Chapter 14 Semiconductor Electronics

NCERT Solutions Class 12 Physics Chapter 14 - Semiconductor Electronics: Materials, Devices and Simple Circuits

Chapter 14 (Semiconductor Electronics) of Physics Class 12 is the conclusive part of Physics and forms the core of today's digital revolution. In this Chapter, students will gain insight into the materials where the conductivity of these materials can be regulated in a controlled manner, as opposed to classical physics of conductors and insulators.

The basic concepts of energy bands, n-type and p-type semiconductor distinctions, and how to describe the functions of a p-n junction are introduced in this chapter. Students must know these fundamental building blocks of electronics so they can understand how the technology works (like smartphones, computers, solar panels). This chapter is significant for performance in CBSE Board Examinations; JEE; and NEET because it has a high emphasis on understanding concepts and constructing graphs that illustrate key information.

To help students develop a thorough understanding of "Logic," NCERT Solutions for Semiconductor Electronics in Chapter 14 of Class 12 Physics provides systematic steps that outline an orderly transition from atomic energy levels to the practical applications of diodes and logic gates.

1.0Class 12 Physics Chapter 14 Semiconductor Electronics: Key Concepts

Class 12 Physics Chapter 14, Semiconductor Electronics, explains the properties of semiconductors and how semiconductor devices are used in electronic circuits. The chapter covers energy bands, doping, p-n junctions, diodes, rectifiers, and logic gates. The main concepts covered in this chapter include:

Classification of Solids: Explains the difference between conductors, insulators, and semiconductors based on their electrical conductivity, resistivity, and energy band structure.
Intrinsic and Extrinsic Semiconductors: Explains pure semiconductors and the process of doping. It covers n-type and p-type semiconductors formed by adding pentavalent and trivalent impurities.
p-n Junction: Explains how a p-n junction is formed and how the depletion region and barrier potential develop when p-type and n-type semiconductors are joined.
Semiconductor Diode: Covers the working of a p-n junction diode under forward bias and reverse bias. It also explains the current-voltage characteristics and breakdown of a diode.
Diode as a Rectifier: Explains how diodes are used to convert alternating current (AC) into direct current (DC) through half-wave and full-wave rectification.
Special Purpose p-n Junction Diodes: Introduces important semiconductor devices such as LEDs, photodiodes, and solar cells, along with their working principles and uses.
Digital Electronics and Logic Gates: Introduces binary numbers and explains the working and truth tables of NOT, OR, AND, NAND, and NOR gates.

2.0NCERT Solutions Class 12 Physics Chapter 14 Semiconductor Electronics : Detailed Solutions

SOLVED EXAMPLES

C, Si and Ge have same lattice structure. Why is C insulator while Si and Ge intrinsic semiconductors? Sol. The 4 bonding electrons of C , Si or Ge lie, respectively, in the second, third and fourth orbit. Hence, energy required to take out an electron from these atoms (i.e., ionisation energy $E_{g}$ ) will be least for Ge, followed by Si and highest for C. Hence, number of free electrons for conduction in Ge and Si are significant but negligibly small for C.
Suppose a pure Si crystal has $5 \times 1 0^{28}$ atoms $m^{- 3}$ . It is doped by 1 ppm concentration of pentavalent As. Calculate the number of electrons and holes. Given that $n_{i} = 1.5 \times 1 0^{16} m^{- 3}$ . Sol. Note that thermally generated electrons $(n_{i} \sim 1 0^{16} m^{- 3})$ are negligibly small as compared to those produced by doping. Therefore,

$n_{e} \approx N_{D} .$

Since $n_{e} n_{h} = n_{i}^{2}$ , The number of holes

$n_{h} = \frac{( 2.25 \times 1 0 ^{32} )}{( 5 \times 1 0 ^{22} )} \sim 4.5 \times 1 0^{9} m^{- 3}$

Can we take one slab of p-type semiconductor and physically join it to another n-type semiconductor to get p-n junction? Sol. No! Any slab, howsoever flat, will have roughness much larger than the inter-atomic crystal spacing (~2 to $3 \overset{˚}{A}$ ) and hence continuous contact at the atomic level will not be possible. The junction will behave as a discontinuity for the flowing charge carriers.
The V-I characteristic of a silicon diode is shown in the Figure. Calculate the resistance of the diode at (a) $I_{D} = 15 mA$ and (b) $V_{D} = - 10 V$ .

