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Original question:

DRV10987: Component selection and PCB layout

DRV10987: 47uH inductor

David Gustavson1
Prodigy 210 points
Part Number: DRV10987
Other Parts Discussed in Thread: , TIDA-01585

There are several posts already about the 47uH inductor needed for the +5 volt regulator buck supply, but none of them adequately address the issue.

Please don't just parrot back to me what is in the datasheet.  I can read the datasheet.  What is in the datasheet is contradictory and unclear, and there cannot be a one size fits all answer.  

The datasheet says to use a 47uH inductor with a 1.15A saturation current.

1.  You don't do this on your own eval board.  You use a 0.11A inductor.

2.  The layout shown in the datasheet contradicts 47uH @ 1.15A.  There is no way that size inductor will fit the footprint shown.  An inductor of that rating will be as big as the chip.

3.  The value in the datasheet may be a bulletproof worst case scenario, but certainly it can be optimized based on how much power will be used for external devices and what the supply voltage for the chip is.  In order to calculate the best inductor value, I would need to know the hysteresis of the buck converter, which is not in the datasheet.

Please provide help in how to better spec this inductor.

Thanks,

Dave Gustavson

  • 0
    TI__Mastermind 24415 points

    Hi David,

    Thanks for posting your question in MD forum.

    I'm not sure if you are looking at the right evaluation board. DRV10987EVM is the evaluation board for DRV10987. The part number of the 47 uH inductor that is used in DRV10987EVM is DR74-470-R. This has 1.15A current rating (rms) and 1.41A saturation current rating which matches the datasheet recommendation. Please follow this link to download the DRV10987EVM design files. 

    Regards,

    Vishnu 

  • 0 in reply to Vishnu Balaraj
    Prodigy 210 points

    Hi Vishnu,

    Thanks for your response.  I looked at the link you sent, and that design does have a very large inductor.  It's as big as the chip. 

    The snippet below is from TIDA-01585:

    In any case, the minimum inductor is going to vary with how much external power is required from the +5.  This is no different than selecting an inductor for any other buck converter.  Most buck converter datasheets will have 1/2 a page or more dedicated to how to calculate the most cost effective inductor for the particular application.  Similar information should be provided for this chip.

    Thanks,

    Dave Gustavson

  • 0 in reply to David Gustavson1
    TI__Mastermind 24415 points

    Hi David,


    I agree with you that there is no information provided in the datasheet about choosing the inductor value. The reason why we do not want customers to choose their own inductor values is because the 5V buck regulator is designed to not only power the external load but also the internal 3.3V LDO which in-turn powers several other critical internal circuitry inside the device such as ADC, 1.8V LDO, band-gap reference etc. Changing the inductor value or choosing a 47 µH inductor with lower current rating will affect the functionality of all the aforementioned internal circuitry.

    As you rightly mentioned, inductor value in buck regulator depends on the output voltage, current ripple and the switching frequency of the regulator. Also, the minimum inductor value for continuous conduction mode depends on the load, duty cycle and switching frequency. Also, the switching frequency of the regulator in DRV10987 is not fixed and varies with load. If you are not loading the 5V Vreg with external load, then you may pick a 47 µH inductor rated for 1A saturation current.

    Regards,

    Vishnu