NCERT Solutions for Class 12 Physics Chapter 14 – Free PDF Download
The NCERT Solutions for Class 12 Physics Chapter 14 Semiconductor Electronics: Materials, Devices and Simple Circuits are important resource materials if students are aiming to score good marks in CBSE Class 12 Physics examination. The NCERT Solutions for Class 12 Physics of this chapter have answers to the textbook questions, exemplary questions, the important questions from previous years’ question papers and sample papers together with exercises, worksheets and MCQs.
The NCERT Solutions for Class 12 Physics help students to sort out their weaknesses and strengths in the topic of Semiconductors. They can also learn the right methods to solve numerical problems using these solutions, which are updated for the CBSE Syllabus (2023-24). Further, the NCERT Solutions for Class 12 Physics Chapter 14 help learners take short notes, which will come in handy for revisions when the board exam nears.
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Class 12 Physics NCERT Solutions Semiconductor Devices Important Questions
Q 1. In an n-type silicon, which of the following statement is true:
(a) Electrons are the 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 the majority carriers, and trivalent atoms are the dopants.
Ans:
Here, (c) is the correct option.
For n-type silicon, the majority carriers are electrons, while the minority carriers are holes. An n-type semiconductor is obtained by dropping pentavalent atoms like phosphorus in silicon atoms.
 Q2. Which of the statements given in Exercise 14.1 is true for p-type semiconductors?
Ans:
Here, (d) is the correct explanation.
For p-type semiconductor, holes are the majority carriers, while electrons are the minority carriers. A p-type semiconductor is obtained by using trivalent atoms like aluminium in silicon atoms.
Q 3. Carbon, silicon and germanium have four valence electrons each. These are characterised by valence and conduction bands separate by energy band gap respectively equal to\(\begin{array}{l}\left (E_{g} \right )_{Si}\end{array} \)
, \(\begin{array}{l}\left (E_{g} \right )_{c}\end{array} \)
and \(\begin{array}{l}\left (E_{g} \right )_{Ge}\end{array} \)
. Which of the following statements is true?
(a)\(\begin{array}{l}\left (E_{g} \right )_{Si} < \left (E_{g} \right )_{Ge} < \left (E_{g} \right )_{C}\end{array} \)
(b)\(\begin{array}{l}\left (E_{g} \right )_{C} < \left (E_{g} \right )_{Ge} > \left (E_{g} \right )_{Si}\end{array} \)
(c)\(\begin{array}{l}\left (E_{g} \right )_{C} > \left (E_{g} \right )_{Si} > \left (E_{g} \right )_{Ge}\end{array} \)
(d)\(\begin{array}{l}\left (E_{g} \right )_{C} = \left (E_{g} \right )_{Si} = \left (E_{g} \right )_{Ge}\end{array} \)
Ans:
(C) is the correct option.
Out of carbon, germanium, and silicon, carbon has the maximum energy bandgap, whereas germanium has the least energy bandgap.
For all these elements, the energy band gap can be related as:
\(\begin{array}{l}\left (E_{g} \right )_{C} > \left (E_{g} \right )_{Si} > \left (E_{g} \right )_{Ge}\end{array} \)
Q 4. In an unbiased p-n junction, holes diffuse to the n-region from the p-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 the p-region is more as compared to the n-region.
(d) All of the above.
 Ans:
(c) is the correct option.
The usual tendency of the charge carriers is to disperse towards the lower concentration region from the higher concentration region. So it can be said that in an unbiased p-n junction, holes disperse from the p-region to the n-region, as the p-region has a greater concentration of holes than in the n-region.
Q 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) potential barrier is reduced.
(d) None of the above
 Ans:
(c) is the correct option
The potential barrier reduces for a p-n junction when a forward bias is applied.
In the above case, the potential barrier across the junction reduces as the applied voltage is opposed by the potential barrier.
Q 6. 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?
Ans:
For a half-wave rectifier, the output frequency is equal to the input frequency. In this case, the input frequency of the half-wave rectifier is 50 Hz.
On the other hand, the output frequency for a full-wave rectifier is twice the input frequency.
Therefore, the output frequency is 2 × 50 = 100 Hz.
