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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.
| 101. |
A p-type semiconductor is obtained by doping silicon withA. germaniumB. galliumC. bismuthD. phosphorus |
| Answer» Correct Answer - B | |
| 102. |
The forbidden energy gap of germanium is 0.72eV. What do you understand by it? |
| Answer» It states that if an energy of 0.72 eV is given to an electron in the valence band of germanium it will jump to the conduction band, crossing an energy gap of 0.72eV. | |
| 103. |
Stonsium (Sn) and silicon (Si) both belong to group IV elements but Sn is a metal whereas Si is a semiconductor. Why? |
| Answer» Sn is a metal because the energy gap in its energy band diagram is zero, whereas Si is a semiconductor because the energy gap in its band diagram is 1.1 eV. | |
| 104. |
Why do Ge and Si are semiconductor? |
| Answer» In the energy band diagram of Ge and Si. The energy gap is `0.72 eV `and 1.1 eV respectively between conduction band and valence band. As a result of it, they behave as semiconductor. | |
| 105. |
Why diamond behaves like an insulator? |
| Answer» In the energy band diamond, there is a large energy gap` =5.54eV`. Due to it, no electron can go from valence band to conduction band. | |
| 106. |
Define transconductance of a transistor. On what factor does it depend? |
| Answer» It depends on the geometry, doping levels and biasing of the transistor. It is defined as the ratio of the change in collector current to the change in base-emitter voltage. | |
| 107. |
What will happen if both, emitter and collector of a transistor are reversed biased? |
| Answer» In this situation, on current will flow through transistor because there is no conduction due to majority carriers across the emitter-base junction or collector-base junction. | |
| 108. |
What is the phase relationship between collector and base voltage in common-emitter configuration. |
| Answer» The output collector voltage is `180^(@)` out of phase with respect to input signal voltage applied to base. | |
| 109. |
What will happen if emitter as well as collector in a transistor are forward biased? |
| Answer» In this aituation, the majority charge carriers will flow in the emitter-base circuit. In this case, the working of transistor will be equivalent to two p-n junction diode with a common base. It means the purpose of transistor will be defeated. | |
| 110. |
What is the phase relationship in the output and input voltage in the common base transistor amplifier. |
| Answer» Output voltage is in phase with the input signal voltage. | |
| 111. |
Define current amplification factor in a common emitter mode of transistor. |
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Answer» Current amplification factor `(beta_(ac))` is defined as the ratio of small change in collector current `(DeltaI_C)` to the corresponding small change in the base current `(DeltaI_B)` when the collector emitter voltage is kept constant, i.e., `beta_(ac)=[(DeltaI_C)/(DeltaI_B)], V_(CE)`is constant. |
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| 112. |
Is transistor a current controlled or temperature controlled device? |
| Answer» Transistor is a current controlled device. | |
| 113. |
Transistor is a temperature-sensitive device. Explain. |
| Answer» In a transistor,free electrons and holes are charge carriers. When a transistor is properly biased i.e. emitter is forward biased and collector is reverse biased, these charge carriers are responsible for the current through the transistor as well as in external circuit. If the temperature of the transistor increases, some of the covalent bonds may be broken, giving rise to additional number of free electrons and holes. As a result of it, the current in a transistor will increase. If the current in a transistor is very strong it may causeb large heat and finally may result complete breakdown of semi-conducting device. That is why transistor is a temperature sensitive device. | |
| 114. |
A transistor is current operated device. Explain. |
| Answer» In a transistor, emitter current= base current+collector current. It means emitter current controls the collector current and base current. For a given emitter current, collector current is controlled by base current. Therefore, the change in collector current is related with the base current and not to the base voltage change. That is why, the transistor is a current operating device. | |
| 115. |
Assertion : In a common emitter transistor amplifier, the input current is much less than output current. Reason : The common-emitter transistor amplifier has a very high input impedance.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 - C The CE amplifier has high output impedance `~~ 10 kOmega` and low input impedance `~~ 1 kOmega`. So assertion is true but reason is false. |
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| 116. |
Distinguish between electrons and holes. |
| Answer» (i) Electron is a negatively charged particle having charge `=1.6 xx 10^(-19)C`. Hole is a seat having positive charge which is produced when an electron breaks away from a covalent bond in a semiconductor. Hole is having the same charge as that of electron. (ii) Energy of a hole is high as compared to that of electron. (iii) The mobility of a holes is smaller than that of electron. | |
