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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.
| 651. |
The mean free path of conduction electorns in copper is about `4xx10^(-8)`m. for a copper block, find the electric field which can give, on an average, 1eV energy to a conduction electron. |
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Answer» Let the electric field be E. the force on an electron is eE. As the electron moves through a distance d, the work done on it is eEd. This is equal to the energy transferred to the electron As the electron travels an average distance of `4xx10^(-8)`m before a collision, the energy transferred is `eE(4xx10^(-8)m)`. To get 2 eV energy from the electric field, `eE(4xx10^(-8)m)=1eV` energy from the electric field, `eE(4xx10^(-8)m)=1eV` or, `E=2.5xx10^(7) V//m`. |
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| 652. |
The majority charge carriers in `P`-type semiconductors areA. ElectronsB. ProtonsC. HolesD. neutrons |
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Answer» Correct Answer - C (3). In P-type semiconductors, holes are the majority charge carriers. |
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| 653. |
In case of a semiconductor, which of the following statement is wrong?A. Doping increases conductivityB. Temperature coefficient of resistance is negativeC. Resistivity is in between that of a conductor and insulatorD. At absolute zero temperature, it behaves like a conductor |
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Answer» Correct Answer - D (d) is wrong because at absolute zero temperature it behaves like an insulator and not as a conductor. |
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| 654. |
Which of the following energy band diagrams shows the `N`-type semiconductor?A. B. C. D. |
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Answer» Correct Answer - B In N-type semiconductor impurity energy level lies just below the conduction band. |
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| 655. |
The energy band diagrams for three semiconductor samples of silicon are as shown. We can then assert that A. Sample X is undoped while samples Y and Z have been doped with a third group and a fifth group impurity respectivelyB. Sample X is undoped while both samples Y and Z have been doped with a fifth group impurityC. Sample X has been doped with equal amounts of third and fifth group impurities while samples Y and Z are undopedD. Sample X is undoped while samples Y and Z have been doped with a fifth group and a third group impurity respectively |
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Answer» Correct Answer - D Sample X is an intrinsic semiconductor. Impurity level is not shown. Hence it is not doped. Sample Y is an n-type semiconductor, because the impurity energy level Iies just below the conduction band. It is doped with fifth group impurity (donor). Sample Z is a p-type semiconductor because the impurity energy level lies just above the valence band and it is doped with a third group impurity (acceptor). |
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| 656. |
Which of the energy band diagrams shown in the figure corresponds to that of a semiconductor?A. B. C. D. |
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Answer» Correct Answer - D In semiconductors the forbidden energy gap between the valence band and conduction band is very small almost equal to kT. Moreover, valence bank is completely filled where as conduction band is empty. |
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| 657. |
In p-type semiconductor, the major charge carriers are:A. holesB. electronsC. protonsD. neutrons |
| Answer» Correct Answer - A | |
| 658. |
Which of the following energy band diagrams shows the `N`-type semiconductor?A. B. C. D. |
| Answer» Correct Answer - B | |
| 659. |
What will be conductivity of pure sillicon crystal at 300 K temp? if electron hole pairs per `cm^(3)` is `1.072xx10^(10)` at this temp. `mu_(n)=1350 cm^(2)//"volt"` sec and `mu_(p)=480 cm^(2)//"volt"` sec-A. `3.14 xx 10^(-6) mho//cm`B. `3 xx 10^(6) mho//cm`C. `10^(-6) mho//cm`D. `10^(6) mho//cm` |
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Answer» Correct Answer - A `sigma=en_(e)(mu_(e)+mu_(h))` `=1.6xx10^(-19)xx1.072xx10^(10)(1350+480)` `=3.14 xx10^(-6) "mho/cm"` |
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| 660. |
GaAs is-A. an elemental semiconductorB. a compound semiconductorC. an insulatorD. a metallic semiconductor |
