I can understand how a buck converter can step down the voltage but only with the formulas. I am struggling to understand intuitively how the inductor stores energy and releases it in a way so that the voltage is stepped down. Same thing for boost, buck-boost and other converters.


6 Answers 6


Intuitively, one way of thinking about it is that you can't instantaneously change the current in an inductor similarly to the way you can't change the voltage on a capacitor. If you were to short out an ideal capacitor, the instantaneous current would be infinite, and in the real world a capacitor is discharged through some resistance, but if the resistance is low, the current is very high.

Most people have no problem with this concept but for some reason have a harder time with inductors. Once you have a current flowing in an ideal inductor, if you were to open the circuit the voltage would be infinite. In the real world, the voltage rises until a current path is found, but the voltage can become very high if the resistance is high.

Keeping this in mind, in a boost converter, a switch initiates a flow of current in inductor. When the switch is opened, the current has to go somewhere, and you have provided only one path. The current must continue to flow in the same direction, so it flows from the supply through your diode and out into your load. The inductor voltage must rise above your load voltage or the diode won't conduct. enter image description here BOOST CONVERTER


In a DCM buck converter, the inductor current starts out at zero. When the switch closes, the current in the inductor starts to increase. Then the switch opens, and the only inductor current path is from ground, through the diode and into the load. This current flows until the inductor's stored energy is spent, and the inductor current is zero. Then the entire process repeats.

In continuous mode, the inductor energy is not allowed to be fully expended, and the switch closes before the inductor current drops all the way to zero. So there is always some current in the inductor. Otherwise, it functions in the same way, drawing current through the diode when the switch is open. enter image description here BUCK CONVERTER https://en.wikipedia.org/wiki/Buck_converter#/media/File:Buck_conventions.svg

  • \$\begingroup\$ I believe you mean "inductor" than "conductor" in the very first phrase. \$\endgroup\$ Apr 21 at 13:49
  • \$\begingroup\$ You are correct! \$\endgroup\$ Apr 21 at 15:36
  • 1
    \$\begingroup\$ I think the problem is that, although electric and magnetic fields in fact coexist in a circuit, most people tend to think "voltage" as a cause and "current" as an effect because most practical power and signal sources behave like voltage sources. This mental image makes back-EMF difficult to understand. Understanding the interchangeability of voltage and current is crucial for beginners. \$\endgroup\$ Apr 22 at 8:48
  • \$\begingroup\$ I think you are also correct. \$\endgroup\$ Apr 22 at 17:35

I am struggling to understand intuitively how the inductor stores energy and releases it in a way so that the voltage is stepped down

A buck converter needs to be controlled by a "system" that measures the output voltage and adjusts the buck converter's duty cycle to maintain that output voltage at the correct level. The basic buck circuit cannot do this on its own; there needs to be a control-system and maybe that's what is confusing you?

But, it's really all about storing and transferring energy to the load

The amount of energy that the inductor stores during the period of one switching cycle is quite small. Many, many switching cycles are needed just to get the output capacitor charged to the desired output voltage and, when this point is reached, all the buck has to continue doing is storing and transferring just enough energy to the load to maintain the output voltage at the correct level. Maybe it was this bit that you didn't understand?

So, if the load consumes 1 watt of power at the required output voltage level and, the converter runs at 100 kHz, the inductor needs to store and transfer 10 μJ each cycle.

So, in short, the buck's inductor stores energy and transfers it to the output every switching cycle. A control system is needed to ensure that once the correct output voltage is reached, the duty cycle is maintained at a value that sustains the output voltage.

Same thing for boost, buck-boost and other converters.

The same principle applies to all of them. Here's a mechanical equivalent: -

enter image description here

  • Instead of voltages we have rotational speed
  • Instead of MOSFETs and diodes you can use on/off clutches
  • When clutch A is connected, clutch B is disconnected and vice versa

As you'd expect it cannot "switch" very fast because it's mechanical but, pretty much exactly the same formula apply and, it transfers energy from a high-speed to a low-speed at theoretically 100% efficiency. It is a mechanical buck converter.

