# Electronic – Convert Noise to dBm

noise

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 ?

$$35\text{ nV}\times \sqrt{100,000} = 11.07\text{ }\mu V$$
(If the bandwidth of interest was 1 MHz then the noise would be $$\35\text{ }\mu V\$$).