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