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LM61495: Inductor value selection

Kei Kuwahara
Expert 7650 points
Part Number: LM61495

Hello team,

My customer is selecting inductor for this device.

The customer design is below:

  • VIN: 24V
  • VOUT: 15V
  • IOUT MAX: 8A
  • FSW: 400kHz
  • K: 0.3

The equation for chosing the L from datasheet is below (equation 5)

So, L = (24V - 15V)/(400kHz*0.3*8A)*(15V/24V) = 5.86uH

The datasheet states that the minimum value of L is calculated by below (equation 6)

In this case, L > 15V/(400kHz*0.6*10A) = 6.25uH

The L chosen by the equation 5 can't meet the minimum L calculated by equation 6.
Also if they consider the variation of L (normally +/-30%), they have to chose the L with size of 10uH. (because 7uH at -30% variation)

Using the 10uH inductor, the inductor ripple current (K) will be 17%.
17% of ripple still fulfill the minimum ripple of 10%, but it will be out of 20% - 40% range.

Can customer ignore the general rule of inductor ripple current of 20% -40%?
If it is not recommended, what value of inductor should customer use?

Best Regards,
Kei Kuwahara

  • +1
    TI__Guru 50120 points

    Hello Kuwahara-san,

    Usually for current mode control devices, subharmonic oscillation happens when steady state operation duty cycle is above 50%.

    In customer's application of 24Vin and 15Vout, the device will have duty 50% so inductor should follow Equation 6.

    Note that Equation 6 specifies the the inductor must be bigger than the calculated value should be greater than 6.25uH and higher. 

    Also note that as a rule-of-thumb, the minimum inductor ripple current must be no less than about 10% of the device maximum rated current under nominal conditions.

    The general rule of thumb inductor range (20% - 40%) can be ignored for Equation 6 in customer's particular use case. 

    Regards,

    Jimmy