RE: LF: RE: VLF - vertical probe measurements

[Rik Strobbe <Rik.Strobbe@fys.kuleuven.be>]

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

 

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Van: owner-rsgb_lf_group@blacksheep.org [owner-rsgb_lf_group@blacksheep.org] namens Chris Osborn [g3xiz@yahoo.co.uk]
Verzonden: vrijdag 15 april 2011 15:22
Aan: rsgb_lf_group@blacksheep.org
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 <Rik.Strobbe@fys.kuleuven.be> wrote:

From: Rik Strobbe <Rik.Strobbe@fys.kuleuven.be>
Subject: LF: RE: VLF - vertical probe measurements
To: "rsgb_lf_group@blacksheep.org" <rsgb_lf_group@blacksheep.org>
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: owner-rsgb_lf_group@blacksheep.org [owner-rsgb_lf_group@blacksheep.org] namens Chris Osborn [g3xiz@yahoo.co.uk]
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