Wikipedia says that power supply rejection ratio (PSRR) is the ratio of output noise referred to the input vs noise at the power supply:
The PSRR is defined as the ratio of the change in supply voltage to the equivalent (differential) input voltage it produces in the op-amp
Good Quality Design of Analog Cmos Integrated Circuits by Razavi seems to say the same thing:
The power supply rejection ratio (PSRR) is defined as the gain from the input to the output divided by the gain from the supply to the output.
So the overall rejection from power supply to output vary with the closed-loop gain of the op-amp?
So an op-amp with +40 dB gain and 100 dB PSRR, with 0 dBV noise at the power supply would have -60 dBV noise at the output? The Wikipedia example seems to say that it would be -120 dBV instead, which I don't understand.
Is there also an output component of PSRR? Like if you lower the gain of the amp, the input referred noise would decrease, right? But is there then a constant component coupled from the power supply through the output stages that starts to dominate?
Analog Devices MT-043, on the other hand, says:
PSRR or PSR can be referred either to the output (RTO) or the input (RTI). The RTI value can be obtained by dividing the RTO value by the amplifier gain. In the case of the traditional op amp, this would be the noise gain. The data sheet should be read carefully, because PSR can be expressed either as an RTO or RTI value.
Is this true? How do you find out from the datasheet which method is used?
Best Answer
Gain is absolutely the ONLY important part of PSRR. Essentially what you are saying is how much can an op-amp when feeding back a signal cancel out any ripples introduced from the power supply, not from the input of the circuit..
Lets take a simple example: an ideal (infinite open loop gain) voltage follower (output tied directly to the inverting input, fed from the non inverting input). The circuit has a closed loop gain of 1, but the feedback (since the overall gain is SOOO high) will mean that any power supply ripple will be canceled due to the feedback forcing the non inverting and inverting inputs to be in perfect lockstep..
But take the SAME example, but make the OPEN loop gain of the opamp 1, still with closed loop gain of 1, then suddenly the op amp can't keep up with the changes between the non-inverting input and the output-inverting input. And hence all ripple from the power supply would be visible on the output (essentially the op-amp would turn into a noise source with the noise being the coupled power supply ripple)
I understand HOW stevenvh could say that the gain is not meaningful, because he meant CLOSED loop gain... But the gain of question is OPEN loop gain, and YES, that is EVERYTHING in PSRR.
EDIT: And to answer your question, just to follow up slightly here, the PSRR is related to open loop gain, but the more closed loop gain you introduce, the more power supply ripple you will get on the output (hence the 60dB you reference above)
Here is why: Same example I give above, except this time you have a REAL op amp, (finite open loop gain), and resistors in your feedback path, meaning you have a closed loop gain of some value, say 6dB. Since the resistors behave as a voltage divider, the op-amp has to OVERCOMPENSATE for the power supply ripple being fed back to the non-inverting input. If it can only compensate for 100dB of power supply ripple, you will only get 94dB of rejection. The more closed loop gain you introduce, the less of the power supply ripple you are able to reject.
The whole conversation stems from the separate meanings of open loop and closed loop gain.
2nd EDIT: And the way that you get 60dB, or I get my 94dB is that you have to realize you have to convert dB BACK so for example you need to use
$$ 20 \log10\left(\frac{10^\frac{100}{20}}{10^\frac{6}{20}}\right) = 94 \mathrm{dB} $$
$$ 20 \log10\left(\frac{10^\frac{100}{20}}{10^\frac{40}{20}}\right) = 60 \mathrm{dB}. $$
And YES the other guy who said it should be 1mV not 1µV on Wikipedia is correct.