A Cree white LED XLampXML is used. According to the LED manufacturer’s data sheet, this LED has 2.9-V forward voltage at 1-A current, so VOUT = 2.9 V × 4 + 0.1 V = 11.7 V. Choose KIND = 0.3, which gives a 36-µH inductance. With this inductance, the ripple current on the inductor is only 0.3-A peak-to-peak, which is too conservative and increases total system cost and size.
For this application, with concerns about system cost and size taken into account, decide the inductance by choosing a larger peak-to-peak inductor ripple current. To choose a proper peak-to-peak inductor ripple, the low-side FET sink current limit should not be exceeded when the converter works in a no-load condition. To meet this requirement, half of the peak-to-peak inductor ripple must be lower than that limit. Another consideration with this larger peak-to-peak ripple current is the increased core loss and copper loss in the inductor, which is also acceptable. Once this peak-to-peak inductor ripple current is chosen, Equation 14 can be used to calculate the required inductance.
Choose 1-A peak-to-peak inductor ripple current, and half of the current is 0.5 A, much lower than the minimum low-side sink current limit of 1.25 A. The calculated inductance is 10.9 µH. Choose a 10-µH inductor with part number 744066100 from Wurth. The ripple, peak, and rms currents of the inductor are 1.09 A, 1.54 A, and 1.05 A, respectively. The chosen inductor has ample margin in this design.