NCERT Solutions for Class 12 Physics Chapter 14 Semiconductor Electronics Materials

NCERT Solutions for Class 12 Physics Chapter 14 Semiconductor Electronics: Materials, Devices and Simple Circuits is an important resource material if you are planning to score good marks in CBSE class 12 physics examination. This solution has answers to the textbook questions, exemplary questions, the important question from previous year question papers, sample papers together with exercises, worksheets and MCQ’S.

NCERT Solutions for Class 12 Physics Chapter 14 helps you to sort out your weaknesses and strengths in the topic Semiconductors. NCERT solution also helps you to take class 12 physics chapter 14 notes which will come handy when 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 careers 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 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 about 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 this 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 knowledge.

Class 12 Physics NCERT Solutions Semiconductor Devices Important Questions


Q 1. Find which of the given statement is true for n-type silicon.

(a) Majority carriers are the electrons and dopants are the trivalent.

(b) Minority carriers are the electrons and dopants are pentavalent.

(c) Minority carriers are the holes and dopants are pentavalent.

(d) Majority carriers are the holes and dopants are the trivalent.

Ans:

Here, (c) is the correct option.

For n-type silicon, the majority carriers are electrons while the minority carriers are holes. n-type semiconductor is obtained by dropping pentavalent atoms like phosphorus in silicon atoms.

 Q2. Find the correct statement in the previous question for a p – type semiconductors.

Ans:

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 is obtained by using trivalent atoms like aluminum in silicon atoms.

Q 3. We have Silicon, carbon and germanium, each of them are having four valence electrons. Here, there are categorized by their conduction and valence band which are separated by energy gap respectively equal to (Eg)Si\left (E_{g} \right )_{Si}, (Eg)c\left (E_{g} \right )_{c} and (Eg)Ge\left (E_{g} \right )_{Ge}. Find out the correct statement.

(a)(Eg)Si<(Eg)Ge<(Eg)C\left (E_{g} \right )_{Si} < \left (E_{g} \right )_{Ge} < \left (E_{g} \right )_{C}

(b)(Eg)C<(Eg)Ge<(Eg)Si\left (E_{g} \right )_{C} < \left (E_{g} \right )_{Ge} < \left (E_{g} \right )_{Si}

(c)(Eg)C<(Eg)Si<(Eg)Ge\left (E_{g} \right )_{C} < \left (E_{g} \right )_{Si} < \left (E_{g} \right )_{Ge}

(d)(Eg)C<(Eg)Si<(Eg)Ge\left (E_{g} \right )_{C} < \left (E_{g} \right )_{Si} < \left (E_{g} \right )_{Ge}

Ans:

(C) is the correct option.

Out of carbon, germanium, and silicon, carbon has the maximum energy band gap whereas germanium has the least energy  band gap.

For all these elements the energy band gap can be related as: (Eg)C<(Eg)Si<(Eg)Ge\left (E_{g} \right )_{C} < \left (E_{g} \right )_{Si} < \left (E_{g} \right )_{Ge}

Q 4. In an unbiased p – n junction, holes diffuses to n – region from p – region because

(a) The electron in the n – region and which of them are free attracts them

(b) They are moved to the junction by the potential difference.

(c) The concentration of the holes at p – region is more as compared to that of the n – region

(d) All of the above.

 Ans:

(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 greater concentration of holes than in n-region.

Q 5. When  a p – n junction is forward biased, its

(a) potential power is increased.

(b) majority carrier current is reduced to zero.

(c) potential barrier is reduced.

(d) None of the above.

 Ans:

(c) is the correct option

The potential barrier reduces for a p-n junction when 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. Find out the correct statements from the following for the case of a transistor action

(a)Emitter, collector and base regions should have same size and doping concentration

(b) The base region must be very thin and lightly doped.

(c) The collector junction is reverse biased and the emitter unction is forward biased.

(d) Both of the emitter and the collector junction are forward biased.

Ans:

(b) and (c) are the correct options.

In case of a transistor action, the base region is thin and lightly doped. The collector junction and the emitter junction are reverse biased and forward biased respectively.

Q 7. In the case of a transistor amplifier, the voltage gain

(a) Does not change for any frequencies.

(b) Will remain constant in the middle frequency and it will be high at high and low frequency.

