[Elecraft] RF Power

[David Robertson]

Greetings,

With all the information on calculating actual RF power and all these formulas being presented I fine it can be very confusing to people who don't have an electronic or mathematic background.  Being a technical instructor and self-made engineer I always find that a little explanation of the basics before springing a formula on someone really eliminates a lot of confusion.

So if those knowledgeable people will excuse my bandwidth let me explain a couple of things.

If you have a 110 volt 100 watt light bulb and you connect it to a 110 volt DC source (like a battery) it will dissipate 100 watts of power.

On AC power the voltage and current is always reversing polarity in a sine wave pattern at a given frequency (60 HZ for our power lines in North America).
Because AC is constantly changing, we must apply 110 volts RMS AC voltage (Root Mean Square). This strange RMS value is simply the actual peak AC voltage that is needed to produce the DC equivalent (100 watts) across the load or bulb or the equivalent of driving the load with DC voltage of 110 volts. I am not going to derive RMS here but I will tell you the value  RMS voltage = Peak voltage/Square root of 2.
Having said that you would have to have a peak value of 155.56 volts (or 311.27 Peak to Peak (maximum positive to negative voltage swing on AC)) to dissipate 100 watts of energy across the 100 watt load or bulb.

Now to RF power.

RF out of your rig is a sine wave (or better be or you got harmonic problems). If you look at your RF output with a scope you will see a peak to peak ac voltage which is being dissipated across your load (which is either your antenna or dummy load). If you use a RF probe that uses a diode to rectify the RF you will see a  peak voltage of the RF minus the voltage dropped across the diode in your probe.

Power (RMS) = E ( the RMS voltage) squared / The resistance R of the load (EG 50 Ohms).

If you are using a scope you have to read the Peak to peak value and divide that by 2 for the peak value. Now to convert this value to an equivalent DC power value ( RMS) you must divide your peak value by the square root of 2 or 1.414 ( this is plenty accurate for our calculations or you can use 1.414213562373 and on and on if you are a accuracy type. Look at the example below.

200 volts peak to peak / 2 = 100 volts peak
100 volts peak / 1.414 = 70.72 volts RMS
Now Square the 70.72 value which will be ( E^2 ) 5110.31 and divide it by the load resistance ( 50 Ohms ) which is 100.02 watts.

If you use a probe, in order to have most accuracy, you should measure the voltage dropped across the diode in the probe. Silicon diodes (1N1458 ) drop about 0.6 volts. Germanium diodes ( 1n34 ) drop about 0.3 volts.

Power (RMS) = ((Probe reading + diode drop) divided by 1.414) squared, divided by the load resistance.

For simplicity, assume the diode drop is .3 volts. your measure 20 volts.  Add .3 volts ( diode drop ) for a total of 20.3 volts. this your peak value.
divide 20.3 volts by 1.414 for the RMS value which is  14.36 volts. Square 14.36 for a value of 206.21 (E^2) and divide the squared value by the load resistance ( 50 Ohms ) (R) for a value of 4.124 watts.

If you don't know your diode drop and always use of 0.3 volts your power calculations will be within 2 to 3% unless you are trying to work in the 1 watt area or less.

I hope this helps.

Thanks and 73
Dave KD1NA
Ke #934