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

Gate I is OR gate, $$Y' = A + B$$

Gate II is AND gate, $$Y = Y' \cdot C$$
$$\therefore A = 1,B = 0,C = 1\,{\text{will}}\,{\text{give}}\,Y = 1$$

$${D_2}$$ is forward biased whereas $${D_1}$$ is reversed biased. So effective resistance of the circuit
$$\eqalign{ & R = 4 + 2 = 6\Omega \cr & \therefore i = \frac{{12}}{6} = 2\,A \cr} $$

Given, current gain $$\beta = 100,{R_c} = 3\,k\Omega ,{R_b} = 2\,k\Omega $$
Voltage gain $$\left( {{A_v}} \right) = \beta \frac{{{R_c}}}{{{R_b}}} = 100\left( {\frac{3}{2}} \right) = 150$$
Power gain $$ = {A_v}\beta = 150\left( {100} \right) = 15000$$

Integrated circuits are miniature electronic circuit produced within a single crystal of a semiconductor such as silicon. They contain a million or so transistors and resistors or capacitors. They are widely used in memory circuits, micro computers, pocket calculators and electronic watches on account of their low cost and bulk, reliability into specific regions of the semiconductor crystals.

$$\frac{{{I_e}}}{{{I_h}}} = \frac{{{n_e}eA{v_e}}}{{{n_h}eA{v_h}}} \Rightarrow \frac{7}{4} = \frac{7}{5} \times \frac{{{v_e}}}{{{v_h}}} \Rightarrow \frac{{{v_e}}}{{{v_h}}} = \frac{5}{4}$$

A	B	C
1	1	1
1	0	1
0	1	1
0	0	0

This truth table follows the boolean algebra $$C = A + B$$   which is for $$OR$$  gate

The current gain $$\left( \alpha \right)$$  in common-base configuration is
$$\eqalign{ & \alpha = \frac{{\Delta {i_c}}}{{\Delta {i_e}}}\,\,\left[ {{\text{at}}\,{\text{constant voltage across}}\,{\text{collector base junction}}} \right] \cr & = 0.98 \cr} $$
The current gain in common-emitter configuration
$$\beta = \frac{\alpha }{{1 - \alpha }} = \frac{{0.98}}{{1 - 0.98}} = \frac{{0.98}}{{0.02}} = 49$$

Input current or base current is given by
$${i_b} = \frac{{\Delta {V_b}}}{{{R_b}}} = \frac{{0.01}}{{1000}} = {10^{ - 5}}A\left[ {_{{R_b} = {\text{ base resistance}}}^{\Delta {V_b} = {\text{ input voltage}}}} \right]$$
Also current gain
$$\beta = \frac{{{i_c}}}{{{i_b}}}$$
$$\eqalign{ & \therefore {i_c} = \beta {i_b} = 50 \times {10^{ - 5}}A \cr & = 500 \times {10^{ - 6}}A \cr & = 500\,\mu A \cr} $$

In common emitter configuration current gain
$${A_i} = \frac{{ - h{f_e}}}{{1 + {b_{oe}}{R_L}}} = \frac{{ - 50}}{{1 + 25 \times {{10}^{ - 6}} \times 1 \times {{10}^3}}} = - 48.78$$

The minimum voltage range of $$DC$$  source is given by
$$\eqalign{ & {V^2} = PR \cr & \because P = 1\,watt, \cr & R = 100\,\Omega = 1 \times 100 \cr & \therefore V = 10\,volt. \cr} $$