Power Electronics: Buck, Boost and Thermal Design
How switching converters work, choosing the inductor and capacitors, feedback dividers, efficiency and losses, heat and layout, with worked designs.
Why switch?
A linear regulator wastes the difference between the input and output voltage as heat: dropping 12 V to 5 V at 2 A burns (12 − 5) × 2 = 14 W. A switching converter turns a transistor on and off hundreds of thousands of times a second and uses an inductor and a capacitor to store and release energy, so almost all of the power is delivered: the same job loses less than 1 W. The price is complexity: ripple, noise and layout sensitivity. This page explains how the common converters work and gives worked designs. The basics of regulators and batteries are in Batteries and Power Supplies.
| Linear (LDO) | Switching | |
|---|---|---|
| Efficiency | Vout ÷ Vin: poor when the drop is large | 85–97 percent, nearly independent of the drop |
| Output noise | Very low | Ripple and spikes at the switching frequency |
| Parts and cost | One chip and two capacitors | Controller, inductor, diode or MOSFET, capacitors |
| Voltage range | Output below input only | Step down, step up or both |
| Layout | Forgiving | Critical: loop area, ground, feedback routing |
| Best for | Low power, small drop, quiet analog rails | Higher power, battery life, large voltage change |
The buck converter
The buck converter steps voltage down. When the switch is on, the input voltage drives the inductor and the current rises, storing energy in its magnetic field. When the switch turns off, the inductor keeps the current flowing through the diode (or a second, synchronous MOSFET) into the output, and the current falls. The capacitor smooths the output. The average output voltage is the input voltage times the duty cycle D (the fraction of the time the switch is on): Vout = D × Vin.
| Quantity | Formula (continuous conduction) | Meaning |
|---|---|---|
| Duty cycle | D = Vout ÷ Vin | Fraction of the period that the switch is on |
| Inductor ripple current | ΔI = (Vin − Vout) × D ÷ (f × L) | The up-and-down swing of the inductor current |
| Peak inductor current | Ipk = Iout + ΔI ÷ 2 | What the inductor and switch must withstand |
| Output ripple voltage | ΔV ≈ ΔI ÷ (8 × f × C) | Ideal capacitor: ESR adds ΔI × ESR |
| Input capacitor ripple current | IRMS = Iout × √(D × (1 − D)) | Largest at D = 50 percent: 0.5 × Iout |
A converter in continuous conduction mode (CCM) never lets the inductor current reach zero; below a certain load it enters discontinuous mode and many chips switch to a pulse-skipping or burst mode (PFM) to keep the light-load efficiency high.
Designing a buck: inductor and capacitors
Worked example: 12 V to 5 V at 2 A, 500 kHz
- Duty cycle: D = 5 ÷ 12 = 0.417.
- Inductor: choose a ripple current of about 30 percent of the load current, ΔI = 0.3 × 2 = 0.6 A. Then L = (12 − 5) × 0.417 ÷ (500 kHz × 0.6 A) = 9.7 µH; use the standard 10 µH.
- With 10 µH the actual ripple is (7 × 0.417) ÷ (500 kHz × 10 µH) = 0.58 A, so the peak current is 2 + 0.29 = 2.29 A. Choose an inductor with a saturation current at least 20–30 percent above that (3 A or more), and a heating current above 2.0 A. The stored energy at the peak is ½ × L × I² = 26.3 µJ.
- Output capacitor: for a ripple of 20 mV, C = ΔI ÷ (8 × f × ΔV) = 0.58 ÷ (8 × 500 kHz × 0.02 V) = 7.3 µF. A 22 µF ceramic is typical, but ceramic capacitors lose a large part of their capacitance under DC bias (about half at 5 V in a small package), so check the datasheet curve and use two in parallel. A bulk electrolytic or polymer part helps the transient response.
- Input capacitor: it must carry Iout × √(D × (1 − D)) = 2 × 0.49 = 0.99 A rms of pulsed current: use low-ESR ceramics (10–22 µF) close to the switch, with rating of at least 1.5 times the maximum input voltage.
| Switching frequency | Smaller inductor and capacitors | Higher switching losses and EMI | Typical use |
|---|---|---|---|
| 100–300 kHz | Larger parts | Low losses; easy layout | High-power converters; wide-input designs |
| 400 kHz–1 MHz | Compact | Moderate | Most point-of-load and battery regulators |
| 1–3 MHz and more | Very small | Higher losses; careful layout needed; avoids the AM radio band | Space-constrained and automotive designs |
Losses and efficiency
Efficiency η = Pout ÷ (Pout + losses). The main losses are the conduction loss in the switches (I² × RDS(on)), the voltage drop of the diode, the switching loss while the transistor changes state, the inductor’s winding and core losses, and the controller’s own supply current.
