Martin wrote:
..."When I said a short vertical, I meant electrically short, i.e. <<
one wavelength (and also << 1/4 wavelength). Such a monopole above a
ground plane always presents a high impedance, and it's mainly
capacitive. If you want to pump a kW into such an antenna, you've got
to work at it.... Voltage breakdown will be the limiting factor of how
much power you can put in."...
Very true, assuming the antenna is not tuned to resonance by an inductor
on the antenna. If it is, then it reduces to the same as the mag loop
example.
The part that is surprising to most is that with a non-resonant antenna
not only is the voltage very high, the current is also very high. The
current however is usually easily handled by most common wire sizes.
But wire size is what started this whole thread, wasn't it.
Here is an example: 10 ft vertical on 7 MHz driven with 1KW. If you
assume perfect ground, the voltage to the antenna will be about 11.4KV
and the current will be about 20.6 A. In the real world it never gets
that high because of ground loss which appears in series with the
radiation resistance. So if you assume 8 ohms ground loss then the
voltage to the antenna is about 5.4 KV and the current is 9.8 amps.
Jerry, K4SAV
Martin Ewing wrote:
>When I said a short vertical, I meant electrically short, i.e. << one
>wavelength (and also << 1/4 wavelength). Such a monopole above a ground
>plane always presents a high impedance, and it's mainly capacitive. If you
>want to pump a kW into such an antenna, you've got to work at it.... Voltage
>breakdown will be the limiting factor of how much power you can put in.
>
>This discussion is difficult when limited to words and no equations.
>
>73
>Martin
>
>
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