The following discussion applies to all the topologies.

If keeping everything else fixed (including D) we double the frequency, the voltseconds will halve, because the durations t _ON and t _OFF have halved. But since ?I is "voltseconds per unit inductance," this too will halve. Further, since I _DC has not changed, r = ?I/I _DC will also halve. So if we started off with r = 0.4, we now have r = 0.2.

If we want to return the converter to the optimum value of r = 0.4, we will now need to somehow double the ?I we were left with at the end of the last step. The way to do that is to halve the inductance.

• Therefore, we can generally state that inductance is inversely proportional to frequency.

Finally, having restored r to 0.4, the peak will still be 20% higher than the dc level. But the dc level has not changed. So the peak value is also unchanged (since r hasn't changed either, eventually). However, the energy-handling requirement (size of inductor) is L I _PK ². So since L has halved, and I _PK is unchanged, the required size of the inductor has halved.

• Therefore, we can generally state that the size of the inductor is inversely proportional to frequency.

• Note also that the required current...