# Electronic – Low input voltage noise nV/√Hz

I am using ADA4937 as the driver circuit to my 14 bit high speed ADC. A term in the datasheet is Low Input Voltage Noise is used which is 2.2nV/√Hz. Please explain the importance of this term and unit used(nV/√Hz)in ADC driver. How lower or higher value of Low Input Voltage Noise will affect the overall performance.

The datasheet says Low input voltage noise: 2.2 nV/√Hz. The parameter is really just input voltage noise, and they are saying that it's low to make the part sound more awesome.

The noise is usually modelled as having a constant spectral density. Thus, the spectral density of noise power would be specified in \$\mathrm{W/Hz}\$. The higher the bandwidth (for an ADC, this depends on your sampling rate), the more spectrum there is with noise in it, so the more total noise power there is in your measurement.

However, this parameter is voltage noise. Power is proportional to the square of voltage:

$$P = \frac{E^2}{R}$$

So if \$R\$ is constant, and in this case it's the input impedance of the ADC, which is approximately constant, and you want just the voltage component of the noise (because the ADC measures voltage, not power), then you need to take the square root of the noise power, and you are left with a measurement in units \$\mathrm{V/\sqrt{Hz}}\$.

With this parameter, you can know how much voltage noise there will be, and thus, how much noise there will be in your measurements. So let's say your input bandwidth is 24 kHz. Take the square root of this, and multiply it by the input voltage noise spectral density to get the RMS noise voltage:

$$\require{cancel} \frac{2.2\:\mathrm{nV}}{\cancel{\sqrt{\mathrm{Hz}}}} \sqrt{24000}\cancel{\sqrt{\mathrm{Hz}}} \approx 341\:\mathrm{nV_{(RMS)}}$$

This means that the measurements you get from the ADC will look like noise with an RMS amplitude of 341 nV was added to your signal if the input bandwidth is 24 kHz. More bandwidth means more noise, less bandwidth, less noise.