NCERT Solutions for Class 12 Physics Chapter 14 Electronics Devices

NCERT Solutions for Class 12 Physics Chapter 14 Electronics Devices

NCERT Solutions for Class 12 Physics Chapter 14 Electronics Devices are part of NCERT Solutions for Class 12 Physics. Here we have given. NCERT Solutions for Class 12 Physics Chapter 14 Electronics Devices

Board	CBSE
Textbook	NCERT
Class	Class 12
Subject	Physics
Chapter	Chapter 14
Chapter Name	Electronics Devices
Number of Questions Solved	19
Category	NCERT Solutions

Question 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 doplants.
(d) Holes are majority carriers and trivalent atoms are the dopants.
Answer:
(c) ‘Holes are minority carriers and pentavalent atoms are the dopants in n type semiconductor.’

Question 2.
Which of the statements given in Exercise 1 is true for p-type semiconductors ?
Answer:
(d) Holes are majority carriers and trivalent atoms are the dopants in p-type semiconductors.

Question 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)(E)c<(Eg)Ge>(Eg)si
(c) (Eg)c > (Eg)si > (Eg)Ge
(d) (Eg)c = (Eg)si = (Eg)Ge
Answer:
(C) (Eg)c > (Eg)Si > (Eg)Ge. Energy band gap is maximum in carbon and least in germanium among the given elements.

Question 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:
(c) hole concentration in p-region is more as compared to n-region because hole diffusion takes place from higher concentration to lower concentration.

Question 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:
(c) lowers the potential barrier by cancelling the depletion layer.

Question 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:
(b) and (c) : The base region must be very thin, lightly doped and the emitter junction is forward biased whereas collector junction is reverse biased to avoid unnecessary diffusion of charge carrier in the base and also for proper amplification.

Question 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:
(c) is low at high and low frequencies and constant at mid frequencies as per frequency response of a transistor.

Question 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:
A half wave rectifier rectifies only one half cycle of input A.C.
.’. frequency of the output A.C.
= frequency of input A.C. = 50 Hz A full wave rectifier rectifies both halve cycles
of the A.C. input
.’. frequency of output A.C. = 2 x frequency of input A.C. = 2 x 50 = 100 Hz

Question 9.
For a CE-transistor amplifier, the audio signal voltage across the collector resistance of 2 kQ 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:


Question 10.
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.
Answer:


Question 11.
A p-n photodiode is fabricated from a semiconductor with bandgap of 2.8 eV. Can it detect a wavelength of 6000 nm ?
Answer:

Since the energy of the light photon is less than the bandgap energy of the p-n diode, it can not be detected.

Question 12.
The number of silicon atoms per m³ is 5 x 10²⁸. This is doped simultaneously with 5 x 10²² atoms per m³ of Arsenic and 5 x 10²⁰ per m³ atoms of Indium. Calculate the number of electrons and holes. Given that n_i = 1.5 x 10¹⁶ m^-3. Is the material n-type or p- type ?
Answer:


Question 13.
In an intrinsic semiconductor, the energy gap E_g is 1.2 eV. Its hole mobility is much smaller than electron mobility and independent of temperature. What are the ratio between conductivity at 600 K and that at 300 K? Assume that the temperature dependence of intrinsic carrier concentration n_i is given by

Answer:

Thus, conductivity of a semiconductor increases with rise in temperature.

Question 14.
In a P-n junction diode, the current I can be expressed as

where I₀ is reverse saturation current. V is the voltage across the diode and is positive for forwarding bias and negative for reverse bias, and I is the current through the diode, K_B is the Boltzmann constant (8.6 x 10^-5 eV/K) and T is the absolute temperature. If for a given diode I₀ = 5 x 10^-12 A and T = 300 K, then
(a) What will be the forward current at a forwarding voltage of 0.6 V?
(b) What will be the increase in the current if the voltage across the diode is increased to 0.7 V?
(c) What is the dynamic resistance?
(d) What will be the current if reverse bias voltage changes from 1 V to 2 V?
Answer:
The statement of the given question is incorrect. The relation should be

Question 15.
You are given the two circuits as shown in Figure. Show that circuit (a) acts as OR gate while circuit (b) acts as AND gate.

Answer:


Question 16.
Write the truth table for a NAND gate connected as given in Fig. Hence identify the exact logic operation carried out by these circuits.

Answer:
The NAND gate shown in the truth table has only one input. Therefore, the truth table is

A	A	y = \( \bar { A.A }\)
0	0	1
1	1	0

Since Y = \( \bar { A }\) in this case, the circuit is actually a NOT gate with the truth table

A	Y
0	1
1	0

Question 17.
You are given two circuits as shown in Fig., which consist of NAND gates. Identify the logic operation carried out by the two circuits.

Answer:

Question 18.
Write the truth table for circuit given in Fig. below consisting of NOR gates and identify the logic operation (OR, AND, NOT) which this circuit is performing.

Answer:
Let y₁ be the output which appears at the first operation of NOR gate.

A	B	Y
0	0	0
1	0	1
0	1	1
1	1	1

Question 19.
Write the truth table for the circuits given in Fig., consisting of NOR gates only. Identify the logic operations (OR, AND, NOT) performed by the two circuits

Answer:

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