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NCERT Exemplar Class 12 Physics Solutions Chapter 14 Semiconductor Electronics Materials Devices and Simple Circuits

NCERT Exemplar Class 12 Physics Solutions Chapter 14 Semiconductor Electronics Materials Devices and Simple Circuits

#CBSE Class 12th
Vishal kumarUpdated on 14 Jul 2025, 07:49 PM IST

Have you ever wondered how simple electronic circuits work or how semiconductors are used in devices like diodes and transistors? Chapter 14: Semiconductor Electronics Materials Devices and Simple Circuitss in Class 12 Physics explains the basics of how these devices work. In this chapter you will learn about semiconductors, diodes, transistors (NPN and PNP) and logic gates (Like - OR gate, AND gate etc) . These topics are very important for understanding how modern electronic gadgets are built. The NCERT Exemplar Solutions for class 12 are prepared by experienced teachers as per the CBSE syllabus.

The NCERT Exemplar Solutions for Class 12 Physics Chapter 14 by Careers360 provide, step-by-step answers to all the exemplar problems. The solutions cover MCQs, Very short answer (VSA) short answer(SA) and long answer(LA) questions. solving these NCERT exemplar questions will help you understand electronics better, improve your application skills, and prepare well for exams like CBSE Board Exams, JEE, and NEET.

This Story also Contains

  1. NCERT Exemplar Class 12 Physics Solutions Chapter 14 MCQ I
  2. NCERT Exemplar Class 14 Physics Solutions Chapter 14 MCQ II
  3. NCERT Exemplar Class 12 Physics Solutions Chapter 14: Very Short Answer
  4. NCERT Exemplar Class 12 Physics Solutions Chapter 14: Short Answer
  5. NCERT Exemplar Class 12 Physics Solutions Chapter 14: Long Answer
  6. Main Subtopics of NCERT Exemplar Class 12 Physics Solutions Chapter 14
  7. NCERT Exemplar Class 12 Physics Chapter Wise Links
NCERT Exemplar Class 12 Physics Solutions Chapter 14 Semiconductor Electronics Materials Devices and Simple Circuits
NCERT Exemplar Class 12 Physics Solutions Chapter 14 Semiconductor Electronics Materials Devices and Simple Circuits

NCERT Exemplar Class 12 Physics Solutions Chapter 14 MCQ I

Question:1

The conductivity of a semiconductor increases with increase in temperature because
A. number density of free current carriers increases.
B. relaxation time increases.
C. both number density of carriers and relaxation time increase.
D. number density of current carriers increases, relaxation time decreases but effect of decrease in relaxation time is much less than increase in number density.

Answer:

The answer is the option (d)
The number density of current carriers increases, relaxation time decreases but the effect of the decrease in relaxation time is much less than the increase in number density.

Question:2

In Fig. 14.1, Vo is the potential barrier across a p-n junction, when no battery is connected across the junction
q-2
A. 1 and 3 both correspond to forward bias of junction
B. 3 corresponds to forward bias of junction and 1 corresponds to reverse bias of junction
C. 1 corresponds to forward bias and 3 corresponds to reverse bias of junction.
D. 3 and 1 both correspond to reverse bias of junction.

Answer:

Answer: The answer is the option (b)
The height of a potential barrier increases when p-n junction is biased forward, and it opposes the potential barrier junction. But if p-n is reversed biased, the potential barrier junction is supported, which results in an increase in the potential barrier.

Question:3

In Fig. 14.2, assuming the diodes to be ideal,
a-3
A. $D_1$ is forward biased and $D_2$ is reverse biased and hence current flows from A to B
B. D2 is forward biased and $D_1$ is reverse biased and hence no current flows from B to A and vice versa.
C. $D_1$ and $D_2$ are both forward biased and hence current flows from A to B.
D. $D_1$ and $D_2$ are both reverse biased and hence no current flows from A to B and vice versa.

Answer:

The answer is the option (b)
$D_2$ is forward biased and $D_1$ is reverse biased and hence no current flows from B to A and vice versa.

