For most applications, choose an inductance such that the inductor ripple current, ΔI_L, is between 30% and 50% of the rated load current at nominal input voltage. Use Equation 26 to calculate the inductance. For this example we assume V_IN = 12 V, F_SW = 500 kHz, and a ripple current of 30% of 0.3A. This gives us an inductance of about 65 μH. Choose the next standard value of 68 μH for this design. Next, we use Equation 27 to calculate the actual inductor ripple current across the input voltage range. Finally, use Equation 28 to determine the peak inductor current at our maximum input voltage and compare with the current limit of the LM5168. Arrive at a peak current of about 0.37 A at V_IN = 115 V, which is less than the current limit of the LM5168.

Ideally, the saturation current rating of the inductor is at least as large as the peak current limit. This ensures that the inductor does not saturate even during a short circuit on the output. When the inductor core material saturates, the inductance falls to a very low value, causing the inductor current to rise very rapidly. Although the valley current limit is designed to reduce the risk of current run-away, a saturated inductor can cause the current to rise to high values very rapidly. This can lead to component damage. Do not allow the inductor to saturate. Inductors with a ferrite core material have very hard saturation characteristics, but usually have lower core losses than powdered iron cores. Powered iron cores exhibit a soft saturation, allowing some relaxation in the current rating of the inductor. However, they have more core losses at frequencies above about 1 MHz. In any case, the inductor saturation current must not be less than the maximum peak inductor current at full load.