| 117. |
Name the category of the integrated circuit which is utilizing the circuit components (i) `le 100` and (ii) `le 1000`. |
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Answer» (i) The integrated circuit utilizing the circuit components `le 100` is called Medium scale integation circuit (MSI). (ii) The integrated circuit utilizing the circuit components `le 1000` is called Large scale integration circuit (LSI). |
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| 118. |
Write the full form of the terms (i) SSI and (ii) VLSI used for different types of integrated circuits. |
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Answer» (i) SSI stands for small Scale Intergration circuits. (ii) VLSI stands for Very Large Scale Integration circuit. |
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| 119. |
The input wave form for NOT gate is shown in Fig.Draw the output wave form of this gate. |
| Answer» For NOT gate, the Boolean expression is `y=barA`. Thus the output wave from is as shown in Fig. | |
| 120. |
What is integrated circuit? |
| Answer» Integrated circuit is that circuit in which the circuit components such as resistors, capacitors, diode and transistors, etc., are automatically parts of a small semiconductor chip. | |
| 121. |
The gain of a common emitter amplifier is given by `A_(V)=-g_(m)R_(L)`. Does it mean that if we keep on increasing indefinitely `R_(L)`, the gain of the amplifier will also increase indefinitely? Explain your answer. |
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Answer» The gain of the amplifier will not increase indefinitely on increasing `R_(L)` indefinitely. As `V_C=V_(CE)-I_C R_(L)` Therefore, as `R_(L)` increases, the value of `V_C` decreases. When the value of `V_C` becomes less than the base voltage, both the junction in transistor get forward biased. Due to it, the current becomes saturated. It means no further increase of current takes place and hence no further gain of the amplifier. |
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| 122. |
How does potential barrier of a semiconductor vary with temperature? |
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Answer» The barrier potential of a semiconductor increases with rise in temperature. The barrier potential `(V_B)` of a semiconductor is directly proportional to temperature in kelvin (T). The relation between barrier potential `V_B` and temperature TK in terms of the hole concentration on either side of the junction is given by `V_(B)=(k_(B)T)/qlog_(e)(p_(p)/P_(n))` where `k_(B)=` Boltzmann constant, `p_(p) =` hole concentration in p-side,` p_(n) =` hole concentration in n-side. The relation between barrier potential and temperature in terms of donor concentration `(N_d)` on the n-side and acceptor concentration `(N_a)` on the p-side is given by `V_(B)=(k_(B)T)/qlog_(e)((N_a N_d)/n_(i)^(2))` where `n_(i)` is the intrinsic carrier concentration. |
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| 123. |
When a p-type semiconductor is brought into a close contact with n-type semiconductor, we get a p-n junction with a barrier potential `0.4V` and the width of depletion region is `4.0 xx 10^(-7)m`. this p-n junction is forwaed biased with a battery of voltage `3V` adn negligible internal resistance, in series with a resistance of resistance R, ideal milliammeter and key K as shown in Fig. If an electron with speed `4.0xx10^(5) ms^(-1)` approaches the p-n junction from the n-side, the speed with which it will enter the p-side isA. `1.39xx10^(5) ms^(-1)`B. `2.78xx10^(5) ms^(-1)`C. `1.39xx10^(6) ms^(-1)`D. `2.78xx10^(6) ms^(-1)` |
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Answer» Correct Answer - A Let `v_(1)` be the speed of electron when enters the depletion region and `v_(2)` is the speed when it comes out of depletion layer. According to principle of conservation of total energy, K.E. of the incident electron =workdone against potential barrier +K.E. of the emerging electron. i.e. `1/2mv_(1)^(2)=eV_+1/2mv_(2)^(2)` or `1/2xx(9.1xx10^(-31))xx(4xx10^(5))^(2)` `=(1.6xx10^(-19)xx(0.4)+1/2xx9.1xx10^(-31)xxv_(2)^(2)` on solving, we get, `v_(2)=1.39xx10^(5) ms^(-1)` |
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| 124. |
When a p-type semiconductor is brought into a close contact with n-type semiconductor, we get a p-n junction with a barrier potential `0.4V` and the width of depletion region is `4.0 xx 10^(-7)m`. this p-n junction is forwaed biased with a battery of voltage `3V` adn negligible internal resistance, in series with a resistance of resistance R, ideal milliammeter and key K as shown in Fig. Wattage of diode isA. 0.060 WB. 0.052 WC. 0.008 WD. 0.048 W |
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Answer» Correct Answer - C Wattage of diode = voltage drop across diode xx current `=0.4xx20xx10^(-3)=0.008 W` |
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| 125. |
