A Semi-Analytical Thermal Elastic Model for Directional Crystal Growth with Weakly Anisotropy

A Semi-Analytical Thermal Elastic Model for Directional Crystal Growth with Weakly Anisotropy School of Mathematical Sciences Peking University with Sean Bohun, Penn State University Huaxiong Huang, York University 25th May 2005

Directional growth techniques such as Czocholski method is frequently used to produce high quality single crystals. anisotropic effects such as facets are often visible on the surface of the crystals grown by Cz method. We extend perturbation model for axisymmetric crystals developed in C. Sean Bojun, Ian Frigaard, Huaxiong Huang & Shuqing Liang, A Semi-Analytical Model for the Cz growth of Type III-V Compounds to anisotropic ones. Both the geometry and elastic parameters anisotropy effect are considered.

Weak geometric anisotropy Axisymmetric shape: j k i z=0 h gs (z) Ω S(t) Γ g C = 2πR A = πr 2 R(z) Γ S

Weak geometric anisotropy (cont.) Weak geometric anisotropy: R depending on the angle φ. Assume R(z, φ, t) = A(z, t)(1 + α m β k cos(n k (φ + δ k )) k=1 where α is small constant (weak anisotropy), and m k=1 β2 k = 1. For [1,0,0] pulling direction, 4-fold symmetric: m=1, β 1 = 1, n 1 = 4. For [1,1,1] pulling direction, 6-fold symmetric: m=1, β 1 = 1, n 1 = 6.

elastic constant anisotropy The effect of elasticity problem with cubic anisotropy The stresses σ = (σ xx, σ yy, σ zz, σ yz, σ xz, σ xy ) are related to the strains e = (e xx, e yy, e zz, 2e yz, 2e xz, 2e xy ) by C 11 C 12 C 12 C 12 C 11 C 12 σ = C 12 C 12 C 11 C 44 e = Ce. C 44 C 44 isotropic case: 2C 44 C 11 C 12 = 0.(C 12 = λ, C 44 = µ, Lamé constants) cubic anisotropy: H = 2C 44 C 11 C 12 > 0.

Basic Equations ( T 2 ) ρ s c s t = κ T s x 2 + 2 T y 2 + 2 T z 2, X = (x, y, z) Ω, t > 0. where ρ s, c s and k s are the density, specific heat and thermal conductivity of the crystal. The boundary conditions are below, κ s T n = h gs(t T g ) + h F (T 4 T 4 b ), X Γ g κ s T z = h ch(t T ch ), z = 0, where h gs and h ch represent the heat transfer coefficients; h F the radiation heat transfer coefficients; T g, T ch and T b denote the ambient gas temperature, the chuck temperature and background temperature respectively.

The crystal-melt interface is denoted Γ S and is where T = T m, the melting temperature. Explicitly we denote the melting isotherm by z S(X, t) = 0, X Γ S. The motion of the interface of the phase transition is governed by the Stefan condition ρ s L S t = κ T s n q l, z S where L is the latent heat and q l is the heat flux from the melt. The speed S t above is the speed at which S moves in the direction of the outward normal n.

Basic equations after non-dimensionlization Define the Biot number by ɛ = h gs R κ s. where h gs is the mean value of h gs. ɛ St Θ t = 1 r (rθ r ) r + 1 r 2 Θ φφ + ɛθ zz, X Ω, t > 0; with boundary conditions Θ r + ɛr z Θ z + Θ φ R φ /R 2 = ɛf (Θ), 1 + ɛ(rz ) 2 + (R φ ) 2 /R2 r = R(φ, z), Θ = 1, z = S(r, Φ, t), Θ z = δ(θ Θ ch ), z = 0.

The right hand side of last equality is F (Θ) = h F (T 4 g T 4 b ) h gs T + Θ(β(z) + 4h F h gs T 3 g ) + h F h gs T Θ 2 (6T 2 g + 4T g T Θ + T 2 Θ 2 )).

Perturbation Solution Basic assumption: the Biot number ɛ( 0.03) for the lateral heat flux is small and the geometric anisotropy is weak. Zeroth order equation Θ Θ 0 (z, t) + ɛθ 1 (r, φ, z, t) + S S 0 (t) + ɛs 1 (r, φ, t) + 1 St Θ 0,t = Θ 0,zz L S F (Θ 0) + Θ 0,zS z, S 0 < z < S 0 (t), Θ 0,z = δ(θ 0 Θ ch ), z = 0, Θ 0 = 1, z = S 0 (t) γ + S 0,t = Θ 0,z z=s0 (t), S 0 (0) = Z seed,

Θ 1 (r, φ, z, t) Θ 0 1(z, t) + r 2 4 ( L S F (Θ 0) + Θ 0,zS z ) S m + α β k r n F (Θ k 0 ) n k A nk 1 (z) cos(n kφ + δ k ). k=1

If the temperature field is known, how to compute the stress (displacement)? Now we assume that the z-component of the displacement is zero. How to decide the equations with cubic anistropy? how to find the (Semi-)Analytic Solution?

3-D General elastic problem with cubic anisotropy For brevity, we write the thermoelastic equation as LU = F, (x, y, z) Ω with the boundary condition σ xx σ xy σ xz σ xyz n = σ xy σ yy σ yz σ xz σ yz σ zz n = g, (x, y, z) Ω where n = (n 1, n 2, n 3 ) is the unit outer normal direction. For the 3-D thermoelastic problem with cubic anisotropic, then F = (C 11 + 2C 12 ) Θ, g = (C 11 + 2C 12 )Θn.

