Measuring power
WA4EFS
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, Lloyd


Charles W. Powell
Lloyd,
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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 peaktopeak voltage, I make my calculation by deriving the RMS voltage, then using RMSsquared/resistance. So with about 13.4 DC volts input, my peak output voltage on the HT20 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 QRPometer, 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. 72, Chas  NK8O


KB9BVN 
If I measure my HT20 into a dummy load with my WM2 power meter, I see close to 7 watts. Perfect situation. If I measure my HT20 into my attic dipole thru my MFJ antenna tuner, I see close to 5 watts. Imperfect situation. My attic dipole is a halfwave 40m dipole...so tuning it up on 20m is not real efficient anyway.
On Wed, Apr 17, 2019 at 10:03 AM Charles W. Powell via Groups.Io <doctorcwp=yahoo.com@groups.io> wrote:

73 de KB9BVN Brian Murrey


chuck adams
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 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. :) FYI chuck, k7qo
On 4/17/19 7:03 AM, Charles W. Powell
via Groups.Io wrote:
Lloyd,


WA0ITP
My personal favorite is (Vpp/2.8)^2/feedline
R This makes sense to me and is easy to remember. Same as yours
Chas.
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72 WAØITP I love this radio stuff. www.wa0itp.com www.4sqrp.com On 4/17/2019 9:03 AM, Charles W. Powell
via Groups.Io wrote:
Lloyd,


WA4EFS
Thanks for the several informative replies re. RF power measurement. With a fully charged 12 volt battery as power source and a 50 ohm resistive load my scope indicates 40 volts PP from the Hilltopper 40, equating to 4 watts, which seems about right. 13.8 volts might produce an additional watt or two out.
73, Lloyd
From: HilltopperKit@4SQRP.groups.io [mailto:HilltopperKit@4SQRP.groups.io] On Behalf Of WA0ITP
Sent: Wednesday, April 17, 2019 2:03 PM To: HilltopperKit@4SQRP.groups.io Subject: Re: [HilltopperKit] Measuring power
My personal favorite is (Vpp/2.8)^2/feedline R This makes sense to me and is easy to remember. Same as yours Chas. 72 WAØITP I love this radio stuff. www.wa0itp.com www.4sqrp.com On 4/17/2019 9:03 AM, Charles W. Powell via Groups.Io wrote:

