Buck Converter Inductor Calculator

How to use the buck inductor calculator

1. Type the input and output voltages, the load current and the switching frequency (on the controller's data sheet: 100 kHz to 2 MHz is typical). Use the highest input you expect: that is where the ripple is largest.
2. Leave the inductor blank and set a ripple target (30 % by default): the inductance is designed and rounded up to the next standard value. Or type an inductor to check it.
3. Add the output capacitor for the voltage ripple and the efficiency for a truer duty. The drawing shows the inductor current with the on and off times; the Inductor values tab compares the standard values around yours.

How a buck works

The switch connects the inductor to the input for a fraction D of each period. While it is on, V_in − V_out sits across the inductor and its current ramps up; when the switch opens the inductor keeps the current flowing through the diode (or a second switch), now with V_out across it the other way, so the current ramps down. Over a cycle the inductor's average voltage must be zero, which forces V_out = D × V_in: the duty sets the output, the inductor carries the current, and the capacitor smooths what is left of the triangle. Nothing in the chain dissipates on purpose, which is why a buck runs at 85–95 % where a linear regulator at the same ratio would burn half the power.

The ripple formula

ΔI = (V_in − V_out) × D / (f × L)    L = (V_in − V_out) × D / (f × ΔI)    D = V_out / V_in

From V = L di/dt during the on time: the current rises by (V_in − V_out) × t_on / L, and t_on is D/f. The ripple does not depend on the load, only on the voltages, the frequency and the inductance. Designers pick ΔI as a fraction of the full load current, 20–40 %: smaller means a bigger, slower inductor and a lazier response to load steps; larger means high peaks, more core and capacitor loss and an early slide into discontinuous conduction. Doubling the frequency halves the inductor for the same ripple, which is why modern converters switch at a megahertz and use inductors of a few microhenries.

Continuous and discontinuous conduction

I_boundary = ΔI / 2    L_crit = (V_in − V_out) × D / (2 f I_out)

Because the ripple is fixed, lowering the load lowers the whole triangle until its valley touches zero, at a load of half the ripple. Below that the inductor current stops each cycle: discontinuous conduction, where the duty no longer equals V_out/V_in, the control loop changes and the output ripple grows. Most controllers handle it (some switch to pulse-skipping for efficiency), but if you need continuous conduction down to a light load, that sets a minimum inductance: the critical value above, evaluated at the lightest load.

Choosing the inductor

Three ratings matter. The saturation current must exceed the peak, I_out + ΔI/2, with margin for start-up and load steps: a saturating core loses its inductance and the current spikes. The RMS (heating) current, √(I_out² + ΔI²/12), is nearly the load current, and the DC resistance times its square is the copper loss. And the inductance should be the data-sheet value at the peak current, not at zero, since soft-saturating powder cores lose 20–30 % under bias. Shielded parts keep the switching field away from sensitive traces; ferrite is efficient at high frequency, powder iron tolerates overload gracefully.

The output capacitor

ΔV_out = ΔI / (8 f C) + ΔI × ESR

The triangle's AC part flows into the output capacitor. The charge it carries over half a cycle makes a voltage ripple of ΔI / (8 f C), and its current across the capacitor's ESR makes ΔI × ESR on top, often the larger term with electrolytics and negligible with ceramics. A few tens of microfarads of ceramic at a megahertz give millivolts of ripple; the same job at 100 kHz with electrolytics needs hundreds of microfarads and low-ESR parts.

Worked example: 12 V to 5 V at 2 A, 500 kHz

The duty is 5 / 12 = 42%, so the switch is on for 833 ns of each 2 µs period. For a ripple of 30 % of 2 A, 0.6 A, the inductor is (12 − 5) × 0.417 / (500 kHz × 0.6 A) = 9.722 µH; the next standard value is 10 µH, which gives a ripple of 583 mA (29%). The inductor current swings between 1.71 A and 2.29 A, with an RMS of 2.01 A: choose a part with a saturation current of 3 A or more and a low DC resistance. The converter stays in continuous conduction down to 292 mA of load; the critical inductance at 2 A would be only 1.46 µH. The core stores 26.3 µJ at the peak, and a 22 µF output capacitor sees a voltage ripple of 6.63 mV from its capacitance, plus the ripple current times its ESR.

Questions

Why 30 % ripple?

A compromise: less means a bigger, slower inductor; more means higher peaks and losses and an early slide into discontinuous mode. Most reference designs sit between 20 and 40 %.

Which input voltage do I design at?

The maximum: the ripple grows with V_in − V_out. Check the minimum input for the duty limit of the controller.

Does the inductor's rated current have to be the load current?

Its saturation rating must exceed the peak, load plus half the ripple, with margin; its heating rating must exceed the RMS, which is about the load current.

Is discontinuous mode bad?

Not necessarily: it is normal at light load and more efficient with controllers that skip pulses. It changes the control loop and raises the ripple, so size the inductor for continuous mode at the loads that matter.