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OPA211: Gain Calculation in TI precision lab

GUOOO
Prodigy 140 points
Part Number: OPA211
Other Parts Discussed in Thread: TINA-TI

When i learn the OPA input offset voltage and bias current at TI precision LAB:

video: https://training.ti.com/ti-precision-labs-op-amps-vos-and-ib-lab?cu=14685

PDF: https://training.ti.com/system/files/docs/1100-L%20-%20Vos%20and%20Ib%20-%20Lab%20-%20slides_0.pdf

They calculated the gain of OPA by its equal resistor in page 4-6 at PDF and 01:45-03:52 int the video.

Fig 1 is the topology of circuit, and Fig 2 is the simulation with ideal device and using current and voltage source replace the bias current and offset voltage by me in TINA-TI.

FIG1  ↑↑

FIG2  ↑↑(IS1 should be 60nA and I also simulated with right IS1, and the result is same to the old one)

And also the result I calculated with pencil is equal to my simulation, so I wonder WHY the video using equal resistor and ignored the second OPA's offset voltage and bias current?

Thanks!

  • 0
    TI__Guru* 93105 points

    Hello Kang,

    It appears that the authors simply decided to use the op amp voltage offset and Ib contributions related offset of the first stage U5 alone, and then multiply them by the total voltage gain of the total circuit, +2100 V/V. The offset contributions of the second stage U6 are only amplified by +21 V/V, and are highly overshadowed by the offset of U5 which is amplified by +2100 V/V. That overlooks the offset errors of the second stage but the total resulting error isn't large.

    To be absolutely correct you would find the offset errors of the first stage and multiply them by +2100 V/V. Then derive the offset errors of the second stage and multiply them by +21 V/V. The total output referred offset for the two op amps would be the sum appearing at the U6 output.

    Regards, Thomas

    Precision Amplifiers Applications Engineering

  • 0 in reply to Thomas Kuehl
    Prodigy 140 points

    Hi Thomas,

    Thanks for your answer! But there are still some issues in that thesis.

    At the analysis of the author of TI Percision Lab, his calculation of the output voltage caused by IB (if we ignore the input offset voltage) is the IB multiply the EQ_Resistor( EQ_Resistor = R1//RF), you can see in the fig 3:

    fig3 calculation of voltage caused by IB(ignore the Vos) of FIG1

    BUT, int the video of https://training.ti.com/ti-precision-labs-op-amps-vos-and-ib?cu=14685

    He calculated the vout caused by IB through IB*Rf, in the fig4, and the resistor is Rf, not EQ_Resistor.

    So the FIG3 &FIG4 that calculate the output voltage caused by IB is not the same! 

    Fig 4 calculation of IB and output voltage

    So, I think the IB calculation is not related to R1, and the FIG3 is wrong!

  • 0 in reply to GUOOO
    Prodigy 40 points

    Hello,

    agree with You, think also using Req for calculation of the output voltage (Vib) of the first Opamp due to Ib is wrong.

    Clarification through the author would be helpful.

    repetition of the problem with effect of the input bias current to the output voltage of the first amplifier:

    In Chapter 2.2 of the course

    („1100-L - Vos and Ib - Lab - slides_0.pdf“, page 6)

    Vib : Output voltage due to ib (input bias current)

    the Vib voltage is calculated

    Vib = ib*Req    with  Req = (Rf*R1)/(Rf+R1)

     

    Where as in Chapter 2.1 of the course
    Vib = ib * Rf

    is used. (which I supposed)

    Please, why in 2.2 Req is used?

    Thanks& Regards

    Michael

  • 0 in reply to Michael Deppert
    Prodigy 140 points

    Hi Michael,

    There are all right above.

    EQ_resistor mulply the Ib means the EQ  voltage caused by ib attached to the input pin.

    Vib=Ib*Rf=Ib*(R1//Rf)*(1+Rf/R1) where R1//Rf means EQ_R and (1+Rf/R1) means Gain of the OPA!

  • 0 in reply to GUOOO
    Prodigy 40 points

    Hallo, thanks all !


    Thank You GUOOO,

    Your explanation made it clear to me.


    It mixed me up that -
    Vib     in the slides, is not OP1(OP_U5)'s output voltage due to it's input bias current ib !



     One thing to add is, that OP2(OP_U6)'s input bias current resulting to an additional (rel. small) output voltage failure

     isn't considered in the total output error calculation, as it was in "1100 - Vos and Ib - exercises-and-solutions.pdf".