root/trunk/matml/transport/problems/dopantdrive/dopantdrive-solution.tex

\documentclass{article}
\usepackage{fullpage}
\newcommand{\PSbox}[3]{\mbox{\rule{0in}{#3}\special{psfile=#1}\hspace{#2}}}
\begin{document}
\begin{enumerate}
\item Drive-in diffusion of semiconductor dopant

  \begin{enumerate}
  \item Because there is a finite amount of material diffusing into a
    semi-infinite solid, the shrinking Gaussian solution is the one to use.
    The ``box'' near the surface (x=0) represents the initial phosphorous
    distribution at $C_{P0}$ over thickness $\Delta x$, the horizontal line
    near the bottom is the uniform boron content, and the curves represent
    phosphorous concentrations during drive-in diffusion.

    \begin{center}
      \PSbox{phosbor.ps}{238pt}{135pt}
    \end{center}

  \item The one-sided semi-infinite timescale is given by:
    $$t=\frac{L^2}{16D} =
    {\rm \frac{(0.1cm)^2}{16\cdot2.49\times10^{-12}\frac{cm^2}{s}} =
      2.5\times10^6 seconds}$$
    About a month!

  \item We want to find at what depth the phosphorous and boron concentrations
    are equal.  The phosphorous concentration is the shrinking Gaussian:
    $$C_P=\frac{C_{P0}\Delta x}{\sqrt{\pi Dt}}
    \exp\left(-\frac{x^2}{4Dt}\right)$$
    Set that to the boron concentration $C_B$ and solve for $x$:
    $$C_B=\frac{C_{P0}\Delta x}{\sqrt{\pi Dt}}
    \exp\left(-\frac{x^2}{4Dt}\right)$$
%    $$\frac{C_B\sqrt{\pi Dt}}{C_{P0}\Delta x} =
%    \exp\left(-\frac{x^2}{4Dt}\right)$$
    $$\ln\left(\frac{C_B\sqrt{\pi Dt}}{C_{P0}\Delta x}\right) =
    -\frac{x^2}{4Dt}$$
    $$x=\sqrt{-4Dt\ln\left(\frac{C_B\sqrt{\pi Dt}}{C_{P0}\Delta x}\right)}$$
    For $t=1800$ seconds, this gives $x=1.95\times10^{-4}$ cm, or about 2
    microns.

    For $t=5400$ seconds, this gives $x=2.92\times10^{-4}$ cm, or about 3
    microns.
  \end{enumerate}
\end{enumerate}
\end{document}