Conventional wisdom has it that a flaw in the Princeton Reverb is that the PI distorts before the power tubes do. I spent some time this afternoon with my oscilloscope but am not sure how to interpret the results.
Here's a Gallery of 7 o'scope shots which you can look at either individually or as a slideshow. [edit 10/4/10 - added 4 shots of the PI discussed below] The first four shots have the phase inverter grid (V4 pin 7) on CH 1 and a power tube grid (V5 pin 5) on CH 2. At first I thought "this proves it," but I'm wondering if the flat top on CH 2 is PI distortion or the power tube shutting off as the amp transitions to Class AB. ...
Maybe I'm missing something... Isn't the power tube grid on the lower trace? I see it distort first.
So my next question is what is the peak voltage when the bottom trace begins to show distortion, and how does that peak voltage compare to the bias voltage. *If* the peak voltage equals the absolute value of the bias voltage (e.g., 34v peak, -34v bias), then you are seeing the 6V6 grid being driven positive and conducting rather than being a super-huge input impedance.
EDIT: The point behind the above paragraph is that if the grid is driven positive and starts to conduct, then distortion viewed at that grid could be the result of grid current, and not actual distortion in the phase inverter. Grid current doesn't necessarily happen all at once either, but once the peak input equals the bias, it becomes very severe very fast. And in a case of the serpent eating its tail, if the output tube draws grid current, the load represented by the grid becomes very small, and the phase inverter could be pushed to distort by trying to drive a load smaller than it can handle.
You could measure what is happening with the output tube current by inserting a 1 ohm cathode resistor and measure millivolts across this resistor to watch the current through the output tube (or use 10 ohms, remembering to multiply voltage readings by 10).