I need to automate my antenna connections. I have a network centric relay board. I need to know the amperage of 500  600 watt RF signal. Does anyone know how to figure that out?
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Posted 3 years ago
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The current is a function of the impedance. One needs to know that before Ohm's law can be applied.
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Assuming the impedance is 50 ohms, the Ohms Law formula says at 600 watts, the current will be 3.46 amps with a voltage of 173.21 volts. Let's assume you don't switch the antenna with the transmitter keyed, and then a relay with 5 amp or better contact rating should get the job done.
Good luck,
Barry (W4TGA)
Good luck,
Barry (W4TGA)
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Divide the power by the impedance and take the square root of the answer.
73, Bill W6WRT
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Youi answer is correct only if the impedance is real. If it is complex, you will get both average real and reactive power.
KY6LA  Howard, Elmer
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Jim is correct. The RMS current of 3.162 A is only valid if your SWR is 1:1.
So you really need to design your system for a worst case SWR.
When you have a SWR greater than 1:1 you have both a resistive current and a reactive current
So please give us your worst case SWR and whether or not it's capacitive or inductive so we can calculate your worst case current
THEN follow the usual Engineering safety factor so that your relay is able to handle at LEAST TWICE your worst case current just to be safe because RMS currents are Only 70.7% Peak currents.
So you really need to design your system for a worst case SWR.
When you have a SWR greater than 1:1 you have both a resistive current and a reactive current
So please give us your worst case SWR and whether or not it's capacitive or inductive so we can calculate your worst case current
THEN follow the usual Engineering safety factor so that your relay is able to handle at LEAST TWICE your worst case current just to be safe because RMS currents are Only 70.7% Peak currents.
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Power seems to be a misunderstood subject. When I hear someone use the term RMS power I want to go somewhere and puke. Power is a squared function and why in the world would one want to square a squared function, takes its average over a period and, then, take the square root. Oh well life goes on.
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