When a p-type semiconductor is brought into a close contact with n-type semiconductor, we get a p-n junction with a barrier potential `0.4V` and the width of depletion region is `4.0 xx 10^(-7)m`. this p-n junction is forwaed biased with a battery of voltage `3V` adn negligible internal resistance, in series with a resistance of resistance R, ideal milliammeter and key K as shown in Fig. The resistance of resistor R isA. `150 Omega`B. `300 Omega`C. `130 Omega`D. `180 Omega` |
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Answer» Correct Answer - C Pot. Diff. across `R=3-0.4=2.6 V` Resistance, `R=("pot. diff.")/("current")=2.6/(20xx10^(-3))=130 Omega` |
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| 126. |
A p-n junction is fabricated from a semiconductor with band gap of `2.8eV`. Can it detect a wavelength of `6000nm`? |
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Answer» Energy, `E=(hc)/lambda=(6.6xx10^(-34)xx3xx10^(8))/(600xx10^(-9)xx1.6xx10^(-19))eV=2.06eVlt2.8eV` As `E lt E_(g)`, so p-n junction cannot detect the radiation of given wavelength. |
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| 127. |
In an intrinsic semiconductor the energy gap `E_(g) is 1.2 eV`. Its hole mobility is much smaller than electron mobility and independent of temperature. What is the ratio between conductivity at `600K` and `300K`? Assume that temperature dependence intristic concentration `n_(i)` is given by `n_(i)=n_(0) exp ((-E_(g))/(2k_T))`, where `n_(0)` is a constant and `k_=8.62xx10^(-5)eV//K`. |
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Answer» Here, `E_(g)/(2k)[1/T_(1)-1/T_(2)]=1.2/(2xx8.62xx10^(-5))[1/300-1/600]=11.6` `n_(600)/n_(300)=e^(E_(g)/(2k)[1/T_(1)-1/T_(2)])=e^(11.6)=(2.718)^(11.6)=1.089xx10^(5)=1.1xx10^(5)` `sigma_(600)/sigma_(300)=n_(600)/n_(300)=1.1xx10^(5)` |
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| 128. |
In Fig . `V_(0)` 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 junctionB. 3 corresponds to forward bias of junction and 1 corresponds to reverse bias of junctionC. 1 corresponds to forward bias and 3 corresponds to reverse bias of junctionD. 3 and 1 both correspond to reverse bias of junction |
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Answer» Correct Answer - B When p-n junction is forward biased, it opposes the potential barrier across junction. When p-n junction is reverse biased, it supports the potential barrier junction, resulting increase in potential barrier across the junction. |
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| 129. |
A semiconductor is known to have an electron concentration of `5xx10^(12) cm^(-3)` and a hole concentration `8xx10^(13)cm^(-3) ` Is the semi-conductor n-type or p-type? What is the resistivity of the sample, if the electron mobility is `23,000 cm^(2) V^(-1)s^(-1)`? Take charge on electron, `e=1.6xx10^(-19) C`. |
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Answer» Here, `n_(e)=5xx10^(12)cm^(-3)=5xx10^(18)m^(-3)`, `n_(h)=8xx10^(13)cm^(-3)=8xx10^(19)m^(-3)`, `mu_(e)=23000cm^(2)V^(-1)s^(-1)=2.3m^(2)V^(-1)s^(-1)`, `mu^(h)=100cm^(2) V^(-1)s^(-1)=0.01m^(2)V^(-1)s^(-1)`. since the semiconductor has greater hole density, hence it is p-type Now, `1/rho=e(n_(e)mu_(e)+n_(h)mu_(h)) =1.6xx10^(-19)[5xx10^(18)xx2.3+8xx10^(19)xx0.01]` `1.6xx10^(-19)[1.15xx10^(19)+0.08xx10^(19)]` `=1.968` or `rho=1/1.968=0.508ohm-m` |
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| 130. |
Figure Shows a diode connected to an external resistance and an e.m.f. Assuming that the barrier potential developed in diode is `0.5 V`, obtain the value of current in the circuit in milliampere. |
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Answer» Here, `E=4.5V, R=100Omega`, voltage drop across p-n junction`=0.5 V`. Effective voltage in circuit, `V=4.5-0.5=4.0V` Current in the circuit, `I=V/R=4.0/100=0.04A` `0.04xx1000mA=40mA` |
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| 131. |
A diode having potential difference `0.5 V` across its junction which does not depend on current, is connected in series with resistance of `20 Omega` across source. If `0.1A` passes through resistance then what is the voltage of the source?A. 1.5VB. 2.0 VC. 2.5VD. 5V |
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Answer» Correct Answer - C Here, `V_(D)=0.5 V, R=20Omega, I=0.1 A` Voltage of the source = Voltage drop across p-n junction + pot. Diff. avross resistance `=0.5+0.1xx20=0.5+2=2.5 V` |
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| 132. |
In the given figure, a diode `D` is connected to an external resistance `R=100 Omega` and an emf of `3.5 V`. If the barrier potential developed across the diode is `0.5 V`, the current in the circuit will be : A. 35mAB. 30mAC. 40mAD. 20mA |
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Answer» Correct Answer - B Effective voltage of circuit, `V=3.5-0.5=3.0V` Current in circuit `I=V/R=3.0/100A=30 mA` |
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| 133. |
Why are elemental dopants for Silicon or Germanium usually chosen from group XIII or group XV? |
| Answer» The size of the dopant atoms should be such that their presence in the pure semiconductor does not distort the semiconductor but easily contribute the charge carriers on forming covalent bonds with Si or Ge atoms. | |
| 134. |