| Answer» Correct Answer - B | |
| 661. |
A P-type sillicon semiconductor is made by adding one atom of indium per `5xx10^(7)` atoms of sillicon is `25xx10^(28) "atom"//m^(3)`. Point the number of acceptor atoms in per cubic cm. of silliconA. `2xx10^(30) "atom"//cm^(3)`B. `5xx10^(15) "atom"//cm^(3)`C. `1xx10^(15) "atom"//cm^(3)`D. `2.5xx10^(36) "atom"//cm^(3)` |
| Answer» Correct Answer - B | |
| 662. |
In a `n`-type semiconductor, which of the following statement is true?A. electrons are minority carriers and pentavalent atoms are dopantsB. holes ar majority carriers and pentavalent atoms are dopantsC. holes are majority carriers aned trivalent atoms are dopantsD. electrons are majority carriers and trivalent atoms are dopants |
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Answer» Correct Answer - B Holes are minority carriers and pentavalent atoms are depends |
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| 663. |
What will be conductance of pure sillicon crystal at 300 K temp? if electron hole pairs per `cm^(3)` is `1.072xx10^(10)` at this temp. `mu_(n)=1350 cm^(2)//"volt"` sec and `mu_(p)=480 cm^(2)//"volt"` sec-A. `3.14xx10^(-6) "mho/cm"`B. `3xx10^(6) "mho/cm"`C. `10^(-6) "mho/cm"`D. `10^(6) "mho/cm"` |
| Answer» Correct Answer - A | |
| 664. |
Freuency of given AC signal is 50 Hz. When it connected to a half - wave rectifier, then what is the number of output pulses given by rectifier within one second ?A. 50B. 25C. 100D. 150 |
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Answer» Correct Answer - A Number of pulses in the output of half wave rectifier in one second is equal to frequency of ac source i.e. 50 |
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| 665. |
Freuency of given AC signal is 50 Hz. When it connected to a half - wave rectifier, then what is the number of output pulses given by rectifier within one second ?A. 50B. 100C. 25D. 150 |
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Answer» Correct Answer - A In half-wave rectification, there is only one output pulse for each complete cycle of the input AC voltage. Therefore output frequency is same as that of AC supply frequency, i.e. `f_("out")=f_("in")` So, pulses obtained in 1 s by half-wave rectifier will be 50. |
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| 666. |
Which of the following is correct, about doping in a transistor ?A. Emitter is lightly doped, collector is heavily doped and base is moderately dopedB. Emitter is lightly doped, collector is moderately doped and base is heavily dopedC. Emitter is heavily doped, collector is lightly doped and base is moderately dopedD. Emitter is heavily doped, collector is moderately doped and base is lightly doped |
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Answer» Correct Answer - D Emitter is always forward biased w.r.t. base, so that it can supply a large number of majority carriers. The emitter is heavily doped so that it can inject a large number of charge carriers (electrons or holes) into the base. The base is lightly doped and very thin, it passes most of the emitter injected charge carriers to the collector. The collector is moderately doped. |
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| 667. |
Which of the following is correct, about doping in a transistor ?A. Emitter is heavily dopped, collector is lightly dopped and base in moderatelyB. Emitter is lightly dopped, collector is heavily dopped and base in moderatelyC. Emitter is heavilyD. Emitter is lightly |
| Answer» Correct Answer - C | |
| 668. |
What is amount of doping tansistor ?A. Emitter is moderately doped, collector is heavily doped and base is lightly dopedB. Emitter is moderately doped, collector is lightly doped and base is heavily dopedC. Emitter is heavily doped, collector is lightly doped and base is moderately dopedD. Emitter is heavily doped, collector is moderatly doped and base is lightly doped |
| Answer» Correct Answer - D | |
| 669. |
To obtain n-type semiconductor , the impurity introduced isA. ArsenicB. AluminiumC. Silicon D. Indium |
| Answer» Correct Answer - A | |
| 670. |
To obtain n-type semiconductor , the impurity introduced isA. ArsenicB. AluminiumC. SiliconD. Indium |
| Answer» Correct Answer - A | |
| 671. |
Which of the following is a typical example of a semiconductor ?A. MicaB. QuartzC. PlatinumD. Germanium |
| Answer» Correct Answer - D | |
| 672. |
A hole in semiconductor is different from an electron, because it isA. masslessB. an antiparticleC. negatively charged vacancyD. positively charged vacancy |
