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BQ25703A: REGN has no power, can't communicate with device

Part Number: BQ25703A

Hello,

My customer has a BQ25703A on their board, they have based their design off of the TIDA-01515.

Situation:

  1. They are trying to communicate to the BQ25703A over I2C.
  2. They don't have a voltage applied to VBUS, but the batteries are connected, 3S1P.
  3. They see 10.3V on VSYS.

Issues:

  1. REGN is at 0V
  2. They can't communicate over I2C to the device

Questions:

  1. Based on the datasheet, REGN can be powered by either VSYS or VBAT, but it looks like the LDO enable is only triggered by VBUS, is that correct?
  2. Can the device be communicated to without REGN enabled?

We are trying to narrow down if the issues is on the BQ25703A or the I2C master device, but we noticed this REGN discrepancy in the process of debugging.

Thanks,
Nick

  • Hi Nick,

    When only battery is connected, the charger by default is in low power mode which should shuts off REGN to achieve minimum quiescent current. If you set REG0x01[7]= 0 to disable low power mode, then REGN should comes up.

    BQ25703A is able to communicate even if REGN=0V. But you need to make sure SCL/SDA pins are pulled up to external 3.3V(10 kohm) to start the communication.

    Regards

  • Hello,

    Thanks for the details on the REGN pin. I started to look into the I2C potential issues and I ran across a discrepancy in the datasheet.

    Section 8.5 of the datasheet says the I2C address is "D6h (1101101_x) where "X" is the read/write bit". The issue is that D6h = 11010110 not 1101101_x. 

    So my questions are:

    1. is the write address actually D6h (and the binary is wrong) or is the address actually DAh (and the binary is correct?
    2. for reads the user would send D7h/DBh, correct?

    Thanks,
    Nick

  • Hi Nick,

        The real address of BQ25703A is 6BH(110 1011), for I2C communication the read(1b)/write(1b) will be added after 7bits address and contribute to 8bits byte (1101 011_X), X is the read/write bit (1 for read and 0 for write). Therefore for read user would send D7h(1101 011_1 b) and for write user would send D6h(1101 011_0 b).

    Regards