Signal to noise ratio Question?

Hi i have attempted answering this question:

The unwanted noise power is measured at the output of an amplifier by measuring the output power with the input shorted to 0v (ground). The measurement is 2mW. An input is then applied to produce full output power. The result is 26W.Calculate the signal to noise ration of the amplifier in db.

this is my answer si it correct?

Sn ratio = 20 log 10 ps/pn

=

20 log 10 26w/0.002w = 180641 db

I think i have miscalculated the end answer as i do not have a scientific calc to hand and am using the windows one...

Can anyone tell me if i have calculated it correctly?

Many thanks - wasnt sure where to post but it has been bothering me

That is a ridiculous number of dBs!

Your formula is wrong, you should use 10 Log10 (Ps/Pn)

I calculate the S/N ratio as 41dB.

If you want to be really nit picky, what is calculated is actually the ratio of (signal + noise)/noise, because the noise will still be there even when the signal is present.

You would use the 20Log10 P1/P2 type formula if you had measured the voltage rather than the power.

JimB

thanks for your response i had th other formula and have worked through it and got the same value of 41db.