View Single Post
  #38   Report Post  
Old March 31st 07, 05:11 PM posted to sci.electronics.design,sci.electronics.basics,rec.radio.amateur.antenna
RST Engineering RST Engineering is offline
external usenet poster
 
First recorded activity by RadioBanter: Jul 2006
Posts: 49
Default VSWR doesn't matter? But how about "mismatch loss"?

Bill ...

I would bet that most of the "confustion" comes from the conditions people
put on their answers to the question.

Some postulate a "tuner" between generator and load.

Some postulate a specific internal impedance of the generator.

Some postulate a specific length of feeder line (either lossless or
resistive, which is another parameter in and of itself).

Some postulate lots of other stuff, almost all of which is valid in the
context of their answer.

What exactly do we mean when we say that we have a "100 watt transmitter"?
What we are actually saying is that the transmitter will cause a resistor of
a specific value to dissipate 100 watts of energy when tied to the
transmitter output port and the transmitter keyed. Let's not muddy the
waters up by asking if we are talking about peak power, PE power, average
power, or whatnot. Let's just assume an unmodulated carrier putting out a
constant power into the resistor that gets just as hot as when 100 dc watts
(E*I) is pumped into it from a battery.

What value resistor? Whatever the designer/engineer/manufacturer specifies.
32 ohms? Sure. 50 ohms? No problem. 300 ohms? Certainly. Any competent
engineer can give you a specified power into a specified resistive load.

The crux of the question becomes, "What happens if my transmitter is
specified into a 50 ohm resistor and I put a 100 ohm resistor as the load?
How much "loss" do I get (or another way of asking that same question is how
much power is dissipated in the 100 ohm resistor)?"

The answer is that it is impossible to tell without making the measurement.
That may seem like a "wiggle" answer, but the truth of it is that the output
stage design of the transmitter will dictate how it handles a "VSWR load".
In some output stages, the output voltage will increase to the point of
nearly driving 100 watts into the 2:1 VSWR resistor. Some will shut
themselves down with a protection circuit. Some will go into parasitic
oscillation. Some will fry the output devices.

Now increase the magnitude (and probably the sign also) of the problem to
toss in a complex impedance instead of a resistive load and the confusion
factor goes up rapidly. What DOES that output stage do when the load has an
inductive component? Or a capacitive component? And both a resistive
component and an imaginary component that varies with frequency?

The simple answer to your question outside the "laboratory environment"
where everything is nicely matched and the internal impedances are set "just
so" is that there IS NO SINGLE RIGHT ANSWER to this simple question.

And that is most probably the cause of your confusion.

Jim



"billcalley" wrote in message
oups.com...


But here again, I'm probably not seeing the
entire picture here. What am I missing??

Confused!

-Bill