[Boatanchors] Dipping the plate current at resonance

[Jim Haynes]

I'll take a stab at the question.

We're mostly talking about the classical Class C amplifier, not the
modern linear.  Ideally the tube in the Class C amplifier is acting
as a switch, thus a short circuit for a small part of the cycle and
an open circuit for the rest of the cycle.  Since it isn't really a
short circuit when conducting there is a limit to the  plate current.

When the L-C circuit is completely unloaded and is at resonance
then its impedance is very high.  It takes very little plate current
to keep an oscillatory current going in the L-C circuit, just enough
to overcome the losses.  When it is tuned off resonance it will
look either inductive or capacitive.  Suppose it is capacitive.
Then the tube runs extra current to charge the capacitor when the
tube is switched on; and when the tube is switched off the
capacitor discharges into the inductor, but because of losses the
oscillation dies out right away.  So the tube is having to recharge
the capacitor every time it turns on, and this adds to the idling
current.

Treating it as a transient circuit rather than as an AC circuit, we
can note that at resonance the tube also has to charge the
capacitor when it is switched on, but it is helped by current
coming from the inductor at just the right time to minimize the
amount the tube has to contribute.  Whereas off resonance the
current from the inductor doesn't come at just the right time;
it is either early or late; and the tube has to work harder to
charge the capacitor.

Then as loading is increased the power that goes out the antenna
is power lost by the tuned circuit, so again the tube has to work
harder to charge the capacitor because there is less energy
stored in the inductor to help out.




jhhaynes at earthlink dot net