NCERT Solutions for Class 12 Physics Chapter 14 Semiconductor Electronics Materials Devices And Simple Circuit

where $I_{0}$ 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, $k_{B}$ is the Boltzmann constant $(8.6\times 10^{-5}eV/K)$ and $T$ is the absolute temperature. If for a given diode $I_{0}=5\times 10^{-12}A$ and $T=300\; K,$ then

(a) What will be the forward current at a forward voltage of $0.6\; V\; ?$

Answer:

As we have

$I=I_{0}\; [exp \frac{eV}{K_{B}T}-1 ]$

Here, $I_{0}=5\times 10^{-12}A$ , $T=300\; K,$ and , $k_{B}$ = Boltzmann constant = $(8.6\times 10^{-5}eV/K)$ $=(1.376*10^{-23}J/K)$

When the forward voltage is 0.6V:

$I=5*10^{-12}\;[ exp \frac{1.6*10^{-19}*0.6}{1.376*10^{-23}*300}-1 ]=0.0625A$

Hence forward current is 0.0625A

Q.14.10 (b) In a p-n junction diode, the current I can be expressed as

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, $k_{B}$ is the Boltzmann constant $(8.6\times 10^{-5}\; eV/K)$ and $T$ is the absolute temperature. If for a given diode $I_{0}=5\times 10^{12}A$ and $T=300\; K,$ then

(b) What will be the increase in the current if the voltage across the diode is increased to $0.7 \; V?$

Answer:

As we have

$I=I_{0}\; [exp \frac{eV}{K_{B}T}-1 ]$

Here, $I_{0}=5\times 10^{-12}A$ , $T=300\; K,$ and , $k_{B}$ = Boltzmann constant = $(8.6\times 10^{-5}eV/K)$ $=(1.376*10^{-23}J/K)$

When the forward voltage is 0.7V:

$I=5*10^{-12}\;[ exp \frac{1.6*10^{-19}*0.7}{1.376*10^{-23}*300}-1 ]=3.029A$

When the forward voltage is 0.6V:

$I=5*10^{-12}\;[ exp \frac{1.6*10^{-19}*0.6}{1.376*10^{-23}*300}-1 ]=0.0625A$

Hence the increase in the forward current is

$I(whenv=0.7) - I(whenv=.6)$ $= 3.029- 0.0625 = 2.967A$

Q. 14.10 (c) In a p-n junction diode, the current I can be expressed as

where $I_{0}$ 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, $k_{B}$ is the Boltzmann constant $(8.6\times 10^{-10}\; eV/K)$ and $T$ is the absolute temperature. If for a given diode $I_{0}=5\times 10^{-12}A$ and $T=300\; K,$ then

Answer:

Dynamic Resistance = $\frac{voltage-change}{ current-change}$

Resistance change = 0.7 - 0.6 = 0.1

Current change = 2.967(calculated in prev question)

Therefore

, $Dynamic Resistance = \frac{0.1}{2.967} = 0.0337\Omega$

Q.14.10 (d) In a p-n junction diode, the current I can be expressed as

Q 14.13 You are given two circuits as shown in Fig. 14.38, which consist of NAND gates. Identify the logic operation carried out by the two circuits.

where $I_{0}$ 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, $k_{B}$ is the Boltzmann constant $(8.6\times 10^{-5}V/K)$ and $T$ is the absolute temperature. If for a given diode $I_{0}=5\times 10^{-12}A$ and $T=300\; K,$ then

(d) What will be the current if reverse bias voltage changes from 1 V to 2 V?

Answer:

As we have

$I=I_{0}\; [exp \frac{eV}{K_{B}T}-1 ]$

Here, $I_{0}=5\times 10^{-12}A$ , $T=300\; K,$ and , $k_{B}$ = Boltzmann constant = $(8.6\times 10^{-5}eV/K)$ $=(1.376*10^{-23}J/K)$

When reverse voltage is 1V, V= -1

$I=5*10^{-12}\;[ exp \frac{1.6*10^{-19}*(-1)}{1.376*10^{-23}*300}-1 ]\approx5\times10^{-12}$

When the reverse voltage is -2V:

$I=5*10^{-12}\;[ exp \frac{1.6*10^{-19}*(-2)}{1.376*10^{-23}*300}-1 ]\approx5\times10^{-12}$

In both case current is very small and approximately equal to the reverse saturation current, hence their difference is negligible which causes dynamic resistance of infinity.