Worked example: the loss budget of the 12 V to 5 V converter
| Loss | Calculation | Power |
|---|---|---|
| High-side MOSFET, conduction | I² × RDS(on) × D = 2² × 0.03 × 0.417 | 0.05 W |
| Switching (both edges) | ½ × Vin × I × tsw × f = 0.5 × 12 × 2 × 20 ns × 500 kHz | 0.12 W |
| Inductor winding | I² × DCR = 2² × 0.015 | 0.06 W |
| Schottky diode | Vf × I × (1 − D) = 0.45 × 2 × 0.583 | 0.53 W |
| Total with a diode | Output power 10 W | 0.76 W: efficiency 93.0 percent |
| Synchronous low-side MOSFET | I² × RDS(on) × (1 − D) = 4 × 0.02 × 0.583 | 0.05 W |
| Total, synchronous | replaces the diode loss | 0.28 W: efficiency 97.3 percent |
Replacing the diode by a second MOSFET (synchronous rectification) removes the largest loss: the diode’s 0.53 W becomes 0.05 W. This matters most at a low output voltage, where the diode drop is a large part of what is delivered.
- Light-load efficiency is set by the controller’s quiescent current and the switching loss: use a chip with PFM or burst mode for battery products.
- Efficiency curves in the datasheet show the real figures: at light load they can fall to 60–70 percent.
- Heat is what limits the output current, not the electrical rating of the parts: see thermal design below.
Feedback divider and the controller
A controller chip compares a fraction of the output with an internal reference Vref (often 0.6–1.25 V) and adjusts the duty cycle to make them equal. A resistor divider R1 (top) and R2 (bottom) sets the output: Vout = Vref × (1 + R1 ÷ R2).
Worked example: the feedback divider
- With Vref = 0.8 V and a wanted output of 5 V: R1 ÷ R2 = 5 ÷ 0.8 − 1 = 5.25. Choose R2 = 10 kΩ, so R1 = 52.5 kΩ; the nearest 1 percent value is 52.3 kΩ.
- The output becomes 0.8 × (1 + 52.3 ÷ 10) = 4.984 V (-0.32 percent). Resistor tolerance (1 percent) and the reference tolerance add to this.
- The divider current is 80.26 µA, small against the load but significant for a battery in sleep: use larger resistors where quiescent current matters, and keep the feedback node short and away from the switch node.
- The Resistor Network and Standard Value Calculator finds standard-value pairs for a target.
- Voltage mode, current mode and constant on-time controllers differ in speed and stability; current-mode chips are the easiest to use.
- Compensation keeps the loop stable: many chips have it built in, others need a resistor and capacitor chosen from the datasheet.
- Integrated modules and simple switchers (LM2596, MP1584, TPS54xxx, module regulators with the inductor included) reduce the design to a few parts.
- Enable, power-good and soft-start pins sequence the supply rails.
Boost, buck-boost and other topologies
| Topology | Output | How it works | Typical use |
|---|---|---|---|
| Buck | Below the input | Switch in series, inductor, freewheel diode | Reducing 12–48 V to 5 V, 3.3 V, 1.8 V |
| Boost | Above the input | The inductor charges from the input when the switch is on; when it opens, the inductor voltage adds to the input | One lithium cell to 5 V; LED strings |
| Buck-boost | Inverted (or above or below) | One inductor, switch and diode; output polarity reversed in the basic form | Negative rails; battery that falls above and below the output |
| SEPIC, Ćuk, four-switch buck-boost | Above or below, not inverted | Two inductors or capacitors, or four switches | A battery that crosses the output voltage (a lithium cell to 3.3 V) |
| Flyback | Any, isolated | A coupled inductor (transformer) stores energy and releases it to the secondary | Mains adaptors, isolated supplies up to 100 W |
| Forward, half- and full-bridge, LLC | Isolated, higher power | Transformer-coupled topologies | Server and industrial supplies |
| Charge pump | Multiple of the input, low current | Capacitors switched in stages; no inductor | Small negative rails, 3.3 V to 5 V at low current |
Worked example: boost from a lithium cell to 5 V at 1 A
- The input is a cell at 3.7 V (falling to 3.0 V when nearly flat). For a 90 percent efficient boost, the input current is Pout ÷ (η × Vin) = 5 × 1 ÷ (0.9 × 3.7) = 1.50 A: the cell must supply 50 percent more current than the load draws, and more as it discharges.
