To All from PA0SE
AO gives the feed point impedance as Z = 0.380 + j60.1 ohm and
the efficiency as 0.03%.
So for 1 kW fed into the loop the current must be SQR(0.380 *
1000) = 19.5A.
The voltage at the feed point becomes 19.5 * 0.380 =
1172V.
Thanks Andy!
I made the same stupid mistake in my message nr. III on this
subject (14-07-01 10.37)
The corrected figures are even more dramatic than for the
loop.
To feed 1kW into the 180m high 15mm copper pipe the current
must be 154A and the voltage on the antenne becomes 2.5MV! Perhaps the pipe
would survive the current but the voltage makes the system completely
unrealistic (corona).
And what would the required 286mH loading coil for that voltage
and current look like ...
73, Dick, PAoSE
