Dilon Earl wrote:

Bill;
Thanks, that all makes sense. Can you consider a Transmitter to
have an internal resistance like the generator that changes with the
plate and tune controls?
If I have a 100 watt transmitter and my wattmeter shows 3 watts
reflected. Is 3 watts actually being dissipated in the tank and final
PA?


No.

If the transmitter output is 100 W and the
reflected power is 3 W, then the 100 W is the
difference between 100+3=103 W (forward power) and
3 W (reflected power).

The question "where does the reflected power go?"
never seems to have an acceptable answer. Very
strange.

A good way to look at is as follows: The junction
of the transmitter output jack and the coax to the
antenna is a "node", which is just a "point" or
"location" where the jack and the coax meet. At
this node the voltage is exactly equal to the
voltage output of the amplifier (VPA) and also the
voltage across the input of the coax (VCOAX). The
voltage VCOAX) across the coax is equal to the
phasor sum of a forward voltage wave that travels
toward the antenna and a reverse voltage wave that
is traveling from the antenna backward toward the
transmitter.

Also, at the node, IPA is the current from the PA
and ICOAX is the phasor sum of a current wave that
travels to the antenna and a return current wave
that travels toward the transmitter.

At the node, the IPA current and the ICOAX current
are exactly equal and in the same direction
(toward the antenna). At the node the IPA current
is equal to the ICOAX coax forward current minus
the ICOAX reflected current. In other words there
is an *EQUILIBRIUM* at the node between VPA
voltage and VCOAX voltage, and an *EQUILIBRIUM*
between IPA current and ICOAX (forward and
reflected) current.

This explanation accounts for everything that is
going on at the node. The answer to the question
"where does the reflected power go?" is the
following: "It is a nonsense question that has
caused nothing but misery". The reflected power
does not actually *GO* anywhere. The correct
answer is that forward and reflected coax waves
always combine precisely and exactly with the
voltage and current that is delivered by the PA.
The voltage and current at the junction are
correctly accounted for. The basic principles here
are Kirchhoff's voltage law and Kirchhoff's
current law, as applied to the node. You can study
Kirchhoff's laws in the textbooks.

If we apply these laws and calculate the 100 W
power out of the PA and the 100 W power that is
dumped into the coax, they are exactly equal. They
cannot possibly be unequal. The power delivered is
the real part of the product of VPA and IPA (100
W), which is identical to the real part of the
product of VCOAX and ICOAX (100 W).

Observe carefully the following: We do not need to
know anything about the PA and its circuitry. The
PA is nothing more than an anonymous "black box".
In other words, any 100 W (output) PA will perform
exactly as I have described.

Bill W0IYH