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LMR33630ARNXEVM: select right inductor for LMR33630AQRNXTQ1

Part Number: LMR33630ARNXEVM
Other Parts Discussed in Thread: LMR33630, LMR33620, LMR33610

Hi 

  I used LMR33630AQRNXTQ1 in my project. There are some questions when choosing inductance.

When I simulated DC / DC using TI's WEBENCH : Input 9-16v(typical 12V), output 6V, 1.5A, and the recommended inductance of L1 is 8.2uH. When I change the output current to 0.6A, The L1 is also 8.2uH. This inductance seems to be independent of the output current.

  But if I used the Equation 4 in datasheet(SNVSB26C – JUNE 2018 – REVISED OCTOBER 2020) page 21, calculated the inductance value is about 16.7uH.

  If my input voltage is 12V, output 6V@0.6A, Can you help me to select which value of inductance?

  Thanks.

Best regards.

Mingli Zhang

  • Hello

    For this device, we base our inductance on a ripple current of 30% of the rated current of the regulator, rather than the output 

    current required in the application.  That is why the inductance did not change when you changed your load current in Webench.

    As shown in the data sheet, the ripple current percentage can vary between 20% and 40% of the rated current of the device.

    In the case of the LMR33630, 30% of 3A would give a ripple current of 0.9A.  This is somewhat large if your load current is only 0.6A,

    so you could use a smaller percentage of ripple.

    However, it would be better to choose a device with a smaller rated current.  You may want to look at the LMR33620, or LMR33610.

    In any case the best way to calculate the inductance, and the other components, is to use the calculator tool on our product folder for

    this device.  You can use the pull down menu to select the other devices in this family.  I have attached a copy here for your convenience.

    Thanks

    LMR336x0_calculator_B1.xlsx