Q 7. A p-n photodiode is fabricated from a semiconductor with a bandgap of 2.8 eV. Can it detect a wavelength of 6000 nm?
Ans:
No, the photodiode cannot detect the wavelength of 6000 nm because of the following reason:
The energy bandgap of the given photodiode, Eg = 2.8 eV
The wavelength is given by λ = 6000 nm = 6000 × 10−9 m
We can find the energy of the signal from the following relation:
E = hc/λ
In the equation, h is Planck’s constant = 6.626 × 10−34 J and c is the speed of light = 3 × 108 m/s
Substituting the values in the equation, we get
E = (6.626 x 10-34 x 3 x 108) / 6000 x 10-9 = 3.313 x 10-20 J
But, 1.6 × 10 −19 J = 1 eV
Therefore, 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.
Q 8. The number of silicon atoms per m3 is 5 × 1028. This is doped simultaneously with 5 × 1022 atoms per m3 of Arsenic and 5 × 1020 per m3 atoms of Indium. Calculate the number of electrons and holes. Given that nI = 1.5 × 1016m–3. Is the material n-type or p-type?
Ans:Â
The following values are given in the question:
Number of silicon atoms, N = 5 × 10 28 atoms/m3
Number of arsenic atoms, nAS =5×1022atoms/m3
Number of indium atoms, nIn=5×1022atoms/m3
ni=1.5×1016electrons/m3
ne=5×1022−1.5×1016=4.99×1022
Let us consider the number of holes to be nh
In the thermal equilibrium, nenh = ni2
Calculating, we get
nh=4.51×109
Here, ne>nh, therefore, the material is an n-type semiconductor.
Q 9: In an intrinsic semiconductor, the energy gap Eg is 1.2eV. 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 ni is given by
\(\begin{array}{l}n_{i}=n_{0}exp\left [ -\frac{E_{g}}{2k_{B}T} \right ]\end{array} \)
where n0 is constant.
Ans:
Energy gap in an intrinsic semiconductor, Eg = 1.2 eV
The temperature dependence of the intrinsic carrier concentration is given as
\(\begin{array}{l}n_{i}=n_{0}exp\left [ -\frac{E_{g}}{2k_{B}T} \right ]\end{array} \)
Here, kB is the Boltzmann constant = 8.62 x 10-5 eV/K
T is the temperature
n0 is a constant
Initial temperature, T1 = 300 K
The intrinsic carrier concentration at the temperature 300 K can be written as
\(\begin{array}{l}n_{i1}=n_{0}exp\left [ -\frac{E_{g}}{2k_{B}\times 300} \right ]\end{array} \)
—–(1)
Final temperature, T2 = 600 K
The intrinsic carrier concentration at the temperature 600 K can be written as
\(\begin{array}{l}n_{i2}=n_{0}exp\left [ -\frac{E_{g}}{2k_{B}\times 600} \right ]\end{array} \)
——-(2)
The ratio between conductivity at 600K and that at 300K is equal to the ratio between the respective carrier concentration at these temperatures.
\(\begin{array}{l}\frac{n_{i2}}{n_{i1}}= \frac{n_{0}exp\left [ -\frac{E_{g}}{2k_{B}\times 600} \right ]}{n_{0}exp\left [ -\frac{E_{g}}{2k_{B}\times 300} \right ]}\end{array} \)
Â
\(\begin{array}{l}= exp \frac{E_{g}}{2k_{B}}\left [ \frac{1}{300}-\frac{1}{600} \right ]\end{array} \)
Â
\(\begin{array}{l}= exp \frac{1.2}{2\times 8.62\times 10^{-5}}\left [ \frac{1}{300}-\frac{1}{600} \right ]\end{array} \)
Â
\(\begin{array}{l}= exp \frac{1.2}{17.24\times 10^{-5}}\left [ \frac{2-1}{600} \right ]\end{array} \)
= exp (11.6) = 1.09 x 105
There, the ratio between the conductivities is 1.09 x 105.
