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LM60440: Ripple current.

Kengo.Y
Guru 11245 points
Part Number: LM60440

Hi team,

I'd like to know about the Ripple current.
My customer wants to use LM60440 for USB PD.
Since USBPD changes the output voltage, I am worried about the value of the output inductor.
Because Ripple current is less than 10% when input/output conditions are severe.

I think 10uH is good for my customer's condition.
However, the ripple current is severe under the following conditions.
Vin=20.4V
Vout=20V

Do I need to adhere to the minimum of 10% stated in the datasheet?
What happens if you don't follow it?

Sincerely.
Kengo.

  • 0
    TI__Guru 50120 points

    Hi Kengo,

    As per the datasheet, it suggests staying at or above the 10% minimum ripple current. The inductor selection usually consists of picking a ripple current percentage between 20%-40% of which 30% ( ie. K=0.3 in the equation 4) is the optimal value. 

    Picking that high of an inductance such that minimum inductor ripple current is below 10% may result in the internal PLL not responding properly and result in unstable design. If the customer expects to see 20.4Vin|20Vout in normal operation, then the inductor selected should account for this condition too. 

    In this case I would suggest decreasing the inductor value to stay within limits.

    Regards,

    Jimmy