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

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

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

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Have you ever questioned yourself how your smartphone, laptop, LED lights or even solar panels operate? It is in semiconductors, upon which all modern electronics devices are based. That is why Chapter 12 Physics- Semiconductor Electronics: Materials, Devices and Simple Circuits is so much important in CBSE board exams and competitive entry tests such as JEE and NEET.

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Semiconductor Electronics Materials Devices And Simple Circuit NCERT Solutions: Download PDF
Semiconductor Electronics Materials Devices And Simple Circuit NCERT Solutions: Exercise Questions
Semiconductor Electronics Materials Devices And Simple Circuit NCERT Solutions: Additional Questions
Class 12 Physics Chapter 14 - Semiconductor Electronics Materials Devices And Simple Circuit: Higher Order Thinking Skills (HOTS) Questions
Class 12 Physics Chapter 14 - Semiconductor Electronics Materials Devices And Simple Circuit: Topics
Class 12 Physics Chapter 14 - Semiconductor Electronics Materials Devices And Simple Circuit: Notes
Approach to Solve Questions of Semiconductor Electronics: Materials, Devices and Simple Circuits
What Extra Should Students Study Beyond NCERT for JEE/NEET?
NCERT solutions for class 12 physics chapter-wise

NCERT Class 12 Physics Chapter 14

The NCERT Solutions for Class 12 Physics Chapter 14 - Semiconductor Electronics Materials Devices And Simple Circuit have been prepared by experts in the subject and would give step by step answers to all the questions in the exercises hence making the students to have an easy time grasping the more complicated topics such as semiconductor materials, p-n junction diodes, rectifiers, transistors and logic gates. These NCERT solutions are not just an answer to the textbook exercises, but also a boost to the analytical and problem-solving skills necessary to be able to succeed in exams. These NCERT Solutions for Class 12 Physics Chapter 14 - Semiconductor Electronics Materials Devices And Simple Circuit will allow students to develop a solid conceptual foundation, revising well on solved examples and gain the confidence required to not only score high on CBSE exams but also to score good in JEE and NEET.

Also read :

Semiconductor Electronics Materials Devices And Simple Circuit NCERT Solutions: Download PDF

The Class 12 Physics Chapter 14 - Semiconductor Electronics Materials Devices And Simple Circuit question answers offer detailed answers to each of the textbook questions so that complicated concepts become easy to understand. These are the solutions formatted according to the current CBSE syllabus and are very useful during revisions, examinations at the board level and even competitive applications like JEE/NEET examinations. They can also print the PDF to study offline and get last-minute reading practice.

Download PDF

Semiconductor Electronics Materials Devices And Simple Circuit NCERT Solutions: Exercise Questions

The Class 12 Physics Chapter 14 - Semiconductor Electronics Materials Devices And Simple Circuit question answers (Exercise questions) offer step-by-step solutions to all questions in the textbook, and that is why it becomes easy to learn and apply the semiconductor concepts among students. Such solutions are accurate, clear and they are used to prepare board exams as well as competitive exams such as JEE/NEET.

Q. 14.1 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.

(d) Holes are majority carriers and trivalent atoms are the dopants.

Answer:

An n-type semiconductor has electrons as majority carriers and holes as minority carriers. It is formed when we dope a pentavalent impurity into a 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 the majority carriers and trivalent atoms are the dopants.

(b) Electrons 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).

Q. 14.3 Carbon, silicon and germanium have four valence electrons each. These are characterised by valence and conduction bands separated by energy bandgap respectively equal to $(E_{g})_{C}, (E_{g})_{S i}$ and $(E_{g})_{G e}$ . Which of the following statements is true?

(a) $(E_{g})_{S i} < (E_{g})_{G e} < (E_{g})_{C}$

(b) $(E_{g})_{C} < (E_{g})_{G e} > (E_{g})_{S i}$

(d) $(E_{g})_{C} = (E_{g})_{S i} = (E_{g})_{G e}$

Answer:

Since carbon is a non-metal, its energy band gap would be the highest and the energy band gap of Ge would be the least, as it is a metalloid.

