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From: email@example.com (Tom Bruhns)
Subject: Re: RF power measurement
Date: 14 Nov 2002 10:28:21 -0800
NNTP-Posting-Date: 14 Nov 2002 18:28:22 GMT
Paul Burridge wrote in message news:...
> I've got several books on RF and they all seem to give slightly
> different variations for the equation for power measurement.
> For example one says Pout = V^2/8R which seems a bit dubious.
> Let's say I get a sine wave of 10v peak-peak on my scope when I
> connect the output of an amp. across a 50 ohm purely resistive load.
> How do I arrive at a figure for a) peak output power and b) RMS output
Instantaneous power delivered to a resistive load is e^2/R, where e is
the voltage drop across the resistor (load) at that instant. Average
power is E(rms)^2/R. It's convenient, in case you are dealing with
reactive loads, to express average power as I(rms)^2*re(Z) =
I(rms)^2*R, or E(rms)^2*re(Y) = E(rms)^2*G. You'll often find thermal
(RMS-reading) RF ammeters at the feedpoint of antennas of known Z.
It's easy, then, to find the power delivered to the antenna. It's
probably best to not use the term, "RMS output power." RMS of what??
Surely not the power!
When you're working with 50-ohm systems, one type of power meter
you'll find presents a 50 ohm load to its input connector. And it may
actually work by letting the input power heat a real resistor, and
measuring the temperature rise of that resistor (usually by comparing
it to the temperature rise of another resistor in a matched
environment, and servoing a DC voltage to that second resistor so the
temperatures match). That sort of meter measures the power your
source actually delivers to a 50 ohm load. As another poster pointed
out, the power you get from a source depends on the load you put on
it, and what you get into 50 ohms will in general be different that
what you get into any other load. Typically power meters like that
operate over a limited power range--if it's a low power range, you'd
put attenuators between the source and the power meter head to get the
power within range, and find the power as the indicated power in dBm
plus the value of attenuation you've put in front.
There are also through-line power meters, often capable of resolving
forward and reverse power...under the assumption of a known impedance
(commonly 50 ohms).
You've probably just scratched the surface of this question...it can
be a challenge to _accurately_ measure power over a wide frequency
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