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. Isquared*R. how can I compute it correctly?
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. Isquared*R. how can I compute it correctly?
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