Q. 14.11 (a) You are given the two circuits as shown in Fig. 14.36. Show that circuit

(a) acts as OR gate while the circuit

1645428242590

Fig. 14.36

Answer:

Here, THE Input = A and B

Output = Y

The left part of the figure acts as a NOR and right part acts as NOT Gate.

The output of NOR gate = $\overline{A+B}$

the output of the NOR gate would be the input of NOT Gate and hence

Y = $\over\overline{A+B}$ = $\overline\overline{A+B}$

Hence the figure functions like an OR Gate.

or compare the truth table by giving different input and observing the output

INPUTS	OUTPUT
A B	Y
0 0	0
0 1	1
1 0	1
1 1	1

Q. 14.11 (b) You are given the two circuits as shown in Fig. 14.36. Show that circuit

(b) acts as AND gate.

1645428384709

Answer:

The output of NOT gate ( left part of the circuit) is the input of the NOR gate

Hence the output of total circuit Y = $\over(\overline A + \overline B)$

= $\overline{\overline A}.\overline{\overline B}$ $\overline{A+B}=\overline A. \overline B$

= $A*B$

Hence the circuit functions as AND gate.

or give the inputs 00,01,10,11 and observe the truth table

INPUTS	OUTPUT
A B	Y
0 0	0
0 1	0
1 0	0
1 1	1

The truth table is the same as that of AND gate

Q. 14.12 Write the truth table for a NAND gate connected as given in Fig. 14.37

1645428401276

Hence identify the exact logic operation carried out by this circuit.

Answer:

Here A is both input of the NAND gate and hence Output Y will be

$Y = \overline {A*A}$

$Y = \overline {A} + \overline A$

$Y = \overline {A}$

Hence circuit functions as a NOT gate.

The truth table for the given figure:

Input	Output
A	Y
0	1
1	0

1645428416591

Answer:

A and B are inputs of a NAND gate and output of this gate is the input of another NAND gate so,

Y = $\over(\overline {A.B})(\overline {A.B})$

Y= $\over(\overline {A.B})$ $+$ $\over(\overline {A.B})$

Y= $AB$

Hence this circuit functions as AND gate.

A is input to the NAND gate output of whose goes to the rightmost NAND gate. Also, B is input to the NAND gate whose output goes to the rightmost NAND gate.

Y = $\over \overline A .\overline B$

Y = $\over\overline A .$ + $\over\overline B.$

Y = A + B

Hence the circuit functions as an OR gate .

Alternative method

fig. a

construct the truth table by giving various input and observe the output

INPUT	INTERMEDIATE OUTPUT	OUTPUT
00	1	0
01	1	0
10	1	0
11	0	1

The above truth table is the same as that of an AND gate

fig. b

INPUTS	OUTPUT
00	0
01	1
10	1
11	1

The above truth table is the same as that of an OR gate

Q. 14.14 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.

14ncrt

(Hint: $A=0,B=1$ then $A$ and $B$ inputs of 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.)

Answer:

A and B are the input od a NOR gate and Output of this NOR gate is the Input of Another NOR gate whose Output is Y. Hence,

Y = $\over(\overline{A+B} + \overline{A+B})$

Y = $\over\overline {A+B}$ . $\over\overline {A+B}$

Y = A + B

Hence Circuit behaves as OR gate.

Truth table

INPUTS	OUTPUT
00	0
01	1
10	1
11	1

Q. 14.15 Write the truth table for the circuits given in Fig. 14.40 consisting of NOR gates only. Identify the logic operations (OR, AND, NOT) performed by the two circuits.

1645428438468

Figure 14.40

Answer:

A is the two input of the NOR gate and Hence Output Y is:

Y = $\overline {A+A}$

Y = $\overline {A}$

Hence circuit functions as a NOT gate.

