Semiconductor Electronics

Given, $R_{\text {out }}=200 \Omega, R_{\text {in }}=100 \mathrm{k} \Omega=10^5 \Omega$

$V_{C C}=3 \mathrm{~V}, V_{B E}=0.7 \mathrm{~V}, V_{C E}=0 \mathrm{~V}, \beta=200$

Applying KVL at output side,

$\begin{aligned} & & V_{C E} & =V_{C C}-I_C R_{\text {out }} \\ & & 0 & =3-I_c \times 200 \\ \Rightarrow & & I_C & =15 \mathrm{~mA} \\ \therefore & & I_\beta & =\frac{I_C}{\beta}=\frac{15 \times 10^{-3}}{200}=75 \times 10^{-6} \mathrm{~A} \\ \therefore & & I_B & =75 \mu \mathrm{A} \end{aligned}$

Again, Applying KVL at input side,

$\begin{aligned} V_{B E} & =V_{B B}-I_B R_{\text {in }} \\ \Rightarrow \quad V_{B B} & =V_{B E}+I_B R_{\text {in }} \\ & =0.7+75 \times 10^{-6} \times 10^5 \\ & =0.7+7.5=8.2 \mathrm{~V} \end{aligned}$

Given, voltage across Zener diode, $V_Z=6 \mathrm{~V}$

Since, Zener diode provide statelised supply of $6 \mathrm{~V}$.

In this case, the value of $R$ should be chosen such that negligible current flows through the Zener diode.

$\begin{aligned} & \therefore \quad I=I_Z+I_L=0+I_L \\ & I=\frac{30}{R+10^3} \\ & \text { But } \quad I=6 \mathrm{~mA}=6 \times 10^{-3} \mathrm{~A} \\ & \therefore \quad 6 \times 10^{-3}=\frac{30}{R+10^3} \\ & \Rightarrow \quad R=4 \times 10^3 \Omega=4 \mathrm{~k} \Omega \\ & \end{aligned}$

A p-n junction photo diode can be operated under photovoltaic conditions similar to that of a solar cell. The current-voltage characteristics of a photodiode under illumination are also similar. When light energy falls on the solar cell, then it converts solar energy into electrical energy. The area of solar is larger so that it could absorbed more amount of sunlight and hence more amount electrical energy is obtained. Hence, assertion and reason both are true but Reason is not the correct explanation of assertion.

Current passing in the circuit is

$I=\frac{P}{V}=\frac{100 \times 10^{-3}}{0.5}=0.2 \mathrm{~A}$

Value of connected series resistance,

$\begin{aligned} & R=\frac{1.5-0.5}{0.2} \Rightarrow R=\frac{1}{0.2} \\\\ \therefore R & =5 \Omega \end{aligned}$

Thickness of depletion layer depends upon the biasing of semiconductor devices and in the depletion layer, there is no free charge carriers are available in depletion layer.

Given, $\quad \beta=49$

Current gain $\beta=\frac{\Delta I_C}{\Delta I_B}$, where $\Delta I_C$ is change in collector current and $\Delta I_B$ is change in base current.

$\begin{aligned} \Delta I_C & =\beta \Delta I_B=49 \times 5 \\ & =245 \mu \mathrm{A} \\ \Delta I_E & =\Delta I_B+\Delta I_C \\ & =(245+5) \mu \mathrm{A} \\ & =250 \mu \mathrm{A} \end{aligned}$

Given for one indium atom to be doped in $5 \times 10^7$ silicon atoms.

Number density of silicon $=5 \times 10^{22}$

Number of acceptor atom$/ \mathrm{cm}^3=\frac{5 \times 10^{22}}{5 \times 10^7}$

$=1 \times 10^{15} \mathrm{~cm}^{-3}$

Hence, number of acceptor atom$/ \mathrm{cm}^3$ is $1 \times 10^{15} \mathrm{~cm}^{-3}$.

The simplified circuit is shown in figure below.

So, output $y=\overline{\bar{A} \cdot \bar{B} \cdot \bar{C}}$

If $\quad A=0, B=1, C=1$

$\therefore \quad y=\overline{\overline{0} \cdot \overline{1} \cdot \overline{1}}=\overline{1 \cdot 0 \cdot 0}=\overline{0}=1$