Forget about Ghost notes, Princeton seems to have a Ghost tube?

[SoulFetish]

This Princeton seems to be able to do the impossible: Take the signal from the input, and drive a push pull output with no phase inverter; I mean the entire tube which drives the vibrato and cathodyne phase inverter is pulled from the amp. Tell me, how does it manage to do that?
So, not only is the circuit interrupted by opening an entire gain stage, but somehow, it manages to take an input signal which goes to nowhere, create 2 signals 180 degrees out of phase, and drive the output tubes to the speaker output.
I recorded some video of this phenomenon in action. I'll uploaded it a bit later on, but chew on that for a bit and give me something so that everything I thought I knew isn't wrong

[Helmholtz]

...create 2 signals 180 degrees out of phase, ..

Did you verify that by scoping the grid signals simultaneously?
Any difference between grid signals will produce a little output.
I suspect a leaky board.

[Chuck H]

This ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

If you're getting "output" only technically, that is the output is there but low, low and distorted. etc. then there is likely signal bleeding to the grids without the signal path being directly connected.


EDIT: No offense intended but another possibility is that this is a Princeton Reverb and not a Princeton and you pulled the reverb tube. Pretty sure that's not the case though.

And yet another possibility is that someone worked on or modified the amp in some failed way and this is the result.

[g1]

Any difference between grid signals will produce a little output.

How about if the signal were on the supply line. I would think if the push and pull were not perfectly balanced you could still get output.

[Helmholtz]

How about if the signal were on the supply line. I would think if the push and pull were not perfectly balanced you could still get output.

More than unlikely. Too late for me to explain.

[Chuck H]

Well I had also considered the "supply line". ie: Errant signal modulation on a higher resistance (impedance) than ideal power supply. It does look unlikely schematically but I've known some small signal bleed to happen there. Ergo my caveat about lower signal and distortion. I think it more likely there's something hinky in the amp circuit. I haven't seen conductive boards cause this sort of thing. And I've seen A LOT of conductive boards. Usually it takes some higher voltage than what is in the signal chain to present significant conduction. But we're early on in this and I may yet learn something.

[glebert]

Which model Princeton? Using the schematic for AA964, it seems like if the PI tube is pulled and the capacitor at node B is not good you get signal leaking to the grid of the upper 6V6 but not the lower one.

https://cdn.shopify.com/s/files/1/23...71360534828114

[Jon Snell]

The AA964, B120, PR770 and 6G2 all have a phase splitter and very efficient it is.
The other Princeton models use a single 6V6.

Some other amplifiers use a different drive circuit; as the first grid rises and the first valve draws current, the second valve sees a reflection and works by relaxing. When the first valve relaxes, the second vale conducts, (in sympathy it would look like), causing the wave form correction and producing a push pull circuit as normal.

Have a look here, this covers the basics of both valve an transistor output stages.

https://www.youtube.com/watch?v=y3P7lAH3F7c


Hope that helps.
Just in case in more depth, the Parraphase invertor;
Old school but works well.

Edit; More information added.

[SoulFetish]

The “reverb tube, Chuck? That’s cute. What the hell is “reverb”?

[SoulFetish]

Yeah, so I apologize for the shorthand. Like any Princeton worth a damn,
It’s a Princeton Reverb.
However, I already explained that the specific tube is responsible for driving both the tremolo, and cathodybe
phase inverter. its a Single triode circuit.
it differs from The long tailed pair for paraphrase PIs in that Certainly wouldn’t be reasonable for some coupling to drive one Of the two triodes And allow some weak output. That’s almost impossible with a cathodyne.
but that’s moot. Anyways, there is no tube there. There’s nothing to draw current through those resistors.

Here’s the craziest part, if I disconnect the wire which connects the preamp to the grid pin on the tube socket, The Output goes away. Make that contact, output comes back.
I would like to reiterate again that there is no tube and no grid/plate,/cathode or heater for that matter In that tube socket.

[Helmholtz]

Using the schematic for AA964, it seems like if the PI tube is pulled and the capacitor at node B is not good you get signal leaking to the grid of the upper 6V6 but not the lower one.

That is indeed a possibility I overlooked last night.
Adding another filter cap at point B will tell.

[Helmholtz]

Here’s the craziest part, if I disconnect the wire which connects the preamp to the grid pin on the tube socket, The Output goes away. Make that contact, output comes back.

It's not clear to me which tube and wire you're speaking of. Can we agree on a specific schematic?

[g1]

It's not clear to me which tube and wire you're speaking of. Can we agree on a specific schematic?

From the description of the circuit, I have 2 matches, AA1164 and AB1270. I'm attaching the AB1270 as it calls out the tube names (V3A and V3B) which was confusing me (the AA1164 schem. does not call them out).