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NCERT Solutions for Class 12 Physics Chapter 14 Semiconductor Electronics Materials Devices And Simple Circuit - This chapter is one of the most important chapter in Class 12 Physics. Students preparing for the upcoming board examination must refer to the NCERT Solutions for Class 12 Physics Chapter 14 Semiconductor Electronics Materials Devices And Simple Circuit. These NCERT Solutions for Class 12 Physics Chapter 14 comprise of the solutions to the questions aksed in the NCERT books for Class 12 Physics.The Semiconductor Class 12 solutions are given in an easy-to-understand language. These NCERT Solutions will be beneficial for academics as well as competitive examinations.
Semiconductor Electronics is one of the important topics that comes under the Unit Modern Physics of Class 12. Semiconductor Class 12 discuss the basics of semiconductors electronics and devices like diode, rectifiers and transistors. Students can use the following materials along with the Semiconductor NCERT solutions. The NCERT Solutions for Class 12 contain solutions provided by subject matter experts, and students can use these solutions to prepare for their board exams or another competitive exam like JEE or NEET.
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Free download chapter 14 physics class 12 ncert solutions pdf for CBSE exam.
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. 14.7 A p-n photodiode is fabricated from a semiconductor with bandgap of Can it detect a wavelength of
Answer:
Given
the energy band gap of photodiode is 2.8eV.
wavelength = = 6000nm =
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, is constant.
Answer:
Energy gap of given intrinsic semiconductor = E g = 1.2eV
temperature dependence of intrinsic carrier concentration is given by
Where is constant, is Boltzmann 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. 14.10 (a) In a p-n junction diode, the current I can be expressed as
where 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 is the current through the diode, is the Boltzmann constant and is the absolute temperature. If for a given diode and then
(a) What will be the forward current at a forward voltage of
Answer:
As we have
Here, , and , = Boltzmann constant =
When the forward voltage is 0.6V:
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, is the voltage across the diode and is positive for forward bias and negative for reverse bias, and is the current through the diode, is the Boltzmann constant and is the absolute temperature. If for a given diode and then
(b) What will be the increase in the current if the voltage across the diode is increased to
Answer:
As we have
Here, , and , = Boltzmann constant =
When the forward voltage is 0.7V:
When the forward voltage is 0.6V:
Hence the increase in the forward current is
Q. 14.10 (c) In a p-n junction diode, the current I can be expressed as
where is called the reverse saturation current, is the voltage across the diode and is positive for forward bias and negative for reverse bias, and is the current through the diode, is the Boltzmann constant and is the absolute temperature. If for a given diode and then
(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.14.10 (d) In a p-n junction diode, the current I can be expressed as
where is called the reverse saturation current, is the voltage across the diode and is positive for forward bias and negative for reverse bias, and is the current through the diode, is the Boltzmann constant and is the absolute temperature. If for a given diode and then
(d) What will be the current if reverse bias voltage changes from 1 V to 2 V?
Answer:
As we have
Here, , and , = Boltzmann constant =
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. 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
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 =
the output of the NOR gate would be the input of NOT Gate and hence
Y = =
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
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. 14.12 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.
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: then and inputs of second NOR gate will be and hence Similarly work out the values of for other combinations of and 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 =
Y = .
Y = A + B
Hence Circuit behaves as OR gate.
Truth table
INPUTS | OUTPUT |
00 | 0 |
01 | 1 |
10 | 1 |
11 | 1 |
Figure 14.40
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 ) is the one input of another NOR gate. B is the two input of NOR gate whose output (which is ) is the input of another NOR gate. Hence,
Y =
Y = .
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.
Intrinsic Semiconductors: ne=ni=nn
Where:
ne represents the free electron density in the conduction band of a semiconductor, ni stands for the intrinsic carrier concentration and nn represents the hole density in the valence band of a semiconductor.
Extrinsic Semiconductors:
N-type: ne≅ Nd>nn
P- type: nn ≅ Nd>ne
Where: Nd is the number density
Mobility
=Vd/E
Where: Vd = the drift velocity and E is the electric field
Action of Transistor
IE=IC+IB
Where:
IE is the emitter current, IC is the collector current and IB is the base current
Some of the important topic of ncert solutions of chapter 14 physics class 12 are listed below:
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
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.
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.
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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.
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