Rapid Computation of thermal stress in crystals with facets and - PowerPoint PPT Presentation

Constrained Growth The Thermal Problem Thermoelastic Equations Results Conclusions Rapid Computation of thermal stress in crystals with facets and allowing for material anisotropy Canada-China Workshop on Industrial Mathematics C. Sean Bohun, Faculty of Science, University of Ontario Institute of Technology Joint work with Jinbiao Wu 1 & Huaxiong Huang 2 1 Peking University, 2 York University BIRS: August 6, 2007

Constrained Growth The Thermal Problem Thermoelastic Equations Results Conclusions Outline Constrained Lateral Growth & Facet Formation 1 Coordination Polyhedra Constrained Growth Equilibrium Crystal Shapes The Thermal Problem 2 Basic Equations Perturbation Solution Thermoelastic Equations 3 Basic Relations Operator Splitting Perturbation Series A Simplified Case: Plane Strain Results 4 Effect of Facets (Geometric) Effect of Material Anisotropy Conclusions 5

Constrained Growth The Thermal Problem Thermoelastic Equations Results Conclusions Coordination Polyhedra The growth rate is based on a coordinate polyhedron model This is capable of naturally explaining the different growth rates between the positive and negative directions in a polar crystal such as the III-V semiconductors If AB is the III-V semiconductor under consideration, then its anion-coordination polyhedra are AB 6 − tetrahedra 4

Constrained Growth The Thermal Problem Thermoelastic Equations Results Conclusions Coordination Polyhedra k j i Shown are the four tetrahedra of an AB unit cell. To the left only the B atoms in the unit cell are shown. B atoms in the unit cell but not included in the four growth units are represented with hollow circles. At the centre of each tetrahedral growth unit is a A atom accounting for all the atoms in the AB unit cell. On the right only the tetrahedra are shown.

Constrained Growth The Thermal Problem Thermoelastic Equations Results Conclusions Coordination Polyhedra k j i For a crystal pulled in the [001] direction, [00¯ 1] is into the melt gives v axial = 1 . 7321 v lateral has four-fold symmetry

Constrained Growth The Thermal Problem Thermoelastic Equations Results Conclusions Constrained Growth Crystal r = R ( z ) v n n ∂ S k ∂ t TP Gas Melt Meniscus v axial ∆ t θ r = R c θ c v lateral ∆ t If not constrained by the meniscus then tan( θ − θ c ) = v lateral v axial For growing a cone θ − θ c is 1 / 2 the opening angle of the cone

Constrained Growth The Thermal Problem Thermoelastic Equations Results Conclusions Pulling in the [001] direction Pull = [0 0 1] Relative growth rate 3 3 Growth rate relative to faces 2 2.5 1 2 0 -1 1.5 -2 1 -3 0 100 200 300 -2 0 2 θ Baserate =1.7321 Constrained rate 3 80 Growing at: θ =5 degrees 2 60 1 Growth angle 0 40 -1 20 -2 0 -3 0 100 200 300 -2 0 2 θ

Constrained Growth The Thermal Problem Thermoelastic Equations Results Conclusions Pulling in the [¯ 1¯ 1¯ 1] direction Pull = [-1 -1 -1] Relative growth rate 3 3 Growth rate relative to faces 2 2.5 1 2 0 -1 1.5 -2 1 -3 0 100 200 300 -2 0 2 θ Baserate =3 Constrained rate 3 80 Growing at: θ =5 degrees 2 60 1 Growth angle 0 40 -1 20 -2 0 -3 0 100 200 300 -2 0 2 θ

Constrained Growth The Thermal Problem Thermoelastic Equations Results Conclusions Pulling in the [¯ 211] direction Pull = [-2 1 1] Relative growth rate 3 3 Growth rate relative to faces 2 2.5 1 2 0 -1 1.5 -2 1 -3 0 100 200 300 -2 0 2 θ Baserate =2.1213 Constrained rate 3 80 Growing at: θ =5 degrees 2 60 1 Growth angle 0 40 -1 20 -2 0 -3 0 100 200 300 -2 0 2 θ

Constrained Growth The Thermal Problem Thermoelastic Equations Results Conclusions Equilibrium Crystal Shapes Constrained v lateral ( n ) Constrained v lateral ( n ) Constrained v lateral ( n ) 3 3 3 p = [001] p = [111] p = [211] 2 2 2 a = [100] a = [211] a = [111] b = [010] b = [011] b = [011] 1 1 1 0 0 0 -1 -1 -1 -2 -2 -2 v axial = 1.7321 v axial = 3 v axial = 2.1213 -3 -3 -3 -2 0 2 -2 0 2 -2 0 2 ECS ECS ECS p = [001] p = [111] p = [211] 0.4 0.4 0.4 0 0 0 -0.4 -0.4 -0.4 -0.4 0 0.4 -0.4 0 0.4 -0.4 0 0.4

Constrained Growth The Thermal Problem Thermoelastic Equations Results Conclusions Equilibrium Crystal Shapes For the purpose of computing thermal stress, we assume the following expression in the case of weak anisotropy ( α small) � m � R ( φ, z ) = ¯ � R ( z ) 1 + α β k cos ( n k φ + δ k ) , k =1 where m , n 1 < n 2 < · · · < n m are positive integers and � m k =1 β 2 k = 1. α is the (small) geometric anisotropy factor 4-fold symmetry ( m = 1 , n 1 = 4) 6-fold symmetry ( m = 1 , n 1 = 6) We assume that the lateral shape of the crystal is in equilibrium

Constrained Growth The Thermal Problem Thermoelastic Equations Results Conclusions Basic Equations Within the crystal Ω, the temperature T ( x , t ) satisfies the heat equation, ∂ T ρ s c s ∂ t = ∇ · ( κ s ∇ T ) , x ∈ 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, ∂ T ∂ n = h gs ( T − T g ) + h F ( T 4 − T 4 − κ s x ∈ Γ g , b ) , ∂ T κ s ∂ 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 coefficient; T g , T ch and T b denote the ambient gas temperature, the chuck temperature and background temperature respectively.

