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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
Thermal conductivity of the composite material PCB consists of the three layers: FR4 dielectric, copper foil and FR4 dielectric. FR4 dielectric Copper foil 0.035 mm 0.5 mm

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

CaseTemperature
difference
Cross-section
area
Path lengthFluxThermal
conductance
Along40°C0.535 mm²1 mm0.268 W12.5 W/K*m
Across40°C1 mm²0.535 mm0.01 W0.133 W/K*m

Thermal conductivity of the composite material