RF Input power

[Amr Wael]

Hello,
I have been always a noob in RF/Microwave. When reading a datasheet for an attenuator/amplifier. There is always a rating for the maximum input power in dBm. I understand that power in watts = 10^((Pdbm-30)/10) but if I want to calculate the maximum voltage based on this power is it calculated based on the 50 ohm impedance?
Does this mean that a maximum input power of 10 dBm is equivalent to 0.707 Volts only?
Thank you very much in advance!

 

[tyassin]

Yes 0.5Vrms

 

[albbg]

Yes 0.5Vrms

No, 0.707 Vrms. The power is referred to the RMS voltage value. If you want, instead, the peak you have to multiply by sqrt(2), that is:
10 dBm (10 mW) ==> 0.707 Vrms ==> 1 Vpeak

 

[stenzer]

power is referred to the RMS voltage

Further that might help:

\[ P ={10}^{\frac{P_{\mathrm{dBm}}}{10}} \cdot {10}^{-3} ~\mathrm{W}\]

\[P = \frac{{V_{\mathrm{RMS}}}^{2 }}{50 ~\Omega}\]

\[V_{\mathrm{RMS}} = \sqrt{P \cdot 50 ~\Omega}\]

BR

 

[KlausST]

Hi,

So
0 dBm = 1 mW
10 dBm = 10 mW
20 dBm = 100 mW
30 dBm = 1000 mW = 1W
...
-10 dBm = 0.1 mW
-20 dBm = 0.01 mW
...
10 dBm = 10mW on 50 Ohms system.

P = U x U / R --> U = sqrt(P x R) = sqrt(10 mW x 50 Ohms) = sqrt(500 m) V RMS = sqrt(0.5) V RMS = 0.707 V RMS

On a clean sinusoidal waveform: 1 V amplitude, 1 V peak, +/- 1V peak = 2V peak to peak

Klaus

 

[tyassin]

Yes 0.5Vrms

Yes sorry for the error and confusion. It is 0.707Vrms.

 

[biff44]

Power in watts = (RMS voltage)^2 / (Zo)

Usually Zo is 50 ohms