NCERT Solutions for Class 12 Science Physics Chapter 6 – Semiconductor Electronics: Materials, Devices And Simple Circuits
NCERT Solutions for Class 12 Science Physics Chapter 6, titled “Semiconductor Electronics: Materials, Devices And Simple Circuits,” offer comprehensive explanations with step-by-step guidance. These solutions are widely favored by Class 12 Science students for Physics, as they prove invaluable for completing assignments and preparing for exams. All the questions and answers from this chapter in the NCERT Book are provided here for free, aiding students in their understanding of Semiconductor Electronics: Materials, Devices And Simple Circuits. Page No 509: Question 14.1: In an n-type 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. ANSWER: The correct statement is (c). In an n-type silicon, the electrons are the majority carriers, while the holes are the minority carriers. An n-type semiconductor is obtained when pentavalent atoms, such as phosphorus, are doped in silicon atoms. Page No 509: Question 14.2: Which of the statements given in Exercise 14.1 is true for p-type semiconductors. ANSWER: The correct statement is (d). In a p-type semiconductor, the holes are the majority carriers, while the electrons are the minority carriers. A p-type semiconductor is obtained when trivalent atoms, such as aluminium, are doped in silicon atoms. Page No 509: Question 14.3: 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 (Eg)C, (Eg)Si and (Eg)Ge. Which of the following statements is true? (a) (Eg)Si < (Eg)Ge < (Eg)C (b) (Eg)C < (Eg)Ge > (Eg)Si (c) (Eg)C > (Eg)Si > (Eg)Ge (d) (Eg)C = (Eg)Si = (Eg)Ge ANSWER: The correct statement is (c). Of the three given elements, the energy band gap of carbon is the maximum and that of germanium is the least. The energy band gap of these elements are related as: (Eg)C > (Eg)Si > (Eg)Ge Page No 509: Question 14.4: In an unbiased p-n junction, holes diffuse from the p-region to n-region because (a) free electrons in the n-region attract them. (b) they move across the junction by the potential difference. (c) hole concentration in p-region is more as compared to n-region. (d) All the above. ANSWER: 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. Page No 510: Question 14.5: When a forward bias is applied to a p-n 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. ANSWER: The correct statement is (c). When a forward bias is applied to a p-n 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. Page No 510: Question 14.6: 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. (d) Both the emitter junction as well as the collector junction are forward biased. ANSWER: 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 forward-biased and collector junction should be reverse-biased. Page No 510: Question 14.7: 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. ANSWER: 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. Page No 510: Question 14.8: In half-wave rectification, what is the output frequency if the input frequency is 50 Hz. What is the output frequency of a full-wave rectifier for the same input frequency. ANSWER: Input frequency = 50 Hz For a half-wave rectifier, the output frequency is equal to the input frequency. ∴Output frequency = 50 Hz For a full-wave rectifier, the output frequency is twice the input frequency. ∴Output frequency = 2 × 50 = 100 Hz Page No 510: Question 14.9: For a CE-transistor 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. ANSWER: Collector resistance, RC = 2 kΩ = 2000 Ω Audio signal voltage across the collector resistance, V = 2 V Current amplification factor of the transistor, β = 100 Base resistance, RB = 1 kΩ = 1000 Ω Input signal voltage = Vi Base current = IB We have the amplification relation as: Voltage amplification Therefore, the input signal voltage of the amplifier is 0.01 V. Base resistance is given by the relation: Therefore, the base current of the amplifier is 10 μA. Page No 510: Question 14.13: In an intrinsic semiconductor the energy gap Egis 1.2 eV. Its hole mobility is much smaller than electron mobility and independent of temperature. What is the ratio between conductivity at 600K and that at 300K? Assume that the temperature dependence of intrinsic carrier concentration niis given by where n0 is a constant. ANSWER: Energy gap of the given intrinsic semiconductor, Eg = 1.2 eV The temperature dependence of the intrinsic carrier-concentration is written as: Where, kB = Boltzmann constant = 8.62 × 10−5 eV/K T = Temperature n0 = Constant Initial temperature, T1 = 300 K The intrinsic carrier-concentration at this temperature can be written as: … (1) Final temperature, T2 = 600 K The intrinsic carrier-concentration at this temperature can be written as: … (2) The ratio between the conductivities at 600 K and at 300 …