True, but only in a linear system representable by a Thevenin source having
a resistive component that is dissipative. But an r-f amplifier is a
non-linear system with a non-dissipationless "internal resistance" and
cannot be modeled with a Thevinin dissipative source.

IF the Thevenin source approach worked, we would have to be content with max
50% efficient amplifiers. We know we can do better than that substantially.

The latest QEX has a revealing examination of impedance and conjugate
matching matters.

Bottom line seems to be that one concentrates upon the specific load
*resistance* specified for all other specs to be met. Such a load permits
the amplifier to deliver maximum power within specifications and as such is
conjugately matched to its load.


--
73, George W5YR
Fairview, TX

http://www.w5yr.com




"Richard Harrison" wrote in message
...
John Woodgate wrote:
"The problem is that people say "output impedance" when they mean "load
impedance".

Quite right. I`ll use "source" and "load".

Current through a load depends on the voltage. Ratio of volts to amps is
the impedance. A source with the same resistance and offsetting
reactance to the load enjoys a Goldilocks relationship with its load.
The source`s volts and amps perfectly match the demands of the load.
It`s just right. There`s no surplus of either volts or amps when source
and load are connected. Its a match. Only a matched source and load
deliver all the power available in a source.

If we have too much resistance in our load, it doesn`t take as much
power as it could.

If we have too little resistance in our load, too much power is lost in
our source.

The perfect match of equal source and load resistances, with the
reactance neutralized, is the only condition permitting maximum power
transfer.

Somme amateurs want all the power they can get from their transmitters.

Best regards, Richard Harrison, KB5WZI