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BQ24266: The Right Chip for my Needs?

Adam Wheeler
Intellectual 270 points
Other Parts Discussed in Thread: BQ24266, BQ25895, BQ25892, BQ25890

Hello,

I am designing an embedded system with a Raspberry Pi and several peripherals (camera, sensors, LCD touchscreen etc), and am looking at the BQ24266 for a power solution. The system load needs between 12.5-15W (5V @ 2.5A-3A) at all times. We are looking for a power system solution that:

  • Charges our 8000mAh LiPo battery (4.2V regulated) and powers our system load simultaneously (TI's PowerPath feature seems to do this)
  • In case of input power loss or disconnection (the device is intended to be powered by a AC/DC converter 99% of the time), the chip will allow the battery to continue to power the load without interruption or a brown out

Our input power supply right now is a 5V at 3A (15W). I need to be able to supply at least 2.5A to the system load at all times, whether through DC input or the battery. The BQ24266 meets our current output requirements, but I have a few questions over the operation of this chip.

  1. If our DC input is being supplied to the chip, does this same voltage and current, minus the current which we allocate for charge current, appear at the output of the BQ24266?
  2. When the DC input supply is cut off from the system, the LiPo battery should immediately begin to supply the system load. However, the output voltage will only be that of the LiPo battery, correct? It seems I would need to add a boost converter between the system load and the BQ24266 to continue to supply the 5V the system requires if I am not mistaken; is this correct?

If this is not correct, what is the right answer to my problem? I've tried several chips and so far I am having trouble supplying the load current from most of these small battery management ICs, but have yet to really find a solution to my problem. It seems that supplying the system current I need is quite a problem...

Thanks for any help,

-Adam

  • 0
    TI__Intellectual 1520 points

    Hi Adam,

    in general our chargers have a buck converter integrated that takes the input voltage and steps it down at the SW-SYS pin. So answering your first question, you will not have the same input voltage at the bq output. This output voltage will be the VSYS voltage that you have selected, like shown in the below picture

    This way you can provide at the SYS pin up to  VBATreg. If you want to provide 5V from a battery, there is no need of an external boost because most of our chargers can provide 5V entering OTG mode. In this case the output would be provided at the PMID pin (and I hope this answered your second question).

    The bq4266 OTG mode can provide 5V with only 1A . So I would rather suggest the bq25890, bq25892 or the bq25895. These chargers can provide up to 2.4A at the PMID pin when in OTG mode. 

    Since you need to enter OTG mode only when VIN is cut off, I would suggest in particular the bq25895 since the OTG mode is enabled by default. With the other 2 chargers you might need to detect when the VIN is disconnected and enable the OTG via I2C.

    Hope this helps

    Let me know if you have further questions.

    Sergio

  • 0 in reply to Sergio Iannace
    Intellectual 270 points

    Hi Sergio,

    Thank you for the recommendation on the BQ25895; I've begun to look into this chip and it does look like it will be a single chip solution (i.e. I will not need an external boost converter between this chip and my system). I had some questions to ask to hopefully solidify my understanding, as this is the first battery powered/charging PCB I've designed...usually I just slap on some nice DC/DC converters and 'set it and forget it' ;)

    • The SYS pin is the system the chip is powering, while the PMID is the output from the internal battery boost converter. In my application, where the system load is the same load the battery will be supplying, can I tie PMID to SYS like shown below?
    • The 'simplified schematic' from the datasheet shows the SYS pin as ranging from 3.5-4.5V; is this just an example voltage, or can I supply 5V on this pin as well?

    • 0 in reply to Adam Wheeler
      TI__Intellectual 1520 points
      Hi Adam,
      The SYS pin is the load for you chip when operating in buck mode. Here it is a normal operation: If the DC supply is connected, the load will be powered by the DC supply; when disconnected, the battery will provide the power to the SYS pin. In Boost mode (or OTG mode) the load is at the PMID pin instead.
      The output voltages shown in the picture are the practical values: this range depends upon the battery voltage, which for a Li-Po battery (when fully charged) can go up to 4.3V.
      So you cannot reach the 5V at the SYS pin but you can instead at the PMID in both of your operative condition: when the DC supply is connected PMID will follow the input (as you can see from the diagram in the datasheet) so you can power your load directly from the input supply. When the supply it is disconnected, the device enters OTG mode, providing 5V at the PMID (=your load) from the battery.
      In other words, you can connect your load at the PMID pin (regardless of the SYS pin that you do not need to use) and not in the way you showed in the picture.
      Let me know if you have further questions.

      Sergio
    • 0 in reply to Sergio Iannace
      Intellectual 270 points

      Hi again Sergio,

      Thank for you that reply, this chip is making more and more sense, and I'm well on my way with the design! I am a little confused as to where in the datasheet it states that I can power my load from PMID when the DC supply is connected? This section of the datasheet states that PMID is only active when these conditions are met; one of which is the input bus voltage is less than the battery voltage (i.e. VBUS is less than 3.8 - 4.3V, which will never happen if I have a 5V DC input).

      Am I missing something?

      Thanks again,

      -Adam

    • 0 in reply to Adam Wheeler
      TI__Intellectual 1520 points
      Hi Adam,
      just bear in mind that this section is referring to the boost mode. When you have the DC supply connected you won't be in boost mode but you will provide the 5V from the DC supply. When you have the DC supply disconnected, the second condition can be met and the device (as discussed in my previous answer) enters boost mode.

      Sergio