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 Wu1 & Huaxiong Huang2

1Peking University, 2York University

BIRS: August 6, 2007

Constrained Growth The Thermal Problem Thermoelastic Equations Results Conclusions

Outline

1

Constrained Lateral Growth & Facet Formation Coordination Polyhedra Constrained Growth Equilibrium Crystal Shapes

2

The Thermal Problem Basic Equations Perturbation Solution

3

Thermoelastic Equations Basic Relations Operator Splitting Perturbation Series A Simplified Case: Plane Strain

4

Results Effect of Facets (Geometric) Effect of Material Anisotropy

5

Conclusions

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 AB6−

4

tetrahedra

Constrained Growth The Thermal Problem Thermoelastic Equations Results Conclusions

Coordination Polyhedra

i j k

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

i j k

For a crystal pulled in the [001] direction, [00¯ 1] is into the melt gives vaxial = 1.7321 vlateral has four-fold symmetry

Constrained Growth The Thermal Problem Thermoelastic Equations Results Conclusions

Constrained Growth

vn n ∂S k ∂t Crystal Melt Gas Meniscus r = R(z) vlateral∆t TP r = Rc θ θc vaxial∆t

If not constrained by the meniscus then tan(θ − θc) = vlateral

vaxial

For growing a cone θ − θc is 1/2 the opening angle of the cone

Constrained Growth The Thermal Problem Thermoelastic Equations Results Conclusions

Pulling in the [001] direction

100 200 300 1 1.5 2 2.5 3 θ Growth rate relative to faces Pull = [0 0 1]

• 2

2

• 3
• 2
• 1

1 2 3 Relative growth rate 100 200 300 20 40 60 80 θ Growth angle Growing at: θ =5 degrees Baserate =1.7321

• 2

2

• 3
• 2
• 1

1 2 3 Constrained rate

Constrained Growth The Thermal Problem Thermoelastic Equations Results Conclusions

Pulling in the [¯ 1¯ 1¯ 1] direction

100 200 300 1 1.5 2 2.5 3 θ Growth rate relative to faces Pull = [-1 -1 -1]

• 2

2

• 3
• 2
• 1

1 2 3 Relative growth rate 100 200 300 20 40 60 80 θ Growth angle Growing at: θ =5 degrees Baserate =3

• 2

2

• 3
• 2
• 1

1 2 3 Constrained rate

Constrained Growth The Thermal Problem Thermoelastic Equations Results Conclusions

Pulling in the [¯ 211] direction

100 200 300 1 1.5 2 2.5 3 θ Growth rate relative to faces Pull = [-2 1 1]

• 2

2

• 3
• 2
• 1

1 2 3 Relative growth rate 100 200 300 20 40 60 80 θ Growth angle Growing at: θ =5 degrees Baserate =2.1213

• 2

2

• 3
• 2
• 1

1 2 3 Constrained rate

Constrained Growth The Thermal Problem Thermoelastic Equations Results Conclusions

Equilibrium Crystal Shapes

• 2

2

• 3
• 2
• 1

1 2 3 Constrained vlateral(n) p = [001] a = [100] b = [010] vaxial = 1.7321

• 2

2

• 3
• 2
• 1

1 2 3 p = [111] a = [211] b = [011] vaxial = 3

• 2

2

• 3
• 2
• 1

1 2 3 p = [211] a = [111] b = [011] vaxial = 2.1213 Constrained vlateral(n) Constrained vlateral(n) ECS ECS ECS p = [001] p = [111] p = [211]

• 0.4

0.4

• 0.4

0.4

• 0.4

0.4

• 0.4

0.4

• 0.4

0.4

• 0.4

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) R(φ, z) = ¯ R(z)

• 1 + α

m

• k=1

βk cos (nkφ + δk)

• ,

where m, n1 < n2 < · · · < nm are positive integers and m

k=1 β2 k = 1.

