Negative Feedback question?

[topbrent]

I have read the Aiken articles on Negative Feedback.  I don't think I completely grasp the technical side of it.  I understand that it is injecting some of the signal back into the amp, and that amount is the ratio of the series resistor and the shunt resistor.

This probably reveals my "paint-by-the numbers" and "follow there standard recipes" level of understanding, but my question is simple.

Which amp has more NFB, the 6g9b tremolux, '63 vibroverb reissue or the AB763 bandmaster?  And a simple Why?

- Tremolux 6g9B-  100k-series, 4.7k shunt  http://www.prowessamplifiers.com/schematics/fender/Tremolux_6g9b-Schematic.html
- 63 Vibroverb reissue-  10k-series, 470ohm-shunt  http://www.prowessamplifiers.com/schematics/images/63RI_Vibroverb.pdf
- AB763 Bandmaster - 820ohm-series, 100ohm-shunt  http://www.prowessamplifiers.com/schematics/fender/Bandmaster_ab763-Schematic.html

All 3 amps are 6L6 amps and all use a 4ohm output transformer.

Why Leo, did you monkey around with stuff sooo much?,....why...Chronic Tinker-itis...

[sluckey]

I would say the AB763 Bandmaster has more NFB because 100/820>470/10,000. Notice that 470/10,000=4,700/100,000=0.047, so those two amps should have about the same feedback.

The tail resistors are slightly different values and that will also have an effect on the level of NFB, but 100/820=0.12 and that is significantly larger than 0.047

[VMS]

Richard Kuehnel has an easy to understand example for NFB at the bottom of this page:

http://www.pentodepress.com/home/classic-circuits/marshall-jcm800-2204-power-amp/

and a LTP feedback calculator here:

http://www.pentodepress.com/home/amplifier-calculators/feedback/

[jjasilli]

I'm thinking that the source voltage for the NFB loop might be a factor here.  Note that the Tremolux has about 65 less plate volts than the Bandmaster.  If we pretend that both amps use the same OT, and the voltage difference is due only to the B+ supply, then the amp with the higher plate voltage will put commensurately more voltage on the OT secondaries.  Therefore, to get the same NFB voltage delivered to the PI in both amps, the Bandmaster would require more attenuation of its NFB voltage.  But as sluckey states, the PI's are not exactly the same; hence it would take a different injection of NFB voltage to get the same final result.

Also:  Maybe Leo didn't want the same NFB result, to help make different amp models with essentially the same general circuitry sound different from one another.

[HotBluePlates]

I'm thinking that the source voltage for the NFB loop might be a factor here.  Note that the Tremolux has about 65 less plate volts than the Bandmaster.  If we pretend that both amps use the same OT, and the voltage difference is due only to the B+ supply, then the amp with the higher plate voltage will put commensurately more voltage on the OT secondaries.

But the assumption with that is the different d.c. voltages result in a different a.c. voltage on the speaker. Notice also the bias voltage is different for each of these amps by the same ratio as the d.c. supply voltage.

I think this is a red herring. The big assumption I make is that given a pair of 6L6's in all 3 amps, essentially the same amount of audio power should be transferred to the speakers of all 3 amps. Call it 40w. A great thing topbrent did to compare apples to apples was to see that all compared amps have the same output tube complement and the same speaker load. 40w across 4 ohms -> 40w/4 = 10, which is voltage-squared, so when the amp is making 40w there is ~3.16v RMS at the speaker terminals. This is the source a.c. which is then divided by the feedback network.

If some amps were a pair of 6V6, or a quad of 6L6, or 8 ohm speaker load, then we'd have to assume an output power level and do the math to find out what the raw audio voltage is at the speaker terminals before considering how the feedback network divides down that voltage. Topbrent's example simplifies things so that looking only at the resistor values gives the right answer.

[jjasilli]

Notice also the bias voltage is different for each of these amps by the same ratio as the d.c. supply voltage.  Yes, a power tube with less plate voltage will want a hotter bias = less bias voltage.  So bias voltage will vary commensurately with plate voltage.

But the assumption with that is the different d.c. voltages result in a different a.c. voltage on the speaker. Yes, that is my assumption.  Is it wrong?  My belief is that IF a 10VAC signal rides atop a 400VDC plate voltage in the OT primary, THEN the OT secondary "sees" a 410VAC signal operating on it; AND it (the secondary winding) steps-down that voltage, say around 14:1, to maybe ~20 some odd VAC.  

