Driver PI question

[trem]

Can you guys explain how this works? I am use to seeing single tube, power amps with no phase inverter, and I know what a PI looks like in the schematic of a "push-pull" amp. But in the schematic below it looks like the driver feeds the grid of one of the power tubes, but also the grid of the other half of the driver itself. Can you guys explain how this works? Is this fairly unique, or is this just something I am personally seeing for the first time ?
Thank You

http://www.angelfire.com/mech/beansa...hems/S6420.pdf

[Bruce / Mission Amps]

Can you guys explain how this works? I am use to seeing single tube, power amps with no phase inverter, and I know what a PI looks like in the schematic of a "push-pull" amp. But in the schematic below it looks like the driver feeds the grid of one of the power tubes, but also the grid of the other half of the driver itself. Can you guys explain how this works? Is this fairly unique, or is this just something I am personally seeing for the first time ?
Thank You

http://www.angelfire.com/mech/beansa...hems/S6420.pdf

It's just drawn a bit different but I think it is a simple paraphase type driver.
Similar to an old Gibson or Fender amp, before the cathodyne driver was used... even some old VOX amps use a similar PI.
Very common in old lower powered cathode biased amps.

[Enzo]

OK, you are aware, I hope, that in a basic triode stage, like the first stage in your amp, the signal entering the grid is inverted at the plate. And of course in a push pull amp, the second power tube needs to have the signal at its grid the inverse of that at the other tube grid.

Now look V2. The signal leaves the preamp via the tone control, and into the grid of V2a. V2a inverts and amplifies the signal which it sends from its plate to one side of the power amp, V3. We need that same signal for the other power tube, V4, but it needs to be inverted. SO we sample the signal from V3 grid, run it into V2b, which inverts it tof the grid of V4. The signal at V3 grid is too strong for the V2b grid, so there is a simple voltage divider made from the 270k and 12k resistors. It knocks the signal down to about 1/20th of what it was. V2b will step it back up though.

You may note that the signal is inverted in V2a and then again in V2b, resulting in the original phase. But that doesn;t matter. what matters is that V2a drives V3, and V2b inverts that for V4.

[ThermionicScott]

Can you guys explain how this works? I am use to seeing single tube, power amps with no phase inverter, and I know what a PI looks like in the schematic of a "push-pull" amp. But in the schematic below it looks like the driver feeds the grid of one of the power tubes, but also the grid of the other half of the driver itself. Can you guys explain how this works? Is this fairly unique, or is this just something I am personally seeing for the first time ?
Thank You

http://www.angelfire.com/mech/beansa...hems/S6420.pdf

That's a "paraphase" PI. It takes advantage of the fact that a gain stage inverts the signal. So a gain stage feeds both a power tube grid and another gain stage, but attenuates the signal by the reciprocal of that stage's gain so the outputs are roughly equal.

Merlin's excellent page on the subject: The Valve Wizard -Paraphase

- Scott

[trem]

Thank you everybody...
Hey Enzo, yes I am "aware" of those facts, but I need to be reminded sometimes I guess. I'm sure this all seems painfully simple to you, but I am still trying to retain the knowledge I learn. Things like phase inversion of a signal entering the grid of a triode and leaving via the plate.
So I will ask another simple one, as long as I have you down at my level... The same triode, but as a cathode follower, I assume the phase is the same, with the signal going grid through cathode. That is to say, there is no "inversion".
Thank You

[ThermionicScott]

Yep, no phase inversion in a cathode follower. (That's how a cathodyne phase inverter works -- the signal off the plate is inverted, but the signal on the cathode isn't.)

[Enzo]

Correct. A cathode follower is so named because the signal at the cathode "follows" or mimics the signal at the grid.

So if V2b were wired as a cathode follower, then the output from its cathode would be in phase with its grid, which is to say with the plate of V2a. That would leave the grid of V4 having the same signal as V3 grid, so the push pull wo0uld no longer work.

[trem]

10-4
Thanks again everybody.