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Original question:
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:
Kind regards,
Zeki
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