Question:4

A 220 V A.C. supply is connected between points A and B (Fig. 14.3). What will be the potential difference V across the resistor?

tgrtgttg
A. 220V
B. 110V
C. 0V
D. $220 \sqrt{2}V$

Answer:

As p-n junction diode will conduct during positive half cycle only during the negative half cycle diode is reverse biases. during this diode will not give any output so potential difference across capacitor C-peak voltage of the given AC voltage

$V_{0}=V_{rms}\sqrt{2}=220 \sqrt{2}V$

Question:5

Hole is
A. an anti-particle of electron.
B. a vacancy created when an electron leaves a covalent bond.
C. absence of free electrons.
D. an artificially created particle.

Answer:

The answer is the option (b)
The hole is a vacancy created when an electron leaves a covalent bond.

Question:6

The output of the given circuit in Fig. 14.4.

q-6
A. would be zero at all times.
B. would be like a half wave rectifier with positive cycles in output.
C. would be like a half wave rectifier with negative cycles in output.
D. would be like that of a full wave rectifier.

Answer:

The answer is the option (c)

would be like a half wave rectifier with negative cycles in output

Question:7

In the circuit shown in Fig. 14.5, if the diode forward voltage drop is 0.3 V, the voltage difference between A and B is

q-7
A. 1.3 V
B. 2.3 V
C. 0
D. 0.5 V

Answer:

The answer is the option (b)
Let us consider the fig (b)given above in the problem suppose the potential difference between A and B is VAB
Then,
$\begin{aligned} & V_{A B}-0.3=\left[\left(r_1+r_2\right) 10^3\right] \times\left(0.2 \times 10^{-3}\right)\left[\text { because } V_{A B}=i r\right] \\ & =\left[(5+5) 10^3\right] \times\left(0.2 \times 10^{-3}\right) \\ & =10 \times 10^3 \times 0.2 \times 10^{-3}-2 \\ & \Rightarrow V_{A B}=2+0.3=2.3 \mathrm{~V}\end{aligned}$

Question:8

Truth table for the given circuit (Fig. 14.6) is
fgttggt4
A.

A
B
C
0
0
1
0
1
0
1
0
1
1
1
0

B.
A
B
C
0
0
1
0
1
0
1
0
0
1
1
1
C.
A
B
C
0
0
0
0
1
1
1
0
0
1
1
1
D.
A
B
C
0
0
0
0
1
1
1
0
1
1
1
0

Answer:

The answer is the option (c)
(c) In this problem the input C of OR get and when which is an output of AND gate . So, "C equals A AND B" or C=A.B and "D equals NOT A AND B" or D=$\bar{A}$.B
and "E equals C AND D "or E=C+D=(A.B)+(A.B) Now we can generate the truth table of this arrangement of gates can be given bt
A
B
$\bar{A}$
C=A.B
d=$\bar{A}$.B
E=(C+D)
0
0
1
0
0
0
0
1
1
0
1
1
1
0
0
0
0
0
1
1
0
1
0
1

NCERT Exemplar Class 14 Physics Solutions Chapter 14 MCQ II

Question:9

When an electric field is applied across a semiconductor
A. electrons move from lower energy level to higher energy level in the conduction band.
B. electrons move from higher energy level to lower energy level in the conduction band.
C. holes in the valence band move from higher energy level to lower energy level.
D. holes in the valence band move from lower energy level to higher energy level.

Answer:

The answer is the option (a, c)
The electrons in the valence band are not able to gain energy from the external electric field. While the electrons in the conduction band can gain energy from the external field. In semiconductors, when an electric field is applied, the electrons in the conduction band are accelerated and gain energy. The movement is from lower energy level to higher energy level. The movement of holes in the valence band is from higher energy level to lower energy level. Here they will have more energy

Question:10

Consider an npn transistor with its base-emitter junction forward biased and collector base junction reverse biased. Which of the following statements are true?.
A. Electrons crossover from emitter to collector.
B. Holes move from base to collector.
C. Electrons move from emitter to base.
D. Electrons from emitter move out of base without going to the collector.

Answer:

The answer is the option (a, c)
Key Concept: Transistor
A junction transistor is formed by sandwiching a them layer of P- type swmiconductor between two N-type semiconductors or by sandwiching a thin layer of N-type semiconductor between two P-type semiconductors
q-101
E- Emitter(emits majority charge carriers)
C- Collects majority charge carriers
B- Base (provide proper interaction between E and C)
q-102

Question:11

Figure 14.7 shows the transfer characteristics of a base biased CE transistor. Which of the following statements are true?

q-11
A. At Vi = 0.4V, transistor is in active state.
B. At Vi = 1V, it can be used as an amplifier.
C. At Vi = 0.5V, it can be used as a switch turned off.
D. At Vi = 2.5V, it can be used as a switch turned on.