$(E_{g})_{C} > (E_{g})_{S i} > (E_{g})_{G e}$

Hence correct option would be (c)

Q14.4 In an unbiased p-n junction, holes diffuse from the p-region to n-region because

(a) free electrons in the n-region attract them,

(b) they move across the junction by the potential difference.

(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 the p-region.

and hence the 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.

(d) none of the above.

Answer:

When a p-n junction is forward biased, the negative voltage repels the electrons toward the junction and gives them the energy to cross the junction and combine with the holes which are also being pushed by a positive voltage. This leads to a reduction in the depletion layer, which means a reduction in the potential barrier across the junction.

Hence correct option would be (c)

Q. 14.6 In half-wave rectification, what is the output frequency if the input frequency is $50 H z .$ What is the output frequency of a full-wave rectifier for the same input frequency

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.

Semiconductor Electronics Materials Devices And Simple Circuit NCERT Solutions: Additional Questions

Class 12 Physics Chapter 14 - Semiconductor Electronics Materials Devices And Simple Circuit additional questions give students extra practice on concepts like p-n junctions, diodes, transistors, and logic gates. These questions go beyond the NCERT exercise, helping in deeper understanding and exam-focused preparation for boards as well as JEE/NEET.

Q.1 A p-n photodiode is fabricated from a semiconductor with bandgap of $2.8 e V .$ Can it detect a wavelength of $6000 n m ?$

Answer:

Given

The energy band gap of the photodiode is 2.8eV.

wavelength = $λ$ = 6000nm = $6000 * 10^{- 9}$

The energy of signal will be $\frac{h c}{λ}$

where c is speed of light(300000000m/s) , h is planks constant ( = $6.626 * 10^{- 34} J s$ )

putting the corresponding value

The energy of signal = $\frac{(6.626 * 10^{- 34} * 3 * 10^{8})}{6000 * 10^{- 9}}$

= $3.313 * 10^{- 20} J$

= $0.207 e V (s i n c e 1.6 * 10^{- 20} = 1 e V)$

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.

Q.2 The number of silicon atoms per m 3 is $5 \times 10^{28} .$ This is doped simultaneously with $5 \times 10^{22} .$ atoms per $m^{3}$ of Arsenic and $5 \times 10^{20}$ per $m^{3}$ atoms of Indium. Calculate the number of electrons and holes. Given that $n_{i} = 1.5 \times 10^{16} m^{- 3} .$ Is the material $n - t y p e$ or $p - t y p e$ ?

Answer:

Given:

number of Silicon atoms per $m^{3}$ = $5 \times 10^{28}$

number of Arsenic atoms per $m^{3}$ = $5 \times 10^{22}$

number of Indium atoms per $m^{3}$ = $5 \times 10^{20}$

number of thermally generated electrons $n_{i} = 1.5 \times 10^{16} m^{- 3}$

Now,

Number of electrons

$n_{e} =$ $5 * 10^{22} - 1.5 * 10^{16}$ = $4.99 * 10^{22} (a p p r o x)$

number of holes is $n_{h}$

in thermal equilibrium

$n_{h} * n_{e} = n_{i}^{2}$

$n_{h} = n_{i}^{2} / n_{e}$

$n_{h} = (1.5 * 10^{16})^{2} / 4.99 * 10^{22}$

$n_{h} = 4.51 * 10^{9}$

Now, since the number of electrons is higher than number of holes, it is an n-type semiconductor.

Q.3 In an intrinsic semiconductor the energy gap $E_{g}$ is $1.2 e V .$ Its hole mobility is much smaller than electron mobility and independent of temperature. What is the ratio between conductivity at $600 K$ and that at $300 K$ Assume that the temperature dependence of intrinsic carrier concentration $n_{i}$ is given by

$n_{i} = n_{0} e x p [- \frac{E_{g}}{2 K_{B} T}]$

Where, $n_{0}$ is constant.

