LM61460-Q1: Inductor Selection: maximum current

Fan Cunhai

Part Number: LM61460-Q1
Other Parts Discussed in Thread: LM61460

Tool/software:

From the datasheet:

"Note that when selecting the ripple current for applications with much smaller maximum load than the maximum available from the device, the maximum device current must still be used. "

How to understand the maximum current.

For Example: Requirement: Input 12V, output 5V/ maximum 4A.

But the LM61460 maximum available current 6A.

When I select the ripple current for applications, should I set the maximum 4A or 6A?

What's that mean the device? DCDC or load?

7 months ago

0 Andrew Kutzler 7 months ago

TI__Mastermind 23815 points

Hi Fan,

I've assigned an expert to your question.

Thanks,

Andrew

+1 Taru Kapoor 7 months ago in reply to Andrew Kutzler

TI__Genius 11815 points

Hi Fan,

There are a couple of things to consider here.

1. Stability: Minimum Inductance required (which comes from Equation 10). You shouldn't use an inductor with an inductance lower than this value.

2. Inductor Current Ripple: Please use the max current of your design (in this case 4A) to select the inductor in Equation 9.

You can use the quick start calculator for this device to quickly select the right passives around the device. https://www.ti.com/tool/download/SNVR479

Thanks,

Best Regards,

Taru

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LM61460-Q1: Inductor Selection: maximum current