Constrained Growth The Thermal Problem Thermoelastic Equations Results Conclusions Basic Equations 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 � ∂ T | v n | = v n = ∂ S � ρ s L | v n | = κ s z → S − − q l , n , ∂ t k · n � ∂ n � where L is the latent heat, | v n | is the speed of the interface in the direction of its outward normal n , and q l , n is the heat flux from the melt normal to the interface. The speed ∂ S /∂ t is the speed of the interface S in the k direction.

Constrained Growth The Thermal Problem Thermoelastic Equations Results Conclusions Rescaled Equations Identify the Biot number ¯ h gs ˜ R ǫ = (1) κ s as a small parameter (small lateral heat flux). Rescaling, St Θ t = 1 ǫ r ( r Θ r ) r + 1 r 2 Θ φφ + ǫ Θ zz , x ∈ Ω , t > 0 , with, � 1 / 2 � R 2 − Θ r + 1 φ R 2 + ǫ R 2 x ∈ Γ g , R 2 R φ Θ φ + ǫ R z Θ z = ǫ F (Θ) 1 + , z Θ z (0 , φ, t ) = δ (Θ(0 , φ, t ) − Θ ch ) , Θ = 1 , x ∈ Γ S , q l ˜ Θ z − 1 ǫ S r Θ r − 1 R ǫ r 2 S φ Θ φ = γ + S t , γ = ǫ 1 / 2 κ s ∆ T .

Constrained Growth The Thermal Problem Thermoelastic Equations Results Conclusions Rescaled Equations β ( z ) = h gs / ¯ h gs , and δ = ǫ 1 / 2 h ch / ¯ h gs and γ ( q l ) is the non-dimensional (dimensional) heat flux in the liquid across the crystal/melt interface in the axial direction. Also, F (Θ) = h F ( T 4 g − T 4 b ) � β ( z ) + 4 h F � T 3 + Θ ¯ ¯ g h gs ∆ T h gs + h F ∆ T (6 T 2 g + 4 T g ∆ T Θ + ∆ T 2 Θ 2 )Θ 2 . ¯ h gs

Constrained Growth The Thermal Problem Thermoelastic Equations Results Conclusions Perturbation Solution The Biot number for the lateral heat flux is small ( ǫ ∼ 0 . 03) and the geometric anisotropy is weak ( α ≪ 1). Expansion: Θ ∼ Θ 0 ( z , t ) + ǫ Θ 1 ( r , φ, z , t ) + ǫ 2 Θ 2 ( r , φ, z , t ) + · · · , S ∼ S 0 ( t ) + ǫ S 1 ( r , φ, t ) + ǫ 2 S 2 ( r , φ, t ) + · · · . Zeroth order model (Fast to compute): St Θ 0 , t − Θ 0 , zz = 2 1 � ¯ � R ′ Θ 0 , z − F (Θ 0 ) , 0 < z < S 0 ( t ) , t > 0 , ¯ R Θ 0 , z (0 , t ) = δ (Θ 0 (0 , t ) − Θ ch ) , t ≥ 0 , Θ 0 ( S 0 ( t ) , t ) = 1 , t ≥ 0 , S ′ 0 ( t ) = Θ 0 , z ( S 0 ( t ) , t ) − γ, S 0 (0) = Z 0 , t > 0 .

Constrained Growth The Thermal Problem Thermoelastic Equations Results Conclusions Perturbation Solution First order model: Θ 1 ( r , φ, z , t ) = Θ a 1 ( z , t ) + r 2 Θ b 1 ( z , t ) + α Θ c 1 ( r , φ, z , t ) + O ( α 2 ) where, keeping only those terms to O ( α ), 1 ( z , t ) = 1 � ¯ Θ b R ′ Θ 0 , z − F (Θ 0 ) � , 2¯ R m � r β k � n k cos( n k φ + δ k ) . Θ c 1 ( r , φ, z , t ) = ¯ � RF (Θ 0 ) ¯ n k R k =1 These last two terms are completely determined by Θ 0 and ¯ R . Θ a 1 does not play a role in the stress.

Constrained Growth The Thermal Problem Thermoelastic Equations Results Conclusions Basic Relations For a crystal with cubic symmetry the stresses σ = ( σ xx , σ yy , σ zz , σ yz , σ xz , σ xy ) T and strains e = ( e xx , e yy , e zz , 2 e yz , 2 e xz , 2 e xy ) T are related through  C 11 C 12 C 12  C 12 C 11 C 12     C 12 C 12 C 11   σ = C rect e , C rect = .   C 44     C 44   C 44 For an anisotropic material the quantity H = 2 C 44 − C 11 + C 12 � = 0. We assume that the z -component of the displacement is zero because of the free surface at the melt.

Constrained Growth The Thermal Problem Thermoelastic Equations Results Conclusions Directional Dependence of the Young’s modulus for an InSb Crystal 6 4 2 0 -2 -4 -6 6 4 6 2 4 0 2 0 -2 -2 -4 -4 -6 -6