Semiconductor Electronic Material Devices and Simple Circuits

Physics

NCERT

1   In an n-type silicon, which of the following statements is true:$\\$

Solution :

The correct statement is (c).$\\$ In an n-type silicon, the electrons are the majority carriers, while the holes are the minority carries. An n-type semiconductor is obtained when pentavalent atoms, such as phosphorus, are doped in silicon atoms.

2   Which of the statements given in Exercise 14.1 is true for p-type semiconductor.

Solution :

The correct statement is (d).$\\$ In a p-type semiconductor, the holes are the majority carriers, while the holes are the minority carries. A p-type semiconductor is obtained when trivalent atoms, such aluminium, are doped in silicon atoms.

3   Carbon, silicon and germanium have four valence electrons each. These are characterized by valence and conduction bands separated by energy band gap respectively equal to $(E_g)_c,(E_g)_{Si}$ and $(E_g)_{Ge}$. Which of the following statements is true?

Solution :

The correct answer is (c).$\\$ Of the three given elements, the energy band gap of carbon is the maximum and that germanium is the least.$\\$ The energy bang gap of the these elements are related as :$\\$ $(E_g)_C > (E_g)_{Si} > (E_g)_{Ge}$

4   In an unbiased p-n junction, holes diffuse from the p-region to n-region because

Solution :

The correct statement is (c).$\\$ The diffusion of charge carriers across a junction takes place from the region of higher concentration to the region of lower concentration. In this case, the p-region has greater concentration of holes than the n-region. Hence, in an unbiased p-n junction, holes diffuse from the p-region to the n-region.

5   When a forward bias is applied to a p-n junction, it

Solution :

The correct statement is (c).$\\$ When forward bias is applied to a p-n junction, it lower the values of potential barrier. In the case of forward bias, the potential barrier is opposed by the applied voltage. Hence, the potential barrier across the junction gets reduced.

6   For transistor action, which of the following statements are correct.$\\$ (a) Base, emitter and collector regions should have similar size and doping concentrations.$\\$ (b) The base region must be very thin and lightly doped.$\\$ (c) The emitter junction is forward biased and collector junction is reverse biased.$\\$ Both the emitter junction as well as the collector junction are forward biased.

Solution :

The correct statement is (b), (c).$\\$ For a transistor action, the base region must be lightly doped so that the base region is very thin. Also, the emitter junction must be forward-biased and collector junction should be reversed-biased.

7   For a transistor amplifier, the voltage gain$\\$ (a) Remains constant for all frequencies.$\\$ (b) Is high at high and low frequencies and constant in the middle frequency range.$\\$ (c) Is low at high and low frequencies and constant at mid frequencies.$\\$ (d) None of the above.

Solution :

The correct statement is (c).$\\$ The voltage gain of a transistor amplifier is constant at mid frequency range only. It is low at high and low frequencies.

8   In half-wave rectifier, what is the output frequency if the input frequency is $50Hz$. What is the output frequency of a full-wave rectifier for the same input frequency

Solution :

Input frequency $= 50Hz$$\\$ For a half-wave rectifier, the output frequency is equal to the input frequency. $\therefore $ Output frequency =$ 50Hz$$\\$ For a full-wave rectifier, the output frequency is twice the input frequency. $\therefore $ Output frequency $= 2 × 50 = 100 Hz$

9   For a CE-transmitter amplifier, the audio signal voltage across the collector resistance of $2 k \Omega $ is $2 V$. Suppose the current amplification factor of the transistor is $100$, find the input signal voltage and base current, if the base resistance is $1 k \Omega $ .

Solution :

Collector resistance, $R _c = 2 k\Omega = 2000 \Omega $$\\$ Audio signal voltage across the collector resistance,$ V = 2 V$$\\$ Current amplification factor of the transistor, $\beta = 100$$\\$ Base resistance, $R_ B = 1 k\Omega = 1000 \Omega $$\\$ Input signal voltage = $V_ i$$\\$ Base current =$ I_ B$$\\$ We have the amplification relation as :$\\$ Voltage amplification=$\dfrac{V}{V_i}=\beta \dfrac{R_C}{R_B}$$\\$ $V_i=\dfrac{V R_B}{\beta R_C}\\ =\dfrac{2*1000}{100*2000}=0.01V$$\\$ Therefore, the input signal voltage of the amplifier is $ 0.01V.$$\\$ Base resistance is given by the relation:$\\$ $R_B=\dfrac{V_i}{V_B}\\ =0.01/1000=10^{-6}A$$\\$ Therefore, the base current of the amplifier is $10\mu A$

10   Two amplifier are connected one after the other in series (cascaded). The first amplifier has a voltage gain of $10$ and the second has a voltage gain of $20$. If the input signal is $0.01 $ volt, calculate the output ac signal.

Solution :

Voltage gain of the first amplifier,$ V_ 1 = 10$$\\$ Voltage gain of the second amplifier, $V _2 =20$$\\$ Input signal voltage, $V_ i = 0.01 V$$\\$ Output AC signal voltage $= V o$$\\$ The total voltage gain of a two-stage cascaded amplifier is given by the product of voltage gains of both the stages, i.e.,$\\$ $V =V_ 1 * V_ 2\\ = 10 * 20 = 200$$\\$ We have the relation:$\\$ $V=\dfrac{V_0}{V_i}\\ V_o=V*V_i\\ =200*0.01=2V$$\\$ Therefore, the output AC signal of the given amplifier is $2V.$