Power Tip: Don't get burned by inductor core losses

Typically, to select an inductor, you'd simply figure out the maximum load current, and establish an inductance by allowing 20% ripple current. It would have a temperature rise similar to the data sheet as the core loss would be insignificant. However, as switching frequencies rise above 500 kHz, core loss and winding ac losses can dramatically reduce the allowable dc current in an inductor. Using 20% ripple current to calculate the inductance sets the same flux excursion in the core material, independent of frequency. The core loss equation takes the general form:

Pcore = K × F^1.3

So if the frequency (F) goes from 100 to 500 kHz, the core loss increases by a factor of eight. This increase is shown in Figure 1, which also depicts the allowable copper loss, which decreases with increasing core loss. At 100 KHz, most of the loss is in the copper and it's possible to utilize the full dc current rating. At higher frequencies, core loss becomes significant. Since the total allowable loss is set by the sum of the core and copper loss, copper loss must be reduced as core loss rise. This continues until the losses become equal. This is an optimum, at higher frequency where the losses are best kept equal and allows the maximum output current to be achieved from the magnetic structure.

Figures 1 and 2 are based on a fixed core volume and winding area, with only the number of turns varied. Figure 2 shows the inductance and allowable dc current for the core loss shown in Figure 1. Below 1.3 MHz, the inductance is inversely proportional to the switching frequency. The inductance reaches a minimum around 1.3 MHz. Above this frequency, the inductance must be increased to limit the core flux and, hence, limit the core loss to 50% of the total. The inductor's resulting current rating is also calculated. At low frequency, where core losses aren't significant, the current rating is set by power losses in the windings.