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Original question:

INA199: not working as expected.

INA199: Output voltage stuck at 0.03V

Michael Browning
Prodigy 210 points
Part Number: INA199

I have an INA199B3DCKR that is measuring the current through a 1 Ohm resistor, the current varies between 0 to 120mA.

I have confirmed that I'm getting the expected voltage drop across the 1 Ohm resistor but the output of the INA199 is always 0.03V.

The IN+ is ~24V and the IN- is ~23.88V. In+ and IN- will both occasionally drop to 0V. Is that my problem?

The reference pin is tied to 0V

The layout of the chip is shown below.

  • 0
    TI__Mastermind 32395 points

    Hey Michael,

    Thank you posting to the forum. I hope we solve the issue so you can get back to engineering.

    It would be best to get multiples measurements of the input pin voltages (IN+ to GND pin, or IN- to GND pins, probed as close to the pins as possible) when the input falls to 0V to make sure the voltages are not dropping below -0.3V, which would cause device damage. Keep the time scale as small as possible to get the best resolution of this fast transient voltage.

    Another thing to check (if the input voltages do not indicate anything) is if there is significant impedance in between INA199 GND and system/board ground. You can simply measure resistance between INA199 GND pin and the system/source ground return. Or you can use a differential voltage probe and measure for any significant voltage drops (< -0.3V) during these occasional input pins falling to 0-V.

    What is happening to system when the input Vcm drops to 0-V?

    Does the output remain stuck even when cycling power?

    How many device or boards have been tested?

    How many failures have you detected?

    Sincerely,

    Peter

  • 0 in reply to Peter Iliya
    Prodigy 210 points

    I measured the Vin+ and - with an oscilloscope and the pins are dropping down to -0.7V. It's an inductive load with a flyback diode so that makes sense.

    I will have to redesign the board so that it is a fixed 24V source, then the sense resistor, then the switch, and then the relay coils as opposed to 24V source, then the switch, then the sense resistor, and then the inductive load.

  • +1 in reply to Michael Browning
    TI__Mastermind 32395 points

    Hey Michael,

    Yes that will have to be one of the options (switching it from low-side to high side).

    The other options are:

    1. Insert clamping schottky diodes at input pins to ground to clamp input pins to to their forward voltage, but this is difficult because it needs to be <0.3V.

    2. Another option is to use current-limiting resistors at input pins to limit the internal ESD turn on current which is what generates damaging heat in these electrical overstress scenarios. Limiting the current to < 5 mA is necessary. Assuming the voltage can drop to -1V, a quick calculation to get the correct Rprotect is 1V/5mA = 200-Ω. In reality the equation should be 1V-Vf_bodyDiode because the internal body diode of the ESD cell will generate some forward voltage, but it is usually small and unpredictable.

    However, inserting these 200-Ohm resistors at input pins will attenuate overall shunt voltage gain. You can refer to section 8.4.1 of datasheet to calculate the new gain.

    Additionally, the gain error (centered around your new attenuated gain) will increase significantly +/ few percent because internal resistors of the device can vary +/-25% due to process variation. You can refer to the video and training pdf on how to calculate the new error here.

    https://www.youtube.com/watch?v=hLGN_wl-xgA

    https://www.ti.com/video/series/precision-labs/ti-precision-labs-current-sense-amplifiers.html

    Sincerely,

    Peter