I am having an issue understanding how the a buck converter circuit actually steps down the input voltage. Specifically, my issue is regarding the capacitor charging and discharging. Let's consider the basic buck converter diagram with an ideal switch:

enter image description here

When the switch is closed an increasing current flows through the inductor, going to the capacitor and the capacitor is charging, right? Now, when the switch is opened, a negative spike appears at the node prior to the inductor such that the diode is forward biased and the polarity of the inductor voltage is reversed. Now, I really haven't figured out what happens with the voltage of the capacitor. Is there a negative current now going to the capacitor which means that the capacitor voltage is decreasing?


4 Answers 4


It may be more useful to focus on the inductor:

First assume it's fairly large and the switching frequency is fast.

Now the current through the inductor will remain fairly close to constant and always in the same direction.

With the switch ON, the current will gradually increase, from: $$V =L \frac{dI}{dt}$$ where $$V = V_s-V_{out}$$ Because \$L\$ is large, \$\frac{dI}{dt}\$ is fairly small.

With the switch OFF, \$V\$ becomes \$0 - V_{out}\$ (negative) or more strictly, \$-0.6-V_{out}\$, so \$\frac{dI}{dt}\$ is negative, and the current gradually reduces.

And that's pretty much it. (for a buck operating in CCM, Continuous Current Mode)

Oh - the capacitor - it just helps Vout remain close to constant, despite the variation in incoming current.

(Switch OFF for too long and the curent eventually falls to \$0\$ : the maths for Discontinous Current Mode, DCM, are different.)

  • \$\begingroup\$ Only thing I didn’t quite get is the fact that when some current flows to a capacitor, the capacitor charges, right ? How is it that it’s voltage remains quite constant ? \$\endgroup\$ Nov 1, 2020 at 1:19
  • \$\begingroup\$ Not perfectly constant > the voltage changes slowly as the cap charges, or discharges into a load. \$\endgroup\$
    – user16324
    Nov 1, 2020 at 11:56
  • \$\begingroup\$ @TeoProtoulis That's what a capacitor does. It tries to keep voltage constant by absorbing and releasing current. dV/dt=I/C and C is large. \$\endgroup\$
    – user253751
    Nov 1, 2020 at 12:02

Imagine the circuit without the capacitor. When the switch is closed, the current will flow from the power supply, through the inductor, through the resistor, back to the power supply.

During this phase, the current through the inductor will gradually increase over time. The inductor is storing energy in its magnetic field.

When the switch opens, the inductor becomes a current source, using the energy it had stored. Current flows from the inductor through the resistor, trough the diode, back to the inductor.

During this phase, the current through the inductor will gradually decrease over time.

This means the current through the load is a triangle-shaped waveform, rising when the switch is closed and falling when the switch is open.

Following ohm's law, the voltage across the resistor will thus also be a triangle waveform.

The capacitor will shunt the AC portion of the current, leaving only the DC current going trough the load, thereby stabilising the voltage on the load.

To be clear, the capacitor (and thus resistor) voltage is not completely stable. It does go up when the capacitor is charging up with the excess current in the switch closed phase and it does go down again when the capacitor is discharging, supplying the deficit current in the switch open phase. With the correct capacitor, this ripple can be made very small though.


Inductors, magnetics, Lenz's law, L di/dt ... all great. But let's answer the original question.

when the switch is opened, a negative spike appears

We must understand this confusing phenomenon before we go into circuit analysis. Try to figure out how a "Joule Thief" (blocking oscillator) works without knowing this.

During the first phase of operation, the circuit effectively looks like this: charging phase

The voltage source is forcing current through the inductor. Current flows from + to - through the inductor.

Once the switch opens, the circuit now looks like:

discharging phase

The inductor has indeed switched polarity! Due to the collapsing magnetic field, it has become a voltage source itself! Current still flows from + to - , but since the inductor is now driving the circuit it flows around the loop to do so.

Since current travels the same direction in both phases of operation, it never tries to discharge the capacitor. Only the load (R) will do that.

  • \$\begingroup\$ Thank you for the nice explanation. I have figured out that generally speaking the capacitor attracts the AC components of the output voltage and only the DC component appears at the load (R). So, the capacitor voltage “oscillates” around some magnitude because the capacitor attracts the AC voltage components (that’s what it has to do since the whole concept is about it being low pass filter and attenuating the AC components of the switching frequency and its harmonics). \$\endgroup\$ Nov 2, 2020 at 0:44

The easiest way is to view the buck converter as a low-impedance square-wave generator delivering a signal toggling between the input voltage \$V_{in}\$ and 0 V and averaged by the \$LC\$ filter:

enter image description here

An integrated circuit usually instructs the power switch to close and open at a pace set by the clock. The time during which the switch is closed is the on-time and is labeled \$t_{on}\$. During \$t_{on}\$ the power switch is closed and applies \$V_{in}\$ across the diode. The inductor is magnetized and a current circulates in the load. When the switch opens, the inductor demagnetizes and the diode starts conducting to keep the current circulating in the same direction. The voltage across the diode ideally drops to 0 V and its lasts until the next clock cycle. In this mode, we consider the current in the inductor as continuous: it never returns to 0 A within a switching cycle. This is called the continuous conduction mode or CCM. Considering this mode, the average voltage delivered to the load by a perfect buck converter (no losses of any kind) is thus calculated as:

enter image description here

This is obviously a very simplified operation and literature abounds with more detailed description (you can have a look at a seminar I taught at an APEC conference here or check the book I wrote on switching converters). So you see that the duty ratio \$D\$ being constrained for a buck converter between 0 and 100%, the formula \$V_{out}=DV_{in}\$ can only lead to an output voltage lower than the input. For instance, with a 10-V source and a buck operated in continuous conduction mode at a 50% duty ratio, the output voltage is \$V_{out}=DV_{in}=10\times 5 = 5\;\mathrm{V}\$


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