A solution of (1.8) for an ideal diffusion system can be obtained given appropriate initial and boundary conditions. In the case of Si diffusors in an Al mediator, the rate of Si deposition at the exposed Al surface may be easily controlled such that the surface concentration, $C_s$, remains constant during the entire diffusion process (Fig. 2.4a),
| $$C(0,t)=C_s.$$ | (2.1) |
The concentration at some infinite distance must go to zero,
| $$C(\infty,t)=0.$$ | (2.2) |
|
Figure 2.4. SMME Diffusion Model. a) Si diffusors are deposited at the surface of the c-Al mediator. Once a sufficient concentration of diffusors is present at the surface, b) the Si atoms begin to diffuse through the c-Al mediator in the $+\hat{x}$ direction until they fall into the low-energy lattice sites of the c) buried c-Al/c-Si interface. A buried homoepitaxial c-Si thin film grows. |
The initial conditions of this system can be constructed as follows. Consider an infinite range $(-\infty < x < \infty)$ with the surface of Al at $x=0$. Given an infinite period of time, half the atoms in the negative range ($x < 0$) will diffuse into the material ($x > 0)$ since at this point the concentration on either side of $x=0$ is equal, i.e.,
| $$C(0^-,\infty)=C(0^+,\infty).$$ | (2.3) |
An equilibrium about the surface is obtained.
If the initial concentration is $2C_s$ ($x \lt 0$), an equilibrium concentration of $C_s$ will result in the material, $x >0$ (Fig. 2.4b). The initial condition to be satisfied then is,
|
$$\begin{eqnarray} C(x,0)=\left\{ \begin{array}{cl} 2C_S& x\lt 0\\ 0& x\gt 0 \end{array} \right. \end{eqnarray}.$$ |
(2.4) |
The solution of (1.8) must satisfy boundary conditions (2.1) and (2.2) and the initial condition (2.4). By the general principle of superposition, one such solution of the diffusion equation is (Appendix A.1)
|
$$C(x) = \int_{-\infty}^{\infty} k(\omega)\textrm{e}^{-i\omega x}\textrm{e}^{D\omega^2}\textrm{d}\omega.$$ |
(2.5) |
The initial condition (2.4) is satisfied if
$$C(x) = \int_{-\infty}^{\infty} k(\omega)\textrm{e}^{-i\omega x}\textrm{d}\omega$$
where $k(\omega)$ is the Fourier transform of $C(x)$,
$$k(\omega)=\frac{1}{2\pi}\int_{-\infty}^{\infty} C(x)\textrm{e}^{i\omega x}\textrm{d}x$$.
Combining these expressions,
$$C(x)=\int_{-\infty}^{\infty} \left[ \frac{1}{2\pi} \int_{-\infty}^{\infty}C(x')\textrm{e}^{i\omega x'}\textrm{d}x' \right]\textrm{e}^{-i\omega x}\textrm{e}^{-D\omega^2 t}\textrm{d}\omega$$
where $x'$ is a dummy variable. Interchanging the order of integration yields the result (Appendix A.2),
$$\begin{array}{rcl} C(x)&=&\frac{1}{2\pi} \int_{-\infty}^{\infty}C(x') \left[ \int_{-\infty}^{\infty} \textrm{e}^{-k\omega^2 t}\textrm{e}^{-i\omega(x-x')}\textrm{d}\omega\right]\textrm{d}x'\\ &=&\frac{1}{2\pi}\int_{-\infty}^{\infty}C(x') f(x)\textrm{d}\omega\\ &=&\int_{-\infty}^{\infty}C(x')\frac{1}{\sqrt{4\pi kt}}\textrm{e}^{-(x-x')^2/4kt}\textrm{d}x' \end{array}.$$
To satisfy (2.4) for $x < 0$,
| $$\begin{array}{rcl} C(x\lt0,t)&=& \frac{2C_S}{\sqrt{4\pi Dt}} \int_{-\infty}^{0}\textrm{e}^{-(x-x')^2/4Dt}\textrm{d}x'\\ &=& \frac{2C_s}{\sqrt{\pi}}\int_{-\infty}^{-x/\sqrt{4Dt}}\textrm{e}^{-z^2}\textrm{d}z\\ &=& \frac{2C_S}{\sqrt{\pi}}\left[ \int_0^{\infty}\textrm{e}^{z^2}\textrm{d}z-\int_0^{x/\sqrt{4Dt}}\textrm{e}^{-z^2}\textrm{d}z \right] \end{array}$$ | (2.6) |
where
$$\begin{array}{ccc} z=\frac{(x'-x)}{\sqrt{4Dt}} & \textrm{ and } &\textrm{d}z=\frac{\textrm{d}x'}{\sqrt{4Dt}} \end{array}$$
has been made. The first integral of (2.6) is well-known (Appendix A.3). The second integral is given by definition of the error function (Appendix A.4) in integral form. Hence,
| $$\begin{array}{rcl} C(x,t)&=&\frac{2C_S}{\sqrt{\pi}}\left[ \frac{\sqrt{\pi}}{2} -\frac{\sqrt{\pi}}{2} \textrm{erf}\left( \frac{x}{\sqrt{4DT}}\right)\right]\\ &=& C_S \left[ 1-\textrm{erf}\left( \frac{x}{\sqrt{4Dt}}\right) \right]\\ &=&C_S \textrm{erfc}\left( \frac{x}{\sqrt{4Dt}}\right) \end{array}.$$ | (2.7) |