Chapter 6
Highlights:
1. Know what diffusion is (material transport by atomic motion) and how its action
during processing affects materials properties (annealing depends on diffusion).
2. Know diffusion mechanisms (vacancy and interstitial).
3. Understand the concept of a diffusion flux (amount of material transported) and
the concept of steady state (no change in flux).
4. Understand the relationship between distance and time in unsteady state diffusion
problems.
5. Understand factors that influence diffusion rate (species, temperature, etc.).
Notes:
Throughout Chapter 6, two repeated
examples of engineering problems for
which diffusion is important are:
1)
“Case hardening,” where C is
diffused into the near-surface
region of a gear, drill tap,
or
other machine part. This makes
the outside of the part harder
(and more
brittle),
while
maintaining a more ductile
interior. This is illustrated in the
Figure at the right.
2)
Diffusion into the surface of a Si wafer to create p-n junctions, and transistors.
Transistors are created from the top surface of a Silicon wafer down. This is
illustrated by the Figure below, where the concentration of an n-type or p-type
dopant increases with time during a diffusion process. We will learn about Si
dopants in Chapter 12.
________________________________________________________________________
Chapter 6 turns chapter 5 into chapter 3.
Diffusion turns crystals with point defects, line defects, and grain boundaries, into
perfect crystals.
Show figures 6.1 and 6.2. This configuration is known as a diffusion couple.
Interdiffusion, or impurity diffusion, may be desired or undesired.
Desirable-
Provides a hard outer coating on cutting edges/tools, allows engineers to
tailor the electrical properties of Si.
Undesirable- Degrades the optical properties of a semiconductor laser, which is composed
of alternating layers of GaAs/AlxGa1-xAs, in a compact disc player.
Self-diffusion also occurs, this is harder to see, but is important during materials processing.
Self-diffusion may reduce the number of defects during heat treatment (annealing).
Imagine two different heat treatment procedures, cooling a polycrystalline Cu sample from
800°C to 20°C in one minute, and cooling an identical sample in on day. Comparing the
two samples, the latter sample (slow cooling) will have far fewer defects, since diffusion
will occur much more rapidly at high temperature.
Diffusion Mechanisms
Must involve stepwise migration of atoms from lattice site to lattice site.
A)
Vacancy diffusion (interdiffusion and self-diffusion)
Show figure 6.3.
This type of diffusion depends on the presence of vacancies and therefore increases
with the vacancy concentration as the temperature increases. Motion of vacancies in
one direction is equivalent to motion of atoms in the opposite direction.
B)
Interstitial diffusion (interdiffusion only)
Show figure 6.3.
Normally this is faster than vacancy diffusion.
Steady-state Diffusion
Diffusion is time-dependent, the amount of matter transferred depends on time and is
characterized by the diffusion flux J.
J=
M Mass(or atoms or moles)
g
=
in 2
At
Area x Time
cm sec
In differential form, the instantaneous flux J is
J(t) =
1 ∂M
A ∂t
Eventually, steady-state conditions may be reached, and the diffusion flux no longer
changes with time. The concept of a steady state is an important one, and we will divide up
diffusion problems into steady state and unsteady state diffusion.
Example
Steady state diffusion of a gas through a plate of metal, where the gas pressure on either side
of the plate is kept constant. Show figure 6.4.
dc
c -c
= slope of line in figure 6.4b = A B
dx
x A - xB
For steady-state diffusion, we have Fick's first law,
J=-D
dC
dx
The negative sign ensures that the flux goes from the high to the low concentration regions.
This concentration gradient can be the driving force for a reaction. The quantity D is the
diffusivity and describes the rapidity with which material A can diffuse into material B.
The configuration in figure 6.4 can be used for hydrogen purification. Take a mixture of H,
O, and N on the left-hand side. DH >> DO, DN, so only hydrogen diffuses through at a
significant rate. Why doesn’t this reach eventual steady state conditions, which is not useful
for hydrogen purification? Discuss this extensively in class. Pd membranes are used
commercially for separating H gas, which diffuses through Pd much faster than other gases.
Example Problem:
A sheet of BCC Fe 1.0 mm thick is exposed to a carburizing gas on
one side and a decarburizing gas on the other at 725°C. After
reaching steady state, the Fe membrane is quenched to room
temperature, and the C concentrations at each side of the membrane
are 0.012 and 0.075 wt%. Calculate the diffusion coefficient if the
diffusion flux is 1.4x10-8 kg/m2-sec.
We want to use the equation above, Fick’s first law of diffusion. However, the units of
diffusion flux and concentration are inconsistent, so we need to use equations (5.12) to
convert from wt% to kg C/m3 Fe.
C1 " =
C1
C1
ρ1
CC =
+
C2
x1000
ρ2
0.012
0.012
99.988
+
7.87 g / cm 3 2.25 g / cm 3
x1000
C C = 0.270 kg / m 3
Similarly, 0.075 wt% C can be converted to 1.688 kg/m3. Now we can substitute into Fick’s
first law above:
J=-D
1.4 x10
−8
dC
dx
(0.270 − 1.688 kg / m )
− sec = − D
(1.0 x10 m )
3
kg / m
2
−3
D = 9.87 x10 −12 m 2 / sec
Unsteady-state Diffusion
Unsteady-state diffusion describes processes where the diffusion flux and the concentrations
change with time. This is encountered, for example, for the diffusion couple shown in
figures 6.1 and 6.2. Unsteady state diffusion is also encountered with diffusion of materials
into a Si surface or the surface of a machine tool. The former allows alteration of the
electrical properties of Si, and the latter allows hardening of the surface of cutting tools.
This is encountered more often than steady-state diffusion. Unsteady-state diffusion is
governed by Fick's second law:
∂C ∂  ∂C 
=

