NEET/JEE Coaching Scholarship
Get up to 90% Scholarship on Offline NEET/JEE coaching from top Institutes
From mobile phones to PCS, LEDS to solar cells — semiconductors are the fundamental building materials of contemporary electronic devices. The knowledge of their operation is necessary for every student who is attempting to excel at physics and has a career linked with technology.
Chapter 14 – Semiconductor Electronics: Materials, Devices and Simple Circuits is a crucial chapter in NCERT Class 12 Physics, especially for those students who are preparing for board-level exams or competitive exams like the JEE and NEET. In order to have a better grasp of the topic, it is highly recommended to study the NCERT Solutions of the chapter under study.
These responses are the correct and proper answers to each question that are located in the NCERT Class 12 Physics book. Authored by subject experts, the content is written in an understandable form that is apt for students. The descriptions are meant to demystify complex concepts, such as semiconductor materials, p-n junction diodes, rectifiers, transistors, and applications. For their optimal achievement, the students must supplement their textbook study with these NCERT solutions for Class 12 and other study materials for better understanding and effective revision.
Also read :
Q. 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:
An N-type semiconductor has electron as majority carriers and holes as minority carriers. It is formed when we dope pentavalent impurity in Silicon atom. Some pentavalent dopants are phosphorus, arsenic, and bismuth.
Hence, the correct option is C.
Q. 14.2 Which of the statements given in Exercise 14.1 is true for p-type semiconductors.
(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:
In a p-type semiconductor, holes are the majority carrier and electrons are the minority carrier. It is formed when a trivalent atom-like aluminium is doped in a silicon atom. Hence correct option for p-type conductor would be (d).
(a)
(b)
(c)
(d)
Answer:
Since carbon is a non-metal, its energy band gap would be highest and energy band gap of Ge would be least as it is a metalloid.
Hence correct option would be (c)
Q14.4 In an unbiased p-n junction, holes diffuse from the p-region to n-region beca
(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:
Charge flows from higher concentration to the lower concentration in a junction. In this case, holes are diffusing from the p-region to n-region and hence the concentration of hole is greater in p region.
and hence correct option would be (c)
Q. 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:
When a p-n junction is forward biased, the negative voltage repels the electron toward junction and give them the energy to cross the junction and combine with the hole which is also being pushed by a positive voltage. This leads to a reduction in the depletion layer which means a reduction in potential barrier across the junction.
Hence correct option would be (c)
Answer:
As we know :
output frequency for half-wave rectifier = input frequency, and hence output frequency in half-wave rectifier will be 50Hz.
also, output frequency for full-wave rectifier = 2*(input frequency) and Hence output frequency in full-wave rectifier will be 2*50 = 100 Hz.
Q.1 A p-n photodiode is fabricated from a semiconductor with bandgap of
Answer:
Given
the energy band gap of photodiode is 2.8eV.
wavelength =
The energy of signal will be
where c is speed of light(300000000m/s) , h is planks constant ( =
putting the corresponding value
The energy of signal =
=
=
The energy of the signal is 0.207eV which is less than 2.8eV ( the energy and gap of photodiode). Hence signal can not be detected by the photodiode.
Answer:
Given:
number of Silicon atoms per
number of Arsenic atoms per
number of Indium atoms per
number of thermally generated electrons
Now,
Number of electrons
number of holes is
in thermal equilibrium
Now, since the number of electrons is higher than number of holes, it is an n-type semiconductor.
Where,
Answer:
Energy gap of given intrinsic semiconductor = E g = 1.2eV
temperature dependence of intrinsic carrier concentration
Where is constant,
T is temperature
Initial temperature = T1 = 300K
the intrinsic carrier concentration at this temperature :
Final temperature = T2 = 600K
the intrinsic carrier concentration at this temperature :
the ratio between the conductivities at 300K and at 600K is equal to the ratio of their intrinsic carrier concentration at these temperatures
Therefore the ratio between the conductivities is
Q.4 In a p-n junction diode, the current I can be expressed as
where
(a) What will be the forward current at a forward voltage of
Answer:
As we have
Here,
When the forward voltage is 0.6V:
Hence forward current is 0.0625A
Q.5 In a p-n junction diode, the current I can be expressed as
where I0 is called the reverse saturation current,
(b) What will be the increase in the current if the voltage across the diode is increased to
Answer:
As we have
Here,
When the forward voltage is 0.7V:
When the forward voltage is 0.6V:
Hence the increase in the forward current is
Q.6 In a p-n junction diode, the current I can be expressed as
where
(c) What is the dynamic resistance?
