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![]() Roy Lewallen wrote: We can use the inductor as an autotransformer. If we tap up on the inductor some fraction k of the whole way up, the impedance we'll see at that tap will be very nearly 1275 * k^2, and it'll be purely resistive (no reactance) because the impedance across the whole coil is purely resistive. For example, half way up the coil we'll see 1275 * (.5)^2 = 1275 / 4 = 319 ohms. So to get 50 ohms, we tap up sqrt(50/1275) = 20% of the way up the coil. Roy Lewallen, W7EL Hi Roy, I'm OK with 319+j0 half way up the coil, but at 20% of the way up I keep getting 50+j50. I did it on the Smith Chart, and on a Spice based circuit analysis program. Too old to do this stuff by hand. That's not a real good match for 50 ohm coax. The are other taps that will provide a better match, but no where did I find 50+j0. In an earlier post I stated that it looked like the real part of the antenna impedance needed to be less than 50 ohms to get a perfect match, using this method. The impedance across the whole coil is not purely resistive, it is at the 50% point. Apparently I am modeling incorrectly, or missing something. Gary N4AST |
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