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

$$i = \frac{V}{R} = \frac{{20}}{{2 \times {{10}^3}}} = 10\,mA.$$

$$DC$$  current gain in common base amplifier is given by
$$\alpha = \frac{{{i_c}}}{{{i_e}}}$$
where, $${{i_c}}$$ is collector current and $${{i_e}}$$ is emitter current.
Given, $$\alpha = 0.96,{i_e} = 7.2\,mA$$
$$\eqalign{ & \therefore {i_c} = 0.96 \times 7.2\,mA \cr & = 6.91\,mA \cr & {\text{As,}}\,\,{i_e} = {i_b} + {i_c} \cr & \therefore {\text{Base}}\,{\text{current}}\,{i_b} = {i_e} - {i_c} \cr & = 7.2 - 6.91 \cr & = 0.29\,mA \cr} $$

$$\eqalign{ & \frac{{{V_O}}}{{{V_{{\text{in}}}}}} = \frac{{{R_O}}}{{{R_{{\text{in}}}}}} \times \beta = \frac{{5 \times {{10}^3} \times 62}}{{500}} = 10 \times 62 = 620 \cr & {V_O} = 620 \times {V_{{\text{in}}}} = 620 \times 0.01 = 6.2\,V \cr & \therefore {V_O} = 6.2\,volt. \cr} $$

Semiconductors doped with acceptor atoms are called $$p$$-type semiconductors. The $$p$$ stands for positive to imply that the holes are introduced into the valence band, which behave like positive charge carriers, greatly in number than the electrons in the conduction band.
In $$p$$-type semiconductors, holes are the majority carriers and electrons are the minority carriers.

$$\eqalign{ & {\text{Voltage gain}} = \beta \times {\text{Impedance gain}} \cr & 50 = \beta \times \frac{{200}}{{100}} = 2\beta \Rightarrow \beta = 25 \cr & {\text{and power gain}} = {\beta ^2} \times \frac{{200}}{{100}} = 1250. \cr} $$

Input signal of a $$CE$$  amplifer, $${V_{{\text{in}}}} = 2\cos \left( {15t + \frac{\pi }{3}} \right)$$
Voltage gain $${A_v} = 150$$
As $$CE$$  amplifier gives phase difference of $$\pi $$ between input and output signals.
$$\eqalign{ & {\text{So,}}\,\,{A_v} = \frac{{{V_0}}}{{{V_{{\text{in}}}}}} \Rightarrow {V_0} = {A_v}{V_{{\text{in}}}} \cr & {V_0} = 150 \times 2\cos \left( {15t + \frac{\pi }{3} + \pi } \right) \cr & V = 300\cos \left( {15t + \frac{{4\pi }}{3}} \right) \cr} $$

Conductivity
$$\eqalign{ & \sigma = {n_i}e{\mu _e} = {10^7} \times \left( {1.6 \times {{10}^{ - 19}}} \right) \times 3800 \cr & = 60.8\,mho/cm \cr} $$

In the triode amplifier, there is a phase difference of $${180^ \circ }$$ between the input (i.e. signal) voltage and output voltage. In other words, the positive half-cycle of the signal appears as amplified negative-half in the output while the negative half-cycle of the signal appears as amplified positive half in the output.
This is known as phase reversal. In other words, as the signal is increasing in the negative half-cycle, the output is increasing in the positive sense.

We know that a single $$p$$-$$n$$ junction diode connected to an $$a$$-$$c$$ source acts as a half wave rectifier [Forward biased in one half cycle and reverse biased in the other half cycle].

$${P_v} = {A_v}\alpha = 4 \times 0.96 = 3.84$$