α is the (small) geometric anisotropy factor 4-fold symmetry (m = 1, n1 = 4) 6-fold symmetry (m = 1, n1 = 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, ρscs ∂T ∂t = ∇ · (κs∇T) , x ∈ Ω, t > 0 where ρs, cs and ks are the density, specific heat, and thermal conductivity of the crystal. The boundary conditions are below, −κs ∂T ∂n = hgs(T − Tg) + hF(T 4 − T 4

b ),

x ∈ Γg, κs ∂T ∂z = hch(T − Tch), z = 0, where hgs and hch represent the heat transfer coefficients; hF the radiation heat transfer coefficient; Tg, Tch and Tb 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 ΓS and is where T = Tm, the melting temperature. Explicitly we denote the melting isotherm by z − S(x, t) = 0, x ∈ ΓS. The motion of the interface of the phase transition is governed by the Stefan condition ρsL|vn| = κs ∂T ∂n

• z→S− − ql,n,

|vn| = vn = ∂S ∂t k · n where L is the latent heat, |vn| is the speed of the interface in the direction of its outward normal n, and ql,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 ǫ = ¯ hgs ˜ R κs (1) as a small parameter (small lateral heat flux). Rescaling, ǫ StΘt = 1 r (rΘr)r + 1 r2 Θφφ + ǫΘzz, x ∈ Ω, t > 0, with, −Θr + 1 R2 RφΘφ + ǫRzΘz = ǫF(Θ)

• 1 +

R2

φ

R2 + ǫR2

z

1/2 , x ∈ Γg, Θz(0, φ, t) = δ (Θ(0, φ, t) − Θch) , Θ = 1, x ∈ ΓS, Θz − 1 ǫ SrΘr − 1 ǫr2 SφΘφ = γ + St, γ = ql ˜ R ǫ1/2κs∆T .

Constrained Growth The Thermal Problem Thermoelastic Equations Results Conclusions

Rescaled Equations

β(z) = hgs/¯ hgs, and δ = ǫ1/2hch/¯ hgs and γ (ql) is the non-dimensional (dimensional) heat flux in the liquid across the crystal/melt interface in the axial direction. Also, F(Θ) = hF(T 4

g − T 4 b )

¯ hgs∆T +

• β(z) + 4hF

¯ hgs T 3

g

• Θ

+ hF ¯ hgs ∆T(6T 2

g + 4Tg∆TΘ + ∆T 2Θ2)Θ2.

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 ∼ Θ0(z, t) + ǫΘ1(r, φ, z, t) + ǫ2Θ2(r, φ, z, t) + · · · , S ∼ S0(t) + ǫS1(r, φ, t) + ǫ2S2(r, φ, t) + · · · . Zeroth order model (Fast to compute): 1 StΘ0,t − Θ0,zz = 2 ¯ R ¯ R′Θ0,z − F(Θ0)

• ,

0 < z < S0(t), t > 0, Θ0,z(0, t) = δ(Θ0(0, t) − Θch), t ≥ 0, Θ0(S0(t), t) = 1, t ≥ 0, S′

0(t) = Θ0,z(S0(t), t) − γ,

S0(0) = Z0, t > 0.

Constrained Growth The Thermal Problem Thermoelastic Equations Results Conclusions

Perturbation Solution

First order model: Θ1(r, φ, z, t) = Θa

1(z, t) + r2Θb 1(z, t) + αΘc 1(r, φ, z, t) + O(α2)

where, keeping only those terms to O(α), Θb

1(z, t) = 1

2¯ R ¯ R′Θ0,z − F(Θ0)

• ,

Θc

1(r, φ, z, t) = ¯

RF(Θ0)

m

• k=1

βk nk r ¯ R nk cos(nkφ + δk). These last two terms are completely determined by Θ0 and ¯

• R. Θa

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 = (σxx, σyy, σzz, σyz, σxz, σxy)T and strains e = (exx, eyy, ezz, 2eyz, 2exz, 2exy)T are related through σ = Crecte, Crect =         C11 C12 C12 C12 C11 C12 C12 C12 C11 C44 C44 C44         . For an anisotropic material the quantity H = 2C44 − C11 + C12 = 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

2 4 6

• 6
• 4
• 2

2 4 6

• 6
• 4
• 2

2 4 6

Constrained Growth The Thermal Problem Thermoelastic Equations Results Conclusions

Operator Splitting

Split Crect into a diagonal anisotropic part and an isotropic part Crect = C0 − Ca,rect, Ca,rect = H/4 × diag(2, 2, 2, −1, −1, −1), and C0 =         C 0

11

C 0

12

C 0

12

C 0

12

C 0

11

C 0

12

C 0

12

C 0

12

C 0

11

C 0

44

C 0

44

C 0

44

        is isotropic. Ca,rect is chosen to minimize ρ(C −1

0 Ca,rect).

E and ν in term of Cij are given by E = (C11 + 2C12 + H/2)(C11 − C12 + H/2) C11 + C12 + H/2 , ν = C12 C11 + C12 + H/2.