Answer:
Dynamic Resistance =
Resistance change = 0.7 - 0.6 = 0.1
Current change = 2.967(calculated in prev question)
Therefore
,
Q.7 In a p-n junction diode, the current I can be expressed as
where
(d) What will be the current if reverse bias voltage changes from 1 V to 2 V?
Answer:
As we have
Here,
When reverse voltage is 1V, V= -1
When the reverse voltage is -2V:
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.8(a) You are given the two circuits as shown in Fig. Show that circuit
(a) acts as OR gate while the circuit
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 =
the output of the NOR gate would be the input of NOT Gate and hence
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. 8(b) You are given the two circuits as shown in Fig. Show that circuit
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 =
=
=
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.9 Write the truth table for a NAND gate connected as given in the figure.
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
Hence circuit functions as a NOT gate.
The truth table for the given figure:
Input | Output |
A | Y |
0 | 1 |
1 | 0 |
Answer:
a)
A and B are inputs of a NAND gate and output of this gate is the input of another NAND gate so,
Y =
Y=
Y=
Hence this circuit functions as AND gate.
b)
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 =
Y =
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
(Hint:
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 =
Y =
Y = A + B
Hence Circuit behaves as OR gate.
Truth table
INPUTS | OUTPUT |
00 | 0 |
01 | 1 |
10 | 1 |
11 | 1 |
Answer:
a)
A is the two input of the NOR gate and Hence Output Y is:
Y =
Y =
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
Y =
Y =
Y = A.B
Hence it functions as AND gate.
TRUTH TABLE:
INPUTS | OUTPUT |
00 | 0 |
01 | 0 |
10 | 0 |
11 | 1 |
Q1:
For an amplifier (NPN), VBC = 0,
Answer:
As we learn
Relation between emitter current, Base current, collector current -
- wherein
Q2:
A Si diode has a saturation current of 10-7 A. Calculate the junction current for a forward bias of 0.7 V and 300 k (y = 2 for Si and V = 26 mV)
Answer:
As we learn,
Relation between current I & Voltage V -
- wherein
K = Boltzmann constant
I0 = reverse saturation current
In forward bias
Then, the forward biasing current is
Q3:
Find the current through the circuit for Si diode`
Given that -
Knee voltage for Ge is 0.3 V
Knee voltage for Si is 0.7 V
Answer:
Knee voltage of P-N junction -
It is defined as that forward voltage at which the current through the junction starts rising rapidly with increase in voltage .
Knee voltage for Ge is 0.3 V
Knee voltage for Si is 0.7 V
Q4:
The truth table for the above logic circuit is the same as that of :
Answer:
As we learn
NOR Gate -
NOT + OR Gate
- wherein
A and B are input
Y is output
The output (y) of two input (A, B) NOR gates is :
Q5:
For the circuit shown current through 1.5K
Answer:
As we learn
Zener diode can operate continuously without being damaged in the region of reverse bias
- wherein
1) It acts as a voltage regulator
2) In forward biasing it acts as an ordinary diode.
As per latest 2024 syllabus. Maths formulas, equations, & theorems of class 11 & 12th chapters
Current gain in a transistor:
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.
Two or three questions can be expected from the chapter. Questions from both analogue and digital electronics part can be expected.
Yes, this chapter is the basics for higher studies in electronics-related branches and for VLSI and Nano Technology.
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.
Changing from the CBSE board to the Odisha CHSE in Class 12 is generally difficult and often not ideal due to differences in syllabi and examination structures. Most boards, including Odisha CHSE , do not recommend switching in the final year of schooling. It is crucial to consult both CBSE and Odisha CHSE authorities for specific policies, but making such a change earlier is advisable to prevent academic complications.
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!
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:
Consider Professional Help:
Explore Alternative Options:
Focus on NEET 2025 Preparation:
Seek Support:
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.
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!
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.
Register for ALLEN Scholarship Test & get up to 90% Scholarship
Get up to 90% Scholarship on Offline NEET/JEE coaching from top Institutes
This ebook serves as a valuable study guide for NEET 2025 exam.
This e-book offers NEET PYQ and serves as an indispensable NEET study material.
As per latest 2024 syllabus. Physics formulas, equations, & laws of class 11 & 12th chapters
As per latest 2024 syllabus. Chemistry formulas, equations, & laws of class 11 & 12th chapters