Ok what I have is a dummy load of 50 ohms attached to a CB radio. On the dummy load, for measuring purposes there is a voltage divider and a diode, the usual. one thing that keep stumping me and I am losing my memory, getting old, need an excuse, is: is this accurate? what I mean is after it goes through the diode do I add back the voltage drop across the diode to get a true reading or is it correct just the way it is? Now for the diagram: Which I have attached.
the output from the coax goes into a 100k in series with a 10k resistor which terminates into ground. The center tap of the two resistors goes to the diode and as I recall maybe to a capacitor. which I suppose might raise the voltage as well. I forgot about that in the diagram too. So now the reading I get is right at .9 volts for one CB and right at 1 volt for another. So I am assuming of course that maybe that is close to 4 watts output. One of these radios was a cobra 148 which was the one showing one volt. What is your calculations on this? I don't have a wattmeter. I tried to compute it say finding the total current through the two resistors then squaring it and multiplying by 110,000 ohm. doesn't compute right. I-squared*R. how can I compute it correctly?

 

First of all, I wouldn't bother with the potential divider - you want as much signal as possible to reduce the effect of the diode's forward voltage drop (which is something of an unknown). As you've said, the diode ought to be used to charge a small capacitor (and you ought to have a high value resistor across the capacitor too).

The capacitor will then be charged on positive half-cycles, to a voltage equal to the peak signal voltage minus the diode's Vf. If we ignore Vf, then the rms power will be V(squared)/2R watts. For R=50 ohms we'll have V(squared)/100 watts.

 

I see your point about the voltage divider. If the voltage was high enough the voltage drop across the diode would be nothing compared to the real voltage and could basically be ignored. This particular divider was already existing not sure if it was factory or not. I won this thing at a Ham Radio Christmas party, white elephant gift, but to me it is more than that. You know another mans junk thing. I did get lost in your final calculation. I thought that e-squared / R would equal watts, but how did you come up with 2R and the 100 watts or 100 and answer in watts?
So I should come straight off the output and directly into the diode then the cap with a resistor across it? Then calculate and ignore the VF of diode?
Is the 2R because of RMS?
Now the resistor across the cap what is the purpose, to bleed off the voltage so that you are constantly getting new readings or to slow it up a bit? What size of resistor would you suggest?

 

Yes, the 2R is all about rms. With a small capacitor the input resistance of your measuring device will probably be sufficient to ensure the capacitor is in a discharged state when there's no signal (you need, as you say, to slowly bleed off charge to keep refreshing the reading). But just to put some figures on it, I would suggest 10nF and 1Mohm.

The 100 figure is just 2 times 50 ohms, so you have power in watts is equal to V(squared)/100.

 

ah ha I see, well that sounds good. Thanks for the advice I will give it a go. I just hate to buy anything when you can make it yourself. I have enough stuff here as I have been saving all my life. The unfortunate thing is I have to get rid of it because I am getting older and will never get to the stuff. crazy isn't it. As you can see I like messing with rf output. The one thing I'd like to see and I haven't checked here on the forum yet is people who mess with the 160 - 190 KHz band. Always wanted to play with that since I was young. I see some info on the web but those guys are serious - serious about it. Anyhow maybe I will start another thread on that. Thanks for the help, greatly appreciated for your time.

 

You're very welcome, pleased to be of assistance!