I know that many questions on operation of dc-dc buck converters have been asked in the forum. I'm really confused with the exact operation of the circuit. I've been reading about the converters and I am unable to find clear explanation.

I understand that buck converters provide regulated output whose value is controlled by the on and off periods of the switch. From what I've read, the inductor and capacitor form a low pass filter and reduce the output voltage ripple. (This supposedly happens because inductor opposes changing current and capacitor opposes changing voltage)

But a Low Pass filter is the one that attenuates frequencies higher than a desired frequency and passes the lower frequencies. So How does this definition fit in this case?

The analysis of the circuit begins with the assumption that output voltage remains constant and hence during the on-time of switch, voltage across inductor is constant resulting in ramp current through inductor. But how is the output voltage maintained at constant level?

What exactly is the need of inductor? As in case of rectifier only a filter capacitor, provides almost constant output voltage.

I've some other questions regarding the analysis of current and voltage waveforms that would not fit into this question. I'd really appreciate some help.

** https://electronics.stackexchange.com/users/20218/andy-aka

I've been reading your answers on questions related to buck converters.


"you are using an inductor and capacitor to form a low pass circuit and providing the inductor's internal resistance isn't too big then there won't be too much of a dc volt drop across L1 and you'll still get 3.3V at the output."

Could you please explain this clearly? What is the filtering action here?

  • 1
    \$\begingroup\$ The inductor is essential to efficint voltage translation. Charging a capacitor directly loses energy in the charging resistance. Doing it with a very low R wire gives a very high current and still loses energy. An ideal L (Inductor) can have a step voltage applied and current will ramp up. Energy is stored in the inductor. When the Vin switch is turned off the inductor input end is grounded and current flows out of the inductor - energy is transferred to the load. \$\endgroup\$
    – Russell McMahon
    Oct 16, 2015 at 14:21
  • \$\begingroup\$ The inductor in a buck converter has multiple functions. First, Storing energy with 1/2*L*I^2. Second, Acting as a part of the LC filter at the end. Third, acts as a current limiter. If you need further theory I would suggest reading Switching power Supplied A to Z (Sanjaya Maniktala) great book to understand this and other topologies. \$\endgroup\$
    – Weaverworm
    Oct 16, 2015 at 14:23

3 Answers 3


But a Low Pass filter is the one that attenuates frequencies higher than a desired frequency and passes the lower frequencies. So How does this definition fit in this case?

The important thing you haven't grasped is that the on-off square wave has an average level that is totally dependent on the duty cycle. This mathematically average level gets converted to a real dc level by the L and C low pass filter action.

Here's a square wave that rises to 3V from 0V (gold trace): -

enter image description here

The blue trace has a little bit of low pass filtering applied - can you see that if a lot more low pass filtering were applied, the blue trace would be almost a constant DC level of 1.5 volts?

  • \$\begingroup\$ I've been reading your answers. Please help me out here. Yes, the average voltage is dependent on duty cycle. But we need an output that is pure DC. How does low pass filter achieve this? As I've mentioned in question, what filtering action is happening here? The input is a desired single frequency squarewave. \$\endgroup\$
    – JGalt
    Oct 16, 2015 at 14:05
  • \$\begingroup\$ Thank you! The waveform you've shown could be achieved with an RC filter as well right? Why inductor? \$\endgroup\$
    – JGalt
    Oct 16, 2015 at 14:07
  • \$\begingroup\$ It could but then you wouldn't have an efficient switching converter and the output would droop under load. \$\endgroup\$
    – Andy aka
    Oct 16, 2015 at 14:10
  • \$\begingroup\$ Could you please explain how the ramp current in inductor is prevented from flowing through load? Why does the ramp current flow only through capacitor? \$\endgroup\$
    – JGalt
    Oct 16, 2015 at 14:28
  • \$\begingroup\$ As a dynamic impedance, the capacitor is hugely numerically smaller than the load resistance and therefore nearly all the ramping current flows into the capacitor leaving just a small amount of ripple voltage. That ripple voltage remaining does create a small ripple current in the load of course. \$\endgroup\$
    – Andy aka
    Oct 16, 2015 at 14:37

For a signal to change quickly, it needs to have higher frequency components. The low pass filter removes high frequency components, so the voltage will only change slowly.

The inductor improves the filter; to get the same effect with just a capacitor, it would have to be a lot bigger (because the charging needs to be slow enough that we can turn off when we reach the target voltage), and we'd get a massive inrush current. The capacitor after the rectifier is always charged to peak voltage, so there is no need to limit the current, except to avoid problems at startup (so an inductor is a good idea here as well).

The target voltage is reached by increasing the voltage until it is too large, then letting it drop until it is too small, and then smoothing out the ripple, so the duty cycle will increase as more current is drawn.


Operation of the buck converter is based on the fact that the average voltage across an inductor is 0:

\$(U_{o}-U_{i})\delta T + U_{o}(1-\delta)T = 0\$,

which simplifies to \$ U_{o} = \delta U_{i}\$

where \$\delta\$ is the duty cycle, \$T\$ the switching period time, \$U_{i}\$ and \$U_{o}\$ are input and average output voltages. The capacitor's function is to smooth \$U_{o}\$.

Similar statements can be made for boost and flyback converters.


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