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Original question:

INA381: Differential input range and reset function

INA381: Differential Input Impedance

Zeki Sisman
Intellectual 320 points
Part Number: INA381

Hi,

I am using INA381A1 (20V/V gain) for a current sensing application. My sense resistance is 1.025R. I also designed a resistor divider at the input of INA381A1 with 200R-200R-200R as shown below. I cannot reduce the sense resistance due to some other reasons so I need the attenuator.

From the datasheet, using gain error factor, and from the internal model given by TI in (https://e2e.ti.com/support/amplifiers/f/14/t/675407) I deducted that input stage of INA381A1 acts like (1250R//25000R)+(1250R//25000R) = 2380Ω differential impedance. 

Then from V(Rsense) to V(INA381A1_out) I am calculating the overall gain as = ((2380R//200R)/((2380R//200R) + 400R)) x 20 = 6.31.

My questions are:

  • In TINA Spice I cannot observe the input resistance voltage attenuation behavior. I see that my divider creates perfect 1/3 division. And from input to output I see 20/3 gain. Is it because the Rint of INA381A1 is not modeled in the spice model?
  • Is the calculated 2380Ω differential input impedance correct? I don't understand why it is not directly given in the datasheet and instead some equations are given for a "gain error factor", so perhaps I am making an incorrect assumption.
  • If 2380Ω calculation is correct, what is the tolerance level of the internal resistances of INA381A1?

Kind regards,

Zeki

 

  • 0
    TI__Mastermind 26660 points

    Hey Zeki,

    You are right to use Equation (2) of INA381 datasheet to derive the effective input differential resistance of the device; however, I believe it is 2.5kΩ nominal with a tolerance of +/-20% over process and temperature. Equation 2 of datasheet incorporates this resistance when determining the new circuit gain after inserting input filter resistors.

    Anyway, the model does not appear to incorporate this behavior as it is a behavioral-based model and not a transistor-level model. The best thing is to compare the model with a discrete version of the internal amplifier configuration to determine the behavior. Below I simulated the INA381A1 with TI model and discrete version of internal amplifier. On left I exclude your attenuating resistor and on the right I include the resistor. 

    INA381_inputFilter_gainChange.TSC

    As you can see, the gain does not change for the model according to Equation 2 when inserting the input resistors (R8 and R10 your schematic). The gain is overall attenuated with the 200Ω attenuation resistor. The top-right circuit is correct with a gain of 64.595858mV/10.231984mV = 6.3131 V/V.

    All the resistors inside the INA381 can vary ±20% over process and temperature, so the overall gain error range will increase as you incorporate input resistors to the front-end of INA381. You can use circuit on top-right and vary the resistors and calculate the gain error to determine what the circuit's gain error range will become.

    Best,

    Peter

  • 0 in reply to Peter Iliya
    Intellectual 320 points

    Hey Peter,

    Thank you for clarification! 

    Just a minor remark, you said that the 2.5kΩ looks like the effective differential input impedance. However right after the 2.5kΩ the following differential amplifier part also has 50kΩ differential input impedance which acts parallel to the 2.5kΩ inside the IC.

    So 2.5kΩ // 50kΩ then becomes the total equivalent input impedance which is 2.38kΩ.

    Kind regards,

    Zeki