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OPA322: Why use the circuit to test PSRR of OPA

wenyue wu
Prodigy 90 points
Part Number: OPA322

Hi，

I am learning the PSRR of OPA.

and Why use the circuit to test PSRR of OPA? 

in lab test, we need to Amplify VOS  to measure

But why need to adjust V1 for Vout=0? to In order to prevent the open loop gain from bringing errors?

Could you give me a detailed explanation?

i tested the psrr in lab, and got the result below:

but the psrr in datasheet is 10uV/V. Is this error normal?

Thanks very much.

Joise Wu

  • 0
    TI__Intellectual 775 points

    Hi Josie,

    You have the correct result and is very close to datasheet value. I am assuming the 88mV in Vos column is 88uV (typo and not mV). 

    PSRR is the ratio of ( Difference in Vos measured at two different supply voltage)/(The difference in power supply voltages). So you have to measure two Vos voltages at two different power supply voltages. The circuit is all about this.

    You change V1 to get Vout to be 0V at certain supply Voltage.  The voltage at V2 gives you a measure of your initial input offset voltage. The ratio of R2 and R3 determines the exact ratio of V2 to Vos. You changed the supply voltage which will cause Vout to have a non-zero value. So You change the V1 again to set Vout to be 0. The new measurement at V2 is different and gives you a measurement of the new offset voltage. These two readings should give change in offset voltage due to power supply. You can calculate PSRR from here.

    Here is the link to TIPL video lecture on PSRR which you might find helpful: https://training.ti.com/ti-precision-labs-op-amps-power-supply-rejection?cu=14685

    I hope this answered your question.

    Sanjeev

  • 0 in reply to Sanjeev Manandhar
    TI__Prodigy 90 points

    Hi Sanjeev,

    Many thanks for your help.

    I understand the reason why choose the circuit.

    Could you explain why need to adjust V1 to 0 to test vos in detail?

    Is it to eliminate the error caused by the open loop gain?

    Thanks very much.

    Best regards

    Joise wu

  • 0 in reply to wenyue wu
    TI__Intellectual 775 points

    Hi Joise,

    You adjust V1 to get 0 at Vout. This adjusted voltage at V1 will cause voltages at the inputs of the op-amp (inverting input and non-inverting input) to be not equal.  The difference between these two voltages is the definition of the input offset voltage. That is what you are measuring at two different power supply. 

    What do you mean by an error caused by open-loop gain?  I do not understand what you are referring to. Do you mean loop gain?

    Open-loop gain is very high 130 dB and the closed-loop gain from non-inverting input is 2 (6dB) hence loop gain is 124dB, which is very high so I do not think there is an error due to low loop-gain if that is what you meant. 

    Regards,

    Sanjeev

  • 0 in reply to Sanjeev Manandhar
    TI__Prodigy 90 points

    Hi Sanjeev，

    Thank you for continued help.

    I have learnt the PSRR in Precision Lab.

    in Precision Lab, simulate the circuit by Tina. VOS= output offset voltage/A,

     so, i think maybe can measure like below. Vos=Vout/A.

    But i think in this circuit, Open-loop Gain is not  ∞, SO it need an offset to support the output

    so the vos measuerd by this circuit has a error.

    in the circuit, we use v1 to make vout=0, have not the vos error from open loop Gain.

    could i think so? right?

    Thanks very much.

    Best regards

    Joise wu

  • 0 in reply to wenyue wu
    TI__Intellectual 775 points

    Hi Joise,

    I agree with limited A will have an effect in gain error but the A is a very high number of 130 dB = 3000000. If you look at the formula for the gain error caused by this limited A for the circuit above of gain (1) is very small in order 1uV/V or less.

    G =A/(1+AB). (G is closed-loop gain, B is feedback factor, A is open-loop gain)

    In addition to that, you are taking two measurements with different voltage supplies. when you subtract these two measurements you cancel the effect of limited A -as long as you do not change the configuration of the circuit. You are accounting for the effect of the change from the power supply only. 

    Sanjeev