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INA210: Adding external series resistor for adjust the gain of INA210?

Red Yen
Expert 4310 points
Part Number: INA210

Hi Support team 

Regarding to the below picture:

If we add 300 ohm resistors, shall the gain of INA210 will be 188.6 (from calculation)?

Gain = Rout/Rin = 1Meg/ (5k + 300 ohm) = 188.6

 

Please advise from TI point of view and let us know your concern.

  • 0
    Guru 140200 points

    Hi Red,

    I wouldn't do that. Have you read section 7.4.1 of datasheet?

    Why not decreasing the shunt resistor? Or put a resistor is parallel to the shunt resistor. Or split the shunt resistor into two shunt resistors.

    Kai

  • 0
    TI__Mastermind 26730 points

    Hello Red,

    The gain will not change according to your equation. The reason is that there really is ~2.5kΩ of input differential resistance from IN+ to IN- and this is not shown in the simplified diagrams of datasheet. The input bias network is shown in Figure 24 and it is accounted for in equations 1 and 2 of the datasheet. So in order to calculate the nominal gain change when determine the gain error factor (GEF) with equation 1 and then multiply this change to the device's gain. So for INA210 and Rs=300Ω, the new gain is 200*(1-GEF) = 153.85 V/V.

    There will be some widening of the gain error because the internal resistors (Rdiff, Rint, RFB) can vary by +/-20% while the input resistos (Rs=300Ω) will remain relatively constant. Rdiff=2.5k, Rint = R3=R4, and RFB=R1=R2 when referring to the figure you have posted. The ratio of RFB/Rint will only vary by the approximately the specified gain error of the device.

    The new gain error will need to be determined by calculating/simulating gain when Rint and RFB are both +20% and then both -20%. Also, Rdiff can vary +/-20% in both of those scenarios as well.

    Best,

    Peter