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
On Sat, Jan 3, 2009 at 11:00 PM, K4SAV <RadioIR@charter.net> wrote:
> ..."I'm not buying that! Current flow depends on impedance, R+jX, not
> radiation
> resistance.".....
>
> Ah, but it's confusing. For a short vertical that is not tuned to
> resonance, if the source is a voltage source, the feedpoint is at a low
> current and the voltage at the feedpoint is equal to the source. If the
> source is a current source, the feedpoint is at a high voltage and the
> current is equal to the source. If the feedpoint is a power source,
> then the feedpoint will be at a high voltage AND at a high current. No,
> you cannot multiply voltage times current and get the power. Remember
> the phase calculations done when figuring real power? When you split
> that vector, the equivalent current going into the antenna at zero phase
> angle is highly dependent on the radiation resistance of the antenna.
> So the lower that resistance, the higher the feedpoint current (the
> feedpoint voltage also goes up). There is a lot of reactive power but
> all that does is drive up the feedpoint voltage.
>
> The mag loop analogy may be confusing to some because it is usually
> tuned to resonance by a cap. That leaves a very small impedance because
> the radiation resistance is very low, and everyone tries to make
> resistive losses as low as possible to make the thing work. So now you
> have to drive maybe 0.5 +j0 ohms. I think you can do the math for that.
>
> Jerry, K4SAV
>
> 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
> >
> >On Sat, Jan 3, 2009 at 4:46 PM, Steve Hunt <steve@karinya.net> wrote:
> >
> >
> >
> >>Nope - the current will be much higher in an electrically short antenna.
> >>What matters is not its high reactance - that will get tuned out by the
> >>matching circuitry - but the low radiation resistance where the power is
> >>being dissipated. If you don't believe it, just picture how much current
> >>is flowing in a small magnetic loop antenna or a mobile whip.
> >>
> >>73,
> >>Steve G3TXQ
> >>
> >>Martin Ewing wrote:
> >>
> >>
> >>>This all assumes you're talking about a full size dipole. If your
> >>>
> >>>
> >>antenna
> >>
> >>
> >>>is more like a short whip, the current in the wire is much less, since
> it
> >>>
> >>>
> >>is
> >>
> >>
> >>>a high impedance antenna. Then, your problem is in the antenna
> >>>tuner/coupler and ground system, where there could be high losses.
> >>>
> >>>
> >>>
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> >>
> >>
> >>
> >
> >
> >
> >
> >
>
> _______________________________________________
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--
Martin Ewing, AA6E
Branford, CT
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