NCERT Solutions for Class 12 Physics Chapter 14 Semiconductor Electronics : Materials, Devices and Simple Circuits

NCERT Solutions for Class 12 Physics Chapter 14 helps you to sort out your weaknesses and strengths in the topic of Semiconductors. NCERT Solutions also helps you to take Class 12 Physics Chapter 14 notes which will come handy when the exam nears.

Class 12 Physics NCERT Solutions for Chapter 14 Semiconductor Electronics

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 are:

  1. Introduction
  2. Classification Of Metals, Conductors, And Semiconductors
  3. Intrinsic Semiconductor
  4. Extrinsic Semiconductor
  5. P-n Junction
    1. p-n Junction formation
  6. Semiconductor Diode
    1. p-n junction diode under forward bias
    2. p-n junction diode under reverse bias
  7. Application Of Junction Diode As A Rectifier
  8. Special Purpose P-n Junction Diodes
    1. Zener diode
    2. Optoelectronic junction devices
  9. Junction Transistor
    1. Transistor: structure and action
    2. Basic transistor circuit configurations and transistor characteristics
    3. Transistor as a device
    4. Transistor as an Amplifier (CE-Configuration)
    5. Feedback amplifier and transistor oscillator
  10. Digital Electronics And Logic Gates
    1. Logic gates
    2. 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 you 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 your questions here.

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 you will be figuring out which of the given circuits comes under these two categories and you will be writing the truth table for the NAND gates used and additionally, you will be seeing questions on the different logical operation. These questions will help you understand everything about the semiconductor devices and the problems related to them. Practising these problems will definitely ensure that you get a good grasp of this subject’s knowledge.

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 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.


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?


Here, (d) is the correct explanation.

For p-type semiconductor, holes are the majority carriers while electrons are the minority carriers. 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(Eg)Sileft (E_{g} ight )_{Si}, (Eg)cleft (E_{g} ight )_{c} and (Eg)Geleft (E_{g} ight )_{Ge}. Which of the following statements is true?.

(a)(Eg)Si<(Eg)Ge<(Eg)Cleft (E_{g} ight )_{Si} < left (E_{g} ight )_{Ge} < left (E_{g} ight )_{C}

(b)(Eg)C<(Eg)Ge<(Eg)Sileft (E_{g} ight )_{C} < left (E_{g} ight )_{Ge} < left (E_{g} ight )_{Si}

(c)(Eg)C<(Eg)Si<(Eg)Geleft (E_{g} ight )_{C} < left (E_{g} ight )_{Si} < left (E_{g} ight )_{Ge}

(d)(Eg)C<(Eg)Si<(Eg)Geleft (E_{g} ight )_{C} < left (E_{g} ight )_{Si} < left (E_{g} ight )_{Ge}


(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: (Eg)C<(Eg)Si<(Eg)Geleft (E_{g} ight )_{C} < left (E_{g} ight )_{Si} < left (E_{g} ight )_{Ge}

Q 4. In an unbiased p – n junction, holes diffuse to n – region from 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 p-region is more as compared to n-region.

(d) All of the above.


(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 p-region to the n-region as the p-region has a greater concentration of holes than in 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.


(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?


For a half-wave rectifier, the output frequency is equal to the input frequency, in this case, the output 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?


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 the 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?

Following values are given in the question:

Number of silicon atoms, N = 5 × 10 28 atom/m3

Number of arsenic atoms, nAS =5×1022atom/m3

Number of indium atoms, nIN=5×1022atom/m3



Let us consider the number of holes to be nh

In the thermal equilibrium, nenh = ni2

Calculating, we get


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 are the ratio between conductivity at 600K and that at 300K? Assume that the temperature dependence of intrinsic carrier concentration ni is given by

ni=n0exp[Eg2kBT]n_{i}=n_{0}expleft [ -frac{E_{g}}{2k_{B}T} ight ]

Here, n0 is constant.


Energy gap in an intrinsic semiconductor, Eg = 1.2 eV

The temperature dependence of the intrinsic carrier-concentration is given as

ni=n0exp[Eg2kBT]n_{i}=n_{0}expleft [ -frac{E_{g}}{2k_{B}T} ight ]

Here, kB is the Boltzman  = 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

ni1=n0exp[Eg2kB×300]n_{i1}=n_{0}expleft [ -frac{E_{g}}{2k_{B} imes 300} ight ] —–(1)

Final temperature, T2 = 600 K

The initrinsic carrier concentration at the temperature 600 K can be written as

ni2=n0exp[Eg2kB×600]n_{i2}=n_{0}expleft [ -frac{E_{g}}{2k_{B} imes 600} ight ] ——-(2)

