title: Tutorial 201: Buck Converter¶

Tutorial 201: Buck Converter¶

Overview¶

The buck converter is the most fundamental step-down DC-DC topology. It efficiently converts a higher DC voltage to a lower DC voltage using PWM control. This tutorial covers buck converter operation, design equations, and simulation analysis.

Level: Intermediate (⅔)

Duration: 30-40 minutes

Series: DC-DC Converters

Learning Objectives¶

By the end of this tutorial, you will: - [ ] Understand buck converter operation in CCM and DCM - [ ] Apply the volt-second balance principle - [ ] Calculate output voltage, inductor current ripple, and capacitor voltage ripple - [ ] Design a buck converter for given specifications - [ ] Simulate and verify converter operation

Prerequisites¶

• Complete Tutorial 103: PWM Basics
• Understanding of inductor and capacitor behavior
• Basic circuit analysis skills

Materials¶

Circuit Description¶

Buck Converter Topology¶

        +Vin ──────┬──[S]──┬──[L]──┬── +Vout
                   │       │       │
                   │      [D]     [C]     [R] Load
                   │       │       │       │
        GND ───────┴───────┴───────┴───────┴── GND

Components: - S: High-side switch (MOSFET/IGBT), controlled by PWM - D: Freewheeling diode (Schottky recommended) - L: Output inductor (energy storage) - C: Output capacitor (voltage smoothing) - R: Load resistance

Operating Modes¶

Switch ON (0 < t < D·Ts): - Current path: Vin → S → L → C/R → GND - Inductor voltage: VL = Vin - Vout (positive, current increases) - Diode: Reverse biased (OFF)

Switch OFF (D·Ts < t < Ts): - Current path: L → C/R → GND → D → L (freewheeling) - Inductor voltage: VL = -Vout (negative, current decreases) - Diode: Forward biased (ON)

Key Equations¶

Output Voltage (Volt-Second Balance)¶

In steady state, the average inductor voltage is zero:

(Vin - Vout)·D·Ts + (-Vout)·(1-D)·Ts = 0

Solving:

Vout = D × Vin

Key insight: Output voltage is directly proportional to duty cycle!

Inductor Current Ripple¶

ΔIL = (Vin - Vout) × D / (fs × L)
    = Vout × (1 - D) / (fs × L)

Output Voltage Ripple¶

Assuming ESR = 0:

ΔVout = ΔIL / (8 × fs × C)

Including ESR:

ΔVout ≈ ΔIL × ESR + ΔIL / (8 × fs × C)

Boundary Condition (CCM/DCM)¶

CCM is maintained when average inductor current > ripple/2:

IL,avg > ΔIL/2

Iout > (Vin × D × (1-D)) / (2 × fs × L)

Critical inductance for CCM:

Lcrit = (Vin × (1-D) × D) / (2 × fs × Iout,min)

Design Parameters¶

Example Design Specifications¶

Step-by-Step Design¶

1. Calculate Duty Cycle:

D = Vout / Vin = 12/48 = 0.25 (25%)

2. Calculate Load Resistance:

R = Vout / Iout = 12/5 = 2.4 Ω

3. Calculate Inductance: For 30% ripple:

ΔIL = 0.3 × Iout = 0.3 × 5 = 1.5 A
L = Vout × (1-D) / (fs × ΔIL)
L = 12 × 0.75 / (100k × 1.5) = 60 μH

4. Calculate Capacitance: For 1% ripple (120 mV):

C = ΔIL / (8 × fs × ΔVout)
C = 1.5 / (8 × 100k × 0.12) = 15.6 μF

Building the Circuit¶

Step 1: Power Stage¶

1. Add voltage source (Vin = 48V DC)
2. Add ideal switch (or MOSFET) - high-side position
3. Add diode - cathode to switch node, anode to ground
4. Add inductor (L = 68 μH)
5. Add capacitor (C = 22 μF) in parallel with load
6. Add resistor (R = 2.4 Ω)

Step 2: PWM Control¶

1. Add PWM signal generator:
2. Frequency: 100 kHz
3. Duty cycle: 0.25
4. Connect PWM output to switch gate

Step 3: Measurements¶

1. Add SCOPE
2. Connect channels to:
3. Output voltage (Vout)
4. Inductor current (IL)
5. Switch voltage (Vds)

Step 4: Simulation Settings¶

• Simulation time: 2 ms (200 switching cycles)
• Time step: 50 ns (or automatic)
• Solver: TRZ

Expected Results¶

Steady-State Waveforms¶

Waveform Characteristics¶

Output Voltage: - DC level at 12V - Small triangular ripple - Frequency = 2×fs (double switching frequency)

Inductor Current: - Triangular waveform - Average = Iout - Ramps up during ON, down during OFF

Exercises¶

Exercise 1: Vary Duty Cycle¶

1. Open buck_simple.ipes
2. Change duty cycle from 0.1 to 0.5 in steps of 0.1
3. Record Vout for each D
4. Verify: Vout = D × Vin

Exercise 2: CCM to DCM Transition¶

1. Set D = 0.25, L = 68 μH
2. Increase R from 2.4Ω to 24Ω (light load)
3. Observe: Does inductor current reach zero?
4. Calculate: At what load does DCM begin?

Exercise 3: Ripple Analysis¶

1. With L = 68 μH, C = 22 μF, measure ripple
2. Double L to 136 μH, measure ripple
3. Double C to 44 μF, measure ripple
4. Compare: Which has more effect on voltage ripple?

Exercise 4: Component Stress¶

1. Measure peak switch current (= IL,max)
2. Measure peak diode voltage (= Vin)
3. Design: Select components with 2× margin

Exercise 5: Efficiency Estimation¶

1. Add realistic component losses:
2. Switch: Ron = 10 mΩ
3. Diode: Vf = 0.5V
4. Inductor: DCR = 20 mΩ
5. Calculate: Pin, Pout, efficiency

Common Issues¶

Efficiency Considerations¶

Typical loss breakdown: | Loss Type | Formula | Example | |-----------|---------|---------| | Switch conduction | Irms² × Ron | 25mW | | Switch switching | Vds × Id × (trise + tfall) × fs | 1W | | Diode conduction | Iavg × Vf + Irms² × Rd | 2.5W | | Inductor copper | Irms² × DCR | 0.5W | | Inductor core | From datasheet | 0.2W |

Related Tutorials¶

• 202 - Boost Converter - Step-up topology
• 203 - Buck-Boost - Inverting topologies
• 501 - Loss Calculation - Thermal analysis

References¶

1. Erickson, R.W., Maksimovic, D. "Fundamentals of Power Electronics" - Chapter 7
2. Mohan, N. "Power Electronics" - DC-DC Converters
3. Texas Instruments SLVA477: "Buck Converter Design"

Tutorial Version: 1.0 Last updated: 2026-02 Compatible with GeckoCIRCUITS v1.0+