[AMPS] High-SWR protection

Tom Rauch W8JI@contesting.com
Tue, 31 Oct 2000 21:29:32 -0500


> You simply mis-read what Ten-Tec said.  It says that the  "SWR reading
> gives no indication of reactive components".  I understood that to mean
> that a 2:1 SWR reading simply does not tell you what combination or
> resistance and reactance creates that voltage.  

Maybe it's just poor wording, but they didn't say it doesn't indicate 
the amount of reactance. They said it gives "no indication of 
reactance". That is untrue, a properly working directional coupler 
will ALWAYS indicate the presence of reactance.
 
> Now, with regard to a short of the output shutting the rig down, I know of
> at least one case and probable second that it did not.  A short in the
> output will change the operating conditions of the amplifier driving the
> circuit and change the efficiency of the networks.  If you had a constant
> power source, the current would increase with a short and voltage would be
> induced into the reflected pickup of the directional coupler.  But, the
> source will change drastically with a short.

It would be a design flaw of the highest order if a SWR shutdown 
circuit failed to shut the PA down safely into a low Z presented 
across the output connector. A low Z could occur there from an 
open or high impedance elsewhere on the line, or even a reactive 
load with the correct line length.

I'm not trying to argue with you, it's just a simple fact that properly 
working directional couplers always have maximum voltage output 
into either a short or open. 

Depending on the phase shift through the output filter, 
transformers, and transmission lines...a shorted load can appear 
as anything from a near-open to a near-short at the output device. It 
is unlikely that a short at the output port appears as a short at the 
PA device after running through that system. 

This isn't meant to be an argument, it's just how the system works. 
I have often used low pass filters along with a PIN diode that shorts 
the output of the filter to present an open circuit to a source feeding 
the filter.

I'm not saying you didn't see what you saw, just that the rig must 
have a design flaw to fail into a short.  
73, Tom W8JI
w8ji@contesting.com

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