Return-Path: Received: from rly-de11.mx.aol.com (rly-de11.mail.aol.com [172.19.170.147]) by air-de08.mail.aol.com (v125.7) with ESMTP id MAILINDE083-5144b0a89cb3de; Mon, 23 Nov 2009 08:10:50 -0500 Received: from post.thorcom.com (post.thorcom.com [193.82.116.20]) by rly-de11.mx.aol.com (v125.7) with ESMTP id MAILRELAYINDE115-5144b0a89cb3de; Mon, 23 Nov 2009 08:10:36 -0500 Received: from majordom by post.thorcom.com with local (Exim 4.14) id 1NCYfq-0000jb-L3 for rs_out_1@blacksheep.org; Mon, 23 Nov 2009 13:09:46 +0000 Received: from [83.244.159.144] (helo=relay3.thorcom.net) by post.thorcom.com with esmtp (Exim 4.14) id 1NCYfq-0000jS-3Y for rsgb_lf_group@blacksheep.org; Mon, 23 Nov 2009 13:09:46 +0000 Received: from smtp831.mail.ukl.yahoo.com ([217.12.13.166]) by relay3.thorcom.net with smtp (Exim 4.63) (envelope-from ) id 1NCYfo-00025V-RU for rsgb_lf_group@blacksheep.org; Mon, 23 Nov 2009 13:09:46 +0000 Received: (qmail 41424 invoked from network); 23 Nov 2009 13:09:39 -0000 DomainKey-Signature: a=rsa-sha1; q=dns; c=nofws; s=s1024; d=btopenworld.com; h=Received:X-Yahoo-SMTP:X-YMail-OSG:X-Yahoo-Newman-Property:Message-ID:From:To:References:In-Reply-To:Subject:Date:MIME-Version:Content-Type:Content-Transfer-Encoding:X-Priority:X-MSMail-Priority:X-Mailer:X-MimeOLE; b=GL+2Sd4C+2uJkif6M1Bp1P84Icqwp556yzETeZ0Zuqav2wwrSdDn/G32KnQ0WV1O0xCixWRkRTiE/RQnlWaY8kaxmYkLq0SDnXbZDTMjj5Z/fjaLxgdtYBlAAZfbysjkOyJRzJ4RD1Dqa3GnjPc87lPhxVnhRKBcm3JRQbGJ5T4= ; Received: from unknown (HELO JimPC) (james.moritz@86.180.119.35 with login) by smtp831.mail.ukl.yahoo.com with SMTP; 23 Nov 2009 13:09:39 -0000 X-Yahoo-SMTP: qKIhhNCswBB2TTHr2BORWcGGR2mpopxhCcunGIxpCKQYiG07Q7UOhNo- X-YMail-OSG: cy.ivQcVM1mNX_V9XdUI95bmXiAhZzY0y69MUEspPN3TNEn4ySbwm9vbSRiyNh5.pbtfDEH2bmptvdW3wcpM8TrrQp1sN8IUoGQ_VsJVDqkyBTYxrxz0O8eTkt6wx1KxYLVGu.FiIFcH2jvF8q7470CsOTfDhhJ.qmNEttuR9ZIvJiY0_t0q0q927Z8neMPuVPxGRwvtukR9hzsxhqY4Z4MB.wfIeH0VCB3pdIFW0Oi0hrlztmA6aykyr_7w_FVqNKiNRLbS6AQXFuLw9QVpfeemjTghxNNXuCf8 X-Yahoo-Newman-Property: ymail-3 Message-ID: <7C21B1DD169746B0967307E2AB9E7FDD@JimPC> From: "James Moritz" To: References: <4B09D502.8030200@toya.net.pl>,<006101ca6bd9$9ab3cb70$0900a8c0@lark> In-Reply-To: Date: Mon, 23 Nov 2009 13:09:38 -0000 MIME-Version: 1.0 X-Priority: 3 X-MSMail-Priority: Normal X-Mailer: Microsoft Windows Mail 6.0.6002.18005 X-MimeOLE: Produced By Microsoft MimeOLE V6.0.6002.18005 DomainKey-Status: good (testing) X-Spam-Score: 0.0 (/) X-Spam-Report: autolearn=disabled,none Subject: LF: Re: N-turn TX Loop Content-Type: text/plain; format=flowed; charset="iso-8859-2"; reply-type=original Content-Transfer-Encoding: 7bit X-Spam-Checker-Version: SpamAssassin 2.63 (2004-01-11) on post.thorcom.com X-Spam-Level: X-Spam-Status: No, hits=0.1 required=5.0 tests=MISSING_OUTLOOK_NAME 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-AOL-IP: 193.82.116.20 Dear Piotr, LF Group, Considering only the conductor losses for the moment, in principle increasing the number of turns of the loop by a factor N increases the radiation resistance by approximately N^2, while the loss resistance of the conductors increases by approximately N times, giving an improvement in efficiency of a factor N. However, if you instead connected the same N turns in parallel, you would have a single-turn loop with N parallel conductors, with a loss