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BQ25570EVM-206: IS IT POSSIBLE TO HAVE 3.3V CONTINUALLY AT VOUT?

Ander Lahidalga
Prodigy 30 points
Part Number: BQ25570EVM-206
Other Parts Discussed in Thread: BQ25570

Hello,

I am designing an application where a micro needs to be supplied continually. The source is a solar harvesting system and the battery is a Li-Ion battery of 400 mAh. This way, I don´t know if the BQ25570 allows me to supply the micro with 3.3V in the following situations:

  1. The source is ON but the battery is discharged.
  2. The source is OFF and the battery is charged.

What I mean is (if it is no well explained),

  1. What would happen if the battery is discharged and the source is ON? 3.3 V will be in Vout and the battery will be charged?
  2. What would happend if the battery is charged and the source is OFF? 3.3 V will be in Vout?

Thank you in advanced.

  • 0
    TI__Genius 10385 points

    Ander,

    The exact behavior of VOUT is dependent on what signal pin you connect VOUT_EN to.

    Regardless of where you you connect it, I can tell you that if your battery is deeply discharged with the source ON, the VOUT will not be able to power up and produce 3.3V. This is because the input of the buck converter is VSTOR. VSTOR is greater than VBAT, the two voltage are effectively shorted across the body diode of the FET connecting them. If this VSTOR voltage is less than your VOUT, the buck converter will not be able to operate.

    However, if the source is off but the battery is charged, then the VOUT will produce 3.3V as the input to the buck converter (VSTOR) will be sufficient to power the VOUT.

    Thanks,

    Ricardo

  • 0
    Intellectual 260 points

    Why don't you charge a super capacitor and draw from that when the micro is sleeping, or even after the input voltage to the regulator drops out? A well designed and good micro should only draw a few mA of current running, and uA when in sleep. A super capacitor would keep it running for quite some time.

  • 0 in reply to Ricardo
    Prodigy 30 points

    Ricardo,

    Thanks for your fast reply and perfect explanation. Could you please confirm this is right?

    In the moment the source is ON, the battery will start charging. When the battey is charged at the minimun voltage for the Buck to start working, a regulated voltage (3.3) will be in VOUT. In the moment the source is OFF, the Buck converter will be working while the battery is discharging until the battery voltage is less than the minimun for the Buck convertert to work. 

    Thank you in advanced.

    Ander.

  • 0 in reply to Ander Lahidalga
    TI__Genius 10385 points

    Ander,

    Your understanding is correct. If you plane to drive VOUT_EN with the VBAT_OK signal, I suggest you look at Figure 21 in the datasheet as it describes the operation of the device across the various voltage thresholds clearly for this configuration.

    Thanks,

    Ricardo