- The ideal duty cycle is D = 1 − Vin ÷ Vout = 1 − 3.7 ÷ 5 = 0.26.
- At 1 MHz with a ripple current of 30 percent of the input current (0.45 A): L = Vin × D ÷ (f × ΔI) = 3.7 × 0.26 ÷ (1 MHz × 0.45) = 2.1 µH; use 2.2 µH. The switch and inductor peak current is about 1.72 A: the chip’s switch-current limit must exceed it.
- The output capacitor supplies the load while the switch is on (a fraction D of the time): for 50 mV ripple C = Iout × D ÷ (f × ΔV) = 5.2 µF, so 22 µF ceramic or more.
- A boost cannot be turned off by its switch alone: the input still connects to the output through the inductor and diode, so a short on the output is a short on the battery. Use a chip with true load disconnect where that matters.
Thermal design
Heat in a part raises its temperature above the surroundings by the dissipated power times the thermal resistance: Tj = Ta + P × θJA, where θJA (junction to ambient, °C/W) depends on the package and the copper area under it. Keep the junction below the maximum (usually 125 °C) with margin, and remember that ambient inside an enclosure is higher than in the room.
Worked example: linear against switching
- Linear: an LDO dropping 12 V to 5 V at 2 A dissipates 14 W. With θJA = 50 °C/W the temperature rise would be 14 × 50 = 700 °C: impossible without a large heat sink or a lower input voltage.
- Switching: the buck above loses 0.28 W, shared between the chip, the inductor and the diode or MOSFETs. If the controller’s share is half of it (0.14 W) in a package with the same θJA, the rise is 6.9 °C: easy.
- At 60 °C ambient, a junction limit of 125 °C leaves 65 °C of rise: the permitted dissipation is 65 ÷ 50 = 1.3 W for a package with that thermal resistance.
- Spread the heat: use a large copper pour under the thermal pad, with vias to inner layers or the other side. A good pad can reduce θJA from 80 to below 30 °C/W.
- Derate with temperature: the output current rating falls at high ambient temperature (read the derating curve).
- Look at the inductor, too: its winding loss and core loss heat it; a saturated inductor is a short.
- Measure. A thermal camera or a thermocouple on the chip case under full load tells more than any calculation.
Layout and noise
A switching converter makes sharp edges: the current in the input loop changes from zero to the full load in a few nanoseconds. The stray inductance of the wiring then produces a voltage spike V = L × di/dt that radiates and rings.
Worked example: a spike from loop inductance
A hot loop (input capacitor, switch and diode) with 10 nH of inductance, switching 2 A in 5 ns, produces a spike of 10 nH × 0 A/ns = 4.0 V on top of the 12 V input. Halving the loop length and area halves the spike; a snubber or a slower gate drive reduces it further.
- Keep the hot loop tiny: the input capacitor, high-side switch and low-side switch (or diode) must be adjacent, with the shortest, widest traces.
- One solid ground plane under the converter, with the input and output grounds returning to it close to the chip.
- Keep the switch node small: it has fast voltage edges and is the main source of radiated noise. Do not route sensitive signals under it.
- Route the feedback trace away from the switch node and the inductor, on a quiet layer, from the output capacitor to the divider.
- Follow the datasheet layout example exactly if there is one.
- Add input filtering (a ferrite bead and capacitor, or an LC filter) when the converter shares a supply with analog or radio circuits.
- Shield and bond the enclosure and the cable shields as described in Cabling, Grounding and Noise.
