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INA240-Q1: About test method

Part Number: INA240-Q1
Other Parts Discussed in Thread: INA194A-Q1, INA240,

Hi All,

I received the following question from the customer.

1. When doing ICT inspection, is there no problem even if applying 10 mA between each terminal ?

2. In case of function check How can I connect IN - IN +? They are connected by shunt resistors.

Does the same answer apply to INA194A-Q1 ?

Best Regards,

Koji Hayashi

  • Hello Koji,
    1. I do not know what a ICT inspection is, but (VIN+)-(VIN-)=10mV or (VIN+)-(VIN-)=-10mV is fine. This can also be done with the INA194A-Q1.
    2. You can connect INA240 input pins such that they are high-side, low-side, or in-line. The INA194-Q1 is a unidirectional device, be sure to orient this part such that VIN+ is at a higher potential than VIN-.
  • Hi Patrick-san

    Thank you for response.

    1.
    ICT stands for In-circuit test.
    We are planning to confirm also soldering in this test.
    Patrick-san's answer is about voltage.
    Is it OK to apply 10mA ?

    2.
    I understand this matter.


    Best Regards,
    Koji Hayashi
  • Hello Koji,

    If I understand correctly, you are doing something like in the figure below.  The differential impedance between the input terminals is about 3.2kohm.  10mA through that create a differential voltage of 32V.  According to the datasheet absolute maximum ratings table, the device can handle -80V to 80V accross the terminal pins.  So 10mA applied here should be ok.

    If I misinterpreted what you said, please provide a diagram of what you are wanting to do.

  • Hi Patrick-san

    Thank you for response.

    I understand that applying 10mA between IN+ and IN- has no problem.
    How about each terminal ?
    If the answer is difficult, could you tell me how to electrically check if solder is attached ?
    My customer's purpose is to see if solder is attached.

    Best Regards,
    Koji Hayashi
  • Hello Koji,

    I do not have specifications on current limits for all pins. However, I know that a forced current will generate a differential voltage between pins. You will need to ensure that these differential voltages fall within the specifications listed in the "Absolute Maximum Ratings" section of the datasheet. For such a large current, this may require that you have diodes between various pins and ground or supply to clamp the voltage.
  • Hi Patrick-san

    Thank you for response.
    I apologize for the late response .

    >I do not have specifications on current limits for all pins.
    Is also unknown about INA194A-Q1 ?
    Is differential impedance between of INA194A-Q1 also 3.2kohm ?

    Best Regards,
    Koji Hayashi
  • Hello Koji,

    Per the absolute maximum ratings table in the datasheet, all pins need to have the current limited to 5mA. The differential impedance is 5k with a tolerance of 30%.
  • Hi Patrick-san

    Thank you for response.

    I understood as follows.

    The differential impedance between the input terminals.

    INA240-Q1 is 3.2kohm.

    INA194A-Q1 is 5kohm.

    Current limit on all pins

    INA240-Q1 is not clear.

    INA194A-Q1 is 5mA.

    Is it right ?

    Best Regards,

    Koji Hayashi.

  • Hello Koji,

    That is correct.
  • Hi Patrick-san

    Thank you for quick response.

    INA194A-Q1 is 5kohm.

    If 5mA is applied, it becomes 25V with 5mA x 5 kohm

    I think that it is necessary to set it to 2.7 mA or less for the differential input.
    Because 18 V / (5 kΩ × 1.3 times) = 2.77 mA

    Best Regards,

    Koji Hayashi

  • Hello Koji,

    The part should handle up to 5mA. I presume the 1.3 times is accounting for the tolerance in the resistance between the device inputs and you are wanting some margin between the max current you will source to the device and the max specification for the device. For worst case, I would use 0.7 instead of 1.3. For an 18V input this will yield a higher current value.
  • Hi Patrick-san

    Thank you for response.

    At 5kohm * 0.7 = 3.5kohm it will be over 5mA.
    18 / (5 * 0.7) = 5.14mA
    This exceeds the absolute maximum rating of Input Current.

    If the resistance is calculated at 1.3 times, it will exceed the absolute maximum rating 18V of the differential input.
    Considering the differential input and the input current, I think that it is necessary to calculate with under 2.7mA.

    What do you think?

    Best Regards,

    Koji Hayashi

  • Hello Koji,

    The device impedance between IN+ and IN- can have a 20 to 30% tolerance. Hence that is why I presumed you were multiplying 5k * 1.3 and that is why I suggested you use 0.7 for your analysis.  Based off what you noted above, you will need to lower your common mode voltage for this test so that in the worst case the device sinks less than 5mA.  Alternatively you could use a resistor in series with the device impedance.  However, this will lead to a gain error. 

    If your 1.3 does not correspond to the device impedance including tolerance or it does not correspond to some total resistance from the sum of the device resistance and an external series resistance, please provide a diagram to indicate what you are wanting to do.

  • Hello Koji,

    I have not heard from you in a while. Therefore I presume your issue is resolved.  However, if you need additional help, please reply below or start a new thread.

  • Hi Patrick-san

    Thank you for concern.

    I am so sorry for the delayed reply.

    I informed the customer of the answer.

    If there are additional questions, I will post new thread.

    Best Regards,

    Koji Hayashi