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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

Edited By Safeer PP | Updated on Sep 14, 2022 01:33 PM IST | #CBSE Class 12th

NCERT Exemplar Class 12 Physics solutions chapter 14 dives deep into the heart of microprocessors and transistors, something we cannot survive a day without. Class 12 Physics NCERT Exemplar solutions chapter 14 explores Semiconductors as devices whose properties lie between an insulator and a conductor. It possesses eccentric electrical properties and is mounted inside many electronics appliances that support our everyday lives. NCERT Exemplar solutions For Class 12 Physics chapter 14 provided here would help a student achieve academic success in both 12 boards and competitive exams. They would also help to understand and practice the concepts both theoretically and practically. Students can use NCERT Exemplar Class 12 Physics solutions chapter 14 PDF download to understand the best way to approach a problem.

Also see - NCERT Solutions for Class 12 Other Subjects

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.

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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. D1 is forward biased and D2 is reverse biased and hence current flows from A to B
B. D2 is forward biased and D1 is reverse biased and hence no current flows from B to A and vice versa.
C. D1and D2 are both forward biased and hence current flows from A to B.
D. D1 and D2 are both reverse biased and hence no current flows from A to B and vice versa.

Answer:

The answer is the option (b)
D2 is forward biased and D1 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_{nms}\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 outpu

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,
\\V_{AB}-0.3= [ ( r_{1} +r_{2}) 10^{3} ]\times ( 0.2 \times 10^{-3} ) [ \ because\ V_{AB}=ir ]\\ = [( 5+5 ) 10^{3}]\times ( 0.2 \times 10^{-3} )\\ =10\times 10^{3}\times 0.2\times 10^{-3}-2\\ \Rightarrow V_{AB}=2+0.3=2.3V


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

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.

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}=l 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.

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: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: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

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
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
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.
1535axaxa1535axaxa

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: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 )
a-40

The topics and subtopics covered in this chapter are as follows:

Main Subtopics of NCERT Exemplar Class 12 Physics Solutions Chapter 14 Semiconductor Electronics: Materials, Devices, and Simple Circuits

  • Introduction
  • Classification Of Metals, Conductors, And Semiconductors
  • Intrinsic Semiconductor
  • Extrinsic Semiconductor
  • P-n Junction
  • P-n Junction formation
  • Semiconductor Diode
  • p-n junction diode under forward bias
  • p-n junction diode under reverse bias
  • Application Of Junction Diode As A Rectifier
  • Special Purpose P-n Junction Diodes
  • Zener diode
  • Optoelectronic junction devices
  • Junction Transistor
  • Transistor: structure and action
  • Basic transistor circuit configurations and transistor characteristics
  • Transistor as a device
  • Transistor as an Amplifier (CE-Configuration)
  • Feedback amplifier and transistor oscillator
  • Digital Electronics And Logic Gates
  • Logic gates
  • Integrated Circuits.
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What will the students learn NCERT Exemplar Class 12 Physics Solutions Chapter 14 Semiconductor Electronics: Materials, Devices, and Simple Circuits?

Semiconductors help in equipment control in a variety of fields, such as contributing to road safety by improving automobile efficiency, operating air conditioners, and smoothly changing temperature and in medical centres for laser treatments. In addition to this, semiconductor technology has successfully achieved efficient techniques of energy-saving, system efficiency, miniaturisation of global warming, which in turn helps to conserve the global environment and to live a safe and comfortable life. A popular semiconductor is silicon. It is an important component of the IC, which is an integrated circuit of transistors. Different types of transistors are used as fundamental working concepts behind logic gates, which in turn are used to draft and design electrical and digital circuits. The transistors in analogue circuits, which use semiconductors as a protagonist in building their structure, work as responders to a range of input and output. Some Analog circuits include amplifiers and oscillators. So Class 12 Physics solutions NCERT Exemplar chapter 14 is an important part of Class 12 curriculum

NCERT Exemplar Class 12 Physics Chapter Wise Links

Important Topics To Cover For Exams From NCERT Exemplar Class 12 Physics Solutions Chapter 14

· The devices that use semiconductors have assembled a wide area of usage because of their reliability, compactness, accuracy, and power efficiency, with considerably cheaper costs. As individual operating components, they have found use in optical sensors, power devices, light emitters, and lasers.

· By using NCERT Exemplar Class 12 Physics solutions chapter 14, students will be able to demonstrate the movement of electrons and holes inside a semiconductor, study the idea of doping, the types of doping, the formation of diodes and their applications, and the setting up of LEDs' in a circuit.

· Through NCERT Exemplar Class 12 Physics chapter 14 solutions, we would learn about the materials used as semiconductors, common types of semiconductors used in a wide array of fields, their working principles, junction formats, transistors, and a variety of digital applications of the same.

NCERT Exemplar Class 12 Solutions

Also, check the NCERT solutions of questions given in book

Also read NCERT Solution subject wise

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Frequently Asked Question (FAQs)

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

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

2. What is the weightage of the chapter in NEET exam?

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.

