🚀 go-pugleaf

In article <1990Aug7.182544.7140@ns.network.com> logajan@ns.network.com (John Logajan) writes:
>peter.saulesleja@f630.n250.z1.fidonet.org (peter saulesleja) writes:
>>Why high current?  There is very little current travelling through 
>>the antenna, so you'd require a higher voltage, wouldnt you?  After 
>>all, P=VI...  And air is a very good insulator.   A high current would 
>>be impossible, at reasonable  (below 1,000,000 Volts) voltages.
>
>The impedance of the antenna/radio wave interface is generally on the
>order of 50-300 ohms.  Given this "resistance" you can come up with
>the voltages needed for a given power.  For simplicity, to drive one
>watt into a 100 ohm load would require 10 volts.

  More specifically, the formula is:

	P = VI

  and:

	I = V/R

  combine the two, you get:

	P = V^2 / R

   Of course, the RF output is usually expressed in RMS Watts, so the
Peak-Peak output must be 1.414 (square root of 2) greater than the RMS
voltage derived above, therefore you can work them around a little:

	V (Peak to Peak) = sqrt(2PR)
	V (RMS) = sqrt(PR)
	P (rms) = V (P-P) / 2R

  Enjoy.


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