resistance 1/N times that of a single conductor loop, but with the same radiation resistance. This would also give rise to the same N times efficiency improvement. So from the conductor loss point of view, increasing the number of conductors in the loop should improve efficiency by much the same amount, it would not matter much whether the turns are connected in series or parallel. However, the multi-turn loop inductance is also approximately proportional to N^2, if the turns are fairly close together. The current in the multi-turn loop will be reduced by a factor of N to achieve the same radiated power. Most of the loop impedance is due to its inductance, so the overall voltage across the loop will increase by a factor of N for the same radiated power.This means a greater electric field strength around the antenna, and so increased environmental losses due to the multiple turns - dielectric loss increases as the square of the applied electric field. So assuming a constant value of environmental loss (1.5ohm in your calculation) is not realistic, if a significant proportion of this resistance is due to dielectric losses, which is probably the case for most practical antennas. Probably the main virtue of the transmitting loop is the reduced environmental loss due to it being a relatively low-impedance, low-voltage device. so to maintain this advantage, it would seem to be a better idea to have multiple parallel turns rather than series turns. Cheers, Jim Moritz 73 de M0BMU > > This sunday evening i decided to do some simple math and it turns out > > that such a N-turn TX loop should work ( at least on the paper) > > The radiation resistance is proportional to the square of so called > > "effective > > height" and this last term can be easily derived for a loop i.e. it is > > equal to > > 2*pi*A*N/lambda where A denotes area closed by a loop a N is the number > > of turns > > so the radiation resistance for a single turn loop reads as > > 320*pi^4*A^2/lambda^4 > > For the N-turns the radiation resistance obtained for a single turn is > > multiplied > > by N^2. Ok, the R (ac) is increased as we increase N but this is linear > > with respect to N > > and therefore we should have gain in the radiated power. > > (there is an implicit assumption made: the loop is "small" i.e. the > > current is constant) > > > > i did some calculations: assumed TX power ( and later, perfect match to > > the loop) 200 Watt > > environmental loss: 1.5 Ohm, diameter of the wire d = 3 mm, rectangular > > shape of the loop > > i.e 10 meters by 20 meters ( less optimal than square or circle ) > > so A = 200 sq.m For N =1 (classical tx loop) we get R(AC) = 0.62 Ohm > > ( Rac formula taken from ARRL Antenna Handbook, f = 137.7kHz) > > radiation resistance RRAD = 55.5 microOhm, total R loss = 2.12 Ohm, > > efficiency is 0.0026% and radiated pwr 5.2 miliWatts, I = 9.7 Amp. > > Next, I took N =3 so the wire length is changed from 60 meters to 180, > > everything else was kept the same and now one gets: R(ac)= 1.85 Ohm so > > R loss = 3.35 Ohm > > RRAD = 499.5 microOhm, efficiency increased to 0.015% and radiated pwr > > abt 30 miliWatt, I=7.7 Amp > > I am sorry bothering you but i simply would like to learn > > where is the 'catch' here - if there is one ... > > 73 de Piotr, sq7mpj > > qth: Lodz /jo91rs/