various times showing that the oxygen concentration will drop to about 6 × 1017 cm3 at 20 μm depth, as required, in about 104 seconds, or about 3 hours.
11.6 FINITE SOURCE SOLUTIONS
11.6 Finite Source Solutions 395
11.6.2 e
xtendeds
ourcesEquation 11.16 holds for any type of extended source, f(x), as well as individual sources. For example, consider f(x) = C0 for − ≤ ≤a x a. A more complex function of x could be chosen other than a constant concentration, C0, but this just makes the algebra involving the integrations more time consuming without adding any additional insight. Therefore, for f(x) = C0, Equation 11.16 becomes
C x t C
Dt dx
x x Dt a a
, e .
( )
( )
= − −∫
−0 4
4 0
02
π (11.18)
The procedure to put these types of equations into a more familiar and calculable form is just the reverse of that used in the Appendix to obtain Equation 11.16, simply make a variable substitution.
An obvious choice is to let y=(x x− 0)/ 4Dt so x0= −x 4Dt y and dx0= − 4Dt dy, and when x0 = −a, y=(x a+ )/ 4Dt and when x0 = a, y=(x a− )/ 4Dt so
Equation 11.18 is now C
C Dt e Dtdy
C
C e
y x a Dt
x a Dt
y
x a D
0 4
4
0 4
1
4 4
1
2
2
= −
= −
− +
−
− +
π
∫
π
( ) ( )
( )
.
/ /
/ tt
x a Dt y
x a Dt y
y
dy e dy
C
C e dy e
0
0 4
0 0
4
0
1
1 1
2
2 2
∫ ∫
∫
−
= −
− −
+ −
−
π
π π
( )
( ) (
/
/ xx a Dt
dy
∫
−)/ 4(ugly but no fancy mathematics) which, of course, is C
C erf x a
Dt erf x a Dt
0
1
2 4
1
2 4
= +
− −
(11.19)
which is nothing but a sum of two numbers for specific values of a and 4Dt. Figure 11.12 shows the solution for some values of 4Dt. An example where such a solution might apply is for a thin sheet of one metal (e.g., gold) sandwiched between two pieces of silver and heated to get interdiffusion and a solid-state bond or weld between the two pieces of silver. Another is an adhesive bond between two polymers or pieces of wood with the diffusion of the solvent (with some of the adhesive) into the pieces being joined.
−3 −2 −1 0 1 2 3
0 2 4
Concentration (mol/cm3)
Distance (x)
−x2 +
4Dt −(x−1)4Dt2
1 2
C(x, t) = e e
4πDt 4πDt
4Dt = 0.1
4Dt = 0.5
4Dt = 3.0
FIGURE 11.11 C(x, t) with infinite boundary conditions from two sources: one at x = 0 of strength Q = 1 and one at x = 1 of strength Q = 2 illustrating how there concentrations blend together.
A little mathematical trick is to put Equation 11.19 into a dimensionless variable form by divid- ing the top and bottom of the parameters in the parentheses by a to give
C
C erf x a
Dt a erf x a Dt a
0 2 2
1 2
1 4
1 2
1
= + 4
− −
( )
( )
( )
( )
/ /
/ /
.
The solution is now in terms of a dimensionless distance variable, x/a, and a dimensionless time variable, (4Dt a/ 2).
11.6.3 m
ethodofI
maGes11.6.3.1 Introduction
For semi-infinite boundary conditions, there are times when the concentration at the boundary of the medium, x = 0, needs to be a fixed value. For example, in Figure 11.11, the influence of the source at x = 1 extends beyond the origin into negative values of x. If this were a semi-infinite medium, clearly the material could not leave the end of the bar at x = 0 but would rather be reflected back to positive x values. The somewhat graphical way to handle this is by the method of images.