(c) Will remain constant at middle frequency and it will be low at high and low frequencies.

(d) None of the above.

 Ans:

(c) is the correct option

During the middle frequency, the voltage gain of a transistor amplifier remains constant and during high and low frequencies, the voltage gain will be low.

Q 8. Calculate the output frequency of a half – wave rectifier, if we will provide an input with frequency 50 Hz. Also calculate the output frequency for a full wave rectifier if the input is given the same.

Ans: 50Hz is the input frequency.

We know that, for a half-wave rectifier the output frequency is equal to the input frequency.

Therefore, output frequency is 50Hz.

Whereas in full-wave rectifier, the output frequency is double the input frequency.

Therefore, output frequency = 2 × 50 = 100 Hz

Q 9 . Find the input signal voltage and base current for a CE – transistor amplifier, the voltage of audio signal across the collector resistance of 2kΩ2k\Omega is 2 V. Assume that the transistor have current amplification factor equals to 100  and the given base resistance is 1kΩ1k\Omega

Ans:

Collector resistance, RC=2kΩ=2000ΩR _{C} = 2k\Omega = 2000 \Omega

Voltage of audio signal across the collector resistance, V = 2V

Current amplification factor of the transistor, β=100\beta = 100

Base resistance, RB=1kΩ=1000ΩR _{B} = 1k\Omega = 1000 \Omega

Input Signal = ViV _{i}

Base current = IBI _{B}

We have amplification relation as:

VVi=βRCRB\frac{V}{V _{i}} = \beta \frac{R _{C}}{R _{B}}

Voltage amplification

Vi=VβRBRCV _{i} = \frac{ V }{ \beta } \frac{R _{B}}{R _{C}} =2×100×10002000=0.01V= \frac{ 2 \times }{ 100 \times } \frac{ 1000 }{ 2000 } = 0.01V

Therefore, the input signal voltage of the amplifier is 0.01 V.

Below is the relation of base resistance:

RB=ViIB    =0.011000=106×10μA\\R _{B} = \frac{V_{i}}{I _{B}} \\ \\ \;\; = \frac{0.01}{1000} = 10^{-6 } \times 10 \mu A

Therefore, the base current of the amplifier is 10 A.

Q 10. Calculate the output ac signal, if two amplifiers are connected in series one after another (cascaded). Where the voltage gain for first amplifier is 20 and for the second it is 40.  And the input signal is provided for 0.01 volt, calculate the output ac signal.

Ans: 

Voltage gain, V1 for first amplifier = 20

Voltage gain, V2 for second amplifier = 40

Input signal voltage, Vi = 0.01V

Output voltage of AC signal = V0

By multiplying the voltage gains of each stage, we can calculate the total voltage gain of the two stage cascade amplifier

V=V1×V2V = V _{1} \times V _{2}

= 20 × 40

= 800

We know that:

V=VoViVo=V×Vi\\V = \frac{V _{o}}{V _{i}} \\ \\ V _{o} = V \times V _{i}

= 800 × 0.01 = 8 V

Therefore, the output AC signal is 8V.

Q 11.Is it possible for a p – n photodiode to detect a wavelength of 5000 nm, if it is fabricated from a semiconductor with a band gap of 3.2 eV?

Ans:

Energy band gap for the photodiode = Eg=3.2eVE _{g} = 3.2\, eV

Wavelength, λ\lambda = 5000 nm = 5000×109m5000 \times 10 ^{-9} m

We know from the relation of energy:

E=hcλE = \frac{hc}{\lambda }

Where, h = Planck’s constant = 6.626×1034Js6.626 \times 10 ^{-34}\, Js

C = Speed of light

=3×108m/sec3 \times 10 ^{8}\, m/sec E=6.626×1034×3×1085000×1093.975×1020J∴ E = \frac{6.626 \times 10 ^{-34} \times 3 \times 10 ^{8}}{5000 \times 10 ^{-9}} \\ \\ 3.975 \times 10^{-20} J

But 1.6×1019J=1  eV1.6 \times 10 ^{-19} J = 1\; eV E=3.975×1020j=3.975×10201.6×1019=0.248  eV∴E = 3.975 \times 10 ^{-20} j \\ \\ = \frac{3.975 \times 10 ^{-20}}{1.6 \times 10 ^{-19} } = 0.248 \;eV

Therefore, the energy of a signal of wavelength 5000nm is 0.248eV which is less than the energy band gap of the photodiode. Hence, the detection of the signal by the photodiode is not possible.

