Remember that the impedance ratio is dependent upon the sizes of the
parallel wires in the folded dipole.
If one wire has a different diameter from one or more of the others,
the impedance ratio will change.
Most of this can be found in older editions of the ARRL Handbook
and/or Antenna Handbook.
Don
N8DE
Quoting Martin Ewing <martin.s.ewing@gmail.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<steve@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.
>>> _______________________________________________
>>>
>>>
>>>
>>> _______________________________________________
>>> 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
>
_______________________________________________
_______________________________________________
TowerTalk mailing list
TowerTalk@contesting.com
http://lists.contesting.com/mailman/listinfo/towertalk
|