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BQ25708: BATDRV (BATFET) state when battery is removed

Part Number: BQ25708

Question: what is the state of BATDRV, and therefore BATFET, when the battery is removed?

Background: we have an instance in which there is a front panel switch to isolate the battery from the circuit, as illustrated below.

If the BATFET is held in the conducting state in this case, while the system is powered from the external AC/DC converter, VSYS will feed through to the switch, where it encounters the open circuit to the battery, but some small current draw (approx 10mA) through the LED.

Is the state of BATDRV and the BATFET predictable in this circumstance?

Thank you for your attention.

  • Hi Steve,

    In this case, the simplest thing to do is to disable charge. Then the BATFET will stop conducting and no current can go that path.

    Please see below from the datasheet:

    "When battery is full or battery is not in good condition to charge, host terminates charge by setting REG0x12[0] to 1, or setting ChargeCurrent() to zero."

    Thanks and I hope this helps,

    Peng

    *Please press "This Resolved My Issue" on the thread if my answer is satisfactory

  • Thank you for the response Peng,

    As I understand the response it will require activity by the host. The circumstance I describe is an electro-mechanical front panel switch. A user could change the state of the switch and not notify the host of the change of state. In this case the host would have to monitor the state of the switch. That is a potential solution. But it is a work-around to the question I asked, as opposed to a direct answer.

    If the BQ25708 opens the BATFET when it detects the battery is not connected then the host does not have to monitor the switch state. Which is a simpler solution.

    I appreciate pointing out the work-around, and it may come to that.

    Have a good day,

    Steve

  • Hi Steve,

    The charger does not detect battery not connected and shut off BATFET by itself. If you remove the battery, the SRN pin is floating and the charger will continue to flow current in that path due to the parasitic capacitance and other decoupling caps on that node for that instance of time.

    Thanks and I hope this helps,

    Peng

    *Please press "This Resolved My Issue" on the thread if my answer is satisfactory

  • Thanks again Peng,

    The response implies once the parasitics and decoupling caps are charged the charger will detect the open condition. Is that a true statement?
    If so, will it shut off the BATFET in that case?

    Additionally, in the diagram I shared there will be a constant current flow (approx 10mA) as long as the BATFET is allowing current to flow, due to the LED. Is that current sufficient to prevent the charger from detecting an 'open battery' condition?

    I recognize these are unusual conditions I am asking about, but I do not have any boards in house to experiment with and will have to change the battery charger or switch , or add additional circuitry, if the BATFET does not open when the switch is opened. And so I am trying to nail it down with a high level of confidence.

    I appreciate your attention.

  • Hi Steve,

    For high level of confidence, you would need to go with the original solution, which was:

    "When battery is full or battery is not in good condition to charge, host terminates charge by setting REG0x12[0] to 1, or setting ChargeCurrent() to zero."

    This part is designed to work with a CPU.

    Thanks and I hope this helps,

    Peng

    *Please press "This Resolved My Issue" on the thread if my answer is satisfactory

  • Thanks again Peng,

    I am curious if you are a knowledgeable user or a TI employee. As this is a public forum I imagine in either case this response does not explicitly connote a formal response from TI. Is that so?

    I will mark this question as Resolved after your response.

    Take care,

    Steve

  • Hi Steve,

    I am a TI application engineer supporting this part.

    Thanks and I hope this helps,

    Peng

    *Please press "This Resolved My Issue" on the thread if my answer is satisfactory