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Original question:

[FAQ] INA193: Measuring Vsense < 20mV with the INA193-INA198 products

INA193: INA193 bias currents

Chris Cooper
Prodigy 50 points
Part Number: INA193

Tool/software:

I have a circuit using a INA193 to sense the current through a 60m ohm sense resistor. My sense supply is +/-15V. My INA193 supply is 7V. Both + and - circuits are identical except where the +IN and -IN on the negative rail are connected in reverse. The +IN on both are the high side.

It appears that the input bias current for +IN is higher than the stated datasheet values for the +15 rail. I have an input filter using matching 100ohm resistors per the datasheet. The resistor on the +IN for +15V drops1.7mv, while the resistor on -IN drops 0.7mv. Both resistors on the +IN and -IN pins for the negative rail drop close to the same value at around 1.0mV and 1.2mV respectively. 

A 350mA load on both +15 and -15 create different results from the two INA193's. I can only assume this difference is due to the additional drop in the bias currents. 

I have removed the filter resistors on the + Rail and the results between the two INA193's match. 

Isense: Both match

Vsense: Both match

My question is, why are the two INA193's yielding different results when they are connected similarly and have identical currents through the sense resistor and is there a way to make them match with the filtering network? 

  • 0
    TI__Genius 13810 points

    Hi Chris, 

    Thank you for your post. 

    I want to point you to the figure illustrating the difference in bias currents between both IN+ and IN- with a supply voltage of 12V:

    There is roughly a 525µV offset between both inputs with series 100Ω resistors as a result of a 5.25µA delta between both inputs at your common mode voltage. So this 1mV to 1.7mV 700uV difference is expected. 

    Whenever precision is required, we typically do not encourage the use of the input filters with the INA193. 

    What are these filters intended to help with? There may be a different network that would be better for your application. 

    One thing you can try for an approximate 10X reduction in the error is to replace your 100Ω resistors with 10Ω resistors. 

    What accuracy ratings are your 100Ω resistors?

    Best regards, 

    Joe