InterviewSolution
Saved Bookmarks
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.
| 601. |
The impurity atoms with which pure silicon should be doped to make a p - type semiconductor are those ofA. PhosphorousB. AntimonyC. BoronD. Aluminium |
| Answer» Correct Answer - C::D | |
| 602. |
A light emitting diode `(LED)` has a voltage drop of `2V` across it and passes a current of `10 mA`. When it operates with a `6V` battery through a limiting resistor `R`. The value of `R` isA. `40kOmega`B. `4kOmega`C. `200Omega`D. `400Omega` |
|
Answer» Correct Answer - D `R=(6-2)/(10xx10^(-3))=400Omega` |
|
| 603. |
In a semiconductor diode p-side is earthed and N-side is applied a potential of -2V, the diode shallA. ConductB. Not conductC. Conduct partiallyD. Breakdown |
| Answer» Correct Answer - A | |
| 604. |
In silicon diode, the reverse current increases from 10A to `20 muA` . When the reverse voltage change from 2V to 4V . Find the reverse a.c. resistance of the diodeA. `3xx10^(-5) Omega`B. `2xx10^(5) Omega`C. `1xx10^(5) Omega`D. `4xx10^(5) Omega` |
|
Answer» Correct Answer - B `r_(r)=(deltaV)/(deltaI)=(2)/(10xx10^(-6))=0.2xx10^(6) Omega`. |
|
| 605. |
What is the value of D.C. voltage in a half wave rectifier in converting AC. voltage V = 100 sin (314 t) into D.C.?A. 100 voltB. 50 voltC. 32 voltD. 0 |
|
Answer» Correct Answer - C V = 100 sin (314 t) In this case, `V_0` = 100 V In the half wave rectifier, the value of d.c. voltage in one cycle is given by `V_"ac" = V_0/pi=100/3.14=31.85` V = 32 V |
|
| 606. |
An n-p-n transistor in a common - emitter mode is used as a simple voltage amplifier with a collector current of 4 mA. The positive terminal of a 8 V battery is connected to the collector through a load resistance ` R_L ` and to the base through a resistance ` R_B ` . The collector - emitter voltage ` V_(CE) = 4 V` , the base - emitter voltage ` V_(BE) = 0.6 V` and the current amplification factor ` beta = 100 ` . Calculate the values of ` R_L ` and `R_B` . |
|
Answer» Correct Answer - A Given , `i_c = 4 mA ` Applying Kirchhoffs second law in loop 1, we have `V_(CE) = 8 - i_c R_L` `:. R_L = 8 - V_(CE) / i_c = (8 - 4)/ (4xx 10^(-3)) ` ` = 1000 Omega ` ` = 1 k Omega ` Further, ` beta = i_c / i_b ` ` :. i_b = i_c / beta = (4 xx 10^(-3)) / 100 A = 40 mu A ` Now, ` V_(BE) = 8 - i_b R_B ` `:. R_B = (8 - V_(BE)) / i_b ` `= (8 -0.6 )/( 40 xx 10^(-6) )` ` = 1.85 xx 10^(5) Omega ` . |
|
| 607. |
For a common emitter amplifier, the voltage gain is 40. Its input and output impedances are 100 `Omega` and 400 `Omega` respectively. The power gain of the C.E. amplifier will beA. 300B. 400C. 450D. 500 |
|
Answer» Correct Answer - B 40=Current gain x `400/100` `therefore ` Current gain = 10 `therefore` Power gain = Voltage gain x Current gain = 400 |
|
| 608. |
A silicon specimen is made into a `P`-type semiconductor by dopping, on an average, one helium atoms per `5xx10^(7)` silicon atoms. If the number density of atoms in the silicon specimen is `5xx10^(28) at om//m^(3)` then the number of acceptor atoms in silicon per cubic centimeter will beA. `2.5xx10^(30)atoms//cm^(3)`B. `1.0xx10^(13)atoms//cm^(3)`C. `1.0xx10^(15)atoms//cm^(3)`D. `2.5xx10^(36)atoms//cm^(3)` |
|
Answer» Correct Answer - C Number of indium atoms=`(5xx10^(28))/(5xx10^(10^(7)))=10^(21)//m^(3)=10^(15)//cm^(3)` |
|
| 609. |
Pure `Si` at `500K` has equal number of electron `(n_(e))` and hole `(n_(h))` concentration of `1.5xx10^(16)m^(-3)`. Dopping by indium. Increases `n_(h)` to `4.5xx10^(22) m^(-3)`. The doped semiconductor is ofA. `n`-typpe with electron concentration `n_(e)=2.5xx10^(23)m^(-3)`B. `p`-type having electron concentration `n_(e)=5xx10^(9)m^(-3)`C. `n`-type with electron concentration `n_(e)=5xx10^(22m^(-3)`D. `p`-type with electron concentration `n_(e)=2.5xx10^(10)m^(-3)` |
|
Answer» Correct Answer - B `n_(i)^(2)=n_(e)n_(h)` `(1.5xx10^(16))^(2)=n_(e)(4.5xx10^(22))` `n_(e)=0.5xx10^(10)=5xx10^(9)` `n_(h)=4.5xx10^(22)` `n_(h)gtn_(e)(p-type)` |
|
| 610. |
In the energy band diagram of a material shown below, the open circles and filled circles denote holes and electrons respectively. The material is a/an A. p-type semiconductorB. insulatorC. metalD. n-type semiconductor |
|
Answer» Correct Answer - A In the given situation the number of holes velence band is greater than number of electrons in conduction band. So it is p-type semiconductor. |
|
| 611. |
A `p-n` photodiode is fabricated from a semiconductor with a band gap of `2.5 eV`. It can detect a signal of wavelengthA. 6000 ÅB. 4000 nmC. 6000 nmD. 4960 Å |
|