  • \$\begingroup\$ Of course, mechanically, it's much easier to do this with gears. But there's no gear-equivalent when it comes to electronics, at least not at DC. \$\endgroup\$
    – Hearth
    Apr 22 at 2:18
  • \$\begingroup\$ The statement about power confuses power and energy. A 1 W converter provides 1 W on average, even if at 100 KHz, each cycle might only push 10 uJ into the output cap. \$\endgroup\$
    – AI0867
    Apr 22 at 7:00
  • \$\begingroup\$ @AI0867 thanks for pointing that error out. \$\endgroup\$
    – Andy aka
    Apr 22 at 8:20

The buck you can imagine as pulsed Vin voltage filtered by LC:

enter image description here

The switch creates a Vin pulses with some D and so outputs some V_average to LC.
In case of continuous conduction mode the V_average is Vout (diode drop is neglected).
The LC doesn’t change anything.

In discontinues mode it’s more complex.


With both bucks and boosts, you're taking energy at one potential, converting it, and storing it on an output capacitor at another potential. You actually convert this energy twice: first, you take the energy stored as charge at a potential (on the input) and store it in the magnetic field of the inductor. If voltage on a capacitor is potential energy, then the inductor is like a flywheel, and it stores the energy for a short duration as kinetic energy.
The second conversion is when you take this kinetic energy in the inductor and transfer it into the output capacitor as potential again. Depending on how you do this, the potential can be lower or higher than the input, and you need a control loop to regulate it. This is generally easier on a buck and harder on a boost; there's lots of architectures to do both.


I am struggling to understand intuitively how the inductor stores energy and releases it in a way so that the voltage is stepped down.

Firstly, don't analyze with energy. Use voltage and current.
Secondly, the inductor can't do it alone. It needs a capacitor to be complete.

It is a four step process:

  1. (Interval 1), A voltage is applied across the inductor allowing the current to increase in the circuit for a time.
  2. The circuit is reformed to allow the charge to flow into the capacitor. (Inductors integrate voltage to produce current).
  3. (Interval 2), The capacitor integrates the current allowing the capacitor's voltage to increase for a time.
  4. The circuit is reformed to apply the voltage across the inductor, disconnecting the capacitor. The steps repeat.

The two inductor circuits, in intervals 1 and 2, must be closed. If the capacitor circuit is open in 1, then its voltage will remain constant for that interval. The capacitor voltage will increase in interval 2. Depending on how the voltage is applied to the inductor, the capacitor voltage can increase without bound.

So the voltage isn't really stepped up or down. It is converted into a current, then the current is converted into the output voltage by the capacitor.

If a resistor (the load) is used instead of a capacitor, the output voltage will be zero during the first interval, then set according to Ohm's law. So the capacitor is needed to maintain constant voltage across the load.

So what determines the output voltage? The load voltage is determined by the average current and the load resistance (Ohm's law).

So the inductor is considered a pulsating current source. The average current is going to the load to create the voltage while the current pulses maintain the capacitor's voltage.


simulate this circuit – Schematic created using CircuitLab

During interval 1, if the buck load voltage equals Vin, then the inductor voltage is zero. For the boost converter, the entire input voltage is applied to the inductor, allowing large currents depending on the duration. Diverting this current to the load can result in very high voltages.

During interval 2, for the buck converter, the entire load voltage is applied in reverse to the inductor discharging the inductor quickly. For the boost, Vout-Vin is reverse applied to the inductor. If Vout is only a little higher, then the inductor will discharge more slowly.

There is a lot more to be said, but I will stop here.

It is the conversion to current that allows the step-up/step/down to work.

  • 1
    \$\begingroup\$ A continuous-operation buck converter wouldn't need the capacitor if driving a constant-resistance load. The current will increase while the source is connected, and decrease while the source is disconnected, but the change in current within each cycle would be slight, and thus the change in load voltage would also be slight. \$\endgroup\$
    – supercat
    Apr 22 at 14:58

If you just connect a capacitor to a voltage source, you'll get a surge of current that charges the capacitor very quickly, and heats up the wire so much that a large part of the energy you put in is wasted as heat.

A Buck converter puts an inductor between the voltage source and the capacitor to control the current and prevent this waste.

The current through the inductor can be precisely controlled by the timing of the switch that connects and disconnects it from the voltage source, and by controlling this current, the Buck convert controls the voltage on the output capacitor.


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