Can someone explain the phase inverter in the Traynor YBA-2 ?

[waspclothes]

I recently fixed up a YBA-2 for myself and now trying to figure out the circuit, which is all very straight forward. But the phase inverter doesn't make sense to me... I've read that this amp is very similar to the Fender Harvard, but the PI isn't close...

The "top" stage looks like a straightforward amplification stage, but the "bottom" stage gets it's input from the node at R9,R10,R11.

On a sort of related note, when I got the amp - the person had the two preamp tubes switched. So a 12AX7 in that lone pre-amp section and the 12AU7 in the PI position.. strangely, reversing them yielded a sound that I didn't prefer. I checked it out on the scope and the cranked distorted signal looks far more interesting compared to the proper way (which essentially turns into asymmetrical square wave).

[mozz]

Some sort of Paraphase. There is a model with a 6av6 preamp, that's where somebody said it's close to a Harvard.

[Chuck H]

It is essentially a paraphase inverter. There may be some attention payed to balancing that resolves itself in the operating voltages and testing, but considering that the cathode and plate resistors are symmetrical typical values and there is no (otherwise) calculated voltage divider for the non inverting grid feed than the 27k 0V shunt, if it does balance with the intended tubes and make a reasonably symmetrical square wave, well, I think it's brilliant. I'll need to experiment with this design because it's one of the more eloquent but non obvious I've seen. I'm greatly intrigued.

But yes, it's a paraphase. That is, the output of the inverting triode is feeding signal to the grid of the non inverting triode.

[Enzo]

The job of the PI is to take a signal and feed it to a power tube, and also to invert that signal and feed it to the other power tube.

The top stage is a basic gain stage, yes. SO is the bottom one really. The top one amps the signal and sends it through C5 to the power tube grid. Then R9 and 11 form a voltage divider across that signal. Very roughly a 10/1 divider. That signal is of course opposite phase from pin 2 of the PI. SO we take the reduced signal there and feed it to the other PI grid to invert again to the lower output tube.

[eschertron]

Is the placement of the second grid signal as the junction of the two grid leaks significant? I'm trying - and failing - to envision what the voltage of that node is actually doing. Is it staying at ground potential because the two grids are being drawn to (theoretically) equal but opposite voltages? If so, there's no signal to the inverting grid. If it's being drawn towards the grid voltage (as in a voltage divider) of the upper pentode, then the lower pentode's grid is being yanked away from ground in the opposite direction from the upper pentode's grid. And so the 6V6's have unequal bias references. How would that work?

[Dave H]

Is the placement of the second grid signal as the junction of the two grid leaks significant? I'm trying - and failing - to envision what the voltage of that node is actually doing.

The bottom triode of the PI is a unity gain virtual earth negative feedback circuit (like an op-amp inverter). R9 is the feed on resistor, R10 is the feedback resistor making the junction of R9,10,11 a virtual earth. R11 is to to provide a grid leak path to ground for the 6V6s. There's no signal voltage across R11 because it's connected between the virtual earth point and ground.

[pdf64]

Exactly:-)
Merlin has a page on it http://www.valvewizard.co.uk/paraphase.html see the last 2 ‘anode follower’ paragraphs.

The old JMI Vox AC50 has a variant https://el34world.com/charts/Schemat...Vox_ac50_2.pdf

[Helmholtz]

The bottom triode of the PI is a unity gain virtual earth negative feedback circuit (like an op-amp inverter)

^^^Excellent explanation.

I would like to add that as the lower triode has a finite "open loop" gain of around 25, it needs a grid signal of 4% of its plate signal. This means that the upper triode's output will always need to be around 4% higher than the lower one's and the signal across R11 will be a couple of volts (while with a high gain op-amp it would be µVs and perfect symmetry). - Not significant, really, but in my first attempt to understand the circuit, I assumed perfectly symmetrical PI output signals. But this would lead to zero grid signal for the lower triode and can't work.

