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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

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    \$\begingroup\$ How much energy do you think the inductor carries? It's often on the order of tens of microJoules. There will be a capacitor to dump it into, too. Suppose you are bucking down from 15 V to 5 V using a typical 100 uH inductor into a typical 100 uF output capacitor using the ever-present-on-ebay 52 kHz LM2576 IC. The inductor's cycle energy will cause the capacitor voltage to rise about 100 mV. That's without the load drawing at the same time. Not scary. \$\endgroup\$ – jonk Jul 9 '20 at 2:13
  • \$\begingroup\$ 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. \$\endgroup\$ – helloguys Jul 9 '20 at 2:22
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    \$\begingroup\$ dV/dt rise= I/C with a single pole switch. skipping cycles can prevent further rise. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jul 9 '20 at 2:30
  • \$\begingroup\$ Worthwhile to note that it will go "anywhere available". If there are output caps then other answers cover it. With no cap voltage will rise until energy is stored in stray capacitance or dissipated in a spark. The definition of an inductor includes the concept that current is continuous at all times - you cannot step change current so it MUST go somewhere when the switch opens. Not giving it somewhere formal to go can be a bad mistake. \$\endgroup\$ – Russell McMahon♦ Jul 9 '20 at 22:12
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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.

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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\$).

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