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BQ27621-G1: GPOUT - Reason for pulling to VDD through 10K resistor

Todd Daynes
Prodigy 45 points
Part Number: BQ27621-G1

Hi,

I am using the bq27621-G1 in an application where the battery is not removable from the system.  This design is also extremely space constrained: 0201 size components used throughout the design.

I am not using GPOUT in the application.  Our programming model will never put the gauge into shutdown power mode.  We are not using GPOUT for detecting gauge status.

In the datasheet, there is note that even if GPOUT is not being used, it SHOULD not be left open and it is recommended to pull it to VDD through a 10K resistor.  If the datasheet said ".... MUST not be left open ...", I would not be asking this question. 

My question is why is this pull-up required?   

I have the bq27621 EVM board and have it configured so that GPOUT is floating.  I don't see any erroneous operation in this configuration.

Regards,

Todd  

  • 0
    TI__Guru 61266 points

    This pin becomes an input in shutdown mode so if you leave it floating there's a possibility that it will automatically exit shutdown mode. If you don't use shutdown mode (e.g. during long time storage where you want as little current through the gauge) then it's a non-issue.

  • 0 in reply to Dominik Hartl11
    Prodigy 45 points
    Thank you Dominik.
    Your explanation makes sense.

    Is there any concern that with GPOUT floating to an intermediate value, that we may have excessive leakage current flowing through the input transistors of the GPOUT pad? If this is a possibility, do you have any idea of the order of magnitude of the leakage current? In my particular application (shutdown mode not used), would I be better tying GPOUT to VSS to prevent GPOUT from floating?

    Regards,
    Todd