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BQ40Z50: BAT Pin - Apply 300 Ohm resistor to trace connected to pin

Joseph DeWitt
Intellectual 320 points
Part Number: BQ40Z50

Hi,

I hope you are doing well.

I would like to install a 300 Ohm resistor to the trace feeding the BAT pin. The purpose of the resistor will be to limit the amount of power that can be seen by the IC under a fault condition for intrinsic safety purposes. 

My current understanding of the IC is that it will consume 15 mA of current. My battery assembly will be comprised of three cells in series (3.6 V x 3 = 10.8 V). The battery assembly will likely be configured to shutoff at around 8.3 V. Therefore, installing a 300 Ohm resistor will result in a voltage drop of 15mA x 300 Ohms = 4.5 V. At the near fully discharged condition the BAT pin will see 8.3 V - 4.5 V = 3.8 V. According to Section 7.3 Recommended Operating Conditions of the datasheet, a minimum voltage of 2.2 V is recommended. Given that the voltage (3.8 V) will be greater than the minimum recommended value (2.2 V), can you please advise / confirm that installing the 300 Ohm resistor on the trace feeding the BAT pin will not interfere with the IC's functionality.

Thanks in advance for your support.

Best regards,

Joe DeWitt 

  • +1
    TI__Mastermind 27321 points

    Hi Joe,

    The gauge does not use BAT pin voltage for gauging. You might want to disable this feature (see below) since the BAT pin voltage would be less than the sum of all cell voltages. 


    In any case, the gauge will be using individual cell voltages and the sense resistor voltage drop to trigger protections. What is the main reason of using the 300Ohm? 

    Regards,
    Jose Couso

  • 0 in reply to Jose Couso
    Intellectual 320 points

    Hi Jose,

    That is super helpful feedback. I was unaware of that shutdown logic. The installation of the 300 ohm resistor would be strictly for intrinsic safety thermal compliance purposes. Essentially the 300 ohm resistor will limit the power that can be seen by the IC under a worst case fault scenario. For this example the power will be limited to 10.8 V^2 / 300 Ohms = 0.388 W. I then multiply 0.388 W by the RthetatJA (C / W) of the IC which is 47.4 C / W to get the thermal rise which will be 18.42 C (0.388 W x 47.4 C = 18.42 C). This will put me in compliance for my application as a function of my anticipated max ambient temperature (40 C). 

    Best regards,

    Joe DeWitt

  • 0 in reply to Joseph DeWitt
    TI__Mastermind 27321 points

    Hi Joseph,

    Understood. Best of luck!

    Regards,
    Jose Couso

  • 0 in reply to Jose Couso
    Intellectual 320 points

    Thanks Jose, much appreciated!

    Best regards,

    Joe DeWitt.