Message-ID: <4DA840AF.6040507@iup.uni-heidelberg.de>
Date: Fri, 15 Apr 2011 14:57:19 +0200
From: =?ISO-8859-1?Q?Stefan_Sch=E4fer?= <Stefan.Schaefer@iup.uni-heidelberg.de>
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To: rsgb_lf_group@blacksheep.org
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Subject: Re: LF: VLF - vertical probe measurements
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Hello Chris,

I would expect a sqare law relationship as well since the probe is a 
capacitive divider and thus the voltage should be directly proportional 
to the antennas voltage. And this rises with sqrt(2) x the power...


73, Stefan


Am 15.04.2011 14:40, schrieb Chris Osborn:
> 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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Hello Chris, <br>
<br>
I would expect a sqare law relationship as well since the probe is a
capacitive divider and thus the voltage should be directly proportional
to the antennas voltage. And this rises with sqrt(2) x the power...<br>
<br>
<br>
73, Stefan<br>
<br>
<br>
Am 15.04.2011 14:40, schrieb Chris Osborn:
<blockquote cite="mid:618180.72059.qm@web28509.mail.ukl.yahoo.com"   type="cite">
  <table border="0" cellpadding="0" cellspacing="0">
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        <div>LF,</div>
        <div>&nbsp;</div>
        <div>Acting on advice I have installed a whip aerial (probe)&nbsp;on
top of a mast at the&nbsp;boundary of my property.</div>
        <div>This is in order to monitor my VLF transmissions and to
provide a reference level for any system improvements.</div>
        <div>&nbsp;</div>
        <div>I noticed that the probe's pick-up voltage is directly
proportional to&nbsp;my PA's supply current.</div>
        <div>Assuming that the PA efficiency doesn't change drastically
with current then would I be correct in assuming that the probe's
voltage is <u>directly proportional</u> to the radiated power?</div>
        <div>&nbsp;</div>
        <div>I was expecting a square law relationship . . </div>
        <div>&nbsp;</div>
        <div>73</div>
        <div>Chris G3XIZ</div>
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