Now what about power output? How do we determine that?
In Radiotron Designer's Handbook (
RDH4), it is shown that the power output of a push-pull pentode output stage can be calculated by by knowing the peak current maximum of one side of the output stage (which happens at the knee of the 0v grid curve), and the supply voltage and plate voltage where the loadline intersects the knee. See the graphic below.
The diagonal line AB is the loadline of the output stage, and (because this is a class AB stage) represents the OT primary plateto-plate impedance divided by 4.
Point B is at "Ebb" which is RDH4's shorthand for "supply voltage". Point A is at the knee of the "Ec1=0" curve, which means the 0v gridline from a set of plate curves, and which is the point reached on the loadline when the peak input signal driving the tube just equals the bias voltage (because 18v peak signal plus -18v bias equals 0v at that instant).
"Ibm" is RDH4's shorthand for "plate current, maximum", which happens at point A on the loadline. "Emin" is shorthand for "minimum plate voltage" which also happens at point A.
The equation for power when you know all these quantities is:
Po = 1/2 * Ibm * (Ebb - Emin)
Now if you stop and think about voltage, current and
resistance, you may realize that Ohm's Law says they're all linked, all the time. In fact, the diagonal line AB represents the resistance of the circuit (in this case, 1/4 OT primary impedance), and its slope is the magnitude of the impedance. More-vertical equals less impedance (which should make sense, because for the same voltage change you get more current change, which is what Ohm's Law also tells you).
We don't have curves for the 6V6 and 6L6 at 290v on the screen, so what do we do?
Let's assume we again use our 5E3-type amp and the same OT exists regardless of whether we plugged in 6V6's or 6L6's. Let's also assume the powe supply voltage doesn't drop with the 22mA extra total plate current the 6L6's are pulling.
Given these assumptions, point B on our diagram doesn't change. Also, because the same OT (and same OT primary impedance) is being used, the
slope of AB doesn't change. The only variable that can change output power, then, is if the 6L6's can pull their plate a little closer to 0v so Emin is a smaller quantity.
If you look at the bottom graph of page 4 of the 6V6 data sheet, and the top graph of page 4 of the 6L6 data sheet, you'll see a knee for G2=250v happens at ~35v for the 6V6 and ~50v for the 6L6. So it would seem the 6V6 is more likely to output more power with G2=290v.
Now that presumed the OT was the right load for the 6V6 to make maximum output power...