On 12/17/18 3:11 PM, Jim Brown wrote: > This overly simplified equation fails to account for proximity effe Proximity effect will effect impedance, which will then effect the loss.
The wire loss equation is dependent on the TL Zo. The previous example was for Zo=100. Using Zo=75, wire loss for #12 is 0.47 dB/100. For the Teflon, that would leave 0.47 dB for dielectric loss. This is sounding closer. John KN5L > On 12/17/2018 12:58 PM, John Oppenheimer wrote: >> On 12/17/18 1:58 PM, Jim Brown wrote: >>> In order of loss at dB/100 ft at 10 MHz from low to high, the >>> Teflon #12 is lowest at 0.94dB, then RG400 at 1.22 dB (about the same as >>> RG58), then #12 THHN at 1.34 dB. #12 or #10 enameled copper had the >>> greatest loss, 2.4 dB/100 ft. >> Hi Jim, >> >> Two parallel transmission line is the easiest of all to evaluate using a >> RF resistance table and knowing the impedance of the line. >> http://ve3efc.ca/wireohms.htm >> >> The wire loss is dB = 10 * log10((Zo + WireR)/Zo) >> >> Assuming Zo=100 for the #12 Teflon and THHN. R at 10 MHz = 4.24/100 = >> 8.48/200. >> >> Wire dB = 10 * log10(108.48 / 100) = 0.35 dB >> >> Using the two values above at 10 MHz, 0.94 and 1.34, the dielectric loss >> is 0.59 dB and 1 dB. The dielectric loss is about two and three times >> the wire loss. >> >> John KN5L ______________________________________________________________ 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]
