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INA296A: Output Voltage

Tamaki Tanji
Prodigy 20 points
Part Number: INA296A

Tool/software:

Hi all

We are currently considering ensuring the accuracy of the sense IC at low current when INA296A5 is connected as shown in Fig.1.

Considering the voltage drop of the shunt resistor, offset voltage, gain error, etc., I think it will look like Table.1.


Specifically, when measuring 5A, the theoretical maximum output voltage of INA296B5 is 85.3mV, but 92.0mV is output.
There is a difference of about 6.0mV.

In another experiment, I mounted INA296A5 on a universal board as shown in Fig. 2, shorted IN-/+ to GND, and turned on a 3.3V power supply, resulting in an output of about 6.5mV.
We think this is the reason why the voltage output is larger than the theoretical value.

Question
Q1. Can an output of about 6.5mV be calculated?
Q2. Is there a way to lower the output to around 6.5mV when there is no input?


Best Regards,
Tamaki




  • 0
    TI__Mastermind 29630 points

    Hello Tamaki,

    Normally the output is higher due to resistance being off or how you are sensing the Rshunt especially when dealing with very small resistances.  I think we discussed the layout in a different post https://e2e.ti.com/support/amplifiers-group/amplifiers/f/amplifiers-forum/1392025/ina296b-about-accuracy.

      Also please verify the Rshunts are not the issue.  In the calculation you assume they are 0% error.  

    Tamaki Tanji said:
    Q1. Can an output of about 6.5mV be calculated?

    You cannot measure with output because it is below the swing low (20mV). 

    We specify Gain error from 50mV to VS -200mV at the output.  

    Tamaki Tanji said:
    Q2. Is there a way to lower the output to around 6.5mV when there is no input?

    Swing low is what dominates this.  There are ways to avoid this.  Placing a voltage on Vref above this and using that as your 0A is one.  Shifting supplies and connecting Vref to GND is another option.

    Regards,

    Javier

  • 0 in reply to Javier Contreras4
    Prodigy 20 points

    Hi, Javier

    Thank you for your reply.

    Q3. If 110mV is input to the REF1/REF2 terminals, what is the variation range of the output voltage when there is no input?

    Best Regards,
    Tamaki

  • 0 in reply to Tamaki Tanji
    TI__Mastermind 29630 points

    Hello Tamaki,

    I want you to be careful with no input vs 0 current input.  No current will still have the IB( 35µ) going into the pins.  If floating this could be an issue.  If not floating then your output at 0 current should be around 110mV.  The variation is due to a couple of things and I will point out the most impact.  Vos, CMRR and PSRR and drift.  You can adjust calculations below based on your conditions.

    Vos = 8µV

    PSRR = PSRR * |Specified VS - Used VS| 

    • (0.5µV/V) * (5V-3.3V) = 0.85µV

    CMRR = CMRR * |Specified VCM - Used VCM|

     150dB = 31.6nV/V,  Assuming VCM is 110 which is largest delta.

    • 31.6nV/V * |48 -110| = 1.96µV

    Drift= dVos/dT = 100nV/°C

    Assume 125°C

    • 100nV/°C * |125°C - 25°C| = 10µV

    Total Offset error = 8µV + 0.85µV + 1.96µV + 10µV = 20.81µV

    20.81µV/83.33µΩ = 0.24972 A of error.

    20.81µV * 500V/V =  10.41mV--> Vout = 110mV + 10.41 = 120.41mV

    Regards,

    Javier 

  • 0 in reply to Javier Contreras4
    Prodigy 20 points

    Hi, Javier

    Thank you for your reply.

    >20.81µV * 500V/V =  10.41mV--> Vout = 110mV + 10.41 = 120.41mV

    When using INA296A3IDDFR, the gain is 200V/V.

    20.81µV * 200V/V =  4.16mV--> Vout = 110mV + 4.16 = 114.16mV
    Is that correct？

    Best Regards,
    Tamaki

  • 0 in reply to Tamaki Tanji
    Prodigy 20 points

    Hi, Javier

    Thank you for your help.

    Regarding the CMRR calculation,
    I understand that the MIN condition of 150dB is stricter, so the calculation is done with a value of 150dB.
    Is it 5.01nV/V when the TYP is 166dB?

    Best Regards,
    Tamaki

  • 0 in reply to Tamaki Tanji
    TI__Mastermind 29630 points

    Hello Tamaki,

    That is not correct for the INA296A3 due to different Specifications.

    Vos = 10µV

    PSRR = PSRR * |Specified VS - Used VS| 

    • (0.5µV/V) * (5V-3.3V) = 0.85µV

    CMRR = CMRR * |Specified VCM - Used VCM|

     150dB = 31.6nV/V,  Assuming VCM is 110 which is largest delta.

    • 31.6nV/V * |48 -110| = 1.96µV

    Drift= dVos/dT = 100nV/°C

    Assume 125°C

    • 100nV/°C * |125°C - 25°C| = 10µV

    Total Offset error = 10µV + 0.85µV + 1.96µV + 10µV = 22.81µV

    22.81µV * 200V/V =  4.56mV--> Vout = 110mV + 4.56mV = 114.56mV

    Not much difference but you have to be aware they are different.

    Regards,

    Javier

  • 0 in reply to Tamaki Tanji
    TI__Mastermind 29630 points
    Tamaki Tanji said:
    o the calculation is done with a value of 150dB

    Tamaki,

    You are correct but I did use the max for all of the specifications.  If you want typical I would change all the specifications not just CMRR.  CMRR has the least impact on the specification as well.

  • 0 in reply to Javier Contreras4
    Prodigy 20 points

    Hi, Javier

    Thank you so much.

    I understood.

    Best Regards,
    Tamaki