Ralph Mowery wrote:
"Sal M. Onella" wrote in message
...
This group has presented members with valuable lessons in antennas and
transmission lines, like how to measure, how to match, etc.

Something I haven't seen is a discussion of the source impedance of
the transmitter. My curiosity was piqued today as I took some baby
steps into EZNEC. A particular antenna had such-and-such VSWR if fed
with a 50-ohm cable and a different value if fed with a 75-ohm cable.

While this is hardly news, it got me wondering whether a 75-ohm cable
will load the transmitter the same. Doesn't seem like it.

My point: Using 75-ohm cable to improve the match at the antenna
won't help me ... IF ... I suffer a corresponding loss due to
mismatch at the back of the radio. My HF radios, all solid state,
specify a 50 ohm load. As necessary, I routinely use an internal
autotuner and either of two external manual tuners. (I'm aware of the
published 1/12 wavelength matching method.)

Wisdom in any form would be appreciated. Thanks.

"Sal"
(KD6VKW)

A transmitter output impedance is designed for maximum power transfer at a
specific impedance. Most of the the older tube transmitters impedance was
tunable within a range.

In simple terms the impedance of the transmitter tube is the plate voltage
devided by the current. This impedance is then transformed to the nominal
50 ohms of the antenna system. If the transmitter has the usual tune and
load controls, the exect impedance will not mater as you adjust for maximum
transmitter output.

Most of the transistor transmitters are not adjustable so the output
impedance is usually fixed at 50 ohms for maximum power transfer. If the
impedance of the antenna system is not 50 ohms, then the output power will
be less than the designed output. You can use the antenna tuner to adjust
for a match.


Not exactly..

A "match" provides the optimum power transfer from generator to load,
but that is NOT the maximum load power, nor is it either the maximum or
minimum power dissipated in the source.

Say I have a zero output impedance on my source and I'm putting out 7
Volts RMS into a 50 ohm load. That's about 1 watt into the load. Now..
if I reduce the load impedance to 25 ohms, and since I've got zero
output impedance, I'm now putting out 2 Watts. The source impedance is
zero, so I'm not dissipating any extra power in the source, either.

It is true that a "matched load" to my zero ohm source would, in fact,
be zero ohms, and would have infinite power. Any other load impedance
would have less power into the load, so the Thevenin theorem is satisfied.


Now.. if my generator had a fixed output impedance, it's true that the
load impedance that will get the most power out is the conjugate of the
output Z. For resistive sources/loads, here's an example..

You also have to be careful about looking at Thevenin equivalent sources
(e.g. a ideal voltage in series with a Z, or a ideal current in parallel
with a Z), because just because *the model* has a resistor in series
does NOT mean that you're actually dissipating any power in the source.
(If I had a very efficient op amp, I could simulate any arbitrary output
impedance, without dissipating any power in the source)


Say my generator is 40 ohms, and I'm putting out 7 Volts into a 40 ohm
load. OK, that means that the imaginary voltage source is putting out
14 V. I'm getting about 1.23 Watts into my load. Now, if I decrease my
load Z to 20 ohms, what do I get? Now, I have 4.67 (=14/3) Volts
instead of 7, and I get 1.1 Watts. Yep, less.. Thevenin works. Let's
try increasing the Z to 60 ohms.. Now the voltage on my load is 8.4 V,
and I'm dissipating 1.18W, again, less than my 1.23.

But here's some weird stuff.. let's look at how much power is dissipated
in that imaginary resistor (i.e. our source *really is* a ideal voltage
source in series with a resistor)

At 40 ohm load, we've got 7 volts on the load and 7 volts across the
resistor, so they both dissipate the same 1.225 Watts. Pload/Pgen = 1

In the 20 ohm load case, we've got 1.1 dissipated in the load and 2.2
dissipated in the generator. Pload/Pgen = 0.5

In the 60 ohm load case, we've got 1.18 dissipated in the load, and 0.78
dissipated in the generator. Pload/Pgen = 1.5 (i.e. we dissipate more
in the load than in the generator... how about that!)

And let's look at "efficiency" of the system (assuming that the total
power in is the sum of what's dissipated in the generator and the load)

20 ohm load, 33%
40 ohm load, 50% (what you'd expect)
60 ohm load, 60% (hey.. it's more efficient, too)

- take home message... a "good match" is sort of an artificial thing
from a power transfer standpoint.. it depends on what you're trying to
optimize for.

- Where you get bitten is when "match" varies with frequency... now,
all of a sudden, you have a system that has a response that varies with
frequency, which is generally undesirable. When you get up into the
microwave region, where a transmission line is often many wavelengths
long, that mismatch can result in remarkably wild fluctuations in gain
with respect to frequency.