Diffusion Diffusion = the spontaneous intermingling of the particles of two or more substances as a result of random thermal motion
Fick s First Law Γ ΔN AΔt Γ = flux ΔN = number of particles crossing cross sectional area A in time Δt Γ = D dc dx C = concentration D = diffusion coefficient Diffusion: Concepts and Solved Problems ( S.O. Kasap, 1990-001: v.1.1) An e-booklet
Diffusion Coefficient D = D o exp E A kt or D Q = D exp diff o RT D = diffusion coefficient D O = constant E A or Q diff = activation energy k = Boltzmann constant, R = gas constant T = temperature [E A ] =ev k = 8.617 10-5 ev K -1 [Q diff ] = kj mol -1 R = 8.314 10-5 J K -1 mol -1
Diffusion of impurities in crystals U = PE(x) U A* E A U A = U B A A* B A A* B X Displacement Fig. 1.8: Diffusion of an interstitial impurity atom in a crystal from one void to a neighboring void. The impurity atom at position A must posses an energy E A to push the host atoms away and move into the neighboring void at B. Boltzmann s distribution n E N E =C exp kt n E = number of impurities with energy E N = total number of particles T = temperature k = Boltzmann s constant From Principles of Electronic Materials and Devices, Second Edition, S.O. Kasap ( McGraw-Hill, 00) http://materials.usask.ca Probability (E>E A ) = Number of impurities with E>E A Total number of impurities υ Frequency of jumps = = = Frequency of attempt along AB Probability (E>E A ) E A n E EA de = Aexp( ) kt E A = activation energy A = dimensionless constant EA = Aν 0 exp( ) kt
Random walk of impurity atom inside crystalline lattice and root mean square (rms) displacement θ = 180 θ = 90 θ = 70 θ = 0 y x a O X L O' After N jumps Fig. 1.9: An impurity atom has four site choices for diffusion to a neighboring interstitial vacancy. After N jumps, the impurity atom would have been displaced from the original position at O. From Principles of Electronic Materials and Devices, Second Edition, S.O. Kasap ( McGraw-Hill, 00) http://materials.usask.ca Y Total square displacement in X-direction after N jumps X X = a = 1 cos a L N ( θ ) + a cos ( θ ) +... + a = 1 X + Y Y = a = 1 N a cos Total square displacement after N jumps However, N = υt Therefore L = a υt Dt N ( θ ) where υ = frequency of jumps N
Mean Square Displacement L = a ϑt = Dt ϑ = Aν exp( 0 EA ) kt L = distance, a = closest void to void separation, ϑ = frequency of jumps, t = time, D = diffusion coefficient = L /t. Diffusion Coefficient D = D o exp E A kt D = diffusion coefficient, D O = constant, E A = activation energy, k = Boltzmann constant, T = temperature
Diffusion of impurities in crystals E sd < E bd < E vd Diffusion: Diffusion and Oxidation ( S.O. Kasap, 1990-001: v.1.1) An e-booklet
Purification of hydrogen D Q = D exp d o RT D 0 = 4.5 10-7 m s -1 Q d = 4.6 kj mol -1 M at = 106.5 g mol -1 ρ = 1 g cm -3 pre-exponential factor activation energy atomic mass of Pd density of Pd
Surface of crystal : passivation with hydrogen
k
Fick s Second Law D C( x, t) C( x, t) = x t C(x,t) = concentration, which depends on time and distance D Q = D exp d o RT = diffusion coefficient, where Q d = activation energy, R = gas constant, T = temperature, D 0 = constant.
Diffusion from unlimited supply C x t C x t D (, ) (, = ) x t C (0, t) = Solution C s C ( x,0) = C C (, t) = C 0 0 at the surface always at t=0 anywhere in the bulk always at far end erf (z) = error function erf ( z) = π z e 0 x dx erfc (z) = 1- erf (z)
Error and complimentary error functions erfc (z) = 1- erf (z)
Carburization of steel T s = 750 0 C t = hours C 0 = 0.15 % C s = 1.5 % C(1mm, t) =? Qdiff D = Do exp RT D = 0 6. 10-7 m s -1 Q diff = 80.3 kj mol -1 pre-exponential factor activation energy
Carburization of steel : solution T s = 750 0 C t = hours C 0 = 0.15 % C s = 1.5 % C(1mm, t) =? D 0 = 6. 10-7 m s -1 Q diff = 80.3 kj mol -1
Error and complimentary error functions erfc(0.84)=0.4 erf(0.84)=1-erfc(0.84) = 0.76
Diffusion from limited supply N 0 = total number of diffusing particles (cm - ) D = diffusion coefficient t = time Diffusion: Concepts and Solved Problems ( S.O. Kasap, 1990-001: v.1.1) An e-booklet
Diffusion doping in semiconductor technology E hole e (a) (b) (c) Fig..5 (a) Thermal vibrations of the atoms rupture a bond and release a free electron into the crystal. A hole is left in the broken bond which has an effective positive charge. (b) An electron in a neighboring bond can jump and repair this bond and thereby create a hole in its original site; the hole has been displaced. (c) When a field is applied both holes and electrons contribute to electrical conduction. From Principles of Electronic Materials and Devices, Second Edition, S.O. Kasap ( McGraw-Hill, 00) http://materials.usask.ca [Ne] s p
