1 EE143 N.Cheung Notes on CVD Kinetics • Deposition depends on the sequence of events: (1) Diffusion of Reactants to surface (2) Absorption of Reactants at surface (3) Chemical reaction at surface (4) Desorption of products from surface (5) Diffusion of products from surface. * (1) and (5) are mass-transfer mechanisms. • The slowest event will be the rate- determing step. (1) Temperature Dependence of Deposition rate. (i) • From the above Si deposition example, for low T, rate R ∝ e- ∆E/kT (Surface Reaction limited) where ∆E = 25 - 100 kcal/mole for most CVD processes ( or 1- 4 eV/atom). (ii) For high temperature, R ∝ Tn (1.5 < n (experimental) < 2.0) (Mass Transport limited)2 For mass-transport limited reactions, R ∝ D (diffusion constant) and D ∝ T3/2/ P for gaseous diffusion. Proof: F = n1v4 - n2v4 = v4 (λ ⋅ dndx) where λ = mean free path of gas collision. F = Ddndx ⇒ D = λ v2 Since λ ∝ kTP and v ∝ T ∴ D ∝ T3/2/P (2) Growth Rate Model [Note] This model is a special case of the Grove-Deal model for thermal oxidation. Here , we don’t have to consider the diffusion througfh the oxide layer. In this model, fluxes of products are ignored (i.e., their mass-transfer coefficients >> those of reactant). Let F1 = flux from bulk of gas to substrate surface. ≡ hG ⋅ (CG - CS) where hG = mass-transfer coefficient. CG, CS = reactant conc. at bulk of gas and substrate surface respectively. Let F2 = flux consumed in film-growth reaction. = kS ⋅ CS where kS = surface-reaction rate coefficient. Steady state ⇒ F1 = F2 = F ∴ CS = CG1 + kS/hG and F = kShGkS + hG ⋅ CG3 Since CG = Y ⋅ CT where Y = mole fraction of reactant. CT = total # of molecules/cm3 in gas mixture. and dydt (thickness growth rate) = Fρ where ρ = atomic density of film. ∴ dydt = kShGkS + hG ⋅ CTρ ⋅ Y = 1(1ks + 1hG) ⋅ CTρ ⋅ Y Comments: (a) dydt is determined by the smaller of hG and kS (i.e., mass-transfer control or surface-reaction control). (b) dydt ∝ Y (mole fraction of reactant in bulk of gas mixture)4 (3) Boundary Layer Theory for Stagnant Gas Layer The boundary layer thickness δ(x) is shown in the figure below and L is the length of the substrate (e.g. substrate or wall of reactor). The gas velocity u is a function of x and y and is equal to zero at plate’s surface and is equal to U in the free gas stream. Let µ = viscosity of gas. Then frictional force / unit area along the x-direction = µ × ∂u∂y Let us consider a volume element of unit depth (i.e., into the paper) , height δ(x) and width dx. Total friction force on element = µ × ∂u∂y × (1 × dx) = µ ∂u∂y dx = decelerating force Total accelerating force on element. = ρ × δ(x)dx × dudt = ρ × δ(x)dx × dudx × dxdt = ρ × δ(x)dx × dudx × u where ρ is the gas mass density Balanced forces : µ ∂u∂y = ρ × δ(x)ududx and u(x,y) can be solved exactly. Approximate Solutions Let u » U ; ∂u∂y ≈ Uδ(x) ; ∂u∂y ≈ Ux then δ(x) ≈ A µxρU1/2 - B “parabolic dependence’ where A,B are constants. ∴ The “average” boundary layer thickness δ = 1L ⌡⌠0 L δ(x) dx = 23LρULµ = 23LReL 5 ReL is called the Reynold Number of the reactor. When ReL is small (≤ 2000) , viscous flow dominates. When ReL very large (≥ 2000) , turbulent flow dominates. The Exact Solutions: The stagnant layer thickness with u= 0.99U is equal to : δ(x) ≈ 5.0 µxU1/2 See H. Blasius, NACA Tech. Mem., 1949, p. 1217. In the CVD growth rate model, it was assumed that mass transport across stagnant layer proceeds by diffusion, then. F1 ≡ DG ⋅ CG - CS δ ⇒ hG = DGδ where DG = diffusivity For mass-transfer limited deposition, model ⇒ dydt ∝ hG ∝ 1 δ ∝
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