OPA445: PSRR

Hi Loris,

PSRR is simply the change in the naturally occurring offset voltage of an op amp in response to a change in the supply voltage. For the OPA445 operating with ±40 V supplies it is specified with a  ±4 uV/V typical PSRR, with the supplies changing ±10V to ±45V.

Imagine the OPA445 supply voltage initially set to ±40 V and the voltage offset is measured and recorded. Then, the supplies are dropped to ±10 V and voltage offset measured again and recorded. The PSRR is calculated from the total change in the voltage offset (usually hundreds of microvolts to millivolts), divided by the total change in the supply voltage (70 V) in this case.

If you are powering the OPA445 from ±12 V supplies, the PSRR won't be substantially different from what you would see with ±40 V supplies. Although not specified in the datasheet it will remain high. Remember during the PSRR testing the supplies are brought as low as ±10 V. If the PSRR were badly degraded it would show up in the PSRR measurement.

Any ac signal, which includes noise, riding on the dc supply lines will exercise the op amp's PSRR capabilities. PSRR is highest at dc and low ac frequencies and then rolls off with increased frequency. You can see this in the OPA445 PSRR vs frequency, Typical Characteristics graph shown below. Note that the PSRR is different for each of the supplies, the positive supply having more rejection compared to the negative supply.

Regards, Thomas

Precision Amplifiers Applications Engineering

Hi Thomas,

thanks for the answer, however I still have some doubts.

first case: intial supply voltage Vs = ±40V, then Vs  is dropped to ±10V, so t the supply variation is 60V. To evaluate the consequent offset voltage I use PSRR = ±4 uV/V .

second case: intial supply voltage Vs = ±12V, then Vs  is dropped to ±10V, so the supply variation that I have to consider is 60V (i.e. ±10V referred to ±40V) or ±4V (i.e.  ±10V referred to ±12V). Then, to evaluate the consequent offset voltage I still have to use PSRR = ±4 uV/V?

AC case:  intial supply voltage Vs = ±12V. I have a noise with amplitude 0.5V at 1kHz on both the supplyies (hp: the two noises are out of  phase of 180°), so null mean but minimum Vs  =  ±11V and max Vs =  ±13V. From the plot I see that I have to consider a PRSS of  80 dB for the positive supply and 50dB for the negative supply. Let's assume that 80 dB is for both the supplies. What is the variation of the supply voltage Vs that I have to consider:

2V (because ±11V and ±13V differ from ±12V by±1V),

4V (because ±11V and ±13V differ by ±2V)

58V (because ±11V differ from ±40V (the reference Vs value in the graph) by ±29V)

54V (because ±13V differ from ±40V (the reference Vs value in the graph) by ±27V)

Thanks,

Loris

Hello Loris,

Concerning your questions:

first case: intial supply voltage Vs = ±40V, then Vs  is dropped to ±10V, so t the supply variation is 60V. To evaluate the consequent offset voltage I use PSRR = ±4 uV/V .

I had the voltage delta at 70 V, but you are correct with 60 V. The PSRR = ±4 uV/V typical is only for the Vs = ±40V case, but I wouldn't expect it to be much different for the ±12V case. The OPA445 is targeted for higher voltage applications and that is why its specifications are based on ±40V. Therefore, the OPA445 is PSRR is not characterized specifically for a ±12V condition. You could run a PSRR test with the particular OPA445 device you are evaluating and move its supplies from ±10 V to ±14 V, and measure how the voltage offset changes for that 8 V supply change. It may come close to the 4 uV/V number, but then again it may not. The OPA445 datasheet makes no assurances about the PSRR performance at any supply level conditions other than the ±40V as discussed.

second case: initial supply voltage Vs = ±12V, then Vs  is dropped to ±10V, so the supply variation that I have to consider is 60V (i.e. ±10V referred to ±40V) or ±4V (i.e.  ±10V referred to ±12V). Then, to evaluate the consequent offset voltage I still have to use PSRR = ±4 uV/V?

The voltage across the OPA445 for Vs = ±12V is 24 V, between V+ to V-. For Vs = ±10V, it is 20 V between V+ to V-. Therefore the total supply change is 4 V. You would divide the corresponding voltage offset change by the 4 V change to derive the PSRR in uV/V.

AC case:  intial supply voltage Vs = ±12V. I have a noise with amplitude 0.5V at 1kHz on both the supplyies (hp: the two noises are out of  phase of 180°), so null mean but minimum Vs  =  ±11V and max Vs =  ±13V. From the plot I see that I have to consider a PRSS of  80 dB for the positive supply and 50dB for the negative supply. Let's assume that 80 dB is for both the supplies. What is the variation of the supply voltage Vs that I have to consider:

2V (because ±11V and ±13V differ from ±12V by±1V),

4V (because ±11V and ±13V differ by ±2V)

58V (because ±11V differ from ±40V (the reference Vs value in the graph) by ±29V)

54V (because ±13V differ from ±40V (the reference Vs value in the graph) by ±27V)

It looks like you have a correct understanding about the voltage deltas used in the PSRR calculation.

PSRR(dB) can be converted to V/V using the following equation:

PSRR V/V = 10^{(-PSRR_dB) / 20 dB)}

For 80 dB PSRR

PSRR V/V = 10^{(-80 dB) / 20 dB)} = 100 uV/V

Keep in mind that the datasheet graph is typical PSRR vs. frequency performance, with the ±40 V supplies.

Regards, Thomas

Precision Amplifiers Applications Engineering

Hello Thomas,

thanks for your answers. However I still have some open point about the AC case. Which of the four supply variations should I consider in the assesment of the offset voltage by PSRR? answer 1, 2, 3 or 4 ?

AC case:  intial supply voltage Vs = ±12V. I have a noise with amplitude 0.5V at 1kHz on both the supplyies (hp: the two noises are out of  phase of 180°), so null mean but minimum Vs  =  ±11V and max Vs =  ±13V. From the plot I see that I have to consider a PRSS of  80 dB for the positive supply and 50dB for the negative supply. Let's assume that 80 dB is for both the supplies. What is the variation of the supply voltage Vs that I have to consider:

1) ΔVs = 2V (because ±11V and ±13V differ from ±12V by±1V),

2) ΔVs = 4V (because ±11V and ±13V differ by ±2V)

3) ΔVs = 58V (because ±11V differ from ±40V (the reference Vs value in the graph) by ±29V)

4) ΔVs = 54V (because ±13V differ from ±40V (the reference Vs value in the graph) by ±27V)

Regards,

Loris