Constrained Growth The Thermal Problem Thermoelastic Equations Results Conclusions

Operator Splitting

Denote the displacement vector as w, the strain by e = S(w) and the stress by σ = CS(w) with C = C0 − Ca. The thermoelastic problem becomes ∇ · CS = (C11 + 2C12)∇Θ, x ∈ Ω, t > 0, CS · n = (C11 + 2C12)Θn, r = R(φ, z)

• r by rescaling

∇ · CS = 1 − ν 1 − 2ν − H 2

• ∇Θ,

x ∈ Ω, t > 0, CS · n = 1 − ν 1 − 2ν − H 2

• Θn,

r = R(φ, z) with n denoting the outward normal of the surface r = R(φ, z).

Constrained Growth The Thermal Problem Thermoelastic Equations Results Conclusions

Operator Splitting

Using the form of C, ∇ · CS = ∇ · C0S − ∇ · CaS = L0 − La, CS · n = C0S · n − CaS · n = B0 − Ba, to solve for w(x) one starts with w0 given by L0(w0) = 1 − ν 1 − 2ν − H 2

• ∇Θ,

x ∈ Ω, t > 0, B0(w0) = 1 − ν 1 − 2ν − H 2

• Θn,

r = R(φ, z). w0 is the isotropic displacement found previously [Bohun et al.], multiplied by a factor of 1 − H

2 1−2ν 1−ν .

Constrained Growth The Thermal Problem Thermoelastic Equations Results Conclusions

Operator Splitting

We know w0 explicitly for a given crystal shape R(φ, z). Having defined w0, we denote by wk+1 = Nwk, with k ≥ 0, the solution to L0(wk+1) = La(wk), x ∈ Ω, t > 0, B0(wk+1) = Ba(wk), r = R(φ, z).

Constrained Growth The Thermal Problem Thermoelastic Equations Results Conclusions

Perturbation Series

Continuing this process we have for w(x) w = w0 + Nw0 + N 2w0 + · · · + N nw0 + · · · . Since N ≤ ω in a suitable norm, where ω = |H|/2 C11 − C12 + H/2 = |2C44 − C11 + C12| 2C44 + C11 − C12 < 1 is an anisotropic factor, the series converges and an error can be estimated when replaced by a finite sum. For typical cubic anisotropic materials ω ∼ 1/3. C11 C12 C44 ω GaAs 12.16 × 104 5.43 × 104 6.18 × 104 0.295 InP 10.76 × 104 6.08 × 104 4.233 × 104 0.288 InSb 6.70 × 104 3.65 × 104 3.02 × 104 0.329

Constrained Growth The Thermal Problem Thermoelastic Equations Results Conclusions

Perturbation Series

For a given pulling direction C0 is invariant however, the explicit form of Ca depends on the crystal orientation Consequently La and Ba depend on the orientation Ca transforms as a fourth rank tensor and includes only trigonometric factors cos mφ and sin mφ where m depends on the orientation of the crystal For example, if (c4, s4) = (cos 4φ, sin 4φ) then C [001]

a,cyc = H

4         1 + c4 1 − c4 −s4 1 − c4 1 + c4 s4 2 −1 −1 −s4 s4 −c4         .

Constrained Growth The Thermal Problem Thermoelastic Equations Results Conclusions

Plane Strain

To illustrate the procedure assume that the displacement is only in the (r, φ) plane. Stress strain relation for the [001] direction becomes   σa,rr σa,φφ σa,rφ   = H 4   1 + c4 1 − c4 −s4 1 − c4 1 + c4 s4 −s4 s4 −c4     err eφφ 2erφ   . For the [¯ 1¯ 1¯ 1] direction   σa,rr σa,φφ σa,rφ   = H 12   2 2 −1     err eφφ 2erφ   .

Constrained Growth The Thermal Problem Thermoelastic Equations Results Conclusions

A Canonical Problem

To find w0 + w1 = w0 + Nw0 the thermoelastic equations L0(w1) = La(w0), x ∈ Ω, t > 0, B0(w1) = Ba(w0), r = R(φ, z) reduce to finding sequence of solutions of the form ∂σrr ∂r + 1 r ∂σrφ ∂φ + σrr − σφφ r = frrk−2 cos(nφ + δ), r < ¯ R(z), ∂σrφ ∂r + 1 r ∂σφφ ∂φ + 2σrφ r = fφrk−2 sin(nφ + δ), r < ¯ R(z), with integers n ≥ 0, k ≥ 1, and σrr = grrk−1 cos(nφ + δ), r = ¯ R(z), σrφ = gφrk−1 sin(nφ + δ), r = ¯ R(z), where fr, fφ, gr, gφ depend on Ca.

