# Thermal Effects in High Power LED Packaging

## Overview

The aim of this test case is to validate the following functions:

• Heat transfer solvers

The simulation results of SimScale were compared to the analytical results presented in [Adam]. The mesh used was created using first order tetrahedral with local refinements meshing algorithm on the SimScale platform.

Import validation project into workspace

## Geometry

The pitch is the Light Emitting Diodes (LEDs) separation distance between each other. Total of 25 LEDs with dimension and thickness of 7 mm x 7 mm and 100 μm respectively mounted on a heat sink of 10 cm x 10 cm were simulated. The pitch of 1 cm, 5 mm and 1 mm were tested. Due to symmetry, only quarter of the model was taken for the analysis. The geometries used for the analysis (with different LED pitches) are shown in the figure below: Geometries used in the analysis with 1 cm, 5 mm and 1 mm of separation distance between LEDs

## Analysis type and Domain

Tool Type : CalculiX, Code_Aster

Analysis Type : Heat transfer

Mesh and Element types :

Information of different tested cases
Case Pitch Mesh algorithm No. of nodes No. of 3D elements Solver
A 1 cm Tetrahedral with local refinements 766903 2826272 CalculiX
B 1 cm Tetrahedral with local refinements 766904 2826273 Code_Aster
C 5 mm Tetrahedral with local refinements 764085 2814528 CalculiX
D 5 mm Tetrahedral with local refinements 764086 2814529 Code_Aster
E 1 mm Tetrahedral with local refinements 760497 2799922 CalculiX
F 1 mm Tetrahedral with local refinements 760498 2799923 Code_Aster

Mesh used for all the cases are shown below. In order to get better results, at least 2 elements were placed over the thickness of TIM (Thermal Interface Material) layer.

## Simulation Setup

Material:

• Aluminum: isotropic: ρ
$\rho$

= 2700 kg/m³, κ

$\kappa$

= 202.4 W/(mK), Specific heat = 897 J/(kg K)

• TIM (Thermal Interface Material): isotropic: ρ
$\rho$

= 7870 kg/m³, κ

$\kappa$

= 0.22 W/(mK), Specific heat = 480 J/(kg K)

Initial Condition:

• Temperature = 300.15 K

Each case contains total of 14 simulation runs with different configuration setups. The applied surface heat flux and convective heat flux boundary condition for all the simulation runs are given below:

Surface heat flux:

Surface heat flux of 1 W, 3 W and 5 W per LED was applied on all the red highlighted surface of different LED packages as shown in figure above under ‘Geometry’ section. The table below shows the run cases with applied surface heat flux:

Applied surface heat flux for all the cases
Case Surface heat flux [W/m²]
1 – 5 20408.2 (1 W/LED)
6 – 10 61224.5 (3 W/LED)
11 – 14 102040.8 (5 W/LED)

Convective heat flux:

Convective heat flux of 10 W/(m²K), 25 W/(m²K), 50 W/(m²K), 75 W/(m²K) and 100 W/(m²K) was applied to all the heat sink surfaces accept the surfaces along symmetry plane and surfaces in contact with TIM. The table below shows the run cases with applied convective heat flux:

Applied convective heat flux for all the cases
Case Convective heat flux [W/(m²K)] Reference temperature [K]
1 & 6 10 300.15
2, 7 & 11 25 300.15
3, 8 & 12 50 300.15
4, 9 & 13 75 300.15
5, 10 & 14 100 300.15

## Reference Solution

Since the overall resistance of 9 K/W was used for TIM (Thermal Interface Material) in [Adam], the resistance was converted to conductance in order to apply this value under material definition. The formulation used to convert this value is shown below:

κ=dR.A

$\kappa =\frac{d}{R.A}$

where,

Kappa, κ

$\kappa$

= conductivity

thickness, d

$d$

= 100 μm

Resistance, R

$R$

= 9 K/W

Cross sectional area, A

$A$

= 4.9e-5 m²

Putting all these values give conductance of:

κ=100e694.9e5=0.22W/(mK)

$\kappa =\frac{100e-6}{9\ast 4.9e-5}=0.22W/\left(mK\right)$

The equation used to convert the resistance was taken from [infineon].

## Results

Comparison of the temperature results for Case A-F with different boundary conditions configuration taken from SimScale with [Adam].