Diffusion in Solids...Solids – vacancy diffusion or interstitial diffusion • Interdiffusion: In...

17 March 2015

Diffusion in Solids

ELEC3215 Fluids and Mechanical Materials

Pt 3 Electromechanical Engineering

• C

changes as we solidify.• Cu-Ni case:

• Fast rate of cooling:Cored structure

• Slow rate of cooling:Equilibrium structure

First

to solidify has C

= 46 wt% Ni.Last

to solidify has C

= 35 wt% Ni.

Cored vs Equilibrium Phases

First to solidify: 46 wt% Ni

Uniform C:35 wt% Ni

Last to solidify: < 35 wt% Ni

Diffusion and Kinetics of Phase Transformation are important!

ISSUES TO ADDRESS...

• 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 in Solids

Diffusion

Diffusion - Mass transport by atomic motion

Mechanisms

Gases & Liquids – random (Brownian) motion

Solids – vacancy diffusion or interstitial diffusion

• Interdiffusion: In an alloy, atoms tend to migrate from regions of high concentration to regions of low concentration

Initially

Diffusion

After some time

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

Label some atoms After some time

Diffusion

A

B

C

DA

B

C

D

Mechanisms:

Diffusion Mechanisms

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

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

increasing elapsed time

Diffusion Mechanisms

Interstitial diffusion – smaller atoms can diffuse between atoms.

More rapid than vacancy diffusion

• Case Hardening:--Diffuse carbon atoms

into the host iron atomsat the surface.

--Example of interstitialdiffusion is a casehardened gear.

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

Processing Using Diffusion

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

3. Result: Dopedsemiconductorregions.

silicon

Processing Using Diffusion

magnified image of a computer chip

0.5mm

light regions: Si atoms

light regions: Al atoms

2. Heat it.

1. Deposit P richlayers on surface.

silicon

Doped region need to be interconnected !

• What material to choose? Cu, Ag, Au, Al ?• Process involves further heat treatments.

Conducting circuit paths

Al is the best choice in spite of lower electric conductivity

Cu can be use but thin barrier layer of Ta has to be deposited first ( cost ! )

Diffusion

How do we quantify the amount or rate of diffusion?

Measured empirically

Make thin film (membrane) of known surface area

Impose concentration gradient

Measure how fast atoms or molecules diffuse through the membrane

smkgor

scmmol

timearea surfacediffusing mass) (or molesFlux 22J

1M dMJAt A dt

M =mass

diffusedtime

J

slope

Geometry of Fick’s first law

Steady-State Diffusion

dxdCDJ

Fick’s first law of diffusionC1

C2

x

C1

C2

x1 x2

D

diffusion coefficient

Rate of diffusion independent of timeFlux proportional to concentration gradient =

dxdC

12

12 linear ifxxCC

xC

dxdC

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 glove?

Data:diffusion coefficient in butyl rubber:

D = 110x10-8 cm2/s surface concentrations:

C2 = 0.02 g/cm3

C1 = 0.44 g/cm3

scmg 10 x 16.1

cm) 04.0()g/cm 44.0g/cm 02.0(/s)cm 10 x 110( 2

5-33

28-

J

Example (cont).

12

12- xxCCD

dxdCDJ

D

tb 6

2

gloveC1

C2

skinpaintremover

x1 x2

• Solution – assuming linear conc. gradient

D = 110x10-8 cm2/s

C2 = 0.02 g/cm3

C1 = 0.44 g/cm3

x2 – x1 = 0.04 cm

Data:

Thermal Activation

Process path showing how an atom must overcome an activation energy, q, to move from one stable position to a similar adjacent position

/( )"jump" rate * Bq k Tconst e

Diffusion and Temperature

• Diffusion coefficient increases with increasing T.

D Do exp

Qd

RT

= pre-exponential [m2/s]= diffusion coefficient [m2/s]

= activation energy [J/mol or eV/atom] = gas constant [8.314 J/mol-K]= absolute temperature [K]

DDo

Qd

RT

Diffusion and Temperature

D has exponential dependence on T

Dinterstitial >> DsubstitutionalC in -FeC in -Fe

Al in AlFe in -FeFe in -Fe

1000K/T

D (m2/s) C in -Fe

C in -Fe

Al in Al

Fe in -Fe

Fe in -Fe

0.5 1.0 1.510-20

10-14

10-8T(C)15

00

1000

600

300

Alternative Diffusion Paths

Self-diffusion coefficients for silver depend on the diffusion path

In general, diffusivity is greater through less- restrictive structural regionsSurfacesGrain boundaries

(Lattice)

Diffusion at Micro scale

Schematic illustration of how a coating of impurity B can penetrate more deeply into grain boundaries and even further along a free surface of polycrystalline A, consistent with the relative values of diffusion coefficients

(Dvolume < Dgrain boundary < Dsurface )

Effect of Grain Boundaries

Represent a grain by a cylinder: diameter d with grain boundary width .

Area of grain =

Area of grain boundary =

Hence ratio of areas is

2lattice gbD D D

d

“Top view”

22d

22 2d 2

d

lattice lattice gb gbJ A J A J A

2lattice gbJ J J

d

Effect of Grain Boundaries

Grain boundaries have a more open structure, so Qgb < Qlattice

Should result in faster diffusion in polycrystalline samples than in single crystals.

However, the grain boundary area forms such a small fraction of the total area

The effect is only significant at low temperatures, when Dlattice << Dgb .

The effect is dependent on grain size.

For smaller grains the effect of grain boundary diffusion is greater.

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?

101

202

1lnln and 1lnlnTR

QDDTR

QDD dd

121

212

11lnlnln TTR

QDDDD d

transform data

D

Temp = T

ln D

1/T

Example (cont.)

K 5731

K 6231

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

2D

1212

11exp TTR

QDD d

T1 = 273 + 300 = 573KT2 = 273 + 350 = 623K

D2 = 15.7 x 10-11 m2/s

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

2

2

xCD

tC

Fick’s Second Law

Profile changes in time!Steady State: Fixed profile

v.s.

Non-steady State Diffusion

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 =

• Copper diffuses into a bar of aluminum.

pre-existing conc., Co of copper atoms

Surface conc., C of Cu atoms bars

Cs

Solution:

CS

Co

C(x,t)

Dtx

CCCt,xC

os

o

2 erf1

dye yz 2

0

2

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

erf (z) = error function

erf(z) values are given in tables

Error FunctionERF

Approximation

0

01

2s

c c xc c Dt

0

00.25

s

c cc c

If

Example: 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 the temperature at which the treatment was carried out.

Solution: use equation given:

Dtx

CCCtxC

os

o

2erf1),(

Solution (cont.)

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

Dtx

CCC)t,x(C

os

o

2erf1

)(erf12

erf120.00.120.035.0),( z

Dtx

CCCtxC

os

o

erf(z) = 0.81250

00.19 0.25

s

c cc c

Solution (cont.)

We must now determine from Table ERF 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

Now solve for D

Dtxz

2

tzxD 2

2

4

/sm 10 x 6.2s 3600

h 1h) 5.49()93.0()4(

m)10 x 4(4

2112

23

2

2

tzxD

To solve for the temperature at which D has above value, we use a rearranged form of Arrhenius equation )lnln( DDR

QTo

d

from tables, for diffusion of C in FCC Fe

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 000,148

21125 T

Solution (cont.)

T = 1300 K = 1027°C

Diffusion FASTER for...

• open crystal structures

• materials with secondarybonding

• smaller diffusing atoms

• lower density materials

Diffusion SLOWER for...

• close-packed structures

• materials with covalentbonding

• larger diffusing atoms

• higher density materials

Summary