Bruce Raymond wrote:

I've seen the equation Po = V^2 / 2R applied to the design
of Class C amplifiers. This doesn't make sense to me and
I'm looking for corroboration, or somebody to tell me I'm
an idiot ;-). The formula makes sense for a Class A amplifier
which has conduction over 360 degrees, but would seem
to overstate the power output for a Class C amplifier with,
say a 120 degree conduction.

An amplifier with a 120 degree conduction angle would
only produce about 47% as much power as one with a
360 degree conduction angle (if I did the math right).
Therefore, I'm assuming that the formula should be
Po = .47 * V^2 / 2R, or Po = V^2 / 4.2R in this case.
Is this correct?


I think the answer depends on what the letter V stands for in the
equation. If it is the peak voltage of a sine wave that rings out of
a tuned circuit (or any other pretty good sine source), then it
doesn't matter how the sine wave was generated. Somehow, the energy
put into the resonator is driving a resistor with positive and
negative peak swings and that dumps
V^2 / 2R watts into the resistor. The instantaneous peak power put
into the resonance must be higher than that, for the class C amplifier
to pump it up to that voltage in a small fraction of a cycle.

If you want a challenge, figure the power out of a class C DC
amplifier.

--
John Popelish