Thermal conductivity of the composite material
QuickField simulation example
PCB consists of two layers: FR4 dielectric and a copper foil. Dielectric and copper feature different thermal conductivity.
Problem Type
Plane-parallel problem of heat transfer.
Geometry
Given
Thermal conductivity of copper is 380 W/K-m;
Thermal conductivity of FR4 dielectric is 0.25 W/K-m;
Task
Calculate effective thermal conductivity of a printed circuit board.
Solution
Thermal conductance = (Heat flux / Temperature difference) * ( Path length / Cross-section area ).
Path length is measured along the heat flux. Cross-section area is measured in the plane across the heat flux.
We apply temperature difference and QuickField calculates the heat flux.
Results
Case | Temperature difference | Cross-section area | Path length | Flux | Thermal conductance |
---|---|---|---|---|---|
Along | 40°C | 0.535 mm² | 1 mm | 0.268 W | 12.5 W/K*m |
Across | 40°C | 1 mm² | 0.535 mm | 0.01 W | 0.133 W/K*m |
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