Chapter 5, Problem 5.1P

To determine

The area and dimensions (width and length) of the given capacitor which is fabricated using MOS technology.

Answer to Problem 5.1P

  $\begin{array}{l}\text{For,}{\left({\text{t}}_{\text{ox}}\right)}_{\text{1}}\text{=1nm}\\ \text{Area}=2.89\times {10}^{-10}{m}^2\\ W=L=17\mu m\\ \text{For,}{\left({\text{t}}_{\text{ox}}\right)}_2\text{=10nm}\\ \text{Area}=0.289\times {10}^{-10}{m}^2\\ W=L=5.37\mu m\end{array}$

Explanation of Solution

Given:

  $\begin{array}{l}$ {(t_{OX})}_1=1nm$$ {(t_{OX})}_2=10nm$C=10pF\end{array}$

Calculation:

Case 1:

  ${\left(t_{ox}\right)}_1=1nm$

Although formula oxide capacitance which is the quotient of permittivity of the silicon dioxide and oxide thickness is given by

  $\begin{array}{l}{C}_{ox}\text{}=\frac{{\epsilon }_{ox}}{t_{ox}}\\ \text{}\text{}{C}_{ox}\text{}=\frac{3.9{\epsilon }_0}{t_{ox}}\\ \text{plugging}\text{values}\\ \text{}{C}_{ox}\text{}=\frac{3.9\times 8.854\times {10}^{-12}}{1\times {10}^{-9}}\\ \text{}{C}_{ox}\text{}=34.53\times {10}^{-3}F/m^2\end{array}$

Although,

  $\begin{array}{l}C=C_{ox\text{}\cdot }\cdot W\cdot L\\ \text{where,}\\ W=\text{ channel width, }\\ L=\text{ channel length, }\\ C=\text{ total capacitance}\\ \text{Hence,}\\ W\cdot L=\frac{C}{C_{ox\text{}\cdot }}\\ \text{so,}\\ W\cdot L=\frac{10\times {10}^{-12}}{34.43\times {10}^{-3}}\\ W\cdot L=2.89\times {10}^{-10}{m}^2\end{array}$

Now, take square root from W · L in order to get channel width and channel length.

  $\begin{array}{l}W=L=\sqrt{2.89\times {10}^{-10}}\\ W=L=1.70\times {10}^{-5}\\ W=L=17\mu m\end{array}$

Case 2:

  ${\left(t_{ox}\right)}_2=10nm$

Although formula oxide capacitance which is the quotient of permittivity of the silicon dioxide and oxide thickness is given by:

  $\begin{array}{l}{C}_{ox}\text{}=\frac{{\epsilon }_{ox}}{t_{ox}}\\ \text{}\text{}{C}_{ox}\text{}=\frac{3.9{\epsilon }_0}{t_{ox}}\\ \text{plugging}\text{values}\\ \text{}{C}_{ox}\text{}=\frac{3.9\times 8.854\times {10}^{-12}}{10\times {10}^{-9}}\\ \text{}{C}_{ox}\text{}=3.453\times {10}^{-3}F/m^2\end{array}$

Although,

  $\begin{array}{l}C=C_{ox\text{}\cdot }\cdot W\cdot L\\ \text{where,}\\ W=\text{ channel width, }\\ L=\text{ channel length, }\\ C=\text{ total capacitance}\\ \text{Hence,}\\ W\cdot L=\frac{C}{C_{ox\text{}\cdot }}\\ \text{so,}\\ W\cdot L=\frac{10\times {10}^{-12}}{3.443\times {10}^{-3}}\\ W\cdot L=0.289\times {10}^{-10}{m}^2\end{array}$

Now take square root from W · L in order to get channel width and channel length

  $\begin{array}{l}W=L=\sqrt{0.289\times {10}^{-10}}\\ W=L=0.537\times {10}^{-5}\\ W=L=5.37\mu m\end{array}$