Semiconductor Electronics Materials Devices And Simple Circuits
Chapter 14
SEMICONDUCTOR ELECTRONICS MATERIALS, DEVICES AND SIMPLE CIRCUITS
MCQ I
14.1 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: (d)14.2 In Fig. 14.1, $V_o$ is the potential barrier across a p-n junction, when no battery is connected across the junction
Fig. 14.1
(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.
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Answer: (b)14.3 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) $D_2$ 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.
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Answer: (b)14.4 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 capacitor?
(a) $220 V$.
(b) $110 V$.
(c) OV.
(d) $220 \sqrt{2} V$.
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Answer: (d)14.5 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 artifically created particle.
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Answer: (b)14.6 The output of the given circuit in Fig. 14.4.
(a) would be zero at all times.
Fig. 14.4
(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.
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Answer: (c)14.7 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) $0.5 V$
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Answer: ** (b)**14.8 Truth table for the given circuit (Fig. 14.6) is
(a)
| $A$ | $B$ | $E$ |
|---|---|---|
| $O$ | 0 | 1 |
| 0 | 1 | 0 |
| 1 | 0 | 1 |
| 1 | 1 | 0 |
(b)
| $A$ | $B$ | $E$ |
|---|---|---|
| 0 | 0 | 1 |
| 0 | 1 | 0 |
| 1 | 0 | 0 |
| 1 | 1 | 1 |
(c)
| $A$ | $B$ | $E$ |
|---|---|---|
| 0 | 0 | 0 |
| 0 | 1 | 1 |
| 1 | 0 | 0 |
| 1 | 1 | 1 |
(d)
| $A$ | $B$ | $E$ |
|---|---|---|
| 0 | 0 | 0 |
| 0 | 1 | 1 |
| 1 | 0 | 1 |
| 1 | 1 | 0 |
MCQ II
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Answer: (c)14.9 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.
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Answer: (a), (c)14.10 Consider an npn transitor 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.
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Answer: (a), (c)14.11 Figure 14.7 shows the transfer characteristics of a base biased CE transistor. Which of the following statements are true?
Fig. 14.7
(a) At $V_i=0.4 V$, transistor is in active state.
(b) At $V_i=1 V$, it can be used as an amplifier.
(c) At $V_i=0.5 V$, it can be used as a switch turned off.
(d) At $V_i=2.5 V$, it can be used as a switch turned on.
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Answer: (b), (c), (d)14.12 In a npn transistor circuit, the collector current is $10 mA$. 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 $2 mA$.
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Answer: (b), (c)14.13 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.
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Answer: (a), (b), (d)14.14 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 $(R_s)$ changes.
(c) The Zener resistance is constant.
(d) The resistance offered by the Zener changes.
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Answer: (b), (d)14.15 To reduce the ripples in a rectifier circuit with capacitor filter
(a) $R_L$ should be increased.
(b) input frequency should be decreased.
(c) input frequency should be increased.
(d) capacitors with high capacitance should be used.
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Answer: (a), (c), (d)14.16 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.
VSA
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Answer: (a), (d)14.17 Why are elemental dopants for Silicon or Germanium usually chosen from group XIII or group XV?
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Answer: The size of dopant atoms should be such as not to distort the pure semiconductor lattice structure and yet easily contribute a charge carrier on forming co-valent bonds with $Si$ or Ge.14.18 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?
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Answer: The energy gap for $Sn$ is $0 eV$, for $C$ is $5.4 eV$, for $Si$ is $1.1 eV$ and for Ge is $0.7 eV$, related to their atomic size.14.19 Can the potential barrier across a p-n junction be measured by simply connecting a voltmeter across the junction?
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Answer: No, because the voltmeter must have a resistance very high compared to the junction resistance, the latter being nearly infinite.14.20 Draw the output waveform across the resistor (Fig.14.8).
Fig. 14.8
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Answer:
14.21 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 $10 V$ ?
(ii) if de supply voltage is $5 V$ ?
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Answer: (i) $10 \times 20 \times 30 \times 10^{-3}=6 V$
(ii) If de supply voltage is $5 V$, the output peak will not exceed $V _{cc}=5 V$.
Hence, $V_0=5 V$.
14.22 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 voilate conservation of energy?
S.A
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Answer: No, the extra power required for amplified output is obtained from the DC source.14.23
(a)
(b)
Fig. 14.9
(i) Name the type of a diode whose characteristics are shown in Fig. 14.9 (A) and Fig. 14.9(B).
(ii) What does the point $P$ in Fig. (A) represent?
(iii) What does the points $P$ and $Q$ in Fig. (B) represent?
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Answer: (i) ZENER junction diode and solar cell.
(ii) Zener breakdown voltage
(iii) Q- short circuit current
P- open circuit voltage.
14.24 Three photo diodes D1, D2 and D3 are made of semiconductors having band gaps of $2.5 eV, 2 eV$ and $3 eV$, respectively. Which ones will be able to detect light of wavelength $6000 _A^{\circ}$ ?