Q 10. In a p-n junction diode, the current I can be expressed as
\(\begin{array}{l}I = I_{0}exp\left ( \frac{eV}{2k_{B}T}-1\right )\end{array} \)
where I0 is called the reverse saturation current, V is the voltage across the diode and is positive for forward bias and negative for reverse bias, and I is the current through the diode, kB is the Boltzmann constant (8.6×10–5 eV/K), and T is the absolute temperature. If for a given diode I0 = 5 × 10–12 A and T = 300 K, then
(a) What will be the forward current at a forward 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?
​Ans:
In a p-n junction diode, the expression for current is given as
\(\begin{array}{l}I = I_{0}exp\left ( \frac{eV}{2k_{B}T}-1\right )\end{array} \)
Here, I0 = 5 × 10–12 A
T = 300 K
kB is the Boltzmann constant = 8.6 x 10-5 eV/k =Â 8.6 x 10-5Â x 1.6 x 10-19 = 1.376 x 10-23 J/K
(a) Forward voltage, V = 0.6 V
\(\begin{array}{l}I = 5\times 10^{-12}exp\left ( \frac{1.6\times 10^{-19}\times 0.6}{2\times 1.376\times 10^{-23}\times 300}-1\right )\end{array} \)
Â
\(\begin{array}{l}I = 5\times 10^{-12}exp (22.36)\end{array} \)
= 0.0256Â A
(b) Voltage across the diode is increased to 0.7 V
\(\begin{array}{l}I’ = 5\times 10^{-12}exp\left ( \frac{1.6\times 10^{-19}\times 0.7}{2\times 1.376\times 10^{-23}\times 300}-1\right )\end{array} \)
Â
\(\begin{array}{l}I’ = 5\times 10^{-12}exp (26.25)\end{array} \)
I’ = 1.257 A
Change in current, ΔI = I’ – I
= 1.257 – 0.0256
= 1.23Â A
Change is voltage = 0.7 – 0.6 = 0.1 V
(c) Dynamic resistance = Change in voltage/ Change in current
= 0.1/1.23
= 0.081 Ω
(d)Â If the reverse bias voltage changes from 1 V to 2 V, the current will almost remain equal to I0 in both cases. Therefore, the dynamic resistance in the reverse bias will be infinite.
Q 11. You are given the two circuits as shown in Fig. 14.36. Show that
circuit (a) acts as OR gate while circuit (b) acts as AND gate.
a)
b)
Ans:
(a) Input for the circuit: A and B
Output for the circuit: Y
Here, the left half of the circuit works as the NOR gate, while if we look at the right half of the circuit, then it is a NOT gate.
We can show it in the figure as:
Now, the output for the NOR gate =
\(\begin{array}{l}\overline{A + B}\end{array} \)
We are giving the output from NOR gate to the input at the NOT gate. So, its output will be
\(\begin{array}{l}\overline{\overline{A + B}}\end{array} \)
= A + B
\(\begin{array}{l}∴ Y = A + B\end{array} \)
Hence, this circuit functions as an OR Gate.
(b) Input for the circuit: A and B
Output for the circuit: Y
We can observe from the figure that the output from the first half of the circuits goes to the input of the next half, which is working as NOR gate.
Hence, the output of the given circuit can be written as
\(\begin{array}{l}Y = \overline{\bar{A} + \bar{B}} = \bar{\bar{A}}.\bar{\bar{B}} = A.B\end{array} \)
Hence, this circuit functions as an AND gate.
Q 12. Write the truth table for a NAND gate connected as given in
Fig. 14.37
Hence, identify the exact logic operation carried out by this circuit.
Ans:
From the figure, we know A is the input, and we are getting an output B.
Hence, the output can be written as
\(\begin{array}{l}Y = \overline{ A . A } = \bar{A} + \bar{A} = \bar{A}\end{array} \)
We can make the truth table for equation (1) as:
| A |
Y= \(\begin{array}{l}\bar{A}\end{array} \) |
| 0 |
1 |
| 1 |
0 |
Below is the symbol of the NOT gate:
Q 13. You are given two circuits, as shown in Figure, which consist of NAND gates. Identify the logic operation carried out by the two circuits.
Ans:
(a) The Boolean expression for NAND gate is
\(\begin{array}{l}Y=\overline{AB}\end{array} \)
Let Y’ be the output of the NAND gate.