Assertion: `NOT` gate is also called inverter circuit. Reason: `NOT` gate inverts the input order.A. Statement-1 is true, Statement-2 is true, Statement-2 is a correct explanation of Statement-1.B. Statement-1 is true, Statement-2 is true, Statement-2 is not a correct explanation of Statement-1.C. Statement-1 is true, Statement-2 is false.D. Statement-1 is false, Statement-2 is true. |
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Answer» Correct Answer - A Here, both Assertion and Reason are correct and Reason is correct explanation of Assertion because a NOT gate puts the input condition in the opposite order i.e. for high input voltage, we get low output voltage and vice versa. |
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| 135. |
In p-n junction, p-side is known as........and n-side is known as.......... |
| Answer» Correct Answer - anode; cathode | |
| 136. |
A p-n junction can be considered to be equivalent to a..........with p and n-regions acting as the plates of a.........and depletion region as the............. |
| Answer» Correct Answer - capacitor, capcitor, dielectric medium | |
| 137. |
In p-n juction, the physical distance from one side of the barrier to the other is know as the ...........barrier and the difference of potential from one side of the barrier to other side is.......known as...........barrier. |
| Answer» Correct Answer - width of the; height of the | |
| 138. |
Those solids which have very low conductivity and very high resistivity are called......... |
| Answer» Correct Answer - insulators | |
| 139. |
Those solids which have high conductivity and low resistivity are called............ |
| Answer» Correct Answer - metal conductors | |
| 140. |
In p-type semiconductors, the..........are majority carries and..........are minority carries. |
| Answer» Correct Answer - holes, electrons | |
| 141. |
The number of minority carriers crossing the junction of a diode depends primarily on theA. concentration of doping impuritiesB. magnitude of potential barrierC. magnitude of the forward bias voltageD. rate of thermal generation of electron-hole pair |
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Answer» Correct Answer - D The number of minority carriers crossing the junction diode depends on rate of thermal generation diode depends on rate of thermal generation of electron-hole pair. |
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| 142. |
In n-type semocionductor, the..........are majority carries and..........are minority carries. |
| Answer» Correct Answer - electron,holes | |
| 143. |
In.............semiconductor, the fermi level lies in the energy gap, very close to valence band. |
| Answer» Correct Answer - p-type | |
| 144. |
Transistor does not work in railway carriage. Why? |
| Answer» The railway carriage works as an electric screen. The electric field inside the carriage is zero and any change from outside in electric field cannot enter the carriage. Hence the electromagnetic signal do not find their entry in the railway carriage. Due to it, the transistor does not work in railway carriage. | |
| 145. |
A semiconductor has equal electron and hole concentration of `6xx10^(8)//m^(3)`. On doping with certain impurity, electron concentration increases to `9xx10^(12)//m^(3)`. (i) Identify the new semiconductor obtained after doping. (ii) Calculate the new hole concentration. (iii) How does the energy gap vary with doping? |
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Answer» Here, `n_(i)=6xx10^(8)m^(-3) , n_(e)=9xx10^(12)m^(-3)` `n_(h)=n_(i)^(2)/n_(e)=((6xx10^(8))^(2))/(9xx10^(12))=4xx10^(4)m^(-3)` As, after doping, `n_(e)gtn_(h)`, so the new semiconductor is n-type. Energy gap decreases with doping. |
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| 146. |
What type of charge carriers are there in n-type semiconductor? |
| Answer» The charge carriers in n-type semiconductor are both electrons (majority carriers) and holes (minority carriers). | |
| 147. |
What type of charge carriers are there in p-type semiconductor? |
| Answer» The charge carriers in p-type semiconductor are both holes (majority carriers) and electrons (minority carriers). | |
| 148. |
Give the ratio of number of holes and the number of conduction electrons in a (i) pure semiconnductor (ii) n-type semiconductor and (iii) p-type semiconductor. |
| Answer» (i) `n_(h)//n_(e)=1` (ii) `n_(h)//n_(e)lt1` (iii) `n_(h)//n_(e)lt1` | |
| 149. |
The resistance of p-n junction is low when forward biased and is high when reverse biased. Explain. |
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Answer» A small increase in forward voltage across p-n junction shows large increase in forward current. Hence the resistance `(="voitage /current")` of p-n junction is low when forward biased. A large increase in reverse voltage across p-n junction shows small increase in reverse current. Hence the resistsnce of p-n junction is high when reverse biased. |
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| 150. |
Write the full form of the terms (i) MSI and (ii) LSI used for different types of integrated circuits. |
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Answer» (i) MSI stands for medium Scale Intergration circuits. (ii) LSI stands for Large Scale Integration circuit. |
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