Subject: Re: RF power measurement
Date: 14 Nov 2002 09:23:36 GMT
User-Agent: tin/1.4.5-20010409 ("One More Nightmare") (UNIX) (CYGWIN_NT-4.0/1.3.12(0.54/3/2) (i686))
In sci.electronics.design Chuck Simmons wrote:
> Paul Burridge wrote:
>> 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.
This isn't a power *measurement* - it's the *definition* of the maximum
power that can be delivered (from the source) into the load if the source
and load are conjugately matched. If you don't know what conjugately
matched means, you'd better read a bit further into those books!
Take a look at this circuit; it's a signal source of internal
impedance Rsource, connected to a load of Rload:
gen ___ V'
/ \ |
V ( ~ ) .-.
\ / | | Rload
| | |
Okay, V is the peak GENERATOR voltage, i.e.
v(t) = V * cos(2*Pi*f*t)
so the RMS voltage AT THE GENERATOR is:
Vrms(gen) = V/sqrt(2)
Rsource and Rload form a resistive divider, and for "maximum power
transfer", we require that Rload = Rsource, so the RMS voltage at V' is:
Vrms(V') = Vrms(gen)/2
power delivered to load is "rms voltage squared over Rload", i.e.
Power = Vrms(V')^2/R
= (V/(sqrt(2)*2))^2 / R
= V^2 / 8 / R
>> 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
a) be careful what you mean by peak power - it more usually means
something like "the highest power observed when the envelope of
a signal is fluctuating in amplitude", which is very different
from measuring the highest voltage in a repetitive sine-wave
(Vpeak, of course) and using that to define power. That's more
like peak instantaneous power.
b) Again, be careful...in RF, "output power" is not generally the
same as "power delivered to load", which is probably what you
mean. In the very specific case of a load conjugately matched
to the source, delivered through a lossless transmission line,
then your 10V P-P is equal to 5V peak, which is 5/sqrt(2) V rms.
square that and divide by the load: 25 / 2 / 50 = 0.25W rms.