Push Pull phase - operation

[lowell]

I just wanna clear something up.. for myself. The more theory I learn the more I realize I assume or take things for granted. Such as this. I've known that a push pull amp works because the 2 power tubes are on opposite ends of the center-tapped OT and each tube is driven with opposite phase signals from a phase inverter, AND that the OT recombines these signals to produce the full cycle and deliver it to the speaker. I just made some sketches to try to clear this up for myself and I think I have but I want to confirm that I'm right. I think I am as this is the only way I can see it working.

As of yet I've thought that 1 power tube amplifies the positive going cycle and the other power tube amplifies the negative going cycle. I now think that this is false. The problem that I'm finding here is that how can your typical power tube plate, sitting almost at B+, amplify a negative going cycle on its grid? I believe that both power tubes are amplifying the positive going cycle (on the grid), but they alternate doing just that. Then it's the OT that converts these same cycles back to a full cycle, because the OT windings are of opposite phase. I almost look at it this way. If one were to scope both power tube plates at the same time, they would see negative pulsating unfiltered DC.

??

[R.G.]

Good question.

What's really happening is that each tube can only pull current through the OT in one direction: from B+ toward ground. A positive going signal on the grid causes more current to flow in the tube, and pulls that tube's end of the primary down from the B+ in the middle towards ground. Notice that the grid is sitting at some negative bias voltage, and a positive going signal on that grid means it heads up from the below-ground bias voltage toward ground; I mention this only to keep the voltage senses straight. It's always "more positive grid, more current through the plate".

If you have two tubes fed from a phase inverter, one is fed a positive going signal and the other grid gets a negative going signal from the phase inverter, and every half cycle, the PI switches which one is getting the positive going signal at any instant. In class AB, the tubes are biased almost off for the no-signal condition. When one of them gets a positive going grid signal, it pulls current in the OT, and the other one gets a negative going signal that turns it more off; at some point its plate current hits zero, and stays zero until its grid voltage comes back. It doesn't matter HOW off it gets turned, it stays off.

So the active tube plate pulls down on its half of the OT primary. The primary being an autotransformer, the active, current carrying side gets pulled down and the inactive, non-carrying side gets pulled UP above B+ by an amount equal to the other side being pulled below B+. It's a little like kids playing on a see-saw, but only able to pull their side down with a rope, never push it up.

The pull down causes the flux in the transformer iron to be increased in one direction on one tube's pull, and then in the other direction when the opposite tube pulls. The magnetic field in the transformer iron is where the tube signals mix. The secondary can see nothing of what the tubes do, only what they cause the magnetic flux in the iron to do.

[Jazz P Bass]

You have hit on the reason for the bias voltage.
Yes the phase inverter turns each tube on & off.
Each output tube is (ideally in class B) on the verge of conduction.
Top of the phase inverter goes high.
Tube #1 conducts.
Tube #2 is waiting.
Bottom phase inverter (hint) turns on, on the negative portion of the sine wave.
Inverts it, turns on tube #2.
Tube #1 is waiting.
Setting the bias voltage , among other things, gets rid of the "crossover" notch when one tube turns off & the next tube turns on.
You really only need enough to get rid of the notch.
(Internet bias setup experts not withstanding)

[lowell]

Ok thanks for the replies... so I'm right. Glad I have that ironed out in my brain for good.

[Wilder Amplification]

It's positive going current through both tubes. The phase inverter just makes the tubes pass current through the OT at opposite times.

The phase inversion of 1/2 the sine wave occurs in the output transformer primary due to 1/2 the primary being wound backwards relative to the center tap, which is the primary current source. This causes the magnetic polarity of the coil to also be inverted from the other. This is why the tubes have to conduct current at OPPOSITE times...if they pass current at the same time, both magnetic fields being of opposite polarity and equal in amplitude would cancel each other (common mode rejection).

Of course there's no such thing as a 100% symmetrical output so they wouldn't completely cancel...you'd end up with the DIFFERENCE at the output, hence the name "Difference Amplifier".

From start to finish it's wound in the same direction...but from center tap to start and center tap to finish you'll find that one side is backwards from the other.

It's just like a power transformer with a grounded center tap and how each end of the winding is of opposite phase from the other relative to the center tap. As Steve Connor once said...a push pull output section is just a full wave center tap rectifier circuit flipped upside down and backwards.

The interesting is that with all this phase reversal going on they STILL manage to get the OT in phase with the speaker.

[MWJB]

The phase inverter doesn't control when either tube turns off/on, it just splits the preamp signal in to two signal 180deg out of phase - the bias condition of the tubes is the primary factor here. You can have push-pull operation without a phase inverter, it's called self split push pull, or SIPP.

[mooreamps]

Of course there's no such thing as a 100% symmetrical output so they wouldn't completely cancel...you'd end up with the DIFFERENCE at the output, hence the name "Difference Amplifier".

Difference Amplifier (Subtractor)

-g

[Wilder Amplification]

The phase inverter doesn't control when either tube turns off/on, it just splits the preamp signal in to two signal 180deg out of phase - the bias condition of the tubes is the primary factor here.

Correct....but driving it with a differential input ensures that they're not passing current at the same time "in phase" (putting aside the "overlap" region of Class AB).

You can have push-pull operation without a phase inverter, it's called self split push pull, or SIPP.

Yes...where you set the input to the power valves up similar to a LTP. However I think your only option with that setup is to cathode bias.

[MWJB]

Jon Wilder wrote: "Yes...where you set the input to the power valves up similar to a LTP. However I think your only option with that setup is to cathode bias." Indeed, but as you know, methode of bias doesn't directly relate to how much of the cycle is handled by each tube. Magnatone did have a screen coupled SIPP amp, which allowed the cathode to be bypassed, I did briefly look into fixed bias SIPP, I wouldn't rule it out as "impossible". Its more that previous designs have been based around entry level amps, with a low parts count, that utilise tubes that run fairly efficiently in SIPP. Final wattage is well down on PP amp with a PI tube, so it's unlikely that a high powered, fixed bias, SIPP amp would be pursued due to high cost & impracticality.

[Wilder Amplification]

Indeed, but as you know, methode of bias doesn't directly relate to how much of the cycle is handled by each tube.

Correct...the margin between cutoff grid voltage and the idle negative voltage at the grid in quiesent state determine that. The tube could care less where it was derived from.