Constrained Growth The Thermal Problem Thermoelastic Equations Results Conclusions

A Canonical Problem

We solve this with a two stage approach.

1

Find a particular solution that does not necessarily satisfy the boundary condition

2

Find a homogeneous solution with a (perhaps) modified boundary condition The point here is that the solution can be written out explicitly for general fr, fφ, gr, gφ so that the problem becomes a bookkeeping problem. Fast

Constrained Growth The Thermal Problem Thermoelastic Equations Results Conclusions

Results - Geometric [001]: Total Resolved Stress

4 40 40 4 40 4 4 40 40 40 40 50 5 5 5 50 5 60 6 6 60 60 60 30 30 30 30 30 30 30 30 50 5 70 70 70 60 60 8 8 80 70 70 80 80 90 9 9 90 90 20 20 20 20 −1 −0.5 0.5 1 −1 −0.8 −0.6 −0.4 −0.2 0.2 0.4 0.6 0.8 1 40 40 40 40 40 40 4 40 40 40 40 50 50 50 5 50 5 60 60 60 6 6 60 3 30 30 30 30 30 30 30 50 50 7 70 70 6 6 8 80 8 70 70 80 80 90 90 9 9 9 20 20 2 2 −1 −0.5 0.5 1 −1 −0.8 −0.6 −0.4 −0.2 0.2 0.4 0.6 0.8 1

α = 0: max |σtot

rs | = 9.23

α = 0: max |σtot

rs | = 9.23

Constrained Growth The Thermal Problem Thermoelastic Equations Results Conclusions

Results - Geometric [¯ 1¯ 1¯ 1]: Total Resolved Stress

2 20 20 20 20 20 20 20 20 30 30 30 3 40 40 40 40 40 50 50 50 50 50 50 50 60 60 60 60 60 60 60 −1 −0.5 0.5 1 −1 −0.8 −0.6 −0.4 −0.2 0.2 0.4 0.6 0.8 1 20 20 20 20 2 20 30 30 3 30 30 30 3 30 3 3 30 40 40 40 40 40 40 4 4 40 40 50 50 50 50 50 50 50 50 5 50 60 60 60 60 60 6 60 60 60 60 20 2 20 20 2 2 2 20 20 3 3 3 30 70 70 70 70 70 70 3 30 80 80 80 80 70 70 70 70 90 90 90 90 100 100 100 100 80 8 90 90 1 1 110 1 1 110 110 110 1 1 120 120 120 120 120 120 1 3 130 130 1 3 130 130 −1 −0.5 0.5 1 1.5 −1 −0.8 −0.6 −0.4 −0.2 0.2 0.4 0.6 0.8 1

α = 0: max |σtot

rs | = 6.07

α = 0.123: max |σtot

rs | = 13.4

Constrained Growth The Thermal Problem Thermoelastic Equations Results Conclusions

Results - Geometric [¯ 211]: Total Resolved Stress

20 30 30 30 30 30 30 30 30 30 30 40 40 40 40 40 5 50 50 5 50 50 50 5 6 60 60 60 70 7 70 70 2 20 20 80 80 80 8 −1 −0.5 0.5 1 −1 −0.8 −0.6 −0.4 −0.2 0.2 0.4 0.6 0.8 1 20 20 20 20 20 20 30 30 30 30 30 30 3 3 30 30 3 30 4 40 40 40 40 40 40 40 40 40 4 40 50 50 50 50 50 50 3 30 30 60 60 60 60 6 60 50 50 50 20 20 20 20 70 70 70 70 80 80 80 80 6 60 70 70 90 90 −1 −0.5 0.5 1 −1 −0.8 −0.6 −0.4 −0.2 0.2 0.4 0.6 0.8 1

α = 0: max |σtot

rs | = 8.19

α = 0.089: max |σtot

rs | = 8.78

Constrained Growth The Thermal Problem Thermoelastic Equations Results Conclusions