This is bolstered by the fact that the two amps should not have the same wattage rating:  For Push-Pull:  Power = (Idle Plate Voltage)² / OT's Primary Impedance. Let's say the 6L6_p-p impedance = the OT's Primary Impedance = 3000 Ohm's.  Then 365² / 3000 =  133,225 / 3000 = 44.4 Watts.   440² / 3000 =  193,600 / 3000 = 64.5 Watts.

But I'm no expert.  If this is wrong please let us know.  

[HotBluePlates]

My belief is that IF a 10VAC signal rides atop a 400VDC plate voltage in the OT primary, THEN the OT secondary "sees" a 410VAC signal operating on it ...

Yep, that's the error.

Ultimately, the only thing that causes an output on the secondary is a changing voltage/current on the primary. If d.c. resulted in an impact, then there would be a d.c. voltage present on the secondary with no signal input on the primary (just B+). If that was the case, the speaker would be very unhappy! The speaker cone would be trying to jump out of the magnet gap all the time.

So only the signal a.c. voltage at the primary causes an output voltage on the secondary. Becuase it's somewhat hard to figure the output tube plate voltage swing (read as extra steps), the easy method is to recognize that power is voltage-squared divided by resistance (impedance), and use an assumed power output to figure voltage at the speaker terminals.

You could say that any signal level less than full power output results in less voltage at the speaker terminals, which means less feedback. But that would also be in error, because there would be a smaller signal input to the phase inverter to create that lower power output; that means that reduction in feedback voltage is proportional to the reduction in phase inverter input voltage, so one tracks the other up and down, as determined by the feedback voltage divider.

For Push-Pull:  Power = (Idle Plate Voltage)2 / OT's Primary Impedance. Let's say the 6L6p-p impedance = the OT's Primary Impedance = 3000 Ohm's.  Then 3652 / 3000 =  133,225 / 3000 = 44.4 Watts.   4402 / 3000 =  193,600 / 3000 = 64.5 Watts.

But what 6L6 amp running in class AB outputs 64w? Recall that when Psychonoodler and PRR were bouncing this idea around that there was a proposed fudge-factor of less-than-one to multiply along with these numbers.

Also know that this approach was acknowledged as approximate. The output tubes cannot swing their plates down to zero volts, so the voltage used must be somewhat less. And it's been a while since I saw this formula in a definitive explanation... are we sure that the numbers don't have to be converted from peak volts to RMS volts?

That question too may be meaningless, for the reason I pointed out above. If the supply voltage is lower, and the bias voltage is lower, then the signal input needed to make full output power is lower. Maybe output power is smaller and feedback voltage is therefore smaller, but then that feedback voltage is also acting against a smaller signal voltage. I guess I should brush up on my output stage math to put real numbers to this, or measure an actual amp. My gut-feeling is that the proportion of feedback to signal will be determined in all cases by the feedback voltage divider, and that the audible differences (if any) will also follow that ratio.

[jjasilli]

Some thoughts & comments, or points of confusion:

Yep, that's the error. . . only the signal a.c. voltage at the primary causes an output voltage on the secondary  OK, thanks.

Also know that this approach was acknowledged as approximate.  Even though the method is approximate, it implies that there are different voltages on the secondary.  

it's been a while since I saw this formula in a definitive explanation... are we sure that the numbers don't have to be converted from peak volts to RMS volts?  The formula uses only DC volts.  I think peak & RMS concepts do not apply to DC.

Maybe output power is smaller and feedback voltage is therefore smaller, but then that feedback voltage is also acting against a smaller signal voltage.  Now I see your point.  Very elegant!

measure an actual amp  Where the rubber hits the road!  

[HotBluePlates]

it's been a while since I saw this formula in a definitive explanation... are we sure that the numbers don't have to be converted from peak volts to RMS volts?  The formula uses only DC volts.  I think peak & RMS concepts do not apply to DC.

Ahhh.... but audio is not d.c. 

Here's the thing: when figuring output stages, we can do it right or we can do it quick and "right enough". Both rarely coincide, especially if you read some of the old books on figuring tube parameters graphically (or even mathematically).

So we make some approximations that do the job fast and get us close enough. One cheat is to assume that the plate curves of a pentode/beam power tube are perfectly horizontal and are parallel and evenly spaced. The horizontal part is pretty close to true in beam power tubes, and we don't have horrible error regarding the spacing of the curves. Exactly horizontal lines would only happen if the tube plate resistance were infinite, though beam power tubes have high enough plate resistance to make this fudge close to the truth.

The d.c. supply voltage is used because if the tube were operated in class A and were "ideal" it could swing from the idle supply voltage to twice B+ and back, then down to 0v. You would specify that a.c. swing in peak volts as the change from idle to twice B+, which is equal to the value of the idle supply voltage. So that's where the reference to the d.c. supply voltage comes from.

