Date: Thu, 03 Mar 2005 11:45:20 +0100
From: Dick Rollema <dickrollema@casema.nl>
In-reply-to: <000001c51f54$fb9902b0$e6a4c593@RD40002>
To: rsgb_lf_group@blacksheep.org
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 <000001c51f54$fb9902b0$e6a4c593@RD40002>
Subject: RE: LF: Re: Current "lost" in loading coil
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<font size=3>Thanks Jim for providing the final answer!<br><br>
73, Dick, PA0SE<br><br>
At 19:23 2-3-05, you wrote:<br><br>
</font><blockquote type=cite class=cite cite=""><font face="Times New Roman, Times" size=3>I
do like G3GVB's kind of reasoning. But I think the situation is
simplified a bit too much.&nbsp; Assume as an example that the coil is
wound on a toroid of high permeability material&nbsp; (e.g. for use with
a QRP transmitter....).<br>
The magnetic flux in the core is proportional to the current in the
winding.&nbsp; When the current increases from one end of the winding to
the other end this would mean thatt he flux in the core would rise as
well&nbsp; along the core. But this is impossible. The flux must be
uniform.&nbsp; So current in must be equql to current out.<br><br>
&nbsp;73, Dick, PA0SE<br>
</font><font face="Times New Roman, Times" size=3 color="#000080">&nbsp;<br>
Dear Dick, LF Group,<br>
&nbsp;<br>
This is not actually true – there is no requirement for the current to be
equal in all parts of the winding; if the magnetic flux is the same in
all parts of the winding, the voltage across each turn will be identical
( = d(phi)/dt), but the only requirement on the current is that the
vector sum of all the (current x turns) products flowing in all the
windings will be that required to generate the magnetic flux phi. As an
example, if you divide the winding on your high permeability core into
two halves, and connect an AC source across one half while the other half
is left open circuit, obviously no current will flow in the open circuit
half, while the other half will carry a current determined by the source
voltage and the winding inductance. The voltage across each half of the
winding will be equal, but the current in each half will be
different.<br>
&nbsp;<br>
</font><font face="Times New Roman, Times" size=3>The inductance and
capacitance of the loading coil depend a lot on how they are measured and
at what frequency – as a rough approximation think of the total
distributed capacitance as small, equal, capacitors between each turn and
ground. If you apply a signal far below the self-resonant frequency of
the coil to one end of the coil, and leave the other end open circuit,
the impedance at that point will nearly be the sum of the small
capacitors in parallel. But as the frequency increases, the inductive
reactance of the coil partly cancels the reactance of the capacitors,
reducing the overall reactance, so the overall effective capacitance is
larger. Because the capacitors at the “hot” open end of the coil have
more inductance in series with them, their effective reactance is reduced
more, and so more current flows in them than the capacitors at the driven
end of the coil – this also means that the voltage across the hot end
capacitors is higher than at the driven end. The limiting case is where
the coil and the distributed capacitance becomes self resonant, and most
of the current flows through the capacitance at the hot end, where the
voltage is only limited by the Q. Of course in practice the distributed
capacitance along the coil will not normally be uniform, and there will
be extra capacitance attached to the “hot” end in the form of the
antenna, but qualitatively the same thing will happen – with most of the
displacement current flowing to ground from the high-voltage end of the
coil. The difference in current measured at the two ends of the coil is
due to the presence of this displacement current. The smaller the antenna
is, (and so the bigger the coil and its distributed capacitance is) the
greater the fraction of the total current flows through the distributed
capacitance of the coil.<br>
&nbsp;<br>
Cheers, Jim Moritz<br>
73 de M0BMU<br>
</font><font face="arial" size=2 color="#000080">&nbsp;<br>
&nbsp;</font></blockquote></body>
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