[ Home ]

# Emulation and Simulation of Microelectronic Fabrication Processes

#### 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:

$$p = \frac {\hat {n} \cdot \vec {P}}{n_z} - \frac {n_x x + n_y y}{n_z} = q - \frac {n_x x + n_y y}{n_z} \quad .$$

In the x-direction, the integral is bound by $$a$$ and $$b$$, which are defined as

$$a = \frac {n_z}{n_x}q = \frac {\hat {n} \cdot \vec {P}}{n_x} \quad ,$$

$$b = \frac {n_z}{n_x}q - \frac {n_y}{n_x} = \frac {\hat {n} \cdot \vec {P} - n_y}{n_x}$$

On the top plane of the cube, the integral is bound by $$c$$ and $$d$$:

$$c = \frac {\hat {n} \cdot \vec {P}}{n_x} - \frac {n_z}{n_x}$$

$$d = \frac {\hat {n} \cdot \vec {P}}{n_x} - \frac {n_z}{n_x} - \frac {n_y}{n_x}$$

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

$$f(x) = \frac {\hat {n} \cdot \vec {P}}{n_y} - \frac {n_x}{n_y} x \quad .$$

The intersect with the side of the cube at $$p=1$$ is given by

$$g(x) = \frac {\hat {n} \cdot \vec {P} - n_z}{n_y} - \frac {n_x}{n_y} x \quad .$$

The volume $$V$$ under the plane is then given by the sum of volumes:

$$V = V_1(b) + V_2(a, b) - V_3(d) - V_4(c, d)$$

Explicitly, these volumes are:

$$\begin {split} V &= \int _{x=0}^{b} \int _{y=0}^{1} p ~ dy dx + \int _{x=b}^{a} \int _{y=0}^{f(x)} p ~ dy dx \\ &- \int _{x=0}^{d} \int _{y=0}^{1} p ~ dy dx - \int _{x=d}^{c} \int _{y=0}^{g(x)} p ~ dy dx \end {split}$$

The solutions to the integrals are:

$$V_1(b) = -\frac {n_x}{2n_z} b^2 + \left ( \frac {\hat {n} \cdot \vec {P}}{n_z} - \frac {n_y}{2n_z} \right ) b$$

$$V_2(a, b) = \frac {1}{2n_z n_y} \left [ \frac {n_x^2}{3} \left ( a^3 - b^3 \right ) - n_x \left (\hat {n} \cdot \vec {P}\right )\left (a^2 - b^2\right ) + \left (\hat {n} \cdot \vec {P}\right )^2 \left (a - b\right ) \right ]$$

$$V_3(d) = V_1(d)$$

$$V_4(c, d) = V_2(c, d) - \frac {n_z}{2n_y} \left (c - d\right )$$