TRUTH TABLE:

INPUT	OUTPUT
0	1
1	0

b) A is the two input of a NOR gate whose output(which is $\overline {A}$ ) is the one input of another NOR gate. B is the two input of NOR gate whose output (which is $\overline {B}$ ) is the input of another NOR gate. Hence,

Y = $\over\overline {A} + \overline {B}$

Y = $\over\overline {A}$ . $\over\overline {B}$

Y = A.B

Hence it functions as AND gate.

TRUTH TABLE:

INPUTS	OUTPUT
00	0
01	0
10	0
11	1

Semiconductors ncert solutions holds significant importance for students preparing for board exams, as well as competitive ones like JEE and NEET. Its relevance lies in the practical applications of semiconductor devices in modern technology. Scoring in this ncert solutions class 12 physics chapter 14 can be relatively manageable, provided students have a solid grasp of semiconductor concepts. Therefore, mastering this chapter not only ensures success in board exams but also aids in tackling competitive exams with confidence, making it a valuable part of the syllabus.

NCERT solutions for class 12 physics chapter-wise

NCERT solutions for class 12 physics chapter 1 Electric Charges and Fields

NCERT solutions for class 12 physics chapter 2 Electrostatic Potential and Capacitance

NCERT solutions for class 12 physics chapter 3 Current Electricity

NCERT solutions for class 12 physics chapter 4 Moving Charges and Magnetism

NCERT solutions for class 12 physics chapter 5 Magnetism and Matter

NCERT solutions for class 12 physics chapter 6 Electromagnetic Induction

NCERT solutions for class 12 physics chapter 7 Alternating Current

NCERT solutions for class 12 physics chapter8 Electromagnetic Waves

NCERT solutions for class 12 physics chapter 9 Ray Optics and Optical Instruments

NCERT solutions for class 12 physics chapter 10 Wave Optics Solutions

NCERT solutions for class 12 physics chapter 11 Dual nature of radiation and matter

NCERT solutions for class 12 physics chapter 12 Atoms

NCERT solutions for class 12 physics chapter 13 Nuclei

NCERT solutions for class 12 physics chapter 14 Semiconductor Electronics Materials Devices and Simple Circuits

Semiconductor Class 12 NCERT Solutions: Important Formulas and Diagrams

Intrinsic Semiconductors: n_e=n_i=n_n

JEE Main Highest Scoring Chapters & Topics

Just Study 40% Syllabus and Score upto 100%

Download EBook

Where:

n_e represents the free electron density in the conduction band of a semiconductor, n_i stands for the intrinsic carrier concentration and n_nrepresents the hole density in the valence band of a semiconductor.

Extrinsic Semiconductors:

N-type: n_e≅ N_d>n_n

P- type: n_n ≅ N_d>n_e

Where: N_d is the number density

Mobility

=V_d/E

Where: V_d = the drift velocity and E is the electric field

Action of Transistor

I_E=I_C+I_B

Where:

I_E is the emitter current, I_C is the collector current and I_B is the base current

Important Topic Covered In Semiconductor Class 12

Some of the important topic of ncert solutions of chapter 14 physics class 12 are listed below:

Basics of Semiconductors and Valance Band Theory
Types of Semiconductors
Diode and its Chara
Rectifiers, Photo Diodes, LED and Solar Cell
Logic Gates

Significance of NCERT Solutions for Class 12 Physics Chapter 14 Semiconductor Electronics Materials Devices And Simple Circuit

As far as the CBSE board exam is considered you can expect more theoretical questions from Class 12 Physics Chapter 14 NCERT solutions.
With the help of NCERT solutions for class 12 physics chapter 14 semiconductor electronics materials devices and simple circuits, you can prepare well and it is easy to score in this chapter.
NCERT Solutions for class 12 help in preparations of exams like NEET and JEE Mains. For NEET exams one or two questions and for JEE Mains one question is expected from this chapter

How to use NCERT class 12 physics chapter 14 semiconductor electronics materials devices and simple circuits:

Go through the NCERT Syllabus for Class 12 Physics first. Therein, check all the important topics and subtopics.
Try to solve all the questions by your own first. Afterwards check the NCERT Solutions for Class 12 Physics Chapter 14 Semiconductor Electronics Materials Devices And Simple Circuit.
Along with the NCERT solutions, also practice from the CBSE previous year question papers and sample papers to score well in the examination.
NCERT Solutions for Class 12 Physics Chapter 14 PDF Download will also be available soon.

