Diffusion in Solids

ISSUES TO ADDRESS...

• How does diffusion occur?• How does diffusion occur?

• Why is it an important part of processing?

• How can the rate of diffusion be predicted forsome simple cases?

• How does diffusion depend on structureand temperature?

Diffusion MechanismsVacancy Diffusion:

• atoms exchange with vacancies• atoms exchange with vacancies• applies to substitutional impurities atoms • rate depends on:

number of vacancies--number of vacancies--activation energy to exchange.

increasing elapsed time

Diffusion Mechanisms• Interstitial diffusion – smaller atoms

can diffuse between atoms.can diffuse between atoms.

Adapted from Fig. 5.3 (b), Callister 7e.

More rapid than vacancy diffusion

Processing Using Diffusion• Doping silicon with phosphorus for n-type semiconductors:• Process:

0.5mm 1. Deposit P rich

layers on surface.

magnified image of a computer chip

silicon

3. Result: Dopedsemiconductor

light regions: Si atoms

2. Heat it.

regions.

light regions: Al atomssilicon

light regions: Al atoms

Adapted from chapter-opening photograph, Chapter 18, Callister 7e.

Example: Chemical Protective Clothing (CPC)• Methylene chloride is a common ingredient of paint

removers. Besides being an irritant, it also may be g yabsorbed through skin. When using this paint remover, protective gloves should be worn.

• If butyl rubber gloves (0.04 cm thick) are used, what is the diffusive flux of methylene chloride through the

l ?glove?• Data:

diff i ffi i t i b t l bb– diffusion coefficient in butyl rubber: D = 110x10-8 cm2/s

– surface concentrations: C1 = 0.44 g/cm3

C2 = 0.02 g/cm3C1 0.44 g/cm

Example (cont).

glove

• Solution – assuming linear conc. gradient

12- CCDdCDJ −−≅=2l

gC1

skinpaintremover

12-

xxD

dxDJ

−≅=

Dtb 6

l=

C2

remover

x1 x2

D = 110x10-8 cm2/s

C = 0 02 g/cm3

C1 = 0.44 g/cm3Data:

33

C2 = 0.02 g/cm3

x2 – x1 = 0.04 cm

scmg 10 x 16.1

cm) 04.0()g/cm 44.0g/cm02.0(/s)cm 10 x 110( 2

5-33

28- =−

−=J

Diffusion and Temperature

• Diffusion coefficient increases with increasing T.

D = D exp⎛

⎝⎜

⎞

⎠⎟−

QdD Do exp⎝

⎜ ⎠ ⎟

RT

= diffusion coefficient [m2/s]D= pre-exponential [m2/s]= activation energy [J/mol or eV/atom]

Do

Qd

= gas constant [8.314 J/mol-K]= absolute temperature [K]

RT

Diffusion and TemperatureD has exponential dependence on T

0 0

D ( 2/ )

10-8T(°C)15

00

1000

600

300

Dinterstitial >> DsubstitutionalC in α-FeC in γ-Fe

Al in AlFe in α-Fe

D (m2/s)

10-14 C in γ Fe Fe in α FeFe in γ-Fe

10 20

Adapted from Fig. 5.7, Callister 7e. (Date for Fig. 5.7 taken from E.A.

1000K/T0.5 1.0 1.510-20

Brandes and G.B. Brook (Ed.) Smithells Metals Reference Book, 7th ed., Butterworth-Heinemann, Oxford, 1992.)

Example: At 300ºC the diffusion coefficient and activation energy for Cu in Si areactivation energy for Cu in Si are

D(300ºC) = 7.8 x 10-11 m2/sQd = 41.5 kJ/molQd 41.5 kJ/mol

What is the diffusion coefficient at 350ºC?

transform data

D ln D

⎞⎛⎞⎛ 11 QQ

Temp = T 1/T

⎟⎟⎠

⎞⎜⎜⎝

⎛−=⎟⎟

⎠

⎞⎜⎜⎝

⎛−=

101

202

1lnln and 1lnlnTR

QDDTR

QDD dd

⎟⎞

⎜⎛2 11lll QDDD d

⎟⎟⎠

⎞⎜⎜⎝

⎛−−==−∴

121

212

11lnlnln TTR

QDDDD d

Example (cont.)

⎥⎦

⎤⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−−= 12

11exp TTR

QDD d

⎦⎣ ⎠⎝ 12 TTR

T1 = 273 + 300 = 573K1

T2 = 273 + 350 = 623K

⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛ −

−= −

K 5731

K 6231

K-J/mol 314.8J/mol 500,41exp /s)m 10 x 8.7( 211

2D

D2 = 15.7 x 10-11 m2/s2

Non-steady State Diffusiony• 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

2CDC ∂∂Fick’s Second Law2x

Dt ∂

=∂

Fick s Second Law

Non-steady State Diffusion• Copper diffuses into a bar of aluminum.

pre existing conc C of copper atoms

Surface conc., C of Cu atoms bars pre-existing conc., Co of copper atoms s

Cs

Adapted from Fig. 5.5, Callister 7e.

