Bruce Raymond wrote:

John,

Thanks for the reply. I'm actually interested in the version
of the formula (Rl = Vcc^2 / 2Pout) listed on page 2.33
of "Experimental Methods in RF Design", by Wes Hayward, ...
(excellent book).

In the example Vcc is 12 volts and Po is 1.5 watts. Rl
works out to 48 ohms. Wouldn't the peak current be
12 volts/48 ohms = 250 ma? If this were a sine wave then
the RMS power would be 1.5 watts. However, the amplifier
is run as Class C and only produces output less than half
of the time, so the output would then be less than 0.75 watts.

I guess this assumes that with a 12 volt supply, a resonant load could
produce almost a 24 volt peak to peak sine wave. A fair
approximation.

When the class C amplifier comes on, it sees an impedance a lot lower
than the 48 ohms on the tank. The amp loads the tank with all the
energy it will lose in the next cycle, but the tank meters this energy
to the load in sinusoidal form. A piston in an internal combustion
engine puts out more peak torque than the flywheel delivers to the
transmission by the same mechanism. But the average power put out by
the piston is the same as the the average power delivered by the
flywheel to the transmission.

It seems that the formula should be adjusted to account
for the conduction angle, e.g. Rl should be smaller by at
least a factor of 2 to compensate for the conduction
angle. What am I missing?

Energy storage in the tank.

--
John Popelish