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BQ24780S: Minimum VCC for correct opperation

Mark Cullen
Prodigy 180 points
Part Number: BQ24780S

From the datasheet it says the charger is good for 1 to 4 cell designs

I have followed the design guidelines from the data sheet using a diode OR to power the device either from the input supply or from the battery.

However with a single cell Li-Ion battery with nominal voltage of 3.65, minus the voltage drop of the diode we will be well below the minimum VCC requirement of 4.5 from the data sheet when the system is running from the battery.

So do I need to include a boost converter on the battery before powering the charger or can the charger run at less than 4.5, if it can what is the lower limit of VCC for correct operation.

  • 0
    TI__Genius 11850 points

    Hi Mark,

    VCC must be above UVLO in order for the BQ24780S to power up and begin SMBus communication.

    However, please note that the BQ24780S is a buck charger, so there will be no charging unless the adapter voltage is sufficiently above the battery voltage.

    Best regards,

    Angelo

  • 0 in reply to Angelo Zhang87
    Prodigy 180 points

    Hi Angelo

    Thanks for that.

    I'm interested in this for the situation where the normal supply is off, I'm running from a 1 cell battery and the battery is becoming discharged.

    I need to be sure that I can still access the device via I2C and that the BATDRV pin can keep the associated FET on.

    So if the worst case for the lockout is 2.8V but my voltage is falling and there is a 200mV hysteresis my cutoff would be around 2.6V is that correct

    Cheers

    Mark

  • 0 in reply to Mark Cullen
    TI__Genius 11850 points

    Hi Mark,

    Yes. In the situation you're describing, the worst case falling threshold would be 2.6 V. However, a 1s Li-ion battery typically should not be allowed to discharge much below 3 V anyway due to safety reasons and potential damage to the battery.

    Best regards,

    Angelo

  • 0 in reply to Angelo Zhang87
    Prodigy 180 points

    Hi Angelo

    I'll set my cut off at 3V on the battery, with the diode drop that would be about 2.7 on the device.

    Thanks for your help

    Cheers

    Mark