ISSUES TO ADDRESS...

• How does diffusion occur?

• Why is it an important part of processing?

Chapter 6: Diffusion in Solids

Chapter 6 - 1

• 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

Diffusion - Mass transport by atomic motion

Mechanisms• Gases & Liquids – random (Brownian) motion

Chapter 6 - 2

• Gases & Liquids – random (Brownian) motion• Solids – vacancy diffusion or interstitial diffusion

• Interdiffusion: In an alloy, atoms tend to migratefrom regions of high conc. to regions of low conc.

Initially

Diffusion

After some time

Chapter 6 - 3From Figs. 6.1 and 6.2Callister’s Materials Science and Engineering,Adapted Version.

• Self-diffusion: In an elemental solid, atomsalso migrate.

Label some atoms After some time

Diffusion

A

C

A

C

D

Chapter 6 - 4

B

DA

B

Diffusion Mechanisms

Vacancy Diffusion:

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

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

Chapter 6 - 5

--activation energy to exchange.

increasing elapsed time

Diffusion Mechanisms

• Interstitial diffusion – smaller atoms can diffuse between atoms.

Chapter 6 - 6

More rapid than vacancy diffusionFrom Fig. 6.3 (b)Callister’s Materials Science and Engineering, Adapted Version.

From chapter-opening photograph, Chapter 6, Callister’s Materials Science and Engineering,Adapted Version.

(Courtesy of

• Case Hardening:--Diffuse carbon atoms

into the host iron atomsat the surface.

--Example of interstitialdiffusion is a casehardened gear.

Processing Using Diffusion

Chapter 6 - 7

Surface Division, Midland-Ross.)

hardened gear.

• Result: The presence of C atoms makes iron (steel) harder.

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

Processing Using Diffusion

magnified image of a computer chip

0.5mm

1. Deposit P richlayers on surface.

silicon

Chapter 6 - 8

3. Result: Dopedsemiconductorregions.

silicon

light regions: Si atoms

light regions: Al atoms

2. Heat it.

silicon

From chapter-opening photograph, Chapter 17Callister’s Materials Science and Engineering,Adapted Version..

Diffusion• How do we quantify the amount or rate of diffusion?

• Measured empirically– Make thin film (membrane) of known surface area– Impose concentration gradient

( )( ) smkg

orscm

gtimearea surface

diffusing mass)(or molesFlux 22=ººJ

Chapter 6 - 9

– Impose concentration gradient– Measure how fast atoms or molecules diffuse through the

membrane

dtdM

Al

AtM

J ==

M =mass

diffused

time

J µ slope

Steady-State Diffusion

dC-=

Fick’s first law of diffusionC1C1

Rate of diffusion independent of time

Flux proportional to concentration gradient =dxdC

Chapter 6 - 10

dxdC

DJ -=C2

x

C2

x1 x2

D º diffusion coefficient

12

12 linear ifxxCC

xC

dxdC

--

=DD

@

Example: Chemical Protective 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.

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

Chapter 6 - 11

is the diffusive flux of methylene chloride through the glove?

• Data:

– diffusion coefficient in butyl rubber: D = 110x10-8 cm2/s

– surface concentrations:

C2 = 0.02 g/cm3

C1 = 0.44 g/cm3

Example (cont).

12

12- xxCC

DdxdC

DJ--

-@=D

tb 6

2l=

gloveC1

C2

skinpaintremover

x x

• Solution – assuming linear conc. gradient

D = 110x10-8 cm2/s

C = 0.44 g/cm3

Data:

Chapter 6 - 12

scm

g 10 x 16.1

cm) 04.0()g/cm 44.0g/cm 02.0(

/s)cm 10 x 110( 2

5-33

28- =-

-=J

x1 x2

C2 = 0.02 g/cm3

C1 = 0.44 g/cm3

x2 – x1 = 0.04 cm

Diffusion and Temperature

• Diffusion coefficient increases with increasing T.

D = Do expæ

è ç

ö

ø ÷-

Qd

RT

= diffusion coefficient [m2/s]D

Chapter 6 - 13

= pre-exponential [m2/s]

= diffusion coefficient [m /s]

= activation energy [J/mol or eV/atom]

= gas constant [8.314 J/mol-K]

= absolute temperature [K]

D

Do

Qd

R

T

• The activation energy may be thought of as the energy required to produce the diffusive motion of one mole of atom

Diffusion and TemperatureD has exponential dependence on T

Dinterstitial >> Dsubstitutional

a

D (m2/s)

10-8

T(°C)

1500

1000

600

300

Chapter 6 - 14

From Fig. 6.7Callister’s Materials Science and Engineering, Adapted Version.(Date for Fig. 6.7 taken from E.A. Brandes and G.B. Brook (Ed.) Smithells Metals Reference Book, 7th ed., Butterworth-Heinemann, Oxford, 1992.)