B.C. at t = 0, C = Co for 0 ≤ x ≤ ∞

at t > 0, C = CS for x = 0 (const. surf. conc.)

C = Co for x = ∞

Solution: ( )

⎟⎠

⎞⎜⎝

⎛−=−

−Dtx

CCCt,xC

os

o

2 erf1

C(x,t) = Conc. at point xat

CS

⎠⎝ DtCC os 2

at time t

erf (z) = error function C(x t)erf (z) error function

Co

C(x,t)

dye yz 2

0

2 −∫π=

erf(z) values are given in T bl 5 1Table 5.1

How much is erf(0 24)? =>erf(z)?How much is erf(0.24)? =>erf(z)?erf (?) is 0.85? =>erf inverse(0.85)?

Non-steady State Diffusiony• Sample Problem: An FCC iron-carbon alloy

initially containing 0 20 wt% C is carburized atinitially containing 0.20 wt% C is carburized at an elevated temperature and in an atmosphere that gives a surface carbon concentrationthat 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

i d tcarried out.

⎞⎛− xCtxC )(• Solution: use Eqn. 5.5 ⎟⎠

⎞⎜⎝

⎛−=−

−Dtx

CCCtxC

os

o

2erf1),(

Solution (cont ): ⎟⎞

⎜⎛−=

− xC)t,x(C o erf1Solution (cont.): ⎟⎠

⎜⎝

−=− DtCC os 2

erf1

– t = 49.5 h x = 4 x 10-3 m– Cx = 0.35 wt% Cs = 1.0 wt%– Co = 0.20 wt%

⎞⎛ )(erf12

erf120.00.120.035.0),( z

Dtx

CCCtxC

os

o −=⎟⎠

⎞⎜⎝

⎛−=

−−

=−

−

∴ erf(z) = 0.8125

Solution (cont.):We must now determine from Table 5.1 the value of z for which the error function is 0.8125. An interpolation is necessary as follows

z erf(z)0.90 0.7970 7970.08209.0

7970.08125.090.095.0

90.0−−

=−

−z

z 0.81250.95 0.8209

z = 0.93

Now solve for D

Dtxz

2=

tzxD 2

2

4=

/sm10x 6.23600

h 1m)10 x 4( 2112

23

2

2−

−==⎟

⎟⎞

⎜⎜⎛

=∴xD

s 3600h)5.49()93.0()4(4 22 ⎟⎠

⎜⎝ tz

Solution (cont.):• To solve for the temperature

at which D has above value, )lnln( DDRQTo

d−

=

we use a rearranged form of Equation (5.9a);

f T bl 5 2 f diff i f C i FCC Ffrom Table 5.2, for diffusion of C in FCC Fe

Do = 2.3 x 10-5 m2/s Qd = 148,000 J/mol

/s)m10x6.2ln/sm10x3.2K)(ln-J/mol 314.8(J/mol 000,148

21125 −− −=T∴

))((

T = 1300 K = 1027°CT 1300 K 1027 C

Example: Chemical Protective Cl hi (CPC)Clothing (CPC)

• Methylene chloride is a common ingredient of paint removers. Besides being an irritant, it also may be absorbed through skin. When using this paint remover, protective gloves should be worn.protective gloves should be worn.

• If butyl rubber gloves (0.04 cm thick) are used, what is the breakthrough time (tb), i.e., how long could the gloves be used before methylene chloride reaches the hand?

• Data (from Table 22.5)Data (from Table 22.5)– diffusion coefficient in butyl rubber:

D = 110x10-8 cm2/s

Example (cont).

glove

• Solution – assuming linear conc. gradientg

C1

skinpaintremover 2l

Equation 22.24

C2

remover

x1 x2 Dtb 6

2l=

i4240cm)04.0( 2t

Time required for breakthrough ca 4 min

min4 s 240/s)cm 10 x 110)(6(

cm)04.0(28- ===bt

Time required for breakthrough ca. 4 min

SummaryDiffusion FASTER for... Diffusion SLOWER for...

y

• open crystal structures

• materials w/secondary

• close-packed structures

• materials w/covalent• materials w/secondarybonding

• smaller diffusing atoms

• materials w/covalentbonding

• larger diffusing atoms• smaller diffusing atoms

• lower density materials

• larger diffusing atoms

• higher density materials