Tainwreck master volume question

[Randall]

I installed this PPI master volume in a 100 watt Marshall type amp. I have done it before using a 1MA pot, but I liked the taper of a 500KA pot this time. After sealing back up, I find myself wondering, when the pot is all the way open and there is then 500K ohms between the two sides of the PI, does this have any effect? I wasn't looking specifically for this when I tested it after installation, but I didn't notice anything off either.

[Dave H]

I don't think it will be audible. The 500k loading will just cause a slight attenuation of the signal.

[Jon Snell]

I must say, I am not keen on this circuitry as it literally strangles the signal to the output section by adding them so the output valves work hard but tend to work against each other instead.
As the pot reduces in resistance, the inverted drive becomes less in differential, causing the valves to still work but in phase with each other instead of being push pull, they both work the same phase.

[Chuck H]

I must say, I am not keen on this circuitry as it literally strangles the signal to the output section by adding them so the output valves work hard but tend to work against each other instead.
As the pot reduces in resistance, the inverted drive becomes less in differential, causing the valves to still work but in phase with each other instead of being push pull, they both work the same phase.

Huh?

The like phase characteristics of the PI output is somewhat cancelled with a resistance coupling between them. It's the differences between them that is less affected by the circuit. This should leave the remaining signal to the power tubes with more differential, not less with only like phase characteristics cancelled before the output tube grids. How can that make the output tubes more alike in phase?

[Helmholtz]

Common mode/in phase PI output signal don't produce a voltage difference between the ends of the MV, so no current and no loading.

This type 3 MV mainly works by loading the differential outputs of the PI, thus making the LTPI triodes work against each other.

I don't think the remaining common mode signal can be large, though. It should mainly be a consequence of asymmetry.

[pdf64]

I must say, I am not keen on this circuitry as it literally strangles the signal to the output section by adding them so the output valves work hard but tend to work against each other instead.
As the pot reduces in resistance, the inverted drive becomes less in differential, causing the valves to still work but in phase with each other instead of being push pull, they both work the same phase.

I don't think you've got that right. Have you tried monitoring the cathode or plate current of the power valves as the type 3 master is turned down? If you start out at full load, you should find that it falls to idle level. As the signal Vac at the power valve control grids (both differential and with respect to 0V) will reduce to near 0Vac.

[Dave H]

I think it works by loading down the outputs. Imagine the master vol pot (rheostat) is two equal resistors in series. The PI outputs are out of phase so the junction of the two resistors is always at 0V and could be grounded. i.e. it's the same as having individual resistors from each output to ground.

[Jon Snell]

If you connect the +ve signal of a balanced line to the -ve signal, (short the signal, pins 2 & 3 from a microphone XLR cable for instance) ... what do you think you get?
Yes no signal or the stronger phase influences the weaker phase in the case of the PI.
Obviously if it is truly balanced then it works as a volume control but I don't ever make any of my amplifiers use that system.

[Chuck H]

Common mode/in phase PI output signal don't produce a voltage difference between the ends of the MV, so no current and no loading.

This type 3 MV mainly works by loading the differential outputs of the PI, thus making the LTPI triodes work against each other.

I don't think the remaining common mode signal can be large, though.

I'm not assuming each PI output to be a perfect mirror image of the other. Which they certainly aren't when the PI is clipping. I think that would usually be the case when the master volume is being used. I would expect it's the differences that don't get canceled. Otherwise I think most of us recognize that the "type 3 MV" works by phase cancellation before the power tube grids.

[Helmholtz]

I would expect it's the differences that don't get canceled.

Yes, its these differences which will cause a partial common mode drive.

Otherwise I think most of us recognize that the "type 3 MV" works by phase cancellation before the power tube grids.

The term "phase cancellation" doesn't make much sense for different sign/out-of-phase voltages applied to opposite ends of a resistor. Any AC voltage across a resistor means out-of-phase voltages at both ends (wrt a virtual center point) - otherwise its voltage would be zero.

[Chuck H]

The term "phase cancellation" doesn't make much sense for different sign/out-of-phase voltages applied to opposite ends of a resistor. Any AC voltage across a resistor means out-of-phase voltages at both ends - otherwise voltage would be zero.

Well I'm still working with simpler concepts in my understanding of circuits. "phase cancellation" get's tossed around in these circumstances enough that I've fallen into it.

[Enzo]

In any case, it is lost on me how shorting opposing power tube grids together would make the power tubes work harder.

[Jon Snell]

In any case, it is lost on me how shorting opposing power tube grids together would make the power tubes work harder.

If both power valves conduct at the same time, the output transformer will not be able to induce the current from both sides of the primary at the same time as they will oppose each other, just like two transformer windings connected together, they make a virtual short circuit if the same polarity of energy is sent to them and any drive to the transformer primary will show a lower impedance causing a mismatch for the power valves.
I did not mean to imply that turning the volume down increases the current flow but any 'in phase' power will stop being induced into the secondaries.
Consider drive of 10volts on each anode of the phase inverter, the drive is still there in this circuit wherever the control is set.
Hope that clarifies it better.

[Helmholtz]

If both power valves conduct at the same time, the output transformer will not be able to induce the current from both sides of the primary at the same time as they will oppose each other, just like two transformer windings connected together, they make a virtual short circuit if the same polarity of energy is sent to them and any drive to the transformer primary will show a lower impedance causing a mismatch for the power valves.
I did not mean to imply that turning the volume down increases the current flow but any 'in phase' power will stop being induced into the secondaries.
Consider drive of 10volts on each anode of the phase inverter, the drive is still there in this circuit wherever the control is set.
Hope that clarifies it better.

I agree in principle.
The question remains, how high common mode drive signals could actually be. Are there any measurements?

[Jon Snell]

I only know the theory as I have not recorded any results.
I will next time I have one in for repair and then publish the results.
Hopefully we will all find that useful.