flemingmras wrote:
Aiken, this brings up some interesting thoughts here. Some math I was playing around with earlier...not sure if it's correct and if it isn't please feel free to correct it -
Assuming a 100% efficient zero loss output transformer and single frequency test signal conditions into a resistive load -
With each 1/2 of the OT primary being 400 ohms (roughly), you would need 500mA draw through that 400 ohm load, which would in turn create a 200 volt drop across the 400 ohm load, to get 100 watts on 1/2 the OT primary, which then would get transferred to the secondary.
However, no transformer is even close to 100% efficient, which means that you'd have to be feeding MORE THAN 100 watts of power in to get 100 watts out, which would require more current and therefore a higher voltage drop across each 1/2 of the OT primary.
And of course, as current flow through the tube & the load increases, so does the voltage drop across the load, which causes the voltage drop across the tube (from cathode to plate) to decrease as current flow increases. Effectively, the PI grid drive is modulating the cathode - plate resistance of the tube at input signal frequency. As plate - cathode resistance decreases, plate/load current increases and hence the decrease in plate voltage, which is what puts your plate voltage/current out of phase. Effectively you have the exact same thing happening in the bias supply right at the bias pot as you adjust it.
Am I correct in thinking this?
First, I think your math is a bit wrong because you didn't take into account RMS for power, and you're using the wrong impedance. The way to look at power is this:
The plate signal must swing along the loadline, so the plate voltage swing will run from roughtly 480V at cutoff, down to around 50V at the point where it clips at the zero volt grid curve. This gives a voltage swing of 430V peak (not peak-to-peak, because we are only looking at one half of the sine wave in a class B, or cold-biased class AB amp).
Next, you look at the current - we have a current swing of roughly 500mA for that voltage swing.
Now, these voltages and currents are peak, so we have to convert them to RMS, which means dividing by a factor of sqrt(2), or 1.414.
You could multiply the RMS voltage by the RMS current to get power (the square roots cancel, so you just divide by two): P = (430V*500mA)/2 = 107W
Since we can determine power using P=I^2*R, the power for one tube pair will be (500mA/1.414)^2*850 = 106W.
Of course, this is assuming an ideal tube, ideal transformer, etc., so you won't get that in real life.
RA