3. Whether the NCERT Exemplar Class 12 Physics Solutions Chapter 14 is useful for exam.

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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Questions related to CBSE Class 12th

Have a question related to CBSE Class 12th ?

Here are some options you can explore to get admission in a good school even though admissions might be closed for many:

  • Waitlist: Many schools maintain waitlists after their initial application rounds close.  If a student who secured a seat decides not to join, the school might reach out to students on the waitlist.  So, even if the application deadline has passed,  it might be worth inquiring with schools you're interested in if they have a waitlist and if they would consider adding you to it.

  • Schools with ongoing admissions: Some schools, due to transfers or other reasons, might still have seats available even after the main admission rush.  Reach out to the schools directly to see if they have any open seats in 10th grade.

  • Consider other good schools: There might be other schools in your area that have a good reputation but weren't on your initial list. Research these schools and see if they have any seats available.

  • Focus on excelling in your current school: If you can't find a new school this year, focus on doing well in your current school. Maintain good grades and get involved in extracurricular activities. This will strengthen your application for next year if you decide to try transferring again.


Best CBSE schools in Delhi: Click Here


In India, the design and coding fields offer exciting career options that can leverage your interest in both. Here's how you can navigate this path:

Choosing Your Stream:

  • Graphic Design Focus: Consider a Bachelor's degree in Graphic Design or a design diploma. Build a strong portfolio showcasing your creative skills. Learn the basics of HTML, CSS, and JavaScript to understand web development better. Many online resources and bootcamps offer these introductory courses.

  • Coding Focus: Pursue a Computer Science degree or a coding bootcamp in India. These programs are intensive but can equip you with strong coding skills quickly. While building your coding prowess, take online courses in graphic design principles and UI/UX design.

Engineering Subjects (for a Degree):

  • Information Technology (IT): This offers a good mix of web development, networking, and database management, all valuable for web design/development roles.

  • Human-Computer Interaction (HCI): This is a specialized field that bridges the gap between design and computer science, focusing on how users interact with technology. It's a perfect choice if you're interested in both aspects.

  • Passing NIOS in October 2024 will make you eligible for NIT admissions in 2025 . NIT admissions are based on your performance in entrance exams like JEE Main, which typically happen in January and April. These exams consider the previous year's Class 12th board results (or equivalent exams like NIOS).

Here's why 2025 is more likely:

  • JEE Main 2024 Admissions: The application process for NITs through JEE Main 2024 is likely complete by now (May 2024). They consider your 2023 Class 12th marks (CBSE in this case).
  • NIOS Results: Since NIOS results typically come out after the NIT admission process, your October 2024 NIOS marks wouldn't be available for JEE Main 2024.

Looking Ahead (2025 Admissions):

  • Focus on JEE Main: Since you have a computer science background, focus on preparing for JEE Main 2025. This exam tests your knowledge in Physics, Chemistry, and Mathematics, crucial for engineering programs at NITs.
  • NIOS Preparation: Utilize the time between now and October 2024 to prepare for your NIOS exams.
  • Eligibility Criteria: Remember, NITs typically require a minimum of 75% marks in Class 12th (or equivalent) for general category students (65% for SC/ST). Ensure you meet this criteria in your NIOS exams.

Yes, scoring above 99.9 percentile in CAT significantly increases your chances of getting a call from IIM Bangalore,  with your academic background. Here's why:

  • High CAT Score: A score exceeding  99.9 percentile is exceptional and puts you amongst the top candidates vying for admission. IIM Bangalore prioritizes  CAT scores heavily in the shortlisting process.

  • Strong Academics: Your 96% in CBSE 12th and a B.Tech degree demonstrate a solid academic foundation, which IIM Bangalore also considers during shortlisting.

However, the shortlisting process is multifaceted:

  • Other Factors: IIM Bangalore considers other factors beyond CAT scores, such as your work experience (if any), XAT score (if you appear for it), academic diversity, gender diversity, and performance in the interview and Written Ability Test (WAT) stages (if shortlisted).

Here's what you can do to strengthen your application:

  • Focus on WAT and PI: If you receive a shortlist, prepare extensively for the Written Ability Test (WAT) and Personal Interview (PI). These stages assess your communication, soft skills, leadership potential, and suitability for the program.

  • Work Experience (if applicable): If you have work experience, highlight your achievements and how they align with your chosen IIM Bangalore program.

Overall, with a stellar CAT score and a strong academic background, you have a very good chance of getting a call from IIM Bangalore. But remember to prepare comprehensively for the other stages of the selection process.

hello,

Yes you can appear for the compartment paper again since CBSE gives three chances to a candidate to clear his/her exams so you still have two more attempts. However, you can appear for your improvement paper for all subjects but you cannot appear for the ones in which you have failed.

I hope this was helpful!

Good Luck

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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

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

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

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 .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

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.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

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

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

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

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