Phaez gives away EL84 design

[Dave H]

That would give best balance before FB is applied which should be your aim.

Does it make a difference that the NFB is applied at the cathode coupled, non inverting triode rather than at the tail? It seems to me that it would.

I've tried that on an AC15 type output stage. Without NFB it was well balanced but when NFB was applied just to the non inverting triode's grid the balance was way off. Increasing the NFB made the balance worse. If the NFB is applied to both the non inverting triode's grid and the the tail then the balance is fine.

[Helmholtz]

This leads to mutual gain degeneration.

This assumption of mine was wrong. Actually the mutual interaction of the cathode currents reduces the degeneration effect and increases gain compared to single triode with the same cathode resistor.

[nickb]

Thanks nickb,

I thought I was close.

I used this as a guide.
http://www.valvewizard.co.uk/acltp.html

As Helmholtz says, the circuit has to be treated as a differential amplifier where V3a grid is the +ve input and V3b grid is the -ve input.

What I understand (or possibly miss-understand) is:
With any differential amplifier (SS or tube or whatever)
When balance is perfect:
- 2nd harmonic distortion is cancelled
- what distortion remains is mostly odd harmonics and is a result of non-linearities in the devices (triodes in this case) themselves.

Deliberatly unbalacing a differential amp is a way to allow more of the naturally resulting 2nd harmonic distortion (from the diff amp) to remain (not be cancelled).

The Push Pull output stage is a differential amp too, if it is perfectly balanced then the same 2nd harmonic distortion (from the output stage) cancellation occurs.

Feeding the output stage unbalanced inputs causes increased 2nd and other even harmonic distortions to occur.

AND (a point which is often missed)
The mechanism which produces Harmonic Distortion also produced Intermodulation Distortion, you simply can not have one without the other. It is often the Intermodulation Distortion products which can sound nasty.

Cheers,
Ian

This would have been really close on if it were not some for something going wonky in the calculation. The formula is

1 + (Ra+rp)/(Rk(mu+1))

which comes to 1.136 with mu=100, rp=70k and Ra=68k. That would make other Ra = 77.3k, really close to the sim.

Thinking about the operation of the feedback, if you simplify and call it class B then on the positive going half cycle of the plate where the bigger Ra is located, the NFB gain will be 1.136^2 times that of the other plate. This has the effect of reducing the gain on that side and so explains why they went all the way to 120K to unbalance it again.