Answer:

Energy gap of given intrinsic semiconductor = E g = 1.2eV

temperature dependence of intrinsic carrier concentration $n_{i}$ is given by

$n_{i} = n_{0} e x p [- \frac{E_{g}}{2 K_{B} T}]$

Where is constant, $K_{B}$ is Boltzmann constant = $8.862 * 10^{- 5} e V / K$ ,

T is temperature

Initial temperature = T1 = 300K

The intrinsic carrier concentration at this temperature :

$n_{i 1} = n_{0} e x p [\frac{- E_{g}}{2 K_{B} * 300}]$

Final temperature = T2 = 600K

The intrinsic carrier concentration at this temperature :

$n_{i 2} = n_{0} e x p [\frac{- E_{g}}{2 K_{B} * 600}]$

The ratio between the conductivities at 300K and at 600K is equal to the ratio of their intrinsic carrier concentration at these temperatures

$\frac{n_{i 2}}{n_{i 2}} = \frac{n_{0} e x p [\frac{- E_{g}}{2 K_{B} * 600}]}{n_{0} e x p [\frac{- E_{g}}{2 K_{B} * 300}]}$

$= e x p \frac{E_{g}}{2 K_{B}} [\frac{1}{300} - \frac{1}{600}] = e x p [\frac{1.2}{2 * 8.62 * 10^{- 5}} * \frac{2 - 1}{600}]$

$= e x p [11.6] = 1.09 * 10^{5}$

Therefore, the ratio between the conductivities is $1.09 * 10^{5}$ .

Q.4 In a p-n junction diode, the current I can be expressed as

$I = I_{0} [e x p \frac{e V}{K_{B} T} - 1]$

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

What will be the forward current at a forward voltage of $0.6 V ?$

Answer:

As we have

$I = I_{0} [e x p \frac{e V}{K_{B} T} - 1]$

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

When the forward voltage is 0.6V:

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

Hence forward current is 0.0625A

Q.5 In a p-n junction diode, the current I can be expressed as

$I = I_{0} [e x p \frac{e V}{K_{B} T} - 1]$

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

What will be the increase in the current if the voltage across the diode is increased to $0.7 V ?$

Answer:

As we have

$I = I_{0} [e x p \frac{e V}{K_{B} T} - 1]$

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

When the forward voltage is 0.7V:

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

When the forward voltage is 0.6V:

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

Hence the increase in the forward current is

$I (w h e n v = 0.7) - I (w h e n v = .6)$ $= 3.029 - 0.0625 = 2.967 A$

Q.6 In a p-n junction diode, the current I can be expressed as

$I = I_{0} [e x p \frac{e V}{K_{B} T} - 1]$

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

What is the dynamic resistance?

Answer:

Dynamic Resistance = $\frac{v o l t a g e - c h a n g e}{c u r r e n t - c h a n g e}$

Resistance change = 0.7 - 0.6 = 0.1

Current change = 2.967(calculated in prev question)

Therefore

, $D y n a m i c R e s i s t a n c e = \frac{0.1}{2.967} = 0.0337 Ω$

Q.7 In a p-n junction diode, the current I can be expressed as

$I = I_{0} [e x p \frac{e V}{K_{B} T} - 1]$

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

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

Answer:

As we have

$I = I_{0} [e x p \frac{e V}{K_{B} T} - 1]$

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

When reverse voltage is 1V, V= -1

$I = 5 * 10^{- 12} [e x p \frac{1.6 * 10^{- 19} * (- 1)}{1.376 * 10^{- 23} * 300} - 1] \approx 5 \times 10^{- 12}$

When the reverse voltage is -2V:

$I = 5 * 10^{- 12} [e x p \frac{1.6 * 10^{- 19} * (- 2)}{1.376 * 10^{- 23} * 300} - 1] \approx 5 \times 10^{- 12}$

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.8(a) You are given the two circuits as shown in Fig. Show that circuit acts as OR gate while the circuit

1645428242590

Answer:

Here, THE Input = A and B

Output = Y

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

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

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

$Y = \overset{―}{\overset{―}{A + B}} = A + B$

Hence the figure functions like an OR Gate.