Reality is that output tubes are not ideal, and might swing their plate as low as 40-90v (wherever the slope below the knee of the plate curve is), but it won't get to 0v. Depending on the actual load, the swing might not even be supply voltage minus this number; it could be somewhat less.

So I'm saying that this rough formula is very hopeful, and I vaguely recall PRR proposing a fudge factor to multiply this by to account for errors. The net effect was that the rough formula estimated output power too optimistically. My only point in talking about all that is that if that formula is optimistic, the degree of difference between each output stage would similarly be overstated.

[PRR]

Find some basic (not specific) NFB theory.

You figure the forward gain without feedback.

Figure the feedback network ratio (including loading).

Find the ratio of difference. Or the "excess gain".

Example. A '741 chip opamp has open-loop gain of a million. We use 9K and 1K (or 90K and 10K) NFB resistors. The closed-loop gain will be slightly less than 10. This 10 is 100,000 times less than the open-loop gain of a million. We have excess gain of 100,000. The error from "exactly 10" is 1/100,000 or 0.001%.

So what is the open-loop gain of a Fender tube-amp output stage? Sometimes hard to guess.

But the DC voltage has very little effect on signal gain. Occasionally you can reverse-guess the gain from DC conditions. And very over-volted power stages have lower voltage gain.

The "similar" amps cited give roughly these gains from one 6L6 grid to 4 ohm load:

Grid=36Vpk Power=35W Output 17Vpk 1:0.47

Grid=52Vpk Power=60W Output 22Vpk 1:0.42

But (as shown in Kuehnel'a 5F6A book, gain varies a lot from small to large signal. That's the nature of hard-worked power stages, and why we may want some NFB.

We may often assume that gain of self-bias 6L6 in 4 ohms is "around 0.4"; any greater accuracy is a waste of time.

Remember that this is for _normal_ load. The gain will rise for higher loads. Loudspeaker impedance it NOT constant; often we use NFB to control the hi-Z resonances not the midband where impedance is near nominal. Power pentodes have plate resistance 10 times the typical nominal load, so gain will increase almost 10 times for loads more than 10 times higher than nominal.

The long-tail with usual values has gain of 50 for one side. Because we drive one side and take output from both sides, the useful gain is 25.

So the open-loop gain is about 50*0.4= 20 for 4 ohms, and perhaps over 100 at speaker resonance.

> the ratio of the series resistor and the shunt resistor.

Actually, since it is a voltage-divider, there is a "plus one" in the math. For the examples you cite, this is lost in tolerance. There's an amp with 810:100 NFB network: this looks like 1:8.2 but is really 1:9.2, just enough difference to maybe matter a bit.

> 100k-series, 4.7k shunt

100K and 5K is 1:20. Or doing all the math, 1:22.3. (Ignoring loading.)

At nominal load, we have forward gain of about 20 and NFB network of 1:22. We have "no" excess gain. We have "nearly no" NFB.

At bass resonance we may have gain of 100, still NFB factor of 1:22. We have 100/22= 4.5:1 of excess gain, and "significant" NFB.

> 10k-series, 470ohm-shunt

Still 1:22.3.

> 820ohm-series, 100ohm-shunt

9.2. This is different. Forward gain of 20, NFB factor of 1:9, we have about 2:1 of excess gain and "some" NFB at nominal load. At bass resonance, 100/9= 11, very significant NFB.

Loading matters. If Leo had returned NFB to the idle grid, as it "should" be, loading would be negligible. If you bring NFB back to a cathode, ~~1K impedance, then the 4.7K shunt is very heavily loaded, the 100 ohm barely notices. The way Fender "mis"-wires the long-tail defies easy understanding.

But why think so hard? Do NOT "paint-by-the numbers"! OK, start with the suggested values, then CHANGE them and use your EARS. The "optimum" NFB depends heavily on the speaker system, the style of music, and the player. The "classic" amps work, but may not be YOUR optimum.

Particular case: the AB763 Bandmaster is a fashion-swing. Tube amps may be soft and sloppy or hard and clean.

> Why Leo, did you monkey around with stuff sooo much?

Salemen need a "groovy NEW sound!" every year. The 820:100 is a swing to "clean". The early Marshall took this further (by moving the NFB from 4 to 16 ohms). Also early Sunn and the big UltraLinear amps. We had the BIG CLEAN amps. But after a few years, transistors were doing BIG CLEAN better, lighter, and cheaper. That may be where you want to go. OTOH for the last few decades we have re-discivered the joy of a near-naked (low NFB) power pentode. You may like less NFB. Or if you play a range of music, you may want a sloppy-to-tight NFB control.