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.
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