Answer:

The answer is the option (b, c, d)
According to the graph, the transfer characteristics of base biased common emitter transistor are followed by in the following options.

Question:12

In a npn transistor circuit, the collector current is 10mA. If 95 per cent of the electrons emitted reach the collector, which of the following statements are true?
A. The emitter current will be 8 mA.
B. The emitter current will be 10.53 mA.
C. The base current will be 0.53 mA.
D. The base current will be 2 mA.

Answer:

The answer is the option (b, c)
It is given here that; the collector current is 95% of the electrons reaching the collector after the emission. And Ic = 10 mA
$I_{E}$=emiter current Also, $I_{C}=\frac{95}{100}I_{E}\\$
$ \Rightarrow I_{E}=\frac{10 \times 100}{95}=10.53mA\\$
Also,$I_{B}=I_{E}-I_{C}=10.53-10=0.53mA$

Question:13

In the depletion region of a diode
A. there are no mobile charges
B. equal number of holes and electrons exist, making the region neutral.
C. recombination of holes and electrons has taken place.
D. immobile charged ions exist.

Answer:

The answer is the option (a, b, d)
The concentration of charge carriers in N-region and P-region is different. Therefore, due to diffusion, the neutrality of both p-type and n-type semiconductor is disturbed. Hence, a layer of negatively charged ions appears at the p junction while positive ions at a junction in N-crystal.
The thickness of layer (depletion) is 1 micron = 10-6 m.
Width of depletion layer ∞ 1/Dopping
And the depletion layer is directly proportional to the temperature.
a-13

Question:14

What happens during regulation action of a Zener diode?
A. The current in and voltage across the Zenor remains fixed.
B. The current through the series Resistance (Rs) changes.
C. The Zener resistance is constant.
D. The resistance offered by the Zener changes.

Answer:

The answer is the option (b,d)
In forward biased mode, the zener diode act as a voltage regulator and used as an ordinary diode.
a-14
The zener diode in the reverse biased mode it offers constant voltage drop across the terminals as the unregulated voltage is applied. And during the regulation action of the zener diode, the current across the series resistance Rs is changed, therefore, resistance offered by the zener diode also changes. The current in the zener diode changes, but the voltage remains the same and constant.

Question:15

To reduce the ripples in a rectifier circuit with capacitor filter
A. RL should be increased.
B. input frequency should be decreased.
C. input frequency should be increased.
D. capacitors with high capacitance should be used.

Answer:

The answer is the option (a, c, d)
Ripple factor is defined as the ratio of r.m.s. value of the ripple to the absolute value of the direct current voltage of the output voltage. It is usually denoted in percentage. The ripple voltage is also expressed as peak to peak value. The ripple factor of a full-wave rectifier for a capacitor filter is given by:
$r=\frac{0.236R}{\omega L}$
Where L is inductance of the coil and $\omega$ is the angular frequency.
or Ripple factor can also be given by
$r=\frac{I}{4\sqrt{3}VR_{L}C_{Y}}\\ i.e \: r\infty \frac{I}{R_{L}}\Rightarrow r\infty \frac{I}{C}.r\infty \frac{I}{V}$

Question:16

The breakdown in a reverse biased p–n junction diode is more likely to occur due to
A. large velocity of the minority charge carriers if the doping concentration is small.
B. large velocity of the minority charge carriers if the doping concentration is large.
C. strong electric field in a depletion region if the doping concentration is small.
D. strong electric field in the depletion region if the doping concentration is large.

Answer:

The answer is the option (a, d)
Reverse biasing is when the positive terminal of the battery is connected to the N-crystal and negative terminal is of the battery is connected to P-crystal.
a-16
In reverse biasing, ionization takes place because the minority charge carriers get accelerated due to reverse biasing. They strike with the electrons which in turn increase the number of charge carriers. And when the doping region is large, there will be a large number of ions in the depletion region. This will give rise to a strong electric field.

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NCERT Exemplar Class 12 Physics Solutions Chapter 14: Very Short Answer

Question:17

Why are elemental dopants for Silicon or Germanium usually chosen from group XIII or group XV?

