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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 51. |
Find the output y of the logic gate whose symbol Is shown in Fig. and give a truth table for the same. |
| Answer» It is an AND gate with three input A, B and C. Thus the output y is given by `y=A.(B.C)=(A.B).C`. The truth table for the same is shown in Fig. | |
| 52. |
State the rule used in the operation of AND gate. |
| Answer» The output of an AND gate assumes 1 (in level) if one all the inputs assume 1 (in level). | |
| 53. |
State the rule used in the operation of NOT gate. |
| Answer» The output of an NOT gate assumes 1 (in level) if input is 0 (in level) and vice versa. | |
| 54. |
State the rule used in the operation of OR gate. |
| Answer» The output of an OR gate assumes 1 in (level) if one or more inputs assume 1 (in level). | |
| 55. |
Name the 2-input logic gate, whose truth table is given here, If this logic gate is connected to a NOT gate. What will be the output when (i) `A=1, B=1` and (ii) `A=0, B=1`? `|{:(A,B,"Output"y),(0,0,1),(0,1,1),(1,0,1),(1,1,0):}:|` `|{:(A,B,"Out put"y,"Output of Not gate"),(1,1,0,1),(0,1,1,0):}|` |
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Answer» For the output y, the Boolean expression applicable is `y=bar(A.B)` , Hence, the table is of NAND gate. When output of NAND gate is connected to input of NOT gate, the output of NOT gate `=bar(bar(A.B))=A.B` (i.e. AND gate). |
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| 56. |
In which region transistor with `h_(fe)=80` operates as shown in Fig. Also find `R_(B)`. |
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Answer» Since `V_(CB) lt V_(c c )` and not equal to zero, therefore, Transistor operates in avtive region: Current in the collector circuit is `I_(c)=((9-5)V)/(2kOmega)=2mA` `h_(fe)=I_c/I_b` or `I_b=I_c/H_(fe)=(2mA)/80=2.5xx10^(-5)A` we know that `V_(BE)` in active region is `0.7V`, therfore, `R_(B)=((3.2-0.7)V)/((2.5xx10^(-5))A)=92xx10^(3)Omega=92kOmega` |
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| 57. |
The width of depletion region in a P-N junction diode is 500 nm and an intense electric field of `5xx10^(5)Vm^(-1)` is also found to exit init. Determine the height of the potential barrier. Also calculate the kinetic energy which a conduction electron must have inorder to diffuse from the n-side to p-side. |
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Answer» `V=Ed=(5xx10^(5))xx(500xx10^(-9))=0.25V` Max. K.E. of an electron to cross the potential barrier `V=eV=1.6xx10^(-19)xx0.25J` `=(1.6xx10^(-19)xx0.25)/(1.6xx10^(-19))eV=0.25 eV` |
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| 58. |
What is meant by depletion region in a junction diode? How is this region formed? |
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Answer» Depletion region is a region created around the p-n junction which is devoid of free charge carriers and has immobile ions. Regarding the formation of depletion region, refer to |
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| 59. |
Name the type in which the electronic circuit have been classified. |
| Answer» Analogue circuit and Digital circuit. | |
| 60. |
Find the Boolean expression of the output y in terms of the input A and B for the circuit shown in Fig.Name the gate formed. |
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Answer» The output from the upper AND gate `=(barA.B)`. The output from the upper AND gate `=(A.barB)` The output y of the OR gate is `y=(barA.B)+(A.barB)=Ao+B` In this gate if `A=0` and `B=1` or vice versa, the output `y=1` and if both A and B are identical, then the output `y=0`. This operation is also called Exclusive OR gate, designated EXOR gate. |
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| 61. |
Assertion: Two `P-N` junction diodes placed back to back, will work as a `NPN` transistor. Reason: The `P-N` junction of two `PN` junction diodes back to back will form the base of `NPN` transistor.A. if both the Assertion and Reason are true and the Reason is the correct explanation of Assertion.B. if both the Assertion and Reason are true but the reason in not a correct explanation of the Assertion.C. if the Assertion is true but the Reason is false.D. if both Assertion and Reason are false. |
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Answer» Correct Answer - D Two p-n junction placed back to back, cannot work as n-p-n transistor. Hence Assertion is wrong. Here, Reason is also wrong. |
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| 62. |
