Electronic – Convert Noise to dBm


I have a noise output of 35nv / (sqroot (hz))

So in dbm this is just 20*log(35nv) = -149.1 dbm/hz ?

Is that correct ?

Best Answer

I have a noise output of 35nv / (sqroot (hz))

No, what you have is a noise spectral density. To convert that to a voltage you have to consider the bandwidth that is of interest to you. So, if the bandwidth of interest is (say) 100 kHz, then the noise voltage you get in a 100 kHz of bandwidth is this: -

$$35\text{ nV}\times \sqrt{100,000} = 11.07\text{ }\mu V$$

(If the bandwidth of interest was 1 MHz then the noise would be \$35\text{ }\mu V\$).

To convert that to dBm requires knowledge of the nominal impedance of the system (usually 50 Ω or 600 Ω) so, if we assume 50 Ω then a voltage of 11.07 μV produces a noise power of 2.45 pW and, 2.45 pW is -86.11 dBm.