C in a-FeC in g-Fe

Al in AlFe in a-FeFe in g-Fe

1000K/T0.5 1.0 1.5

10-20

10-14

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

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

What is the diffusion coefficient at 350ºC?

transform D ln D

Chapter 6 - 15

÷÷ø

öççè

æ-=÷÷

ø

öççè

æ-=

101

202

1lnln and

1lnln

TRQ

DDTR

QDD dd

÷÷ø

öççè

æ--==-\

121

212

11lnlnln

TTRQ

DD

DD d

data

Temp = T 1/T

Example (cont.)

úû

ùêë

é÷÷ø

öççè

æ--=

1212

11exp

TTRQ

DD d

T1 = 273 + 300 = 573K

T2 = 273 + 350 = 623K

Chapter 6 - 16

úû

ùêë

é÷øö

çèæ -

-= -

K 5731

K 6231

K-J/mol 314.8J/mol 500,41

exp /s)m 10 x 8.7( 2112D

T2 = 273 + 350 = 623K

D2 = 15.7 x 10-11 m2/s

Question: Rank the magnitudes of the diffusion coefficients from greatest to least for the following systems:

N in Fe at 700°CCr in Fe at 700°C

Chapter 6 - 17

Answer: Nitrogen is an interstitial impurity in Fe (on the basis of its atomic radius), whereas Cr is a substitutional impurity. Since interstitial diffusion occurs more rapidly than substitutionalimpurity diffusion, DN > DCr.

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

Chapter 6 - 18

2

2

xC

DtC

¶¶

=¶¶Fick’s Second Law

Non-steady State Diffusion

Adapted from Fig. 5.5,

• Copper diffuses into a bar of aluminum.

pre-existing conc., Co of copper atoms

Surface conc., C of Cu atoms bar

s

Cs

Chapter 6 - 19

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:

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

erf (z) = error function

CS

( )÷ø

öçè

æ-=-

-Dt

xCC

Ct,xC

os

o

2 erf1

Chapter 6 - 20

erf (z) = error function

Co

C(x,t)dye yz 2

0

2 -òp=

Non-steady State Diffusion

• Sample Problem: An FCC iron-carbon alloy initially containing 0.20 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

Chapter 6 - 21

at a position 4.0 mm below the surface, determine the temperature at which the treatment was carried out.

• Solution: ÷ø

öçè

æ-=-

-Dt

xCC

CtxC

os

o

2erf1

),(

Solution (cont.):

– t = 49.5 h x = 4 x 10-3 m

– Cx = 0.35 wt% Cs = 1.0 wt%

– Co = 0.20 wt%

÷ø

öçè

æ-=-

-Dt

xCC

C)t,x(C

os

o

2erf1

Chapter 6 - 22

)(erf12

erf120.00.120.035.0),(

zDt

xCC

CtxC

os

o -=÷ø

öçè

æ-=

--

=-

-

\ erf(z) = 0.8125

Solution (cont.):

We must now determine from the Table below the value of z for which the error function is 0.8125. An interpolation is necessary as follows

z erf(z)

0.90 0.7970z 0.81250.95 0.8209

7970.08209.07970.08125.0

90.095.090.0

--

=-

-z

z = 0.93

Chapter 6 - 23

0.95 0.8209

Now solve for D

Dt

xz

2=

tz

xD

2

2

4=

/sm 10 x 6.2s 3600

h 1

h) 5.49()93.0()4(

m)10 x 4(

4211

2

23

2

2-

-==÷

÷ø

öççè

æ=\

tz

xD

• To solve for the temperature at which D has above value, we use a rearranged form of this equation:

)lnln( DDRQ

To

d-

=

It is given that for diffusion of C in FCC Fe

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

Solution (cont.):

Chapter 6 - 24

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

/s)m 10x6.2 ln/sm 10x3.2 K)(ln-J/mol 314.8(

J/mol 000,14821125 -- -

=T\

T = 1300 K = 1027°C

Diffusion FASTER for...

• open crystal structures

• materials w/secondarybonding

Diffusion SLOWER for...

• close-packed structures

• materials w/covalentbonding

Summary

Chapter 6 - 25

• smaller diffusing atoms

• lower density materials

• larger diffusing atoms

• higher density materials