11.6.3.2 Concentration Confined to Positive Values of x
The top graphic in Figure 11.13 illustrates how this is done by the simple graphical/mathematical trick of applying an image source. For example, the shaded portion near the origin is the amount of material from the source at x0 that is reflected back from the origin. This is the same C(x, t) as the tail of a similar source placed at −x0 because its tail in the positive x direction will add to the source at x = x0. To get the solution, the real source and its image simply have to be added and evaluated for positive values of x. Therefore, the solution for a single source with semi-infinite boundary condi- tions in which material is not allowed to leave the end of the region at x = 0 is
C x t
Dt Dt
x x
Dt Dt
, e e
( ) (x x )
( )
= 1 − − + − +4
1 4
02
02
4 4
π π (11.20)
where the second term is the image source at −x0. Figure 11.14 shows the resulting concentration as a function of positive x for various values of 4Dt with a source of Q = 1.0 at x0 = 1. Note that, at x = 0, the C(0, t) concentrations approach a constant value or dC( , )0t dx/ =0, a derivative boundary condition presented in Chapter 8, because there can be no flux in the negative x direction at x = 0.
−5 −4 −3 −2 −1 0 1 2 3 4 5
0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.1
C(x, t)/C0
Distance
4Dt = 0
4Dt = 2 4Dt = 5 4Dt = 1 4Dt = 0.5
FIGURE 11.12 C(x, t) for infinite boundary conditions for an extended source of C(x, 0) = C0 for −1 ≤ x ≤ +1.
11.6 Finite Source Solutions 397
11.6.3.3 C(0, t) = 0
In some applications, the problem requires that the concentration remains zero—or some other fixed value—at the interface of the semi-infinite medium, C(0, t) = 0. The lower graphic in Figure 11.13 shows that this requires a negative image source, so that the concentration for the source at x = 1 is just can- celed by the overlapping concentration from the source of Q = −1 at x = −x0. The resulting concentra- tion, C(x, t: t > 0), is shown in Figure 11.15 as a single source at x0 = 1 and Q = 1:
C x t
Dt Dt
Dt Dt
, e e .
(x x ) (x x )
( )
= 1 − − − − +4
1 4
02
02
4 4
π π (11.21)
0 1 2 3 4
0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9
Concentration
Distance from surface 4Dt = 0.5
4Dt = 1
4Dt = 2
FIGURE 11.14 Semi-infinite boundary conditions with the application of a positive image (at x =−1.0) of a source at x = 1.0 of strength Q = 1 showing that dC/dx(0, t) = 0, or the material diffusing to the left is reflected at the boundary back into the positive direction.
C(x,t)
0 ∞
x
−x0
−x0
x0 C(x,t)
0
∞ x
x0
(x+x0)2
− 4Dt
1 e 4πDt
(x−x0)2
− 4Dt
1 e 4πDt
(x−x0)2
− 4Dt
1 e 4πDt
(x+x0)2
− 4Dt
−1 e 4πDt (a)
(b)
FIGURE 11.13 Illustration of the method of images for semi-infinite boundary conditions with a single source and image. (a) shows how the addition of an image source leads to a derivative boundary condition at x = 0: all of the material is reflected back to positive x at x = 0; (b) shows that with a negative image the concentration C(0, t) = 0: material leaves the surface.
11.6.4 s
urfaces
ource andm
easurement ofd
A very important experimental approach to measuring diffusion coefficients in solids is to place a single source of a radioactive tracer element on the surface of a bar and heat-treat it for a given time.
The method of images is very useful for measuring the resultant diffusion coefficient. For a single source in Figure 11.14 placed at the origin, one-half of the material is reflected back to positive values of x and the solution becomes
C x t Q
Dt
Q
, e Dte
x Dt
x
( )
=2 − = − Dt4
2 2
4 4
π π (11.22)
and taking the log10 of both sides gives
log , log
10 10 .
2
2 303 4
C x t Q
Dt
x
( )
= Dt
− ×
π (11.23)
which provides a simple, and very commonly used, method of determining diffusion coefficients in solids. As an example, the diffusion of nickel in NiO is to be measured. Deposit a thin surface layer of amount Q of radioactive nickel (e.g., 63Ni t, 1 2/ =101year) at the surface of a NiO bar then heat for a time (t) at some temperature. When cooled, take thin sections, Δx, and measure the radioactivity in each Δx-thick section as a function of x and then plot the concentration as a function of x2 on a semi- log plot, Equation 11.23, as shown in Figure 11.16. The diffusion coefficient can be determined from the slope of the line and, once determined, can be used to calculate Q from the intercept. If there are rapid diffusion paths, such as dislocations or grain boundaries, then their contributions will appear with less steep slopes at low concentrations and their diffusion coefficients can also be measured.
