Theory behind PI problem(?)

[Geezer]

Hey guys! I've been busy with other stuff lately & not had any time for fun, but will soon be back to "normal" & get once again into the amp swing!

However, a friend amp tech had a question that I thought I'd throw out to you guys for opinions.

Here's his notes:

I did a successful repair on a Bandmaster AB763 the other day which left me a little confused about the theory of what was wrong. Didn't know if you might have interest in shedding some light on this. Attached is a drawing of the phase inverter, showing symptoms and fix.

When the amp came in, it sounded awful. A quick signal check showed
that no signal was being delivered to one of the output tubes. Next a
voltage check on the plates of the phase inverter. DC voltage on one
plate was way high.

My reckoning was that the plate voltage was high because there was no
complete circuit through the triode, hence no current flow. Replacing
the .1 capacitor brought voltages back where they belong and the amp
sounded perfectly fine.

Here's where I'm confused: The capacitor that was replaced would
apparently affect DC idle current in the triode circuit. But why, or
how? Voltage on the plate is DC and the capacitor doesn't have much
effect on the DC current flow. Or does it?

It seems like the lack of idle current would indicate no voltage on
the grid, and that was the case. So, with no voltage on the grid, why
was the triode in cutoff?

Any help appreciated, Geezer

[sluckey]

There is normally 108.5Vdc on that grid and 110Vdc on the cathode. That leaves the grid about 1.5Vdc negative with respect to the cathode. That's a typical bias voltage for that tube.

The bad cap must have been shorted causing the grid to now be at 0Vdc. It may have been a few millivolts just because of the 100 or 47 ohm NFB resistor. Anyway, the cathode is still at some big positive voltage due to current flowing thru the other half of the tube. But with 0Vdc on the grid with the shorted cap, the grid is now many volts negative with respect to the cathode. This will bias the triode well into cutoff, no current will flow, and the plate voltage will rise to B+ supply level. And no signal will pass. The other triode will continue operating, although it may not be optimum.

[Geezer]

Thanks Steve!

That makes perfect sense to me.....excellent explanation!