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Re: [TowerTalk] Dipole fed with balanced line?

To: towertalk@contesting.com
Subject: Re: [TowerTalk] Dipole fed with balanced line?
From: Jim Lux <jimlux@earthlink.net>
Date: Tue, 17 Jan 2012 06:35:44 -0800
List-post: <towertalk@contesting.com">mailto:towertalk@contesting.com>
On 1/17/12 5:34 AM, Ken wrote:
> Steve,
>
> The issue is not about I^2R losses and feedline cannot compensate for
> missing antenna wire.   It's the antenna wire that radiates, not a
> properly balanced feedline.
>
> How do you "increase the feedpoint current...by 19%" other than by
> increasing power?

A shorter wire means the feedpoint impedance (resistive part) gets
lower, so for the same power, you have more current.   That comes with
the inconvenience of a bigger reactive part, which you need to cancel
with a matching network.

The change is roughly as length squared, that is, a 1/4 wavelength long
dipole (half length) has roughly 1/4 the resistance at feedpoint Z
(this works down to about 1/10th wavelength), or 18-20 ohms.  So your
matching network will need to match 18 ohms plus a big capacitive reactance.


As others have noted, you can shorten quite a ways before the losses in
the matching network and IR losses in the antenna start to add up. Start
getting down to 1/10 wavelength or 1/20 wavelength, and it's getting
pretty ugly.

Now, if you don't put the matching network at the feed, so the
circulating power between reactance in the antenna and reactance in
matching network has to flow through a feedline, you get yet another
source of IR loss.  Feeding a short dipole through 100 ft of coax with a
tuner in the shack is a pretty non-optimal way to do things.

the other issue (although a small one) is that short dipoles have less
gain that full size dipoles. An infinitesimally small lossless dipole
has 1.76 dBi gain, compared to 2.15dBi for a half wavelength.  (that is,
the pattern is more omnidirectional)


   BTW, increasing the feedpoint current by 19% is a 40+% increase in
power (P=I^2*R).


Only if R is constant.  The R of the short dipole drops, so for the same
power, you have more current.


  That's a significant increase, not "only a fraction of a db".
>
> If the solution was that simple, we could have efficient "limited
> space" antennas by putting a 600 ohm resistor at the end of an open
> wire feedline and not need the antenna at all!   ;-)



No, but you could do it by having a very low loss matching network and a 
physically small radiator.

At some point, with simple matching networks, the "good SWR bandwidth" 
of the combination might get too narrow for practical use.  If you had 
idealized matching components (superconducting inductors, lossless 
dielectrics, etc) you could conceivably  flatten the SWR curve out with 
a more complex matching network.

There are compact antenna schemes which essentially do the latter by 
coupling the amplifier directly to the antenna, and simulating the 
appropriate impedances, but since amplifiers aren't 100% efficient, this 
isn't necessarily a better scheme.  (Imagine a perfect current source 
amplifier)
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