Define H = 2C 44 C 11 + C 12, split C = C 0 C a, where the isotropic part C 0 = C 11 + H 2 C 12 C 12 C 12 C 11 + H 2 C 12 C 12 C 12 C 11 + H 2 C 44 H 4 C 44 H 4. Corresponding to the splitting, we have L = L 0 L a, σ xyz = σ 0,xyz σ a,xyz Define V = N G if V satisfies L 0 V = L a G, with boundary condition σ 0,xyz (V )n = σ a,xyz (G)n. C 44 H 4

We assume that U 0 is the solution to with boundary condition Define the anisotropic factor ω = L 0 U 0 = F, σ 0,xyz = g H/2 C 11 C 12 + H/2 = 2C 44 C 11 + C 12 2C 44 + C 11 C 12. We can get that N ω with some norm and the following expansion U = U 0 + N U 0 + N 2 U 0 + + N n U 0 +.

Set S n = n i=0 N i U 0, U S n ω n+1 U C 11 C 12 C 44 ω GaAs 12.16e4 5.43e4 6.18e4 0.295 InP 10.76e4 6.08e4 4.233e4 0.288 InSb 6.70e4 3.65e4 3.02e4 0.329

The splitting in (r, φ, z) coordinates The stress-stain relationship: C rφz = C 0 + C a,rφz For example [0,0,1] pulling direction, we choose the z-direction is [0,0,1], [1,0,0] and [0,1,0] are the directions related to φ = 0 and φ = π/2. C a,rφz is given by 1 4 H 1 + cos(4φ) 1 cos(4φ) 0 0 0 sin(4φ) 1 cos(4φ) 1 + cos(4φ) 0 0 0 sin(4φ) 0 0 2 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 sin(4φ) sin(4φ) 0 0 0 cos(4φ)

For [1,1,1] pulling direction, it is nature to choose the z-direction is [1,1,1], [1,1,-2] and [-1,1,0] are the directions related to φ = 0 and φ = π/2. C a,rφz is H 1 1 0 2 6 3 1 1 6 0 3 2 6 s 2 1 1 3 12 0 6 s 2 6 c 0 6 c 0 3 1 6 0 0 0 2 6 s 2 6 s 0 1 2 6 c 2 0 0 0 6 c 0 0 1 where c = cos(3φ) and s = sin(3φ) 12 2 6 c 2 6 s 2 6 c 2 6 s 1 12

Semi-analytic solution for 2-D problem Now we assume that the displacement is only in the (r, φ) plane. The previous 3-D problem reduce to 2-D problem. For [0,0,1] pulling direction σ a,rr σ a,φφ σ a,rφ = H 4 1 + cos(4φ) 1 cos(4φ) sin(4φ) 1 cos(4φ) 1 + cos(4φ) sin(4φ) sin(4φ) sin(4φ) cos(4φ) e rr e φφ 2e rφ For [1,1,1] pulling direction σ a,rr σ a,φφ σ a,rφ = H 1 0 6 0 1 6 0 0 0 0 1 12 e rr e φφ 2e rφ

Solution technique If we want to know the solution U U 0 + N U 0, we need to know the analytic solution to σ rr r + 1 r σ rφ φ + σ rr σ φφ r = d 1 r k 2 cos(nφ + δ), r < A(z), σ rφ r + 1 r σ φφ φ + 2σ rφ r with the boundary condition = d 2 r k 2 sin(nφ + δ), r < A(z), σ rr = d 3 cos(nφ + δ), σ rφ = d 4 sin(nφ + δ), r = A(z), r = A(z).

To find the solution to the equations with the boundary conditions, first we need to find a particular solution to the equations, like W = (W r, W φ ) = (c 1 r k cos(nφ + δ), c 2 r k sin(nφ + δ)) (in the case that k = n + 1 or k = n 1, need another particular solution like c 3 log r((r k cos(nφ + δ), γ 2 r k sin(nφ + δ)) where ((r k cos(nφ + δ), γ 2 r k sin(nφ + δ)) is a solution to homogeneous elasticity equation). Then consider the homogeneous equations with modified boundary conditions.

Resolved stress for [0,0,1] pulling direction Thermal stress with cubic anisotropy: [0,0,1], R(z)=A(z) x 10 3 6 1 0.8 5 0.6 0.4 4 0.2 0 3 0.2 0.4 2 0.6 0.8 1 1 1 0.5 0 0.5 1 0

Thermal stress with cubic anisotropic: [0,0,1], R(z,φ )=A(z)(1 0.05 cos (4φ )) 1 0.8 0.6 0.4 0.2 0 0.2 0.4 0.6 0.8 1 1.5 1 0.5 0 0.5 1 1.5 x 10 3 5 4.5 4 3.5 3 2.5 2 1.5 1 0.5 0

Resolved stress for [1,1,1] pulling direction Thermal stress with cubic anisotropy: [1,1,1], R(z)=A(z) x 10 3 4 1 0.8 3.5 0.6 3 0.4 0.2 2.5 0 2 0.2 0.4 0.6 1.5 1 0.8 0.5 1 1 0.5 0 0.5 1 0

x 10 3 1 0.8 0.6 0.4 0.2 0 0.2 0.4 0.6 Thermal stress with cubic anisotropic: [1,1,1], R(z,φ )=A(z)(1+0.05 cos (6φ )) 4 3.5 3 2.5 2 1.5 0.8 1 1.5 1 0.5 0 0.5 1 1.5 1 0.5 0

Conclusion and future work Conclusion: The anisotropy may not be ignored for accurate simulation. The closest isotropic part for elastic problem with cubic anisotropy is setting λ = C 12 and µ = 2C 44+C 11 C 12 4. The anisotropic factor ω evaluates the anisotropic effect from elastic constants. Future work: use the asymtopic sharp. The effect the other coefficients in the model.

Thank You!