On Jun 6, 4:12�pm, "Antonio Vernucci" wrote:
Someone may regard the following question a bit OT, but as it deals with
impedances I have considered that the antenna newsgroup could be the most
appropriate one where to post it.

Let us regard a transmitter as an ideal RF generator with a resistance in
series. It is well known that, for maximum power transfer, the load resistance
must be equal to the generator resistance. Under such conditions efficiency is
50% (half power dissipated in the generator, half delivered to the load).

To achieve a higher efficiency, the load resistance should be made higher than
the generator resistance, although this would turn into a lower power delivered
to the load (the maximum power transfer condition is now no longer met). This
can be verified in practice: by decreasing the antenna coupling in a
transmitter, one obtains a higher efficiency though with a lower output power.

That said, now the question.

Usually, when a transmitter is tuned for maximum output power, efficiency
results to be higher than 50% (typically 60% for class-B, 70% for class-C).
This would seem to contradict the above cited fact that, under maximum power
transfer condition, efficiency is 50%.

Pertinent comments are welcome.

73

Tony I0JX - Rome, Italy

In addition to Tom's comments, an RF Power Amplifier's efficiency is
defined as (Pout/Pin)X100%. Pout is RF and Pin is usually DC. So if
you pump 1000 watts DC in to a class B RF amp and get 600 watts rms
out you are 60% efficient. Your model of a Voltage Source in series
with an internal resistance does not apply here.

The maximum power theorem gives conditions where power in the load, is
equal to internal power in the generator. Not always a good idea. A
50HZ generator capable of Megawatts of power would dissiapate 1/2 in
the generator and 1/2 in our houses if they designed them to conform
to the MPT. The 50HZ generators would melt. Utilities design their
Generators to have nearly 0.0 ohms internal impedance.

Good question.
Gary N4AST