[Chuck H]

Ok. But what I like about the Traynor circuit is how the power tube grid loads are serving double duty as the virtual earth balance resistors.

And I think there must be signal across R11? The implication of no signal would seem to mean R11 could be made 0 ohms. But that would ground the grid of the VE stage. So, to my simple mind, R11 must be elevating signal from 0V as part of it's function so there must be signal across it. I know that there is an anti phase on the ungrounded end of R11 (ergo local NFB that makes it a VE circuit) but there still has to be a dominant signal modulation on the grid (I guess this would qualify the "virtual" part of the name ) or that triode wouldn't pass signal at all.

[Helmholtz]

And I think there must be signal across R11?

Agree, please see my post #8.
Bypassing R11 would stop the lower triode from producing output.

[Dave H]

I would like to add that as the lower triode has a finite gain of around 25, it needs a grid signal of 4% of its plate signal. This means that the upper triode's output will always need to be around 4% higher than the lower one's and the signal across R11 will be a couple of volts. - Not significant, really, but in my first attempt to understand the circuit, I assumed perfectly symmetrical PI output signals. But this would lead to zero grid signal for the lower triode and can't work.

I omitted the fact that its finite open loop gain makes an imperfect virtual earth because I didn't want to complicate the explanation. The outputs could be balanced by making R10 4% bigger but there's not much point if 10% resistors are being used. 4% could be reduced to 2% by bypassing R6, or the cathodes of V2 could be joined together with just a single 750R to ground (no capacitor)

[Chuck H]

I thought the same thing. Wondering why the cathode of the VE isn't bypassed. And if you're going to fully bypass then why not share the circuit on both cathodes. I saw the rest of the circuit as eloquent enough that I assumed there was a reason for the unbypassed cathode on the VE and it just went over my head.

If we assume R5 as a shared 750R resistor and C3 as a shared bypass cap for both cathodes, could a small, unbypassed resistance be added between R5 and 0V as an insertion point for global NFB? Well, of course it could I suppose, but would there be any detriment to the VE stage function, since it already operates with considerable local NFB.?. I'm thinking it wouldn't be a problem since all other phase inverters also have ample local NFB. But I'm bouncing this off the higher minds

[Helmholtz]

I think 4% controlled asymmetry (meaning that it will be the same with matched or unmatched triode systems) is good enough for a guitar amp (and probably better than most other PIs without selected tubes and/or manual balancing) with typical resistor tolerances of 10%.
Aging of the upper triode doesn't influence symmetry and the "open loop" gain of the lower triode (determining asymmetry) is stabilized against tube variations and aging effects by cathode degeneration.

[Dave H]

If we assume R5 as a shared 750R resistor and C3 as a shared bypass cap for both cathodes, could a small, unbypassed resistance be added between R5 and 0V as an insertion point for global NFB?

It only needs a shared 750R. It doesn't need a shared bypass cap because the cathodes are also a virtual ground. As the current in one triode increases the current in the other triode decreases by the same amount so the current in the 750R is constant as is the voltage at the cathodes.

In the normal LTP circuit the NFB signal is applied to the tail and the inverting grid of the LTP. If it's only applied to the tail it's common mode and will be rejected. If you wanted NFB you could go back to the original circuit, remove C3 and insert the NFB at R5.

[Chuck H]

It only needs a shared 750R. It doesn't need a shared bypass cap because the cathodes are also a virtual ground. As the current in one triode increases the current in the other triode decreases by the same amount so the current in the 750R is constant as is the voltage at the cathodes.

I hadn't even considered that. Makes sense now that I'm looking at it. Neat-o.

Of course, removing the bypass cap from the cathode of the input triode (which is operating as a typical gain stage) will reduce gain of the overall system considerably. If one wanted some gain advantage AND global NFB then use the original circuit with a smaller resistance between R5/C3 and 0V and insert NFB there then. When I say "small resistance" I'm considering something like the typical Fender value of 47 ohms. So not much influence on stage gain other than the NFB applied.