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Loop antenna matching question
Richard Clark wrote in
: On Sat, 30 May 2009 07:19:26 GMT, Owen Duffy wrote: The main loop isn't anything like 2 ohms at resonance, it is much much higher, and the voltage is quite high for a given power. Hi Owen, By your latest comment, it appears you have jumped off the rails completely. Let's look at the spec: On Wed, 27 May 2009 13:36:53 -0400, Michael Coslo wrote: It's Octagonal, .75 inch copper tubing, around 8 feet in diameter. It's designed to just reach into the 80 meter voice segment with it's capacitor setup. which through any of several methods reveals radiation resistance equals 0.27 Ohms. The need for large conductors, high voltage components, and loop (which, for 80M would be half a meter) coupling to obtain the proper match are all found in their proper relationships. My reckoning: I ran up a model in NEC of a vertical circular loop of 2.4m diameter, made of 19mm dia copper conductor, and located with its centre 3m above average ground. The feedpoint impedance is 0.13+j168 at 3.5MHz. If that was resonated with a lossless capacitor, the parallel load impedance would be about 220k+j0. To transform that to 50+j0, the impedance ratio is 4400:1, the voltage ratio is 4400^0.5:1 =66:1, the areas need to be in a ratio of 1:66, the diameters (for circular feed loop) 1:66^0.5 =1:8.1. The feed loop would be about 12% of the diameter of the main loop or about 0.3m. .... Reggie covered all of this years ago. Unfortunately, he buried the best of his legacy in a drift of bantering making it an extreme chore to mine Google for the relevant postings. Reg did leave us a calculator for this problem (MAGLOOP4) but without (AFAIK) documentation of the method used. MAGLOOP4 comes up with slightly different figures to my NEC model, slightly higher loss, slightly lower main loop // Z, slightly larger feed loop. Owen |
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