A lot of regulator datasheets give a figure of Power Supply Rejection Ration in decibals. I understand that a high PSRR is good, but how and where does it help me?

For instance, suppose I have a regulated wall-wart power supply that outputs 5V but with about 20 mV noise imposed upon it - let's further assume that the noise is most significant at around 50 kHz.

Would a regulator with 90 db PSRR from 20kHz to 1MHz help me reject that noise?

Let's further assume that I'm measuring voltage with an Instrumenation Amp + ADC (i.e. the In. Amp is driving the ADC). Would a high PSRR generally help me? (I know it probably depends upon a lot of factors but I'm talking about the general situation).


2 Answers 2


The ideal regulator keeps the output voltage constant as long as:

  1. The input voltage is within the valid range specified for the device,

  2. and the output current draw is within the allowable range.

Of course no regulator is perfect. For voltage regulators, there are two main specs that tell you how much the output voltage varies as a result of operating conditions. Power supply rejection is actually a unusual term applied to voltage regulators. This term makes more sense for a part with some analog output, like a opamp. However, a voltage regulator can be viewed that way so it's not wrong. More commonly though you'd see the term input rejection ratio for voltage regulators.

In any case, this is telling you how much variations on the input of the regulator get onto the output. Ideally none of them would, but in the real world some fraction of input voltage variation is going to appear on the output voltage. Let's say that the input voltage has 1 Vpp ripple on it. If the resulting output voltage has 1 mVpp ripple on it, then the gain from input to output is 1/1000, and the rejection ratio is 1000.

This rejection ratio is often expessed in dB. Keep in mind that dB expresses a ratio of powers, and that power is proportional to the square of the voltage. A dB is 10Log10(power ratio), which is 10Log10((voltage ratio)²), which is more easily expressed as 20Log10(voltage ratio). Therefore, the 1000:1 rejection ratio from the example above could be expressed as 60 dB.

I mentioned there are two main specs used to describe the dynamic performance of a voltage regulator. So far we have talked about the input rejection ratio, which is what you asked about. Sooner or later you'll bump in to some kind of output rejection spec, although that can have different names. This is a measure of how much the output voltage changes for changes in the output current. If you work out the units, you will see this is in Ohms, although it is often not expressed as such explicitly. You can think of this as the resistance in series with the output of a regulator that does not vary its output at all as a function of current.

  • \$\begingroup\$ As always, an absolutely superb answer. Most of the examples, such as the one you give yourself, involve 'large' voltages (i.e. measured in Volts and not mV). Does PSRR also apply to (very) small variations in the input voltage (on the order 10mV?) \$\endgroup\$
    – Saad
    Commented Sep 19, 2012 at 17:52
  • 1
    \$\begingroup\$ @Saad: Actually 1 Vpp ripple into a regulator is "small". The parameters you mention are specified for small voltages, which scale to all other small voltages. In this case, small means not hitting any clipping limit, slew rate limit, etc. So yes, 60 dB input rejection applies to 10 mVpp in as well as 2 Vpp in. At low output noise levels though, other effects dominate. At uV level you have to consider local ground offset, etc. \$\endgroup\$ Commented Sep 19, 2012 at 19:20

From what I understand the PSSR determines how much power fluctuations effect the output of the amplifier.

Say you have a 20mV power supply fluctuation in your instrumentation amplifier with a gain of 40dB (I'm not sure what your gain is, but you can run these calculations yourself), and your amplifier has a PSRR of 90dB.

The total output fluctuation is then:

$$20 mA * 10^{(40-90) / 20} = 0.06325mV$$

You can determine if this fluctuation can be picked up by your ADC by determining the minimum step size the ADC can detect. For a 12-bit ADC with Vpp = 1V,

$$V_{min} = 1V / 2^{12} = 0.24414 mV$$

So the PSRR of the Instr. Amplifier in this case is sufficient at rejecting 20mV power fluctuations.

source: Wikipedia - Power Supply Rejection Ratio

  • \$\begingroup\$ mA should be mV \$\endgroup\$
    – JohannesB
    Commented Dec 22, 2016 at 1:45

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