On Thu, 25 Dec 2003 19:18:17 GMT, Robert Casey
wrote:

Let's ignore, for the moment, the losses involved in the conversion of
DC power to
RF power.

Let's ignore the R resistor, and then we can prove it doesn't exist?
:-)

Now lets say we have 100 watts of RF. If the antenna is a
perfect
load (50 ohms resistive) and if the Thevenin impedance of the transmitter is
50 ohms, then yes, you got 50 watts of extra heat in the transmitter.

See? Your logic failed at the gate. You ALREADY have 100W RF as a
premise. :-)

How did you arrive at 100W without its measure AT the Load? To force
the speculation that the internal resistance (neglected but evident by
such logic) drops it? You are separating those things that are
inseparable and this is the common fault of all speculations that look
beneath the hood of the Thevenin Model.

Now if the
transmiter has a very low Thevenin impedance, then more power is delivered
to the antenna and less waste in the transmitter. I'm not trying to do
the "transfer
the max power to the load and I don't care how much waste in the source"

This is another forced argument. No where, until now, was it offered
that maximum power was being delivered or even demanded. We are
simply observing what IS. With the efficiency sitting at 40% it is
painfully obvious from the beginning that the road of maximum transfer
has not been tread upon - EVER.

Thevenin
thing we had in EE101. If I did that, and the amp is up to it, I could
transmit
even more power to the antenna, but I'd waste more in the source.

Brush up on your EE101 to discover that the Thevenin Model is derived
from observables. I have already described ALL observables that are
consistent with opening the model to look under the hood.

There have been many writers here, over the years, who have offered
bench tests that prove this by observation. As of yet, none has been
toppled with work equal in quality (i.e. demonstration or
measurement). It is no more simple that plunking the load into a
bucket of water, plunking the transmitter into a bucket of water,
hitting the transmit switch and measuring temperature. The
transmitter will release more heat than the load.

To save yourself the issue of submerging the source, common practice
(as described by Thevenin) would have you measure the voltage across
the source, and the current into it. Yields identical results. I've
employed the HP Caloric Wattmeter for years at the Metrologist's bench
to faithfully validate this concept over and over.

In any event Merry Xmas.

Season's Greetings
Richard Clark, KB7QHC