Putta 100K shunt across the coax line input .... letsee what that
looks like: 100K in parallel with a 50 ohm load looks like a 49.98
ohm load, so don't look like much of a load problem. If you run
a full KW the voltage across the 49.98 ohm load will be found from
V=SQRT(1000*49.98)= 223.56 volts. And so 223.56 volts across the
100K will get you I=223.56/100K = 0.0022 amps = 2.2 ma. If you
have a mismatch SWR=50 at the input of a tuner, then the voltage
across the 100K when running 1KW will be 223.56*SQRT(50)=1580 volts
and a corresponding current of 1580/100k = 0.016 = 16 ma. In
the matched case the resistor dissipation needs to be more than
(0.0022^2) * 100K = 0.48 watts and in the bigo mismatch the dissipation
requirement becomes at least (0.016^2)*100K = 25.6 watts.
So, since I have been doing this for years with success, I ask the
question
just asked me by one wiser than I (I never thunk of it!) ...
If this is such a great idea, then WHY don't those who make ant
tuners putta 100K shunt across the ant inputs for precip static
reduction?
Huh?
..... de Dave @ KD7AEE
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