[TowerTalk] How a folded dipole works

[Martin Ewing]

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<steve at karinya.net> 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.
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