Chapter 5: Diffusion
ISSUES TO ADDRESS...
• How does diffusion occur?
• Why is it an important part of processing?
• How can the rate of diffusion be predicted for
some simple cases?
• How does diffusion depend on structure
and temperature?
Chapter 7 - 1
Diffusion
Mechanisms
• Gases & Liquids – random (Brownian) motion
• Solids – vacancy diffusion or interstitial diffusion
Definition
• Mass transport by atomic motion
Conditions
• Enough energy
• A vacant adjacent site
Chapter 7 - 2
Diffusion
• Interdiffusion or impurity diffusion: In an alloy, atoms tend to
migrate from regions of high conc. to regions of low conc.
Initially
After some time
Figs. 7.1 & 7.2,
Callister &
Rethwisch 9e.
Chapter 7 - 3
Diffusion
• Self-diffusion: In an elemental solid, atoms also migrate.
Label some atoms
C
A
D
B
After some time
C
D
A
B
Chapter 7 - 4
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
Chapter 7 - 5
Diffusion Simulation
• Simulation of
interdiffusion
across an interface:
• Rate of vacancy diffusion
depends on:
-- vacancy concentration
-- frequency of jumping
-- diffusing species
(Courtesy P.M. Anderson)
Chapter 7 - 6
Diffusion Mechanisms
• Interstitial diffusion – smaller atoms can
diffuse between atoms.
Fig. 7.3 (b), Callister & Rethwisch 9e.
More rapid than vacancy diffusion
Chapter 7 - 7
Processing Using Diffusion
• Case Hardening:
-- Diffuse carbon atoms
into the host iron atoms
at the surface (carburization)
-- Example of interstitial
diffusion is a case
hardened gear.
Chapter-opening
photograph,
Chapter 7,
Callister &
Rethwisch 9e.
(Courtesy of
Surface Division,
Midland-Ross.)
• Result: The presence of C
atoms makes iron (steel) harder and stronger.
Chapter 7 - 8
Processing Using Diffusion
• Doping silicon with phosphorus for n-type semiconductors:
0.5 mm
• Process:
1. Deposit P rich
layers on surface.
magnified image of a computer chip
silicon
2. Heat it.
3. Result: Doped
semiconductor
regions.
silicon
light regions: Si atoms
light regions: Al atoms
Adapted from Figure 19.27, Callister &
Rethwisch 9e.
Chapter 7 - 9
Diffusion
• How do we quantify the amount or rate of diffusion?
moles (or mass) diffusing mol
kg
=
or
(area)(time)
cm2s m2s
• Measured empirically
J ≡ Flux ≡
– Make thin film (membrane) of known cross-sectional area
– Impose concentration gradient
– Measure how fast atoms or molecules diffuse through the
membrane
M
l dM
J=
=
At A dt
M=
mass
diffused
J µ slope
time
Chapter 7 - 10
Steady-State Diffusion
Rate of diffusion independent of time
Flux proportional to concentration gradient =
C1 C1
dC
dx
Fick’s first law of diffusion
C2
x1
x
C2
dC
J = −D
dx
x2
dC ΔC C2 − C1
if linear
≅
=
dx Δx x2 − x1
D º diffusion coefficient
Chapter 7 - 11
Steady-State Diffusion
•
One practical example of
steady state diffusion is
found in the purification of
hydrogen gas. One side of a
thin sheet of palladium metal
is exposed to the impure gas
composed of hydrogen and
other gaseous species such
as nitrogen, oxygen and
water vapor. The hydrogen
selectively diffuses through
the sheet to the opposite
side, which is maintained at
a constant and lower
hydrogen pressure.
Chapter 7 - 12
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 = 110 x10-8 cm2/s
– surface concentrations: C1 = 0.44 g/cm3
C2 = 0.02 g/cm3
Chapter 7 - 13
Example (cont).
• Solution – assuming linear conc. gradient
glove
C1
paint
remover
C2 − C1
dC
J = −D
≅ −D
dx
x 2 − x1
2
tb =
6D
skin
Data:
D = 110 x 10-8 cm2/s
C1 = 0.44 g/cm3
C2 = 0.02 g/cm3
x2 – x1 = 0.04 cm
C2
x1 x2
(0.02 g/cm3 − 0.44 g/cm3 )
g
J = − (110 x 10 cm /s)
= 1.16 x 10-5
(0.04 cm)
cm2s
-8
2
Chapter 7 - 14
Diffusion and Temperature
• Diffusion coefficient increases with increasing T
D = Do exp -
Qd
RT
D = diffusion coefficient [m2/s]
Do = pre-exponential [m2/s]
Qd = activation energy [J/mol or eV/atom]
R = gas constant [8.314 J/mol-K]
T = absolute temperature [K]
Chapter 7 - 15
Diffusion and Temperature
Chapter 7 - 16
Diffusion and Temperature
300
600
T(°C)
C
10-14
1.0
Dinterstitial >> Dsubstitutional
C in α-Fe
C in γ-Fe
Al
0.5
Ci
nα
-Fe
in
Al
Fe
αn
Fe e i
e
F
gγ-F
in
in
Fe
D (m2/s)
10-20
1000
10-8
1500
D has exponential dependence on T
1.5
Al in Al
Fe in α-Fe
Fe in γ-Fe
1000 K/T
Adapted from Fig. 7.7, Callister & Rethwisch 9e.
(Data for Fig. 7.7 taken from E.A. Brandes and G.B. Brook (Ed.) Smithells Metals
Reference Book, 7th ed., Butterworth-Heinemann, Oxford, 1992.)
