Semiconductor Microchip Non-Uniform Surface Charge Accumulation (Electrical & Computer Engineering) Solution

An integrated circuit (IC) hardware design engineer at the UNLV Howard R. Hughes College of Engineering is analyzing electrostatic charge accumulation on a high-speed microprocessor silicon die. The microscopic planar capacitor plate occupies a rectangular footprint:

During chemical-mechanical planarization (CMP) and high-density ion implantation, dopant dispersion produces an oscillatory non-uniform surface charge density across the die:

where $x$ and $y$ are spatial coordinates on the die face.

Step-by-Step Solution:

1. State the Governing Double Integral Formulation: State the Surface Charge Accumulation Formula (defined in Double Integrals over a Rectangular Region):

   $$Q = \iint_R \sigma(x, y) \, dA$$

2. Select the Strategically Optimal Order of Integration: Under Fubini's Theorem, the integral can be formulated in either order:

   • If integrating $dy \, dx$, the inner integral is $\int 5xy \cos(x^2 y) \, dy$, requiring integration by parts because the derivative of the argument $x^2 y$ with respect to $y$ is $x^2$, which does not match the exterior factor $5xy$.
   • If integrating $dx \, dy$, the inner integral is $\int 5xy \cos(x^2 y) \, dx$. Here, $\frac{\partial}{\partial x}(x^2 y) = 2xy$, which matches the exterior multiplier $xy$ up to a scalar constant, allowing direct single-step $u$-substitution! Therefore, we set up the iterated integral as $dx \, dy$:
     $$Q = \int_{0}^{2} \int_{0}^{3} 5xy \cos(x^2 y) \, dx \, dy$$

3. Evaluate the Inner Integral ($dx$): Hold $y$ constant and use substitution:

   $$\text{Let } u = x^2 y \implies \frac{\partial u}{\partial x} = 2xy \implies xy \, dx = \frac{du}{2}$$
   Convert limits for $x$:
   $$\text{At } x = 0 \implies u = 0$$
   $$\text{At } x = 3 \implies u = (3)^2 y = 9y$$
   Substitute into the inner integral:
   $$\int_{0}^{3} 5xy \cos(x^2 y) \, dx = \int_{0}^{9y} 5 \cos(u) \frac{du}{2} = \frac{5}{2} [\sin(u)]_{0}^{9y} = \frac{5}{2} \sin(9y)$$

4. Evaluate the Outer Integral ($dy$):

   $$Q = \int_{0}^{2} \frac{5}{2} \sin(9y) \, dy = \frac{5}{2} \left[ -\frac{1}{9} \cos(9y) \right]_{0}^{2} = -\frac{5}{18} [\cos(9y)]_{0}^{2}$$
   $$= -\frac{5}{18} \Big(\cos(18) - \cos(0)\Big) = -\frac{5}{18} (\cos(18) - 1) = \frac{5}{18} (1 - \cos(18))$$
   Note that the argument $18$ is in radians. Using $\cos(18\text{ rad}) \approx 0.660317$:
   $$Q = \frac{5}{18}(1 - 0.660317) = \frac{5}{18}(0.339683) \approx 0.09436\text{ nC}$$

Meaning of the Answer & Real-Life Application:

• Quantitative Interpretation: The total net stored electrostatic charge is $Q = \frac{5}{18}(1 - \cos(18))\text{ nC} \approx 0.0944\text{ nC}$ ($94.4\text{ pC}$). The alternating harmonic behavior of $\cos(x^2 y)$ produces localized positive and negative charge pockets across the die surface that partially cancel, leaving a modest residual positive net charge.
• Engineering Application: In sub-micron semiconductor manufacturing, parasitic capacitive charge changes the gate switching delay $\tau = R C_{\text{parasitic}}$. Computer and electrical engineers model surface charge accumulations via double integrals to design on-chip electrostatic discharge (ESD) protection clamp circuits and verify signal timing margins in clock distribution trees.