Why is this Princeton style amp not loud?

[Sonny ReVerb]

I built this amp with blackface Princeton and brownface Vibrolux channels and a 7591 output section. This is the second version of this amp I've built. The first one is a great amp. This one has a bigger power transformer and a 2x10 cab. It's supposed to be louder than the first one, but it's not. Both channels are relatively low volume until you get to about 7-8 on the volume knob. It starts to get loud at that point, but it's no longer clean. This amp should be 35-40W. I have a Vibrolux Reverb that is very loud at 3-4 on the volume. I know a LTP PI would provide more drive, but the Princeton PI should provide plenty of drive for 7591's. I need another set of eyes to check my work.

[Sonny ReVerb]

Some pics

[sluckey]

The pic that shows the actual PI circuitry does not match the schematic PI circuitry. Do you have a layout drawing?

[Sonny ReVerb]

Thank you, Sir. I tried this idea I found on TDPRI for a fixed bias cathodyne phase inverter and forgot to document it in the schematic. Could be part of the problem...

https://www.tdpri.com/threads/fender-cathodyne-phase-inverters.755580/post-7796034

[tubeswell]

Thank you, Sir. I tried this idea I found on TDPRI for a fixed bias cathodyne phase inverter and forgot to document it in the schematic. Could be part of the problem...

https://www.tdpri.com/threads/fender-cathodyne-phase-inverters.755580/post-7796034

^that^ cathodyne grid is set at 42.5V, which will significantly reduce clean output voltage and cause early clipping (the cathode voltage will not stay at 56V - it will automatically drop to follow the grid and this will cause clipping in this stage as the amp is cranked) because it will muck up the B load-line.

[wsscott]

Tubeswell-how did you calculate that number? I’d like to learn something. Thanks

[sluckey]

Tubeswell-how did you calculate that number? I’d like to learn something. Thanks

It's a simple voltage divider that can be solved with Ohm's Law. You have 250V applied across a 3.3M in series with a 680K. Solve for the voltage dropped across the 680K and you have the grid voltage.

R_t = 3,300,000 + 680K = 3,980,000Ω
I_t = 250/3,980,000 = .000063A
Voltage across the 680K is .000063 X 680,000 = 42.8V

There is no grid current so no voltage is dropped across the 1M or 470K so that 42.8V will also appear on the grid.

[wsscott]

Sluckey-Thanks.  I had thought that the grids on the preamp tubes were supposed to be at or near 0 VDC, and maybe similarly for the cathodes.  So seeing these values of 42.8VDC on the grid and 56VDC on the cathode has gotten me confused.

I know I'm missing a lot about operation and theory, but is it different for the grids and cathode on the phase inverter part of the circuit?  Or is it something else that's going on?

So what's the fix for his problem that you describe?

[sluckey]

The actual voltage on the grid is not that important. It's the voltage ***BETWEEN*** the grid and cathode that's important. Go read some Merlin pages.

So what's the fix for his problem that you describe?

The fix is to revert back to the original bootstrap bias circuit used in the original PR.

[wsscott]

Thanks.  I'll check out his website.