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

Given : Resistance $$R = 100\,kilo\,ohm\, = 100 \times {10^3}\,\Omega $$
Capacitance $$C$$ = 250 picofarad = $$250 \times {10^{ - 12}}\,F$$
$$\eqalign{ & \tau = RC = 100 \times {10^3} \times 250 \times {10^{ - 12}}\sec \cr & = 2.5 \times {10^7} \times {10^{ - 12}}\sec \cr & = 2.5 \times {10^{ - 5}}\sec \cr} $$
The higher frequency whcih can be detected with tolerable distortion is
$$\eqalign{ & f = \frac{1}{{2\pi {m_a}RC}} = \frac{1}{{2\pi \times 0.6 \times 2.5 \times {{10}^{ - 5}}}}Hz \cr & = \frac{{100 \times {{10}^4}}}{{25 \times 1.2\pi }}Hz = 10.61\,KHz \cr} $$
This condition is obtained by applying the condition that rate of decay of capacitor voltage must be equal or less than the rate of decay modulated singnal voltage for proper detection of mdoulated signal.

On applying forward bias to a $$p-n$$  junction diode, it increases number of donor on the $$n$$-side and decreases potential barrier. It also decreases electric field of depletion layer.

Voltage gain,
$$\eqalign{ & \frac{{{V_0}}}{{{V_i}}} = \beta \frac{{{R_2}}}{{{R_1}}}\,\,{\text{or}}\,\,{V_0} = \left[ {\beta \frac{{{R_2}}}{{{R_1}}}} \right] \times {V_i} \cr & = \left[ {100 \times \frac{{10}}{1}} \right] \times 1 \times {10^{ - 3}} = 1\,V. \cr} $$

$$\eqalign{ & \lambda = \frac{{hc}}{E} \cr & = \frac{{6.62 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{\left( {60 \times {{10}^{ - 3}} \times 1.6 \times {{10}^{ - 19}}} \right)}} \cr & = 2.07 \times {10^{ - 5}}m \cr} $$

In $$n$$-type semiconductors, electrons are the majority charge carriers.

$$\eqalign{ & \beta = \frac{{\Delta {i_c}}}{{\Delta {i_b}}},{\text{so}}\,\,\Delta {i_b} = \frac{{1 \times {{10}^{ - 3}}}}{{100}} = 0.01\,mA \cr & \Delta {i_e} = \Delta {i_b} + \Delta {i_c} = 0.01 + 1 = 1.01\,mA \cr} $$

In the circuit, diode $${D_1}$$ is forward biased, while $${D_2}$$ is reverse biased. Therefore, current $$i$$ (through $${D_1}$$ and $$100\Omega $$  resistance) will be
$$i = \frac{6}{{50 + 100 + 150}} = 0.02A$$

Here, $$50\Omega $$  is the resistance of $${D_1}$$ in forward biasing.

$$AC$$  current gain $$\beta $$ is defined as the ratio of the collector to the base current at constant collector voltage,
$$\beta = {\left( {\frac{{\Delta {i_c}}}{{\Delta {i_b}}}} \right)_{{V_c}}}$$
Given, $$\Delta {i_c} = 10\,mA - 5\,mA = 5\,mA$$
$$\eqalign{ & \Delta {i_b} = 150\,\mu A - 100\,\mu A = 50\,\mu A \cr & \therefore \beta = \frac{{5\,mA}}{{50 \times {{10}^{ - 3}}mA}} = 100 \cr} $$
NOTE
In common emitter amplifier, the output voltage signal is $${180^ \circ }$$ out of phase with the input voltage signal

The current will flow through diode $${D_1}$$ and so
$$i = \frac{6}{{\left( {150 + 50 + 100} \right)}} = 0.02\,A$$

In $$n-p-n$$   transistor, electrons are majority carriers in emitter ($$n$$-type semiconductor) and are repelled towards base by negative potential of $${V_{BB}}$$  as emitter base junction is forward biased $${i_e}.$$ The base being thin and lightly doped ($$p$$-type semiconductor) has low number density of holes.
When electrons enter the base region, then only a few holes (say $$5\% $$ ) get neutralised by the electron-hole combination, resulting base current $$\left( {{i_b} = 5\% \,{i_e} = 0.05\,{i_e}} \right).$$     The remaining $$95\% $$  electrons pass over to the collector, on account of high positive potential of collector due to battery $${V_{CC}},$$  resulting collector current $$\left( {{i_c} = 95\% \,{i_e} = 0.95\,{i_e}} \right).$$