Topper’s Package Physics - XII Unit SEMICONDUCTOR … · 2020. 8. 8. · Two junction diodes one of Germanium (Ge) and other of Silicon (Si) are connected as shown in fig. to a

Topper’s Package Physics - XII Permutation and CombinationsSemiconductor Devices

103

1. SEMICONDUCTOR & PN JUNCTION

1. Conduction electrons have more mobility thanholes because they(a) are lighter(b) have negative charge(c) need less energy to move them(d) experience collisionsless frequency

2. Fermi energy is the maximum energypossessed by any electron at(a) 273°K (b) 0°K(c) 300°K (d) none of these

3. At absolute zero of temperature, theelectrical conductivity of a pure semiconductoris(a) zero (b) about 3 110 ( )m (c) about 7 110 ( )m (d) infinity

4. With increase in temperature, the electricalconductivity of intrinsic semiconductor(a) increases(b) decreases(c) first increases and then decreases(d) first decreases and then increases

5. A piece of copper and another of germaniumare cooled from room temperature of 80°K.The resistance of(a) each of them increases(b) each of them decreases(c) copper increases and germanium

decreases(d) copper decreases and germanium

increases6. Two identical P–N junction may be connected

in series with a battery in three ways. Thepotential drops across the two P–N junctionsare equal in

P N

CIRCUIT 1

N P

CIRCUIT 2

P NP N

CIRCUIT 3

N PN P

(a) circuit 1 and circuit 2(b) circuit 2 and circuit 3(c) circuit 3 and circuit 1(d) circuit 1 only

7. The electrical conductivity of asemiconductor increases when a radiation ofwavelength shorter than 2480 nm is incidenton it. The band gap (in eV) for thesemiconductor is(a) 0.9 (b) 0.7(c) 0.5 (d) 1.1

8. Which of the following statements is NOT true(a) the resistance of intrinsic

semiconductors decrease with increase oftemperature

(b) Doping pure Si with trivalent impuritiesgive P-type semiconductors

(c) The majority carries is N-typesemiconductors are holes

(d) A P–N junction can act as a semiconductordiode

9. The probability of electrons to be found in theconduction band of an intrinsic semiconductorat a finite temperature:(a) decreases exponentially with increasing

band gap(b) increases exponentially with increasing

band gap(c) decreases with increasing temperature(d) is independent of the temperature and the

band gap10. When a pn junction diode is forward biased, the

flow of current across the junction is mainlydue to :(a) drift of charges(b) diffusion of charges(c) both drift and diffusion of charges(d) depends on the nature of the material

SEMICONDUCTOR DEVICES

Unit 8


104

11. In the ratio of the concentration of electronsthat of holes in a semiconductor is 7/5 and theratio of currents is 7/4 then what is the ratioof their dift velocities(a) 4/7 (b) 5/8(c) 4/5 (d) 5/4

12. Carbon, silicon and germanium have fourvalence electrons each. At room temperaturewhich one of the following statements in mostappropriate ?(a) The number of free electrons for conduction

is significant in Si and Ge but small in C(b) The number of free conduction electrons

is significant in C but small in Si and Ge(c) The number of free conduction elections

is negigibly small in all the three(d) The number of free electrons for conduction

is dignificant in all the three13. If no external voltage is applied across p-n

junction, there would be(a) no electric field across the junction(b) an electric field pointing from n-type to p-

type side across the junction(c) an electric field pointing from p-type to n-

type side across the junction(d) a temporary electric field during formation

of p-n junction that would subsequentlydisppear

14. A transistor has three impurity regions. All thethree regions have different doping level. Inorder of increasing doping level, the regions are(a) emitter, base and collector(b) collector, base and emitter(c) base, emitter and collector(d) base, collector and emitter

15. The V- I characteristic ofa diode is shown in thefigure. The ratio offorward to reverse biasresistance is(a) 100(b) 105

(c) 10(d) 10–5

16. In an unbiased n-p junction electrons diffusefrom n-region to p-region becasue(a) holes in p-region because(b) electrons travel across the junction due to

potential difference(c) electron concentration in n-region is more

as compared to that in p-region

(d) only electrons move from n to p region andnot the vice-versa

17. In the depletion region of an unbiased P-Njunction diode there are :(a) only electrons(b) only holes(c) both electrons and holes(d) only fixed ions

