Where does the inductor energy go in a buck converter when the load is suddenly disconnected?

Question

It is known that inductors resist a change in current. So, my question is that in a buck converter, what prevents the output voltage from going all the way up due to inductor resisting a current change when there is a load transient? For example, when you go from 3A to 0A. I understand there is some voltage regulation going on, but the inductor still has to discharge all those energy right?

Edit: An additional question question. Right now, I see that the main functions of the output capactiors are:

1. supply current when load increase to prevent voltage undershoot

2. takes energy when load decreases to prevent voltage overshoot

3. filters the AC output voltage ripple due switching

Did I miss anything? I am trying to put all the parts in a switching regulator together.

Thanks

Answer 1

It goes into the output capacitor. The pulse is short enough and the capacitor is large enough that it can be engineered not to significantly make the output voltage rise, maybe by few percent, as long as it is within limits so it's not too high. The output waveform also depends on how fast the regulator reacts to this via feedback loop and compensation as well.

Answer 2

The inductor energy goes into the output capacitors, causing some amount of unavoidable overshoot in the output voltage. Worst case is a load release at the peak inductor current, and you can calculate the overshoot from the deltas in the inductor energy (\$1/2\ LI^2\$) and capacitor energy (\$1/2\ CV^2\$).