Reg Edwards wrote:
From where do you get your "thousands of volts" - the old wive's monthly
magazines?

Try this, Reg. A dipole is a standing-wave antenna. Most people know
that the voltage 1/4WL away from a current maximum is pretty high.
Since the current is zero at the end of the dipole, all the energy
is contained in the E-field. So what is the voltage when all the
energy is in the E-field?

A 1/2WL dipole is a lot like a lossy piece of 600 ohm transmission
line. 600 ohms is in the ballpark of the natural Z0 of a dipole if
there were no reflections on it, i.e. if it were terminated such
that reflections were eliminated thus turning it into a traveling-
wave antenna.

If one assumes that at the dipole feedpoint, (VF+VR)/(IF+IR) = 50 ohms,
and if the traveling-wave impedance of a dipole is 600 ohms, one can
calculate the ratio of VR to VF. Turns out to be about 0.9. So VF is
about ten times the feedpoint voltage. At the open-circuit at the end
of a dipole, VF adds in phase with VR so the voltage at the open-circuit
end of the dipole is about 20 times the feedpoint voltage.

The feedpoint voltage at 100W is about 70.7V. Therefore, the voltage
at the ends of the dipole is about 1414V RMS. Multiply by 2.8 to get
peak to peak at about 4kv.

When I worked for Schlumberger in the oil fields, we could easily
draw a 4 inch arc if someone got their aluminum hard hat too close
to the mobile radio whip during transmit. How much voltage does it
take to draw a 4 inch arc?
--
73, Cecil, W5DXP



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