On Thu, 13 May 2004 23:43:44 +0200, "Marc Battyani"
wrote:
This last is more important because it subsumes the radiation
resistance that must be known to perform any efficiency computation.

The loop is 65x25mm

Hi Marc,

Yes, this appears to be extremely problematic. The radiation
resistance of this loop is on the order of 30 nano Ohms. For you to
achieve a 50% efficiency in radiating the input power, no component in
the resonant loop must have an Ohmic resistance greater than this.

In other words, you don't stand a chance. What coupling that you are
getting is probably more capacitive or inductive than radiative.

The radiation resistance of a loop rises or falls by the 4th power of
its ratio to the wavelength of excitation. Double the radius and you
will multiply the radiation resistance 16 fold.n What this says, is
that the radiation resistance is overwhelmed by conductor loss (even
if only micro Ohms) that is turning your power into heat.

The two circuits are very, very different from each other to be
achieving the same purpose. The 50 Ohm resistor, as an equivalent
load appears to be grossly in error.

Yes. This is the problem.

An understatement with the loop dimension given.

73's
Richard Clark, KB7QHC