In the depletion region of an unbiased p-n junction diode, what are the charge carriers? |
| Answer» In the depletion region of an unbiased p-n junction diode, there are only fixed ions but no free charge carries. | |
| 63. |
How does the voltage gain vary with the frequency of the input signal for a transistor amplifier? |
| Answer» For a transistor amplifier, the voltage gain is low at low and high frequencies and constant high at mid frequencies. | |
| 64. |
For an intrinsic semiconductor, the statement//statements that hold good are:A. an intrinsic semiconductor is a perfect insulator at 0KB. the number of charge carriers varies with temperature in an exponential wayC. the number density of electrons in always more than the number density of holesD. the mobility of electrons is more than that of holes. |
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Answer» Correct Answer - A::B::D The choices , and are correct, whereas the choice is incorrect because in an intrinsic semiconductor, the number density of electrons is equal to that of holes. |
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| 65. |
Identify the gate represented by the block diagram of Fig. Write its boolean expression, the truth table and name of the gate, it works. `:|{:(A,B,A+B,bar(A+B)),(0,0,0,1),(1,0,0,1),(0,1,1,0),(1,1,1,0):}:|` |
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Answer» The output of OR gate is made as input for NAND gate. Hence the Boolean expression for final output, `y=bar((A+B).(A+B))=bar((A+B))+bar((A+B))` `=bar((A+B))`. It is NOR gate. |
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| 66. |
Give Boolean expression and Truth table for NOR gate. |
| Answer» The Boolean expression for NOR gate is `bar(A+B)=y`, indicates that y equals A OR B negated. For truth table of NOR gate refer to Fig. | |
| 67. |
Why is the transistor called a junction transistor? |
| Answer» A junction transistor is so called because it is a device which can transfer the resistance by interchanging the biasing across the junctions. | |
| 68. |
Which of the following statements concerning the depletion zone of an unbiased p-n junction is (are) true?A. The width of the zone is independent of the densities of the dopants (impurities)B. The width of the zone is dependent on the densities of the dopants (impurities)The widthC. The electric field in the zone is provided by the electrons in the conduction band and holes in the valence band.D. The electric field in the zone is produced by the ionized dopant atoms. |
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Answer» Correct Answer - A::D The width of the depletion zone of p-n junction is independent of the densities of dopant. The electric field across the depletion zone is due to ionized atoms present on the two sides of p-n junction. |
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| 69. |
Give Boolean expression and Truth table for NAND gate. |
| Answer» The Boolean expression for NAND gate is `bar(A.B)=y`, indicates that y equals A AND B, negated. For truth table of NAND gate, refer to Fig. | |
| 70. |
Name the junction diode whose I-V characteristics are drawn below: |
| Answer» Correct Answer - Photodiode. | |
| 71. |
Three photodiodes `D_1 , D_2 ` and `D_3` are made of semiconductors having band gaps of ` 2.5 eV , 2 eV` and 3 eV , respectively . Which one will be able to detect light of wavelength ` 6000 Å ` ? |
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Answer» Energy of the incident photon, `E=(hc)/lambda=((6.6xx10^(-34))xx(3xx10^(8)))/((6xx10^(-7))xx1.6xx10^(-19))=2.06eV` The incident radiation can be detected by a photodiode if energy of incident radiation photon is greater than the band gap. This is true for `D_(2)(=2eV)`. Therefore, only `D_(2)` will detect these radiations. |
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| 72. |
Dicuss how the OR gate is realised from the NAND gate.A. Only two NAND gatesB. Two NOT gate obtained from NAND gates and one NAND gateC. Four NAND gates and two AND gates obtained from NAND gatesD. Three NAND gates, i.e., three NOT gates obtained from NAND gates |
| Answer» Correct Answer - B | |
| 73. |
The light emitting diode (LED),A. is made from the semiconducting compound gallium arsenide phosphideB. emit light when forward biasedC. is made from one of the two basic semiconducting materials, silicon or germaniumD. emit light when reverse biased. |
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Answer» Correct Answer - A::B The light emitting diode (LED) is made from gallium arsenide phosphide which has the property of emitting light (due to the recombination of electrons and holes) when forward biased. |
