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title: Tutorial 202: Boost Converter

Tutorial 202: Boost Converter

Overview

The boost converter is a step-up DC-DC topology that produces an output voltage higher than its input. It's widely used in battery-powered applications, solar MPPT, and power factor correction. This tutorial covers boost converter operation, design, and the unique challenges of step-up conversion.

Level: Intermediate (⅔)

Duration: 30-40 minutes

Series: DC-DC Converters

Learning Objectives

By the end of this tutorial, you will: - [ ] Understand boost converter operation and volt-second balance - [ ] Calculate output voltage as a function of duty cycle - [ ] Design for CCM operation and calculate component values - [ ] Understand right-half-plane zero and control challenges - [ ] Simulate and analyze boost converter performance

Prerequisites

Materials

File Description
boost_simple.ipes Simple open-loop boost converter
B_Boost.ipes Complete boost converter example

Circuit Description

Boost Converter Topology

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

Components: - L: Input inductor (energy storage) - S: Low-side switch, controlled by PWM - D: Boost diode (high-side) - C: Output capacitor - R: Load resistance

Operating Modes

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

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

Key Equations

Output Voltage (Volt-Second Balance)

In steady state:

Vin·D·Ts + (Vin - Vout)·(1-D)·Ts = 0

Solving:

Vout = Vin / (1 - D)

Voltage Gain:

M = Vout/Vin = 1 / (1 - D)

D M = Vout/Vin
0 1.00
0.25 1.33
0.50 2.00
0.75 4.00
0.90 10.00

Warning: As D → 1, gain → ∞, but efficiency drops rapidly!

Inductor Current

Average inductor current:

IL,avg = Iout / (1 - D) = Pin / Vin

Inductor current ripple:

ΔIL = (Vin × D) / (fs × L)

Output Voltage Ripple

ΔVout = (Iout × D) / (fs × C)

Note: Ripple is higher than buck because current is pulsating!

CCM Boundary

Critical inductance:

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

Design Parameters

Example Design Specifications

Parameter Value Unit
Input Voltage (Vin) 12 V
Output Voltage (Vout) 48 V
Output Current (Iout) 2 A
Output Power (Pout) 96 W
Switching Frequency (fs) 100 kHz
Max Voltage Ripple 2% of Vout
Max Current Ripple 40% of IL,avg

Step-by-Step Design

1. Calculate Duty Cycle:

D = 1 - Vin/Vout = 1 - 12/48 = 0.75 (75%)

2. Calculate Input (Inductor) Current:

IL,avg = Iout / (1-D) = 2 / 0.25 = 8 A

3. Calculate Inductance: For 40% ripple:

ΔIL = 0.4 × IL,avg = 0.4 × 8 = 3.2 A
L = (Vin × D) / (fs × ΔIL)
L = (12 × 0.75) / (100k × 3.2) = 28 μH
Choose: L = 33 μH (standard value)

4. Calculate Capacitance: For 2% ripple (0.96V):

C = (Iout × D) / (fs × ΔVout)
C = (2 × 0.75) / (100k × 0.96) = 15.6 μF
Choose: C = 22 μF (low ESR)

5. Component Stress:

Switch current: IL,peak = IL,avg + ΔIL/2 = 9.6 A
Switch voltage: Vout = 48 V → use 80V+ MOSFET
Diode current: IL,avg = 8 A (pulsed)
Diode voltage: Vout = 48 V → use 60V+ Schottky

Control Challenges

Right-Half-Plane Zero (RHPZ)

The boost converter has an inherent RHPZ in its transfer function:

fz,RHP = (1-D)² × R / (2π × L)

Effects: - Limits control bandwidth - Causes initial wrong-way response - Requires slower feedback loop

Mitigation: - Keep D < 0.7-0.8 if possible - Use current-mode control - Design for lower bandwidth

Input vs. Output Current

Unlike buck, the boost draws more current from input than output:

Iin = Iout / (1-D) > Iout

This affects: - Input capacitor sizing - Wire gauge selection - EMI filter design

