diffusion: chapter 6

Download Diffusion: Chapter 6

Post on 06-Feb-2017




1 download

Embed Size (px)


  • 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 800C to 20C 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.


    Timex Area moles) or atoms Mass(or=

    AtM= J 2

    In differential form, the instantaneous flux J is


    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 =




    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 doesnt 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 725C. 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, Ficks 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.






    11 xCC













    3/270.0 mkgCC =

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

    dxdCD= - J

    ( )( )mx

    mkgD= mkgx 33



    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:




    Assuming D is constant (bad assumption in practice):





    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=




    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-




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

    ... + 7x

    3!1 -


    2!1 +

    3x x -= erf(x)


  • 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 Ficks 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



    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


    -D= D d0lnln



    -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 mDKKT ===

    sec/104.60266.11000 218

    212 mDKKT ===

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

    11 loglog 2.303RT

    Q -D= D d0

    22 loglog 2.303RT

    Q -D= D d0

    Subtracting the 2nd from the 1st yields:


    1 11logTT2.303R


    DD d

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

    ( )


    = mm d



    /31.8sec/10sec/10log 218


    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:

  • (


View more >