| Answer» Correct Answer - D | |
| 673. |
The amplitude of electric field in a parallel light beam of intensity `4Wm^(-2)` isA. `35.5NC^(-1)`B. `45.5NC^(-1)`C. `49.5NC^-1`D. `55.5NC^(-1)` |
| Answer» Correct Answer - D | |
| 674. |
Statement-I : For the same antenna length, power radiated by short wavelength signals would be large. Statement-II : Because power radiated `prop (1)/(lamda^(2))`A. Statetment-I is True, Statement-II is True , Statement-II is a correct explanation for Statement-IB. Statement-I is True, Statement-II is True , Statement-II is NOT a correct explanation for Statement-IC. Statement-I is True, Statement-II is FalseD. Statement-I is False, Statement-II is True |
| Answer» Correct Answer - A | |
| 675. |
Which of these statements correctly describes the orientation of the electric field `(vec(E))` the magnetic field `(vec(B))` and velocity of propagation `(vec(v))` of an electromagnetic wave?A. `vecE` is perpendicular to `vecB`, `vecv` may have any orientation relative to `vecE`.B. `vecE` is perpendicular to `vecB`, `vecv` may have any orientation perpendicular to `vecE`C. `vecE` is parallel to `vecB,vecv` is perpendicular to both `vecE` and `vecB`D. Each of the three vactor is perpendicular to the other two. |
| Answer» Correct Answer - D | |
| 676. |
A vertical electric dipole antennaA. radiates uniformly in all directionB. radiates uniformly in all horizontal direction but more strongly in the vertical direction.C. radiates most strongly and uniformly in the horizontal directions.D. Does not radiate in the horizontal directions |
| Answer» Correct Answer - C | |
| 677. |
A dipole radio transmitter has its rod-shaped antenna oriented vertically. At a point due south of the transmitter, the radio waves have their magnetic field.A. oriented north-southB. oriented east-westC. oriented verticallyD. oriented in any horizontal direction. |
| Answer» Correct Answer - B | |
| 678. |
Statement-1 : Diode lasers are used as optical sources in capital communication. Statement-2 : diode lasers consume less energy.A. Statetment-I is True, Statement-II is True , Statement-II is a correct explanation for Statement-IB. Statement-I is True, Statement-II is True , Statement-II is NOT a correct explanation for Statement-IC. Statement-I is True, Statement-II is FalseD. Statement-I is False, Statement-II is True |
| Answer» Correct Answer - B | |
| 679. |
Would there be any advantage to adding `n`-type or `p`-type impurities to copperA. YesB. NoC. May beD. Data insufficient |
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Answer» Correct Answer - B Pure Cu is already an excellent conductor, since it has a partially filled conduction band, further more, Cu forms a metallic crystal as opposed to the covalent crystals of silicon or germanium, so the scheme of using an impurity of donate or accept an electron does not work for copper. Infact adding impurities to copper decreases the conductivity because an impurity tends to scatter electrons, impeding the flow of current. |
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| 680. |
To get NOT gate from NAND gate, we needA. one NAND gateB. two NOT gates obtained from NAND gatesC. one NAND gate and one NOT gate obtained from NAND gateD. 3 NAND gates and one NOT gate obtained from NAND gate |
| Answer» Correct Answer - A | |
| 681. |
Rectification is the process for the conversion ofA. a.c into d.c.B. d.c. into a.c.C. low a.c. into high a.c.D. low d.c. into high d.c. |
| Answer» Correct Answer - A | |
| 682. |
Transistor switches have the advantage ofA. high speed operationB. low speed operationC. low costD. low working life |
| Answer» Correct Answer - A | |
| 683. |
The major constituent of transistor areA. saltsB. transistorC. conductorsD. semiconductors |
| Answer» Correct Answer - D | |
| 684. |
The mobility of free electrons is greater then that of free holes becauseA. they carry negative chargeB. they are lightC. they mutually collide lessD. they require low energy to continue their motion |
| Answer» Correct Answer - D | |
| 685. |
How many electrodes are there in a transistor ?A. 2B. 3C. 4D. 5 |
| Answer» Correct Answer - B | |
| 686. |
In the working of n-p transistor, the number of free electrons which recombine with holes in the base layer is about.A. 97% of the number injected into the baseB. 50% of the number injected into the baseC. 3% of the number injected into the baseD. 25% of the number injected into the base |
| Answer» Correct Answer - C | |
| 687. |
