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

Question:2

In Fig. 14.1, Vo is the potential barrier across a p-n junction, when no battery is connected across the junction

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 option (b)
The height of a potential barrier increases when a 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. $D_1$ 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 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?


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

Answer:

As a p-n junction diode will conduct during the positive half cycle, only during the negative half cycle diode is reverse-biased. During this diode will not give any output, so the potential difference across the capacitor is the 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 an 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 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.


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 option (c)

would be like a half-wave rectifier with negative cycles in the 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


A. 1.3 V
B. 2.3 V
C. 0
D. 0.5 V

Answer:

The answer is option (b)
Let us consider the figure (b)given above in the problem. Suppose the potential difference between A and B is V_AB
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

A.

B. C. D.

Answer:

The answer is option (c)
(c) In this problem, the input C of the OR gate is an output of the 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 by

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 semiconductor between two N-type semiconductors or by sandwiching a thin layer of N-type semiconductor between two P-type semiconductors

E- Emitter(emits majority charge carriers)
C- Collects the majority charge carriers
B- Base (provide proper interaction between E and C)

Question:11

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


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 a base-biased common emitter transistor are followed by 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 the N-region and the P-region is different. Therefore, due to diffusion, the neutrality of both p-type and n-type semiconductors is disturbed. Hence, a layer of negatively charged ions appears at the p junction, while positive ions appear at the junction in the N-crystal.
The thickness of the layer (depletion) is 1 micron = 10-6 m.
Width of depletion layer ∞ 1/Dopping
And the depletion layer is directly proportional to the temperature.

Question:14

What happens during the 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 acts as a voltage regulator and is used as an ordinary diode.

The zener diode in the reverse-biased mode offers a 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 R_s is changed; therefore, the 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 a 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 the 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 the negative terminal of the battery is connected to the P-crystal.

In reverse biasing, ionisation 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: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 energy gaps of 1.1 eV and 0.7 eV, which makes them semiconductors. The gaps in energy are related to their individual atomic size, which is responsible for making one an insulator, a conductor or a 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).

Answer:

The diode acts as a half-wave rectifier; it offers low resistance when forward-biased and high resistance when reverse-biased.

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 the DC supply voltage is 10V?
(ii) If the DC supply voltage is 5V?

Answer:

The ratio of the output signal voltage to the input signal voltage is known as the total voltage amplification.
According to the problem, voltage gain in X, V_x =10
Voltage gain un Y; V_y=20
Voltage gain un Z; V_Z=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 the DC supply voltage is 10 V, then the output is 6 V since the theoretical gain is equal to the practical gain, i.e output can never be greater than 6V
(ii) If DC supply voltage is 5 V, i.e, V_cc=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 a CE transistor amplifier, the power gain is extremely high. Here, the extra power required for the amplification is gained from a DC source. Therefore, the circuit does not violate the law of conservation.

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:

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?

Answer:

I_bR₁ + V_be = V_bb

Base current = I_b = V_bb – V_be/R₁

I_b is inversely proportional to R₁

Hence, if R₁ is increased, then I_b 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:

So the OR gate gives the desired output

Question:27

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

Answer:

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.

Answer:

This is 'AND' Gate and its characteristics are as follows:

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 fluctuates between 3V and 7V, what should be the value of Rs for safe operation (Fig.14.12)?

Answer:

According to the problem, Power = 1 Watt
Zener breakdown voltage V_z=5V
Minimum voltage V_min=3V
Maximum voltage V_max=7V
We know, P=VI
So current $I_{z_{max}}=\frac{P}{V_{z}}=\frac{1}{5}=0.2A$
For safe operation, R_s 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?

Answer:

According to the problem forward biases resistance = $25 \Omega$ and reverse biases resistance=$\infty$
As shown in the figure, the diode in branch CD is in reverse bias, which has 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 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 are the 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, V_be is zero and V_ce is also zero. Find the values of I_b, I_c and $\beta$

Answer:

According to the problem V₁ =10V, Resistance R_B=400k$\Omega$, V_BE=0, V_CE=0 and R_c=3k$\Omega$
V_i-V_BE=R_BI_{B
$I_{B}=\frac{Voltage\: across\: R_{B}}{R_{B}}\\ =\frac{10}{400 \times 10^{3}}=25 \times 10^{-6}A=25\mu A$}
Voltage across R_C=10V
V_CC-V_CE=I_CR_C
$I_{C}=\frac{Voltage\: across\: R_{C}}{R_{C}}\\ =\frac{10}{3\times 10^{3}}=3.33 \times 10^{-3}A=3.33mA\\$
$ \beta =\frac{I_{C}}{I_{B}}=\frac{3.33 \times 10^{-3}}{25 \times 10^{-6}}\\ =1.33 \times 10^{2}=133$

Question: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.


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).

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 the 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.

Answer:

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

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:

$n_{e}=N_{D}=10^{-6}\times 5 \times 10^{28}atoms/m^{3}\\ =5 \times 10^{22}/m^{3}$
The number of minority carriers (holes) in an n-type wafer is
$n_{h}=\frac{n_{i}^{2}}{n_{e}}=\frac{\left ( 1.5 \times 10^{16} \right )^{2}}{5 \times 10^{22}}=0.45 \times 10^{10}/m^{3}$
A p-type wafer is created with a number of holes, when Boron is implanted in the 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 the p-type wafer are
$n_{h}=\frac{n_{i}^{2}}{n_{e}}=\frac{\left ( 1.5 \times 10^{16} \right )^{2}}{1 \times 10^{25}}=2.25 \times 10^{7}/m^{3}$
(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:

It is represented by the following logical 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 logical 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 can be obtained as the 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 the output of AND gate II, for which one input is of A, and the other input is of B through NOT gate.
Now, Y can be obtained as output from OR, where Y₁ and Y₂ are the inputs of the OR gate
Thus, the logic circuit of this relation is given below

Question:38

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


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).

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

(ii) when A is negative and B is positive

(iii) When B is negative and C is positive

(iv) When B is positive and C is negative

(v) When A is positive and C is negative

(vi) When A is negative and C is positive

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

Question:39

For the transistor circuit shown in Fig.14.19, evaluate V_E, R_B, R_E given I_C = 1 mA, V_CE = 3V, V_BE = 0.5 V and V_CC = 12 V, $\beta$ = 100.

Answer:

As we know, the base current is very small,

$
\begin{aligned}
& \quad I_C=I_E \\
& R_C=7.8 \mathrm{k} \Omega \\
& \text { from the 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 R_C.

Answer:

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



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