Yes, you can use it without the Nano.
Assume the transmitter is driving the 50 ohm dummy load with a 20 volt peak to peak sine wave.
Since the other end of the 50 ohm dummy load is grounded, the 20 volts peak to peak is centered on ground,
so -10 volts and +10 volts at the peaks of the sine wave.
The meter will measure around 10 volts.
When computing power for a sine wave like this, we use the rms voltage, in this case it is 10 volts * 0.707 = 7.07 volts rms.
Now we can use your formula:? ? power = V^2/R = 7.07vrms^2/50ohms = 50/50 = 1 Watt.
In general, measure the dc voltage across the cap, multiply by 0.707, then use that as the voltage in your V^2/R formula.
You will get slightly more exact results if you add 0.6 volts to the meter reading to account for the drop across the diode.
So in the example above where the meter read 10vdc across the cap,? we would use (10v+0.6v)*0.707 = 7.49vrms
You do want something like Jacks 58k ohms of resistance across the cap, value is not critical.
Higher resistance makes the peak voltage stick around longer for you to read with a meter.
Diode is not critical, a 1n4148 or 1n914 should be fine at moderate power levels.
Maximum reverse voltage for these is 100 volts, so should be good up to a peak-to-peak max of 100 volts,
so a peak to ground of 50 volts (as seen on the meter), or 50*0.707=35.35vrms, or 25 Watts.?
Jerry, KE7ER
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On Sat, Dec 30, 2017 at 06:22 pm, Gene Nitschke wrote:
Can I just use the circuit w/o the nano,?I assume the resistor divider is for sampling by the nano???If?I connect the rig across the 50 ohm load and measure the voltage across the capacitor, what is the relationship of voltage to power (is it simply V^2/R)?