D
∂t ∂x  ∂x 
Assuming D is constant (bad assumption in practice):
2
∂C
∂ C
=D
∂t
∂ x2
Since many of you have not had differential equations yet, you may not be able to follow
this discussion, but you should still be able to use the final equations. We need to employ
the following boundary conditions
For t = 0, C = C 0 at 0 ≤ x ≤ ∞
For t = 0 , C = C s , the constant concentration at x = 0
C = C 0 at x = ∞
Including these three boundary conditions Fick's second law (equation 6.4b) can be solved
to yield
 x 
C x - C0
= 1 - erf 

C s - C0
 2 Dt 
The error function(erf) is tabulated in Table 6.1, it is just a mathematical function that can
only be represented by an integral, you can use it just by looking up values in a table and
interpolating. You will not need to calculate error functions numerically, but for your
curiosity erf(x) is:
2 x - y2
erf(x) =
∫ e dy
Π0
The error function erf(x) can also be calculated from the infinite series,
3
1 x 5 1 x7
x
erf(x) = x - +
+ ...
3 2! 5 3! 7
However, many problems in unsteady-state diffusion can be solved without the
complication of error function calculation. For certain problems, one can employ a simple
relationship between the time and distance at which a certain concentration will occur.
Looking at the solution to Fick’s 2nd law that is given above, if we are interested in a certain
concentration C1, then the left-hand-side of the equation is a constant. Therefore the righthand-side of the equation must also be a constant, so
 x 
1 - erf 
 = constant ,
 2 Dt 
2
or
x
= constant
Dt
Factos that Influence Diffusion Rate
A)
Both the diffusing species and the host material affect D (Show table 6.2, figure 6.7).
B)
Temperature
D = D0 e -Q d /RT
ln D = ln D0 -
Qd
RT
or
log D = log D0 -
Qd
2.303RT
Therefore, a plot of ln D versus 1/T should yield a straight line with slope -Qd/R and
intercept ln D0.
Example problem:
From figure 6.7, determine the activation energy and preexponential factor for Al diffusion in Al.
Easiest approach:
Select two points on the straight line, as far apart as possible for
maximum accuracy, and call their coordinates (D1, T1) and (D2, T2). I read two points as
(1000/T, D) = (1.10 K-1, 10-12 m2/sec) and (1.65 K-1, 10-18 m2/sec). Since 1/T is usually
graphed instead as 1000/T, you need to read the graph carefully.
T1 =
1000
= 909.1 K
1.10 K −1
D1 = 10 −12 m 2 / sec
T2 =
1000
= 602.4 K
1.66 K −1
D2 = 10 −18 m 2 / sec
Substitute into the last of the three equations above for each point:
log D1 = log D0 -
Qd
2.303RT1
log D2 = log D0 -
Qd
2.303RT2
Subtracting the 2nd from the 1st yields:
D 
Qd  1 1 
 − 
log  1  =
2.303R
D
 2
 T2 T1 
Now you can substitute values for D1, T1, D2, and T2 to determine the activation energy
Qd.
 10 −12 m 2 / sec 
Qd

