91影视

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

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,$C=\frac{k{蔚}_{0}A}{d}$ (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 ${貜}_0$ is the permittivity of free space.

Area of the square gate is given by:

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

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

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

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

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

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

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