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![]() "nick" wrote in message news ![]() If a (let's say 12v) relay is rated for 10 amps @ 110 volts, it will take 1100 watts. Will it take 1000 watts of RF? I would think not, but I don't know why. There are two parameters specified, but you are reducing them to one. The first is the current, the second is the voltage. With a non-reactive 50-ohm load, the rf current at 1000 watts is about 4.5 amps, and the rf voltage is 70.7 V. These formulae are in every edition of the ARRL Handbook. See also the Handbook discussion of skin effect. At higher frequencies, ac tends to travel near the surface of a conductor. If the relay contacts are silver plated, this may not be a problem. The contacts in the old Dow-Key coax relays weren't very large; it doesn't take a fat relay to switch non-reactive 50-ohm loads. If you want to switch high-impedance or badly mismatched antennas, you may exceed the capability of the relay. Another consideration is whether you will actually be hot-switching the rf. If the contacts are closed before rf is applied, there is less likelihood of arcing and damaging the contacts. Relays designed for rf use are mechanically constructed to minimize impedance variation through the contacts, to maintain the non-reactive 50-ohm impedance of the system. At HF, a slight impedance 'bump' is not a big deal. For amateur HF use, this relay will probably do the job. "PM" |
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