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LM5168-Q1: Inductor selection for standard buck regulator?

Part Number: LM5168-Q1


We’re developing an internal power supply for a PEV application that uses 10S-16S standard Li-Ion packs.

Input voltage to the supply will be 30V-67V.

Output voltage is 3.3V.

Load current is 0.5A-40mA, typically around 10mA when not in Shutdown Mode.

The packs are user replaceable so the contacts will be subjected to ESD and we will be using a 70V TVS to protect the power input terminals for the supply. This results in a worst case TVS clamping voltage of 113V. The LM5168-Q1’s 120V rating and its low quiescent current is perfect for us.

We will be using it in a standard buck regulator configuration though, no secondary fly-buck output. For selecting our inductor can we just assume that Ipri = Iout1 (equations in section of the datasheet, Transformer Selection) and choose our inductor that way?

Any other concerns unique to a standard buck configuration that might not be covered in the datasheet? Thank you!

Regards, John M.

  • Hi John,

    Your understanding is correct. Actually the equations in section are derived from a regular buck converter. So for the inductance value calculation, you can just refer to equation (19), while input your actual maximum output current into IPRI

    In this equation, Vout1 should be your actual Vout. K is the ripple current factor with recommended value of 20% to 40%. Using this equation, you can ensure that  a ripple current of between 20% and 40% of the load current is used.



  • Thank you so much Xiaoying for the quick response! Please consider my issue resolved.

    Regards, John M