18. A pure Si crystal 5 × 1022 atoms m–3. It id dopedby 1 ppm concentration of pentavelent As. Thenumber of holes i

2i p en n n (Take ni = 1.5 ×

1016 m–3)(a) 9 34.5 10 m (b) 6 34.5 10 m(c) 9 32.5 10 m (d) 6 32.5 10 m

19. In n-type semiconductor when all donor statesare filled, then the net charge density in thedonor states becomes(a) 1 (b) > 1(c) < 1, but not zero (d) zero

20. The number density of electrons and holes inpure silicon at 27°C are equal and its value is2.0 × 1016 m–3. On doping with indium the holedensity increases to 4 .5 × 1022 m–3, theelectron density in doped silcon is(a) 10 × 109 m–3

(b) 8.89 × 109 m–3

(c) 11 × 109 m–3

(d) 16.78 × 109 m–3

2. DIODES & APPLICATION

21. In a full wave rectifier with input frequency50 Hz the ripple in the output is mainly ofthe frequency (in Hz)(a) 25 (b) 50(c) 100 (d) none of these

22. A full wave rectifier along with the output isshown in figure. The contribution(s) from thediode 1 is/are

V

O t

Upper

Lower-1

Output

V A B C D D C B A Output

(a) C (b) A, C(c) B, D (d) A, B, C, D


105

23. The output current I versus time (t) curve ofa rectifier is shown in the figure. The averagevalue of the output current in this case is(a) zero

(b) 02I

I0 t

I

(c) 02 I

(d) 0I

24. Which of the junction diode shown in fig. isforward biased?

(a)

R

– 5V

– 10V

(b)

R

+ 5V

+ 10V

(c)

R

– 10V (d)

R

+ 10V

25. Two junction diodes one of Germanium (Ge)and other of Silicon (Si) are connected asshown in fig. to a battery of emf 12 V and aload resistance of 10 kW. The germaniumdiode conducts at 0.3 V and silicon diode at 0.7V. When a current flows in the circuit thepotential of terminal y will be

10k

Ge

Si

y

12V

(a) 12 V (b) 11 V(c) 11.3 V (d) 11.7 V

26. A sinusoidal voltage of peak value 200 voltis connected to a diode and resistor R in thecircuit shown so that half wave rectificationoccurs. If the forward resistance of the diodeis negligible compared to R, the rms voltageacross R is approximately

R Diode

200 volt peak value

(a) 100 (b) 283

(c) 200 (d) 2100

27. In the network shown in figure

(a) the potential difference across 2D is 5V

(b) Current through diode 1D is 1.25 A

(c) Current through diode 2D is 1.25 A(d) Current through resistor equals 1.25 A

28. In the diagram, the input is across theterminals A and C and the output is across Band D. Then the output is

B

C

D

A

(a) zero(b) same as the input(c) full wave rectifier(d) half wave rectifier

29. When a P–N junction diode is reverse biased,the flow of current across the junction is mainlydue to(a) diffusion of charges(b) depends upon the nature of material(c) drift of charges(d) both drift and diffusion of charges

30. Select the correct statements from thefollowing(a) a diode can be used as rectifier(b) a triode can be used as rectifier(c) the current in a diode is always

proportional to the applied voltage(d) the linear portion of the I–V characteristic

of a triode is used for amplification withoutdistortion


106

31. Figure shows a full wave bridge rectifier circuit.The input a.c. is connected across :

A

D

C

B

(a) A and C (b) B and D(c) A and B (d) C and D

32. In the arrangement shown is figure, thecurrent through diode is :

–4V –1V

P N R = 300

(a) 10 mA (b) 1 mA(c) 20 mA (d) Zero

33. The circuit has two oppositely connected idealdiodes in parallel. What is the current flowingin the circuit,

D1

D2

12V

(a) 1.33 A (b) 1.71 A(c) 2.00 A (d) 2.31 A

34. If in a p-n junction diode, a square input signalof 10 V is applied as shown

DRl

5V

–5V

Then the output signal across lR will be

(a)+5V

(b)10V

(c)–10V

(d)–5V

35. A sinusoidal voltage of 200 volt, 50 Hz is appliedto diode and capacitor C in the circuit shown.The final potential difference across capacitorC is(a) 200 V(b) 283 V

Diode

~ 200 V50 Hz

C

(c) 141.4 V(d) 400 V

36.10 V

yx

The potential difference between x and y is(a) 10 V (b) 5 V(c) 2.5 V (d) 0 V

37. If in a p-n junction diode a square voltage isapplied as shown

D

R

+5V

–5V

RP

Q

A

B

O O

Then the output voltage signal between P andQ is

(a)+2.5V

(b)–2.5V

(c)

