To further illustrate what is happening here, I took my "quarter-wave"
dipole example, matched it with a simple L network at 30 MHz (series
4.1uH inductor and shunt 180pF capacitor) and looked at the voltages and
currents around the network at the 1KW level. Assuming lossless matching
components:
Voltage at input of matching network = 224 Volts rms
Current into matching network = 4.47 A rms
Power into matching network = 224*4.47=1KW
Impedance looking into matching network = 224/4.47=50 ohms
Antenna feedpoint impedance = 13-j750
Voltage across antenna terminals = 6,578 Volts rms
Current into antenna terminals = 8.77 A rms
Radiated power = 8.77*8.77*13 = 1KW
Notice that you can't multiply 6,578 Volts by 8.77 A to get the radiated
power, because the voltage and current are almost in phase quadrature at
89 degrees.
Described in simplistic terms, the matching network has transformed the
transmitter voltage to a sufficiently high level to overcome the large
capacitive reactance and force enough current into the antenna to
dissipate 1KW in the 13 ohm radiation resistance.
Whichever way you look at it, for equal radiated power, more current
flows into the shorter antenna!
73,
Steve G3TXQ
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.
>
> 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
>
>
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