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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 P-P from the Hilltopper 40, equating to 4 watts, which seems about right. 13.8 volts might produce an additional watt or two out.
From: HilltopperKit@4SQRP.groups.io [mailto:HilltopperKit@4SQRP.groups.io] On Behalf Of WA0ITP
Sent: Wednesday, April 17, 2019 2:03 PM
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.
I love this radio stuff.
On 4/17/2019 9:03 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.