Or compare the truth table by giving different inputs 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 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 = $(\overset{―}{A} + \overset{―}{B})$

= $\overset{―}{\overset{―}{A}} . \overset{―}{\overset{―}{B}}$ $\overset{―}{A + B} = \overset{―}{A} . \overset{―}{B}$

= $A * B$

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.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 boththe input of the NAND gate and hence Output Y will be

$Y = \overset{―}{A * A}$

$Y = \overset{―}{A} + \overset{―}{A}$

$Y = \overset{―}{A}$

Hence circuit functions as a NOT gate.

The truth table for the given figure:

Input	Output
A	Y
0	1
1	0

Q 10 You are given two circuits as shown in Fig, which consist of NAND gates. Identify the logic operation carried out by the two circuits.

Answer:

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

Y = $(\overset{―}{A . B}) (\overset{―}{A . B})$

Y= $(\overset{―}{A . B})$ $+$ $(\overset{―}{A . B})$

Y= $A B$

Hence this circuit functions as AND gate.

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

Y = $\overset{―}{A} . \overset{―}{B}$

Y = $\overset{―}{A} .$ + $\overset{―}{B} .$

Y = A + B

Hence, the circuit functions as an OR gate.

Alternative method

fig. a

Construct the truth table by giving various inputs and observing 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.11 Write the truth table for the circuit given in Fig below consisting of NOR gates and identify the logic operation (OR, AND, NOT) that this circuit is performing.

(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 = $(\overset{―}{A + B} + \overset{―}{A + B})$

Y = $\overset{―}{A + B}$ . $\overset{―}{A + B}$

Y = A + B

Hence Circuit behaves as OR gate.

Truth table

INPUTS	OUTPUT
00	0
01	1
10	1
11	1

Q.12 Write the truth table for the circuits given in Fig., consisting of NOR gates only. Identify the logic operations (OR, AND, NOT) performed by the two circuits.

1645428438468

Answer:

A is the two inputs of the NOR gate and hence, Output Y is:

Y = $\overset{―}{A + A}$

Y = $\overset{―}{A}$

Hence circuit functions as a NOT gate.

TRUTH TABLE:

INPUT	OUTPUT
0	1
1	0

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

Y = $\overset{―}{A} + \overset{―}{B}$

Y = $\overset{―}{A}$ . $\overset{―}{B}$

Y = A.B

Hence it functions as AND gate.

TRUTH TABLE:

INPUTS	OUTPUT
00	0
01	0
10	0
11	1

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Class 12 Physics Chapter 14 - Semiconductor Electronics Materials Devices And Simple Circuit: Higher Order Thinking Skills (HOTS) Questions

Class 12 Physics NCERT Chapter 14 HOTS (Higher Order Thinking Skills) Questions challenge students to apply semiconductor concepts like diode characteristics, transistor action, rectifiers, and logic circuits in tricky problem-solving. These questions enhance critical thinking and analytical skills, making students better prepared for competitive exams and advanced applications.

Q1:

For an amplifier (NPN), V_BC = 0, $β = 50$ , I_C = 2.475 mA. Then the value of IB =?

Answer:

As we learn
Relation between emitter current, Base current, and collector current -
$I_{E} = I_{B} + I_{C}$
- wherein
$I_{E} =$ Emitter Current
$I_{B} =$ Base Current
$I_{C} =$ Collector Current
$I_{B} = \frac{I_{C}}{β} = \frac{2.475}{50} = 49.5 mA$

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 -

$I = I_{0} (e^{\frac{c v}{K T}} - 1)$

- wherein

K = Boltzmann constant

I₀ = reverse saturation current

In forward bias

$e^{\frac{r v}{K T}} >> 1$

Then, the forward biasing current is

$I = I_{0} \cdot e^{\frac{c v}{K T}}$

$I = I_{0} (e^{\frac{cv}{K T}} - 1) = I_{0} (e^{\frac{v}{y + V_{T}}} - 1) = 10^{- 7} (\frac{0.7}{2 \times 26} - 1) = 70 mA$

Q3:

Find the current through the circuit for the 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 the forward voltage at which the current through the junction starts rising rapidly with an increase in voltage.