Answer:

Silicon and germanium are chosen as a dopants because their size is compatible with the gaps in semiconductors. And they even are capable of forming covalent bonds.

Question:18

Sn, C, and Si, Ge are all group XIV elements. Yet, Sn is a conductor, C is an insulator while Si and Ge are semiconductors. Why?

Answer:

Sn is a conductor as it has an energy gap of 0 EV while C has an energy gap of 5.4 eV therefore, it is an insulator. Si and Ge have an energy gap of 1.1 eV and 0.7 eV, which makes them a semiconductor. The gaps in energy are related to their individual atomic size, which is responsible for making one insulator, conductor and semiconductor.

Question:19

Can the potential barrier across a p-n junction be measured by simply connecting a voltmeter across the junction?

Answer:

We cannot measure the potential barrier of a PN-junction by connecting a sensitive voltmeter across its terminals because in the depletion region, there are no free electrons and holes and in the absence of forward biasing, PN- junction offers infinite resistance.

Question:20

Draw the output waveform across the resistor (Fig.14.8).
q-20

Answer:

The diode act as a half wave rectifier, it offers low resistance when forward biased and high resistance when reverse biased.
ergg

Question:21

The amplifiers X, Y and Z are connected in series. If the voltage gains of X, Y and Z are 10, 20 and 30, respectively and the input signal is 1 mV peak value, then what is the output signal voltage (peak value)

(i) if dc supply voltage is 10V?
(ii) if dc supply voltage is 5V?

Answer:

The ratio of output signal voltage to the input signal voltage is known as the total voltage amplification.
According to the problem, voltage gain in X, Vx =10
Voltage gain un Y; Vy=20
Voltage gain un Z; VZ=30
$\triangle V_{i}=1 m V=10^{-3}V$
And total voltage amplification=
$V_{x}\times V_{y}\times V_{z}$
$\triangle V_{0}=V_{x}\times V_{y}\times V_{z}\times V_{i}\\ =10\times 20\times 30\times 10^{-3} V=6V$
(i)If DC supply voltage is 10 V, then output is 6 V since the theoretical gain is equal to practical gain i.e output can never be greater than 6V
(ii) If DC supply voltage is 5 V i.e, Vcc=5V. Then the output peak will not exceed 5V. hence V0=5V.

Question:22

In a CE transistor amplifier there is a current and voltage gain associated with the circuit. In other words there is a power gain. Considering power a measure of energy, does the circuit violate conservation of energy?

Answer:

$(i) dc \: current\: gain: \beta_{dc}=\frac{i_{c}}{i_{b}}$
$(ii) Voltage \: gain \: A_{v}=\frac{\triangle V_{0}}{\triangle V_{i}}=\beta _{ac}\times Resistance \: gain$
$(iii) Power \: gain \: =\frac{\triangle P_{0}}{\triangle P_{i}}=\beta^{2} _{ac}\times Resistance \: gain$
In CE transistor amplifier, the power gain is extremely high. Here, the extra power required for the amplification is gained from DC source. Therefore, the circuit does not violate the law of conservation.

NCERT Exemplar Class 12 Physics Solutions Chapter 14: Short Answer

Question:23

(a)
q-23a
(b)
q-23b
(i) Name the type of a diode whose characteristics are shown in Fig. 14.9 (A) and Fig. 14.9(B).

Answer:

i) Figure a) it represents the characteristics of Zener diode and figure b) represents a solar cell.
ii) Figure a) the point P represents zener breakdown voltage.
iii) Figure b) the point Q represents negative current and zero voltage.

Question:24

Three photo diodes $D_1$, $D_2$ and $D_3$ are made of semiconductors having band gaps of 2.5eV, 2eV and 3eV, respectively. Which ones will be able to detect light of wavelength $A^{\circ}$

Answer:

a-24
According to the problem,
Wavelength of light $\lambda = 6000 A^{\circ}-6000\times 10^{10}m$
Energy of the light photon
$E=\frac{h_{c}}{\lambda}=\frac{6.62\times10^{-34}\times3 \times 10^{8}}{6000\times10^{10}\times1.6\times10^{19}}eV=2.06eV$

Question:25

If the resistance R1 is increased (Fig.14.10), how will the readings of the ammeter and voltmeter change?

q-25

Answer:

IbR1 + Vbe = Vbb

Base current = Ib = Vbb – Vbe/R1

Ib is inversely proportional to R1

Hence, if R1 is increased, then Ib gets reduced.

