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BQ77915: CBI pin to GND resistor value - for lower most IC in a stack - 10K or zero - For internal Balancing

Part Number: BQ77915

Hi,

We are using internal  cell Balancing .

a) Using single IC in  our 4 cell BMS.

b)Using 4 ICs  cascaded in our 18 cell BMS.

What should be the CBI pin to  GND resistor in a)   and also in lower most IC in b)   above.

In Datasheet Fig 9-13, 9-14 , 10-9  , at 2 places 10K  is  used.  In one case directly shorted to ground.

Please clarify.

Thanks in advance.

Regards,

Anil

  • Hi Anil,

    BQ77915 CBI has an internal pull up with limited current.  It can be connected directly to VSS on the lower part if you always want balancing enabled in designs like your circuit a), or the bottom part on a design like your b).   If you want to control when balancing is enabled you can pull CBI down with a transistor when desired.  You should not drive CBI high since you can push current into the pin and raise the voltage of the AVDD as shown in figure 8-5 of the data sheet.  While CBI can be connected to VSS it can also be connected through RCB to VSS as shown in figures 9-13 and the bottom part of figure 9-14.  If CBI sources 1 uA (the current value is not shown in the data sheet other than figure 8-5),1 uA through the 10k recommended RCB resistor is 10 mV, well below the maximum VCBIL of 0.5 V.  So while using the resistor is ok, connecting to VSS directly as shown in figure 10-9 will save the resistor.

    When using a stacked design such as your circuit b) the CBI of an upper part should be connected through a 10k resistor RCB to the CBO pin of the next lower part.  This resistor is shown between the parts in figure 9-14.  

  • Thank you,

    So we can use 10K or directly connect to GND.

    Marking as Resolved

    Regards,

    Anil