How Inductance and Inductor Size Depend on Frequency

If keeping everything else fixed (including D) we double the frequency, the voltseconds will halve, because the durations toN and toFF have halved. But since AI is voltseconds per unit inductance, this too will halve. Further, since Idc has not changed, r = AI/Idc 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 AI we were left with at the end of the last step. The way to do that is to 

So since L has halved, and Ipk is unchanged, the required size of the inductor has halved. 

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