NEET Practice Questions (MCQs) with Answers & Solutions

(c) Collector current $$\begin{gathered} i_c=\frac{V}{R} \\ =\frac{3}{3\times {10}^3}={10}^{-3}A \end{gathered}$$

Now,base current $i_B=\frac{i_c}{B}=\frac{{10}^{-3}}{100}={10}^{-5}A$

As, voltage      $V_{in}=i_B{R}_B$ 

$\therefore V_{in}={10}^{-5}\times 2\times {10}^3$

              =$2\times {10}^{-2} volts$

So,voltage gain 

        $A_V=\frac{V_{out}}{V_{in}}=\frac{3}{2\times {10}^{-2}}=150$

Power gain = $A_{V\times \beta }$

                $=150\times 100=15000$

(b) Key idea Voltage amplification is

          $A_V=\beta \frac{R_{out}}{R_{in}}=\frac{{\left(V_{out}\right)}_{AC}}{{\left(V_{in}\right)}_{AC}}$

Given, collector resistance= $R_{out}=2k\Omega$

Current amplification factor,$\beta =100$

Base resistance, $R_{in}=1k \Omega$

output signal voltage=4 V

Putting all the values in given equation, we get

        $A_v=\beta \frac{R_{out}}{R_{in}}=100\times \frac{2 k\Omega }{1 k\Omega }\Rightarrow A_v=200$

$Now, A_v=\frac{{\left(V_{out}\right)}_{AC}}{{\left(V_{in}\right)}_{AC}}=200$

$\Rightarrow {\left(V_{in}\right)}_{AC }=\frac{4}{200}=20mV$

(d) Given, resistance across load, R_L=800Ω Voltage drop across load, V_L=0.8V Input resistance of circuit,R_i=192Ω Collector current is given by,

I_C=V_L/R_L=0.8/800=8/8000=1mA 
∴ Current amplification=Output current/Input current
=I_C/I_B=0.96=>I_B=1mA/0.96

$$\begin{gathered} \therefore Voltage gain A_V=V_L/V_{in}=V_L/I_B{R}_i=0.8x0.96/{10}^{-3}x192 \\ {A}_v=4 \end{gathered}$$
and power gain 
A_p=I²_CR_L/I²_BR_i=(I_C/I_B)² x R_L/R_i=(0.96)² x 800/192= 3.84

Input signal of a CE amplifier,V_in=2cos(15t+π/3) Voltage gain A_v=150

As CE amplifier gives phase difference of π between input and output signals.

So, A_v=V_o/V_in=>V_o=A_vV_in

V_o=150x2cos(15t+π/3+π)

V=300cos(15t+4π/3)

(c) The n-type semi-conductor can be produced by doping an impurity atom of valence 5 i.e., pentavalent atoms. i.e., phosphorus.

We know

           $$\begin{gathered} {\mathrm{V}}_{\mathrm{g}}=\mathrm{\beta }\frac{{\mathrm{R}}_{\mathrm{L}}}{{\mathrm{R}}_{\mathrm{i}}} \\ =\frac{∆{\mathrm{I}}_{\mathrm{C}}}{∆{\mathrm{I}}_{\mathrm{B}}}\frac{{\mathrm{R}}_{\mathrm{L}}}{{\mathrm{R}}_{\mathrm{i}}} \left(\mathrm{so}, \mathrm{\beta }=\frac{∆{\mathrm{I}}_{\mathrm{C}}}{∆{\mathrm{I}}_{\mathrm{B}}}\right) \end{gathered}$$

On putting given values 

      $$\begin{gathered} {\mathrm{V}}_{\mathrm{g}}=\frac{2\times {10}^{-3}}{40\times {10}^{-6}}\times \frac{4\times {10}^3}{100} \\ {\mathrm{V}}_{\mathrm{g}}=2000 \end{gathered}$$

When a small amount of antimony is added to

germanium crystal, the crystal becomes n-type

semiconductor because antimoney is a 

pentavalent substrate. It excess free electrons.

In forward biasing of p-n junction the positive terminal of the battery is connected to p-side and the depletion region becomes thin.

 $$\begin{gathered} {\mathrm{n}}_{\mathrm{i}}^2={\mathrm{n}}_{\mathrm{e}}{\mathrm{n}}_{\mathrm{h}} \\ {\mathrm{n}}_{\mathrm{e}}=\frac{({\mathrm{n}}_{\mathrm{i}}{)}^2}{{\mathrm{n}}_{\mathrm{h}}} \\ {\mathrm{n}}_{\mathrm{e}}=\frac{(1.5\times {10}^{16}{)}^2}{(4.5\times {10}^{22})} \\ {\mathrm{n}}_{\mathrm{e}}=5\times {10}^9 {\mathrm{m}}^{-3} \end{gathered}$$

 So, ${\mathrm{n}}_{\mathrm{h}}\gg {\mathrm{n}}_{\mathrm{e}}$ semiconductor is p-type.

p-type semiconductor are obtained by adding a small amount of trivalent impurity to a pure sample of semiconductor(Ge).

Majority charge carriers-holes

Minority charge carries-electrons

n-type semiconductor are obtained by adding a small amount of pentavalent impurity to a pure sample of semiconductor (Ge).

Majority charge carries-electrons

The resistance of intrinsic semiconductors decreases with increase of temperature.