6L6 amp with unique phase splitter - check my math

[kenpeter]

Showcasing unique new PNP concertina with equal impedance phase split.
Collector voltage feedback via Schade = Emitter voltage feedback via Mu.
Also showcasing tight local feedbacks only, no global loop.

Already I forgot a grid leak resistor on V1a. Just pretend its value 100K?
Is 56H the correct plate to plate primary inductance for 5K 100W CXPP?
See any other problems with these numbers?

[DF96]

V1 is biased at about -0.7V, so is likely to take grid current. This will be very non-linear, so the amp will have to be driven from a low impedance source unless you want distortion before the signal even reaches the feedback loop.

[Yvesm]

In other words, plate and cathode load for V1 are far too low !
Anyway, you're trying to solve a non existing problem: concertina remains symetric as long as both sides have equal loads.

Yves.

[DF96]

I have spotted another snag. The 12AX7 has at most a voltage gain of 100, in reality you will get less than this. The negative feedback to the cathode tries to establish a gain of 100. The net result will be a gain a bit below 50. This is 6dB of feedback, exactly the right amount to create problems. It is enough to create high order terms, but not enough to reduce them. The output stage has no feedback so you get whatever it does - important to get the bias right to reduce distortion.

My guess is that this amp will not sound very nice.

[revintage]

Hey Ken,
Think we can expect a lot more inductance from the Edcor. They spec the 25W SE to over 40H.

How does this one sim?

[smoking-amp]

The upper 2.2K plate load is impedance multiplied by the transistor Beta and the 220 Ohm pullup (Base to emitter) resistor is bootstrapped. So it's nearly a CCS load on the plate.
If the load were a true CCS, then the cathode feedback could equal the Mu, since no change in plate current would be needed. But clearly some current drive is needed to operate the splitter loads due to finite Beta. Putting a P Mosfet in place of the bipolar could work with constant plate current if the 220 Ohm got CCS'd.

Thinking backwards from the Mosfet case, a tiny amount of required plate current change for the bipolar case would seem to lead to some solution where the cathode feedback would have to be just slightly less than Mu in order to leave some Vgk delta for operation at that new current. Not sure how to calculate that other than using just the tube gm as a first order approx. Or simulate to fully calc.

Lets see. Some delta Vin gets multiplied by approx. Mu, that Mu*deltaVin then activates the Bipolar (/Beta*Rload) for some current change, which leads to subtracting out a deltaVfdbk = Mu*deltaVin*/(Beta*Rload)*(Beta*Rload)*Rattenuator = Mu*deltaVin*Rattenuator from the original Vgk. Whats left of the original deltaVin-deltaVfdbk (*gm1) should then be what is needed to operate the small plate current change for controlling the bipolar Beta. Making sense? I'm not sure myself. And what would the feedback level be then?

Some fixup needed for the tube bias for sure unless tiny input signals (<0.7 V) expected.

[smoking-amp]

OOPs that "deltaVin-deltaVfdbk (*gm1)" above, should read: (deltaVin-deltaVfdbk)*gm1

After consulting some String Theory and Quantum Field Theory Books, looks like it comes out:
Mu*Attenuator = 1 - Mu/(gm1*Beta*Rload)
or Attenuator = 1/Mu -1/(gm1*Beta*Rload)
on a sunny day maybe

So the bottom resistive attenuator needs to attenuate a little beyond 1/Mu to still get approx. Mu gain overall due to finite Beta. I think. So the feedback available must be from using up Beta rather than Mu. Leading me to suspect that a high Beta transistor should give low distortion. Or us a Mosfet like that split-P splitter. So you callin this thang "P soup"?

[kenpeter]

Cathode swings 50% in phase with the input signal, just as it would in long tail pair.
1.4V peaks would be needed to make the 1st grid forward conduct.
Way more than necessary to drive this gain+splitter to full clip.

Conventional concertinas remain symmetric if both loads are equal impedance on
both rise and fall. Never a problem in AB1, but this amp drives a few milliamps into
Pentode AB2! This was done on purpose to show the improved splitter function.
Misbehavior in A2 is symmetrical about the splitter, not favoring the follower end.
It behaves and misbehaves much more like a long tail pair...

[smoking-amp]

Arrgh... I forgot the feedback voltage also reduces the tube voltage gain effectively too, since Vgk is reduced. ( Mu*(Vin-Vfdb) ) With 1/Mu feedback it will settle on half. So looks like only Mu/2 with that setup. DF96 has it right.

[kenpeter]

Yes Mu/2 per single end. Still Mu Voltage gain if you look at it differentially.
Again I am reminded the behavior of Long Tail Pairs...

Mu/2 is still plenty enough gain to swing the entire concertina headroom
before reasonable danger of V1 grid to cathode forward conduction.

I'm not sure DF96's bit about -6dB being enough to cause high order terms
but not enough to fix anything??? Here its all about faking the cathode
feedback seen by a LTP. I never seen LTP having issue on account of -6dB.

Output stage has local cathode feedback per ARC ST-70-C3 style.