Key features of NCERT Solutions Class 12 Physics Chapter 14

Comprehensive Coverage: These semiconductors ncert solutions encompass all the topics and questions found in Chapter 14, ensuring a thorough understanding of semiconductors and electronic devices.
Detailed Explanations: Each solution offers comprehensive, step-by-step explanations, making complex semiconductor concepts accessible to students.
Clarity and Simplicity: The chapter 14 physics class 12 ncert solutions pdf are presented in clear and straightforward language, ensuring ease of understanding.
Practice Questions: Exercise questions are included for practice and self-assessment, enhancing students' problem-solving skills.
Exam Preparation: These ncert solutions class 12 physics chapter 14 are essential for board exam preparation and provide valuable support for competitive exams.
Real-Life Applications: Concepts covered in this semiconductor class 12 ncert solutions have practical applications in the world of electronics and technology, enhancing students' real-world knowledge.
Free Access: These ncert solutions of chapter 14 physics class 12 are available for free, ensuring accessibility to all students.

Also Check NCERT Books and NCERT Syllabus here:

NCERT solutions subject wise

Frequently Asked Questions (FAQs)

1. How many questions are asked for JEE main from the chapter semiconductor electronics?

1 or maximum 2 questions can be expected for JEE Main from NCERT chapter Semiconductor electronics. The questions may be from any topics of either analog electronics or digital electronics(Logic Gates). Cover the topics from JEE Main syllabus to get good score. Students should practice enough questions to crack an exam like JEE Main or NEET. To pratice students can refer to NCERT questions, NCERT Exemplar questions and JEE Main previous year papers.

2. What is the weightage of class 12 chapter semiconductor electronics materials devices and simple circuits for NEET exam?

Two or three questions can be expected from the chapter. Questions from both analogue and digital electronics part can be expected.

3. Is the chapter semiconductor devices useful for higher studies?

Yes, this chapter is the basics for higher studies in electronics-related branches and for VLSI and Nano Technology.

4. What types of questions are asked from the NCERT chapter semiconductor electronics materials devices and simple circuits for board exams?

Mostly theoretical questions are asked from the chapter semiconductor electronics for board exams. These covers mainly the analog electronics part that has topics like diode and its working, intrinsic and extrinsic semiconductors, how pn junction is formed, concepts of rectifiers, photo diodes, leds and solar cells etc.

Also Read

NCERT Class 12 Maths Notes - Download Chapter wise Free PDFs

NCERT Class 12 Physics Chapter 9 Notes Ray Optics and Optical Instruments - Download PDF

NCERT Class 12 Physics Chapter 8 Notes Electromagnetic Waves - Download PDF

NCERT Class 12 Physics Chapter 7 Notes Alternating Current - Download PDF

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quartely question paper for mathematics cbse

Hello there! Thanks for reaching out to us at Careers360.

Ah, you're looking for CBSE quarterly question papers for mathematics, right? Those can be super helpful for exam prep.

Unfortunately, CBSE doesn't officially release quarterly papers - they mainly put out sample papers and previous years' board exam papers. But don't worry, there are still some good options to help you practice!

Have you checked out the CBSE sample papers on their official website? Those are usually pretty close to the actual exam format. You could also look into previous years' board exam papers - they're great for getting a feel for the types of questions that might come up.

If you're after more practice material, some textbook publishers release their own mock papers which can be useful too.

Let me know if you need any other tips for your math prep. Good luck with your studies!

i have taken admission in clg after my campartment exam in chemistry along with the neet 2025 preperation but unfortunately i again got campartment after doing all this preparation i got 15 marks only now I can't able to think what should I do next.