Question:26

Two car garages have a common gate which needs to open automatically when a car enters either of the garages or cars enter both. Devise a circuit that resembles this situation using diodes for this situation.

Answer:

Here, the OR gate is used to explain the situation:
a-26
So OR gate gives the desired output
A
B
Y=A+B
0
0
0
0
1
1
1
0
1
1
1
1

Question:27

How would you set up a circuit to obtain NOT gate using a transistor?

Answer:

notgateusing-transistor

Question:28

Explain why elemental semiconductor cannot be used to make visible LEDs.

Answer:

In an elemental semiconductor, the bandgap is such that the emission is in the infrared region and not in the visible region.
$\lambda=\frac{hc}{E_{g}}=\frac{1242eVnm}{E_{g}}$
for Si; $E_{g}=1.1eV, \lambda=\frac{1242}{1.1}=1129nm$
for Ge; $E_{g}=0.7eV, \lambda=\frac{1242}{0.7}=1725nm$

Question:29

Write the truth table for the circuit shown in Fig.14.11. Name the gate that the circuit resembles.

q-29

Answer:

a-291
This is 'AND' Gate and its characteristics are as follows:
A
B
V0=A.B
0
0
0
0
1
0
1
0
0
1
1
1

Question:30

A Zener of power rating 1 W is to be used as a voltage regulator. If zener has a breakdown of 5V and it has to regulate voltage which fluctuated between 3V and 7V, what should be the value of Rs for safe operation (Fig.14.12)?

q-30

Answer:

According to the problem Power=1 Watt
Zener breakdown voltage Vz=5V
Minimum voltage Vmin=3V
Maximum voltage Vmax=7V
We know, P=VI
So current $I_{z_{max}}=\frac{P}{V_{z}}=\frac{1}{5}=0.2A$
for safe operation, Rs will be equal to
$R_{s}=\frac{V_{max}-V_{z}}{I_{z_{max}}}=\frac{7-5}{0.2}=\frac{2}{0.2}=10\Omega$

Question:31

If each diode in Fig. 14.13 has a forward bias resistance of $25\Omega$ and infinite resistance in reverse bias, what will be the values of the current I1, I2, I3 and I4?

q-31

Answer:

According to the problem forward biases resistance = $25 \Omega$ and reverse biases resistance=$\infty$
As show in figure the diode in brsnch CD is in reverse biases which having infinite resistance
So current in that branch is zero i.e $I_{3}=\theta$
AB is parallel to EF
So effective resistance
$\frac{1}{R'}=\frac{1}{R_{1}}+\frac{1}{R_{2}}=\frac{1}{150}+\frac{1}{150}=\frac{2}{150}\\\Rightarrow R'=75\Omega$
Total resistance R of the circuit=R'=25=75+25=100$\Omega$
current
$I_{1}=\frac{V}{R}=\frac{5}{100}=0.05A$
According to the Kirchhoff's, current law (KCL)
$I_{1}=I_{4}+I_{2}+I_{3}\: \: \: \: \: \: \: \left ( I_{3}=0 \right )\\ SoI_{1}=I_{4}+I_{2}$
Here the resistances R1 and R2 is same
i.e
$I_{4}=I_{2}\\ \therefore I_{1}=2I_{2}\\ \Rightarrow I_{2}=\frac{I_{1}}{2}=\frac{0.05}{2}=0.025A$
And I4 =0.025A
Therefore we get I1 =0.05A,I2 =0.025A,I3 =0 and I4 =0.025A

Question:32

In the circuit shown in Fig.14.14, when the input voltage of the base resistance is 10V, Vbe is zero and Vce is also zero. Find the values of Ib, Ic and $\beta$

q-32

Answer:

According to the problem V1 =10V, Resistance RB=400k$\Omega$, VBE=0, VCE=0 and Rc=3k$\Omega$
Vi-VBE=RBIB
$I_{B}=\frac{Voltage\: across\: R_{B}}{R_{B}}\\ =\frac{10}{400 \times 10^{3}}=25 \times 10^{-6}A=25\mu A$
Voltage across RC=10V
VCC-VCE=ICRC
$I_{C}=\frac{Voltage\: across\: R_{C}}{R_{C}}\\ =\frac{10}{3\times 10^{3}}=3.33 \times 10^{-3}A=3.33mA\\$
$ \beta =\frac{I_{C}}{I_{B}}=\frac{3.33 \times 10^{-3}}{25 \times 10^{-6}}\\ =1.33 \times 10^{2}=133$