Knee voltage for Ge is 0.3 V

Knee voltage for Si is 0.7 V

$I = \frac{3.4 - 0.7}{600} = \frac{2.7}{600} = 4.5 mA$

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

$Y = \overset{―}{A + B}$

A and B are input

Y is output

The output (y) of two input (A, B) NOR gates is :

$Y = \overset{―}{A + B}$

Q5:

For the circuit shown current through 1.5K $Ω$ is:

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.

$I_{L} = \frac{V_{Z}}{1.5 K} = \frac{6}{1.5 \times 10^{3}} = 4 mA$

CBSE Class 12th Syllabus: Subjects & Chapters

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Class 12 Physics Chapter 14 - Semiconductor Electronics Materials Devices And Simple Circuit: Topics

Class 12 Physics Chapter 14 - Semiconductor Electronics Materials Devices And Simple Circuit introduces the working principles of semiconductors, diodes, transistors, rectifiers, and logic gates that form the backbone of modern electronics. Understanding these topics helps students connect theory with real-world applications in devices like mobiles, LEDs, and computers.

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Class 12 Physics Chapter 14 - Semiconductor Electronics Materials Devices And Simple Circuit: Notes

Class 12 Physics Chapter 14 - Semiconductor Electronics Materials Devices And Simple Circuit Notes offer a neatly organized introduction to semiconductor electronics with all essential definitions, formulas and concepts being simplified. Such notes enable the students to revise and improve their basic knowledge fast to succeed in the CBSE board exams, JEE, and NEET.

Metals, Conductors & Semiconductors

Why do some materials light up your circuits while others block the buzz? Metals conduct electricity well. Conductors let current pass easily. Semiconductors—like silicon—are in-between, and insulators stop the flow completely.

Intrinsic Semiconductor

Pure semiconductors like silicon have equal electrons and holes. They don’t conduct much—until heat gets them moving!

Extrinsic Semiconductor

Add a pinch of impurity, and things change!
n-type: Extra electrons (via elements like phosphorus).
p-type: Extra holes (via elements like boron).
Doping boosts conductivity.

P-N Junction

A magical meeting of p-type and n-type. It only lets current flow one way—thanks to the mysterious depletion region!

Semiconductor Diode

A p-n junction in action. It’s a one-way gate for current—perfect for letting power flow forward and blocking the backflow.

Junction Diode as Rectifier

Want to turn AC into DC? Diodes do the trick!
Half-wave: Uses one side of AC.
Full-wave: Uses both sides.
It’s how your gadgets get steady power.

Approach to Solve Questions of Semiconductor Electronics: Materials, Devices and Simple Circuits

To answer the questions on Semiconductor Electronics, it is essential to have a good understanding of such concepts as p-n junctions, diodes, transistor and logic gates. Using a well-defined step-by-step approach, students will be able to work both theoretically and numerically with a sense of confidence. This approach is useful in achieving good scores in board exams and also JEE/NEET where precision and usage of concepts is important.

Master the basic fundamentals of semiconductors.

Know what conductors, insulators, and semiconductors are.
There are two categories of semiconductors: intrinsic (pure) and extrinsic (doped).

Know the types of doping

n-type: Introduction of pentavalent impurity → Additional electrons
p-type: Introduction of trivalent impurity → Increased holes

Learn important parts:

Diode: Recognise forward and reverse bias.
Zener diode: Used for voltage control
Transistors (n-p-n, p-n-p): Current amplification and switching

Learn circuit behaviour

Explain how voltage and current act in forward/reverse-biased diodes
Study rectifier circuits: Half-wave and full-wave.

Apply proper formulas-

Current gain in a transistor: $β = \frac{I_{C}}{I_{B}}$
$I_{E} = I_{B} + I_{C} (Emitter current = Base + Collector)$

Study I-V characteristics- Draw and label graphs for diodes and transistors.
Learn logic gates (brief introduction)- AND, OR, NOT (if included) – Know the truth tables and how they apply.
Practice circuit numericals- Recognise biasing, compute currents, and perceive the role of the power supply
Know applications-

Diodes: Rectification, switching
Transistors: Logic circuits, amplifiers

Study the theory and practice NCERT questions- Most questions are about numbers or ideas, so put both.