Jefferson 25 September,2024

It's understandable to feel disheartened after facing a compartment exam, especially when you've invested significant effort. However, it's important to remember that setbacks are a part of life, and they can be opportunities for growth.

Possible steps:

Re-evaluate Your Study Strategies:
- Identify Weak Areas: Pinpoint the specific topics or concepts that caused difficulties.
- Seek Clarification: Reach out to teachers, tutors, or online resources for additional explanations.
- Practice Regularly: Consistent practice is key to mastering chemistry.
Consider Professional Help:
- Tutoring: A tutor can provide personalized guidance and support.
- Counseling: If you're feeling overwhelmed or unsure about your path, counseling can help.
Explore Alternative Options:
- Retake the Exam: If you're confident in your ability to improve, consider retaking the chemistry compartment exam.
- Change Course: If you're not interested in pursuing chemistry further, explore other academic options that align with your interests.
Focus on NEET 2025 Preparation:
- Stay Dedicated: Continue your NEET preparation with renewed determination.
- Utilize Resources: Make use of study materials, online courses, and mock tests.
Seek Support:
- Talk to Friends and Family: Sharing your feelings can provide comfort and encouragement.
- Join Study Groups: Collaborating with peers can create a supportive learning environment.

Remember: This is a temporary setback. With the right approach and perseverance, you can overcome this challenge and achieve your goals.

I hope this information helps you.

i scored 84 percent in cbse 2 years ago.Am i eligible for medhavi scholarship if give 12th again after 2 drop years and will got 75+ in mpboard

Kanishka kaushikii 17 September,2024

Hi,

Qualifications:
Age: As of the last registration date, you must be between the ages of 16 and 40.
Qualification: You must have graduated from an accredited board or at least passed the tenth grade. Higher qualifications are also accepted, such as a diploma, postgraduate degree, graduation, or 11th or 12th grade.
How to Apply:
Get the Medhavi app by visiting the Google Play Store.
Register: In the app, create an account.
Examine Notification: Examine the comprehensive notification on the scholarship examination.
Sign up to Take the Test: Finish the app's registration process.
Examine: The Medhavi app allows you to take the exam from the comfort of your home.
Get Results: In just two days, the results are made public.
Verification of Documents: Provide the required paperwork and bank account information for validation.
Get Scholarship: Following a successful verification process, the scholarship will be given. You need to have at least passed the 10th grade/matriculation scholarship amount will be transferred directly to your bank account.

Scholarship Details:

Type A: For candidates scoring 60% or above in the exam.

Type B: For candidates scoring between 50% and 60%.

Type C: For candidates scoring between 40% and 50%.

Cash Scholarship:

Scholarships can range from Rs. 2,000 to Rs. 18,000 per month, depending on the marks obtained and the type of scholarship exam (SAKSHAM, SWABHIMAN, SAMADHAN, etc.).

Since you already have a 12th grade qualification with 84%, you meet the qualification criteria and are eligible to apply for the Medhavi Scholarship exam. Make sure to prepare well for the exam to maximize your chances of receiving a higher scholarship.

Hope you find this useful!

i upload 12th board result screenshot in ncweb form from the cbse site so it is valid

Yuvan 30 August,2024

hello mahima,

If you have uploaded screenshot of your 12th board result taken from CBSE official website,there won,t be a problem with that.If the screenshot that you have uploaded is clear and legible. It should display your name, roll number, marks obtained, and any other relevant details in a readable forma.ALSO, the screenshot clearly show it is from the official CBSE results portal.

hope this helps.

Kya mujhe class 12 ka important question milega jo exam mai per year puche jate hai

Ashvadha 12 July,2024

Hello Akash,

If you are looking for important questions of class 12th then I would like to suggest you to go with previous year questions of that particular board. You can go with last 5-10 years of PYQs so and after going through all the questions you will have a clear idea about the type and level of questions that are being asked and it will help you to boost your class 12th board preparation.

You can get the Previous Year Questions (PYQs) on the official website of the respective board.

I hope this answer helps you. If you have more queries then feel free to share your questions with us we will be happy to assist you.

Thank you and wishing you all the best for your bright future.