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| 74. |
For a CE- transistor amplifier , the audio signal voltage across the collector resistance of ` 2 k Omega ` is 2 V . Suppose the current amplifiaction factor of the transistor is 100 . Find the input signal voltage and base current , if the base resistance is `1 k Omega` . |
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Answer» Here, `R_(0)=2000Omega, V_(0)=2V, beta_(ac)=100, V_(i)=?I_=?, R_(i)=1000Omega` As, `A_(V)=V_(0)/V_(i)=beta_(ac)R_(0)/R_(i) or V_(i)V_(0)/(beta_(ac).(R_(0)//R_(i)))=2/(100(2000//1000))=0.01V` `I_=V_(i)/R_(i)=(0.01V)/(1000Omega)=10muA` |
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| 75. |
A CE amplifier has a voltage gain 50, an input impedance of 1000 ohm, and an output impendance of 200 ohm. The power gain of the amplifier will beA. 24 dBB. 41 dBC. 250 dBD. 12500 dB |
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Answer» Correct Answer - B Power gain in `dB =10 log.P_(0)/P_(i)=10log .(V_(0)^(2)//R_(0))/(V_(i)^(2)//R_(i))` `=10 log.(V_(0)/V_(i))^(2)xxR_(i)/R_(0) =10 log .(50)^(2)xx1000/200` `=10 log 12500=10xx4.0969~~41 dB` |
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| 76. |
The Boolean experssion for NAND gate is............ |
| Answer» Correct Answer - `y =bar(A.B)` | |
| 77. |
In a p-n junction with open ends,A. there is no net charge transfer between the two sidesB. there is a constant electric field near the junctionC. the holes and conduction electrons systematically go from the p-side to n-side and from n-side to p-side respectivelyD. there is no systematic motion of charge carriers. |
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Answer» Correct Answer - A::B::D In a p-n junction with open ends, the net cgarge transferred between the two sides is zero. A depletion between the two sides is zero. A depletion layer is created at the junction due to diffusing of majority carriers from one side to another and a constant electric field is set up across the junction. |
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| 78. |
Which of the following is/are correct equations in Boolean algebra?A. `A+B+C=A+(B+C)`B. `A+0=A`C. `A+1=A`D. `A+A=A` |
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Answer» Correct Answer - A::B::D Boolean Algebra follows associative laws, and option `b, d` are basic OR functions. |
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| 79. |
An n-p-n transistor having a.c. current gain of `50` to be used to make to make an amplifier of power gain of `300`. What will be the voltage gain of the amplifier?A. 8.5B. 6C. 4D. 3 |
| Answer» Correct Answer - B | |
| 80. |
In a CE transistor amplifier there is a current and voltage and gain associated with the circuit. In other words there is a power gain. Considerind power a measure of energy, does the circuit violate conservation of energy? |
| Answer» In CE transistor amplifier, the power gain is very high. The circuit used does not violate the law of conservation of energy because in this circuit the extra power required for amplified output is obtained from d.c. source. | |
| 81. |
In boolean algebra, if `A=B=1`, then the value of `(A.B+A)` isA. `A`B. `B`C. `A+B`D. `B.A+A` |
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Answer» Correct Answer - A::B::C::D Given, `A=B=1`, then, `A.B+A=1.1+1=1+1=1`, Therefore, `A+B=1+1=1`, and `B.A+A=1.1+1=1+1=1`. |
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| 82. |
For a transistor amplifier, the voltage gainA. remain constant for all frequenciesB. is high at high and low frequencies and constant in the middle frequency rangeC. is low at high and low frequencies and constant at mid frequenciesD. none if the above |
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Answer» Correct Answer - C For a transistor amplifier, the frequency-response curve indicates that the voltage gain is low at high and low frequencies and constant and high in the middle frequency range. |
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| 83. |
The current transfer ratio `beta` of a transistor is 50. The input resistance of the transistor when used in common emitter mode is 1 kilo ohm. The peak value of the collector alternating current for an input peak voltage of 0.01 volt isA. `0.01 muA`B. `0.25 muA`C. `100 muA`D. `500 muA` |
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Answer» Correct Answer - D Here `R_(i)=10^(3) Omega, beta=50, V_(i)=0.01 V`, `I_(b)=V_(i)/R_(i)=0.01/10^(3)=10^(-5)A` As `I_(C)=betaxxI_(b)` `=50xx10^(-5)=500xx10^(-6)A=500 muA` |