What Extra Should Students Study Beyond NCERT for JEE/NEET?

Although the basics of semiconductors, diodes, and transistors are taught in NCERT, students who are going to take on the JEE and NEET exams, must take it a step further. Competitive exams require a more in-depth problem-solving, application-based numerals, and exposure to the conceptual questions that are tricky. Comparison of NCERT vs. JEE/NEET requirements can be used to determine the additional topics and practice to achieve success in the exams.

Concept Name	JEE	NCERT
Band Theory Of Solids	✅	✅
Types Of Semiconductor: Intrinsic And Extrinsic Semiconductor	✅	✅
Electric Conductivity	✅
P-N Junction	✅	✅
Semiconductor Diode - Forward Bias And Reverse Bias	✅	✅
P-N Junction As A Rectifier	✅	✅
Zener Diode	✅	❌
Special Purpose P-N Junction Diodes	✅	❌
Bipolar Junction Transistor (N-P-N And P-N-P Transistor)	✅	❌
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NCERT solutions for class 12 physics chapter-wise

NCERT Solutions of Class 12 Physics contain step-by-step explanations on each chapter that makes the complicated theories simple. These solutions have been prepared according to the new CBSE syllabus and are quite beneficial in preparation of board exams; JEE and NEET. The following links lead to chapter-wise solutions that the students can download and revise to access knowledge.

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 Chapter 8 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

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 Circuit

Also Check NCERT Books and NCERT Syllabus here:

NCERT solutions subject wise

Frequently Asked Questions (FAQs)

Q: Why are NCERT Solutions Class 12 Physics necessary to students?

They give step-by-step explanations to make the not-so-easy concepts easier and to solve questions in exercises in a straightforward manner.

Q: Are NCERT Solutions sufficient in achieving good marks in Class 12 board exams?

Yes, NCERT Solutions is enough to pass boards yet however having practice and doing sample papers will increase your mark.

Q: Are these NCERT Solutions useful in JEE and NEET preparation?

Absolutely! Most of the questions in JEE and NEET are directly and indirectly linked to the NCERT concepts; hence, these solutions assist in building a great foundation.

Q: Is it possible to download NCERT Solutions of Class 12 Physics chapter-wise in PDF?

Yes, all NCERT Solutions in each Chapter can be downloaded as PDFs to study and revise.

Q: What is the best way to use NCERT Solutions Class 12 Physics?

You should read the NCERT textbook, go over the NCERT Solutions to clear any confusion, then do previous year papers and mock tests.

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Questions related to CBSE Class 12th

On Question asked by student community

Have a question related to CBSE Class 12th ?

12th Exam 2025Kitin tareekh ko hoga

Hello,

The date of 12 exam is depends on which board you belongs to . You should check the exact date of your exam by visiting the official website of your respective board.

Hope this information is useful to you.

Read Complete Answer

Nidhi Raghuwanshi

21 Sep'25

12th biology question paper 2023-2025

Hello,

Class 12 biology questions papers 2023-2025 are available on cbseacademic.nic.in , and other educational website. You can download PDFs of questions papers with solution for practice. For state boards, visit the official board site or trusted education portal.

Hope this information is useful to you.

Read Complete Answer

Nidhi Raghuwanshi

21 Sep'25

In March 2025, I passed the CBSE 12th board and gave KCET. I am taking a drop year. I will reappear in the CBSE board exam in the upcoming 2026 exam in February as a private candidate to increase my PCM aggregate, as KCET gives 50% weightage to PCM aggregate, and I will also give the KCET exam. So, to calculate rank, do they use the board marksheet from 2025 or 2026? Also for each subject will best out of two marks accross both the marksheets be considered? #kcet

Hello Pruthvi,

Taking a drop year to reappear for the Karnataka Common Entrance Test (KCET) is a well-defined process. As a repeater, you are fully eligible to take the exam again to improve your score and secure a better rank for admissions.