NCERT Exemplar Class 12 Physics Solutions Chapter 14: Long Answer

Question:33

Draw the output signals C1 and C2 in the given combination of gates (Fig. 14.15).
q33

Answer:

a-331
$C_{1}=\bar{A}.\bar{B}=\bar{A}+\bar{B}=\overline{A+B}$
A
B
C
D
E
F
G
H
I
C1
0
0
0
0
1
1
1
0
0
1
1
0
1
0
0
1
0
1
1
0
0
1
0
1
1
0
0
1
1
0
1
1
1
1
0
0
0
1
1
0
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a-332
$C_{2}=\bar{A}.\bar{B}=\bar{A}.\bar{B}=A.B$
A
B
C
D
E
F
G
C2
0
0
0
0
1
1
1
0
1
0
1
0
0
1
1
0
0
1
0
1
1
0
1
0
1
1
1
1
0
0
0
1
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a-333

Question:34

Consider the circuit arrangement shown in Fig 14.16 (a) for studying input and output characteristics of n-p-n transistor in CE configuration.

q-34
Select the values of RB and RC for a transistor whose VBE = 0.7 V, so that the transistor is operating at point Q as shown in the characteristics shown in Fig. 14.16 (b).
q-341
Given that the input impedance of the transistor is very small and VCC = VBB = 16 V, also find the voltage gain and power gain of circuit making appropriate assumptions.

Answer:

According to the problem at point Q, from graph $V_{BE}=0.7V,V_{CC}=V_{BB}=16V$ and $V_{CE}=8V$
$I_{C}=4mA=4\times 10^{-3}A\\ I_{B}=30\mu A=30\times 10^{-6}A\\$
$ Since\: V_{CC}=I_{C}R_{C}+V_{CE}\\ R_{C}=\frac{V_{CC}-V_{CE}}{I_{C}}=\frac{16-8}{4 \times 10^{-3}}=\frac{8 \times 1000}{4}=2k\Omega\\$
Similarly, $V_{BB}=I_{B}R_{B}+V_{BE}\\ R_{B}=\frac{V_{BB}-V_{BE}}{I_{B}}=\frac{16-0.7}{30 \times 10^{-6}}=510 \times 10^{3}\Omega=510k\Omega\\$
Current $\: gain=\beta =\frac{I_{C}}{I_{B}}=\frac{4 \times 10^{-3}}{30 \times 10^{-6}}=133.3\\$
$Voltage \: gain=\beta =\frac{R_{C}}{R_{B}}=\frac{133 \times 2 \times 10^{3}}{510 \times 10^{3}}=0.52\\ $
$Power \: gain=\beta \times voltage\: gain=133 \times 0.52=69$

Question:35

Assuming the ideal diode, draw the output waveform for the circuit given in Fig. 14.17. Explain the waveform.
q-35

Answer:

The waveform obtained from the circuit will be a sine wave with a little dip in the input wave.
1535axaxa

Question:36

Suppose a ‘n’-type wafer is created by doping Si crystal having $5 \times 10^{28}$ atoms/m3 with 1ppm concentration of As. On the surface 200 ppm Boron is added to create ‘P’ region in this wafer. Considering $n_{i} = 1.5 \times 10^{16} m^{-3}$, (i) Calculate the densities of the charge carriers in the n & p regions. (ii) Comment which charge carriers would contribute largely for the reverse saturation current when diode is reverse biased.

Answer:

$n_{e}=N_{D}=10^{-6}\times 5 \times 10^{28}atoms/m^{3}\\ =5 \times 10^{22}/m^{3}$
Number of miniority carriers (holes) in n-type wafer is
$n_{h}=\frac{n_{i}^{2}}{n_{e}}=\frac{\left ( 1.5 \times 10^{16} \right )^{2}}{5 \times 10^{22}}=0.45 \times 10^{10}/m^{3}$
p-type wafer is created with number of holes, when Boron is implanted in Si crystal,
$n_{h}=N_{A}=200 \times 10^{-6} \times \left ( 5 \times 10^{28} \right )=1 \times 10^{25}/m^{3}$
Minority carriers (electrons) created in p-type wafer is
$n_{h}=\frac{n_{i}^{2}}{n_{e}}=\frac{\left ( 1.5 \times 10^{16} \right )^{2}}{1 \times 10^{25}}=2.25 \times 10^{7}/m^{3}$
(ii) The minority carrier holes of n-region wafer $\left (n_{h}=0.45 \times 10^{10}/m^{3} \right )$ would contribute more to the reverse saturation current than minority carrier electrons $\left (n_{e}=2.25 \times 10^{7}/m^{3} \right )$ of p-region wafer when p-n junction is reverse biased

