You can save a few steps on the calculator by using P = V_pk^2/100
for a 50 ohm load.
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Always saves my bacon on a calculator. Formula works only for 50
You can get the above formula from yours below by replacing 0.707/2
thus (0.707/2)^2 = 1/2 and (1/2)/50 = 1/100. For those that missed
that day of
On 4/17/19 7:03 AM, Charles W. Powell
via Groups.Io wrote:
I don’t know about anyone else, but I am measuring
power on mine, key down, using a 50 ohm dummy load, and
connecting my high impedance oscilloscope leads across the load.
The scope is a a Siglent 200 mHz scope, so it is measuring far
away from the rolloff. Using the peak-to-peak voltage, I make
my calculation by deriving the RMS voltage, then using
RMS-squared/resistance. So with about 13.4 DC volts input, my
peak output voltage on the HT-20 was 52 volts. ((52 *
0.707/2)^2)/50 = 6.8 watts into 50 ohms. I get approximately
the same readings with my MFJ tuner’s power readout and also my
QRP-o-meter, although I don’t expect those to be nearly as
Caveats: I might be entirely wrong in my
calculations and I might be missing some subtlety in making this
Chas - NK8O
On Apr 17, 2019, at 8:29 AM, WA4EFS <himself@...
posts have referred to the RF power output of QRP rigs
including the Hilltopper, the relation to amplifier
class, or to final transistor quality, etc. I am
curious as to how power is being measured or assessed
in relation to various reports.