Re: Measuring power

chuck adams <chuck.adams.k7qo@...>

You can save a few steps on the calculator by using  P = V_pk^2/100 for a 50 ohm load.
Always saves my bacon on a calculator.  Formula works only for 50 ohm load.

You can get the above formula from yours below by replacing  0.707/2 by  1/sqrt(2),
thus (0.707/2)^2 = 1/2 and (1/2)/50 = 1/100.  For those that missed that day of
class.  :-)


chuck, k7qo

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 accurate.

Caveats:  I might be entirely wrong in my calculations and I might be missing some subtlety in making this calculation.


Chas - NK8O

On Apr 17, 2019, at 8:29 AM, WA4EFS <himself@...> wrote:

Several 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.
Thanks and 73, 

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