Boost Converter Calculator – Free Calculator

Understanding the boost converter calculator — concept and principle

A boost converter is a switching regulator that raises a DC voltage, for example from a 5 V USB rail or a single lithium cell to 12 V. When the switch closes, energy is stored in an inductor; when it opens, the inductor's voltage adds to the input and pushes current through a diode into the output capacitor. This calculator estimates the duty cycle, the input current the source must supply, and the inductance for a chosen ripple.

Duty 1 − 5 ÷ 12 = 58.33%. Raising the voltage costs current: 0.5 A out needs 1.2 A in.

How it is calculated

Duty cycle: D = 1 − V_in / V_out.
Input current: I_in = I_out × V_out / V_in (power in equals power out at ideal efficiency).
Ripple current: ΔI = (ripple %) × I_in. In a boost converter the inductor sits on the input side, so its average current is I_in.
Inductor: L = V_in × D / (f_s × ΔI), with f_s in hertz (the input is in kHz) and L shown in µH.

The principle

Volt-second balance again sets the duty cycle: the inductor's average voltage over a period must be zero. With the switch on (time D·T), the inductor sees +V_in; with it off (time (1 − D)·T), it sees V_in − V_out, which is negative. Setting V_in·D = (V_out − V_in)·(1 − D) gives V_out = V_in / (1 − D). During the on-time the current rises at V_in/L for D/f_s seconds, which is the ripple ΔI; solving for L gives the formula.

Worked example

Inputs: V_in = 5 V, V_out = 12 V, I_out = 0.5 A, f_s = 500 kHz, ripple 30%.
D = 1 − 5/12 = 58.33%.
I_in = 0.5 × 12 / 5 = 1.2 A.
ΔI = 0.30 × 1.2 = 0.36 A peak to peak.
L = 5 × 0.58333 / (500,000 × 0.36) = 16.2037037 µH.

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

Output (V)	Duty cycle
3	-66.67%
6	16.67%
9	44.44%
12	58.33%
15	66.67%
18	72.22%
24	79.17%

How to read the result

Choose the next standard value up, such as 22 µH, which lowers the ripple slightly. The inductor and switch must handle the peak current I_in + ΔI/2 = 1.38 A here, and the source must supply about 1.2 A, more once losses are included. As V_out/V_in grows, D approaches 1 and the input current and stresses rise sharply; ideal-formula ratios beyond about 5–6 times are usually impractical in a simple boost.

Common mistakes

Using output current instead of input current for ripple and inductor ratings.
Assuming the source only needs to supply I_out; it must supply I_out × V_out/V_in plus losses.
Forgetting that a basic boost cannot turn its output off: with the switch off, input still reaches the output through the inductor and diode.

Keep in mind: These are ideal, continuous-conduction formulas with 100% efficiency. Real converters have losses that raise the duty cycle and input current. Use the controller datasheet and testing, and get qualified review for any safety-related supply.

Frequently asked questions

Can a boost converter step down voltage?

No. Its output is always at least the input minus a diode drop. Use a buck-boost or SEPIC when the input can be above or below the output.

How does efficiency change the input current?

Divide by efficiency: at 90% efficiency the input current here becomes about 1.2 / 0.9 = 1.33 A.

Why does the inductance depend on V_in and not V_out alone?

During the on-time only V_in is across the inductor, so V_in sets how fast the current rises.

Sources: R. W. Erickson and D. Maksimović, Fundamentals of Power Electronics, Springer, Chapter 2 · Texas Instruments, Basic Calculation of a Boost Converter's Power Stage, Application Report SLVA372

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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