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Answer: Energy of incident light photon
$h v=\frac{6.6 \times 10^{-34} \times 3 \times 10^{8}}{6 \times 10^{-7} \times 1.6 \times 10^{-19}}=2.06 e V$
For the incident radiation to be detected by the photodiode, energy of incident radiation photon should be greater than the band gap. This is true only for D2. Therefore, only D2 will detect this radiation.
14.25 If the resistance $R_1$ is increased (Fig. 14.10), how will the readings of the ammeter and voltmeter change?
Fig. 14.10
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Answer: $I_B=\frac{V _{B B}-V _{B E}}{R_1}$. If $R_1$ is increased, $I_B$ will decrease. Since $I_c=\beta I_b$, it will result in decrease in $I_C$ i.e decrease in ammeter and voltmeter readings.14.26 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.
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Answer: 
OR gate gives output according to the truth table.
| $A$ | $B$ | $C$ |
|---|---|---|
| 0 | 0 | 0 |
| 0 | 1 | 1 |
| 1 | 0 | 1 |
| 1 | 1 | 1 |
14.27 How would you set up a circuit to obtain NOT gate using a transistor?
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Answer: 
| Input | Output |
|---|---|
| $A$ | $A$ |
| 0 | 1 |
| 1 | 0 |
14.28 Explain why elemental semiconductor cannot be used to make visible LEDs.
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Answer: Elemental semiconductor’s band-gap is such that emissions are in IR region.14.29 Write the truth table for the circuit shown in Fig.14.11. Name the gate that the circuit resembles.
Fig. 14.11
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Answer: Truth table
| $A$ | $B$ | $Y$ |
|---|---|---|
| 0 | 0 | 0 |
| 0 | 1 | 0 |
| 1 | 0 | 0 |
| 1 | 1 | 1 |
AND Gate
14.30 A Zener of power rating $1 W$ is to be used as a voltage regulator. If zener has a breakdown of $5 V$ and it has to regulate voltage which fluctuated between $3 V$ and $7 V$, what should be the value of Rs for safe operation (Fig.14.12)?
Fig. 14.12
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Answer: $\quad I _{Z \max }=\frac{P}{V_Z}=0.2 A=200 mA$
$R_S=\frac{V_s-V_Z}{I _{Z \max }}=\frac{2}{0.2}=10 \Omega$
14.31 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 $I_1, I_2, I_3$ and $I_4$ ?
Fig. 14.13
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Answer: $I_3$ is zero as the diode in that branch is reverse bised. Resistance in the branch $AB$ and $EF$ are each $(125+25) \Omega=150 \Omega$.
As $AB$ and EF are identical parallel branches, their effective resistance is $\frac{150}{2}=75 \Omega$
$\therefore$ Net resistance in the circuit $=(75+25) \Omega=100 \Omega$.
$\therefore$ Current $I_1=\frac{5}{100}=0.05 A$.
As resistances of $AB$ and $EF$ are equal, and $I_1=I_2+I_3+I_4, I_3=0$
$\therefore I_2=I_4=\frac{0.05}{2}=0.025 A$
14.32 In the circuit shown in Fig. 14.14, when the input voltage of the base resistance is $10 V, V _{b e}$ is zero and $V _{ce}$ is also zero. Find the values of $I_b, I_c$ and $\beta$.
Fig. 14.14
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Answer: As $V _{\text{be }}=0$, potential drop across $R_b$ is $10 V$.
$\therefore I_b=\frac{10}{400 \times 10^{3}}=25 \mu A$
Since $V _{ce}=0$, potential drop across $R_c$, i.e. $I_c R_c$ is $10 V$.
$\therefore I_c=\frac{10}{3 \times 10^{3}}=3.33 \times 10^{-3}=3.33 mA$.
$\therefore \beta=\frac{I_c}{I_b}=\frac{3.33 \times 10^{-3}}{25 \times 10^{-6}}=1.33 \times 10^{2}=133$.
14.33 Draw the output signals $C_1$ and $C_2$ in the given combination of gates (Fig. 14.15).
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Answer:
14.34 Consider the circuit arrangement shown in Fig 14.16 (a) for studying input and output characteristics of npn transistor in CE configuration.
Select the values of $R_B$ and $R_C$ for a transistor whose $V _{B E}=0.7 V$, 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 $V _{C C}=V _{B B}=16 V$, also find the voltage gain and power gain of circuit making appropriate assumptions.
