VSWR doesn't matter? But how about "mismatch loss"?
Roy Lewallen wrote:
An understanding of "mismatch loss" doesn't require SWR, reflections,
power waves, "reflected power", "reflected energy", or other real or
imagined complexities other than simple impedances. Here's what it means:
If you have a generator with a fixed output impedance such as a signal
generator, and connect it to a conjugately matched load, the power
dissipated in that load is the most you can get in any load connected to
the generator. For example, if your generator produces 10 volts RMS open
circuit and has a 50 ohm resistive output impedance, it can deliver 0.5
watt to a 50 ohm resistive load. If you connect any other load impedance
to the generator, you'll get less power to the load. You can calculate
exactly how much with simple circuit theory.
"Mismatch loss" is simply a way of expressing the reduction in power you
get due to the load being mismatched, compared to how much you'd get
with a matched load. For example, if you connect a 100 ohm resistor to
the output of the generator, it would dissipate 0.44 watt instead of
0.5, so the mismatch loss is 10 log 0.5/0.44 = 0.51 dB(*). If you
connect a 25 ohm resistor to the output, you also get 0.44 watt in the
load resistor, again a "mismatch loss" of 0.51 dB. These numbers are
calculated using nothing more complicated than simple lumped circuit
principles.
Mismatch loss is a useful concept when connecting fixed-impedance
circuits together, such as in a laboratory environment. But it doesn't
apply to either antennas or to VSWR. All you have to do to reduce the
"mismatch loss" to zero is to insert a tuner or other matching network
between the generator and the load. Presto, the generator sees 50 ohms
resistive, the load dissipates 0.5 watt, and the mismatch loss is zero.
Bravo. And a great deal simpler to understand than most handwaving on
these threads.
Roy Lewallen, W7EL
Best, Dan.
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