Sol. Considering the diode characteristics as a straight line between $I = 10 mA$ to $I = 20 mA$ passing through the origin, we can calculate the resistance using Ohm's law. (a) From the curve, at $I = 20 mA$

$V = 0.8; I - 10 mA = V = 0.7 V$

Dynamic Resistance,

$r_{fb} = \frac{Δ V}{Δ I} = \frac{( 0.8 - 0.7 ) V}{( 20 - 10 ) mA} = \frac{0.1 V}{10 mA} = 10Ω$

(b) From the curve at

$V = - 10 V, I = - 1 μ A$

Therefore, Static resistance,

$r_{rb} = \frac{10 V}{1 μ A} = 1.0 \times 1 0^{7} Ω$

EXERCISE QUESTIONS WITH SOLUTIONS

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.

Sol. (c)

Which of the statements given in Q. 1 is true for p-type semiconductos.

Sol. (d)

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) $(Eg)_{C} = (Eg)_{Si} = (Eg)_{Ge}$

Sol. (c)

In an unbiased p-n junction, holes diffuse from the p-region to n-region because - (b) From the curve at (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.

Sol. (c)

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.

Sol. (c)

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. Sol. If half wave rectifier, output freq. $= f_{in} = 50 Hz$ If full wave rectifier, output freq. $= 2 f_{in} = 100 Hz$

3.0Class 12 Physics NCERT Solutions – Chapter-wise Links

Access NCERT Solutions for Class 12 Physics chapter-wise, with exercise answers, useful formulas, and solved questions to help students study each topic and practise Physics questions.

Chapter Number	Chapterwise NCERT Solutions
Chapter 1	Electric Charges and Fields
Chapter 2	Electrostatic Potential and Capacitance
Chapter 3	Current Electricity
Chapter 4	Moving Charges and Magnetism
Chapter 5	Magnetism and Matter
Chapter 6	Electromagnetic Induction
Chapter 7	Alternating Current
Chapter 8	Electromagnetic Waves
Chapter 9	Ray Optics and Optical Instruments
Chapter 10	Wave Optics
Chapter 11	Dual Nature of Radiation and Matter
Chapter 12	Atoms
Chapter 13	Nuclei

4.0Key Features of NCERT Solutions for Class 12 Physics Chapter 14

Easy V-I Characteristics: Understand diode V-I graphs, knee voltage, forward bias, reverse bias, and Zener breakdown with clear explanations and step-by-step solutions.
Simple Logic Gate Solutions: Learn how to solve questions based on AND, OR, NOT, NAND, and NOR gates using truth tables and easy steps.
Clear Energy Band Concepts: Understand energy bands in conductors, insulators, and semiconductors, along with the effect of temperature on semiconductor conductivity.
Step-by-Step Rectifier Solutions: Learn how rectifiers convert AC into DC and understand the role of filters in reducing variations in the rectifier output.
Important Semiconductor Terms: Get clear explanations of terms such as doping, intrinsic semiconductor, extrinsic semiconductor, majority carriers, minority carriers, and p-n junction.

NCERT Solutions Class 12 Physics Chapter 14 - Semiconductor Electronics: Materials, Devices and Simple Circuits

Chapter 14 (Semiconductor Electronics) of Physics Class 12 is the conclusive part of Physics and forms the core of today's digital revolution. In this Chapter, students will gain insight into the materials where the conductivity of these materials can be regulated in a controlled manner, as opposed to classical physics of conductors and insulators.

The basic concepts of energy bands, n-type and p-type semiconductor distinctions, and how to describe the functions of a p-n junction are introduced in this chapter. Students must know these fundamental building blocks of electronics so they can understand how the technology works (like smartphones, computers, solar panels). This chapter is significant for performance in CBSE Board Examinations; JEE; and NEET because it has a high emphasis on understanding concepts and constructing graphs that illustrate key information.

To help students develop a thorough understanding of "Logic," NCERT Solutions for Semiconductor Electronics in Chapter 14 of Class 12 Physics provides systematic steps that outline an orderly transition from atomic energy levels to the practical applications of diodes and logic gates.

