Richard, your method of measuring internal impedance of an amplifier sounds
interesting. (o/c volts divided by s/c current)

At present I have no facilities to make such measurements.

Perhaps a few curious readers, who do have facilities, might make crude
measurements on their HF rigs and report approximate impedance values on one
or two bands on this newsgroup. Some sort of average could be obtained.
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Reg, G4FGQ
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"Richard Harrison" wrote in message
...
Asimov wrote:
"Let`s look at it from the dynamic point of view (loss in a Class A
amplifier)."

The no-signal loss of a Class A amplifier is 100%. It equals volts x
amps and appears in the amplifier. Now feed a signal to the ideal
amplifier set just below the clipping level. Average d-c power is
unchanged from the unloaded and no-signal conditions.

Connect a matched load resistor to the amplifier output. If physically
small, the resistor may become warm with heat that were it not for the
load would be otherwise dissipated in the amplifier. Input power to the
Class A amplifier is unchanging.

Finding the internal resistance theoretically is simple. It is simply
the open-circuit output voltage divided by the short-circuit current.

Open-circuit voltage at full output and short-circuit current may be
severe. Internal resistance can be found under less stressful
conditions. Internal resistance will drop the voltage to any load
reasistance. Use the voltage-divider formula to calculate the internal
resistance.

With pure resistances, half the open-circuit volts are dropped by the
internal resistance when the load is a match.

A power amplifier`s internal impedance can be determined.

Power output from a Class A amplifier cools it.

Best regards, Richard Harrison, KB5WZI