mr2racer,

I see others have chimed in, before I wrote this:

Lets find out about how much power we might get from your amp, based on some measurements, assumptions, estimations, and calculations:

You have 10V into 260 Ohms; 10/260 = 0.038 A cathode current. Most of the 38 mA is plate current, not screen current.

Lets swing the plate current from quiescent (38 mA) down to 0 mA, and then up to 2 times quiescent (76 mA). Yes, the current can and will go larger than 76 mA, but it can not go less than 0 mA, so those two extremes are all bounded by clipping and/or distortion.

The plate load is 5k Ohms. (0.038A squared) x 5,000 Ohms = 7.22 Peak Watts 7.22/2 = 3.61 rms watts. But we need to check further . . .

0.038A x 5000 Ohms = 190V peak swing We have 10V cathode self bias, the control grid is at 0 Volts. We could swing the control grid up to 10V before drawing grid current (distorting the signal, and or changing the effective grid to cathode bias, the coupling cap will start charging with signal).

190V / 10 V = 19V. That means the output stage with negative feedback applied needs to have a gain of 19. Therefore, the driver stage needs to swing + 19 V and - 19 V from quiescent (with the shade feedback applied). You could easily check to see if your driver swings that far, if you have a scope.

If you can not make that measurement . . . In order to estimate how far it can swing, I need some more information: The driver tube type (and if it is a paralleled pair) The driver tube cathode resistor, and bias voltage across the resistor, and whether or not the resistor has a bypass capacitor. The driver tube plate load resistor. The Schade feedback resistor. The output tube control resistor (control grid to ground).

Then I should be able to estimate the gain, and the dynamic swing, and the signal required at the driver stage control grid.