[SOLVED] Units conversion 1 dB compression

[tabascorez]

For a load of 50 ohm, can somebody explain to me how if P(1-dB) = -30 dBm, this is equivalent to 10mV(peak)?

 

[johnjoe]

It's simple conversion: just remember P[W]=U(max)^2/(2*R)

 

[albbg]

Convert first dBm to watts.

P(Watts)=10^[P(dBm)/10-3]

so a power of -30 dBm in watt is:

P=10^(-6) watts (that is 1 microwatt)

but since P = V^2/R we can calculate the voltage as

V = sqrt(P*R)

numerically

V=sqrt[10^(-6)*50]=7.07 mV

the above calculation returns a RMS value so we need to multiply by sqrt(2) to have the peak value:

Vpeak=7.07*sqrt(2) = 10 mVpeak

This calculation always apply to convert from dBm to volt (given the load), not only to P1dB measurements

 

[johnjoe]

Mhm, now you took him the chance to find it out on himself ;-)

 

[tabascorez]

Fantastic, thanks for your detailed answer guys