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bq769x0 drive current

Jessica Riley1
Intellectual 280 points
Other Parts Discussed in Thread: BQ76940, BQ76930

Can you tell me what the turn-on and turn-off drive currents are for the bq769x0 DSG and CHG outputs?

  • 0
    TI__Guru 54715 points

    See the "Charge and Discharge Driver" section of the datasheet.  The outputs have different resistances for on and off.  

    Turn on can be calculated from the rise time and should give ~5k ohm, so nominal maximum drive current is when the outputs are low and would be 12/5k or 2.4 mA.  

    Turn off is different between the outputs and the pull down resistance is shown in the datasheet, 1M nominal for CHG and 2.5k for DSG

  • 0 in reply to WM5295
    Intellectual 280 points
    Thank you for the quick response. From the datasheet, it says that it takes about 200us to charge up a load of 10nF from 10% to 90% of V_FET_ON. Wouldn't that give 480uA drive current (0.8*12V*10nF / 200us)?

    I need to know the drive current because I am trying to figure out how quickly the DSG FET will turn on a very large capacitive load (~2mF). I want to make sure that it doesn't trip the BQ SCD due to the large in-rush current. Any suggestions for solving that problem?

    Thanks again,
  • 0 in reply to Jessica Riley1
    TI__Guru 54715 points

    The turn on will be an exponential charging the load capacitance (FET gate capacitance) through the (DSG) pin drive output source resistance.  The current is the "high" 2.4 mA when the voltage is at 0 and very small when the voltage is near final voltage.

    For a high capacitance gate or a FET array with high capacitance you will need a FET driver external to the bq769x0.  You might look at www.ti.com/lit/ml/slup169/slup169.pdf or information from your preferred FET supplier. Be aware that the bq76930 and bq76940 generally do not tolerate long propogation delays for the driver to act since the power pins can be pulled down by a very heavy load such as short circuit.  The SC delays are relatively short, so these parts will work best if your FET switches moderately fast after a short propogation delay, or if your system turn on has a current limit or soft start function. If it is suitable, your system might also be able allow a SCD trip or more at turn on to "charge" the system, but be careful of FET heating and differentiate between a normal turn on and a turn on into a short.

    Of course also be cautious of switching too fast and creating an inductive response in the cells

     

  • 0 in reply to WM5295
    Intellectual 280 points

    Thanks again for the help!  Very useful advice.