11.6.5 I
onI
mplantatIonIon implantation is a process in which high energy (>0.2 keV and <2 MeV) ions are implanted into a silicon substrate (or any other material) in an ion accelerator to replace doping by diffusion from the surface. Some advantages of ion implantation are: the depth of penetration can be controlled by the ion energy; it can be done at room temperature; and the doping location can be controlled by the pattern in the polymer photoresist material (Wolf and Tauber 2000). However, the concen- tration profile generated by ion implantation is usually too steep and narrow and must be given a post-deposition drive-in diffusion heat treatment. Normally, the concentration profile generated by
0 1 2 3 4
0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9
Concentration
Distance from surface 4Dt = 0.5
4Dt = 1
4Dt = 2
FIGURE 11.15 Semi-infinite boundary conditions with the application of a negative image (at x =−1.0) of a source at x = 1.0 of strength Q = 1 showing that C(0, t) = 0, or the material diffusing to the left leaves the material.
11.6 Finite Source Solutions 399
ion implantation is Gaussian with some mean penetration depth on the order of a few tenths of a micrometer. After a relatively short diffusion time, the shape of the original as-implanted profile does not have much effect on the concentration profile, C(x, t). To simplify the problem, yet get rea- sonably accurate values, Figure 11.17 shows concentration profiles as a function of 4Dt starting with the rectangular profile at 4Dt = 0 as the as-deposited ion implantation profile. Then with the method of images described in Section 11.6.2.2, the concentration profiles are given by Equation 11.16:
C x t C
Dt dx
C x t C Dt
x x Dt
x x Dt
, e
, e
( )
( )
( )
=( )
=− −
−∞
∞
− +
−
∫
0 4
0
0 4
4
4
02
02
π
π bb
a x x
Dt a
b
dx C
Dt dx
− − −
∫
0+ 40∫
4 002
π e
( )
where the first term in the second equation is simply the image of the second term. Substitution of y=(x x− 0)/ 4Dt and dx0= − 4Dt dy results in four terms, two each for the source and its image,
C x t e dyy e dy e
x b Dt
x a Dt y
y x a
,
( )
( )
( )
( )
− − − −+
+ −
−
∫ ∫
+1 2 1 2 1 2
4 0
0 4
π / π π
/
//
/ 4
0
0
1 4 2
Dt
x b Dt y
dy e dy
∫
− π∫
+ −( )
and from the definition of erf(z) C x t erf x b
Dt erf x a
Dt erf x a , Dt
( )
= +
− +
+ −
1
2 4
1
2 4
1
2 4 − −
1
2erf x b4
Dt . (11.24)
Admittedly, Equation 11.24 looks pretty ugly! But a, b, and 4Dt are known and the Equation 11.24 is simply a sum of four numbers and easily numerically evaluated. These are plotted in Figure 11.17 for
0 5 10 15 20
10−6 10−5 10−4 10−3 10−2 10−1 100
Concentration
(Distance)2 πDt −
Q x2
log10 C = log10
2.303 × 4Dt
Grain boundary {
Q Δx
x
FIGURE 11.16 Schematic illustrating the technique of a surface source to measure diffusion coefficients in solids. The inset shows a solid with an amount Q of radioactive isotope on the surface and allowed to diffuse into the bar. After a given time, the bar is sectioned parallel to the surface into slices of thickness Δx and the radioactivity (concentration) measured and plotted (the data points) on a semilog plot of concentra- tion versus depth, x2. From both the slope and intercept the diffusion coefficient at that temperature can be calculated. If there are fast diffusion paths present such as grain boundaries or dislocations, these will appear at low concentrations as shown.
a = 0.5 μm, b = 1 μm, and various values of 4Dt. Again, at 4Dt ≅ 1, the diffusion profile is reasonably flat for over 1 μm making it ideal for forming a p-n junction.