The ratio between conductivity at 600K and that at 300K is equal to the ratio between the respective carrier concentration at these temperatures.

ni2ni1=n0exp[Eg2kB×600]n0exp[Eg2kB×300]frac{n_{i2}}{n_{i1}}= frac{n_{0}expleft [ -frac{E_{g}}{2k_{B} imes 600} ight ]}{n_{0}expleft [ -frac{E_{g}}{2k_{B} imes 300} ight ]} =expEg2kB[13001600]= exp frac{E_{g}}{2k_{B}}left [ frac{1}{300}-frac{1}{600} ight ] =exp1.22×8.62×105[13001600]= exp frac{1.2}{2 imes 8.62 imes 10^{-5}}left [ frac{1}{300}-frac{1}{600} ight ] =exp1.217.24×105[12]= exp frac{1.2}{17.24 imes 10^{-5}}left [ frac{1}{2} ight ]

= 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

I=I0exp(eV2kBT1)I = I_{0}expleft ( frac{eV}{2k_{B}T}-1 ight )

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, kis 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?


In a p-n junction diode, the expression for current is given as

I=I0exp(eV2kBT1)I = I_{0}expleft ( frac{eV}{2k_{B}T}-1 ight )

Here, I0 = 5 × 10–12 A

T = 300 K

kB = 8.6 x 10-5 eV/k =  8.6 x 10-5 x 1.6 x 10-19 = 1.376 x 10-23 J/K

(a) Forward voltage, V = 0.6 V

I=5×1012exp(1.6×1019×0.62×1.376×1023×3001)I = 5 imes 10^{-12}expleft ( frac{1.6 imes 10^{-19} imes 0.6}{2 imes 1.376 imes 10^{-23} imes 300}-1 ight ) I=5×1012exp(10.6)I = 5 imes 10^{-12}exp (10.6)

= 2006 x 10-10 A

(b) Voltage across the diode is increased to 0.7 V

I=5×1012exp(1.6×1019×0.72×1.376×1023×3001)I’ = 5 imes 10^{-12}expleft ( frac{1.6 imes 10^{-19} imes 0.7}{2 imes 1.376 imes 10^{-23} imes 300}-1 ight ) I=5×1012exp(12.56)I’ = 5 imes 10^{-12}exp (12.56)

I’ = 14247 x 10-10 A

Change in current, ΔI = I’ – I

= 14247 x 10-10 – 2006 x 10-10 

= 12241 x 10-10 A

Change is voltage = 0.7 – 0.6 = 0.1 V

(c) Dynamic resistance = Change in voltage/ Change in current

= 0.1/(12241 x 10-10)

= (0.1 x 105)/0.1224

= 81699 Ω

(d) 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 the circuit (b) acts as AND gate.



(a) The Boolean expression for NAND gate is Y=A.BˉY = ar{A.B}

Let Y’ be the output of the NAND gate.

Therefore, Final output of the combination,  Y=Y.YˉY = ar{Y’.Y’}


Y=A.BA.Bmathbf{Y}=overline{overline{mathbf{A . B}} cdot overline{mathbf{A . B}}}


Y=A.B+A.BY = A.B+A.B

Y = A.B

The given circuit acts as a AND gate.

(b) Boolean expression for NOT gate is Y=AˉY = ar{A}

The output Y1 of the first NOT gate is Y1=AˉY_{1} = ar{A}


The output Y2 of the second NOT gate is Y2=BˉY_{2} = ar{B}

The output of the combination


Y=Y1.Y2ˉY = ar{Y_{1}.Y_{2}}


Y=Aˉ.BˉˉY = ar{ar{A}.ar{B}}


Y=Aˉˉ+BˉˉY = ar{ar{A}}+ar{ar{B}}

Y = A + B

This is the Boolean expression for OR gate.


Q 14. Write the truth table for 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 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.)



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 A+Aoverline{mathrm{A}+mathrm{A}}

So, the output Y=AY= overline{mathrm{A}}

Therefore, the gate is a NOT gate.

The truth table is given as

A Y=AY= overline{mathrm{A}}

(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 NOT gates are Aoverline{mathrm{A}} and Boverline{mathrm{B}}. This is the input of the NOR gate. Therefore, the output of the circuit is Y=A+B=A.Bmathbf{Y}=overline{overline{mathbf{A}}+overline{mathbf{B}}}=mathbf{A} . mathbf{B}. So, this acts as a AND gate. The truth table is given below



The NCERT Class 12 Physics Solutions for Chapter 14 are created by subject experts according to the latest CBSE syllabus 2020-21. Students must practice the solutions regularly to prepare effectively for their examination.
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Explain intrinsic and extrinsic semiconductor 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 as intrinsic semiconductors.