Building the Circuit

Step 1: Power Stage

  1. Add voltage source (Vin = 12V DC)
  2. Add inductor (L = 33 μH) in series with input
  3. Add ideal switch (low-side, to ground)
  4. Add diode (cathode to output)
  5. Add capacitor (C = 22 μF) at output
  6. Add resistor (R = 24 Ω for 2A at 48V)

Step 2: PWM Control

  1. Add PWM signal generator:
  2. Frequency: 100 kHz
  3. Duty cycle: 0.75
  4. Connect PWM output to switch gate

Step 3: Measurements

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

Step 4: Simulation Settings

  • Simulation time: 5-10 ms (allow settling)
  • Time step: 50 ns (or automatic)
  • Solver: TRZ

Expected Results

Steady-State Waveforms

Signal Expected Value
Vout (average) 48 V
Vout (ripple) ~1 V p-p
IL (average) 8 A
IL (ripple) ~3 A p-p
Switch Vds (OFF) 48 V

Waveform Characteristics

Inductor Current: - Triangular waveform - Never goes negative (CCM) - Average = Iin = Iout/(1-D)

Switch Voltage: - 0 during ON - Vout during OFF - May show ringing at transitions

Output Voltage: - Higher ripple than buck (discontinuous diode current) - DC level at Vout = Vin/(1-D)

Exercises

Exercise 1: Voltage Gain

  1. Open boost_simple.ipes
  2. Set Vin = 12V, vary D from 0.2 to 0.8
  3. Record Vout for each D
  4. Verify: Vout = Vin/(1-D)
  5. Note: What happens at high D?

Exercise 2: Efficiency vs. Duty Cycle

  1. Add component losses (Ron, Vf, DCR)
  2. Measure efficiency at D = 0.5, 0.67, 0.75, 0.85
  3. Plot: Efficiency vs D
  4. Explain: Why does efficiency drop at high D?

Exercise 3: CCM vs DCM

  1. With L = 33 μH, set light load (R = 240Ω)
  2. Observe: Does IL reach zero?
  3. Calculate: Critical load for CCM boundary
  4. Measure: How does DCM affect Vout regulation?

Exercise 4: Transient Response

  1. Apply step load change (R: 24Ω → 12Ω)
  2. Measure: Vout undershoot, recovery time
  3. Compare: Response with different C values
  4. Challenge: Add simple voltage feedback

Exercise 5: Current Mode Control

  1. Add current sense on inductor
  2. Implement peak current-mode PWM
  3. Compare: Transient response vs. voltage-mode

Common Issues

Issue Cause Solution
Output lower than expected High diode drop, losses Use Schottky, lower D
Excessive ripple C too small, high ESR Increase C, use MLCC
DCM at light load L too small Increase L
Switch failure Voltage spike Add snubber, check ratings
Unstable control RHPZ, high bandwidth Slow down controller

Efficiency Analysis

Loss Breakdown

Component Loss Mechanism Typical %
Switch Conduction (I²R) 1-3%
Switch Switching (CV²f) 2-5%
Diode Conduction (Vf×I) 2-5%
Diode Reverse recovery 1-3%
Inductor Copper (I²R) 1-2%
Inductor Core loss 0.5-1%
Capacitor ESR loss 0.2-0.5%

Improving Efficiency

  1. Synchronous rectification: Replace diode with MOSFET
  2. Interleaving: Multiple parallel phases
  3. Lower frequency: Reduce switching losses (trade-off: larger L, C)
  4. Lower D: Limit step-up ratio (cascade if needed)

Boost vs. Buck Comparison

Aspect Buck Boost
Voltage ratio Step-down Step-up
Input current Pulsating Continuous
Output current Continuous Pulsating
RHPZ No Yes
Control Easier Harder
Efficiency at extreme D Good Poor

References

  1. Erickson, R.W., Maksimovic, D. "Fundamentals of Power Electronics" - Chapter 7
  2. Vorpérian, V. "Fast Analytical Techniques for Electrical and Electronic Circuits"
  3. Texas Instruments SLVA372: "Boost Converter Design"

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