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BQ27421-G1: Battery discharge current increases when GPIO goes low

Juergen Demuth
Prodigy 40 points
Part Number: BQ27421-G1

Hello,

I would like to use the GPOUT pin for indication if the battery SOC is below 50% (by setting BATLOWEN). The datasheet recommends a pullup resistor of 10kOhm at this pin which will result in an additional discharge current of approx. 370µA as soon as GPOUT goes low. This is 4 times the supply current in NORMAL mode or 18 times the SLEEP current !!  I want to avoid a faster discharge when battery is low. Is there any risk if the pullup resistor is increased to 47k or 100k (or even more) ?

The easiest way is to use th internal weak pullup of my µC and only activate the pullup during reading. What happens if GPIO is floating between the readings (SHUTDOWN mode is not used)?

Is it possible to deactivate the output during SLEEP?

The battery voltage (LiPo 750mAh) is between 3V and 4.2V

Another thing I do not understand is the SHUTDOWN mode: To exit SHUTDOWN mode, GPOUT must be raised from low to high. GPOUT is an open-drain output. Before raising it to high, it must be held low externally. Must the CPU pull this pin low during SHUTDOWN mode? In this case the pullup resistor draws additional 370µA all the time. This does not make sense as the shutdown current is below 1µA.

I think that I have not yet understood the principle of operation of the GPOUT pin - hope anyone can help me (Datasheet and Reference Manual did not).

Thanks and best regards

Juergen

  • 0
    TI__Guru* 76230 points
    Juergen,
    I believe you can use a larger value, just verify that doesn't make the pin go into a tri-state because of the larger value of this resistor.
    The device has the potential to get into a shutdown state due to esd. If this happens, the host has to be able to toggle the gpout pin, i.e pull it low and then release it back high to cause it to exit shut down.

    thanks
    Onyx
  • 0 in reply to Onyx Ahiakwo
    Prodigy 40 points
    Dear Onyx,
    thanks for your input.
    Is there a maximum value for the pullup resistor specified? I think the risk of ESD is rather low.
    BUt I still would like to understand how SHUTDOWN should work (at 0.6µA) with a pullup resistor attached that draws 370µA. For me that does not make sense. In any case the pullup current is much higher than the supply current of the device.
    Is it possible to leave the pin floating at least during SLEEP mode und just attach a pullup when reading the state of GPOUT ?

    BR Juergen
  • 0 in reply to Juergen Demuth
    TI__Guru* 76230 points
    Hi Juergen,
    No there is no maximum value specified. 10k is just a recommended value.

    At shutdown, the pull up does not draw 370uA as the pin which is an open drain will not be connected to ground. When the device alerts and becomes the pin becomes active low, then yes the current will be high. i do not believe the pin will stay in a perpetually low state after indicating bat low, but it is worth testing to see if that is the case. Incase that is the case, you should be able to use a larger resistor to reduce current draw.
    thanks
    Onyx