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

\documentclass{article}
\usepackage{fullpage}
\newcommand{\PSbox}[3]{\mbox{\rule{0in}{#3}\special{psfile=#1}\hspace{#2}}}
\begin{document}
\begin{enumerate}
\item Chemical vapor deposition

  \begin{enumerate}
  \item The mass transfer Prandtl number is given by:
    $${\rm Pr}=\frac{\mu}{\rho D};\ \rho=\frac{MP}{RT},$$
    where $M$ is the molar mass.  Since 1atm=101300 Pa, this pressure is 10130
    Pa; Pascals are equivalent to Joules per cubic meter.  This therefore
    evaluates to:
    $$\rm\rho = \frac{0.040\frac{kg}{mol}\cdot10130\frac{J}{m^3}}
    {8.314\frac{J}{K\cdot mol}\cdot 500K} = 0.097\frac{kg}{m^3};\ 
    Pr=\frac{3\times10^{-5}\frac{N\cdot s}{m^2}}{0.097\frac{kg}{m^3}\cdot
      2\times10^{-4}\frac{m^2}{s}} = 1.54.$$
    This is not far from one, which is not atypical for a gas (yes, that was a
    double-negative).

  \item With a flow rate of $0.3\rm\frac{m^3}{s}$, and box width and height of
    2m$\times$0.75m, the average velocity is 0.2m/s.  The Reynolds number is
    then:
    $${\rm Re} = \frac{\rho u_{av}H}{\mu} =
    \frac{\rm 0.097\frac{kg}{m^3}\cdot0.2\frac{m}{s}\cdot0.75m}
    {3\times10^{-5}\frac{N\cdot s}{m^2}} = 485.$$
    Looks like flow will remain laminar.

  \item The velocity boundary layer thickness will grow as:
    $$\delta_u=5.0\sqrt{\frac{\nu x}{U_\infty}},$$
    for which we can use $u_{av}$ as $U_\infty$; this gives a maximum thickness
    of 0.27m at $x=2m$.  The velocity boundary layers will just meet at the end
    of the reactor.

    With a Prandtl number just above one, the concentration boundary layer will
    be somewhat smaller than the velocity boundary layer, but our expression
    for relative boundary layer thickness won't quite be right.  So we can
    estimate that the velocity and concentration boundary layers will look
    something like:
    \begin{center}
      \PSbox{cvdboxBL.ps}{420pt}{176pt}
    \end{center}

  \item Though the concentration boundary layer thickness correlation is not
    quite right, the high-Pr Nusselt number correlation is accurate down to a
    Prandtl number of 0.5, so we can use that:
    $${\rm Nu}_x = 0.332{\rm Re}_x^{1/2}{\rm Pr}^{1/3},$$
    along with expressions for the local mass transfer coefficient $h_{Dx}$ and
    diffusive flux:
    $${\rm Nu}_x = \frac{h_{Dx}x}{D_{fl}} \Rightarrow
    h_{Dx} = \frac{{\rm Nu}_xD_{fl}}{x};\ 
    J_y|_{y=0} = h_{Dx}(C_\infty-C_s).$$
    The molar density of the gas mixture is $P/RT$, so with 1 mol\% silane, we
    can use one percent of the overall pressure as the silane partial pressure
    and set $C_\infty=P_{\rm SiH_4}/RT$.  The problem states that we can
    estimate $C_s=0$.  The results are summarized as follows:
    \begin{center}
      \begin{tabular}{c|cccc|}
        Distance $x$ & Re$_x$ & Nu$_x$ & $h_{Dx}$ & $J_y|_{y=0}$ \\ \hline
        0.10m        & 65     & 3.08   & $6.17\times10^{-3}$m/s &
        $1.5\times10^{-4}$mol/m$^2$sec \\
        0.30m        & 195    & 5.34   & $3.56\times10^{-3}$m/s &
        $8.68\times10^{-5}$mol/m$^2$sec \\ \hline
      \end{tabular}
    \end{center}
    Note that this close to the entrance, we're really pushing the envelope of
    validity of this solution (since $\delta_C\ll x$ is not so valid).  Call it
    a rough estimate.

  \item We can just divide the flux by molar density of the solid to calculate
    the deposition rate:
    $\frac{dY}{dt} = \frac{J\cdot M_{\rm Si}}{\rho_{solid}}.$
    At 10 cm, this will be 1.7 nm/sec; at 30 cm, 0.97 nm/sec.  This means a 20
    cm wafer placed 10 cm from the entrance (from $x=10$cm to $x=30$cm) will
    have a 70\% variation in deposition rate across it, which is a huge
    variation!

  \item There are various ways to make the deposition more uniform.  A common
    one is to place the substrates on an incline tilting upward about
    30$^\circ$, so the velocity increases with $x$, flattening out the
    concentration boundary layer a bit.  One can also run under conditions more
    likely to be reaction-limited ({\em e.g.} at lower temperature), so the
    silane concentration in the gas will be roughly uniform throughout the
    chamber and right up to the wafers, making deposition roughly uniform.

    But the real answer is to ditch this reactor design, and go to something
    very different.  Many modern reactors use stagnation flow produced by
    something like a shower head directing reactant gas at an individual wafer,
    or individual wafer rotation, to achieve near-uniform velocity and
    concentration boundary layers.  (In response to a similar question on last
    year's final, a student suggested ``Rotisserie CVD'' with wafers moving
    around like chickens in an oven...)  Because it processes just a single
    wafer per chamber, this equipment is more expensive, but is worth it to
    deliver uniformity to within a few percent across each wafer.
  \end{enumerate}
\end{enumerate}
\end{document}