Therefore, the final output of the combination,Â
\(\begin{array}{l}Y = \overline{Y’.Y’}\end{array} \)
Â
\(\begin{array}{l}\mathbf{Y}=\overline{\overline{\mathbf{A . B}} \cdot \overline{\mathbf{A . B}}}\end{array} \)
Â
\(\begin{array}{l}Y = \overline{A.B}+\overline{A.B}\end{array} \)
Y = A.B
The given circuit acts as an AND gate.
(b) Boolean expression for NOT gate isÂ
\(\begin{array}{l}Y = \bar{A}\end{array} \)
The output Y1 of the first NOT gate isÂ
\(\begin{array}{l}Y_{1} = \bar{A}\end{array} \)
Â
The output Y2 of the second NOT gate isÂ
\(\begin{array}{l}Y_{2} = \bar{B}\end{array} \)
The output of the combination
Â
\(\begin{array}{l}Y = \overline{Y_{1}.Y_{2}}\end{array} \)
Â
\(\begin{array}{l}Y = \overline{\bar{A}.\bar{B}}\end{array} \)
Â
\(\begin{array}{l}Y = \bar{\bar{A}}+\bar{\bar{B}}\end{array} \)
Y = A + B
This is the Boolean expression for the OR gate.
Â
Q 14. Write the truth table for the circuit given in Fig. 14.39 below, consisting
of NOR gates and identify the logic operation (OR, AND, NOT) which
this circuit is performing
 (Hint: A = 0, B = 1, then A and B inputs of the second NOR gate will be 0 and hence, Y=1. Similarly, work out the values of Y for other combinations of A and B. Compare with the truth table of OR, AND, NOT gates and find the correct one.)
Ans
Input for the circuit: A and B
Output from the first circuit (i.e., NOR gate) is
\(\begin{array}{l}\overline{A + B}\end{array} \)
.
From the figure, we can observe that the input for the second gate is automatically the output from the first one.
Therefore, the output obtained from this combination is given as:
\(\begin{array}{l}Y = \overline{\overline{A + B} + \overline{A + B} } = \overline{ \bar{ A } . \bar{ B }} + \overline{ \bar{ A } . \bar{ B }} \\ \\ = \overline{ \bar{ A } . \bar{ B }} = \bar{ \bar{ A } } + \bar{ \bar{ B } } = A + B\end{array} \)
Following is the truth table for the given operation:
| A |
B |
Y (= A + B) |
| 0 |
0 |
0 |
| 0 |
1 |
1 |
| 1 |
0 |
1 |
| 1 |
1 |
1 |
This is the truth table of an OR gate.
Therefore, this circuit functions as an OR gate.
Q 15. Write the truth table for the circuits given in the Figure consisting of NOR gates only. Identify the logic operations (OR, AND, NOT) performed by the two circuits.
Ans:
In figure (a), A acts as the two inputs of the NOR gate, and Y is the output. Hence, the output of the circuit is
\(\begin{array}{l}\overline{\mathrm{A}+\mathrm{A}}\end{array} \)
So, the outputÂ
\(\begin{array}{l}Y= \overline{\mathrm{A}}\end{array} \)
Therefore, the gate is a NOT gate.
The truth table is given as
| A |
\(\begin{array}{l}Y= \overline{\mathrm{A}}\end{array} \) |
| 0 |
1 |
| 1 |
0 |
(b) In figure (b), A and B are the inputs, and Y is the output of the given circuit. By using the result obtained in solution (a), we can infer that the outputs of the first two NOR gates are
\(\begin{array}{l}\overline{\mathrm{A}}\end{array} \)
and \(\begin{array}{l}\overline{\mathrm{B}}\end{array} \)
. This is the input of the NOR gate. Therefore, the output of the circuit is \(\begin{array}{l}\mathbf{Y}=\overline{\overline{\mathbf{A}}+\overline{\mathbf{B}}}=\mathbf{A} . \mathbf{B}\end{array} \)
. So, this acts as an AND gate. The truth table is given below:
| A |
B |
Y= A.B |
| 0 |
0 |
0 |
| 0 |
1 |
0 |
| 1 |
0 |
0 |
| 1 |
1 |
1 |
Class 12 Physics NCERT Solutions for Chapter 14 Semiconductor Electronics
Chapter 14 Semiconductor Electronics of Class 12 Physics, is designed as per the current CBSE Syllabus 2023-24. In this chapter’s NCERT Solutions for Class 12 Physics, most of the questions will consist of an n-type semiconductor device and the majority of carriers and dopants. We will be seeing true-false questions related to semiconductors made out of silicon, carbon and germanium and finding out the energy gap in them. We will be seeing what will happen when a p-n junction is forward-biased and reverse-biased.