Diffusion doping in semiconductor technology E hole e (a) (b) (c) Fig..5 (a) Thermal vibrations of the atoms rupture a bond and release a free electron into the crystal. A hole is left in the broken bond which has an effective positive charge. (b) An electron in a neighboring bond can jump and repair this bond and thereby create a hole in its original site; the hole has been displaced. (c) When a field is applied both holes and electrons contribute to electrical conduction. From Principles of Electronic Materials and Devices, Second Edition, S.O. Kasap ( McGraw-Hill, 00) http://materials.usask.ca [Ne] s p 3 n - doping with P
Diffusion doping in semiconductor technology E hole e (a) (b) (c) Fig..5 (a) Thermal vibrations of the atoms rupture a bond and release a free electron into the crystal. A hole is left in the broken bond which has an effective positive charge. (b) An electron in a neighboring bond can jump and repair this bond and thereby create a hole in its original site; the hole has been displaced. (c) When a field is applied both holes and electrons contribute to electrical conduction. From Principles of Electronic Materials and Devices, Second Edition, S.O. Kasap ( McGraw-Hill, 00) http://materials.usask.ca [He] s p 1 p - doping with B
Diffusion doping in semiconductor technology
Diffusion doping in semiconductor technology 900 0 C 900 0 C 1.04 10-15 7.6 10-74
Diffusion from limited supply positioned in the middle of sample N 0 = total number of diffusing particles D = diffusion coefficient t = time Diffusion: Concepts and Solved Problems ( S.O. Kasap, 1990-001: v.1.1) An e-booklet
MOSFET (Metal-Oxide-Semiconductor Field-Effect-Transistor) Al OXYDATION SiO PHOTOLITHOGRAPHY Si Resist SiO Mask Negative resist DIFFUSION
Diffusion of boron through the opening in SiO Diffusion: Diffusion and Oxidation ( S.O. Kasap, 1990-001: v.1.1) An e-booklet
Oxidation of silicon dry oxidation wet oxidation Diffusion: Diffusion and Oxidation ( S.O. Kasap, 1990-001: v.1.1) An e-booklet
Stages of oxide growth O SiO Si crystal Si crystal Si crystal T I M E Initial linear growth x ~ t Parabolic growth x ~ t x -oxide thickness 0.45 μm Si 1 μm SiO Diffusion: Diffusion and Oxidation ( S.O. Kasap, 1990-001: v.1.1) An e-booklet
Stages of oxide growth x = oxide thickness dx/dt = oxide growth rate C 0 C I = oxidant concentration on the surface = oxidant concentration on the interface Diffusion: Diffusion and Oxidation ( S.O. Kasap, 1990-001: v.1.1) An e-booklet
Initial linear growth Si + O SiO - dry oxidation dx ~ rate of SiO volume growth ~ k[si][o dt ]~ k 1 C 0 C 0 [O ] Mass Action Law rate of reaction A+B C [ C] d dt = k[ A][ B] [A] = concentration of A k = constant x B = k1c0t + xi t A ( +τ ) where B/A = linear rate constant x i = initial oxide thickness and B 1 A E = K1 exp kt rate limiting factor is concentration of oxidant with E 1 ev / atom compare with energy of Si-Si bond breaking 1.8 ev / bond Diffusion: Concepts and Solved Problems ( S.O. Kasap, 1990-001: v.1.1) An e-booklet
Parabolic growth in diffusion limited oxidation What is the rate of growth limiting factor? Diffusion of O to interface! Diffusion: Concepts and Solved Problems ( S.O. Kasap, 1990-001: v.1.1) An e-booklet
Parabolic growth in diffusion limited oxidation Calculations of O diffusion flux to interface dx dc = KD dt dx xdx = KD( C 0 C1 )dt dc Γ = D dx x = x + KD C C ) t 0 ( 0 1 E = K exp kt B E (O ) = 1.4 ev/atom E (HO ) = 0.79 ev/atom Diffusion: Concepts and Solved Problems ( S.O. Kasap, 1990-001: v.1.1) An e-booklet
Overall growth kinetics x 1/ A t + τ = 1+ 1 A / 4B where τ = x i + Ax B i t+τ << A / 4B t+τ >> A / 4B x B = t A ( +τ ) initial linear growth x x + = 0 Bt parabolic growth Diffusion: Concepts and Solved Problems ( S.O. Kasap, 1990-001: v.1.1) An e-booklet
Oxidation rates for dry and wet oxidations dry oxidation wet oxidation
Oxidation rates for dry and wet oxidations linear = rate limiting factor is concentration of oxidant at the interface parabolic = rate limiting factor is diffusion of O to interface dry oxidation wet oxidation
Calculation of oxidation time (at pressure 1 atm.) x = 0.1 μm T = 900+73 =1173 K t wet 0.6 hour t dry 13 hours
Calculations of oxidation time x t 1/ A t + τ = 1+ 1 A / 4B 1 A = x + x τ B B E = K exp kt B B 1 E = K A exp 1 kt τ = x i + t Ax B Experiment i Wet oxidation μm B = 0.1554 hr B A μm = 0.1513 hr τ = 0 0.75 hr t wet 0.6 hour Dry oxidation μm B = 0.004 hr B A μm = 0.00947 hr τ =.13hrs 10.8 hrs t dry 13 hours Diffusion: Concepts and Solved Problems ( S.O. Kasap, 1990-001: v.1.1) An e-booklet