+2.5V

–5V

(d)–2.5V

+5V

38.3A

BAR

6V

The potential difference between the points Aand B is(a) – 21 V (b) + 21 V(c) 15 V (d) – 15 V

39. Assuming that the junction diode is ideal, thecurrent in the circuit is(a) zero(b) 1 mA

1V 2V

(c) 10 mA(d) 30 mA

40. In the given circuit consisting of resistor anddiode the potential difference between A and Bis

30V

A

B


107

(a) 10 V (b) 20 V(c) 30 V (d) 0 V

41. The current gain of a common emitter amplifieris 69. If the emitter current is 7.0 mA, collectorcurrent is(a) 0.69 mA (b) 6.9 mA(c) 69 mA (d) 9.6 mA

42. The value of the resistor, Rs, needed in the dcvoltage regulator circuit shown here, equals

(a) (Vi – VL)/nIL (b) (Vi + VL )/nIL

(c) (Vi – VI )/(n+1)IL (d) (Vi + VL )/(n+1)IL

43. A 2 V battery connected across AB as shown inthe figure. The value of the current supplied bythe battery when in one case battery’s postiveterminal is connected to A and in other casewhen positive terminal of battery is connectedto B will respectively be(a) 0.2 A and 0.1 A

(b) 0.4 A and 0.2 A (c) 0.1 A and 0.2 A(d) 0.2 A and 0.4 A

44. Of the diodes shown in the following figures,which one is reverse biased ?

(a) (b)

(c) (d)

45. The V-I characteristic of a silicon diode is shwonin figure. The resistance of the diode at ID = 15mA is

(a) 5 (b) 10(c) 2 (d) 20

46. Which of the junction diodes shown below areforward biased?

(a) (b)

(c) (d)

47. The circuit shown in the figure contains twodiodes each with a forward resistance of

30 and with infinite backward resistance. Ifthe battery is 3 V, the current through the 50resistance (in ampere) is

(a) zero (b) 0.01(c) 0.02 (d) 0.03

48. A forward biased diode is

(a)

(b)

(c)

(d)

3. JUNCTION TRANSISTORS & APPLICATION

49. In an n-p-n transistor the collector current is10 mA. If 98% of the electrons emitted reachthe collector, the base current is

(a) 0.8 mA (b) 0.2 mA(c) 0.1 mA (d) 0.05 mA

50. In a n-p-n transistor circuit, the collectorcurrents is 10 mA. If 90 % of the electronsemitted reach the collector, the emittercurrent ( )EI and base current ( )BI are givenby

(a) 9 ; 1E BI mA I mA

(b) 1 ; 9E BI mA I mA

(c) 1 ; 1E BI mA I mA

(d) 1 ; 11E BI mA I mA


108

51. The current gain in a common emittertransistor is(a) equal to (b) more than 1(c) more than 1 (d) zero

52. In a transistor circuit, the collector current is50 mA and the base current is 1mA. Thecurrent gain is(a) 0.51 (b) 0.82(c) 0.98 (d) 1.02

53. In a transistor amplifier, the two ac current

gains and are defined as e

E

II

and

e

B

II

. The relation between and is

(a)

1

(b)

1

(c)

1 (d)

154. The LED, or the light emitting diode,

(a) is made from the semiconductingcompound gallium arsenide phosphide

(b) emits light when forward biased(c) is made from one of the two basic

semiconducting materials, silicon orgermanium

(d) geometry of the transistor55. For a transistor, the current amplification

factor is 0.8. The transistor is connected incommon emitter configuration. The change inthe collector current when the base currentchanges by 6 mA is(a) 6 mA (b) 4.8 mA(c) 24 mA (d) 8 mA

56. The difference in the working of an amplifierand a step up transformer is :(a) Amplifier also increases power which is

not possible with transformer(b) Amplifier decreases the power whereas

the transformer increases the power(c) Amplifier keeps the power constant

whereas the transformer decreases thepower

(d) Amplifier keeps the power constantwhereas the transformer increases thepower

57. For detecting light intensity we use :(a) photodiode in reverse bias(b) photodiode in forward bias

(c) LED in reverse bias(d) LED in forward bias

58. A common emitter amplifier is operated with ad.c. supply ccV 20 V . The gain of theamplifier is 100. If a.c. voltage of 2 V peak topeak is applied at the input, the maximumoutput voltage peak to peak will be :(a) 20 V (b) 200 V(c) 0 V (d) 0.02 V

