Chapter 5, Problem 5.1P

To determine

The area required per 1pF of capacitance for oxide thickness ranging from 2nm to 10nm.

The required dimensions for a square plate capacitor of 10pF.

Answer to Problem 5.1P

Area required per $\text{1pF}$ of capacitance for oxide thickness of $\text{2nm}$ is $5.797\times {10}^{-10}{\text{m}}^{\text{2}}$ , and the dimensions needed for square plate capacitor of $10\text{pF}$ is $24.08\text{μm}$ for its channel length and the channel width.

Area required per $\text{1pF}$ of capacitance for oxide thickness of $\text{10nm}$ is $2.9\times {10}^{-9}{\text{m}}^{\text{2}}$ , and the dimensions needed for square plate capacitor of $\text{10pF}$ is $53.8\text{5μm}$ for its channel length and the channel width.

Explanation of Solution

Given Information:

Fabricating a capacitor, utilizing the gate metallization and the substrate as the capacitor electrodes.

  $1pF$ of capacitance.

Oxide thickness ranging from $2nm$ to $10nm$ .

Square plate capacitor of $10pF$ .

Initially determining the oxide capacitance which is the quotient of permittivity of silicon dioxide and oxide thickness of $2nm$ :

  $C_{ox}=\frac{{ϵ}_{ox}}{t_{ox}}=\frac{3.9\times {ϵ}_o}{t_{ox}}=\frac{3.9\times (8.854\times {10}^{-12})}{2\times {10}^{-9}}=0.01725{\text{Fm}}^{\text{-2}}$

Calculating the area required per $1pF$ capacitance using the formula:

  $C=C_{ox}\times W\times L$

Where $W\times L$ is the area required, and $W$ is the channel width and $L$ is the channel length.

  $A_1=\frac{C}{C_{ox}}=\frac{1\times {10}^{-12}}{0.01725}=57.97{\text{μm}}^{\text{2}}$

Since it’s a square plate capacitor, the square root of $57.97{\text{μm}}^{\text{2}}$ will give the channel width and the channel length.

  $W=L=\sqrt{5.797\times {10}^{-10}}=24.08\times {10}^{-6}\text{m}=24.08\text{μm}$

Determining the oxide capacitance which is the quotient of permittivity of silicon dioxide and oxide thickness of $10nm$ :

  $C_{ox}=\frac{{ϵ}_{ox}}{t_{ox}}=\frac{3.9\times {ϵ}_o}{t_{ox}}=\frac{3.9\times (8.854\times {10}^{-12})}{10\times {10}^{-9}}=3.45\times {10}^{-3}F{m}^{-2}$

Calculating the area required per $1pF$ capacitance using the formula:

  $C=C_{ox}\times W\times L$

Where $W\times L$ is the area required, and $W$ is the channel width and $L$ is the channel length.

  $\begin{array}{l}{A}_2=\frac{C}{C_{ox}}=\frac{{10}^{-12}}{3.45\times {10}^{-3}}=289.86\times {10}^{-12}{\text{m}}^{\text{2}}\\ A_2=289.86{\text{μm}}^{\text{2}}\end{array}$

Since it’s a square plate capacitor, the square root of $2.9\times {10}^{-9}{m}^2$ will give the channel width and the channel length.

  $W=L=\sqrt{2.9\times {10}^{-9}}=53.85\times {10}^{-6}\text{m}=53.85\text{μm}$

Conclusion:

Area required per $1pF$ of capacitance for oxide thickness of $2nm$ is $5.797\times {10}^{-10}{m}^2$ , and the dimensions needed for square plate capacitor of $10pF$ is $24.08\mu m$ for its channel length and the channel width.

Area required per $1pF$ of capacitance for oxide thickness of $10nm$ is $2.9\times {10}^{-9}{m}^2$ , and the dimensions needed for square plate capacitor of $10pF$ is $53.85\mu m$ for its channel length and the channel width.