Question

A silicon-based MOSFET has a square gate $\mathbf{0}\mathbf{.}\mathbf{50}\mathbf{\mu m}$on edge. The insulating silicon oxide layer that separates the gate from the p-type substrate is $\mathbf{0}\mathbf{.}\mathbf{20}\mathbf{\mu m}$thick and has a dielectric constant of 4.5 . (a) What is the equivalent gate – substrate capacitance (treating the gate as one plate and the substrate as the other plate)? (b) Approximately how many elementary charges eappear in the gate when there is a gate – source potential difference of 1.0V ?

Short answer

1. The equivalent gate-substrate capacitance is $5.0\times {10}^{-17}\mathrm{F}$.
2. The number of elementary charges appearing in the gate when there is a gate-source potential difference of is $31\times {10}^2$.

Step-by-step solution

The given data

1. The dimension of the square gate of the MOSFET,$a=0.50\mu m$
2. The thickness of the insulating layer separating the gate from the p-type substrate,$d=0.20\mu m$
3. Dielectric constant of the layer, $k=4.5$
4. Source potential difference,$V=1.0\mathrm{V}$

Understanding the concept of capacitance and charge

Capacitance is the ability of any conductor to store charge. It is described as the amount of charge stored per unit change in potential difference.

Formulae:

The capacitance between two plates of the capacitor,$C=\frac{k{\epsilon }_{0}A}{d}$ (i)

The charge between the capacitor plates,$q=CV$ (ii)

The number of elementary charges in the layer,$N=q/e$ (iii)

Here, A is the cross-sectional area of the plate, d is the distance between the plates, V is the potential difference between the plates, q is the charge stored and ${؏}_0$ is the permittivity of free space.

a) Calculation of the value of the capacitance

Area of the square gate is given by:

$$\begin{gathered} A={\left(a\right)}^2 \\ ={\left(0.50\mu m\right)}^2 \end{gathered}$$

Using the given data in equation (i), we can get the capacitance value of the equivalent gate-substrate as follows:

$$\begin{gathered} C=\frac{\left(4.5\right)\left(8.85\times {10}^{-12}\mathrm{F}/\mathrm{m}\right){\left(0.50\mathrm{\mu m}\right)}^2}{\left(0.20\mathrm{\mu m}\right)} \\ =5.0\times {10}^{-17}\mathrm{F} \end{gathered}$$

Hence, the value of the capacitance is $5.0\times {10}^{-17}\mathrm{F}$.

b) Calculation of the number of elementary charges

Using equation (ii) and equation (ii), we can get the number of the elementary charge particles appearing in the gate, as follows:

$$\begin{gathered} N=\frac{CV}{e} \\ =\frac{\left(5.0\times {10}^{-17}\mathrm{F}\right)\left(1.0\mathrm{V}\right)}{\left(1.6\times {10}^{-19}\mathrm{C}\right)} \\ =3.1\times {10}^2 \end{gathered}$$

Hence, the number of the charges is $3.1\times {10}^2$.