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.

seccmgin

Timex Area moles) or atoms Mass(or=

AtM= J 2

In differential form, the instantaneous flux J is

tM

A1= J(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.

x -xc -c= .4bfigure inline ofslope =

dxdc

BA

BA6

For steady-state diffusion, we have Fick's first law,

dxdCD= - J

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.

1000"

2

2

1

1

11 xCC

CC

ρρ+

=

1000

/25.2988.99

/87.7012.0

012.0

33

x

cmgcmg

CC

+=

3/270.0 mkgCC =

Similarly, 0.075 wt% C can be converted to 1.688 kg/m3. Now we can substitute into Fick’s first law above:

dxdCD= - J

( )( )mx

mkgD= mkgx 3

328

100.1/688.1270.0sec/104.1 −

− −−−

sec/1087.9 212 mxD −= 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:

∂∂

∂∂

∂∂

xCD

x=

tC

Assuming D is constant (bad assumption in practice):

xCD=

tC

2

2

∂∂

∂∂

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

∞≤≤ x0 at C=C 0,=t For 0

0=x at ionconcentrat constantthe ,C=C,0 =t For s

∞=x at C=C 0 Including these three boundary conditions Fick's second law (equation 6.4b) can be solved to yield

Dt2xerf -1=

C-CC-C

0s

0x

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:

dye2= erf(x) y-

x

0

2

∫Π

The error function erf(x) can also be calculated from the infinite series,

... + 7x

3!1 -

5x

2!1 +

3x x -= erf(x)

753

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 right-hand-side of the equation must also be a constant, so

constant= Dtxorconstant

Dt2xerf -1

2

,=

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

e D= D /RTQ-0

d

RTQ

-D= D d0lnln

or

2.303RTQ

-D= D d0loglog

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 pre-exponential 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.

sec/101.90910.11000 212

111 mDKK

T −− ===

sec/104.60266.11000 218

212 mDKK

T −− ===

Substitute into the last of the three equations above for each point:

11 loglog

2.303RTQ

-D= D d0

22 loglog

2.303RTQ

-D= D d0

Subtracting the 2nd from the 1st yields:

−

122

1 11logTT2.303R

Q=

DD d

Now you can substitute values for D1, T1, D2, and T2 to determine the activation energy Qd.

( )

−

−

−

−

KKKmolJ2.303Q

= mm d

1.9091

4.6021

/31.8sec/10sec/10log 218

212

molkJQd /205= 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:

( ) ( )( )( )KKmolJ2.303

molJ -D= m 0 1.909/31.8/000,205logsec/10log 212

−−

217.0log 0 −=D

sec/61.0 20 mD =

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:

Dt2xerf -1=

C-CC-C

0s

0x

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):

e D= D /RTQ-0

d

( ) sec/1084.3sec/108.1)1400( 211)1400)(/314.8()/000,152(

25 mxe mx= KD KKmolJmolJ

−°°−

− =°

( ) sec/1035.4sec/108.1)1200( 212)1200)(/314.8()/000,152(

25 mxe mx= KD KKmolJmolJ

−°°−

− =°

Substituting:

−

−

sec]360030][sec/1084.3[105.1

002.3

211

3

xmx2mxerf -1=

-C-

s

( )368.02.3 erf -1= Cs

Interpolating:

3794.04284.03794.0

35.04.035.0368.0

−−

=−− y

397.0)368.0(,397.0 == erfory

397.02.3 -1= Cs

%31.5 wtCs =

Now we have to substitute back in again:

− sec]360030sec][/1035.4[031.5

02.3212 xmx2xerf -1=

--

( )xerf -1= 5.729603.0

( ) 397.05.729 =xerf

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:

368.05.729 =x

mmmxx 50.0100.5 4 == − Clever solution: Start again with the solution to Fick’s 2nd law:

Dt2xerf -1=

C-CC-C

0s

0x

Now note that for both heat treatments, Cs and Cx and C0 are the same, so the entire left-hand 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,

22

2

11

1

tD2x

erf -1= tD2

xerf -1

22

22

11

21

tDx

tDx

=

Substituting into this equation, using the D values calculated above:

( ) ( ) )30(sec/1035.4)30(sec/1084.3)5.1(

212

22

211

2

hrmxx

hrmxmm

−− =

mmx 50.02 =

We obtained the same answer without doing the erf and erf-1 calculation.