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INA191: INA191 - current limited Max voltage?

Roy Tebbe1
Prodigy 50 points
Part Number: INA191
Other Parts Discussed in Thread: INA240, INA280, INA290, INA281

HI,

Looking at the INA191 datasheet in the Absolute Maximum ratings section, Note 3 states - (3) Input voltage at any pin may exceed the voltage shown if the current at that pin is limited to 5 mA.

In my application the current will be limited to 1mA.  Greater than this will likely damage other components in the circuit.

Is it possible for the Common mode voltage to be even higher than 42V? 

If 42V is the maximum under all conditions.  Will the INA191 function properly over temperature (-40 to 85C) ?

Regards,

Roy

  • 0
    TI__Mastermind 32205 points

    Hello Roy,

    Welcome to the forum and thank you for the question.

    As the Absolute Maximum Ratings Tables specifies, the maximum survivable voltage at the input pins is +42V and this is true for the entire survivable temperature range which is-55°C to 150°C. 

    The device can only survive a higher input voltage if there are current-limiting input resistors that keep current into the device to less than 5mA.

    Hope this all makes sense.

    Best,

    Peter

  • 0 in reply to Peter Iliya
    Prodigy 50 points

    HI Peter,

    The maximum current flowing through the sense resistor is limited to 1mA in my application.  So well under the 5mA limit. There is a current limiter circuit before the sense resistor.  The INA191 is well protected from seeing any current higher than ~1.5mA worst case. 

    So in this case what is the Max voltage ?   I only need 45V,  but 50V would be ideal. 

    Assuming 42V Absolute Max is the only value you can specify  (which I understand)  Can you comment on the voltage divider circuit below?

    I need to measure from 7uA  to 1mA.   Spice simulations seem to indicate the INA191 will work correctly.  But WEBench 'error analysis'  indicates 500%+ error at 100uA which would of course not be acceptable.

    Thank you,

    ROy

  • 0 in reply to Roy Tebbe1
    TI__Mastermind 32205 points

    Hello Roy,

    I believe there is a large error showing up because of the device input bias currents (1nA max) and differential impedance (4.6 MΩ). See the simulations below the shunt voltage is attenuated due to voltage drops across the input resistors. This error is discussed in the datasheet. There could be some calibration possible but still with some gain error variance that is beyond the datasheet specification limits. 

    Sincerely,

    Peter

  • 0 in reply to Peter Iliya
    Prodigy 50 points

    HI Peter,

    Ok ,  So is there a way to configure the INA191 to do what I need?   If the series input resistance is reduce and the 8Meg resistor to GND is removed, is that sufficient for the circuit you show to work?

    Regards,

    Roy

  • +1 in reply to Roy Tebbe1
    TI__Mastermind 32205 points

    Hey Roy,

    You could certainly drop the absolute values of the resistors from MΩ to kΩ, but this will increase power loss and overall there will still be large gain error increase. Here is one possible resistor divider combination that minimizes power loss and error, but still keeps input voltage below 42V.

    Another option is to remove the pull-down resistors and limit input current into the device to less than 5mA when VCM > 42V; however, the device will not be operating correctly anymore when VCM>42V and thus no accurate measurements can be taken. So if you have a 20kOhm input resistor for each pin, then input current at 50V would at worst be 50V/20kΩ = 2.5mA.

    Overall, you could make your original circuit work, but you will have to balance gain error vs. power loss as the device's input differential impedance is limited to 4.6 MΩ (+/-20%).

    You could also consider using higher-voltage rated current-sense amplifiers (e.g., INA240, INA290, INA280, INA281) so you do not have to worry about any of these voltage dividers.

    Hope this makes sense.

    Best,

    Peter