that way you’ll have 10% that is 30k on 270+30=300k.
The “percentage” depends on the kind of tube you use. EL84 likes lower values than EL34 or KT88, as per UL.
My preference goes towards 10-12% for EL84 and up to 25-30% for 6V6GT.
Just consider that an higher percentage represents a lower load for the driver, and so it reduces the gain of the driver and can cause it to run out of current before swinging to full power.
The “percentage” depends on the kind of tube you use. EL84 likes lower values than EL34 or KT88, as per UL.
My preference goes towards 10-12% for EL84 and up to 25-30% for 6V6GT.
Just consider that an higher percentage represents a lower load for the driver, and so it reduces the gain of the driver and can cause it to run out of current before swinging to full power.
Yes, it is 10% (or whatever you want to use) of the total, not of the feedback resistor alone.
Jan
Jan