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Original question:

[FAQ] INA199: Why can I exceed the absolute maximum voltage ratings if I am limiting the input current

INA199: INA199C1

Markus Kaestner
Prodigy 10 points
Part Number: INA199


Hello,

our problem is similar. We would like to do some current measurements in a car application. There must be a solution for batterie reversal. If this occours, we have to much current into the IN+ and IN- pins flowing thru the recommended series resistors. To ge maximum performance this series resistors should be as low as possible - 10Ohms are recommended in the datasheet. Our question is, what happens if we have 2.7k resistors instead of the 10 Ohms to protect the inputs against excessive current flowing into these pins. So we have less amplification because of the 20k (INA199C1) internal resistors. If we calculate this in our software, i would say that we can deal with it - or are there other issues beside this behavior?

  • 0
    TI__Mastermind 26335 points

    Hello Markus,

    Thanks for considering to use Texas Instruments. I presume you have already considered eFuses for your scenario, but if not they might be worth looking into.

    As for the 2.7K resistor value, I assume that was based upon VCM_max_expected/5mA≤2.7k. If you decide to use that you will see a gain error and it likely will not be consistent as the 20k resistors have a large tolerance and its actually the ratio of internal resistors that is tightly trimmed for the default device gain.  I would expect the output to be typically the gain error factor (found in equation 1 of the datasheet) multiplied by the ideal output.  However, RINT has a tolerance of ±15% and your RS resistor is only as good as the tolerance you choose.  Consequently you will likely need to calibrate the gain for each device.  Aside from gain error, you Rs tolerance will also create an offset error due to input bias currents.