I’m studying buck converters and I have found contradicting sources claiming either of these statements:

  • The inductor stores and releases the bulk of the current, the capacitor is there just for smoothing its output in an auxiliary role.
  • The capacitor stores and releases the bulk of the current, the inductor is there in an auxiliary role to slow down the capacitor's charge time so that there is no huge current spike when it's charging up at the start of each cycle.

I'm confused, I did a simulation and I seem to be able to omit either the capacitor or the inductor and I'm still getting a decent output, this only deepens my confusion.

enter image description here enter image description here

Which component is supposed to provide the bulk of the current to the load and which component is supposed to perform an auxiliary role? And why is it not the other way around?

EDIT: This link provides aditional explaination about why an RC filter is not a great power source and elaborates on the reason why the inductor is the main current provider.

  • \$\begingroup\$ What is the horizontal time axis time scale ? No ripples are visible in the output. Please mention the scale of the voltage axis also. There should have been some ripple even if it was quite small. \$\endgroup\$
    – AJN
    May 14, 2021 at 5:57
  • \$\begingroup\$ there is, that's why I show the min and max values with arrows. If that's not sufficient I can provide a smaller scale where the ripple is visible. \$\endgroup\$ May 14, 2021 at 6:03
  • \$\begingroup\$ Not necessary, i think. I visited the link in the question. \$\endgroup\$
    – AJN
    May 14, 2021 at 6:03
  • \$\begingroup\$ It's because you are using massive energy storage components, I think. \$\endgroup\$
    – DKNguyen
    May 14, 2021 at 6:12
  • \$\begingroup\$ Probably, maybe realistic component constrains have something to do with the answer? \$\endgroup\$ May 14, 2021 at 6:13

4 Answers 4


tl; dr: the inductor is the primary energy transfer element. The capacitor lowers the power rail impedance for both phases of switching and thus reduces output ripple.

A buck regulator with an inductor has two phases:

  • Charging: current is being poured into the inductor, flux is increasing. Current flows from the power source, through the upper switch, through the inductor and into the load and filter cap.

  • Discharging: inductor is dumping current, flux is decreasing. Current flows from the inductor, into the load and back through ground and through low side switch or catch diode back to the inductor.

So it’s pretty safe to say that the inductor is being used as the energy transfer medium. Its stored energy is in the form of flux, and it’s being alternately refreshed from the source and dumped into the load. It’s doing the bulk of the work.

The filter cap has far less dynamic current variation than the inductor. It's bypassing the triangular ripple current from the inductor, which in a properly tuned buck should be about 30% of the max current. If anything, it’s the input cap before the high side switch that’s really getting the workout: it has a very large dI/dt when the high side FET turns on.

In your sim, try using more reasonable values like say 10uH and 100uF for the filter cap. Also, model the cap and inductor with series resistances to more closely reflect actual components. Finally, add that input cap, again with series resistance. Also, you can set the FET ‘beta’ in Falstad, use about 15 for the switch.

BONUS: here's a constant-on-time type buck that delivers 6A at over 90% efficiency (at least as far as Falstad is concerned.)

enter image description here

Simulate it here


Technically you could omit both and have a rectangle wave power output. It's a chopper and it works. In fact 99% of the PWM DC brushed motor drives could be seen as buck converter and they work (the fact that a DC motor is a huge inductor helps, of course). The same hold when PWM driving leds or lamp, if you drive them from the high side is the same topology.

Of course a regulator should give a 'good' output and some crucial metrics are load regulation (how much it varies with changes of load, more or less like output impedance) and voltage ripple.

They both contribute to the regulation and frankly I don't see one of them as primary and the other as auxiliary. The role gradually shift with the duty cycle and the load condition. There is some difference between DCM and CCM for the inductor, of course. In CCM you could really say that the inductor is the main power provider.

They both charge up when the switch closes and they both discharge and supply the load during the passive phase of the converter. However the capacitor is also responsible for transients, while the current from the inductor is usually considered 'constant' during analysis (it's the base of current mode regulation, which is really popular). If I had to choose I'd say the inductor is the main provider and the capacitor is the auxiliary on but I feel that it's not correct to say so.

From the designer point of view the inductor is the main component to consider because of the varying operating point of the converter. A bigger enough inductor could give you sufficiently low ripple (but no transient response!). However big capacitor are cheaper than big inductors so in practice you have a 'suitable' inductor (for current handling), ceramic capacitors and sometimes bulk capacitor to supplement the filtering (depending on the output quality you need).

Also, by the way, in practice omitting the inductor doesn't work since a discharged capacitor is seen as a short circuit and overstresses the switch.


All current comes through the inductor. Some of it can temporarily be stored into the capacitor.

Leaving out the capacitor can work if the load is constant. If the load suddenly changes its current the voltage jumps or drops radically. In theory the inductor current bulldozes its way generating just as much voltage as needed to find a way if you for example try to disconnect the load. If the load was a transistor which is driven to stop the current the inductor would kill it and the current continues decaying gradually as the inductor (=induction law) decides.

Leaving out the inductor causes the cap will be connected to the input voltage, when the main switch starts to conduct. In practice that that causes a current surge which easily kills the switch. In simulators there's often no smoke programmed to the component models, so the circuit can well look functional. But the current peak at the start of the conducting state of the switch is as big as the drive of the switch allows.


So the concrete answer ended up being it depends

SITUATION 1: If you have a small load connected to the buck converter, it will draw a very small current. If this is the case the capacitor will be the main current provider while the transistor is OFF.

This is because the buck converter will work in DCM or Discontinious conduction mode. This means that because the load is small

  1. It draws very little current from the buck converter
  2. Then the current through the inductor is small when the transistor is ON
  3. That small current cant completely charge up the magnetic field of the inductor when the transistor is ON.
  4. This means that when the transistor turns off, the inductor does not have much (or almost any) energy to provide, so it cant provide a current to the load.
  5. At this point the capacitor provides the current needed, since the amount of current needed is small and it needs to provide it for a small amount of time (beccause the frecuency is high), there is no problem and the output is continious.

At this mode, the inductor is just limiting the speed at which the capacitor recharges, while the capacitor is the main current provider.

The following simulation illustrates this situation enter image description here

SITUATION 2: If you have a big load connected to that very same buck converter, that load will draw a larger amount of current. In this case the inductor will be the main current provider while the transistor is OFF and the capacitor will just be there to smooth the output

This is because the buck converter will work in CCM or Continious Conduction Mode, this means that because the load is large

  1. It draws more current from the buck converter
  2. then the current through the inductor in big when the transistor is ON
  3. That bigger current charges up the magnetic field in the inductor when the transistor is ON
  4. This means that when the transistor turns off, the inductor has a lot of energy to provide, so it provides current to the load while also mantaining the charge of the capacitor
  5. The capacitor then just smooths the output.

At this mode, the inductor is the main power provider, while the capacitor just smooths it's output.

The following simulation illustrates this situation, using the same buck converter with a larger load.

enter image description here

SITUATION 3: The load is somewhere in the middle, the inductor charges up some, but decays quickly during the transistor OFF time and the capacitor takes over, this is still DCM, however here the role of main current provider is distributed in both elements up to some degree.


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