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

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

Van der Waal's bonding is attributed to the attractive forces between molecules of a liquid. The conductivity of semiconductors (covalent bonding) and insulators (ionic bonding) increases with increase in temperature while that of metals (metallic bonding) decreases.

For a transistor $${I_E} = {I_B} + {I_C}$$
where $${I_E} =$$  emitter current
$${I_B} =$$  base current
$${I_C} =$$  collector current
and current gain $$\beta = \frac{{\Delta {I_C}}}{{\Delta {I_B}}}$$
$$\eqalign{ & {\text{Here,}}\,\,\Delta {I_C} = 10 \times {10^{ - 3}} - 5 \times {10^{ - 3}} \cr & = 5 \times {10^{ - 3}}\,A \cr & \Delta {I_B} = 200 \times {10^{ - 6}} - 100 \times {10^{ - 6}} \cr & = 100 \times {10^{ - 6}}A \cr & \because \beta = \frac{5}{{100}} \times 1000 = 50 \cr} $$

For amplifying action of a transistor, emitter-base junction is always forward biased while base-collector junction is reverse biased. For forward biasing of emitter-base junction in $$n-p-n$$   transistor, left $$n$$-side i.e. emitter should be connected to negative terminal and $$p$$-side (base) should be connected to positive terminal of the battery. On the other hand, in right side, $$n$$-side (collector) should be connected to positive terminal of the battery to make base-collector junction reverse biased. The whole situation is shown in the figure for $$n-p-n$$   transistor

$$\eqalign{ & E \to {\text{Emitter}} \cr & B \to {\text{Base}} \cr & C \to {\text{Collector}} \cr} $$

Forward bias resistance $$ = \frac{{\Delta V}}{{\Delta I}}$$
$$ = \frac{{\left( {0.7 - 0.6} \right)V}}{{\left( {15 - 5} \right)mA}} = \frac{{0.1}}{{10 \times {{10}^{ - 3}}}} = 10\,\Omega .$$

$$\eqalign{ & {I_C} = 5.488\,mA,{I_e} = 5.6\,mA \cr & \alpha = \frac{{5.488}}{{5.6}},\beta = \frac{\alpha }{{1 - \alpha }} = 49 \cr} $$

AC power gain $$ = \frac{{{\text{ change in output power }}}}{{{\text{ change in input power }}}}$$
$$\eqalign{ & = \frac{{\Delta {V_c} \times \Delta {i_c}}}{{\Delta {V_i} \times \Delta {i_b}}} \cr & = \left( {\frac{{\Delta {V_c}}}{{\Delta {V_i}}}} \right) \times \left( {\frac{{\Delta {i_c}}}{{\Delta {i_b}}}} \right) = {A_V} \times {\beta _{AC}} \cr} $$
where, $${A_V}$$ is voltage gain and $${\left( \beta \right)_{AC}}$$  is AC current gain. Also,
$$\eqalign{ & {A_V} = {\beta _{AC}} \times {\text{ resistance gain}}\left( { = \frac{{{R_o}}}{{{R_i}}}} \right) \cr & {\text{Voltage}}\,{\text{gain}} = \beta \times {\text{impedance}}\,{\text{gain}} \cr & \Rightarrow 50 = \beta \times \frac{{200}}{{100}} \Rightarrow \beta = 25 \cr & {\text{Also,}}\,{\text{power}}\,{\text{gain}} = {\beta ^2} \times {\text{impedance}}\,{\text{gain}} \cr & = {25^2} \times \frac{{200}}{{100}} = 1250 \cr} $$

Electrons move from base to emmitter.

Number of atoms per unit cell is given by
$$N = {N_b} + \frac{{{N_f}}}{2} + \frac{{{N_c}}}{8}$$
where $${N_b} =$$  number of atoms centred in the body
$${{N_f}} =$$  number of atoms centred in the face
$${{N_c}} =$$  number of atoms centred at the corners
For bcc structure,
$$\eqalign{ & {N_b} = 1,\,{N_f} = 0 \cr & {\text{and}}\,{N_c} = 8 \cr & \therefore N = 1 + \frac{0}{2} + \frac{8}{8} = 2 \cr} $$

Clearly from fig. given in question, Silicon diode is in forward bias.
∴ Potential barrier across diode
$$\Delta V = 0.7\,volts$$
Current, $$I = \frac{{V - \Delta V}}{R} = \frac{{3 - 0.7}}{{200}} = \frac{{2.3}}{{200}} = 11.5\,mA$$