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Author Topic: Wilkinson power divider  (Read 434 times)
LA9XNA
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Posts: 189




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« on: November 24, 2017, 03:09:04 AM »

Can anybody tell me the wattage of the 100ohm resistor in a Wilkinson divider.
I'm planning to use two 100W 2m power modules to get a 200w amp.
I have 100ohm 30W RF resistors meant for divider but are unsure if it is enough.
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AC7ZN
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Posts: 83




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« Reply #1 on: November 24, 2017, 03:53:44 AM »

The power dissipated in the resistor depends on the load mismatch, and source power and phase mismatch.  If your modules are matched in amplitude and phase and your antenna is exactly 50 ohms resistive, no power is dissipated in the Wilkinson resistor, which is what you want (that's how you can get 200 watts to the antenna with low loss).  A severe mismatch or transmitter phase difference will cause dissipation power near what is fed into the network, but that will also probably cause your transmitters to fault or reduce power.   

As long as you are careful to match everything, thirty watts for that resistor seems more than adequate to me.

Good luck and 73,

Glenn AC7ZN

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