Measuring Real World PSRR Figures

Let’s say, for the sake of example, that I drop a non-descript box onto your workbench that contains an unidentified gain block. If I say to you ‘measure the PSRR of this device’, how would you go about doing so, i.e. what measurement would you take and how would you derive the dB rejection figure?

I’ve noticed with certain older solid-state items that you can see a ghost of the audio path’s waveform on the DC rail. I assume you could measure this ‘ghost’ and express it as a function of the input amplitude and available dynamic range – correct?

Justin

I've noticed with certain older solid-state items that you can see a ghost of the audio path’s waveform on the DC rail. I assume you could measure this ‘ghost’ and express it as a function of the input amplitude and available dynamic range--correct?

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No, that's the other direction. The 'ghost' is a function of the power supply impedance (and current draw, i.e. load impedance) and not of the amplifier PSRR.

Measuring AC PSRR is probably best done by running the amplifier at high noise gain (say 40 dB to 80 dB depending on frequency and GBW) with inputs grounded, modulating the according supply and measuring the output voltage.

Samuel

The 'ghost' is a function of the power supply impedance (and current draw, i.e. load impedance) and not of the amplifier PSRR.

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In instances that demand a fair amount of current that are strictly 'Class B', I've observed how you see a 'ghost' of a Vreg's triangular correction waveform when you 'scope the DC rail feeding the Vreg. This 'ghosting' gets smaller as you increase the size of bypass capacitors before the Vreg, and clearly illustrates the impedance of the supply feeding the Vreg is at play.

However, I don't totally understand how this is still the case in the example of a pure Class A stage, passing only a few mV of signal, supplied by a PSU that is able to supply many times more current than it needs - as the Class A stage will represent a steady load?


Justin

Class A doesn't equate to constant supply current---just that no device has its current go to zero while amplifying the signal.*

On the original question, what Samuel said---you have to apply a variation riding on the supply rail and look at the unit's output signal. This can be nontrivial if there is a large local bypass cap within the unit.


EDIT: *bridge-tied load class A can be a pretty good approximation to constant PS current though.

to add to what Brad and Samuel said, accurately quantifying PSRR can be difficult as impedance of feedback networks, and perhaps even topology can make a difference.

What is useful for practical purposes is how does PSRR of your typical circuit behave? To that end a typical high gain non inverting stage, or perhaps (high noise gain) inverting stage to model a summing bus amp, can give useful readings.

Dropping different opamps into a given circuit will give comparative information. You can also play with feedback networks to look for sensitivity to values.
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A secondary effect of PSRR is that a sloppy high impedance PS rail will generate a voltage at the PS terminals from load current that will then be injected relative to the input. For the sake of argument assume a 100 ohm resistor in series with both PS leads. Only if the PSRR is exactly identical for both rails will there be any hope of first order cancellation of this term. Note: any nonlinearity in the load will also dump nonlinear current draw into the rail impedance that can't cancel out.

Brute force is the typical approach for managing PS rails, but knowing the relative CMRR may suggest more brute force on one rail vs. the other. IIRC there was a notable difference in 5532 between +/- PSRR.

JR

Here is an article including a drive circuit for testing PSRR-. They talk about using expensive NWAs but you can substitute generators to produce the AC inputs. On the output, you are going to need a fairly sensitve AC voltmeter to make the readings since the typical range of values of the impressed AC on the amp output are -100 to -120 db although it rises fairly fast with frequency.

http://www.edn.com/article/CA6459062.html?industryid=47046

The nice thing about this circuit is that the power pin is driven from the fairly low impedance output pin of an opamp rather than putting a higher impedance generator output in series with the DC bias.

Thanks for the link. I doubt though that measuring open-loop as shown is a particularly clever idea for typical audio opamps.

You are going to need a fairly sensitve AC voltmeter to make the readings since the typical range of values of the impressed AC on the amp output are -100 to -120 dB.

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Likely not much of a problem if you are measuring open-loop or with high noise gain. Amplifier noise is perhaps more of a problem, a bandpass filter may be needed.

Samuel