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Old March 13th 07, 06:00 AM posted to sci.electronics.design,sci.electronics.basics,rec.radio.amateur.antenna
Bob[_2_] Bob[_2_] is offline
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First recorded activity by RadioBanter: Mar 2007
Posts: 3
Default VSWR doesn't matter?

On Mar 12, 4:56 am, Richard Clark wrote:
On 11 Mar 2007 20:39:46 -0700, "Bob" wrote:

The active part of the transmitter output isn't 50 ohm.
That would cause half the power to be lost as heat in
the output stage.


Hi Bob,

Well, aside from the initial misunderstanding of how power gets to the
load (much less back, and then to the load again); I will put to you
a question that has NEVER been answered by those who know what the
transmitter output Z ISN'T:
"What Z is it?"

73's
Richard Clark, KB7QHC


As Tim Williams alludes, it depends on the transmitter design.
It will often be complex rarther than resistive. Since the active
device changes impedance during a single cycle of the RF
signal it may not even be adequately described by a single
value in ohms for a paticular frequency if you wish to
analyse the case of forward and reflected power.

Consider a class C or class E output stage with an
output transistor that is low impedance during
most of the positive half of a cycle of signal and mostly
somewhere near open circuit for the negative half
of the cycle. It seems to me that the effect of reflected
power is going to be different depending its phase
relative to the forward power.
I think this also applys to a lesser extent to a class
A PA with a nice hi-Q tank circuit.

As usually whan this topic comes up, It don't feel
like we have arrived at a usefull and convincing model
of what happens, possibly because simple
descriptions don't cover everything.

Bob