Calculating RF Power: dBm-Volt-Watt Conversion

[cyclone24]

Hello

I have a basic question about the decibel -voltage - watt conversion.

I am calculating pulsed RF power on the scope. The setup involved an RF transmitter connected to one end of an attenuator. The other end is connected to the scope. Sure I used 100 dB attenuator first and decreased the attenuation from 100 dB to 60 dB (or applied 40 dB) until I see the RF peaks on the scope.

Now I measured the pk-pk RF wave on the scope and it is 130 millivolts. So from here how should I calculate the power in watts?

HInt: 60 dBm = 1000 watts; P = V^2/R; R = 50 ohms...?

 

[waht]

If the scope doesn't have a 50 ohm input impedance, you should terminate at a 50 ohm load and then measure voltage across it.

 

[cyclone24]

Thank you.

I already did that. So how do I convert the voltage value (in mV) in watts? Or how do I define the RF power here? guess I am putting it in a crude way...

 

[waht]

Once you measure the rms voltage across the load, you can use this formula: P = V^2/R to calculate power in watts.

If you want that in dBm, note that 0 dBm is referenced to 1 mW

P_dbm = 10 log(base 10) (P/1mW)

 

[vk6kro]

Start with the voltage and convert it to RMS

So, that is (130 mV p-p / 2*√2) or 0.046 volts RMS

That is E ^ 2 / R or (0.046* 0.046 / 50) or 42.5 microwatts.

You have 60 dB of attenuation ( I think that's what you said).

So, 60 dB = 10 log ( actual power / 0.0000425)

actual power / 0.0000425 = 1000000

actual power = 42.5 watts

 

[cyclone24]

vk6kro and waht, thank you for that explanation. It is now clear.

Just curious, Vrms is usually = Vpk/(sqrt(2)). Why did you divide by another 2 in the equation?

The RF here is a sinc wave.

Thanks

 

[waht]

Just curious, Vrms is usually = Vpk/(sqrt(2)). Why did you divide by another 2 in the equation?

Because initially you measured voltage in Vpp. To get it in Vp you have to divide it by two. Then you use Vp to convert to Vrms.

 

[cyclone24]

That was dumb. Thanks!