The objective here is first to develop a relationship between the diffusion coefficient and the distance diffused, λ, that leads to λ2= 4Dt based on a less-than-mathematically rigorous—but intuitive—
model. This result suggests that a dimensionless variable of the form y x= / 4Dt might be used to transform the partial differential equation into a solvable ordinary differential equation. From this, it follows that some semi-infinite and infinite boundary condition problems of real-world processes can indeed be solved by substitution of the dimensionless variable, y x= / 4Dt.
11.2.2 d
ImensIonlessV
arIableFigure 11.1 schematically shows the diffusion of boron into silicon from a gaseous diborane, B2H, source that establishes a constant surface activity or concentration of boron, CS. The concentration of boron, C(x, t), is a function of both x and t since the boundary conditions are semi-infinite in this case.* The diffusion of boron into the silicon forms a p-n junction 1.0 μm below the surface of a silicon wafer that might be 25 mils†—635 μm—thick, so that semi-infinite boundary conditions clearly apply: 635 μm is essentially infinitely far away from 1 μm and the concentration there will not change during the diffusion. It is assumed that the initial concentration C(x, 0) = 0. This not only makes the problem slightly easier to solve but also is usually true in practice.
The solution to the one-dimensional Fick’s second law, ∂ ∂ =C t D/ (∂2C x/∂ 2), gives C(x, t), and the goal here is to find that solution. But first, with a linear approximation to the diffusion profiles—
C(x, t)—shown in Figure 11.2, a very suggestive result can be obtained that is extremely useful even if the procedure used to obtain the result is less than mathematically rigorous
(Mayer and Lau 1990). The procedure is to find a substitute variable that is a function of both time and distance and, when substituted into the partial differential equation, yields an easily solvable ordinary differential equation.
Assume that the C(x, t) profile for some time of diffusion is approxi- mately linear as shown in Figure 11.2 and the boron has diffused to a depth of x = λ. As usual,
J DdC
= − dx
and the steady-state solution and the concentration gradient, dC/dx, are given by
* As described in Chapter 8, semi-infinite boundary conditions in this case are C(0, t) = CS and C(∞, t) = 0.
† 1 mil = 0.001 in.
CS
x
0 ∞
Silicon B2H6
gas
Boron concentration
C(x, t2)
C(x, t1)
t2> t1
FIGURE 11.1 Schematic showing the C(x, t) of boron into silicon with a fixed surface concentra- tion, determined by the thermodynamics in the gas phase. This is an illustration of diffusion with semi- infinite boundary conditions: 0 ≤ x ≤ ∞.
C x C x dC
dx C
S
S
( )= −
= − 1 λ
λ
where CS is the surface concentration. Therefore, J D C= ( 0/λ), as has been seen several times before in Chapter 10. Now the number of moles of boron per unit length along the surface of the crystal is just the area, A, of the triangle in Figure 11.2, A = 1/2 C0λ. The flux of boron is just the time rate of change of the total amount of boron or the rate of change of the area
J d dt
C C d
= dt
=
0 0
2 2
λ λ
(11.1)
equating the two values for the flux,
J DC C d
= 0 = 0 dt λ 2
λ
gives
2D d
= λdtλ which when integrated gives the desired result
λ2=4Dt. (11.2)
Equation 11.2 basically says that the diffusion distance is proportional to 4Dt and that diffusion time and diffusion distance are not independent. This suggests a single dimensionless variable of (distance2)/(4Dt) or, more precisely, its square root, y x= / 4Dt, might be used to solve the partial differential equation. Indeed, if this dimensionless variable is substituted into the partial differ- ential equation of Fick’s second law, it will yield an ordinary differential equation in y that can be solved by usual straightforward techniques.
11.2.3 s
olutIonoff
Ick’
ss
econdl
awbyV
arIables
ubstItutIonIt is now demonstrated that the substitution of this dimensionless variable into Fick’s second law indeed leads to a solution of
∂
∂ = ∂
∂ C
t D C
x
2 2
in the form C(x, t). The following contains a number of equations simply because each step is carried out in detail; that is, no black boxes—and no new or fancy mathematics! Let the dimensionless vari- able be y2=x2/ 4Dt or y x= / 4Dt. Then, substituting first for ∂ ∂C t/ .
Concentration (mole/cc)
C0
x 0
λ
dλ ∞
FIGURE 11.2 Linear approximation of diffusion in a semi-infinite medium with a penetration depth of λ in time t and additional penetration of dλ after dt.
11.2 Solution with a Dimensionless Variable 385
∂
∂ = ∂
∂
∂
∂
∂
∂ =∂
( )
∂ = − = −
= −
−
C t
C y
y t y
t
x Dt t
xt
D t
x Dt / 4 1 y
2 4
1
2 4
3 2/
22t so that
∂
∂ = − ∂
∂ C
t y
t C y
2 . (11.3)
Similarly, on the right-hand side of the partial differential equation, substituting for ∂2C/∂2t,
∂
∂ = ∂
∂
∂
∂
∂
∂ =∂
( )
∂ =
∂
∂ = ∂
∂ C
x C y
y x y
x
x Dt
x Dt
C
x Dt
C y
/ 4 1
4 1
4 and differentiating one more time,
∂
∂ = ∂
∂
∂
∂
= ∂
∂
∂
∂
= ∂∂
∂
∂
∂
2 2
1 4 1
4 C
x x
C
x x Dt
C y
y Dt C
y y
∂∂ = ∂
∂
∂
∂
= ∂
∂
∂
∂
= ∂
∂
x y Dt
C
y Dt
Dt y C
y Dt
C y
1 4
1 4 1
4
1 4
2 2
so that
∂
∂ = ∂
∂
2 2
2 2
1 4 C
x Dt
C
y . (11.4)
Making the two substitutions of Equations 11.3 and 11.4 for 𝜕C/𝜕t and 𝜕2C/𝜕x2 in Fick’s second law
∂
∂ = ∂
∂
− ∂
∂ = ∂
∂ C
t D C
x y
t C y
D Dt
C y
2 2
2
2 4 2
shows that the substitution was successful because the partial differential equation in x and t is now an ordinary differential equation in y,
d C
dy ydC
dy
2
2 = −2 . (11.5)
To solve this ordinary differential equation, let z = dC/dy, so
dz
dy yz
z dz
dy y
= −
= − 2
1 2
which is easily integrated to give
ln
.
z y A
z Ae y
= − + ′
= −
2
2
Moreover, integrate z one more time to get C(y) z dC
dy Ae C y A e dw B
y
y w
= =
( )
= +−
∫
− 22 0
where w is just a dummy variable of integration. Putting in y x= / 4Dt leads to the final important result
C x t( , )=A
∫
0x Dte−w2dw B+ .4
/ (11.6)
Therefore, the partial differential equation can be solved as an ordinary differential equation, Equation 11.6, with the dimensionless variable y x= / 4Dt, and it gives a specific solution for two integration constants A and B. It should be noted that the partial differential equation was solved simply by making a variable substitution—a similarity variable—to give an easily integrated ordi- nary differential equation: no advanced mathematics beyond calculus! Of course, the right choice of the variable y was critical.