PCB Copper Thermal Resistance

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.