Chapter 5: Diffusion Outline Introduction Diffusion mechanisms Steady-state diffusion Nonsteady-state diffusion Factors that influence diffusion Introduction Diffusion: the phenomenon of material transport by atomic motion Interdiffusion: In an alloy, atoms tend to migrate from regions of high concentration. Initially After some time Adapted from Figs. 5.1 and 5., Callister 6e. 100% Cu Ni 100% 0 Concentration Profiles 0 Concentration Profiles 1
Diffusion phenomenon () Self-diffusion: In an elemental solid, atoms also migrate. Label some atoms C A D B After some time C D A B Diffusion mechanisms Vacancy diffusion Interstitial diffusion small atoms Interstitial is more rapid than vacancy diffusion
Processing using diffusion Case Hardening: --Diffuse carbon atoms into the host iron atoms at the surface. --Example of interstitial diffusion is a case hardened gear. The presence of C atoms makes iron (steel) harder. Processing using diffusion Doping silicon with phosphorus for n-type semiconductors: Process: 0.5mm 1. Deposit P rich layers on surface.. Heat it. silicon 3. Result: Doped semiconductor regions. magnified image of a computer chip Dark regions: Si atoms silicon light regions: Al atoms 3
Steady-state diffusion Diffusion flux J Flux moles (or mass) diffusing mol or kg ( surface area)( time) cm s m s = J = M At Steady-state diffusion: the diffusion does not change with time Concentration profile Flux vs concentration profile 4
Steady-state diffusion (continue) Concentration gradient=dc/dx Linear concentration gradient Driving force = Δ C Δx = CA C x x A B B Examples (steady state) Steel plate at 700 0 C with geometry shown: Q: How much carbon transfers from the rich to the deficient side? Carbon rich gas 0 C 1 = 1.kg/m3 x1 x 5mm C = 0.8kg/m 3 10mm Steady State = straight line! Carbon deficient gas D=3x10-11 m /s J = D C C 1 x x 1 =.4 10 9 kg m s 5
Nonsteady-state diffusion Fick s second law Non-steady state diffusion The concentration of diffucing species is a function of both time and position C = C (x,t) In this case Fick s Second Law is used Fick s Second Law C t C = D x 6
Non-steady state diffusion Copper diffuses into a bar of aluminum. Surface conc., C of Cu atoms bar s pre-existing conc., C o of copper atoms Cs B.C. at t = 0, C = C o for 0 x at t > 0, C = C S for x = 0 (const. surf. conc.) C = C o for x = Solution ( x,t ) C C s C C o o = 1 erf x Dt C(x,t ) = Conc. at point x at time t erf (z) = error function = e y 0 π z dy erf(z) values are given in Table 5.1 C S C(x,t) C o 7
Nonsteady-state diffusion Examples (non-steady state) Sample Problem: An FCC iron-carbon alloy initially containing 0.0 wt% C is carburized at an elevated temperature and in an atmosphere that gives a surface carbon concentration constant at 1.0 wt%. If after 49.5 h the concentration of carbon is 0.35 wt% at a position 4.0 mm below the surface, determine the temperature at which the treatment was carried out. Solution: use Eqn. 5.5 C( x, t) C C C s o o = 1 erf x Dt 8
Solution (cont.) C( x,t ) C C C s o o = 1 erf x Dt t = 49.5 h Cx = 0.35 wt% Co = 0.0 wt% x = 4 x 10-3 m Cs = 1.0 wt% C( x, t) Co Cs Co 0.35 0.0 = = 1 erf 1.0 0.0 x Dt = 1 erf( z) erf(z) = 0.815 Solution (cont.) We must now determine from Table 5.1 the value of z for which the error function is 0.815. An interpolation is necessary as follows z erf(z) 0.90 0.7970 z 0.815 0.95 0.809 z 0.90 0.815 0.7970 = 0.95 0.90 0.809 0.7970 z = 0.93 Now solve for D z = x Dt x D = 4z t 3 x (4 x 10 m) D = = 4 z t (4)(0.93) (49.5 h) 1h 3600 s 11 =.6 x 10 m /s 9
Factors that influence diffusion Diffusing species Diffusion and temperature Diffusion coefficient increases with increasing T D = D o exp Q d RT D D o Q d R T = diffusion coefficient [m /s] = pre-exponential [m /s] = activation energy [J/mol or ev/atom] = gas constant [8.314 J/mol-K] = absolute temperature [K] 10
Diffusion and temperature D has exponential dependence on T D (m /s) 10-8 T( C) 10-14 1500 1000 C in γ-fe Fe in γ-fe Fe in α-fe 600 C in α-fe Al in Al 300 Dinterstitial >> Dsubstitutional C in α-fe C in γ-fe Al in Al Fe in α-fe Fe in γ-fe 10-0 0.5 1.0 1.5 1000 K/T Example: At 300ºC the diffusion coefficient and activation energy for Cu in Si are D(300ºC) = 7.8 x 10-11 m /s Q d = 41.5 kj/mol What is the diffusion coefficient at 350ºC? D transform data ln D Temp = T 1/T Q d 1 lnd = lnd0 and lnd1 = lnd0 R T D Qd 1 1 lnd = = lnd1 ln D1 R T T1 Qd R 1 T1 11
Example (cont.) D = D 1 Qd exp R 1 T 1 T1 T 1 = 73 + 300 = 573K T = 73 + 350 = 63K D 11 41,500 J/mol = (7.8 x 10 m /s) exp 8.314 J/mol - K 1 63 K 1 573 K D = 15.7 x 10-11 m /s Summary Diffusion FASTER for... open crystal structures materials w/secondary bonding smaller diffusing atoms lower density materials Diffusion SLOWER for... close-packed structures materials w/covalent bonding larger diffusing atoms higher density materials 1