Question:37

An X-OR gate has following truth table:

A
B
Y
0
0
0
0
1
1
1
0
1
1
1
0
It is represented by following logic relation
$Y=\bar{A}.B+A.\bar{B}$
Build this gate using AND, OR and NOT gates.

Answer:

XOR can be obtained by combining two NOT gates, two AND gates and one OR gate. The logic relation for the given table is as follows:
$Y=\bar{A}.B+A.\bar{B}=Y_{1}+Y_{2}\\ $
$when\: \: Y_{1}=\bar{A}.B \: \: and\: \: Y_{2}=A.\bar{B}$
Y1 cna be obtained as output of AND gate I for which one input is of A through NOT gate and another input is of B. Y2 can be obtained as output of AND gate II for which one input is of A and other input id of B through NOT gate.
Now Y can be obtained as output from OR where Y1 and Y2 are inputs of OR gate
Thus the logic circuit of this relation is given below
a-37

Question:38

Consider a box with three terminals on top of it as shown in Fig.14.18 (a):

q-381
Three components namely, two germanium diodes and one resistor are connected across these three terminals in some arrangement.
A student performs an experiment in which any two of these three terminals are connected in the circuit shown in Fig. 14.18 (b).
q-382
The student obtains graphs of current-voltage characteristics for unknown combination of components between the two terminals connected in the circuit.
The graphs are
(i) when A is positive and B is negative
q-383
(ii) when A is negative and B is positive
q-384
(iii) When B is negative and C is positive
q-385
(iv) When B is positive and C is negative
q-386
(v) When A is positive and C is negative
q-387
(vi) When A is negative and C is positive
q-388
From these graphs of current – voltage characteristic shown in Fig. 14.18 (c) to (h), determine the arrangement of components between A, B and C.

Answer:

a) n-side of the PN junction is connected to the A terminals while B is connected to the top-side of PN junction.
b) the knee voltage is 0.7V.
c) C is connected to the PN junction and n-side of the junction is connected to the B terminal, and knee voltage is 0.7V.
d) This above condition explains the connection between pn-junction with I and II with the resistance.

Question:39

For the transistor circuit shown in Fig.14.19, evaluate VE, RB, RE given IC = 1 mA, VCE = 3V, VBE = 0.5 V and VCC = 12 V, $\beta$ = 100.

q-39

Answer:

a-39

As we know the base current is very small So,

$
\begin{aligned}
& \quad I_C=I_E \\
& R_C=7.8 \mathrm{k} \Omega \\
& \text { from th fig. } I_C\left(R_c+R_E\right)+V_{C E}=12 \\
& \left(R_E+R_C\right) \times 1 \times 10^{-3}+3=12 \\
& \left(R_E+R_C\right)=9 \times 10^3=9 \mathrm{k} \Omega \\
& R_E=9-7.8=1.2 \mathrm{k} \Omega \\
& V_E=I_E \times R_E \\
& =1 \times 10^{-3} \times 1.2 \times 10^3=1.2 \mathrm{~V} \\
& \text { VoltageV } B=V_E+V_{B E}=1.2+0.5=1.7 \mathrm{~V} \\
& \text { Current } I=\frac{V_B}{20 \times 10^3}=\frac{1.7}{20 \times 10^3}=0.085 \mathrm{~mA} \\
& \text { Resistance } R_B=\frac{12-1.7}{\frac{I_C}{\beta}+0.085} \times 10^3=\frac{10.3}{0.01+0.085} \quad \text { [Given } \beta=100 \text { ] } \\
& =108 \mathrm{k} \Omega
\end{aligned}
$

Question:40

In the circuit shown in Fig.14.20, find the value of RC.
q-40

Answer:

