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Re: [TowerTalk] How a folded dipole works

 To: Steve Hunt Re: [TowerTalk] How a folded dipole works Martin Ewing Wed, 26 Aug 2009 10:03:11 -0400 mailto:towertalk@contesting.com>
 ```Steve, Thanks for pointing this out. Adding more elements to a system adds degrees of freedom that can add complexities to the analysis. My claim that the currents in both wires are equal was obviously a little too simplistic! But it's still a good approximation if you're not using zip cord. Cheers / 73 Martin AA6E On Wed, Aug 26, 2009 at 6:47 AM, Steve Hunt wrote: > Martin, > > Just one point of detail. If you construct a  folded dipole of wire > which has a low(ish) differential-mode characteristic impedance, and > which has a significant difference between its differential-mode and > common-mode Velocity Factors, the connection between the two wires has > to be somewhere other than at the ends. Take a look at my detailed > analysis here: > > http://www.karinya.net/g3txq/folded_dipole/ > > This is not just a "theoretical distraction" - if you try to construct a > folded dipole with "Figure-8" Zip cord you'll find the shorting links > have to be way towards the centre of the antenna. The ARRL Antenna Book > used to address this issue, but they've removed it from the latest edition. > > 73, > Steve G3TXQ > > Martin Ewing wrote: >> In an off-line discussion, we were talking about building a folded >> dipole out of twin lead or ladder line.  Some folks may have the idea >> that if you build an FD out of 450 ohm ladder line, you get a 450 ohm >> feed point impedance, but not so.  It will still be 300 ohms.  (It's >> an accident that an FD made of 300 ohm twin lead has a 300 ohm feed >> impedance.) >> >> I offer a physicist's explanation, based on a symmetry argument. >> >> A 2-wire half-wave folded dipole, radiating 100 W (say), has a certain >> net current flow on the two wires (at the center point).  Both wires >> carry equal current*, i.e. each wire has half of the total current. >> So your FD feed point, in series with only one wire, will be providing >> 1/2 the current of a simple dipole radiating the same power.  But all >> the power comes from your transmitter and P = V x I, so if I is 1/2, V >> has to be 2x the voltage you would have had in a simple dipole. >> Furthermore, the impedance is R = V/I, which will be (2)/(1/2) = 4x >> the simple dipole impedance, so ~70 ohms --> ~280 ohms.  It's more or >> less independent of the spacing of the FD wires.  (Neglecting real >> world issues like wire loss, earth, etc.) >> >> You can extend the argument with N equal size wires in your FD, and >> get an impedance multiplication of N**2.  I suppose you could get >> other interesting impedances by feeding, say, 2 wires out of a 3 wire >> FD, or by using different size wires. >> >> Or that's the way I remember it. >> >> 73 Martin AA6E >> >> * Why equal, you ask?  Because we connected the end points together. >> _______________________________________________ >> >> >> >> _______________________________________________ >> TowerTalk mailing list >> TowerTalk@contesting.com >> http://lists.contesting.com/mailman/listinfo/towertalk >> >> >> > > _______________________________________________ > > > > _______________________________________________ > TowerTalk mailing list > TowerTalk@contesting.com > http://lists.contesting.com/mailman/listinfo/towertalk > _______________________________________________ _______________________________________________ TowerTalk mailing list TowerTalk@contesting.com http://lists.contesting.com/mailman/listinfo/towertalk ```
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