William E. Sabin wrote:
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


This discussion assumes that we want the 100 W PA
to actually deliver 100 W to the coax. Assume a
Bird model 43 wattmeter in the line. If the PA is
actually delivering 100 W to the coax, then the
forward power must be 103 W and the reflected
power must be 3 W. The PA output power is
103-3=100 W. This is how we use the Bird
wattmeter. The Bird instruction manual tells us this.

Keeping everything simple and not getting into
peripheral issues is desirable at this point. For
example, a circulator will dissipate the 3 W, but
the above discussion does not assume a circulator.
The circulator's job is to force a 50 ohm load on
the PA, despite the fact that the coax input
impedance is not 50 ohms.

Bill W0IYH