
Q1 In an ntype silicon, which of the following statement is true:
(a) Electrons are majority carriers and trivalent atoms are the dopants.
(b) Electrons are minority carriers and pentavalent atoms are the dopants.
(c) Holes are minority carriers and pentavalent atoms are the dopants.
(d) Holes are majority carriers and trivalent atoms are the dopants.Ans: The correct statement is (c).
In an ntype silicon, the electrons are the majority carriers, while the holes are the minority carriers. An ntype semiconductor is obtained when pentavalent atoms, such as phosphorus, are doped in silicon atoms.
Q2 Which of the statements given in Exercise 14.1 is true for ptype semiconductors.
Ans: The statement is (d) true.
It is because in a ptype semiconductor, the holes are the majority carriers, while the electrons are the minority carriers. A ptype semiconductor is obtained when trivalent atoms, such as aluminium, are doped in silicon atoms.
Q3 Carbon, silicon and germanium have four valence electrons each. These are characterised 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?
(a) (E _{g }) _{Si }< (E _{g} ) _{Ge} < (E _{g} )_{ C }
(b) (E _{g} ) _{C }< (E_{ g} ) _{Ge} > (E _{g} ) _{Si }
(c) (E _{g} ) _{C} > (E _{g} )_{ Si} > (E _{g} ) _{Ge}
(d) (E _{g} ) _{C} = (E _{g} ) _{Si} = (E _{g} ) _{Ge}
Ans: The energy band gap is maximum for carbon, less for silicon and least for germanium out of the given three elements.
Hence answer is (c) is correct.
Q4 In an unbiased pn junction, holes diffuse from the pregion to nregion because
(a) free electrons in the nregion attract them.
(b) they move across the junction by the potential difference.
(c) hole concentration in pregion is more as compared to nregion.
(d) All the above.
Ans: In an unbiased pn junction, the diffusion of charge carriers across the junction takes place from higher concentration to lower concentration.
Thus answer (c) is correct.
Q5 When a forward bias is applied to a pn junction, it
(a) raises the potential barrier.
(b) reduces the majority carrier current to zero.
(c) lowers the potential barrier.
(d) None of the above.
Ans: When a forward bias is applied to a pn junction, it lowers the value of potential barrier. In the case of a forward bias, the potential barrier opposes the applied voltage. Hence, the potential barrier across the junction gets reduced.
Hence, The correct statement is (c).
Q6 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.
(d) Both the emitter junction as well as the collector junction are forward biased.
Ans: The correct statement is (b), (c).
For a transistor action, the junction must be lightly doped so that the base region is very thin. Also, the emitter junction must be forwardbiased and collector junction should be reversebiased.
Q7 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.
Ans: The voltage gain of a transistor amplifier is constant at mid frequency range only. It is low at high and low frequencies.
Hence, the correct statement is (c).
Q8 In halfwave rectification, what is the output frequency if the input frequency is 50 Hz. What is the output frequency of a fullwave rectifier for the same input frequency.
Ans: Given, input frequency = 50 Hz
For a halfwave rectifier, the output frequency is equal to the input frequency.
∴ Output frequency = 50 Hz
For a fullwave rectifier, the output frequency is twice the input frequency.
∴ Output frequency = 2 × 50 = 100 Hz
Q9 For a CEtransistor amplifier, the audio signal voltage across the collected resistance of 2 kΩ 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 kΩ.
Ans: Collector resistance, R_{ C} = 2 kΩ = 2000 Ω
Audio signal voltage across the collector resistance, V = 2 V
Current amplification factor of the transistor, β = 100
Base resistance, R_{ B} = 1 kΩ = 1000 Ω
Input signal voltage = V_{ i}
Base current = I _{B}
We have the amplification relation as: V/V_{i }= β R_{c}/R_{b}
Voltage amplification, V_{i }= V R_{b} / β R_{c}
_{= 2x1000 / 100x2000 = 0.01 V}
Therefore, the input signal voltage of the amplifier is 0.01 V.
Base resistance is given by the relation: R_{B = }V_{i }/ I_{B}
_{= }_{0.01 / 1000 = 10} μA
Therefore, the base current of the amplifier is 10 μA.
Q10 Two amplifiers 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.
Ans: Given that,
Voltage gain of the first amplifier, V _{1} = 10
Voltage gain of the second amplifier, V _{2} = 20
Voltage of input signal, V _{i }= 0.01 V
Voltage of output AC signal = V _{o}
The total voltage gain of a twostage cascaded amplifier is given by the product of voltage gains of both the stages,
i.e., V = V _{1 }× V _{2 }= 10 × 20 = 200
It can be calculated by the relation: V = V_{o}/V_{i}
V_{ 0} = V × V _{i} = 200 × 0.01 = 2 V
Therefore, the output AC signal of the given amplifier is 2 V.
Q11 A pn photodiode is fabricated from a semiconductor with band gap of 2.8 eV. Can it detect a wavelength of 6000 nm?
Ans: Given that,
Energy band gap of the given photodiode, E _{g} = 2.8 eV
Wavelength, λ = 6000 nm = 6000 × 10 ^{−9} m
The energy of a signal is given by the relation: E = hc/λ
Where, h = Planck’s constant = 6.626 × 10 ^{−34} Js
c = Speed of light = 3 × 10^{ 8} m/s
E = 6.626 x 10^{34} x 3 x 10^{8} / 6000 x 10^{9} = 3.313 x 10^{20} J
But 1.6 × 10^{ −19} J = 1 eV
E = 3.313 × 10 ^{−20} J
∴E = 3.313 × 10 ^{−20} J = 3.313 x 10^{20} / 1.6 x 10^{19} = 0.207 eV
The energy of a signal of wavelength 6000 nm is 0.207 eV, which is less than 2.8 eV − the energy band gap of a photodiode. Hence, the photodiode cannot detect the signal.
Q12 The number of silicon atoms per m^{ 3} is 5 × 10^{ 28} . This is doped simultaneously with 5 × 10^{ 22} atoms per m^{ 3} of Arsenic and 5 × 10^{ 20} per m^{ 3} atoms of Indium. Calculate the number of electrons and holes. Given that n_{ i} = 1.5 × 10^{ 16} m^{ −3} . Is the material ntype or ptype?
Ans: Number of silicon atoms, N = 5 × 10^{ 28} atoms/m^{ 3}
Number of arsenic atoms, n_{ As} = 5 × 10^{ 22} atoms/m^{ 3}
Number of indium atoms, n _{In} = 5 × 10^{ 20} atoms/m^{ 3}
Number of thermallygenerated electrons, n _{i} = 1.5 × 10^{ 16} electrons/m^{ 3}
Number of electrons, n _{e} = 5 × 10 ^{22} − 1.5 × 10^{ 16} ≈ 4.99 × 10 ^{22}
Number of holes = n _{h}
In thermal equilibrium, the concentrations of electrons and holes in a semiconductor are related as: n _{e} n_{ h} = n _{i}^{2}
Therefore, n_{h} = n _{i}^{2 }/ n_{e }= (1.5x10^{16})^{2 }/ 4.99 x 10^{22 }≈ 4.51 x 10^{9}
Therefore, the number of electrons is approximately 4.99 × 10 ^{22} and the number of holes is about 4.51 × 10 ^{9} . Since the number of electrons is more than the number of holes, the material is an ntype semiconductor.