Additional Exercises

Q 12. We have 5×1028  per  m35 \times 10^{28} \; per\; m^{3} silicon atoms with us. Which is doped with 5×1022  per  m3 5\times 10^{22} \; per\; m^{3} of Indium and 5×1020  per  m35 \times 10^{20} \; per\; m^{3} of Arsenic simultaneously. Find out the total number of holes and the electrons. We are given,ni=1.5×1016m3n _{i} = 1.5 \times 10^{16} \, m ^{-3}. Also find if the material n – type or p – type?

 Ans:

No.of Atoms (Silicon), N = 5×1028  atoms/m35 \times 10^{28} \; atoms/ m^{3}

No.of  Atoms (Arsenic), nAs=5×1028  atoms/m3 n_{As} = 5 \times 10^{28} \; atoms/ m^{3}

No.of  Atoms (indium), nIn=5×1020  atoms/m3 n_{In} = 5 \times 10^{20} \; atoms/ m^{3}

No.of thermally generated electrons, ni=1.5×1016  electrons/m3 n_{i} = 1.5 \times 10^{16} \; electrons/ m^{3}

No.of  electrons, ne=5×1022 – 1.5×10164.99×1022 n_{e} = 5 \times 10^{22}  –  1.5 \times 10^{16}\approx 4.99 \times 10^{22}

No.of  holes = nhn _{h}

In the case of thermal equilibrium, the concentration of electrons and holes in a semiconductor are related as:  nenh=ni2nh=ni2 ne=(1.5×1016)24.99×10224.51×109\ n _{e}n _{h} = n _{i}^{2}\\ ∴ n _{h} = \frac{n _{i}^{2}}{\ n _{e}} \\ \\ = \frac{\left ( 1.5 \times 10^{16} \right )^{2}}{4.99 \times 10^{22}} \approx 4.51 \times 10^{9}

Therefore, there are approximately 4.99×10224.99 \times 10 ^{22} and holes are around 4.51×1094.51 \times 10 ^{9}. Since the number of electrons here is more than that of the holes. So, the material we have is an n – type semiconductor.

Q 13. Calculate the ratio between conductivity at 600 k and that at 300 k when an intrinsic semiconductor having energy gap EgE _{g} eV. Whose whole is much smaller than electron mobility and independent of temperature. Assume that the temperature dependence of intrinsic carrier concentration is given by

ni=n0exp[Eg2kBT]n _{i} = n_{0}\,exp \left [ – \frac{E_{g}}{2k_{B} T } \right ]

Where n0n _{0} is a constant.

Ans:Energy gap of the given intrinsic semiconductor, Eg=1.2  eVE _{g} = 1.2 \; eV

We can write the temperature dependence intrinsic carrier – concentration:

ni=n0exp[Eg2KBT]n _{i} = n_{0}\,exp \left [ – \frac{E_{g}}{2K_{B}T} \right ]

Where kBk _{B} = Boltzmann constant = 8.62×105eV/K8.62 \times 10 ^{-5} eV/K

T = temperature

n0n _{0} = Constant

Initial temperature, T1T _{1} = 300 K

We can write the intrinsic – concentration at this temperature as:

ni1=n0exp[Eg2KB×300]...(1)n _{i1} = n_{0}\,exp \left [ – \frac{E_{g}}{2K_{B}\times 300} \right ] . . . (1)

Final temperature, T2T _{2} = 600 K

We can write the intrinsic – concentration at this temperature as:

ni2=n0exp[Eg2KB×600]...(2)n _{i2} = n_{0}\,exp \left [ – \frac{E_{g}}{2K_{B}\times 600} \right ] . . . (2)

The ratio between the conductivity at 600 K and at 300 K is equal to the ratio between the respective intrinsic carrier-concentrations at these temperatures.

ni2ni1=n0exp[Eg2KB×600]n0exp[Eg2KB×300]\frac{n _{i2}}{n _{i1}} = \frac{n_{0}\,exp \left [ – \frac{E_{g}}{2K_{B}\times 600} \right ]}{n_{0}\,exp \left [ – \frac{E_{g}}{2K_{B}\times 300} \right ]} =expEg2kB[13001600]=exp[1.22×8.62×105×21600]=exp[11.6]=1.09×105\\= exp\,\frac{E_{g}}{2k_{B}} \left [ \frac{1}{300} – \frac{1}{600} \right ] \\ \\ = exp\, \left [ \frac{1.2}{2 \times 8.62 \times 10 ^{-5}} \times \frac{2 – 1}{600} \right ] \\ \\ = exp \left [ 11.6 \right ] = 1.09 \times 10^{5}

Therefore, the ratio between the conductivity is 1.09×1051.09 \times 10^{5}.

