• State
  • Locked Locked
  • Replies 4 replies
  • Subscribers 64 subscribers
  • Views 166 views
  • Users 0 members are here
Support feedback
Options
Options
Similar topics

This thread has been locked.

If you have a related question, please click the "Ask a related question" button in the top right corner. The newly created question will be automatically linked to this question.

BQ2973: Design for Primary Battery

Jerry Hsu
Prodigy 10 points
Part Number: BQ2973

Tool/software:

Hi Sir,

I'm working on a battery-protected design using the BQ29733.

The goal is to prevent primary battery passivation, so we need to incorporate a parallel HLC (Hybrid Layer Capacitor).

In our scenario, the user can remove the battery, but the HLC remains embedded inside the DUT.

The reason for using the BQ29733 is to prevent the HLC from powering the system when the battery is removed.

I've attached the schematic, and I observed that the BATT+ voltage(primary battery) remains at the same level as the HLC(BATT-3.6V), keeping the system alive even after the battery is removed until the voltage drops to 2.8V.

Does this meet your expectations?


  • 0
    Prodigy 10 points

    The same situation if change R208 to BATT+

  • 0 in reply to Jerry Hsu
    TI__Mastermind 40090 points

    Hello Jerry,

    I think that sounds about right. Once it hits 2.8-V the part is entering CUV mode, which would disable the DSG MOSFET. 

    Before that, the part will continue to function normally.

    Best Regards,

    Luis Hernandez Salomon

  • 0 in reply to Luis H. S.
    Prodigy 10 points

    Hi Luis

    Thanks for your reply.

    I understand that the DUT will enter CUV mode with my original schematic, but I'm confused about changing R208 to BATT+.

    BATT+ should be 0V if the battery is removed but the voltage didn't drop. Can you also check this scenario?

    Additionally, do you have any suggestions on how to modify the schematic (including component changes) to achieve my goal of cutting off the HLC when the battery is removed?

  • 0 in reply to Jerry Hsu
    TI__Mastermind 40090 points

    Hello Jerry,

    That may be from the capacitance of the C252 capacitor, it will retain some charge even after the battery is removed. Our device is very low power, so it may take sometime before it discharges enough. That would be my first thought. 

    Maybe you could have an external circuit that could pull-down the pin when the battery is removed.

    Best Regards,

    Luis Hernandez Salomon