PCB Copper Thermal Resistance
Rth = d / (k × A)
Quick answer
This calculator solves Rth = d / (k_Cu × A), k_Cu = 385 W/(m·K) for Thermal Resistance. With Conduction Path Length 10 mm, Cross-Section Area 10 mm², the result is 2.597 °C/W. It solves in 3 directions, so you can enter the values you have and get the one you need.
Calculator
Formula
Description
Heat conduction through copper traces and planes on a PCB follows Fourier law. The thermal resistance depends on the path length, cross-sectional area, and thermal conductivity of copper (385 W/(m·K)). Copper planes and wide traces act as effective heat spreaders, conducting heat away from hot components to cooler areas of the board. The thermal conductivity of FR-4 is much lower (0.25-0.3 W/(m·K)), so heat primarily spreads through the copper layers. Thermal vias connecting copper planes on different layers significantly improve vertical heat transfer.
Variables
- Rth — Thermal resistance (°C/W)
- d — Length of the conduction path (m)
- A — Cross-sectional area of the copper (m²)
Practical Notes
For a 1 oz (35 µm) copper plane, 1 cm wide and 1 cm long: A = 0.035 mm × 10 mm = 0.35 mm² = 3.5 × 10⁻⁷ m², d = 0.01 m, Rth = 0.01/(385 × 3.5e-7) = 74 °C/W. Wider and thicker copper dramatically reduces thermal resistance. 2 oz copper halves the resistance. Inner layers conduct heat better than outer layers because they are sandwiched between FR-4 which provides some insulation from ambient, keeping the copper temperature more uniform.
Symbols and units
| Symbol | Meaning | Default unit | Accepted units |
|---|---|---|---|
| Rth | Thermal Resistance | °C/W | °C/W |
| d | Conduction Path Length | mm | mm, cm, m, km |
| A | Cross-Section Area | mm² | mm², cm², m² |
Step by step
- 1. What you know: Conduction Path Length 10 mm, Cross-Section Area 10 mm²
- 2. The relation: Rth = d / (k_Cu × A), k_Cu = 385 W/(m·K)
- 3. Substitute: d = 10 mm, A = 10 mm²
- 4. Result: Thermal Resistance = 2.597 °C/W
Worked examples
Minimum range
Given: Conduction Path Length 0.1 mm, Cross-Section Area 0.1 mm²
Result: Thermal Resistance = 2.597 °C/W
Low range
Given: Conduction Path Length 1 mm, Cross-Section Area 1 mm²
Result: Thermal Resistance = 2.597 °C/W
Typical
Given: Conduction Path Length 10 mm, Cross-Section Area 10 mm²
Result: Thermal Resistance = 2.597 °C/W
High range
Given: Conduction Path Length 10 cm, Cross-Section Area 1 cm²
Result: Thermal Resistance = 2.597 °C/W
Maximum range
Given: Conduction Path Length 1 m, Cross-Section Area 10 cm²
Result: Thermal Resistance = 2.597 °C/W
Solving for each variable
Solve for Thermal Resistance
Conduction Path Length 0.1 mm, Cross-Section Area 0.1 mm² → 2.597 °C/W
Solve for Conduction Path Length
Thermal Resistance 0.1 °C/W, Cross-Section Area 0.1 mm² → 0.00385 mm
Solve for Cross-Section Area
Thermal Resistance 0.1 °C/W, Conduction Path Length 0.1 mm → 2.597 mm²
How Thermal Resistance changes with Conduction Path Length
| Conduction Path Length (input) | Thermal Resistance (result) |
|---|---|
| 1 mm | 0.2597 °C/W |
| 2.5 mm | 0.6494 °C/W |
| 5 mm | 1.299 °C/W |
| 7.5 mm | 1.948 °C/W |
| 10 mm | 2.597 °C/W |
| 15 mm | 3.896 °C/W |
| 2 cm | 5.195 °C/W |
| 4 cm | 10.39 °C/W |
| 7 cm | 18.18 °C/W |
| 10 cm | 25.97 °C/W |
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<p><a href="https://vbatpower.com/formulas/pcb-copper-thermal-r/">PCB Copper Thermal Resistance</a> Calculator by vbatpower.com</p>Frequently asked questions
How do you calculate PCB Copper Thermal Resistance?
Use Rth = d / (k_Cu × A), k_Cu = 385 W/(m·K). With Conduction Path Length 10 mm, Cross-Section Area 10 mm², Thermal Resistance works out to 2.597 °C/W.
What is the formula for PCB Copper Thermal Resistance?
The formula is Rth = d / (k_Cu × A), k_Cu = 385 W/(m·K). Heat conduction through copper traces and planes on a PCB follows Fourier law.
How do you find Conduction Path Length when you know the other values?
Rearrange for Conduction Path Length: with Thermal Resistance 0.1 °C/W, Cross-Section Area 0.1 mm², Conduction Path Length is 0.00385 mm.
How do you find Cross-Section Area when you know the other values?
Rearrange for Cross-Section Area: with Thermal Resistance 0.1 °C/W, Conduction Path Length 0.1 mm, Cross-Section Area is 2.597 mm².
Related Concepts
All Thermal formulas →Need more features?
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