1.0Class 12 Physics Chapter 14 Semiconductor Electronics: Key Concepts

Class 12 Physics Chapter 14, Semiconductor Electronics, explains the properties of semiconductors and how semiconductor devices are used in electronic circuits. The chapter covers energy bands, doping, p-n junctions, diodes, rectifiers, and logic gates. The main concepts covered in this chapter include:

Classification of Solids: Explains the difference between conductors, insulators, and semiconductors based on their electrical conductivity, resistivity, and energy band structure.
Intrinsic and Extrinsic Semiconductors: Explains pure semiconductors and the process of doping. It covers n-type and p-type semiconductors formed by adding pentavalent and trivalent impurities.
p-n Junction: Explains how a p-n junction is formed and how the depletion region and barrier potential develop when p-type and n-type semiconductors are joined.
Semiconductor Diode: Covers the working of a p-n junction diode under forward bias and reverse bias. It also explains the current-voltage characteristics and breakdown of a diode.
Diode as a Rectifier: Explains how diodes are used to convert alternating current (AC) into direct current (DC) through half-wave and full-wave rectification.
Special Purpose p-n Junction Diodes: Introduces important semiconductor devices such as LEDs, photodiodes, and solar cells, along with their working principles and uses.
Digital Electronics and Logic Gates: Introduces binary numbers and explains the working and truth tables of NOT, OR, AND, NAND, and NOR gates.

2.0NCERT Solutions Class 12 Physics Chapter 14 Semiconductor Electronics : Detailed Solutions

SOLVED EXAMPLES

C, Si and Ge have same lattice structure. Why is C insulator while Si and Ge intrinsic semiconductors? Sol. The 4 bonding electrons of C , Si or Ge lie, respectively, in the second, third and fourth orbit. Hence, energy required to take out an electron from these atoms (i.e., ionisation energy $E_{g}$ ) will be least for Ge, followed by Si and highest for C. Hence, number of free electrons for conduction in Ge and Si are significant but negligibly small for C.
Suppose a pure Si crystal has $5 \times 1 0^{28}$ atoms $m^{- 3}$ . It is doped by 1 ppm concentration of pentavalent As. Calculate the number of electrons and holes. Given that $n_{i} = 1.5 \times 1 0^{16} m^{- 3}$ . Sol. Note that thermally generated electrons $(n_{i} \sim 1 0^{16} m^{- 3})$ are negligibly small as compared to those produced by doping. Therefore,

$n_{e} \approx N_{D} .$

Since $n_{e} n_{h} = n_{i}^{2}$ , The number of holes

$n_{h} = \frac{( 2.25 \times 1 0 ^{32} )}{( 5 \times 1 0 ^{22} )} \sim 4.5 \times 1 0^{9} m^{- 3}$

Can we take one slab of p-type semiconductor and physically join it to another n-type semiconductor to get p-n junction? Sol. No! Any slab, howsoever flat, will have roughness much larger than the inter-atomic crystal spacing (~2 to $3 \overset{˚}{A}$ ) and hence continuous contact at the atomic level will not be possible. The junction will behave as a discontinuity for the flowing charge carriers.
The V-I characteristic of a silicon diode is shown in the Figure. Calculate the resistance of the diode at (a) $I_{D} = 15 mA$ and (b) $V_{D} = - 10 V$ .

Sol. Considering the diode characteristics as a straight line between $I = 10 mA$ to $I = 20 mA$ passing through the origin, we can calculate the resistance using Ohm's law. (a) From the curve, at $I = 20 mA$

$V = 0.8; I - 10 mA = V = 0.7 V$

Dynamic Resistance,

$r_{fb} = \frac{Δ V}{Δ I} = \frac{( 0.8 - 0.7 ) V}{( 20 - 10 ) mA} = \frac{0.1 V}{10 mA} = 10Ω$

(b) From the curve at

$V = - 10 V, I = - 1 μ A$

Therefore, Static resistance,

$r_{rb} = \frac{10 V}{1 μ A} = 1.0 \times 1 0^{7} Ω$

EXERCISE QUESTIONS WITH SOLUTIONS

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.

Sol. (c)

Which of the statements given in Q. 1 is true for p-type semiconductos.

Sol. (d)

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) $(Eg)_{C} = (Eg)_{Si} = (Eg)_{Ge}$

Sol. (c)

In an unbiased p-n junction, holes diffuse from the p-region to n-region because - (b) From the curve at (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.

Sol. (c)

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.

Sol. (c)

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. Sol. If half wave rectifier, output freq. $= f_{in} = 50 Hz$ If full wave rectifier, output freq. $= 2 f_{in} = 100 Hz$

3.0Class 12 Physics NCERT Solutions – Chapter-wise Links

Access NCERT Solutions for Class 12 Physics chapter-wise, with exercise answers, useful formulas, and solved questions to help students study each topic and practise Physics questions.

