Return-Path: Received: (qmail 86368 invoked from network); 3 Mar 2005 11:04:51 -0000 Received: from unknown (HELO ptb-spamcore02.plus.net) (192.168.71.3) by ptb-mailstore01.plus.net with SMTP; 3 Mar 2005 11:04:51 -0000 Received: from mailnull by ptb-spamcore02.plus.net with spamcore-l-b (Exim 4.32; FreeBSD) id 1D6o7X-000Eb0-6Y for dave@picks.force9.co.uk; Thu, 03 Mar 2005 11:04:41 +0000 Received: from [192.168.67.2] (helo=ptb-mxcore02.plus.net) by ptb-spamcore02.plus.net with esmtp (Exim 4.32; FreeBSD) id 1D6o6e-000EMy-Kc for dave@picks.force9.co.uk; Thu, 03 Mar 2005 11:03:00 +0000 Received: from post.thorcom.com ([193.82.116.20]) by ptb-mxcore02.plus.net with esmtp (Exim) id 1D6o4Y-000KUd-Eg for dave@picks.force9.co.uk; Thu, 03 Mar 2005 11:00:50 +0000 Received: from majordom by post.thorcom.com with local (Exim 4.14) id 1D6o3t-0006Hf-9M 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-0006AF-17 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 1D6o3g-00011V-6o 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 11:59:50 +0100 (CET) Date: Thu, 03 Mar 2005 11:39:31 +0100 From: Dick Rollema In-reply-to: <1109779478.4225e41606d62@webmail2.kuleuven.be> X-Sender: dickrollema@casema.nl@mail.casema.nl (Unverified) To: rsgb_lf_group@blacksheep.org Cc: "W.F. Oorschot" Message-id: <6.1.0.6.2.20050303113502.02cf20b0@mail.casema.nl> MIME-version: 1.0 X-Mailer: QUALCOMM Windows Eudora Version 6.1.0.6 References: <4225CA7F.15619.BBCBA7@localhost> <6.1.0.6.2.20050302162804.02a13ea0@mail.casema.nl> <1109779478.4225e41606d62@webmail2.kuleuven.be> X-SPF-Result: relay.thorcom.net: 83.80.1.28 is neither permitted nor denied by domain of casema.nl X-Spam-Score: 0.6 (/) X-Spam-Report: autolearn=no,FORGED_RCVD_HELO=0.05,HTML_20_30=0.504,HTML_MESSAGE=0.001 Subject: Re: LF: Lost current in a 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=0.0 required=5.0 tests=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 To All from PA0SE

Rik is right of course. I had overlooked the fact that the voltage over the coil is almost 90 degrees out of phase with the voltage out of the transmitter. So the two voltages cannot be simply added. It must be done as vectors: SQRT(70^2+8541^2) = 8451,3 V.

73, Dick, PA0SE

At 17:04 2-3-05, you wrote:

Hello Dick, Mike

it is indeed a "phase thing".
If a 100 W TX causes a 2 A current in a 5mH coil (XL = 4270 Ohm) and
thus causes a voltage of 8540 V over the coil then at the "hot" end
the phase difference between U and I is 89.7 degrees :
P = U * I * cos(x) thus x = arccos(P/(U*I))
(assuming no coil losses)
At the "cold" end voltage and current are in phase.

73, Rik

> To Mike and All
>
> At 15:15 2-3-05, G3XDV wrote:
> >A useful test to see whether radiation causes the loss, would be to
> >completely screen the loading coil.
> Because the size of the coil is so small, expressed in
> wavelength, radiation is negligible.
>
> >I have never understood why, if the current is thought to stay the
> >same from top to bottom of a coil, the voltage is much bigger at
> the
> >top? Is this a phase thing?
>
> The voltage at the bottom is equal to the output voltage of the
> transmitter; at e.g. 100W this is 70V over 50 ohm.
>
> As an example let the inductance of the coil be 5mH. Then its
> reactance at
> 136kHz is 4270 ohm.
> With an aerial current of for instance 2A flowing through the coil
> the
> voltage at the top of the coil will be
> 2 * 4270 = 8541V higher than at the bottom.
> So total voltage at the top will be 70 + 8541 = 8610V.
>
> 73, Dick, PA0SE
>