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

$$\eqalign{ & {\text{Current gain}} = \frac{{\Delta {I_C}}}{{\Delta {I_B}}}\,{\text{when}}\,{V_{CE}}\,{\text{is}}\,{\text{constant}}. \cr & = \frac{{2.5 \times {{10}^{ - 3}}}}{{25 \times {{10}^{ - 6}}}} = 0.1 \times {10^3} = 100 \cr} $$
\[\left[ \begin{gathered} \Delta {I_B} = 125\,\mu A - 100\,\mu A = 25\,\mu A \hfill \\ \Delta {I_C} = 7.5\,mA - 5\,mA = 2.5\,mA \hfill \\ \end{gathered} \right]\]

In a series $$LC$$ -circuit, frequency of $$LC$$ -oscillations is given by
$$f = \frac{1}{{2\pi \sqrt {LC} }}\,\,{\text{or}}\,\,f \propto \frac{1}{{\sqrt {LC} }}$$
Considering two cases of $$L$$ and $$C,$$
$$\frac{{{f_1}}}{{{f_2}}} = \sqrt {\frac{{{L_2}{C_2}}}{{{L_1}{C_1}}}} $$
Given, $${L_1} = L,{C_1} = C,{L_2} = 2\,L,{C_2} = 4C,{f_1} = f$$
$$\therefore \frac{f}{{{f_2}}} = \sqrt {\frac{{2L \times 4C}}{{LC}}} = \sqrt 8 \Rightarrow {f_2} = \frac{f}{{2\sqrt 2 }}$$

When $$n$$-type semiconductor is heated a few hole-electron pairs generate. When a free electron is produced then simultaneously a hole is also produced. Thus no. of electrons and holes both increases equally.

$$\eqalign{ & {\text{Here,}}\,\,R = 4\,k\Omega = 4 \times {10^3}\Omega \,{V_i} = 60\,V \cr & {\text{Zener voltage}}\,{V_Z} = 10\,V\,{R_L} = 2\,k\Omega = 2 \times {10^3}\Omega \cr & {\text{Load current,}}\,{I_L} = \frac{{{V_Z}}}{{{R_L}}} = \frac{{10}}{{2 \times {{10}^3}}} = 5\,mA \cr & {\text{Current through }}R,I = \frac{{{V_i} - {V_Z}}}{R} = \frac{{60 - 10}}{{4 \times {{10}^3}}} = 12.5\,mA \cr} $$
From circuit diagram,
$$\eqalign{ & I = {I_Z} + {I_L} \cr & \Rightarrow 12.5 = {I_Z} + 5 \Rightarrow {I_Z} = 12.5 - 5 = 7.5\,mA \cr} $$

Let us assume that current through the diode is $$I.$$ From the given condition
$$\because I = \frac{{{V_A} - {V_B}}}{R} = \frac{{4 - \left( { - 6} \right)}}{{1\,K\Omega }} = \frac{{10}}{{1 \times {{10}^3}}} = {10^{ - 2}}A$$

$${I_C} = {I_E} - {I_B} = 90 - 1 = 89\,mA$$

When external voltage is applied to the junction in such a direction that it cancels the potential barrier, or decreases potential barrier thus permitting current flow, it is called forward biasing. To apply forward bias, connect positive terminal of the battery to $$p$$-type and negative terminal to $$n$$-type as shown in figure.

Semiconductors have negative temperature coefficient of resistance, i.e. the resistance of a semiconductor decreases with the increase in temperature and vice-versa. Silicon is actually an insulator at absolute zero of temperature but it becomes a good conductor at high temperatures. Because on giving temperatures to semiconductor some of the electron jumps from valence band to conduction band.

We know that a diode only conducts in forward biased condition. In the given circuit, the diode $${D_1}$$ will be in reverse bias, so it will block the current and diode $${D_2}$$ will be in forward bias, so it will pass the current
$$i = \frac{V}{{{R_1} + {R_3}}} = \frac{{10}}{{2 + 2}} = 2.5\,A$$

In $$'n'$$ region; number of $${e^ - }$$ is due to $$As$$ :
$$\eqalign{ & {n_e} = {N_D} = 1 \times {10^{ - 6}} \times 5 \times {10^{28}}\,atoms/{m^3} \cr & {n_e} = 5 \times {10^{22}}/{m^3} \cr} $$
The minority carriers (hole) is
$$\eqalign{ & {n_h} = \frac{{n_i^2}}{{{n_e}}} = \frac{{{{\left( {1.5 \times {{10}^{16}}} \right)}^2}}}{{5 \times {{10}^{22}}}} = \frac{{2.25 \times {{10}^{32}}}}{{5 \times {{10}^{22}}}} \cr & {n_h} = 0.45 \times 10/{m^3} \cr} $$
Similarly, when Boron is implanted a $$'p'$$ type is created with holes
$${n_h} = {N_A} = 200 \times {10^{ - 6}} \times 5 \times {10^{28}} = 1 \times {10^{25}}/{m^3}$$
This is far greater than $${e^ - }$$ that existed in $$'n'$$ type wafer on which Boron was diffused. Therefore, minority carriers in created $$'p'$$ region.
$${n_e} = \frac{{n_1^2}}{{{n_h}}} = \frac{{2.25 \times {{10}^{32}}}}{{1 \times {{10}^{25}}}} = 2.25 \times {10^7}/{m^3}$$