Protection and start-up
| Feature | What it does | Notes |
|---|---|---|
| Under-voltage lockout (UVLO) | Keeps the converter off until the input is high enough | Prevents chattering from a sagging battery; set with a divider on the enable pin |
| Over-current protection | Limits the peak (or average) switch current | Hiccup mode retries after a short; cycle-by-cycle limits protect the inductor |
| Thermal shutdown | Turns off above about 150 °C | A last resort, not a design aim |
| Soft start | Ramps the output slowly | Limits the inrush current into the output capacitors |
| Over-voltage protection | Clamps or shuts down on a high output | Protects the load from a feedback fault |
| Input protection | A fuse, a TVS diode, reverse-polarity protection | See Switches, Connectors and Protection |
| Input inrush | The input capacitors charge at the plug-in moment | A hot-plug onto a long cable can ring to twice the supply: use a bulk electrolytic or a TVS |
Choosing a regulator
| Need | Choice | Why |
|---|---|---|
| Quiet analog rail, small drop, under 100 mA | LDO | Low noise, no inductor |
| 12–24 V to 3.3 V or 5 V, 0.5 A or more | Buck converter or module | Low heat |
| One lithium cell to 3.3 V | Buck-boost or low-dropout buck | The cell crosses 3.3 V |
| Cell to 5 V | Boost | Output above input |
| Mains to low voltage | Buy a certified adapter or module | Safety and approvals |
| Noise-sensitive load on a high-input rail | Buck followed by an LDO | Efficiency and a clean output |
| Prototype in a hurry | A ready-made buck module (MP1584, LM2596) | Tested layout; check the specifications |
Common faults
| Symptom | Likely cause | Try |
|---|---|---|
| No output | Enable low, UVLO not satisfied, wrong feedback divider, blown input | Check the enable pin and the input voltage at the chip |
| Output too high or too low | Wrong feedback resistors, a floating feedback pin | Check the divider values and the connection |
| Excessive ripple or noise | Too small or too high-ESR output capacitor, long hot loop, an inductor that saturates | Add ceramic capacitors; shorten loops; check the inductor’s saturation current |
| Converter hot or in thermal shutdown | Overload, a diode instead of a synchronous switch, poor copper, a high ambient | Measure the current; add copper; improve airflow |
| Hiccups or squeals under light load | Pulse skipping or burst mode, an audible ceramic capacitor | Force PWM mode, or change the capacitor type |
| Output droops when the load steps | Not enough output capacitance, slow loop | Increase the capacitance or use a faster controller |
| Chip destroyed at power-up | Input spike above the rating (hot-plug ringing), reversed supply | Add a TVS or a bulk capacitor; check the polarity protection |
| Radio or sensor interference | Switch-node noise, a long cable acting as an antenna | Layout, a snubber, a filter, a different frequency |
| Battery life shorter than expected | Quiescent current, light-load inefficiency, feedback divider drain | Measure the no-load current; pick a low-quiescent part |
Related
- Power and Motor Electronics: the overview of this guide.
- Batteries and Power Supplies: cells, linear regulators and supply basics.
- Inductors, Transformers and Relays and Capacitors: the parts a converter is built from.
- Diodes and LEDs and Transistors: the switch and the rectifier.
- Driving Motors and Actuators: MOSFET switching and flyback in motor drives.
- PCB Design and Layout: trace width, copper area and layout.
- Resistor Network and Standard Value Calculator.
FAQ
How do I choose between a linear and a switching regulator?
If the voltage drop is small, the current is low or the rail must be very quiet, use an LDO. If the drop is large or the current is above a few hundred milliamperes, use a switcher. The loss of a linear regulator is (Vin − Vout) × I.
What inductor value should I use?
Choose it for a ripple current of 20–40 percent of the load current: L = (Vin − Vout) × D ÷ (f × ΔI). Then check the saturation current is above the peak and the heating current above the load current.
Why do ceramic capacitors lose capacitance?
High-permittivity ceramics (X5R, X7R) lose a large part of their capacitance when a DC voltage is applied, and more in small packages. Use the datasheet’s DC-bias curve and allow for the loss.
Do I need a synchronous converter?
At low output voltages and high currents the diode loss is large, so synchronous is worthwhile (and most modern chips are). At light load a diode converter can be more efficient than a synchronous one in forced-PWM mode.
Can I power a microcontroller straight from a buck converter?
Yes, with a clean layout and output capacitors. For an analog circuit or ADC reference, filter the output with an LC filter or follow the converter with an LDO.
Is it safe to design a mains converter?
Not without the safety standards, creepage and clearance rules, and approvals. Use a certified module or adaptor; do not work on live mains equipment.
References
- Erickson, R. W. and Maksimovic, D., Fundamentals of Power Electronics, 3rd ed., Springer.
- Texas Instruments, Basic Calculation of a Buck Converter’s Power Stage (SLVA477) and Thermal Design by Insight, Not Hindsight (SNVA419).
- Analog Devices and onsemi application notes on buck and boost design, layout and EMI; controller datasheets (TPS5430, LM2596, MP1584, TPS61xxx).
- Horowitz, P. and Hill, W., The Art of Electronics, 3rd ed., Cambridge University Press: chapter 9, power supplies.
Last reviewed: October 2026. Values are typical: check the datasheets of the parts you use. Do not work on live mains equipment unless you are qualified.