Q 14. The current I for a p – n junction diode can be expressed as:

I=I0exp(eV2KBT1)I = I _{0}\, exp \left ( \frac{eV}{2K _{B} T} – 1 \right )

Where I0 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, kB k_{B} is the Boltzmann constant (8.62×105eV/K8.62 \times 10 ^{-5} eV/K ) and T is the absolute temperature. It is for a given diode I0=5×1012 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?

(b) What will be the increase in the current if the voltage across the diode is increasedto 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?

 Ans:

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

I=I0exp(eV2KBT1)I = I _{0}\, exp \left ( \frac{eV}{2K _{B} T} – 1 \right ) I0 I_{0} = Reverse saturation current =5×1012 5 \times 10 ^{-12} kB k _{B} = Boltzmann constant = 8.62×105eV/K8.62 \times 10 ^{-5} eV/K = 1.376×1023JK11.376 \times 10 ^{-23} JK^{-1}

V = Voltage across the diode.

(a) Forward voltage, V = 0.6 V

Current  I=5×1012[exp(1.6×1019×0.61.376×1023×300)1]=5×1012×esp[22.36]=0.0256  A∴ Current\; I = 5 \times 10^{-12}\left [ exp\left ( \frac{1.6 \times 10^{-19} \times 0.6 }{1.376 \times 10^{-23} \times 300} \right ) – 1 \right ] \\ \\ = 5 \times 10 ^{-12} \times esp\left [ 22.36 \right ] = 0.0256 \; A

Therefore, the forward current is about 0.0256 A.

(b) For forward voltage, V = 0.7 V, we can write:

I=5×1012[exp(1.6×1019×0.71.376×1023×3001)]=5×1012×exp[26.25]=1.257  A\\ I^{‘} = 5 \times 10 ^{-12}\left [ exp\left ( \frac{1.6 \times 10 ^{-19} \times 0.7}{1.376 \times 10 ^{-23} \times 300} – 1 \right ) \right ] \\ \\ = 5 \times 10^{-12} \times exp\left [ 26.25 \right ] = 1.257\; A

Hence, the increase in current, ΔI=II\Delta I = I^{‘} – I

= 1.257 – 0.0256 = 1.23 A

(c) Dynamic Resistance = Change  in  voltageChange  in  current=0.70.61.23=0.11.23=0.081Ω\frac{Change \;\, in \;\, voltage}{Change \;\, in \;\, current} \\ \\ = \frac{0.7 – 0.6}{1.23} = \frac{0.1}{1.23} = 0.081 \Omega

(d) If the reverse bias voltage changes from 1 V to 2 V, then the current (I) will almost remain equal to I0 I _{0} in both cases. Therefore, the dynamic resistance in the reverse bias will be infinite.

Q 15. From the provided two circuits, show that circuit (a) and circuit (b) acts as OR gate and AND gate respectively.

a)
1
b)
34
Ans:

(a) Input for the circuit : A and B

Output for the circuit: Y

Here, the left half of the circuits works as the NOR gate, while if we look at the right half of the circuit then it is a NOT gate.

We can show it in figure as:

NOR gate and NOT gate

Now, the output for the NOR gate = A+B\overline{A + B}

We are giving the output from NOR gate to the input at NOT gate.So, its output will be:A+B\overline{\overline{A + B}} = A + B

Y=A+B∴ Y = A + B

Hence, this circuit functions as an OR Gate.

(b) Input for the circuit : A and B

Output for the circuit: Y

We can observe from the figure that the output from first half of the circuits goes to the input of the next half which is working as NOR gate.

