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

The truth table for the given circuit is

A	B	$${Y_1}$$	$${Y_2}$$	$$Y = \overrightarrow {{Y_1} + {Y_2}} $$
0	0	1	1	0
0	1	1	0	0
1	0	0	1	0
1	1	0	0	1

The truth table shows that both the inputs are high, then we are getting high value of output otherwise zero. Hence, the combination represents AND gate.

Zener diode is a silicon crystal diode having an unusual reverse current characteristic which is particularly suitable for voltage regulating purposes or voltage stabilisation purposes.

The accumulation of electric charges of opposite polarities in the two regions of the junction gives rise to an electric field between these region due to the diffusion of charge carriers. This electric field opposes further flow of electrons from the $$n$$-region to the $$p$$ -region and that of holes from the $$p$$ -region to $$n$$ -region. This electric field sets a potential barrier $${V_B}$$ at the junction which opposes further diffusion of free charge carriers into opposite regions. In the vicinity of the junction, a region is created, which is devoid of free charge carriers and has immobile ions. This region in which no free charge carriers are available is called a depletion region. This is shown in figure.

Current in circuit $$i = \frac{P}{{{V_d}}} = \frac{{100 \times {{10}^{ - 3}}}}{{0.5}}\,\,\left[ {_{ = 200 \times {{10}^{ - 3}}\,A}^{{V_d}\, = \,\,{\text{voltage drop across diode}}}} \right]$$
Voltage across resistance $$R,$$
$$V' = 1.5 - 0.5 = 1.0\,V$$
Thus, resistance $$R = \frac{{V'}}{i}$$
$$\eqalign{ & = \frac{1}{{200 \times {{10}^{ - 3}}}} \cr & = 5\,\Omega \cr} $$

$$E = \frac{V}{d} = \frac{{0.50}}{{5 \times {{10}^{ - 7}}}}\,V/m = 1.0 \times {10^6}V/m$$

Let the energy of radiation falling on the $$p-n$$  photodiode be $$E = h\nu $$
The minimum energy required $$= 2\,eV$$
$$\eqalign{ & \therefore 2eV = h\nu \cr & \therefore \nu = \frac{{2eV}}{h} \cr & = \frac{{2 \times 1.6 \times {{10}^{ - 19}}}}{{6.6 \times {{10}^{ - 34}}}} \cr & = 5 \times {10^{14}}Hz \cr} $$

Both the depletion region and barrier height is reduced.

Given that the current gain in common base emitter is $$\frac{{{i_c}}}{{{i_e}}} = {\text{Current}}\,{\text{gain}}\left( \alpha \right) = 0.96$$
So, current gain in common emitter configuration is $$\beta = \frac{\alpha }{{1 - \alpha }} = \frac{{0.96}}{{1 - 0.96}} = \frac{{0.96}}{{0.04}} = 24$$
NOTE
The current gain $$\beta $$ in common emitter mode is very large as compared to the current gain $$\alpha $$ in common base mode. This is why a transistor is always used as an amplifier in the CE mode.

$$p$$-side connected to low potential and $$n$$-side is connected to high potential.

$$\eqalign{ & n_i^2 = {n_e}{n_h} \cr & {\left( {1.5 \times {{10}^{16}}} \right)^2} = {n_e}\left( {4.5 \times {{10}^{22}}} \right) \cr & \Rightarrow {n_e} = 0.5 \times {10^{10}}\,\,{\text{or}}\,\,{n_e} = 5 \times {10^9} \cr & {\text{Given,}}\,\,{n_h} = 4.5 \times {10^{22}} \Rightarrow {n_h} > > {n_e} \cr} $$
∴ Semiconductor is $$p$$-type and
$${n_e} = 5 \times {10^9}{m^{ - 3}}.$$