Anisotropic Thermal Conductivity

This page shows the results of incorporating anisotropic thermal conductivity into the top layer of the 14 mil VCSEL die model. That is, the thermal conductivity k could take on different values in the radial (lateral) direction, k_r, and the z (vertical) direction, k_z. The motivation for studying the effect of anisotropic conductivity was the discrepancy between the previous steady-state thermal model and experimental values. It was observed empirically that the distribution on Face1 of the die resembled that on Face2, but the steady-state model showed very different pictures on the face of the die. The geometry of the die was modified slightly by dividing the die into four horizontal slices.

The VCSEL cylinder rests in the active region, near the top surface of the die. Experimental values are from a research article by Aframowitz, which gives the thermal conductivities as a function of x in the composition of Al_xGa_1-xAs (M. Aframowitz, J.Appl.Phys. 44, 1292 (1974)). The k values are shown in the table below for the proton implanted (PI) and oxide cases. Note the GaAs substrate retains isotropic thermal conductivity for both cases.

Since FEMLAB is unable to mesh very thin layers in three dimensions, we started by looking at a 2D version of the model. We took the cross-sectional slice of the die passing through the center of the VCSEL cylinder. We used the 15 mA value of 0.02472 for the heat source (Q=heat/volume). The temperature values will be much higher because the heat has only 2 dimensions to dissipate through rather than 3. The gold pad on top was omitted. The picture below is a 2D version of the model with isotropic thermal conductivities throughout. Note this is just a cross-sectional view of the steady-state thermal model developed previously.

Next, we look at a 2D version of the PI case with the anisotropic thermal conductivities described in the table.

Third, we see the effect of the anisotropic thermal conductivities in the oxide case.

The colormaps and scales are the same for all three pictures. The three temperature distributions are almost indistinguishable. So we can conclude that the anisotropic thermal conductivities have little effect on heat flow, at least in 2 dimensions.

We had to simplify the 3D geometry in order for FEMLAB to successfully mesh the VCSEL die. So we omitted the thin layer for the active region. Note this cuts the number of layers from four to two, since the top and bottom mirror layers have the same thermal conductivities for both the PI and oxide cases. The VCSEL is placed near the top surface of the cube. Even with this simplification, FEMLAB had difficulty meshing the thin layer at the top, as the dark top surface in the picture below shows. This mesh used 45,497 nodes and 240,162 elements, which is about four times the number of nodes required for the previous steady-state model. This geometry took roughly 45 minutes to mesh.

The steady-state solution with the mesh using the PI case conductivities is shown below. Since it is virtually identical to the steady-state solution obtained previously, we can conclude that the specified anisotropic thermal conductivities do not significantly affect the temperature distribution.

Since the conductivity values specified by Aframowitz do not change the temperature distribution, we tried to determine what values were necessary to make Face1 remsemble Face2. Since the VCSEL cylinder is off-center, we need conductivity values that will channel heat laterally to the two faces. All changes were made to the top layer, while the substrate retained its isotropic 0.46 W/cmK value. When the k_z value was decreased by a factor of 10 to 0.023 W/cmK, the resulting solution still did not show the desired symmetry between the faces.

Even when the K_z value was decreased by a factor of 10²⁰, the picture looked the same. Since k_z only acts on the top layer of the die (less than 3% of the total height), changing this value has little effect. So we started changing the k_r value in an attempt to increase lateral heat flow. The picture below shows the solution when k_z was increased by a factor of 10 to 4.5 W/cmK.

The heat seems to be flowing to the two faces more evenly. The value of k_r has a greater effect than k_z because k_r acts across the whole lateral surface of the die. Carrying this one step further, we increased k_r by another factor of 10 to 45 W/cmK. The distributions on the two faces are now very similar.

Unfortunately, a thermal conductivity of 45 W/cmK for GaAs is very unrealistic. So we conclude that anisotropic thermal conductivity alone is not enough to account for the observed symmetry between Face1 and Face2 of the die. Perhaps the heat needs to be waveguided to the surface is some other manner.