I forget the math?
P=IxE and E=IxR so P=IxIxR? 7KW=IxIx50 7000/50=IxI 140=IxI
140=me twice
Damn now I've got a headache!


Dave VanHorn wrote:

7kW at 50 ohms isn't THAT much current.

What's the feedline connector?

I forget the math?
P=IxE and E=IxR so P=IxIxR? 7KW=IxIx50 7000/50=IxI 140=IxI
140=me twice
Damn now I've got a headache!

P=I^2R so assuming a properly matched load, P/50=I^2 and I is about 12A.

--
KC6ETE Dave's Engineering Page, www.dvanhorn.org
Microcontroller Consultant, specializing in Atmel AVR

On Mon, 18 Oct 2004 10:39:02 -0500, "Dave VanHorn"
wrote in :

I forget the math?
P=IxE and E=IxR so P=IxIxR? 7KW=IxIx50 7000/50=IxI 140=IxI
140=me twice
Damn now I've got a headache!

P=I^2R so assuming a properly matched load, P/50=I^2 and I is about 12A.


That's 12 amps @ xxMHz -- don't forget about skin effect. At 30 MHz
with 12 AWG copper you will end up radiating about 22 watts per foot
in the form of heat. And to put that into perspective, most small
soldering irons are 15-30 watts.






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