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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