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Tutorial 201: Buck Converter

The buck converter is the fundamental step-down DC-DC topology, efficiently converting a higher voltage to a lower voltage using PWM control.

Overview
Difficulty Intermediate
Duration 30-40 minutes
Prerequisites PWM Basics

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

Circuit Topology

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

Components:

  • S - High-side switch (MOSFET/IGBT)
  • D - Freewheeling diode (Schottky recommended)
  • L - Output inductor
  • C - Output capacitor
  • R - Load resistance

Operating Principle

  • Current path: Vin → S → L → C/R → GND
  • Inductor voltage: \(V_L = V_{in} - V_{out}\) (positive)
  • Current increases: \(\frac{di_L}{dt} = \frac{V_{in} - V_{out}}{L}\)
  • Current path: L → C/R → GND → D → L
  • Inductor voltage: \(V_L = -V_{out}\) (negative)
  • Current decreases: \(\frac{di_L}{dt} = \frac{-V_{out}}{L}\)

Key Equations

Output Voltage (Volt-Second Balance)

In steady state, the average inductor voltage is zero:

\[ V_{out} = D \times V_{in} \]

Key Insight

Output voltage is directly proportional to duty cycle!

Inductor Current Ripple

\[ \Delta I_L = \frac{V_{out} \times (1 - D)}{f_s \times L} \]

Output Voltage Ripple

\[ \Delta V_{out} = \frac{\Delta I_L}{8 \times f_s \times C} \]

Critical Inductance (CCM Boundary)

\[ L_{crit} = \frac{V_{in} \times (1-D) \times D}{2 \times f_s \times I_{out,min}} \]

Design Example

Specifications

Parameter Value Unit
Input Voltage 48 V
Output Voltage 12 V
Output Current 5 A
Switching Frequency 100 kHz
Max Voltage Ripple 1%
Max Current Ripple 30%

Step-by-Step Design

1. Calculate Duty Cycle: \[ D = \frac{V_{out}}{V_{in}} = \frac{12}{48} = 0.25 \]

2. Calculate Load Resistance: \[ R = \frac{V_{out}}{I_{out}} = \frac{12}{5} = 2.4\,\Omega \]

3. Calculate Inductance (for 30% ripple): \[ \Delta I_L = 0.3 \times 5 = 1.5\,\text{A} \] \[ L = \frac{V_{out} \times (1-D)}{f_s \times \Delta I_L} = \frac{12 \times 0.75}{100k \times 1.5} = 60\,\mu\text{H} \]

4. Calculate Capacitance (for 1% ripple): \[ C = \frac{\Delta I_L}{8 \times f_s \times \Delta V_{out}} = \frac{1.5}{8 \times 100k \times 0.12} = 15.6\,\mu\text{F} \]

Simulation

Building the Circuit

  1. Add voltage source (Vin = 48V)
  2. Add switch with PWM gate signal (D = 0.25, fs = 100kHz)
  3. Add freewheeling diode
  4. Add inductor (L = 68µH)
  5. Add capacitor (C = 22µF)
  6. Add load resistor (R = 2.4Ω)
  7. Connect SCOPE to measure Vout and IL

Expected Results

Signal Expected Value
Vout (average) 12 V
Vout (ripple) ~100 mV p-p
IL (average) 5 A
IL (ripple) ~1.5 A p-p

Exercises

Exercise 1: Duty Cycle Variation

  1. Vary D from 0.1 to 0.5 in steps of 0.1
  2. Record Vout for each D
  3. Verify: \(V_{out} = D \times V_{in}\)

Exercise 2: CCM to DCM Transition

  1. Increase load resistance to 24Ω
  2. Observe inductor current waveform
  3. Does IL reach zero?

Exercise 3: Component Sizing

  1. Double the inductance to 136µH
  2. Measure the new current ripple
  3. Compare with calculated value

Common Issues

Issue Cause Solution
Output too low Wrong duty cycle Check D = Vout/Vin
High ripple L or C too small Increase component values
Simulation diverges Time step too large Reduce dt

Download

Buck Converter Circuit (buck_simple.ipes)

Next Steps