On Sun, Jan 29, 2017 at 1:01 PM, Lynn W. Taylor, WB6UUT < [email protected]> wrote:
> If the wire is near 1/2 wavelength on a band, the impedance at the end > will be very high, and the tuner may not be able to match it. > Why is that? A quick heuristic to help understand. The current at the far end of a random wire is zero. One-quarter wavelength from the far end, the current is maximum. Another one-quarter wavelength along (total of one-half wavelength from far end) the current is zero. I = E / Z >> Z = E / I So, one-half wavelength from the far end, Z is going to be whatever E is divided by a very small number >> VERY HIGH and hard to match. This also explains why the impedance at the center of a half-wave dipole is reasonable. Current at either end of the dipole is zero and high at the center. Voltage at either end of the dipole is high and low at the center. Z = E / I At half-wave dipole center, low voltage at center divided by high current at center gives a low impedance, easy to match. It's all basic physics. There is no magic magic number or formula. Despite what some antenna manufactures will tell you. Ham Exuberantly, Hank, W6SX ______________________________________________________________ Elecraft mailing list Home: http://mailman.qth.net/mailman/listinfo/elecraft Help: http://mailman.qth.net/mmfaq.htm Post: mailto:[email protected] This list hosted by: http://www.qsl.net Please help support this email list: http://www.qsl.net/donate.html Message delivered to [email protected]