Chapter 7 - 17
Diffusion and diffusing species
The diffusing species as well as the host material
influence the diffusion coefficient. For example, there is a
significant difference in magnitude between self-diffusion
and carbon interdiffusion in α iron at 500 ◦ C, the D value
being greater for the carbon interdiffusion (3.0x10-21 vs.
2.4x10-12 m2/s)
Chapter 7 - 18
Example: At 300°C the diffusion coefficient and
activation energy for Cu in Si are
D(300°C) = 7.8 x 10-11 m2/s
Qd = 41.5 kJ/mol
What is the diffusion coefficient at 350°C?
transform
data
D
Temp = T
ln D
1/T
Qd " 1 %
Qd " 1 %
lnD2 = lnD0 −
$$ '' and lnD1 = lnD0 −
$$ ''
R # T2 &
R # T1 &
Qd # 1 1 &
D2
∴ lnD2 − lnD1 = ln
= − %% − ((
D1
R $ T2 T1 '
Chapter 7 - 19
Example (cont.)
( Q " 1 1 %+
D2 = D1 exp *− d $$ − ''*) R # T2 T1 &-,
T1 = 273 + 300 = 573K
T2 = 273 + 350 = 623K
D2 = (7.8 x 10
−11
( −41,500 J/mol " 1
1 %+
m /s) exp *
−
$
') 8.314 J/mol-K # 623 K 573 K &,
2
D2 = 15.7 x 10-11 m2/s
Chapter 7 - 20
Non-steady State Diffusion
• The concentration of diffusing 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
∂2C
=D 2
∂t
∂x
Chapter 7 - 21
Non-steady State Diffusion
• Copper diffuses into a bar of aluminum.
Surface conc.,
Cs of Cu atoms
bar
pre-existing conc., Co of copper atoms
Cs
Fig. 7.5,
Callister &
Rethwisch 9e.
B.C.
at t = 0, C = Co for 0 £ x £ ∞
at t > 0, C = CS for x = 0 (constant surface conc.)
C = Co for x = ∞
Chapter 7 - 22
Solution:
( )
C x,t − Co
C s − Co
" x %
= 1 − erf $
'
# 2 Dt &
C(x,t) = Conc. at point x at
time t
erf(z) = error function
=
2
π
z
∫ e
0
− y2
dy
erf(z) values are given in a
table
CS
C(x,t)
Co
Fig. 7.5, Callister & Rethwisch 9e.
Chapter 7 - 23
Non-steady State Diffusion
Chapter 7 - 24
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.
Chapter 7 - 25
Solution (cont.):
– t = 49.5 h
– Cx = 0.35 wt%
– Co = 0.20 wt%
" x %
C( x,t ) − Co
= 1 − erf $
'
C s − Co
# 2 Dt &
x = 4 x 10-3 m
Cs = 1.0 wt%
" x %
C( x,t ) − Co 0.35 − 0.20
=
= 1 − erf $
' = 1 − erf(z)
C s − Co
1.0 − 0.20
# 2 Dt &
\ erf(z) = 0.8125
Chapter 7 - 26
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
z
0.95
0.7970
0.8125
0.8209
Now solve for D
z − 0.90
0.8125 − 0.7970
=
0.95 − 0.90 0.8209 − 0.7970
z = 0.93
z=
x
2 Dt
x2
D= 2
4z t
" x2 %
(4 x 10−3 m)2
1h
−11
2
∴D = $ 2 ' =
=
2.6
x
10
m
/s
2
# 4z t & (4)(0.93) (49.5 h) 3600 s
Chapter 7 - 27
Solution (cont.):
Qd
• To solve for the temperature at
T=
which D has the above value,
R(lnDo − lnD)
we use a rearranged form of
Equation;
from the data table, for diffusion of C in FCC Fe
Do = 2.3 x 10-5 m2/s Qd = 148,000 J/mol
\
T=
148,000 J/mol
(8.314 J/mol-K)(ln 2.3x10−5 m2 /s − ln 2.6x10−11 m2 /s)
T = 1300 K = 1027°C
Chapter 7 - 28
Non-steady State Diffusion
• Sample Problem: check design problem 5.1
Diffusion Temperature–Time Heat Treatment
Specification
Chapter 7 - 29
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
breakthrough time (tb), i.e., how long could the gloves be used
before methylene chloride reaches the hand?
• Data
– diffusion coefficient in butyl rubber:
D = 110 x 10-8 cm2/s
Chapter 7 - 30
CPC Example (cont.)
• Solution – assuming linear conc. gradient
Breakthrough time = tb
glove
C1
2
paint
remover
skin
C2
x1 x2

tb =
6D
Equation from online CPC
Case Study 5 at the Student
Companion Site for Callister &
Rethwisch 9e (www.wiley.com/
college/callister)
 = x2 − x1 = 0.04 cm
D = 110 x 10-8 cm2/s
(0.04 cm)2
tb =
= 240 s = 4 min
-8
2
(6)(110 x 10 cm /s)
Time required for breakthrough is 4 min
Chapter 7 - 31
Summary
Diffusion FASTER for...
Diffusion SLOWER for...
• open crystal structures
• close-packed structures
• materials w/secondary
bonding
• materials w/covalent
bonding
• smaller diffusing atoms
• larger diffusing atoms
• lower density materials
• higher density materials
Chapter 7 - 32
Further Reading
Reading:
Materials Science and Engineering: an Introduction", by
W. D. Callister - Chapter 5
Core Problems:
Self-help Problems:
Chapter 7 - 33