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BQ76200: Current drive capability of DSG and CHG

Nan Zhang
Intellectual 1895 points
Part Number: BQ76200

Hi team,

     Please help confirm the current drive capability of the MOSFET drive pin of BQ76200, including instantaneous drive capability and steady-state drive capability.

     According to the datasheet, if i use the dv/dt*C=I, the output current from the DSG would be 4mA. Am i right?

     I read the bq76200 Beyond the Simple Application Schematic, bq76200 would use the energy stored in Cvddcp,(which is charged by internal charge pump) to drive the gate of mos via DSG or CHG. Meanwhile, the internal charge pump still charging the gate of mos.  Am i right? Further more,  the quick and larger charging current from the cap would be called as instantaneous driving capability. And the charging current from the internal charging pump would be called as steady-state drive capability.

Q1.could you please tell me how to calculate the charging current from the cap?

Q2.And how much current could the internal charging pump output?

    

    yours

nan

  • 0
    TI__Mastermind 42060 points

    Hello Nan, 

    The exact current capabilities for the charge pump were not characterized and it may vary based on load. It can be estimated based on the rise time/fall time and load used.

    Best Regards, 

    Luis Hernandez Salomon

  • 0 in reply to Luis H. S.
    TI__Intellectual 1895 points

    Hi Luis,

          To make it clear.

         Q1, could you please tell me how much current the charger pump could output?

         Q2.We want to calculate how much the current is from Cvddcp. So the Vddcp equals Vbat plus 9-12V. So the driving current via DSG is Vddcp/(Ri1+Ri3+Rdsg).So the question is how much the Ri3 and Ri1.

    yours

    nan

  • 0 in reply to Nan Zhang
    TI__Mastermind 42060 points

    Hello Nan,

    Q1, could you please tell me how much current the charger pump could output?
    Although it has not been characterized, the current that the charge pump can output has been estimated to be about ~40-uA. 

    Q2.We want to calculate how much the current is from Cvddcp. So the Vddcp equals Vbat plus 9-12V. So the driving current via DSG is Vddcp/(Ri1+Ri3+Rdsg).So the question is how much the Ri3 and Ri1.
    The DSG/CHG on/off resistances are in the EC table:

    Best Regards,

    Luis Hernandez Salomon

  • 0 in reply to Luis H. S.
    TI__Intellectual 1895 points

    Hi Luis,

         Thanks for your reply. 

    For question 1, customers want to know the exact range， which could have influence on their device selection.  Could you pleas help to provide such data?

    yours

    nan

  • 0 in reply to Nan Zhang
    TI__Mastermind 42060 points

    Hello Nan,

    What is their application? Have they considered using the BQ769x2 family instead?

    We cannot. This was not fully characterized. 

    Best Regards,

    Luis Hernandez Salomon

  • 0 in reply to Luis H. S.
    TI__Intellectual 1895 points

    Hi Luis,

          In this picture, the customers are concerned that the leakage current through the Rsg would provide a voltage which could open the MOS_DSG.So they want to reduce the value of Rsg. In our reference design, we recommend Rsg as 10MR. Could you please help me to figure out that how much the customers could reduce the value?

    Nan Zhang said:
        

        I will verify their application where they use our bq76200 after figuring out this issue.

    YOURS

    NAN

  • 0 in reply to Nan Zhang
    TI__Mastermind 42060 points

    Hello Nan,

    Unintended turn-on is unlikely. When the CHG FET is OFF, it connects to BAT+, so the voltage different between the gate and source of the CHG FET will always be ~0-V. The DSG FET when OFF will connect to PACK+, so the voltage difference here also is ~0-V.

    So leakage current would have to be significant enough to overcome these, which is unlikely. These are also N-Channel FETs, so the Vgs voltage is what would determine if it turns on or not.

    Regarding the 10-MOhm, I would not recommend to lower this at all. As I mentioned previously, the charge-pump can output roughly ~40-uA. So 10-MOhm was chosen to keep the current consumption low and to not overload the charge-pump.

    Best Regards,

    Luis Hernandez Salomon