Table of Contents
ToggleA boost converter is a DC-DC power supply that steps up an input voltage to a higher output voltage. It uses an inductor, a switch, and a diode.
This guide explains the switching cycle, the output voltage formula, component selection, a fully worked design example, and a live calculator.
A battery discharges from 4.2 V to 3.0 V as it drains. The circuit may need a stable 5 V.
A boost converter steps up the battery voltage continuously, even as it falls. This is one of its most common uses in portable electronics and industrial sensors.

How a Boost Converter Works: The 2-Phase Switching Cycle
A boost converter stores energy in an inductor during the ON phase and releases that energy at a higher voltage during the OFF phase.
The buck converter working principle explains the complementary step-down topology. Together, buck and boost cover the two fundamental DC-DC conversion directions.
Boost Converter Output Voltage Formula
Output voltage (V). Always higher than Vin in a boost converter.
Input voltage (V). The supply being stepped up.
Duty cycle (0 to 1). Fraction of each period the MOSFET is ON.
OFF-time fraction. As D approaches 1, Vout approaches infinity (in theory). In practice, losses limit the maximum achievable gain.
The duty cycle is rearranged to find D from a known Vin and Vout: D = 1 minus (Vin / Vout). For example, to boost 3.3 V to 12 V, the duty cycle is D = 1 minus (3.3/12) = 1 minus 0.275 = 0.725, or 72.5 percent. The MOSFET is on for 72.5 percent of each switching period.
In a real converter, MOSFET and diode losses reduce the actual output voltage slightly below the ideal formula. The controller compensates using its feedback loop. Learn how MOSFET switching characteristics affect this.
As duty cycle approaches 1.0, the output gain increases but efficiency collapses because the MOSFET carries all the input current for a very long fraction of each period. Conduction losses increase as D squared. Most boost converter controllers also have a maximum duty cycle limit of 80 to 90 percent. A duty cycle above 0.85 is a signal to reconsider the input voltage range or the topology.
4 Key Components in a Boost Converter
Live Boost Converter Calculator
Worked Example: 3.3 V to 12 V at 1 A, 300 kHz
Vin = 3.3 V, Vout = 12 V, Iout = 1 A, f = 300 kHz, ripple ratio = 0.30
Step 1: Duty cycle
D = 1 minus (Vin / Vout) = 1 minus (3.3 / 12) = 1 minus 0.275 = 0.725 (72.5%)
Step 2: Average input current (inductor average current)
Iin = Iout / (1 minus D) = 1 / (1 minus 0.725) = 1 / 0.275 = 3.64 A
(Input current is much higher than output current in a the converter)
Step 3: Ripple current
ΔIL = ripple ratio x Iin = 0.30 x 3.64 = 1.09 A peak-to-peak
Step 4: Minimum inductance
L = Vin x D / (f x ΔIL) = 3.3 x 0.725 / (300000 x 1.09)
L = 2.393 / 327000 = 7.32 µH
Select standard value: 10 µH (gives lower ripple than spec)
Step 5: Peak inductor current (saturation current rating)
Ipeak = Iin + ΔIL/2 = 3.64 + 0.545 = 4.19 A
Select inductor with saturation current rating above 4.19 A (e.g. 5 A rated)
Step 6: MOSFET voltage rating
MOSFET sees Vout when off: need rating above Vout x 1.3 = 12 x 1.3 = 15.6 V
Select a 20 V rated MOSFET with low RDS(on) at 12 V gate drive
Boost Converter Design Checklist
Calculate duty cycle and verify it is below 0.85
Use D = 1 minus (Vin / Vout). Use the minimum expected Vin to find D max. If D max exceeds 0.85, either raise Vin or reconsider the topology.
Calculate average input current it is much higher than output current
Use Iin = Iout / (1 minus D). This is the average inductor current. At D = 0.725 and Iout = 1 A, Iin = 3.64 A.
The inductor and MOSFET must be rated for this current, not Iout.
Select inductor: saturation current above Ipeak = Iin + ΔIL/2
The inductor saturation current rating must exceed the peak current at full load. An inductor that saturates loses its inductance and the peak current spikes. See the inductor guide for selection details.
Select MOSFET with voltage rating above Vout x 1.3
The MOSFET in a boost converter sees the full output voltage Vout when it switches off, not Vin. Switching transients add ringing on top of Vout.
Use a MOSFET with a VDS rating at least 30 percent above Vout to provide safe margin.
Select output capacitor for low ESR and sufficient capacitance
The output capacitor supplies the load for D x T every cycle. The output ripple voltage is ΔVout = Iout x D / (f x Co).
For 50 mV ripple at 1 A, 0.725 duty cycle, 300 kHz: Co = 1 x 0.725 / (300000 x 0.050) = 48.3 µF. Select 100 µF for margin.
A the converter has a right-half-plane (RHP) zero in its control-to-output transfer function. This means that increasing the duty cycle initially decreases the output current before increasing it, causing a phase shift that limits how fast the feedback loop can respond. the converter control loops are generally designed with a lower bandwidth than equivalent buck converter loops. Use your controller IC's reference design and compensation guidelines for the feedback loop component values.
Boost vs Buck vs Buck-Boost Converters
| Parameter | Boost Converter | Buck Converter | Buck-Boost Converter |
|---|---|---|---|
| Output vs Input | Vout always greater than Vin | Vout always less than Vin | Vout can be above or below Vin |
| Voltage formula | Vout = Vin / (1 minus D) | Vout = Vin x D | Vout = Vin x D / (1 minus D) (inverting) |
| Inductor position | Between Vin and switch (series with input) | Between switch and output (series with output) | Both input and output sides |
| Input current | Pulsating. Much higher than output current at high duty cycle. | Continuous (ripple only). Equal to output current on average. | Pulsating at both input and output |
| Control difficulty | Harder. RHP zero in control transfer function limits loop bandwidth. | Easier. No RHP zero. Wider control bandwidth achievable. | Hardest. RHP zero plus more complex topology. |
| Typical applications | Battery-to-rail boost, LED drivers, USB power delivery | DC bus regulation, MCU power, point-of-load regulation | Battery-operated systems with wide input voltage range |
Watch: Boost Converter Working Principle Explained
Boost Converter Questions Engineers Ask
Related Articles on This Site
- Buck Converter Working Principle: Duty Cycle Formula and Design Guide
- What Is a MOSFET? Working Principle Explained
- Inductor Working Principle
- How Decoupling Capacitors Work
- Active, Reactive and Apparent Power Explained
External References
- Basic Calculation of a the converter's Power Stage | Texas Instruments SLVA372C
- Boost Regulator Design Guide | Analog Devices
What We Learn Today
- A boost converter steps up DC voltage using an inductor and a switch. The output voltage formula is Vout = Vin / (1 minus D). To boost 3.3 V to 12 V, the duty cycle is 72.5 percent.
- The input current in a the converter is always higher than the output current by the factor 1/(1 minus D). This is critical for inductor and MOSFET selection, where peak current Ipeak = Iin + ΔIL/2 must be used, not Iout.
- Keep duty cycle below 0.85 in any the converter design. The right-half-plane zero in the control transfer function limits feedback loop bandwidth and requires careful compensation design.
