Return-Path: Received: (qmail 8793 invoked from network); 3 Mar 2005 11:01:00 -0000 Received: from unknown (HELO ptb-spamcore01.plus.net) (192.168.71.1) by ptb-mailstore02.plus.net with SMTP; 3 Mar 2005 11:01:00 -0000 Received: from mailnull by ptb-spamcore01.plus.net with spamcore-l-b (Exim 4.32; FreeBSD) id 1D6o6V-0006R9-KK for dave@picks.force9.co.uk; Thu, 03 Mar 2005 11:02:53 +0000 Received: from [192.168.67.1] (helo=ptb-mxcore01.plus.net) by ptb-spamcore01.plus.net with esmtp (Exim 4.32; FreeBSD) id 1D6o6U-0006QM-Le for dave@picks.force9.co.uk; Thu, 03 Mar 2005 11:02:50 +0000 Received: from post.thorcom.com ([193.82.116.20]) by ptb-mxcore01.plus.net with esmtp (Exim) id 1D6o6l-0006Vb-VK for dave@picks.force9.co.uk; Thu, 03 Mar 2005 11:03:09 +0000 Received: from majordom by post.thorcom.com with local (Exim 4.14) id 1D6o3t-0006Hg-69 for rs_out_1@blacksheep.org; Thu, 03 Mar 2005 11:00:09 +0000 Received: from [193.82.116.30] (helo=relay.thorcom.net) by post.thorcom.com with esmtp (Exim 4.14) id 1D6o3i-0006AU-Dj for rsgb_lf_group@blacksheep.org; Thu, 03 Mar 2005 10:59:58 +0000 Received: from cas-mta3-fe.casema.nl ([83.80.1.28] helo=mta.casema.nl) by relay.thorcom.net with esmtp (Exim 4.43) id 1D6o3h-00011V-7u for rsgb_lf_group@blacksheep.org; Thu, 03 Mar 2005 10:59:57 +0000 Received: from s9z5i6.casema.nl ([83.85.106.243]) by cas-mta3.mgmt.casema.nl (Sun Java System Messaging Server 6.1 HotFix 0.11 (built Jan 28 2005)) with ESMTP id <0ICR00ALQVVLDD90@cas-mta3.mgmt.casema.nl> for rsgb_lf_group@blacksheep.org; Thu, 03 Mar 2005 11:59:51 +0100 (CET) Date: Thu, 03 Mar 2005 11:45:20 +0100 From: Dick Rollema In-reply-to: <000001c51f54$fb9902b0$e6a4c593@RD40002> X-Sender: dickrollema@casema.nl@mail.casema.nl (Unverified) To: rsgb_lf_group@blacksheep.org Message-id: <6.1.0.6.2.20050303114428.02cf00f0@mail.casema.nl> MIME-version: 1.0 X-Mailer: QUALCOMM Windows Eudora Version 6.1.0.6 References: <6.1.0.6.2.20050302101906.03115e40@mail.casema.nl> <000001c51f54$fb9902b0$e6a4c593@RD40002> X-SPF-Result: relay.thorcom.net: 83.80.1.28 is neither permitted nor denied by domain of casema.nl X-Spam-Score: 0.3 (/) X-Spam-Report: autolearn=no,FORGED_RCVD_HELO=0.05,HTML_10_20=0.295,HTML_MESSAGE=0.001 Subject: RE: LF: Re: Current "lost" in loading coil Content-type: text/html; charset=windows-1252 X-Spam-Checker-Version: SpamAssassin 2.63 (2004-01-11) on post.thorcom.com X-Spam-Status: No, hits=0.1 required=5.0 tests=HTML_FONTCOLOR_BLUE, HTML_MESSAGE autolearn=no version=2.63 X-SA-Exim-Scanned: Yes Sender: owner-rsgb_lf_group@blacksheep.org Precedence: bulk Reply-To: rsgb_lf_group@blacksheep.org X-Listname: rsgb_lf_group X-SA-Exim-Rcpt-To: rs_out_1@blacksheep.org X-SA-Exim-Scanned: No; SAEximRunCond expanded to false X-Spam-Filtered: by PlusNet SpamCORE (v3.00) Content-transfer-encoding: 8bit Thanks Jim for providing the final answer!

73, Dick, PA0SE

At 19:23 2-3-05, you wrote:

I do like G3GVB's kind of reasoning. But I think the situation is simplified a bit too much. Assume as an example that the coil is wound on a toroid of high permeability material (e.g. for use with a QRP transmitter....).
The magnetic flux in the core is proportional to the current in the winding. 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 along the core. But this is impossible. The flux must be uniform. So current in must be equql to current out.

73, Dick, PA0SE

Dear Dick, LF Group,

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.

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.

Cheers, Jim Moritz
73 de M0BMU