Thermal Calculations & Design

Power electronics thermal analysis and cooling solutions

Essential Thermal Design for Power Electronics

Thermal design is critical for the reliable operation and longevity of power semiconductor devices. Proper thermal management ensures that junction temperatures remain within safe operating limits, preventing degradation and failure of power modules. Semikron provides comprehensive thermal analysis tools and guidelines to help engineers calculate junction temperatures under various operating conditions and design effective cooling solutions. Our thermal models account for conduction, convection, and radiation heat transfer mechanisms, as well as the thermal characteristics of different package types and mounting methods.

Accurate Modeling

Precise thermal models based on actual device characteristics

Multiple Scenarios

Analysis for steady-state, transient, and cyclic conditions

Design Optimization

Tools to optimize cooling solution cost and performance

Thermal Calculation Principles

Fundamental concepts for thermal design

Junction Temperature Calculation

The junction temperature (TJ) is calculated using the thermal resistance model:

TJ = TA + Ptot × Rth(J-A)

Where:

  • TJ = Junction temperature (°C)
  • TA = Ambient temperature (°C)
  • Ptot = Total power dissipation (W)
  • Rth(J-A) = Thermal resistance from junction to ambient (K/W)

Thermal Resistance Path

The total thermal resistance consists of multiple series resistances:

Rth(J-A) = Rth(J-C) + Rth(C-H) + Rth(H-A)

Where:

  • Rth(J-C) = Junction to case thermal resistance
  • Rth(C-H) = Case to heatsink thermal resistance
  • Rth(H-A) = Heatsink to ambient thermal resistance

Power Dissipation

Total power dissipation includes both switching and conduction losses:

Ptot = Pcond + Psw

Where:

  • Pcond = Conduction losses (depend on current and VCEsat)
  • Psw = Switching losses (depend on switching frequency)

Thermal Calculation Tools

Interactive tools for thermal design analysis

Junction Temperature Calculator

Calculate junction temperature based on power dissipation and thermal resistance values.

Junction Temperature: -- °C

Heatsink Sizing Tool

Determine required heatsink thermal resistance based on power dissipation and maximum junction temperature.

Required Heatsink Resistance: -- K/W

Package Thermal Characteristics

Thermal resistance values for common Semikron packages

IGBT Module Thermal Resistance

Package Configuration Rth(J-C) [K/W] Rth(J-H) [K/W] Notes
SEMITRANS Phase Leg 0.07 - 0.12 0.25 - 0.40 Standard industrial package
SEMiX Phase Leg 0.04 - 0.07 0.15 - 0.25 High power density
SEMITOP Phase Leg 0.05 - 0.08 0.18 - 0.30 Top-side cooling
MiniSKiiP Phase Leg 0.03 - 0.05 0.12 - 0.20 Integrated driver
SKiiP Phase Leg 0.03 - 0.06 0.08 - 0.15 Press-fit technology

Thyristor/Diode Module Thermal Resistance

Package Configuration Rth(J-C) [K/W] Rth(J-H) [K/W] Notes
SEMIPACK 4 Single Thyristor 0.08 - 0.10 0.28 - 0.35 Standard generation
SEMIPACK 6 Single Thyristor 0.06 - 0.08 0.22 - 0.30 Latest generation
Semipont Single Diode 0.07 - 0.12 0.25 - 0.40 Rectifier applications

Cooling Methods

Comparison of different cooling approaches

Natural Convection

Passive cooling relying on natural air circulation:

  • Simple and reliable
  • Low cost
  • Limited cooling capacity (typically <50W)
  • Sensitive to ambient temperature
  • Requires large heatsinks

Typical thermal resistance: 10-50 K/W depending on heatsink size and orientation.

Forced Air Cooling

Active cooling with fan-forced air circulation:

  • Good cooling capacity
  • Relatively low cost
  • Requires fan maintenance
  • Sensitive to dust accumulation
  • Typical thermal resistance: 0.5-5.0 K/W

Liquid Cooling

Active cooling with liquid heat transfer medium:

  • Excellent cooling capacity
  • Compact heatsinks
  • Higher complexity and cost
  • Requires pump and heat exchanger
  • Typical thermal resistance: 0.1-0.5 K/W

Thermal Interface Materials

Characteristics and selection guidelines for TIMs

Thermal Resistance of Interface Materials

Material Type Thermal Conductivity [W/m·K] Interface Resistance [K·cm²/W] Applications
Thermal Grease (Standard) 2.0 - 4.0 0.5 - 1.0 General purpose
Thermal Grease (High Performance) 4.0 - 7.0 0.3 - 0.6 High power applications
Thermal Pad 1.0 - 5.0 0.7 - 1.5 Assembly convenience
Phase Change Material 3.0 - 6.0 0.4 - 0.8 Temperature cycling
Solder 20 - 80 0.05 - 0.1 Permanent applications

Selection Guidelines

  • For high power density applications: Use high-performance thermal grease with thermal conductivity >5 W/m·K
  • For assembly convenience: Thermal pads simplify manufacturing but have higher thermal resistance
  • For temperature cycling: Phase change materials perform better under thermal cycling than standard greases
  • For permanent installations: Solder provides the lowest thermal resistance but is not serviceable

FAE Technical Commentary

Expert insights on thermal design

For high-power applications, always calculate the required thermal resistance at the maximum expected ambient temperature and power dissipation. Many designs work well at room temperature but fail at maximum operating conditions. As a rule of thumb, derate the thermal performance by 20-30% for safety margins.

When designing for temperature cycling, consider the impact of thermomechanical stress on both the power module and the thermal interface material. Power cycling can cause degradation in thermal performance over time, particularly for thermal grease.

For press-fit power modules (like SKiiP), the mounting pressure significantly affects thermal performance. Always follow Semikron's recommended torque values for base plate mounting to ensure proper thermal contact.

Thermal Design Examples

Practical examples for common applications

Solar Inverter Thermal Design

Scenario: 10kW string inverter using SEMiX 1200V/200A module

Parameters:

  • Power dissipation: 300W (worst case)
  • Ambient temperature: 50°C (inside enclosure)
  • Max junction temperature: 150°C
  • Module Rth(J-C): 0.05 K/W

Calculation:

Rth(H-A) required = (TJ - TA - P × Rth(J-C)) / P

Rth(H-A) = (150 - 50 - 300 × 0.05) / 300 = (150 - 50 - 15) / 300 = 0.28 K/W

Therefore, a heatsink with thermal resistance ≤ 0.28 K/W is required at the specified ambient temperature.

Motor Drive Thermal Design

Scenario: 50kW industrial drive using SKiiP 3 system

Parameters:

  • Power dissipation: 1200W (total for system)
  • Ambient temperature: 40°C
  • Max junction temperature: 125°C
  • System Rth(J-H): 0.08 K/W

Calculation:

TA required = TJ - P × Rth(J-A)

For Rth(H-A) = 0.05 K/W: TA = 125 - 1200 × (0.08 + 0.05) = 125 - 156 = -31°C

This is impossible, so we need forced cooling with Rth(H-A) ≤ 0.02 K/W

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