A transistor can be used practically inA. only one configurationB. only two configurationsC. three possible configurationsD. four possible configurations |
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Answer» Correct Answer - C The transistor can be connected in either of the following three configurations: Common Emitter (CE), Common Base (CB), Common Collector (CC). The transistor is most widely used in the CE configuration. |
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| 688. |
In the `CB` mode of a transistor, when the collector voltage is changed by `0.5` volt. The collector current changes by `0.05 mA`. The output resistance will beA. `10k Omega`B. `20k Omega`C. `5k Omega`D. `2.5k Omega` |
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Answer» Correct Answer - A Here, `DeltaV_(c)=0.5V and DeltaI_(C)=0.05 mA=0.05 xx 10^(-3)A` Output resistance is given by, `R_("out") =(DeltaV_(C))/(DeltaI_(C))=(0.5)/(0.05xx10^(-3))` `=10^(4)Omega=10Omega` |
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| 689. |
For a transistor, `beta` = 50. To change the collector current by 350 `mu`A, the base current should be changed byA. `(50/350)muA`B. `(350-50)muA`C. `(350+50)muA`D. `(350/50)muA` |
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Answer» Correct Answer - D `beta=50 = (DeltaI_C)/(DeltaI_B)` `therefore DeltaI_B= (DeltaI_C)/beta=(350/50)muA` |
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| 690. |
The current of transistor in common emitter mode is 49. The change in collector current and emitter current corresponding to the change in the base current by `5.0 muA`m will be :-A. `245mu A, 250mu A`B. `240mu A, 235mu A`C. `260mu A, 255mu A`D. None of these |
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Answer» Correct Answer - A Current gain, `beta=(DeltaI_(C))/(DeltaI_(B))`, where `DeltaI_(C )` is change in collector current and `DeltaI_(B)` is change in base current. `rArr DeltaI_(C)=betaDeltaI_(B)=49xx5=245 muA` Also, `DeltaI_(E)=DeltaI_(B)+DeltaI_(C)` `=(245+5)muA=250muA` |
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| 691. |
In a transistor, the value of `alpha=0.9`. Then the value of `beta` is:A. 4B. 5C. 6D. 9 |
| Answer» Correct Answer - D | |
| 692. |
If for a transistor, `beta` = 49, then the value of `alpha` isA. 1B. 0.49C. 0.98D. 15 |
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Answer» Correct Answer - C `beta=alpha/(1-alpha) " " therefore 49=alpha/(1-alpha)` `49=50 alpha " " therefore alpha = 49/50` = 0.98 |
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| 693. |
A zener diode has a contract potential of `0.8 V` in the absence of biasing. It undergoes zener breakdown for an electric field of `10^(6) Vm^(-1)` at the depletion region of p-n junction. If the width of the depletion region is 2.4mu m, what should be the reverse biased potential for the zener breakdown to occur?A. 3.5 VB. 2.5 VC. 1.5 VD. 0.5 V |
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Answer» Correct Answer - B `E="dV"/"dx"` `therefore dV=Edx = 10^6 V/m xx2.5xx10^(-6)` `therefore dV=2.5` V `therefore ` Breakdown voltage = 2.5 V |
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| 694. |
Zener diode is used forA. rectification of voltageB. stabilisation of voltageC. amplification of voltageD. producing electromagnetic oscillation |
| Answer» Correct Answer - B | |
| 695. |
Zener breakdown occursA. mostly in intrinsic semiconductorsB. due to repture of co-valent bondsC. in lightly doped junctionsD. due to thermally generated majority carries |
| Answer» Correct Answer - B | |
| 696. |
Zener breakdown takes place ifA. impurity level is highB. impurity level is lowC. impurity is less in n sideD. impurity is less n p side |
| Answer» Correct Answer - B | |
| 697. |
Once a zener diode is taken in its breakdown region, there is not much change in itsA. currentB. resistanceC. voltageD. capacitance |
| Answer» Correct Answer - C | |
| 698. |
Zener breakdown occurs only whenA. it is lightly dopedB. the temperature is increasedC. it is forward biasedD. it is reverse biased |
| Answer» Correct Answer - D | |
| 699. |
The value of the zener currentA. is determined by the zener currentB. is always in the a microampere rangeC. does not depends upon the temperatureD. is limited by the external circuit resistance |
| Answer» Correct Answer - D | |
| 700. |
For the proper functioning of a zener diode as a voltage stabiliser it should be alwaysA. forward biasedB. reverse biasedC. lightly biasedD. connected in series with the load resistance |
| Answer» Correct Answer - B | |