Semiconductors and Electronic Devices MCQ Questions & Answers in Modern Physics | Physics

The Boolean expression which satisfies the output of this logic gate is $$C = A \cdot B,$$   which is for AND gate.

In the circuit, diode $${D_1}$$ is forward biased and diode $${D_2}$$ is reverse biased. Therefore, no current flows in the arm containing $${D_2}$$ and all of the current flows through arm containing $${D_1}.$$
Thus, current flowing through the circuit is given by
$$I = \frac{V}{{{R_{eq}}}} = \frac{5}{{20 + 30}} = \frac{5}{{50}}\,A\,\,\left[ {\because {R_{eq}} = 20\,\Omega + 30\,\Omega } \right]$$

When the temperature increases, certain bounded electrons become free which tend to promote conductivity. Simultaneously number of collisions between electrons and positive kernels increases

(A) Gate $$A$$ is NAND gate, its output will be
$${Y_A} = \overline {1 \cdot 1} = \overline 1 = 0$$
(B) Gate $$B$$ is NOR gate, its output will be
$${Y_B} = \overline {1 + 1} = \overline 1 = 0$$
(C) Gate $$C$$ is NAND gate, its output will be
$${Y_C} = \overline {0 \cdot 1} = \overline 0 = 1$$
(D) Gate $$D$$ is XOR gate, its output will be
$${Y_D} = 0 \oplus 0 = 0 \cdot \overline 0 + \overline 0 \cdot 0 = 0$$

Thus, gate (C) will give an output of 1.

Collector current $${i_C} = \frac{V}{R} = \frac{3}{{3 \times {{10}^3}}} = {10^{ - 3}}A$$
Now base current, $${i_B} = \frac{{{i_C}}}{B} = \frac{{{{10}^{ - 3}}}}{{100}} = {10^{ - 5}}A$$
As, voltage $${V_{{\text{in}}}} = {i_B}{R_B}$$
$$\eqalign{ & \therefore {V_{{\text{in}}}} = {10^{ - 5}} \times 2 \times {10^3} \cr & = 2 \times {10^{ - 2}}volts \cr} $$
So, voltage gain $${A_V} = \frac{{{V_{{\text{out}}}}}}{{{V_{{\text{in}}}}}} = \frac{3}{{2 \times {{10}^{ - 2}}}} = 150$$
Power gain $$ = {A_V} \times \beta = 150 \times 100 = 15000$$

$$\eqalign{ & I = nA\,e{v_d}\,\,{\text{or}}\,\,I \propto n{v_d} \cr & \therefore \frac{{{I_e}}}{{{I_h}}} = \frac{{{n_e}{v_e}}}{{{n_h}{v_h}}}\,\,{\text{or}}\,\,\frac{{{n_e}}}{{{n_h}}} = \frac{{{I_e}}}{{{I_h}}} \times \frac{{{v_h}}}{{{v_e}}} = \frac{7}{4} \times \frac{4}{5} = \frac{7}{5} \cr} $$

Using $${U_{av}} = \frac{1}{2}{e_0}{E^2}$$
But $${U_{av}} = \frac{P}{{4p{r^2}'c}}$$
$$\eqalign{ & \therefore \frac{P}{{4p{r^2}}} = \frac{1}{2}{e_0}{E^2}'c \cr & E_0^2 = \frac{{2P}}{{4p{r^2}{e_0}c}} = \frac{{2'0.1'9'{{10}^9}}}{{1'3'{{10}^8}}} \cr & \therefore {E_0} = \sqrt 6 = 2.45\;V/m \cr} $$

If a trivalent impurity is mixed in a pure (intrinsic) semiconductor, then it becomes a $$p$$-type semiconductor. As given indium is trivalent impurity so it must be doped to $$Ge$$  or $$Si$$  to make it $$p$$-type semiconductor.

$$\eqalign{ & {\text{Voltage again}}\left( {{A_V}} \right) = \frac{{{V_{{\text{out}}}}}}{{{V_{{\text{in}}}}}} = \frac{{{I_{{\text{out}}}}}}{{{I_{{\text{in}}}}}} \times \frac{{{R_{{\text{out}}}}}}{{{R_{{\text{in}}}}}} \cr & {A_V} = \frac{{2 \times {{10}^{ - 3}}}}{{40 \times {{10}^{ - 6}}}} \times \frac{{4 \times {{10}^3}}}{{100}} = 2 \times 100 = 2000 \cr} $$

$$\eqalign{ & {\text{As}}\,\,{A_v} = \beta \frac{{{R_L}}}{{{R_i}}}\,\,{\text{or}}\,\,G = \left( {\frac{\beta }{{{R_i}}}} \right){R_L} \cr & \Rightarrow G = {g_m}{R_L} \Rightarrow G \propto {g_m}\,\,\left[ {\because {g_m} = \frac{{\Delta {I_C}}}{{\Delta {V_B}}} = \frac{{\Delta {I_C}}}{{\Delta {I_B}{R_i}}}} \right]\,\left[ {\because {g_m} = \frac{\beta }{{{R_i}}}} \right] \cr & \therefore \frac{{{G_2}}}{{{G_1}}} = \frac{{{g_{{m_2}}}}}{{{g_{{m_1}}}}} \Rightarrow {G_2} = \frac{{0.02}}{{0.03}} \times G \cr} $$
So, voltage gain, $${G_2} = \frac{2}{3}G$$