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| 84. |
In boolean algebra, if `A=1` and `B=0` then the value of `A+barB` isA. `A`B. `B`C. `A+B`D. `A.B` |
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Answer» Correct Answer - A::C Given, `A=1, B=0`, then `A+barB=1+bar(0)=1+1=1` Therefore, `A+B=1+0=1`, But `B=0` and `A.B=1.0=0` |
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| 85. |
A transistor is used in common emitter mode as an amplifier. Then (1) the base-emitter junction is forward biased (2) the base emitter junction is reverse biased (3) the input signal is connected in series with the voltage applied to the base-emitter junction. (4) the input signal is connected in series with the voltage applied to the base collector junction.A. 1,2 and 3B. 1 and 2C. 2 and 4D. 1 and 3 |
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Answer» Correct Answer - D While using transistor as a common emitter amplifier the emitter base junction is forward biased and the signal to be amplified is connected in emitter base circuit. |
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| 86. |
A piece of pure semiconductor of silicn of size `1cm xx 1 cm xx 1 mm` is having `5 xx 10^(28)` number of atoms per cubic metre. It is doped simultaneously with `5 xx 10^(22)` atoms per `m^(3)` of aresenic adn `5 xx 10^(20)` per `m^(3)` atoms of indium. The number density of intrisic current carrier (electrons adn holes) in the pure silicon semiconductor is `1.5 xx 10^(16) m^(-3)`. Mobility of electron is `3800 cm^(2) V^(-)S^(-1)`The number of holes in this semiconductor areA. `5.0xx10^(20)`B. `4.54xx10^(13)`C. `4.54xx10^(9)`D. `4.54xx10^(2)` |
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Answer» Correct Answer - D Number density of holes, `n_(h)=n_(i)^(2)/n_(e)=((1.5xx10^(16))^(2))/((4.95xx10^(22)))=4.54xx10^(9)m^(-3)` Number of holes in the given semiconductor `=n_(e)xxvolume` `=(4.54xx10^(9))xx(1/100xx1/100xx1/1000)` `=4.54xx10^(2)` |
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| 87. |
A piece of pure semiconductor of silicn of size `1cm xx 1 cm xx 1 mm` is having `5 xx 10^(28)` number of atoms per cubic metre. It is doped simultaneously with `5 xx 10^(22)` atoms per `m^(3)` of aresenic adn `5 xx 10^(20)` per `m^(3)` atoms of indium. The number density of intrisic current carrier (electrons adn holes) in the pure silicon semiconductor is `1.5 xx 10^(16) m^(-3)`. Mobility of electron is `3800 cm^(2) V^(-)S^(-1)` The number of electrons in this semiconductor areA. `5.0xx10^(15)`B. `4.95xx10^(15)`C. `4.95xx10^(22)`D. `25xx10^(22)` |
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Answer» Correct Answer - B Number density of electrons, `n_(e)=5xx10^(22)-5xx10^(20)=4.95xx10^(22)m^(-1)` Number of electrons in the given semiconductor `=n_(e)xxvolume` `=4.95xx10^(22)xx(1/100xx1/100xx1/1000)` `=4.95xx10^(15)` |
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| 88. |
In a pure semiconductor, the number of conduction electron is `6xx10^(19)` per cubic metre of size `1 cm xx 1 cmxx2mm`? |
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Answer» Number denity of electron, `n_(e)=6xx10^(19)m^(-3),` Volume of sample, `V=1cmxx1cmxx2mm` `=10^(-2)mxx10^(-2)mxx2xx10^(-3)m` `=2xx10^(-7)m^(3)` `:.` No. of electrons in the given sample `=n_(e)xxV=6xx10^(19)xx2xx10^(-7)` `=12xx10^(12).` In a pure semiconductor , no. density of electron is equal to no. density of hole. Therefore number of holes in the given sample`=12xx10^(12)` |
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| 89. |
A piece of pure semiconductor of silicn of size `1cm xx 1 cm xx 1 mm` is having `5 xx 10^(28)` number of atoms per cubic metre. It is doped simultaneously with `5 xx 10^(22)` atoms per `m^(3)` of aresenic adn `5 xx 10^(20)` per `m^(3)` atoms of indium. The number density of intrisic current carrier (electrons adn holes) in the pure silicon semiconductor is `1.5 xx 10^(16) m^(-3)`. Mobility of electron is `3800 cm^(2) V^(-)S^(-1)`The conductivity of doped semiconductor (in `Sm^(-1))` isA. `2xx10^(3)`B. `3xx10^(3)`C. `4xx10^(3)`D. `1xx10^(3)` |
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Answer» Correct Answer - B The conductivity of doped semiconductor is `sigma=n_(e) e mu_(e)` `=(4.95xx10^(22))xx(1.6xx10^(-19))xx(3800xx10^(-4))` `=3xx10^(3) Sm^(-1)` |
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| 90. |
