Selecting an inductor for a buck regulator

From this StackExchange answer.

It might help to understand what happens if you select an inductor of the wrong value.

If you select an inductor with too low a value, the current through it will change too much in each switching period. The current might grow so much in a switching period that it exceeds the current capability of the circuitry driving the inductor. This high ripple current also isn’t nice to the capacitor on the output side. ESR losses in the capacitor will be high, or ripple current will exceed the capacitor’s rating and it will fail.

If you select an inductor with too large a value, you will be paying for a lot of inductor you don’t need. Inductors with a core have a saturation current. This is the current at which the core can not take any more magnetic flux, and the inductor stops being an inductor with a core, and starts being almost a wire. For a given core of a given size and material, you can make an inductor with higher inductance simply by putting more turns of wire around it. But, each of these turns contributes more magnetic flux, so by adding more turns, you are also decreasing the saturation current of the inductor, since your current will be multiplied by the number of turns of wire to arrive at the magnetic flux through the inductor. Thus, if you want a higher inductance at the same saturation current, you need a physically larger core.

User gsills also had a great answer:

Choosing an inductor value for a buck regulator comes directly from V = (L x di) / dt . Where V is the voltage across the inductor, and i is the current through it. First, you want to design for the case where the inductor is in continuous conduction mode (CCM). This means that energy in the inductor doesn’t run out during the switching cycle. So, there are two states, one where the switch is on, and another where the switch is off (and the rectifier is on). Voltage across the inductor during each state is essentially a constant (although it is a different value for each state). Anyway since the voltage is a constant, the inductor equation can be linearized (and rearranged to give L).

L = VΔt / ΔI this is the basis for the equation you saw in the app-note.

ΔI is something you define, not determine.

You will want to maintain CCM operation, so define ΔI as some small fraction of inductor current (I). A good choice is 10% of I. So, for your case ΔI would be 0.24A. This will also define the ripple current in the output capacitors, and less ripple current means less ripple voltage on the output.

Now you can choose an optimal value of L using Vin and Vo (and hence the duty cycle D = Vo/Vin). But you can also make a quick over estimate for the inductance where you don’t consider Vin using L ~ (10 x Vo) / (Io x Fsw )(for more on this look at How to choose a inductor for a buck regulator circuit? ). An over estimate can be worthwhile, especially if you are early in development or uncertain exactly how much the output current will be (output current tends to end up higher than expected usually).

that link leads to this answer, also from gsills:

Here is a quick and somewhat dirty way to calculate an inductor value for buck regulators operating in constant conduction mode (CCM). It will result in an inductance that will be close to what you would get with a more exact calculation, and will not get you into trouble.

What you need to know to calculate inductance:

Output Voltage, Vo
Output Current, Io
Switching Frequency, Fsw

L = ΔtVo / ΔI

Make a couple of assumptions:

ΔI=Io / 10
Δt=1 / Fsw


L = 10 x Vo x Io x Fsw

for Io = 1A and Fsw = 2.2 MHz

L = 22.7 μH

When choosing the inductor:

Find one that is rated for 1.4 to 2 times the output current. In this case 1.4A to 2A. Most standard inductors are specified for 40C heat rise with rated current, which is kind of hot. Conductive losses scale by the square of the current. Using a current rating of 1.4 Io will reduce that heat rise by half, and a current rating of 2 Io will reduce heat rise to 1/4.

Make sure the series resonant frequency (SRF) is at least a decade higher than the switching frequency.

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