[TowerTalk] Relays in RF usage UPDATE

[Jim Lux]

I'd also encourage anyone with interest in this sort of thing to take a look
at the XLZIZL spreadsheet from Dan, AC6LA
http://www.qsl.net/ac6la/xlzizl.html
Very cool.. handles every kind of transmission line, transformers, lumped
components, and calculates all you might want to know, as well as plotting
smith charts, etc.
Jim, W6RMK
----- Original Message -----
From: "Chuck Counselman" <ccc at space.mit.edu>
To: <Towertalk at contesting.com>
Sent: Wednesday, August 13, 2003 8:43 PM
Subject: Re: [TowerTalk] Relays in RF usage UPDATE


> 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 50-Ohm 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 transmission-line 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
> standing-wave pattern on a transmission-line 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
> 50-ohm 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
>
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