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INA282 / Offset voltage TINA-TI simulate.

Sho Ogane
Genius 5575 points

Other Parts Discussed in Thread: INA282, TINA-TI

Hi all.

Please see the attached file as TINA-TI simulation result of INA282.
In this result , output voltage is 7.47mV at Vin=0V.
Why does this simulation become this result?
Could you please let me know the reason of this result 7.47mV? 

INA282 specification is Gain=50V/V and Vos=20uV(typ),70uV(max).
So I calculated the Voutput(@Vin=0V) =20uV*50V/V=1mV.
                                  Voutput(@Vin=0V)=70uV*50V/V=3.5mV.


Best Regards,
Ogane Sho

  • 0
    TI__Genius 14825 points

    Sho,


    There are more sources of offset error than those you listed; see pages 21-22 in the data sheet.  The term you are missing is likely the Rcmr term, which is responsible for the offset contributed to the error when the VREF isn't 2.048V.  In the TINA simulation you sent, the VREF lines are tied to GND.  This causes the Rcmr term to get larger.  Try taking your circuit and biasing the VREF to 2.048V.  When I did this and swept V2 from 0 to 10mV, I got 0.97mV of error on Vout when Vinput = 0.  Let me know how else I can assist.

  • 0 in reply to Jason Bridgmon
    Genius 5575 points

    Hi Jason san.

    Thank you for the reply.
    I understood we should bias the Vref to 2.048V.

    I saw page 21 in the data sheet.
    <Vos_ref=Rcmr*(2.048V-Vref)>
    Could you let me know the method of calculation of Rcmr?
    I don't know how to change the value of Rcmr.

    And if possible, could you give us the simulation circuit and the data?

    Best Regards,

    Sho Ogane

     

  • 0 in reply to Sho Ogane
    TI__Genius 14825 points
    Sho,

    The Reference Voltage Rejection Ratio for the different INA28x gains is found on page 3 of the data sheet. For the INA282, you'll find it is ±25μV/V typical, ±75μV/V max, and 0.055μV/V/°C vs temperature.

    It is very easy to change the TINA circuit you have to a different reference value, attach a voltage supply set to 2.048V to REF1 and REF2, instead of GND. If you still need an example sim file, let me know and I can provide one.
  • 0 in reply to Jason Bridgmon
    Genius 5575 points

    Hi Jason-san.

    Thank you for the reply.
    I change the TINA circuit following your advice.
    Please see  the attached file as TINA simulation result.
    Is this circuit right?

    I would like to simulate linearity of Vout from close 0V vicinity.
    Is this simulation the possible?

    Sorry for asking many questions to you.

    Best Regards,
    Sho Ogane


  • 0 in reply to Sho Ogane
    TI__Genius 14825 points

    Sho,


    This circuit looks like I suggested.

    One other thing to know is that the INA282 has a spec on page 4 of the data sheet that defines how close the output can go to GND.  It is in the VOLTAGE OUTPUT box labeled "Swing to GND" and can be as high as 40mV from GND.  With your VREF=2.048V, this Swing to GND spec is no longer a problem.

    You are asking about the linearity of the INA282 near 0V.  I assume you mean (VIN+) - (VIN-) = 0V, and not VOUT = 0V, because with the reference, VOUT /= 0V.  I would like to know more about your application and what kind of results you are looking for.  Will this be used in a system with a shunt resistor?  What size of shunt resistor will you use?  What will be the common mode voltage on the shunt resistor?  How much current will you measure, and what will you drive with the output of the INA282?  What kind of linearity would be acceptable and over how large an input range?  Will you need bipolar measurement or unipolar measurement?  What kind of accuracy will you need?


    Knowing the answers to these questions will help me understand what you are trying to do and I can assist you better.

  • 0 in reply to Jason Bridgmon
    Genius 5575 points

    Hi Jason san.

    Thank you for the reply.
    And thank you for your advices for me.
    I am checking your questions.
    Could you wait a little?

    Best Regards,
    Sho Ogane

  • 0 in reply to Sho Ogane
    TI__Genius 14825 points
    Sho,

    Of course. Let u sknow if you need anything.