DomainKey-Signature: a=rsa-sha1; c=nofws;
        d=googlemail.com; s=gamma;
        h=mime-version:in-reply-to:references:date:message-id:subject:from:to
         :content-type;
        b=W/x81eAzcq+OylmnyRbw9KJHi/2CfHpebOqfeb83RQN1SmTe5lZJFupNGOLK6s+KEg
         uhrO8WQ39GLpPNamjR8sWd3nPc2qBvJvNPqpKJCgjZfG5OpzlYquUoR7RBxEiKkS1j7h
         jyT0X3laRbcPVOE8nEW92d7FebKEB+jVnMbiQ=
MIME-Version: 1.0
In-Reply-To: <001d01cad190$aa92e590$0901a8c0@lark>
References: <o2kfc7eccec1004010402n7f00e1c3l8b88ea090c0fb8bb@mail.gmail.com>
	 <001d01cad190$aa92e590$0901a8c0@lark>
Date: Thu, 1 Apr 2010 14:05:21 +0100
Message-ID: <n2kfc7eccec1004010605n95ffca10ia8bacd2ef2049820@mail.gmail.com>
From: Andy Talbot <andy.g4jnt@googlemail.com>
To: rsgb_lf_group@blacksheep.org
DomainKey-Status: good (testing) 
Subject: Re: LF: Re: E-Field Probe calibration question
Content-Type: multipart/alternative; boundary=00032555a97ee1d48704832c8305
Sender: owner-rsgb_lf_group@blacksheep.org
Precedence: bulk
Reply-To: rsgb_lf_group@blacksheep.org


--00032555a97ee1d48704832c8305
Content-Type: text/plain; charset=ISO-8859-1

Hi Alan + All...

Actually its a bit more complicated on its own...   The physical length of
the element is 0.75m, which is appreciably longer than the electrical
effective length.   And I'm betting a chunk of that difference is made up by
having a series resistor in the output circuitry to give a 50 ohm source
impedance and help stability but reducing voltage gain to below unity.  If
50R in series, there is an immediate -6dB  halving the length.    If my
original dipole concept *were* to apply, then the other half of the leg
caused by the car would also be affected by the attenuation.

But it appears from your comments so far the argument doesn't apply in which
case the 0.15m length still holds.   How do the EMC measurement community do
it, as their measurements have to be valid in law and could be held up to
scrutiny?

Now, what was the voltage from an untuned loop, again ;-?

Andy
www.g4jnt.com
Mostly retired inguneer


On 1 April 2010 12:44, Alan Melia <alan.melia@btinternet.com> wrote:

> Hi Andy maybe I dont read it right, but if that is an E-field probe it is
> not a "dipole" The input capacitance senses the voltage at two points on
> the
> wavefront. The ground plane is, I believe, irrelevant. You just need
> another
> capacitance separated from the "probe" to reference to voltage to. See
> Renato's balanced probe (www.vlf.it)  In your case this is the car.....but
> it could be the size of a dinky toy.The area of the wavefront intercepted
> by
> the probe or the ground reference does not affect the terminal
> voltage.....as I see it.................of course I might just be wrong cos
> i am really a scientist not an enjuneir!!
>  :-))
> Alan
>

--00032555a97ee1d48704832c8305
Content-Type: text/html; charset=ISO-8859-1
Content-Transfer-Encoding: quoted-printable

<div>Hi Alan + All...</div>
<div>=A0</div>
<div>Actually its a bit more complicated on its own...=A0=A0 The physical=
 length of the element is 0.75m, which is appreciably longer than the elec=
trical effective length.=A0=A0 And I&#39;m betting a chunk of that differe=
nce is made up by having a series resistor in the output circuitry to give=
 a 50 ohm source impedance and help stability but reducing voltage gain to=
 below unity.=A0 If 50R in series, there is an immediate -6dB=A0 halving=
 the length.=A0=A0=A0 If my original dipole concept <em>were</em> to apply=
, then the other half of the leg caused by the car would also be affected=
 by the attenuation.</div>

<div>=A0</div>
<div>But it appears from your comments so far the argument doesn&#39;t app=
ly in which case the 0.15m length still holds.=A0=A0 How do the EMC measur=
ement community do it, as their measurements have to be valid in law and=
 could be held up to scrutiny?</div>

<div>=A0</div>
<div>Now, what was the voltage from an untuned loop, again ;-?</div>
<div>=A0</div>
<div>Andy</div>
<div><a href=3D"http://www.g4jnt.com/">www.g4jnt.com</a></div>
<div>Mostly retired inguneer</div>
<div>=A0</div>
<div>=A0</div>
<div class=3D"gmail_quote">On 1 April 2010 12:44, Alan Melia <span dir=3D"=
ltr">&lt;<a href=3D"mailto:alan.melia@btinternet.com">alan.melia@btinterne=
t.com</a>&gt;</span> wrote:<br>
<blockquote style=3D"BORDER-LEFT: #ccc 1px solid; MARGIN: 0px 0px 0px 0.8e=
x; PADDING-LEFT: 1ex" class=3D"gmail_quote">Hi Andy maybe I dont read it=
 right, but if that is an E-field probe it is<br>not a &quot;dipole&quot;=
 The input capacitance senses the voltage at two points on the<br>
wavefront. The ground plane is, I believe, irrelevant. You just need anoth=
er<br>capacitance separated from the &quot;probe&quot; to reference to vol=
tage to. See<br>Renato&#39;s balanced probe (<a href=3D"http://www.vlf.it/=
" target=3D"_blank">www.vlf.it</a>) =A0In your case this is the car.....bu=
t<br>
it could be the size of a dinky toy.The area of the wavefront intercepted=
 by<br>the probe or the ground reference does not affect the terminal<br>v=
oltage.....as I see it.................of course I might just be wrong cos=
<br>
i am really a scientist not an enjuneir!!<br>=A0:-))<br>Alan<br></blockquo=
te></div>

--00032555a97ee1d48704832c8305--