Emulation and Simulation of
Microelectronic Fabrication Processes
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A.2 3D Problem
The three dimensional problem can be approached in exactly the same way. The plane which intersects a cube is visualised in Fig. A.2 and is given by:
\( \seteqsection {A} \) \( \seteqnumber {8} \)
\begin{equation} p = \frac {\hat {n} \cdot \vec {P}}{n_z} - \frac {n_x x + n_y y}{n_z} = q - \frac {n_x x + n_y y}{n_z} \quad . \end{equation}
In the x-direction, the integral is bound by \(a\) and \(b\), which are defined as
\( \seteqsection {A} \) \( \seteqnumber {9} \)
\begin{equation} a = \frac {n_z}{n_x}q = \frac {\hat {n} \cdot \vec {P}}{n_x} \quad , \end{equation}
\( \seteqsection {A} \) \( \seteqnumber {10} \)
\begin{equation} b = \frac {n_z}{n_x}q - \frac {n_y}{n_x} = \frac {\hat {n} \cdot \vec {P} - n_y}{n_x} \end{equation}
On the top plane of the cube, the integral is bound by \(c\) and \(d\):
\( \seteqsection {A} \) \( \seteqnumber {11} \)
\begin{equation} c = \frac {\hat {n} \cdot \vec {P}}{n_x} - \frac {n_z}{n_x} \end{equation}
\( \seteqsection {A} \) \( \seteqnumber {12} \)
\begin{equation} d = \frac {\hat {n} \cdot \vec {P}}{n_x} - \frac {n_z}{n_x} - \frac {n_y}{n_x} \end{equation}
In addition to the bounds on the x-axis, the bounds for the y-axis also need to be defined, so \(p\) does not contribute to the integral if it is outside of the cube. These are the lines describing the intersection of \(p\) with the sides
of the cube parallel to the x-y plane. \(f(x)\) is the intersect with \(p=0\), given by
\( \seteqsection {A} \) \( \seteqnumber {13} \)
\begin{equation} f(x) = \frac {\hat {n} \cdot \vec {P}}{n_y} - \frac {n_x}{n_y} x \quad . \end{equation}
The intersect with the side of the cube at \(p=1\) is given by
\( \seteqsection {A} \) \( \seteqnumber {14} \)
\begin{equation} g(x) = \frac {\hat {n} \cdot \vec {P} - n_z}{n_y} - \frac {n_x}{n_y} x \quad . \end{equation}
The volume \(V\) under the plane is then given by the sum of volumes:
\( \seteqsection {A} \) \( \seteqnumber {15} \)
\begin{equation} V = V_1(b) + V_2(a, b) - V_3(d) - V_4(c, d) \end{equation}
Explicitly, these volumes are:
\( \seteqsection {A} \) \( \seteqnumber {16} \)
\begin{equation} \begin {split} V &= \int _{x=0}^{b} \int _{y=0}^{1} p ~ dy dx + \int _{x=b}^{a} \int _{y=0}^{f(x)} p ~ dy dx \\ &- \int _{x=0}^{d} \int _{y=0}^{1} p ~ dy dx - \int _{x=d}^{c} \int
_{y=0}^{g(x)} p ~ dy dx \end {split} \end{equation}
The solutions to the integrals are:
\( \seteqsection {A} \) \( \seteqnumber {17} \)
\begin{equation} V_1(b) = -\frac {n_x}{2n_z} b^2 + \left ( \frac {\hat {n} \cdot \vec {P}}{n_z} - \frac {n_y}{2n_z} \right ) b \end{equation}
\( \seteqsection {A} \) \( \seteqnumber {18} \)
\begin{equation} V_2(a, b) = \frac {1}{2n_z n_y} \left [ \frac {n_x^2}{3} \left ( a^3 - b^3 \right ) - n_x \left (\hat {n} \cdot \vec {P}\right )\left (a^2 - b^2\right ) + \left (\hat {n} \cdot \vec {P}\right )^2 \left (a -
b\right ) \right ] \end{equation}
\( \seteqsection {A} \) \( \seteqnumber {19} \)
\begin{equation} V_3(d) = V_1(d) \end{equation}
\( \seteqsection {A} \) \( \seteqnumber {20} \)
\begin{equation} V_4(c, d) = V_2(c, d) - \frac {n_z}{2n_y} \left (c - d\right ) \end{equation}