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

$$p$$-type semiconductor are obtained by adding a small amount of trivalent impurity to a pure sample of semiconductor $$\left( {Ge} \right).$$
Majority charge carriers—holes
Minority charge carriers—electrons
$$n$$-type semiconductor are obtained by adding a small amount of pentavalent impurity to a pure sample of semiconductor $$\left( {Ge} \right).$$
Majority charge carriers—electrons
The resistance of intrinsic semiconductors decreases with increase of temperature.

The energy band scheme of semiconductors is shown here.
In semiconductors, valence band and conduction band are separated by an energy gap called the forbidden energy gap. It is very small. At room temperature some electrons in valence band acquire thermal energy.
This energy is more than forbidden energy gap $${E_g},$$  thus they jump into the conduction band and leave their vacancy in the valence band which act as holes.
Hence, at room temperature valence band is partially empty and conduction band is partially filled.

$$\eqalign{ & \frac{{{J_1}}}{{{J_2}}} = \frac{{AT_1^2{e^{ - \frac{{{W_1}}}{{k{T_1}}}}}}}{{AT_2^2{e^{ - \frac{{{W_2}}}{{k{T_2}}}}}}} \cr & = {\left( {\frac{{{T_1}}}{{{T_2}}}} \right)^2}{e^{ - \frac{{{W_1}}}{{k{T_1}}} + \frac{{{W_2}}}{{k{T_2}}}}} \cr & = {\left( {\frac{1}{2}} \right)^2}{e^0} \cr & = \frac{1}{4} \cr} $$

$$\eqalign{ & \frac{{{I_c}}}{{{I_e}}} = 0.96 \Rightarrow {I_c} = 0.96\,{I_e} \cr & {\text{But}}\,\,{I_e} = {I_c} + {I_b} = 0.96{I_e} + {I_b} \cr & \Rightarrow {I_b} = 0.04{I_e} \cr & \therefore {\text{Current gain}}\,\beta = \frac{{{I_c}}}{{{I_b}}} = \frac{{0.96{I_e}}}{{0.4{I_e}}} = 24 \cr} $$

Voltage amplification is $${A_V} = \beta \frac{{{R_{{\text{out}}}}}}{{{R_{{\text{in}}}}}} = \frac{{{{\left( {{V_{{\text{out}}}}} \right)}_{AC}}}}{{{{\left( {{V_{{\text{in}}}}} \right)}_{AC}}}}$$
Given, collector resistance $$ = {R_{{\text{out}}}} = 2\,k\Omega $$
Current amplification factor, $$\beta = 100$$
Base resistance, $${R_{{\text{in}}}} = 1\,k\Omega $$
Output signal voltage $$= 4\,V$$
Putting all the values in given equation, we get
$$\eqalign{ & {A_V} = \beta \frac{{{R_{{\text{out}}}}}}{{{R_{{\text{in}}}}}} = 100 \times \frac{{2\,k\Omega }}{{1\,k\Omega }} \Rightarrow {A_V} = 200 \cr & {\text{Now,}}\,\,{A_V} = \frac{{{{\left( {{V_{{\text{out}}}}} \right)}_{AC}}}}{{{{\left( {{V_{{\text{in}}}}} \right)}_{AC}}}} = 200 \cr & \Rightarrow {\left( {{V_{{\text{in}}}}} \right)_{AC}} = \frac{4}{{200}} = 20\,mV \cr} $$

Current amplification factor $$\beta = \frac{{\Delta {I_C}}}{{\Delta {I_B}}}$$
Collector resistance $$\Delta {I_C} = \frac{{2V}}{{2 \times {{10}^3}\Omega }} = 1 \times {10^{ - 3}}A$$
Base current
$$\Delta {I_B} = \frac{{{V_B}}}{{{R_B}}} = \frac{{{V_B}}}{{1 \times {{10}^3}}} = {V_B} \times {10^{ - 3}}$$
Given, $$\beta = 100$$
Now, $$100 = \frac{{{{10}^{ - 3}}}}{{{V_B} \times {{10}^{ - 3}}}}$$
$${V_B} = \frac{1}{{100}}V = 10\,mV$$

The intensity of electric field
$$E = \frac{V}{d} = \frac{{0.3}}{{1 \times {{10}^{ - 6}}}} = 3 \times {10^5}\,V/m$$

The electric field retarded the electron. The retardation
$$a = \frac{{Ee}}{m}$$
The speed of the electron on entering $$p$$-side
$$\eqalign{ & {v^2} = {u^2} - 2as = {u^2} - 2\frac{{eE}}{m}.d \cr & = {\left( {5 \times {{10}^5}} \right)^2} - \frac{{2 \times 1.6 \times {{10}^{ - 19}} \times 3 \times {{10}^5}}}{{9.1 \times {{10}^{ - 31}}}} \times \left( {1 \times {{10}^{ - 6}}} \right) \cr & = 3.8 \times {10^5}\,m/s \cr} $$

Pure silicon, at absolute zero, will contain all the electrons in bounded state. The conduction band will be empty. So there will be no free electrons (in conduction band) and holes (in valence band) due to thermal agitation. Pure silicon will act as insulator.

Current gain, $$\alpha = \frac{{{A_V}}}{{{A_R}}} = \frac{{2800}}{{3000}} = 0.93$$

By the definition
$${\beta _{dc}} = \frac{{{i_C}}}{{{i_B}}}$$

$$\eqalign{ & \therefore {i_B} = \frac{{{i_C}}}{{{\beta _{dc}}}} = \frac{{4\,mA}}{{100}} = 0.4\,mA \cr & {\text{From}}\,{\text{figure}} \cr & {i_B}{R_B} + {V_{BE}} = {V_{CC}} \cr & \therefore {R_B} = \frac{{{V_{CC}} - {V_{BE}}}}{{{i_B}}} = \frac{{\left( {8 - 0.6} \right)}}{{0.04 \times {{10}^{ - 3}}}} = 185\,k\Omega \cr & {\text{Also}}\,{R_L}{i_C} + {V_{CE}} = {V_{CC}} \cr & \therefore {R_L} = \frac{{{V_{CC}} - {V_{CE}}}}{{{i_C}}} = \frac{{8 - 4}}{{8 \times {{10}^{ - 3}}}} = 1\,k\Omega \cr} $$