My 2 cents (wont even buy a lolipop any more!) From the old VHF,er
magazine, couple of thoughts: 1) Very lossey coax will show a prefect
termination impedence even if open, or shorted, if sufficiant length is used
(for the coax's attenuation ) . For that reason, still keep a coil of RG-8
size coax (for power dissipation), with RUBBER DIALECTRIC (was origianlly
made for attenuators at the Microwave Region) Even at 80 meters, with no
load , has very low SWR! At 10 meters, and up, with a UHF connector on one
end, and a N connector, on the other, with NO TERMINATION, makes a great
dummy load to 2-300 watts!. and, 2) Just because you think you have a great
match, can be a random event. to test , place a 1/4 wavelength piece of
coax (figuring with velocity factor). If the SWR stays flat, you can assume
a good match, if, howerer, the SWR climbs, then you need to check for a
proper match ( this will throw the worst condition into your coax , with
this extra length, kinda like the opposite of trimming coax for the best
match legends of old CB lore)! as info, Jim NN7K
"Tom Bruhns" wrote in message
m...
(Dr. Slick) wrote in message
. com...
...
What do you mean by "calibrated to the line"?
The SWR meter should read zero reflected power when connected to a
load whose impedance is equal to the line's. Does it? If not, it's
not properly calibrated. Putting it another way, what's the
directionality of the bridge?
...
Not so much a surprise as a disappointment! A difference of 70
watts incident power is totally unacceptable with only 8 feet of coax
length added.
But the load presented to the amplifier is totally different in the
two cases, most likely. Only if the amplifier's output impedance were
the complex conjugate of the line's, and the line were lossless, and
the amplifier behaved as a linear time-invariant system would you
(should you) expect the power to remain unchanged.
The transmitter does not have a mismatch sensor on it for reducing
power at high SWR. It naturally does output less wattage as the
transconductance is reduced with higher temperatures, but my
measurements were done in close proximity to each other.
Exactly how would you measure the output impedance (S22) of a PA?
Drive it with a small signal (class A, linear mode) in one frequency
(F1), and then inject another frequency (F2) into the output, and then
measure the reflected F2 power? If i remember correctly, that's how
someone was trying to measure S22 at my former place of employment.
But all bets are off as soon as you go non-linear, or non-class A.
S-parameters are supposed to be all small signal.
I don't think you can properly measure the S22 of a class C PA.
Tam brought up the suggestion that i try a really long piece of
RG-58 from meter to dummy load, to make the Cantenna more like 50
Ohms. Gonna try it.
Or just tune the load to zero reflected power after you're sure the
meter is properly calibrated...but that can be a catch-22 situation
that the really long RG-58 can help with. Be aware, though, that
"50-ohm" line seldom is -- it can be off 5 ohms or more. It's a cause
of some consternation to those of us involved in calibration of
precision RF test equipment.
Load-pull techniques are commonly used to characterize RF source
impedances. You make known incremental changes to the load, and
deduce from the change in output power what the source impedance is,
assuming it's a linear time-invariant system. Note that adding length
to a mismatched line is one way to make an incremental change to the
load... I wouldn't necessarily say that the output impedance of a
class C amplifier is meaningless, but it may well not be constant for
all loads. and may depend on parameters you'd have trouble controlling
from day to day.
Cheers,
Tom