Buck Converter Calculator (duty, inductor, ripple)

Understanding the buck converter calculator — concept and principle

A buck converter is a switching regulator that steps a DC voltage down, for example 12 V to 5 V, far more efficiently than a linear regulator. This calculator gives the first numbers you need when choosing parts: the duty cycle, the inductance for a chosen ripple current, and the peak current the inductor must carry.

Duty cycle 5 ÷ 12 = 41.67%: the coil smooths the 12 V pulses into 5 V with only 0.6 A of ripple.

How it is calculated

Duty cycle: D = V_out / V_in (the fraction of each period the switch is on).
Ripple current: ΔI = (ripple %) × I_out, peak to peak.
Inductor: L = (V_in − V_out) × D / (f_s × ΔI), with f_s in hertz (the input is in kHz, so it is multiplied by 1000) and L shown in µH.
Peak inductor current: I_peak = I_out + ΔI/2 = I_out × (1 + ripple%/200).

The principle

In steady state the inductor's average voltage over one period must be zero, otherwise its current would grow without limit. This is called volt-second balance. While the switch is on, the inductor sees V_in − V_out for time D·T; while it is off, it sees −V_out for (1 − D)·T. Setting (V_in − V_out)·D = V_out·(1 − D) gives D = V_out/V_in. During the on-time the current rises at a slope of (V_in − V_out)/L, so the rise over D/f_s seconds is the ripple ΔI; solving for L gives the formula above.

Worked example

Inputs: V_in = 12 V, V_out = 5 V, I_out = 2 A, f_s = 500 kHz, ripple 30%.
D = 5 / 12 = 41.67%.
ΔI = 0.30 × 2 = 0.6 A peak to peak.
L = (12 − 5) × 0.41667 / (500,000 × 0.6) = 2.9167 / 300,000 = 9.722 × 10⁻⁶ H = 9.722222222 µH.
I_peak = 2 × (1 + 30/200) = 2.3 A.

Each point is the calculator re-run with only “Input (V)” changed; the gold point is the default example. Change the inputs above and this chart follows your numbers.

Input (V)	Duty cycle
3	166.67%
6	83.33%
9	55.56%
12	41.67%
15	33.33%
18	27.78%
24	20.83%

How to read the result

Pick the nearest standard inductor at or slightly above the result (10 µH here), which keeps the ripple a little below 30%. Its saturation current rating should exceed the peak current with margin, and its RMS rating should exceed the load current. A ripple target of about 20–40% of the load current is a common starting point: lower ripple needs a bigger inductor and slows transient response, higher ripple raises output ripple voltage and core losses.

Common mistakes

Entering the switching frequency in Hz in the kHz field, which makes L 1000 times too small.
Choosing an inductor by inductance only and ignoring its saturation current.
Forgetting that the real duty cycle is somewhat higher than V_out/V_in because of switch and diode drops and losses.

Keep in mind: These are ideal formulas for continuous conduction mode. At light loads (I_out below ΔI/2) the converter enters discontinuous mode and the equations change. Real designs need the controller datasheet, thermal checks and testing; mains-connected or safety-critical supplies need a qualified engineer.

Frequently asked questions

What happens if I double the switching frequency?

The required inductance halves for the same ripple, allowing a smaller inductor, but switching losses in the transistor rise.

Why is a buck converter more efficient than an LDO?

The switch is either fully on or fully off, so it dissipates little power, whereas a linear regulator burns (V_in − V_out) × I as heat.

Sources: R. W. Erickson and D. Maksimović, Fundamentals of Power Electronics, Springer, Chapters 2 and 5 · Texas Instruments, Basic Calculation of a Buck Converter's Power Stage, Application Report SLVA477

Checked 8 Nov 2026. Standard textbook formulas in SI units, computed in your browser. For learning and estimates — real designs must follow your local codes and be checked by a qualified engineer.

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