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Answer: From the output characteristics at point $Q, V _{C E}=8 V & I_C=4 mA$
$V _{C C}=I_C R C+V _{C E}$
$R_c=\frac{V _{C C}-V _{C E}}{I_C}$
$R_c=\frac{16-8}{4 \times 10^{-3}}=2 K \Omega$
Since,
$V _{B B}=I_B R_B+V _{B E}$
$R_B=\frac{16-0.7}{30 \times 10^{-6}}=510 K \Omega$
Now, $\quad \beta=\frac{I_C}{I_B}=\frac{4 \times 10^{-3}}{30 \times 10^{-6}}=133$
Voltage gain $=A_V=-\beta \frac{R_C}{R_B}$
$ \begin{aligned} & =-133 \times \frac{2 \times 10^{3}}{510 \times 10^{3}} \\ & =0.52 \end{aligned} $
Power Gain $=A_p=\beta \times A_V$
$=-\beta^{2} \frac{R_C}{R_B}$
$=(133)^{2} \times \frac{2 \times 10^{3}}{510 \times 10^{3}}=69$
14.35 Assuming the ideal diode, draw the output waveform for the circuit given in Fig. 14.17. Explain the waveform.
Fig. 14.17
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Answer: When input voltage is greater than $5 V$, diode is conducting
When input is less than $5 V$, diode is open circuit
14.36 Suppose a ’ $n$ ‘-type wafer is created by doping Si crystal having $5 \times 10^{28}$ atoms $/ m^{3}$ with $1 ppm$ 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.
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Answer: (i) In ’ $n$ ’ region; number of $e^{-}$is due to As:
$n_e=N_D=1 \times 10^{-6} \times 5 \times 10^{28}$ atoms $/ m^{3}$
$n_e=5 \times 10^{22} / m^{3}$
The minority carriers (hole) is $n_h=\frac{n_i^{2}}{n_e}=\frac{(1.5 \times 10^{16})^{2}}{5 \times 10^{22}}=\frac{2.25 \times 10^{32}}{5 \times 10^{22}}$
$n_h=0.45 \times 10 / m^{3}$
Similarly, when Boron is implanted a ‘p’ type is created with holes
$n_h=N_A=200 \times 10^{-6} \times 5 \times 10^{28}$
$=1 \times 10^{25} / m^{3}$
This is far greater than $e^{-}$that existed in ’ $n$ ’ type wafer on which Boron was diffused.
Therefore, minority carriers in created ‘p’ region
$ \begin{aligned} n_e=\frac{n_i^{2}}{n_h} & =\frac{2.25 \times 10^{32}}{1 \times 10^{25}} \\ & =2.25 \times 10^{7} / m^{3} \end{aligned} $
(ii) Thus, when reverse biased $0.45 \times 10^{10} / m^{3}$, holes of ’ $n$ ’ region would contribute more to the reverse saturation current than 2.25 $\times 10^{7} / m^{3}$ minority $e^{-}$of $p$ type region.
14.37 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=\overline{A} \cdot B+A \cdot \overline{B} $
Build this gate using AND, OR and NOT gates.
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Answer:
14.38 Consider a box with three terminals on top of it as shown in Fig. 14.18 (a):
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).
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
Fig. 14.18 (c)
(ii) when $A$ is negative and $B$ is positive
Fig. 14.18 (d)
(iii) When $B$ is negative and $C$ is positive
Fig. 14.18 (e)
(iv) When $B$ is positive and $C$ is negative
Fig. 14.18 (f)
(v) When $A$ is positive and $C$ is negative
Fig. 14.18 (g)
(vi) When $A$ is negative and $C$ is positive
Fig. 14.18 (h)
From these graphs of current - voltage characteristic shown in Fig. 14.18 (c) to (h), determine the arrangement of components between A, B and $C$.
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Answer:
14.39 For the transistor circuit shown in Fig.14.19, evaluate $V_E, R_B, R_E$ given $I_C=1 mA, V _{C E}=3 V, V _{B E}$ $=0.5 V$ and $V _{C C}=12 V$, $\beta=100$.
Fig. 14.19
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Answer: $I_C \approx I_E \therefore I_C(R_C+R_E)+V _{CE}=12 V$
$R_E=9-R_C=1.2 K \Omega$
Exemplar Problems-Physics
$\therefore V_E=1.2 V$
$V_B=V_E+V _{B E}=1.7 V$
$I=\frac{V_B}{20 K}=0.085 mA$
$R_B=\frac{12-1.7}{I_C / \beta+0.085}=\frac{10.3}{0.01+1.085}=108 K \Omega$
14.40 In the circuit shown in Fig. 14.20, find the value of $R_C$.
Fig. 14.20
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Answer: $I_E=I_C+I_B \quad I_C=\beta I_B$
$I_C R_C+V _{C E}+I_E R_E=V _{C C}$
$R I_B+V _{B E}+I_E R_E=V _{C C}$
From (3) $I_e \approx I_C=\beta I_B$
$(R+\beta R_E)=V _{C C}-V _{B E}, \quad I_B=\frac{V _{C C}-V _{B E}}{R+\beta R_E}=\frac{11.5}{200} mA$
From (2)
$R_C+R_E=\frac{V _{C C}-V _{C E}}{I_C}=\frac{V _{C C}-V _{C E}}{\beta I_B}=\frac{2}{11.5}(12-3) K \Omega=1.56 K \Omega$
$R_C=1.56-1=0.56 K \Omega$
