How Inductance and Inductor Size Depend on Load Current

How Inductance and Inductor Size Depend on Load Current 

For all topologies, if we double the load current (keeping input/output voltages and D fixed), r will tend to halve since AI has not changed but Idc has doubled. Therefore to restore r to its optimum value of 0.4, we need to get AI to double too. But we know that AI is simply voltseconds per unit inductance, and in this case the voltseconds has not changed. So the only way to get AI to double is to halve the inductance. 

What about the size Since we doubled the load current, but still kept r at 0.4, the peak current Idc(1 + r/2) has also doubled. But the inductance has halved. So the energy-handling 

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