VK2PRC Calculating your Transmitter P.E.P. RF Power using a dummy load.

A simple method of measuring your transmitter output power using a dummy load and a few basic components.
Note: The dummy load on the Ipad shows 55 ohms, this should show 50 ohms. My dummy load in the test is a 50 ohm dummy load. The 0.64 volts drop across the diode has not been taken into consideration for simplicity.

[@rickb.6068]

I really like the way you used a ring of wire instead of a copper plate for the resistors on the dummy load,. I will be doing that on my next build.

[@yorkshirebikerbitsnbobs]

You have "flocks" of "parrots" just flying by? Amazing!

[@vk6op648]

Good stuff mate. Very handy. Cheap as chips to put together too.

[@rodolfobonzonjr7490]

Lovely video, very informative .

[@timhenderson2643]

Can you provide the values of the components used in the jumper box for the multimeter??

[@paul.alarner6410]

have you looked at the 320's harmonic output on a rectum paralyser ?

[@jagowadkins]

Hi, Lovely video thank you. What are the components and their values in the circuit please? Thanks

[@paul.alarner6410]

why no resistor values to the right of the dummy load on the schematics?

[@ab9957]

I don't think you mentioned the frequency of your transmitter?

[@vk2prc978]

20M band.

[@vidasvv]

Nice job ! TNX 4 the upload 73 N8AUM

[@hrf1007]

Can it use for fm band too? Or need some changes in circuit or formula?!

[@G7VFY]

Should not make any difference and that method accurate to 30MHZ , maybe higher.

[@peterbold4297]

Hi, I'm piero, I'm in Italy and I'm watching your video. Is the voltage you use for the calculation the peak voltage?

[@vk2prc978]

Yes PEP= PV squared divided by 50 ohm load.

[@peterbold4297]

@@vk2prc978 Hallo, thanks for the reply. For what type of modulation do you consider P.E.P. ? But then, apart from everything, it seems to me that the meter shows not the Vp but the Vpp = 2Vp: Or am I taking a big mistake?

[@simonroberts6490]

@@peterbold4297 His recitfier is half wave. Just single diode, so Vp, not Vpp I believe.

[@peterbold4297]

@@simonroberts6490 Hi. Let's call the 68pF C1 and C2 the 20 nF in parallel to the instrument. C1 charges to Vp during the negative peak through D1. During the positive peak D1 is open and the voltage present on C1 is added to the positive peak in phase with it, so that C1 is charged by the voltage Vpin (pos) + Vc1 = 2Vp or to Vpp. In this way the power obtained from the calculation is obviously four times the real one.

[@simonroberts6490]

@@peterbold4297 Well, when I run this simulation in LTSpice, it shows the voltage on the capacitor to build to the peak voltage minus a diode drop, not the peak to peak. Of course, I can't show you what I did since there are no pictures or attachments in this medium, so that doesn't move us forward very far :)

[@colincolin30]

Must do a test one day as well. 73

[@richylad]

does anyone know how to do this other way, ie, if you already have say 10w PEP, what voltage would you expect to see on your meter ? thanks in advance

[@vk2prc978]

Simple mathematics, If P=V squared divided by 50 then V= square root of P x R

[@dxexplorer]

Thank you for this... it was just on time for me to make some final adjustments for a ten minutes transmitter I found on SolderSmoke blog... and I didn't know how to measure the output voltage ))) So now that I do... I ended up with 7v output voltage... after the calculations I ended up with 0.98.... I'm a little dumb as I don't know what that means... 980mW ? Or 0.98mW ? Hahaha... I'm confused. Thank you one more time. 73

[@vk2prc978]

Yes thats just under a watt. 980mw.

[@dxexplorer]

Thank you... 73

[@joerowland7350]

Nice I have one about like this
I use baby oil from the dollar store
N my resisters are 5 watt

[@user-rn7ft8qd8t]

Your math should be Power = 10 * (Vp)^2 in Volts and mW.. Your meter shows peak voltage Vp (not the rms voltage).. You may add the 0.64V drop to the Vp to be more precise..

[@robinbrown7019]

Audio up and down hard to follow