PI Balance

[martin manning]

It's academic, but I'll argue that with an infinite tail resistance (i.e. a current source) the balance in a LTP depends on the plate loads and not the tubes themselves. The tubes could be mismatched, but if the plate resistors are equal then the balance will be there. Admittedly, the tail resistance in a typical Fender or Marshall LTP is not infinite, but it's higher than might be thought at first because the unusual NFB arrangement bootstraps the tail to some degree. I'm not sure how to even go about calculating the degree of bootstrapping, but I understand the concept to be correct. In any case, the effective tail resistance is very large compared to the resistance looking up into either cathode in the LTP. With absolutely no measured data (why have I never measured this?) I'm willing to bet lunch money that the 82k/100k plate resistors do more to hurt than help the balance in Fender/Marshall PIs.

To look into this question I ran some LTP models. The 82k/100k PI loads date back at least to the 5F6-A Bassman, which had total tail resistance of 12k. In that case, the outputs are seen to balance pretty well, within 0.35 dB, with identical 12AX7 triodes as used in the model. Making the plate loads equal causes the unbalance to increase to 0.87 dB, so the 82k is definitely an improvement. To balance the outputs exactly requires 86.7k on the inverting side. Given E24 values, a 91k would have been a little closer, but with resistor tolerance and tube variation, this seems close enough.

Forward to the AB763, which has a 12AT7 PI tube and the tail resistance increased to 22.6k, the balance would definitely be better with a 91k, yet the 82k is still there. The reason is probably long lost.

Some manufacturers have included a trimmer to dial in PI balance, but in general this practice is rejected due to the time (= money) required to adjust it.

[Tony Bones]

Thanks for taking the time, Martin. I guess I owe you lunch!

It's interesting that Bruce Zinky did use 91k on the VibroKing. Though that was with a 12AX7 and 39k tail, and no NFB to bootstrap the tail.

Which reminds me: did you include NFB at the base of the tail to bootstrap it?

[martin manning]

No, so lunch is still in doubt! Let me try to work that in, but it seems difficult to do without having the rest of the power amp included. In a whole-amp sim, disconnecting the NFB makes little difference to the PI balance, about 0.1 dB difference (PI output/PI input).

[Ten Over]

In the real world with 100K/100K load resistors and an imbalance, the imbalance can be significantly reduced by including the NFB path in the tail.

[martin manning]

Blencowe has a bit on this, and has an example using standard values for ra and mu to calculate a plate load for the inverting triode at 84k, assuming a 10k tail. He mentions that if the tail is above about 18k, the balance is better with equal plate loads, but there is no mention of any effect of NFB. The ra and mu depend on the operating point, so the supply voltage and bias resistor value are involved for any particular case.

[Tony Bones]

In the real world with 100K/100K load resistors and an imbalance, the imbalance can be significantly reduced by including the NFB path in the tail.

To express it another way, looking at fig 5A, the 20k tail resistor has 0.73VAC at one end and 0.54VAC at the other. (I'm ignoring the 470 ohm resistor as it's only 2.3% of 20k - probably smaller than the tolerance on the 20k resistor.) Assuming the AC voltages are the same and in phase, the current through the 20k rezistor is i = v/R = (0.73 - 0.54) / 20k = 9.5uA.

The question is, what size would the tail resistor need to be to have only 9.5uA through it without NFB? The 4.7k resistor between it and ground complicate things, but we can easily calculate what the combined resistance would be (tail resistor + 4.7k NFB resistor): R = v/i = 0.73VAC/9.5E-6A = 77k ohms. 4.7 of that is accounted for by the so-called NFB resistor, so the bootstrapping makes the 20k tail resistor look like 77k - 4.7k = 72.3k.

Bottom line, without NFB we have 20k + 4.7k = 24.7k tail resistance. With NFB we have what is effectively a 77k tail. That's about 3 times as great. I don't know if that translates to 1/3 the imbalance with equal plate resistors, but the effect must certainly be significant.

[martin manning]

Worked some more on this. The model now includes the PI, a power stage, and NFB.

- By far the most significant factor in AC balance for small signals is the plate loads, with the ratio of 0.91 (Ra inverting/non-inverting) required to match the mid-band gain at the outputs (measured at the positive peaks, 24k Rtail in the PI).
- Disconnecting the NFB makes little difference to the matching.
- A miss-matched tube makes little difference to the AC balance for small signals, but when driven to clipping there will be more asymmetry in the positive peaks, and therefore more asymmetry positive to negative in the output.
- A miss-matched tube will upset the DC balance of the PI, but this does not matter unless a DC measurement is used to assess the balance.

Table of results (Inverting/Non-Inverting):

Balanced 100k/110k Ra, NFB: ........Gain 9.367/9.309 (-0.05) dB
100k/110k, NFB disconnected: .......Gain 25.017/25.327 (+0.11) dB
Matched Ra 105k/105k, NFB: .........Gain 9.639/9.044 (-0.55) dB

[Tony Bones]

Martin, what circuit is that? (5F6, BF, ...?)

I've been using the tern 'NFB', but really of interest to this thread, to me anyway, is the influence of bootstrapping the tail in the PI. I don't see any way that NFB in the usual sense of the term can correct for imbalance in the PI. Maybe reduce some distortion that comes as a result of the imbalance, but the Pi is gonna do what it does.

[martin manning]

It's Dumble ODS. Certainly the input impedance is bootstrapped by degenerative FB in the tail, but its the inverting/non-inverting balance I'm interested in. Recall it was the question of the value of matched tubes for the PI that started this. No worries, I definitely learned something from this exercise :^)

[Roe]

this calculator is nicehttps://www.ampbooks.com/mobile/amplifi ... alculator/

[martin manning]

Yes, I've seen that. It's calculating open loop gain, and has some unknown assumption for ra and mu. Compared with the ‘no NFB’ results above, it's pretty accurate as far as balance goes, a little different in gain, probably due to the ra and mu assumption.

[roberto]

It's interesting that Bruce Zinky did use 91k on the VibroKing.

Mesa Dual Rectifier has 82 and 90k.

[martin manning]

82/90 = 0.91 ;^)

[roberto]

Some Diezel amps 120k and 100k.
100 / 120 k = 0,83 not that far.