A general recommendation for the inductor in the LM21305 application is to keep the peak-to-peak ripple current between 20% and 40% of the maximum DC load current (5 A), 30% is desired. The inductor also should have a high enough current rating and a DCR as small as possible.

The peak-to-peak current ripple can be calculated by:

Equation 3.

The current ripple is larger with smaller inductance and/or lower switching frequency. In general, with a fixed output voltage, the higher the PVIN, the higher the inductor current ripple. If PVIN is kept constant, inductor current ripple is highest at 50% duty cycle. It is recommended to choose L such that:

Equation 4.

The inductor should be rated to handle the maximum load current plus the ripple current.

I_L(MAX) = I_LOAD(MAX) + Δi_L(MAX) / 2

An inductor with saturation current higher than the overcurrent protection limit is a safe choice. It is desired to have small inductance in switching power supplies, because it usually means faster transient response, smaller DCR, and smaller size for more compact design. But too low of an inductance will generate too large of an inductor current ripple and it can falsely trigger overcurrent protection at maximum load. It also generates more conduction loss, since the RMS current is slightly higher relative to that with lower ripple current with the same DC load current. Larger inductor current ripple also implies higher output voltage ripple with the same output capacitors. With peak current-mode control, it is recommended not to have too small of an inductor current ripple so that the peak current comparator has enough signal-to-noise ratio.