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LMR43610: Inductor selection

Daeseon Kim
Prodigy 50 points
Part Number: LMR43610


Hi,

I am planning to design a Step-Down circuit using the LMR43610MB3RPER.

Vin: 4.8~36V

Vout: 3.3V

Iout: 0.3A

Ripple Current Ratio K = 0.3

Using the inductor calculation formula provided in the datasheet, I found that when Vin is 4.8V, it's approximately 11uH, and when Vin is 36V, it's about 33uH.

  1. In this case, which value should I choose between 11uH and 33uH? (For reference, designing with WEBENCH yielded a value of 8.2uH.)

  2. Looking at the inductor specs selected by WEBENCH, it seems over-specified compared to the current consumption (0.3A) of the application I intend to use. Is there a reason for this?