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
EfficiencyVout ÷ Vin: poor when the drop is large85–97 percent, nearly independent of the drop
Output noiseVery lowRipple and spikes at the switching frequency
Parts and costOne chip and two capacitorsController, inductor, diode or MOSFET, capacitors
Voltage rangeOutput below input onlyStep down, step up or both
LayoutForgivingCritical: loop area, ground, feedback routing
Best forLow power, small drop, quiet analog railsHigher 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.

Buck converter: 12 V to 5 V at 2 A, 500 kHz12 VSMOSFETswitch nodeDLCload5 VSwitch node voltageInductor current12 V0 Vavg 5 V2 A avgS on (42 percent of the time): the inductor current rises,storing energy. S off: it falls through the diode D.Average output = duty cycle × input = 0.417 × 12 V = 5 V.
A buck converter, 12 V to 5 V at 2 A: the switch node, the inductor current and the output.
QuantityFormula (continuous conduction)Meaning
Duty cycleD = Vout ÷ VinFraction of the period that the switch is on
Inductor ripple currentΔI = (Vin − Vout) × D ÷ (f × L)The up-and-down swing of the inductor current
Peak inductor currentIpk = Iout + ΔI ÷ 2What the inductor and switch must withstand
Output ripple voltageΔV ≈ ΔI ÷ (8 × f × C)Ideal capacitor: ESR adds ΔI × ESR
Input capacitor ripple currentIRMS = 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 frequencySmaller inductor and capacitorsHigher switching losses and EMITypical use
100–300 kHzLarger partsLow losses; easy layoutHigh-power converters; wide-input designs
400 kHz–1 MHzCompactModerateMost point-of-load and battery regulators
1–3 MHz and moreVery smallHigher losses; careful layout needed; avoids the AM radio bandSpace-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

LossCalculationPower
High-side MOSFET, conductionI² × RDS(on) × D = 2² × 0.03 × 0.4170.05 W
Switching (both edges)½ × Vin × I × tsw × f = 0.5 × 12 × 2 × 20 ns × 500 kHz0.12 W
Inductor windingI² × DCR = 2² × 0.0150.06 W
Schottky diodeVf × I × (1 − D) = 0.45 × 2 × 0.5830.53 W
Total with a diodeOutput power 10 W0.76 W: efficiency 93.0 percent
Synchronous low-side MOSFETI² × RDS(on) × (1 − D) = 4 × 0.02 × 0.5830.05 W
Total, synchronousreplaces the diode loss0.28 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

TopologyOutputHow it worksTypical use
BuckBelow the inputSwitch in series, inductor, freewheel diodeReducing 12–48 V to 5 V, 3.3 V, 1.8 V
BoostAbove the inputThe inductor charges from the input when the switch is on; when it opens, the inductor voltage adds to the inputOne lithium cell to 5 V; LED strings
Buck-boostInverted (or above or below)One inductor, switch and diode; output polarity reversed in the basic formNegative rails; battery that falls above and below the output
SEPIC, Ćuk, four-switch buck-boostAbove or below, not invertedTwo inductors or capacitors, or four switchesA battery that crosses the output voltage (a lithium cell to 3.3 V)
FlybackAny, isolatedA coupled inductor (transformer) stores energy and releases it to the secondaryMains adaptors, isolated supplies up to 100 W
Forward, half- and full-bridge, LLCIsolated, higher powerTransformer-coupled topologiesServer and industrial supplies
Charge pumpMultiple of the input, low currentCapacitors switched in stages; no inductorSmall 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

FeatureWhat it doesNotes
Under-voltage lockout (UVLO)Keeps the converter off until the input is high enoughPrevents chattering from a sagging battery; set with a divider on the enable pin
Over-current protectionLimits the peak (or average) switch currentHiccup mode retries after a short; cycle-by-cycle limits protect the inductor
Thermal shutdownTurns off above about 150 °CA last resort, not a design aim
Soft startRamps the output slowlyLimits the inrush current into the output capacitors
Over-voltage protectionClamps or shuts down on a high outputProtects the load from a feedback fault
Input protectionA fuse, a TVS diode, reverse-polarity protectionSee Switches, Connectors and Protection
Input inrushThe input capacitors charge at the plug-in momentA hot-plug onto a long cable can ring to twice the supply: use a bulk electrolytic or a TVS

Choosing a regulator

NeedChoiceWhy
Quiet analog rail, small drop, under 100 mALDOLow noise, no inductor
12–24 V to 3.3 V or 5 V, 0.5 A or moreBuck converter or moduleLow heat
One lithium cell to 3.3 VBuck-boost or low-dropout buckThe cell crosses 3.3 V
Cell to 5 VBoostOutput above input
Mains to low voltageBuy a certified adapter or moduleSafety and approvals
Noise-sensitive load on a high-input railBuck followed by an LDOEfficiency and a clean output
Prototype in a hurryA ready-made buck module (MP1584, LM2596)Tested layout; check the specifications

Common faults

SymptomLikely causeTry
No outputEnable low, UVLO not satisfied, wrong feedback divider, blown inputCheck the enable pin and the input voltage at the chip
Output too high or too lowWrong feedback resistors, a floating feedback pinCheck the divider values and the connection
Excessive ripple or noiseToo small or too high-ESR output capacitor, long hot loop, an inductor that saturatesAdd ceramic capacitors; shorten loops; check the inductor’s saturation current
Converter hot or in thermal shutdownOverload, a diode instead of a synchronous switch, poor copper, a high ambientMeasure the current; add copper; improve airflow
Hiccups or squeals under light loadPulse skipping or burst mode, an audible ceramic capacitorForce PWM mode, or change the capacitor type
Output droops when the load stepsNot enough output capacitance, slow loopIncrease the capacitance or use a faster controller
Chip destroyed at power-upInput spike above the rating (hot-plug ringing), reversed supplyAdd a TVS or a bulk capacitor; check the polarity protection
Radio or sensor interferenceSwitch-node noise, a long cable acting as an antennaLayout, a snubber, a filter, a different frequency
Battery life shorter than expectedQuiescent current, light-load inefficiency, feedback divider drainMeasure the no-load current; pick a low-quiescent part

Related

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.