A piece of pure semiconductor of silicn of size `1cm xx 1 cm xx 1 mm` is having `5 xx 10^(28)` number of atoms per cubic metre. It is doped simultaneously with `5 xx 10^(22)` atoms per `m^(3)` of aresenic adn `5 xx 10^(20)` per `m^(3)` atoms of indium. The number density of intrisic current carrier (electrons adn holes) in the pure silicon semiconductor is `1.5 xx 10^(16) m^(-3)`. Mobility of electron is `3800 cm^(2) V^(-)S^(-1)`Ratio of conductivity of doped silicon and pure silicon semiconductor isA. `2.2xx10^(6)`B. `3.3xx10^(6)`C. `2.2xx10^(8)`D. `3.3xx10^(8)` |
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Answer» Correct Answer - B Ratio of the conductivity of doped silicon and pure silicon semiconductor is `(sigmad)/sigma_(i)=n/n_(i)=(4.95xx10^(22))/(1.5xx10^(16))=3.3xx10^(6)` |
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| 91. |
The diagram Fig.12 shown a piece of pure semiconductor S in series with a variable resistor R, and a source of constant voltage V. Would you increase or decrease the value of R to keep the reading of ammeter constant, when semi-conductor S is heated? Give reason. |
| Answer» when a semiconductor is heated its resistance decreases and hence current in circuit increases. To keep the current in circuit constant, the value of resistance is to be increased. | |
| 92. |
In the following diagram, which bulb out of `B_(1)` and` B_(2)` will glow and why? Draw a diagram of an illuminated p-n junction solar cell Explain briefly the three processes due to which generation of emf takes place in a solar cell. |
| Answer» Here, `D_(1)` is forward biased and `D_(2)` is reverse biased, so bulb `B_(1)` will glow. | |
| 93. |
In the middle of the depletion layer of a reverse - biased `p - n ` junction , theA. electric field is zeroB. potential is zeroC. electric field is maximumD. potential is maximum |
| Answer» Correct Answer - A | |
| 94. |
In an `NPN` transistor the collector current is `24 mA`. If `80%` of electrons reach collector it base current in `mA` isA. 36B. 26C. 16D. 6 |
| Answer» Correct Answer - D | |
| 95. |
In an `NPN` transistor the collector current is `24 mA`. If `80%` of electrons reach collector it base current in `mA` isA. 3B. 6C. 10D. 20 |
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Answer» Correct Answer - B `I_(c)=80/100I_(e)` or `I_(e)=(100I_c)/80=100/80xx24 mA=30 mA` `I_b=I_(e)-I_c=30-24=6 mA` |
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| 96. |
Why doping is done in semiconductor? |
| Answer» The doping is done in semiconductor to increase the number of mobile electrons//holes and hence to increase the conductivity of semiconductor. | |
| 97. |
p-n junction is a semiconductor diode. It is obtained by bringing p-type semiconductor inclose contact with n-type semiconductor. A thin layer is developed at the p-n junction which is devoid of any charge carrier but has immobile ions. It is called depletion layer. At the junction a potential barrier appears, which does not allow the movement of majority charge carriers across the junction in the absence of any biasing of the junction. p-n junction offer low resistance when forward biased and high resistance when reverse biased. Read the above paragaph and answer the following question: (i) Can we measure the potential barrier of p-n junction by putting a sensitive voltmeter across its terminals? (ii) What practical lesson do you draw from the above study? |
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Answer» (i) We cannot measure the potential barrier of p-n junction by putting a sensitive volmeter across its terminals because there are no free electrons or holes in the depletion layer. And in the absence of forward biasing, the depletion layer offers infinite resistance. (ii) From the above study, we find that when some move supports the tendency of majority of the people, the resistance/(opposition) is lowered and output current (desired result) is large (appreciable). However, when some move is against the tendency of majority of people, it suffers rough weather. |
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| 98. |
n-type semiconductor is obtained whenA. germanium is doped with arsenicB. germanium is doped with indiumC. germanium is doped with aluminiumD. silicon is doped with indium |
| Answer» Correct Answer - A | |
| 99. |
Why is n-type semiconductor of Ge so called? |
| Answer» Because in n-type semicondutor, electrons (having negative charge) are majority carriers which are responsible for the conduction. | |
| 100. |
Which type of semiconductor is obtained by mixing arsenic with silicon?A. n-typeB. p-typeC. BothD. None |
| Answer» Correct Answer - A | |