Re: Power output tests at 24v PA

Jerry Gaffke

That's correct, assuming a forward voltage of 0.4 Volts across the diode.
According to fig 7 of the datasheet, that happens at a current of 20 uA when the temp is 25 C.

Dividing the peak voltage (what you measure with a diode detector) by 1.414 
converts it to an RMS voltage, assuming we are dealing with a sine wave.
We must use that RMS voltage when calculating power.


On Sat, Nov 9, 2019 at 12:42 PM, Mike Kilpatrick wrote:
I could be way wrong but I measure DC volts, add 0.4V then divided by 1.414.
Multiply the result by itself (square it) and divide that result by the 50 ohms load. Result is power in watts. 


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