3

Hence, the output of the given circuit can be written as:

Y=Aˉ+Bˉ=Aˉˉ.Bˉˉ=A.BY = \overline{\bar{A} + \bar{B}} = \bar{\bar{A}}.\bar{\bar{B}} = A.B

Hence, this circuit functions as an AND gate.

Q 16. Write the truth table for a NAND gate connected as shown in the figure.

4

And then find out the exact login operation carried out by the given circuit.

Ans:

From the figure, we know A is the input and we are getting an output B.

5

Hence, the output can be written as:

Y=A.A=Aˉ+Aˉ=AˉY = \overline{ A . A } = \bar{A} + \bar{A} = \bar{A}

We can make the truth table for eq(1) as:

A Y
0 1
1 0

Below is the symbol of the NOT gate:

6

 

Q 17. You have two circuits consisting NAND gates arranged to perform different logical operations. Identify the logical operation carried by the two circuits (Shown in figure).

 7 

Ans:

In the given circuits,

Inputs: A and B

Output: Y

(a) Following is the circuit which shows that the output of the left NAND gate will be A.B\overline{ A . B }

8

Therefore, the output for these two NAND gate is given as:

Y=(A.B).(A.B)=AB+AB=AB Y = \overline{\left (\overline{A . B } \right ) . \left (\overline{ A . B } \right )} = \overline{ \overline{A B} } + \overline{ \overline{A B} } = AB

Therefore, this circuit works as an NAND gate.

(b) Aˉ\bar{A} is the output of the upper left of the NAND gate and Bˉ\bar{ B } is the output of the lower half of the NAND gate, as shown in the following figure.

9

Hence, the output from this combination of NAND gates can be shown as:

Y=Aˉ.Bˉ=Aˉˉ+Bˉˉ=A+BY = \bar{A} . \bar{B} = \bar{ \bar{A}} + \bar{ \bar{B}} = A + B

Therefore, we can say that this circuit is working as an OR gate.

Q 18. Draw the truth table for the circuit having NOR gates and find out the logic operation (OR, AND, NOT ) which is formed by this circuit.

(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 thetruth table of OR, AND, NOT gates and find the correct one.)

  10 

Ans

Input for the circuit: A and B

Output from the first circuit (i.e NOR gate) is A+B\overline{A + B}.

From the figure we can observe that the input for the second gate is automatically the output from that first one.

11

Therefore, the output obtained from this combination is given as:

Y=A+B+A+B=Aˉ.Bˉ+Aˉ.Bˉ=Aˉ.Bˉ=Aˉˉ+Bˉˉ=A+BY = \overline{\overline{A + B} + \overline{A + B} } = \overline{ \bar{ A } . \bar{ B }} + \overline{ \bar{ A } . \bar{ B }} \\ \\ = \overline{ \bar{ A } . \bar{ B }} = \bar{ \bar{ A } } + \bar{ \bar{ B } } = A + B

Following is the truth table for the given operation:

A B Y (= A + B)
0 0 0
0 1 1
1 0 1
1 1 1

This is the truth table of an OR gate.

Therefore, this circuit functions as an OR gate.

Q 19.  Find out the truth table forming by combining the OR gates only. (as shown in the following figure). Identify the logic operation (OR, AND, NOT) performed by the two circuits

12

12

Ans:

(a) Here, A is acting as both the outputs for the NOR gate and Y is the output, as shown in the following figure.

So, the output of the circuit is A+A\overline{A + A}.

13

Output, Y = A+A=Aˉ\overline{A + A} = \bar{A}

The truth table for the circuit is given as:

A Y(Aˉ\bar{A})
0 1
1 0

Here, we obtain the truth table for a NOT gate. Hence, the circuits works as a NOT gate.

(b) For the given circuit,

Inputs: A and B

Output: Y

From the solution  obtained from (a) we conclude that the output for the first two NOR gates are Aˉ  and  Bˉ\bar{A} \; and \; \bar{B}

14

Aˉ  and  Bˉ\bar{A} \; and \; \bar{B} are the inputs for the last NOR gate. Therefore, below is the table that has the output for the given circuit:

A B Y (= A! B)
0 0 0
0 1 0
1 0 0
1 1 1

 

The NCERT class 12 physics solutions for chapter 14 is created by subject experts according to the latest CBSE syllabus 2019-20. Students must practice the solutions regularly to prepare effectively for their examination.
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