Subtopics involved in Class 12 Physics Chapter 14 Semiconductor Electronics
- Introduction
- Classification of Metals, Conductors, and Semiconductors
- Intrinsic Semiconductor
- Extrinsic Semiconductor
- P-n Junction
- p-n junction formation
- Semiconductor Diode
- p-n junction diode under forward bias
- p-n junction diode under reverse bias
- Application of Junction Diode as a Rectifier
- Special Purpose P-n Junction Diodes
- Zener diode
- Optoelectronic junction devices
- Junction Transistor
- Transistor: structure and action
- Basic transistor circuit configurations and transistor characteristics
- Transistor as a device
- Transistor as an Amplifier (CE-Configuration)
- Feedback amplifier and transistor oscillator
- Digital Electronics and Logic Gates
- Logic gates
- Integrated Circuits
We will be seeing questions based on calculating the output frequency of a half-wave and full-wave rectifier and even have questions on finding the input signal and base current of a CE-Transistor amplifier. We will even help students find out the output AC signal if two amplifiers are connected in series. If a p-n photodiode is fabricated, will it detect wavelengths? Find out the answers to the questions here with NCERT Solutions.
We have additional exercises containing questions based on finding a number of holes and electrons in atoms of silicon, Indium and arsenic. We have questions pertaining to calculating the ratio between conductivity at two different temperatures.
We even have questions on OR and AND gates where students will be figuring out which of the given circuits comes under these two categories, and they will be able to write the truth table for the NAND gates used. Additionally, students will see questions on the different logical operations. These questions will help them understand everything about semiconductor devices and the problems related to them. Practising these problems will definitely ensure that they get a good grasp of the subject.
The NCERT Solutions are created by subject experts according to the latest CBSE Syllabus and its guidelines. Students must practise the solutions regularly to prepare effectively for their examinations.
BYJU’S presents them with an interactive mode of learning with videos, animations and exercises, along with Exemplar problems.
Disclaimer –Â
Dropped Topics –Â
14.8 Special Purpose p-n junction Diodes
14.9 Digital Electronics and Logic Gates
Exercises 14.7–14.15
Frequently Asked Questions on NCERT Solutions for Class 12 Physics Chapter 14
Q1
What are the benefits of using the NCERT Solutions for Class 12 Physics Chapter 14?
The NCERT Solutions for Class 12 Physics Chapter 14 are the answers to the questions present in the prescribed textbook. The solutions are created by the subject experts at BYJU’S based on the grasping abilities of students. The faculty provides 100% accurate and quality solutions based on the latest CBSE guidelines and the exam pattern. Each and every solution is explained in simple language to boost the confidence among students to score well in the board exam.
Q2
How to get an in-depth knowledge of the concepts covered in Chapter 14 of the NCERT Solutions for Class 12 Physics?
To obtain the right knowledge, understanding the concept is essential for the students of Class 12. Selecting the right study material plays an important role in getting an in-depth knowledge of the concepts. For this purpose, the faculty at BYJU’S have curated solutions to help students improve their academic performance. The solutions can also be used to cross-check their answers and analyse the areas where they are lagging behind.
Q3
Explain intrinsic and extrinsic semiconductors according to the NCERT Solutions for Class 12 Physics Chapter 14.
Extrinsic semiconductors are semiconductors that are doped with specific impurities. The impurity modifies the electrical properties of the semiconductor and makes it more suitable for electronic devices such as diodes and transistors. Semiconductors that are chemically pure, in other words, free from impurities, are termed intrinsic semiconductors.
(Hint: A = 0, B = 1, then A and B inputs of the second NOR gate will be 0 and hence, Y=1. Similarly, work out the values of Y for other combinations of A and B. Compare with the truth table of OR, AND, NOT gates and find the correct one.)
GOOD