At 11:23 PM +0000 8/13/03, 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."
>...Any comments??
Yes. As SWR changes, the ratio of reverse (reflected) power to
forward power changes. So does the ratio of the _net_ forward power
(i.e., the forward power minus the reverse power), to the forward
power. (In the absence of transmissionline loss, the net forward
power anywhere along the line is equal to the power delivered to the
load, or the power delivered by the transmitter.)
Let's assume that, by "voltage," the author means not the voltage at
the load or at the generator/transmitter, but the peak voltage in the
standingwave pattern on a transmissionline sufficiently long to
show the peak. If the load impedance varies, this peak voltage may
or may not vary as the square root of the VSWR, depending on how the
forward power, or the net forward power, varies.
I have an Excel 98 worksheet that calculates the peak current on a
50ohm transmission line, and the net forward power, given the
forward power and the SWR. (The peak voltage is equal to the peak
current multiplied by 50 ohms. Note that peak voltage does not occur
in the same place as peak current.) I will email this worksheet to
anyone who asks.
73 Chuck, W1HIS
