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}