I have a novel idea, just put one up, and see if it works!
73 Chuck KI9A
In a message dated 1/4/2009 5:05:12 P.M. Central Standard Time,
martin.s.ewing@gmail.com writes:
Not to belabor this (and we're stretching the limits of "towertalk" here)...
FWIW, a quarter wave is not "short" in my view. I calculate (using the L <<
lambda approximation in Johnson's Antenna Engineering Handbook, 3rd ed.) the
following for a 1/10 wavelength vertical with L/a = 10000 (we are talking
thin wires!):
Z = 7.9 - j1686 ohms , which happens to be in the same ballpark as your
number.
So that's the impedance of the "antenna". Your job, as a system designer is
to efficiently couple that impedance to 50 ohms resistive. I use TLW to
suggest a typical L match, which dissipates ~66% of the power in resistive
coil losses and presents 13.5 kV to the antenna. Such is the bane of
typical mobile installations. (.05 wavelength leads to 92% loss and 19 kV to
the antenna)
A voltage of 13.5 kV (rms - don't let your kids near this!) into a /Z/ of
~1700 ohms gives a current of 8.1 A. These are similar to your numbers,
except for including some typical circuit losses. So, I guess we don't
disagree - Yes, you need current to radiate, but you need *lots* of voltage
to produce that current.
73 Martin AA6E
On Sun, Jan 4, 2009 at 4:15 AM, Steve Hunt <steve@karinya.net> wrote:
> Martin, with respect this is completely wrong.
>
> Take a simple example of a shortened dipole that's only a
> quarter-wavelength overall length. It has a feedpoint impedance of about
> 13-j750. If this antenna is radiating 1KW the current flowing at the
> feedpoint must be SQRT(1000/13)=8.7A - in other words a lot more current
> than had it been a full half-wave. Only the real part of the feedpoint
> impedance can dissipate power - that's why, for a given radiated power
> the current must increase for a drop in radiation resistance. Matching
> losses will slightly alter this figure, but I wouldn't expect much loss
> matching this particular impedance.
>
> It's a complete misunderstanding to think that, because the total
> feedpoint impedance is high the current must be low. That would only be
> the case if you didn't try to match the antenna. Once properly matched,
> the source impedance at the feedpoint will be the complex conjugate -
> 13+j750 and the reactive component will be cancelled.
>
> Your example of current distribution on a short antenna proves nothing.
> Of course the current is almost zero at the end of the wire, but it can
> be any current you want just a few electrical degrees back from the end
> given sufficient current gradient. In my example, the current at the end
> of the dipole will be close to zero, rising to 8.7A 1/8 wavelength back
> at the feedpoint. If it had been a halfwave dipole the current would
> have been 3.7A 1/4 wavelength from the end.
>
> If you still disagree, could you please explain what the current would
> need to be in my shortened dipole example, assuming it is radiating 1KW.
>
> 73,
> Steve G3TXQ
>
> Martin Ewing wrote:
> > I'm not buying that! Current flow depends on impedance, R+jX, not
> radiation
> > resistance. Your short whip runs at high voltage and low current. There
> is
> > lots of current in the matching network, but little gets into the antenna
> > wire. Another way to see it is that the "boundary condition" on current
> > flow is that it must be zero at the end of the wire (unless there are
> sparks
> > flying!). Transforming back a fraction of a wavelength to the feed
> point,
> > you're still mostly voltage and little current.
> >
> > A small loop antenna is the complement, mostly current and little
> voltage.
> >
> > 73 & Cheers,
> > Martin AA6E
> >
>
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--
Martin Ewing, AA6E
Branford, CT
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