59. In the given transistor circuit, the base currentis 35 µA. The value of bR is

B

7V

RL

A E C

Rb

(a) 100 k (b) 200 k(c) 300 k (d) 400 k

60. In the circuit shown here the transistor usedhas current gain 100 . What should be thebias resistor bR so that CEV 5V ? (neglect

BEV )

RB

E

C

+

–

CE

10V

B

1K

(a) 32 10 (b) 3200 10 (c) 61 10 (d) 500

61. In a n-p-n transistor(a) holes moves from emitter to base(b) holes move from base to collector(c) negative charge moves from emitter to

base(d) negative charge moves from collector to

base62. A common emitter amplifier has voltage gain

of 50, an input unpedance of 100 and anoutput inpedence of 200 . The power gain ofthe amplifier is(a) 1000 (b) 1250(c) 100 (d) 2250


109

63. A transistor is operated in common emitterconfiguration at constant collector voltage

1.5CV V such that a change in the basecurrent from 100 µA to 150 µA produced achange in the collector current from 5 mA to10 mA. The current gain ( ) is(a) 75 (b) 100(c) 50 (d) 67

64. A transistor, when connected in CE mode, has(a) low input resistance and a low output

resistance(b) a high input resistance and a high output

resistance(c) a high input resistance and a low output

resistance(d) a low input resistance and a high output

resistance65. A light emitting diode (LED) has a voltage drop

of 2 V across it and passes a current of 10 mA.When it operates with 6 V battery through alimiting resistor R, the value of R is(a) 40 (b) 4(c) 200 (d) 400

66. In a common base amplifier, the phasedifference between the input voltage signal andoutput voltage is

(a) 0 (b)2

(c)4

(d)

4. LOGIC GATES67. In a 2-input logic circuit both inputs are zero.

If output is 1, the gate would be(a) NAND (b) OR(c) XOR (d) NOT

68. In Boolean Algebra 11 , equals(a) 1 (b) 0(c) 2 (d) none of these

69. What is the boolean expression for the gatecircuit shown in figure?

X

A

B

(a) XB.A (b) XBA

(c) A . B = X (d) XBA 70. The two input terminals of NAND gate are

shorted. This gate is equivalent to a :(a) OR-gate (b) AND-gate(c) NOT-gate (d) XOR-gate

71. Logic gates X and Y have the truth tables shownbelow

XQ

PR YP R

P Q R

0 0 01 0 0

0101 1 1

P R

0 1

1 0

When the output of X is connected to the inplutof Y, the resulting combination is equivalentto a single(a) OR gate (b) NOR gate(c) NAND gate (d) NOT gate

72. Name the gate represented by the combinationof gates shown in the following figure(a) OR(b) AND

A

B

Y

(c) NAND(d) NOR

73. Which of the following is the output of thecombination of the gates shown in the figure ?

YA

B

(a) ( )A A B (b) ( )A B A (c) A A B (d) A A B

74. If input are 0 ; 0A B , 0 ; 1A B then output at y will be

YA

B

(a) 0, 0 (b) 0, 1(c) 1, 0 (d) 1, 1

75. To get an output 1 from the logic circuit of thefollowing figure, the input ABC must be

AB Y

C

(a) 101 (b) 001(c) 110 (d) 010

76. For the given logic circuit 1, 0, 1A B C and 0D , the output y is

AB Y

CD

(a) 1 (b) 0(c) can be both (d) none of these


110

77. If a, b, c d are inputs to a gate and x is its output,then, as per the following time graph, the gateis:

(a) NOT (b) AND

(c) OR

(d) NAND78. The truth table given in figure represents

A B Y0 0 00 1 11 0 11 1 1

(a) OR-Gate (b) NAND- Gate(c) AND - Gate (d) NOR-Gate

79. The combination of NAND gates is shown infigure. The equivalent circuit is(a) ANDgate (b) NOR gate(c) OR gate(d) NOT gate

80. Boolean algebra is essentially based on(a) number (b) truth(c) logic (d) symbol

81. The output of given logic circuit is(a) A · (B + C) (b) A · (B · C)(c) (A + B) · (A + C)(d) A + B + C

82. In the circuit below , A and B represents twoinputs and C represents the output. the circuitrepresents

(a) AND gate (b) NOR gate(c) OR gate (d) NAND gate

83. The symbolic representation of four logic gatesare given below. The logic symbols for OR, NOtand NAND gates are respectively

(a) (iv), (i), (iii) (b) (iv), (ii), (i)(c) (i), (iii), (iv) (d) (iii), (iv), (ii)

84. The given truth table is for which logic gate?(a) NAND A B Y

1 1 00 1 11 0 10 0 1

(b) XOR

(c) NOR

(d) OR85. In Boolean algebra if A = 1 and b = 0, then the

value of A B is(a) A (b) A.B(c) A + B (d) both (a) and (c)

86. The following figure shows a logic gate circuitwith two inputs A and B and the output Y. Thevoltage waveforms of A, B and Y are as given.

