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LM61460: Inductor selection

Takashi Yamaguchi
Genius 13290 points
Part Number: LM61460

Hi team,

I received a question about the inductor selection from my customer.

The d/s states that "The saturation current rating of the inductor must be at least as large as the high-side switch current limit, IL-HS". Therefore, I think customer needs to choose an inductor with a saturation current of at least 11.5A if the duty cycle is less than 35% from Fig.19. I think the definition of saturation current varies among inductor manufacturers. Is it correct that saturation current is <30% from normal inductance value? It seems that the EVM employs Wurth's 74439346047 which rated current is 7.4A and Isat is 14.7A(<30%).

Customer's condition

Vin=24V, Vout=3.3V, Iout=4.5A. L=3.3uH.

Customer cannot use large size inductor due to space limitation. However, when they try to meet 11.5A to support short circuit protection, the inductor size becomes larger. 

Regards,

Yamaguchi

  • 0
    TI__Guru 50120 points

    Hi Yamaguchi,

    The inductor selection should follow Equation 11 if the customer's power conditions are operating at a duty cycle less than 50%. Do you know what frequency they are operating at to confirm that 3.3uH is suitable for their design?

    In any case the inductor's saturation current should be rated for at least 11.5A (max high side switch current limit) as duty cycle approaches 0%. As you mentioned the LMZ61460EVM is using the Wurth's 74439346047 inductor. 

    The saturation current is generally taken to be the current at which the level of inductances falls by a specific amount. I think you are correct to assume that the saturation current of the Wurth inductor 14.7A is taken such that the inductor dropped by 30% of 4.7uH typical. 

    What is the exact space contraint (L x W x H) that the customer is okay? I'm guessing the mentioned worth inductor is too big for their design criteria?

    Regards,

    Jimmy