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To All from PA0SE
As suggested by Rik I have modelled by AO a square loop made of
15mm diameter copper tube.
To make it more or less comparable to the 180m vertical I have
given the loop sides of 180m; the horizontal bottom leg 5m off the ground. The
feedpoint is at the centre of the bottom leg.
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.
Radiated power is 0.3W.
Attached the value of the magnetic field component up to 2km from
the antenna.
The E-field will be sent as a following e-mail.
73, Dick, PA0SE
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