The logic gate is(a) NOR gate (b) OR gate(c) AND gate (d) NAND gate

87. What will be input of A and B for the Boolean

expression (A B).(A ·B) 1?(a) (0, 0) (b) (0, 1)(c) (1, 0) (d) (1, 1)

88. The truth table for the following logic circuit is

(a)

A B Y0 0 00 1 11 0 11 1 0

(b)

A B Y0 0 00 1 11 0 11 1 1

(c)

A B Y0 0 10 1 01 0 11 1 0

(d)

A B Y0 0 10 1 11 0 01 1 1


111

89. The circuit given in figure, is equivalent to

(a) AND gate (b) OR gate(c) NOT gate (d) NAND gate

90. Select the output Y of the combination of gatesshown in figure for inputs A = 1, B = 0; A = 1, B =1 and A = 0, B = 0 respectively.

(a) (0, 1, 1) (b) (1, 0, 1)(c) (1, 1, 1) (d) (1, 0, 0)

91. Which of the following truth tables correspondsto NAND gate ?

(a) (iv) (b) (iii)(c) (ii) (d) (i)

92. The decimal equivalent of the binary number(11010.101)2 is(a) 9.625 (b) 25.265(c) 26.625 (d) 26.265

93. Study the circuit’ shown in the figure. Namethe gate that the given circuit resembles.

(a) NAND (b) AND(c) OR (d) NOR

5. INTEGER TYPE QUESTIONS94. The circuit shown has two diodes each having

resistance 50 and reverse resistance . Abattery (emf = 6V) and a resistance 100 isconnected. The current through 100 resistoris n × 10–2 amp. Then find the value of n.

95. For given combination of gates. If input at A,B, C is A = B = C the output at given circuit is

O

AB G1

C

96. Mass law in semiconductor is given byn.P = ni

a then find the value of a.97. The circuit has two oppositely connected diodes

in parallel what is the current flowing in thecircuit ?

D1 D2

3 2

12V +

4

98. In the circuit shown the current in 5 k,resistor is

+15V

+25V

5k

99. A potential barrier of 0.4 V exists across a p-njunction. A constant electric field of 106 V/mexists in depletion region. The width ofdepletion region is 4 × 10–n, then value of n is

100. In the given circuit the diodes are ideal. Theequivalent resistance between A and B is 10a.Then find the value of a.

20 20

A B

20 20

101. A signal of 20 mV is applied to common emittertransistor amplifier circuit. Due to this, thechange in base current and change in collectorcurrent are 20 µA and 2 mA. The loadresistance is 20 k. The voltage gain is 10x

then calculate the value of x.


112

1. SEMICONDUCTOR & PN JUNCTION1. c 2. b 3. a 4. a 5. d6. b 7. c 8. c 9. a 10. b11. d 12. a 13. b 14. d 15. d16. c 17. d 18. a 19. b 20. b

2. DIODES & APPLICATION21. c 22. c 23. b 24. a 25. d26. a 27. b 28. c 29. c 30. d31. b 32. d 33. c 34. a 35. b36. b 37. d 38. b 39. a 40. a41. b 42. b 43. b 44. c 45. b46. a 47. c 48. a

3. JUNCTION TRANSISTORS & APPLICATION49. b 50. d 51 b 52. b 53. c54. a, b 55. c 56. a 57. b 58. b59. b 60. b 61. c 62. b 63. b64. d 65. d 66. a

4. LOGIC GATES67. a 68. b 69. a 70. c 71. c72. c 73. a 74. a 75. a 76. a77. c 78. a 79. c 80. c 81. c82. c 83. b 84. a 85. d 86. d87. a 88. a 89. a 90. d 91. d92. c 93. b

5. INTEGER ANSWER TYPE94. 4 95. 1 96. 2 97. 2 98. 299. 7 100. 4 101. 3

Documents

Topper’s Package Physics - XII Unit SEMICONDUCTOR … · 2020. 8. 8. · Two junction diodes one of Germanium (Ge) and other of Silicon (Si) are connected as shown in fig. to a