At 11:23 PM 8/13/2003 +0000, RCARIELLO wrote:
>I just received my copy of Worldradio September 2003.
>There is an article by Kurt N. Sterba on page 34, Aerials SWR and Voltage.
>
>Second paragraph "voltage goes up as the square root of the VSWR. So, with
>3:1 VSWR, the voltage and current are only 1.7 times greater then with 1:1.
>A 50Ohm perfectly matched line with 1500 watts power will have a voltage of
>274 volts. With 10:1 SWR the voltage will be 866 volts."
>
>I will have to recalculate my power chart I sent using this new information.
>
>Any comments??
That would be the voltage along a transmission line with that kind of
mismatch, and a resistive one to boot.. What number will you use for the
impedance of the relay, and is there any reactive power circulating.
Thought experiment:
transmission line:Capacitor: 1 wavelength transmission
line:inductor:resistive load
where the C and L are resonant at the frequency of interest. A 1
wavelength TL is perfectly matched (in that the impedance looking into one
end is exactly the impedance connected to the other)
You will have significant circulating current, so the voltage could be
quite high.
Just goes to show that one should carefully examine the constraints on a
simple expression like "voltage goes as square root of VSWR". All that
statement represents is a verbal statement of what the "standing wave
ratio" actually is (peak voltage/minimum voltage) measured along the line.
Another thing to consider is whether 10:1 is actually a worst case? What
if a connector shorts or opens (infinite VSWR)?
Jim, W6RMK
>Rich AA2MF