Chapter Number	Chapterwise NCERT Solutions
Chapter 1	Electric Charges and Fields
Chapter 2	Electrostatic Potential and Capacitance
Chapter 3	Current Electricity
Chapter 4	Moving Charges and Magnetism
Chapter 5	Magnetism and Matter
Chapter 6	Electromagnetic Induction
Chapter 7	Alternating Current
Chapter 8	Electromagnetic Waves
Chapter 9	Ray Optics and Optical Instruments
Chapter 10	Wave Optics
Chapter 11	Dual Nature of Radiation and Matter
Chapter 12	Atoms
Chapter 13	Nuclei

4.0Key Features of NCERT Solutions for Class 12 Physics Chapter 14

Easy V-I Characteristics: Understand diode V-I graphs, knee voltage, forward bias, reverse bias, and Zener breakdown with clear explanations and step-by-step solutions.
Simple Logic Gate Solutions: Learn how to solve questions based on AND, OR, NOT, NAND, and NOR gates using truth tables and easy steps.
Clear Energy Band Concepts: Understand energy bands in conductors, insulators, and semiconductors, along with the effect of temperature on semiconductor conductivity.
Step-by-Step Rectifier Solutions: Learn how rectifiers convert AC into DC and understand the role of filters in reducing variations in the rectifier output.
Important Semiconductor Terms: Get clear explanations of terms such as doping, intrinsic semiconductor, extrinsic semiconductor, majority carriers, minority carriers, and p-n junction.

Which topics should I revise from NCERT Solutions Class 12 Physics Chapter 14?

Why is a semiconductor different from a conductor and an insulator according to NCERT Solutions Class 12 Physics Chapter 14?

According to NCERT Solutions Class 12 Physics Chapter 14, what happens to the depletion region when a p-n junction is forward biased?

A diode is connected in reverse bias. What should I understand from NCERT Solutions Class 12 Physics Chapter 14?

Which V-I characteristics should I study in NCERT Solutions Class 12 Physics Chapter 14?

LED, photodiode and solar cell — what does NCERT Solutions Class 12 Physics Chapter 14 explain about these devices?

How do NCERT Solutions Class 12 Physics Chapter 14 help with logic-gate questions?

Join ALLEN!

NCERT Solutions Class 12 Physics Chapter 14 - Semiconductor Electronics: Materials, Devices and Simple Circuits

1.0Class 12 Physics Chapter 14 Semiconductor Electronics: Key Concepts

2.0NCERT Solutions Class 12 Physics Chapter 14 Semiconductor Electronics : Detailed Solutions

SOLVED EXAMPLES

EXERCISE QUESTIONS WITH SOLUTIONS

3.0Class 12 Physics NCERT Solutions – Chapter-wise Links

4.0Key Features of NCERT Solutions for Class 12 Physics Chapter 14

Table of Contents

Which topics should I revise from NCERT Solutions Class 12 Physics Chapter 14?

Why is a semiconductor different from a conductor and an insulator according to NCERT Solutions Class 12 Physics Chapter 14?

According to NCERT Solutions Class 12 Physics Chapter 14, what happens to the depletion region when a p-n junction is forward biased?

A diode is connected in reverse bias. What should I understand from NCERT Solutions Class 12 Physics Chapter 14?

Which V-I characteristics should I study in NCERT Solutions Class 12 Physics Chapter 14?

LED, photodiode and solar cell — what does NCERT Solutions Class 12 Physics Chapter 14 explain about these devices?

How do NCERT Solutions Class 12 Physics Chapter 14 help with logic-gate questions?

Join ALLEN!

NCERT Solutions Class 12 Physics Chapter 14 - Semiconductor Electronics: Materials, Devices and Simple Circuits

1.0Class 12 Physics Chapter 14 Semiconductor Electronics: Key Concepts

2.0NCERT Solutions Class 12 Physics Chapter 14 Semiconductor Electronics : Detailed Solutions

SOLVED EXAMPLES

EXERCISE QUESTIONS WITH SOLUTIONS

3.0Class 12 Physics NCERT Solutions – Chapter-wise Links

4.0Key Features of NCERT Solutions for Class 12 Physics Chapter 14

Table of Contents