The main procedure involves submitting a new application for the KCET through the official Karnataka Examinations Authority (KEA) website when registrations open for the next academic session. You must pay the required application fee and complete all formalities just like any other candidate. A significant advantage for you is that you do not need to retake your 12th board exams. Your previously secured board marks in the qualifying subjects will be used again. Your new KCET rank will be calculated by combining these existing board marks with your new score from the KCET exam. Therefore, your entire focus during this year should be on preparing thoroughly for the KCET to achieve a higher score.

For more details about the KCET Exam preparation, CLICK HERE.

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

Thank you, and I wish you all the best in your bright future.

Read Complete Answer

Ark Pandey

20 Sep'25

I'm studying scince in karnatakastate board, can i take commerce in cbse for 12th next year?

Yes, you can switch from Science in Karnataka State Board to Commerce in CBSE for 12th. You will need a Transfer Certificate from your current school and meet the CBSE school’s admission requirements. Since you haven’t studied Commerce subjects like Accountancy, Economics, and Business Studies, you may need to catch up before or during 12th. Not all CBSE schools accept direct admission to 12th from another board, so some may ask you to join Class 11 first. Make sure to check the school’s rules and plan your subject preparation.

Read Complete Answer

Abhilasha Agrawal

18 Sep'25

12th class cbse boards most important que ans ans in hindi midum subject hindi impotent question jo exam main aaynge wo btai

Hello

For the 12th CBSE Hindi Medium board exam, important questions usually come from core chapters like “Madhushala”, “Jhansi ki Rani”, and “Bharat ki Khoj”.
Questions often include essay writing, letter writing, and comprehension passages. Grammar topics like Tenses, Voice Change, and Direct-Indirect Speech are frequently asked.
Students should practice poetry questions on themes and meanings. Important questions also cover summary writing and translation from Hindi to English or vice versa.
Previous years’ question papers help identify commonly asked questions.
Focus on writing practice to improve handwriting and presentation. Time management during exams is key to answering all questions effectively.

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Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

Option 1) $2,000 \; J - 5,000\; J$	Option 2) $200 \, \, J - 500 \, \, J$
Option 3) $2\times 10^{5}J-3\times 10^{5}J$	Option 4) $20,000 \, \, J - 50,000 \, \, J$

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

Option 1) $11.2\, L\, H_{2(g)}$ at STP is produced for every mole $HCL_{(aq)}$ consumed	Option 2) $6L\, HCl_{(aq)}$ is consumed for ever $3L\, H_{2(g)}$ produced
Option 3) $33.6 L\, H_{2(g)}$ is produced regardless of temperature and pressure for every mole Al that reacts	Option 4) $67.2\, L\, H_{2(g)}$ at STP is produced for every mole Al that reacts .

Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 2.5 × 10^-2

Option 1) decrease twice	Option 2) increase two fold
Option 3) remain unchanged	Option 4) be a function of the molecular mass of the substance.

Option 1) Molality	Option 2) Weight fraction of solute
Option 3) Fraction of solute present in water	Option 4) Mole fraction.

Option 1) twice that in 60 g carbon	Option 2) 6.023 × 10²²
Option 3) half that in 8 g He	Option 4) 558.5 × 6.023 × 10²³

Option 1) less than 3	Option 2) more than 3 but less than 6
Option 3) more than 6 but less than 9	Option 4) more than 9

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

Semiconductor Electronics Materials Devices And Simple Circuit NCERT Solutions: Download PDF

Semiconductor Electronics Materials Devices And Simple Circuit NCERT Solutions: Exercise Questions

Semiconductor Electronics Materials Devices And Simple Circuit NCERT Solutions: Additional Questions

Related eBooks & Sample Papers

Class 12 Physics Chapter 14 - Semiconductor Electronics Materials Devices And Simple Circuit: Higher Order Thinking Skills (HOTS) Questions

Class 12 Physics Chapter 14 - Semiconductor Electronics Materials Devices And Simple Circuit: Topics

Class 12 Physics Chapter 14 - Semiconductor Electronics Materials Devices And Simple Circuit: Notes

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What Extra Should Students Study Beyond NCERT for JEE/NEET?

NCERT solutions for class 12 physics chapter-wise

Also Check NCERT Books and NCERT Syllabus here:

NCERT solutions subject wise

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