My question is with reference to this answer for this previous question.

The answer states that a buck converter will behave in a boost mode for some cycles during light load or no load condition.

Can someone help me with the exact cycle of operation and why does it happen?

  • 1
    \$\begingroup\$ When it operates in reverse; that is, there is output voltage on the output C and you suddenly remove the input voltage at the wrong moment. So you are boosting FROM the output TO the (now open circuit) input. Verbal Kint's comment there has all the details. \$\endgroup\$
    – user16324
    Dec 20, 2020 at 13:48
  • \$\begingroup\$ Yes. Thank you. But I am not still clear. What is the wrong moment? Aand how does it actually boost from the output ? \$\endgroup\$
    – user220456
    Dec 20, 2020 at 13:49
  • \$\begingroup\$ Just go through his answer bearing in mind that the schematic is now back to front. Imagine the rectifier switch is ON maintaining current through L in your schematic when you disconnect Vin. Now turn the rectifier OFF and turn the FET SW on, connecting L (carrying amps) to the open circuit input. Think through what happens. \$\endgroup\$
    – user16324
    Dec 20, 2020 at 13:51
  • \$\begingroup\$ Yes. I tried reading it. But I am not able to understand because, when the input voltage is OFF, what happens to the internal switching element and why does this behaviour happen only at no load or light load but not on typical load? \$\endgroup\$
    – user220456
    Dec 20, 2020 at 13:57
  • \$\begingroup\$ Well, what is dI/dt in your inductor at the point I described in prev comment? \$\endgroup\$
    – user16324
    Dec 20, 2020 at 14:00

1 Answer 1


Capacitors store voltage. If you charge a capacitor to 9V, remove the supply, and check the capacitor's voltage, it will have (about) 9V across it. The charge stored in the capacitor originates from an electric field. This is stable and wants to stay at the current charge, so degrades slowly and cannot change instantaneously. Once charged, a small cap can supply 9v for a short time, and a large cap for a longer time. This phenomena is why capacitors are said to block DC (they equalize to the DC voltage) but conduct AC (they act like a low-value resistor to alternating current.)

  • Capacitors' charge (voltage) cannot change instantaneously. If you short a charged cap, infinite current flows for a very short time.
  • Capacitors resist instantaneous change of their charge (voltage.)

Inductors are exactly the opposite, in all possible ways. Instead of utilizing an electric field, inductors utilize a magnetic field. This field starts off weak, and grows stronger with time, limited by the inductance value and physical properties such as resistance of the wire. This magnetic field does not want to naturally be there, at all, and is only there because of the current. This phenomena is why inductors are said to conduct DC (they pass DC through) but block AC (they resist alternating current.)

  • Inductors' magnetic field density (current) cannot change instantaneously. If you open an inductor with current flowing through it, an infinite voltage is created for a very short time. This is because an open circuit is a higher impedance - but the same current still wants to flow into a higher impedance now, so the voltage must increase to keep the current constant.
  • Inductors resist change of their current.

Now can you see, that when an inductor has current flowing through it, and the load is suddenly removed, a higher voltage must be made? Such is the nature of inductors. Will this be enough to damage something? Maybe but probably not, because there is usually a filter capacitor after the inductor. Parallel L-C circuits oscillate; so the output voltage will oscillate at the previous DC output voltage - an AC oscillation atop the DC output voltage. So if it were outputting 12VDC before, a sudden load removal might see several peaks of 9-15VDC and gradual relaxation back to 12VDC.

How when the inductor [has] current and the load is removed, the voltage increases?

The answer is the bold bullet-point above.

[What is the] direction of current and the voltages across inductor and output capacitor during this phenomenon?

The capacitor resists change in voltage, so it changes very little. The inductor however, behaves totally opposite. When the load is removed, the inductor current reverses and the voltage rises quickly. That happens because the magnetic field inverts and collapses. This reverse voltage is countered by the capacitor, which doesn't want the voltage to change, so the voltage drops slightly as it absorbs the bulk of that electrical surge. (The capacitor takes current from the inductor.) This now leaves the inductor "depleted" of energy and free to recharge again. The inductor charges from the capacitor (forward current) and the process repeats... For a longer answer, please research LC tank circuit or similar.

And why does this happen at no load or light load condition, but not at typical or maximum load?

At typical or maximum load, the load is there to prevent the inductor from doing anything unusual. It is only when the load is suddenly removed (or reduced significantly) that the inductor tries to maintain current by oscillating with the capacitor.

  • \$\begingroup\$ Thank you very much for your explanation. Just two simple questions I have to get more clarity. Could you please tell , how when the inductor current is increased and the load removed, voltage increases? And 2. Please also tell me the direction of current and the voltages across inductor and output capacitor during this phenomenon , please? \$\endgroup\$
    – user220456
    Dec 21, 2020 at 15:46
  • \$\begingroup\$ And why does this happen at no load or light load condition, but not at typical or maximum load? \$\endgroup\$
    – user220456
    Dec 21, 2020 at 15:46
  • \$\begingroup\$ Thank you very much for the edit. One last request before I accept the answer. I request you to give me some pictures of the circuit detailing the voltage polarity and the current direction during ON and OFF conditions of the internal switches. \$\endgroup\$
    – user220456
    Dec 22, 2020 at 12:53

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