[SOLVED] Units conversion 1 dB compression

Status
Not open for further replies.
T

tabascorez

Junior Member level 2
Joined
Nov 11, 2013
Messages
22
Helped
0
Reputation
0
Reaction score
0
Trophy points
1
Activity points
163
For a load of 50 ohm, can somebody explain to me how if P(1-dB) = -30 dBm, this is equivalent to 10mV(peak)?
 

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

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
 

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

Fantastic, thanks for your detailed answer guys
 

Status
Not open for further replies.

Similar threads

Menu
Log in
Register
Cookies are required to use this site. You must accept them to continue using the site. Learn more…