Estimating opamp output noise due to supply noise

Hi Narek,

as a rough estimation you can divide the noise voltage by the power supppy rejection.

Assume the datasheet specifies a PSRR of 80dB on the positive supply voltage pin at 1kHz and the noise is 1mVrms. Then this noise appears at the input of OPAmp suppressed by 80dB, giving 1mVrms / 10000 = 0.1µVrms. This noise adds to the input noise voltage of OPAmp.

At higher frequencies the PSRR of an OPAmp degrades and you get higher noise levels referred to the input. Assume the datasheet specifies a PSRR of 60dB on the positive supply voltage pin at 10kHz and the noise is 1mVrms. Then this noise appears at the input of OPAmp supressed by 60dB, giving 1mVrms / 1000 = 1µVrms.

What does this mean in practise?

Assume a voltage regulator (like the LM7815 for instance) generating a wideband noise of 100µVrms in the band 10Hz...100kHz. Then this translates to a noise voltage density of 100µV / SQRT(100kHz) = 316nV / SQRT(Hz). With a PSRR of OPAmp of 80dB at 1kHz this gets suppressed to 0.03nV / SQRT(Hz). And with a PSRR of 60dB at 10kHz this gets suppressed to about 0.3nV / SQRT(Hz).

When comparing these noise levels with the input noise voltage of OPAmp you see that the additional noise coming from the voltage regulator is totally negligible usually.

In situations where the PSRR of OPAmp is not sufficient, you can add a RC filter into the supply voltage line of OPAmp. A 10R + 10µF filter will enhance the PSRR above 1.6kHz resulting in a nearly flat overall PSRR curve. Or by other words, the 20dB decrease of PSRR of OPAmp per decade is compensated by the 20dB increase of PSRR per decade of the RC filter.

An even further improvement of PSRR can be achieved when using a "capacitance multiplier" in the supply voltage line.

Kai