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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