Steve Hunt 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.
>
Taking your 13-j750 as the feedpoint impedance (including loss
resistance within the antenna), if you're actually dissipating 1kW
(either by radiation or IR loss), the inphase component of the current
will be 8.7A, as you say, but the quadrature(reactive) current will be
8.7*750/13 or about 500A. The rms current will be sqrt(8.7^2+500^2)..
fairly close to 500A.
The source impedance driving this will be 13+j750. Typically, some sort
of matching network would supply the reactive part, so there's a
circulating current flowing between the reactive antenna and the
reactive matching network.
500A is a healthy current to be flowing, even if it's not contributing
to the radiation. A 0.1 ohm resistance in the wiring would dissipate
25kW. Fortunately, most matching network components have resistances
much lower than 0.1 ohms (AWG 10 wire, ignoring skin effect, is 1mOhm/ft)
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