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

The resistance of metal (like $$Cu$$ ) decreases with decrease in temperature whereas the resistance of a semi-conductor (like $$Ge$$ ) increases with decrease in temperature.

Given, resistance across load, $${R_L} = 800\,\Omega $$
Voltage drop across load, $${V_L} = 0.8\,V$$
Input resistance of circuit, $${R_i} = 192\,\Omega .$$
Collector current is given by,
$$\eqalign{ & {I_C} = \frac{{{V_L}}}{{{R_L}}} = \frac{{0.8}}{{800}} = \frac{8}{{8000}} = 1\,mA \cr & \because {\text{Current amplification}} = \frac{{{\text{Output current}}}}{{{\text{Input current}}}} \cr & = \frac{{{I_C}}}{{{I_B}}} = 0.96 \Rightarrow {I_B} = \frac{{1\,mA}}{{0.96}} \cr & \because {\text{Voltage gain,}}\,\,{A_V} = \frac{{{V_L}}}{{{V_{in}}}} = \frac{{{V_L}}}{{{I_B}{R_i}}} = \frac{{0.8 \times 0.96}}{{{{10}^{ - 3}} \times 192}} = 4 \cr & \Rightarrow {A_V} = 4 \cr} $$
and Power gain,
$$\eqalign{ & {A_P} = \frac{{I_C^2{R_L}}}{{I_B^2{R_i}}} = {\left( {\frac{{{I_C}}}{{{I_B}}}} \right)^2} \cdot \frac{{{R_L}}}{{{R_i}}} \cr & = {\left( {0.96} \right)^2} \times \frac{{800}}{{192}} \cr & {A_P} = 3.84 \cr} $$

In common base amplifier circuit, the input signal voltage and the output collector voltage are in the same phase.

$$\Delta {I_C} = \alpha \,\Delta {I_E} = 0.99 \times 5 = 4.95\,mA$$

Energy, $$E = hv = h\frac{c}{\lambda }$$
$$ \Rightarrow \lambda = \frac{{hc}}{E}$$
Substituting the values of $$h,c$$  and $$E$$ in the above equation
$$\eqalign{ & \lambda = \frac{{6.6 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{2.5 \times 1.6 \times {{10}^{ - 19}}}} = 5000\,\mathop {\text{A}}\limits^ \circ \cr & {\text{As,}}\,\,4000\mathop {\text{A}}\limits^ \circ < 5000\mathop {\text{A}}\limits^ \circ \cr} $$
Signal of wavelength $$4000\mathop {\text{A}}\limits^ \circ $$  can be detected by the photodiode.

Many solids or crystals are composed of neutral atoms or molecules without any transferring, sharing or free movement of electrons. These solids remains bound by much weaker, short range attractive forces, known as 'van der Waals' forces. These forces arise as a result of fluctuations in the charge distributions of nearby molecules.

$$\eqalign{ & {\text{Using}}\,\,{U_{av}} = \frac{1}{2}{\varepsilon _0}{E^2}\,{\text{But}}\,\,{U_{av}} = \frac{P}{{4\pi {r^2} \times c}} \cr & \therefore \frac{P}{{4\pi {r^2}}} = \frac{1}{2}{\varepsilon _0}{E^2} \times c \Rightarrow E = {\left( {\frac{{2P}}{{4\pi {\varepsilon _0}}} \times \frac{1}{{c{r^2}}}} \right)^{\frac{1}{2}}} \cr & \Rightarrow E = {\left( {2 \times 9 \times {{10}^9} \times \frac{{0.1}}{{3 \times {{10}^8} \times 1 \times 1}}} \right)^{\frac{1}{2}}} \cr & = \sqrt 6 = 2.45\,V/m \cr} $$

In transistor the emitter is heavily doped, so that it can inject a large number of charge carriers (electrons or holes) into the base. The base is lightly doped and very thin; it passes most of the emitter injected charge carriers to the collector. The collector is moderately doped.

When a piece of $$p$$-type material is in contact with $$n$$-type material, then to bring Fermi level in a line, electrons in conduction band on $$n$$-type side travel across the junction and leave the positively ionised impurity atoms unneutralized.
Consequently, there is a positively charged region adjacent to the junction in $$n$$-type material.
On $$p$$-type side, the electrons which have traversed the boundary recombine with positive holes in the valence band. Near to the junction on $$p$$-type side, there is a layer of unneutralized negatively ionised trivalent impurity which forms a negatively charged region. This region around the junction is called charge depletion region or space charge region.

Zero; In common base amplifier circuit, input and output voltage are in the same phase.