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On Apr 5, 4:29 pm, Rick wrote:
How do I calculate the voltage rating I'll need for capacitors used to make traps? I am making a portable field antenna for 160 through 40 meters, for use with 100 watts absolute maximum (normal use will be from 5 to 50 watts). One way to put an absolute maximum rating on the trap voltage and current is to assume that ALL your power is dissipated in the trap. Obviously, that would be a real bummer, but it's a limit. In the case of a symmetric dipole, the power would divide, probably fairly close to equally, between the two sides. Remember that you want to be sure the cap doesn't flash over, so take into account peak voltages: peaks of modulation, etc. So if you have a trap that, at resonance, has Xc=Xl, and has some Q, then at power P dissipated in the trap, the voltage will be sqrt(Xl*Q*P). Circulating current will be sqrt(P*Q/Xl). You mentioned in another posting 100 watts absolute max, and 200pF resonant at 80 meters, so Xc=Xl=200 ohms, approximately. You want to make them small, so I'd guess Q300...let's say Q=200. So for the case of 100W in one trap (that coil's gonna get really hot if it really is required to dissipate that for long...), that's Vrms=sqrt(200*200*100)=2000V. Irms=sqrt(100*200/200)=10A. Please note that the CURRENT in the cap may be a bigger problem than the voltage across it! That's absolute worst-case. I'd guess you'll see more like 5 watts max per side dissipated with 100 watts input, the rest being radiated, and if that is indeed the case, Vrms=450V and Irms=2.23A. If there's only 1 watt dissipated, it's still 200V and 1A. At about 30pF/foot, RG-58 could do the job, but 200pF would be almost 7 feet of coax. That would be economical, to be sure, and should be up to the job with no problems. Paralleling some high voltage ceramic discs would probably work...use NPO (C0G) type, maybe 5 or 10 in parallel to avoid current problems. Cheers, Tom |
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