Let us consider the ciruit diagram to solve this problem
$I_{E}=I_{C}+I_{B}\: and\: I_{C}=\beta I_{B}.........(i)\\ I_{C}R_{C}+V_{CE}+I_{E}R_{E}=V_{CC}..........(ii)\\ RI_{B}+V_{BE}+I_{E}R_{E}=V_{CC}.................(iii)\\ \because I_{E}\approx I_{C}=\beta I_{B}\\ from(iii)\\ \left ( R+\beta R_{E} \right )I_{B}=v_{CC}-V_{BE}\\ \Rightarrow I_{B}=\frac{V_{CC}-V_{BE}}{R+\beta .R_{E}}\\ =\frac{12-0.5}{80+1.2\times 100}=\frac{11.5}{200}mA\\ from(ii)\\ \left ( R_{C}+R_{E} \right )=\frac{V_{CE}-V_{BE}}{I_{C}}=\frac{V_{CC}-V_{CE}}{\beta I_{B}}\: \: \: \: \: \: \: \: \: \left ( \because I_{C}=\beta I_{B} \right )$



$\begin{aligned} & \left(R_C+R_E\right)=\frac{2}{11.5}(12-3) \mathrm{k} \Omega=1.56 \mathrm{k} \Omega \\ & R_C+R_E=1.56 \\ & R_C=1.56-1=0.56 \mathrm{k} \Omega \text { or } 560 \Omega\end{aligned}$

Main Subtopics of NCERT Exemplar Class 12 Physics Solutions Chapter 14

The main subtopics of NCERT Exemplar Class 12 Physics Solutions Chapter 14 cover the basics of semiconductor materials, how devices like diodes and transistors work, and their use in simple electronic circuits. These topics help students connect physics concepts to real-world technology used in everyday electronics.

NCERT Exemplar Class 12 Physics Chapter Wise Links

Chapter 1 Electric Charges and Fields

Chapter 2 Electrostatic Potential and Capacitance

Chapter 3 Current Electricity

Chapter 4 Moving Charges and Magnetism

Chapter 5 Magnetism and Matter

Chapter 6 Electromagnetic Induction

Chapter 7 Alternating Current

Chapter 8 Electromagnetic Waves

Chapter 9 Ray Optics and Optical Instruments

Chapter 10 Wave Optics

Chapter 11 Dual Nature of Radiation and Matter

Chapter 12 Atoms

Chapter 13 Nuclei

Chapter 15 Communication Systems

NCERT Exemplar Class 12 Solutions

NCERT Exemplar Class 12 Chemistry Solutions

NCERT Exemplar Class 12 Mathematics Solutions

NCERT Exemplar Class 12 Biology Solutions

Also, check the NCERT solutions of questions given in the book

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, read NCERT Solutions subject-wise

Must read NCERT Notes subject wise

Also, check the NCERT Books and NCERT Syllabus here:

Frequently Asked Questions (FAQs)

Q: What are semiconductors, and why are they important?
A:

Semiconductors are materials that have properties between conductors and insulators. They're super important because they power most modern electronic gadgets.

Q: What devices will I learn about in this chapter?
A:

You’ll study diodes, transistors, logic gates, LEDs, and their working, applications, and how they’re used in electronic circuits.

Q: Does this chapter have practical applications?
A:

Yes! The concepts you learn here are directly linked to real-world devices like smartphones, computers, calculators, and even traffic lights.

Q: What are the topics covered in Semiconductor Electronics: Materials, Devices, and Simple Circuits?
A:

This chapter covers everything related to semiconductors like types, properties, junction transistors, diodes, logic gates, etc.

Q: What is the weightage of the chapter in NEET exam?
A:

In NEET exam 2 to 3 questions are asked every year from the chapter Semiconductor Electronics. Mostly questions from both analog and digital electronics are included.

Q: Whether the NCERT Exemplar Class 12 Physics Solutions Chapter 14 is useful for exam.
A:

Students can get a better understanding of the concepts discussed in the chapter using NCERT Exemplar Solutions For Class 12 Physics Chapter 14, which in turn will be helpful for exams

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After 12th, if you are interested in computer science, the best courses are:

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I am a CBSE class 12 private candidate will my maksheet of CBSE supplement exam will be delivered on main exam form address for supplementary exam form address ,my both addresses are different please tell

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12 arts previous years board paper 2025

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