1
1 
 =


log  −18 2
−
(
)
2.303
J
mol
K
K
K
8
.
31
/
602
.
4
909
.
1
−
m
10
/
sec




Qd = 205 kJ / mol
This differs from the value given in Table 6.2 (144 kJ/mol), and I cannot reconcile the
difference between these two answers. Once you have Qd, D0 can be determined by
substitution into either of the two equations for (D1, T1) or (D2, T2). Using the former:
(
)
log 10 −12 m 2 / sec = log D0 -
(205,000 J / mol )
2.303(8.31 J / mol − K )(909.1 K )
log D0 = − 0.217
D0 = 0.61 m 2 / sec
This disagrees wildly with the typical values for D0 shown in Table 6.2. However, I
cannot find my mistake in this problem.
Example problem:
Consider a diffusion couple between metals A and B. After a 30-hr heat treatment at
1400°K, the concentration of A is 3.2 wt% at a distance of 1.5 mm from the interface. If an
identical diffusion couple is instead heated 1200°K for 30 hr, at what position will the
concentration be 3.2 wt%? Assume that the preexponential factor and activation energy for
diffusion are 1.8x10-5 m2/sec and 152 kJ/mol, respectively, and that the surface
concentration of A is 100 wt%.
Straightforward solution:
Start with the solution to Fick’s 2nd law:
 x 
C x - C0
= 1 - erf 

C s - C0
 2 Dt 
Now substitute in the correct numbers to determine Cs, which will require using the erf table
6.1. We will eventually need to calculate the diffusion coefficient D at each of these
temperatures. From equation (6.8):
D = D0 e - Q
(
−5
2
(
−5
2
/RT
)
−
( 152 , 000 J / mol )
( 8.314 J / mol ° K )(1400° K )
)
−
( 152 , 000 J / mol )
( 8.314 J / mol ° K )(1200° K )
D (1400° K ) = 1.8 x10 m / sec e
D (1200° K ) = 1.8 x10 m / sec e
Substituting:
d
= 3.84 x10 −11 m 2 / sec
= 4.35 x10 −12 m 2 / sec


3.2 - 0
1.5 x10 − 3 m


= 1 - erf
−
11
2


Cs - 0
 2 [3.84 x10 m / sec ][30 x 3600 sec] 
3.2
= 1 - erf (0.368 )
Cs
Interpolating:
0.368 − 0.35
y − 0.3794
=
0.4 − 0.35
0.4284 − 0.3794
y = 0.397, or erf (0.368) = 0.397
3.2
= 1 - 0.397
Cs
C s = 5.31 wt %
Now we have to substitute back in again:


x
3.2 - 0

= 1 - erf 
12
2
−
 2 [4.35 x10 m / sec][30 x 3600 sec] 
5.31 - 0


0.603 = 1 - erf (729.5 x )
erf (729.5 x ) = 0.397
What number has an erf equal to 0.397. Now we need to read Table 6.1 in the reverse
direction. However, if we look above, we see that we just did this problem (in reverse), so
we know that erf(0.368) = 0.397. Therefore we can write:
729.5 x = 0.368
x = 5.0 x10 −4 m = 0.50 mm
Clever solution:
Start again with the solution to Fick’s 2nd law:
 x 
C x - C0
= 1 - erf 

2
Dt
C s - C0


Now note that for both heat treatments, Cs and Cx and C0 are the same, so the entire lefthand side of this equation is the same. For these two heat treatments, the entire right-hand
side of the equation must also be the same. Therefore,

 x1 

 = 1 - erf  x 2 
1 - erf 
2 D t 
2 D t 
1 1 
2 2 


x12
x 22
=
D1 t 1
D2 t 2
Substituting into this equation, using the D values calculated above:
x 22
(1.5 mm ) 2
=
4.35 x10 −12 m 2 / sec ( 30 hr )
3.84 x10 −11 m 2 / sec ( 30 hr )
(
)
(
x 2 = 0.50 mm
We obtained the same answer without doing the erf and erf-1 calculation.
)