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RE: LF: RE: VLF - vertical probe measurements

To: "[email protected]" <[email protected]>
Subject: RE: LF: RE: VLF - vertical probe measurements
From: Rik Strobbe <[email protected]>
Date: Fri, 15 Apr 2011 15:38:39 +0200
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Thread-topic: LF: RE: VLF - vertical probe measurements
Chris,
 
Since you alter the drive level I do assume the PA is linear.
At lower input (and thus output) power the efficiency will drop, due to the bias current (ultimately at zero input level the efficiency will be 0%).
Can you measure the output current of the PA ? That would give a better indication of the output power.
 
73, Rik  ON7YD - OR7T
 

Van: [email protected] [[email protected]] namens Chris Osborn [[email protected]]
Verzonden: vrijdag 15 april 2011 15:22
Aan: [email protected]
Onderwerp: Re: LF: RE: VLF - vertical probe measurements

Hello Stefan / Rik,
 
Thanks for the response.
 
I adjust the PA current by varying the drive level to the FET's so the PA supply voltage stays constant.
I should have made that clear.
 
73 Chris G3XIZ

--- On Fri, 15/4/11, Rik Strobbe <[email protected]> wrote:

From: Rik Strobbe <[email protected]>
Subject: LF: RE: VLF - vertical probe measurements
To: "[email protected]" <[email protected]>
Date: Friday, 15 April, 2011, 14:11

Hello Chris,
 
maybe a stupid question (wouldn't be my first one): how do you alter the PA current ?
If you change the PA current by changing the PA voltage then the output power will change more or less with the square of tht PA current (P = UxI and you change both U and I), what would result in a linear behaviour between PA current and probe voltage.
 
73, Rik  ON7YD - OR7T
 

Van: [email protected] [[email protected]] namens Chris Osborn [[email protected]]
Verzonden: vrijdag 15 april 2011 14:40
Aan: LF Group
Onderwerp: LF: VLF - vertical probe measurements

LF,
 
Acting on advice I have installed a whip aerial (probe) on top of a mast at the boundary of my property.
This is in order to monitor my VLF transmissions and to provide a reference level for any system improvements.
 
I noticed that the probe's pick-up voltage is directly proportional to my PA's supply current.
Assuming that the PA efficiency doesn't change drastically with current then would I be correct in assuming that the probe's voltage is directly proportional to the radiated power?
 
I was expecting a square law relationship . .
 
73
Chris G3XIZ
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