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BQ24192: Inductor at BQ24192's SW pin.

Madhu Wesly
Intellectual 280 points
Part Number: BQ24192

Hello,

The inductor value mentioning in the datasheet is 2.2uH, actually, my application requires around 2.5A load current peak + Battery charging(500mA to 1A is also okay), and the input voltage will be in the range of 4V to 9V.

The 2.2uH inductor suggested is very large in size, what will be the cons if I use 1uH/1.2uH inductor for my application mentioned above??

Thanks in advance.

  • 0
    TI__Genius 10240 points
    Madhu,

    You can use 1uH/1.2uH inductor as shown on the datasheet Page 36. The solution size could be reduced for lower current application. The BQ2419x family has integrated control compensation. To keep the same corner frequency of the LC filter, the output capacitance needs to be doubled to 40uF /10V.

    Regards,

    Eric
  • 0 in reply to ezhao
    Intellectual 280 points

    Hi Eric,
    Thank you.

    One more question: Can I use a single 47uF capacitor at the output, or I have to place 4 X 10uF capacitors as given in datasheet?? Will there be any difference??

    Could you also let me know in general how to decide on capacitor package(size) used at input and output of the regulator ??

    Thanks in advance.

  • 0 in reply to Madhu Wesly
    TI__Genius 10240 points
    Madhu,

    The ESR (equivalent Series Resistor) would be different for the two cases. The output ripple would be different. You can take BQ24192EVM BOM as the guideline for component selection.

    Regards,

    Eric