Results - Anisotropy [001]: Total Resolved Stress

40 40 40 40 40 40 4 40 40 40 40 50 50 50 5 50 5 60 60 60 6 6 60 3 30 30 30 30 30 30 30 50 50 7 70 70 6 6 8 80 8 70 70 80 80 90 90 9 9 9 20 20 2 2 −1 −0.5 0.5 1 −1 −0.8 −0.6 −0.4 −0.2 0.2 0.4 0.6 0.8 1 20 20 20 20 30 3 30 30 30 30 30 30 3 30 30 30 30 30 40 40 40 4 40 40 40 40 40 50 50 50 50 50 50 60 6 60 60 60 6 5 50 60 60 20 20 70 7 70 20 2 70 70 80 80 8 −1 −0.5 0.5 1 −1 −0.8 −0.6 −0.4 −0.2 0.2 0.4 0.6 0.8 1

ω = 0: max |σtot

rs | = 9.23

ω = 0.329: max |σtot

rs | = 7.66

Constrained Growth The Thermal Problem Thermoelastic Equations Results Conclusions

Results - Anisotropy [¯ 1¯ 1¯ 1]: Total Resolved Stress

20 20 20 20 2 20 30 30 3 30 30 30 3 30 3 3 30 40 40 40 40 40 40 4 4 40 40 50 50 50 50 50 50 50 50 5 50 60 60 60 60 60 6 60 60 60 60 20 2 20 20 2 2 2 20 20 3 3 3 30 70 70 70 70 70 70 3 30 80 80 80 80 70 70 70 70 90 90 90 90 100 100 100 100 80 8 90 90 1 1 110 1 1 110 110 110 1 1 120 120 120 120 120 120 1 3 130 130 1 3 130 130 −1 −0.5 0.5 1 1.5 −1 −0.8 −0.6 −0.4 −0.2 0.2 0.4 0.6 0.8 1 20 20 20 20 20 20 30 30 3 30 30 3 30 3 3 30 30 30 40 40 40 40 40 40 40 40 40 40 5 50 50 5 50 50 50 50 50 50 60 60 60 60 60 60 60 60 6 60 20 20 20 20 20 20 70 7 7 7 80 80 80 80 70 70 90 9 90 90 8 80 100 100 100 100 90 9 100 100 70 7 70 70 7 70 110 110 110 110 110 1 1 3 3 −1 −0.5 0.5 1 1.5 −1 −0.8 −0.6 −0.4 −0.2 0.2 0.4 0.6 0.8 1

ω = 0: max |σtot

rs | = 13.4

ω = 0.329: max |σtot

rs | = 12.0

Constrained Growth The Thermal Problem Thermoelastic Equations Results Conclusions

Results - Anisotropy [¯ 211]: Total Resolved Stress

20 20 20 20 20 20 30 30 30 30 30 30 3 3 30 30 3 30 4 40 40 40 40 40 40 40 40 40 4 40 50 50 50 50 50 50 3 30 30 60 60 60 60 6 60 50 50 50 20 20 20 20 70 70 70 70 80 80 80 80 6 60 70 70 90 90 −1 −0.5 0.5 1 −1 −0.8 −0.6 −0.4 −0.2 0.2 0.4 0.6 0.8 1 2 2 20 20 20 20 3 30 3 30 30 30 30 30 30 30 3 40 40 40 40 40 40 40 40 50 50 50 50 5 50 60 60 6 60 60 60 30 3 50 50 50 70 70 70 70 20 2 20 20 80 8 80 80 60 60 90 90 70 7 −1 −0.5 0.5 1 −1 −0.8 −0.6 −0.4 −0.2 0.2 0.4 0.6 0.8 1

ω = 0: max |σtot

rs | = 8.19

ω = 0.329: max |σtot

rs | = 8.78

Constrained Growth The Thermal Problem Thermoelastic Equations Results Conclusions

Conclusions

A simple argument based on the crystal lattice structure predicts facets that depend on both the crystal orientation and growth angle Small opening angles tend to suppress the formation of facets The model naturally incorporates the polarity of III-V semiconductors Facet formation greatly affects the thermal stress distribution Anisotropy has a lesser effect when the crystal has facets The industry preference of the [¯ 211] pulling direction, determined by trial and error, produces facets yet avoids the drastic increase in the stress seen in the [¯ 1¯ 1¯ 1] orientation. Furthermore, effect of the material anisotropy is negligible in this case

Constrained Growth The Thermal Problem Thermoelastic Equations Results Conclusions

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