Answer» Correct Answer - D Only signal having wavelength less than threshold wavelength will be detercted. Energy `E=hv=h(c)/(lamda)implieslamda=(hc)/(E)` Substituting the value of h, c and E in the above equation `lamda=(6.6xx10^(-34)xx3xx10^(8))/(2.5xx1.6xx10^(-19))=5000Å` As `4000Ålt5000Å` Signal of wavelength `4000Å` can be detected by the photodiode |
|
| 612. |
A common emiiter amplifier has voltage gain 50 and current gain is 25. The power gain of the amplifier is:A. 500B. 1000C. 1250D. 100 |
|
Answer» Correct Answer - C AC power gain is ratio of change in output power to the chagne in input power. Ac power gain `=("change in output power")/("change in input power")=(DeltaV_(c)xxDeltai_(c))/(DeltaV_(i)xxDeltai_(b))` `=((DeltaV_(C))/(DeltaV_(i)))xx((Deltai_(C))/(Deltai_(b)))=A_(V)xxbeta_(AC)xx(200)/(100)` `:. beta_(AC)=25` Now, AC power gain `=A_(V)xxbeta_(AC)` `=50xx25=1250` |
|
| 613. |
A `p-n` photodiode is made of a material with a band gap of `2.0 eV`. The minimum frequency of the radiation that can be absorbed by the material is nearlyA. `10xx10^(14)Hz`B. `5xx10^(14)Hz`C. `1xx10^(14)Hz`D. `20xx10^(14)Hz` |
|
Answer» Correct Answer - B p-n photodiode is a semiconductor diode that produces a significant current when illuminated. It is reversed biased but is operated below the breakdown voltage. Energy of radiation `=` band gap energy ie., `hv=2.0eV` or `v=(2.0xx1.6xx10^(-19))/(6.6xx10^(-34))approx5xx10^(14)Hz` |
|
| 614. |
The current gain of a common base transistor circuit is 0.96. On changing the emitter current by 10.0 mA, the change in the base current will beA. 9.6 mAB. 0.4 mAC. 19.6 mAD. 24 mA |
|
Answer» Correct Answer - A Current gain for common base transistor `alpha=((DeltaI_(C))/(DeltaI_(E)))_(VC)` Given, `alpha=0.96, DeltaI_(E)=10.0mA` `0.96=(DeltaI_(C))/(10.0)` `DeltaI_(C)=0.96xx10.0=9.6mA` |
|
| 615. |
The current gain for a transistor working as a common-base amplifier is `0.96`. If the emitter current is `7.2 mA`, the base current will beA. 0.39mAB. 0.43 mAC. 0.35 mAD. 0.29mA |
|
Answer» Correct Answer - D We consider current gain , `alpha` for a common base amplifier . `therefore alpha = I_C/I_E therefore 0.96 = I_C/7.2` `therefore I_C=7.2xx0.96`=6.91 mA The base current `I_B=I_C-I_E`= 7.2-6.91 = 0.29 mA |
|
| 616. |
Transfer characterstics [output voltage `(V_(o))` vs. input voltage `(V_(i))`] for a base biased transistor in `CE` configuration is as shown in the figure. For using transfor as a which, it is used A. ln region IB. Both in region (I) and (III)C. In region IIID. In region II |
|
Answer» Correct Answer - B For using the transistor as a switch, it is used in region I and region III where we get ON and OFF operations for the switch. Region II is the active region , where it is used as an amplifier. |
|
| 617. |
Transfer characterstics [output voltage `(V_(o))` vs. input voltage `(V_(i))`] for a base biased transistor in `CE` configuration is as shown in the figure. For using transfor as a which, it is used A. In region IIIB. both in region (I) and (III)C. in region IID. in region I |
|
Answer» Correct Answer - B `ItoON` `IIItooff` In `II^(nd)` state it is used as a amplifier it is active region. |
|
| 618. |
For a transistor `(I_(C))/(I_(E))=0.96`, then current gain for common emitter configurationA. 6B. 12C. 24D. 48 |
| Answer» Correct Answer - C | |
| 619. |
Output characteristic of n-p-n transistor in CE configuration is shown. From the characteristic curve determine the current gain at `V_(CE)=1V`-A. 30B. 32C. 28D. 40 |
| Answer» Correct Answer - A | |
| 620. |
In which of the transistor configurations, the voltage gain is highest ?A. common baseB. common emitterC. common collectorD. All of these |
| Answer» Correct Answer - C | |
| 621. |
In which of the configuration of a transistor , the power gain is highest ?A. common baseB. common emitterC. common collectorD. same in all of three |
| Answer» Correct Answer - B | |
| 622. |
For a transistor , in a common base arrangement the alternating current gain `alpha` is given by such that , `V_(c)` = constantA. `(DeltaI_(c))/(DeltaI_(b))`B. `(DeltaI_(b))/(DeltaI_(c))`C. `(DeltaI_(c))/(DeltaI_(e))`D. `(DeltaI_(e))/(DeltaI_(c))` |
| Answer» Correct Answer - C | |
| 623. |
Electrimagnetic waves are transverse is nature is evident byA. polarizationB. interference reflectionC. reflectionD. diffraction |
| Answer» Correct Answer - A | |
| 624. |
The combination of gates shown below yields :- A. NAND gateB. OR gateC. NOT gateD. XOR gate |
| Answer» Correct Answer - B | |
| 625. |
A tuned amplifier circuit is used to generate a carrier frequency of 2 MHz for the amplitude modulation. The value of `sqrt(LC)` isA. `(1)/(2pi xx 10^(6))`B. `(1)/(2xx10^(6))`C. `(1)/(3pi xx10^(6))`D. `(1)/(4pi xx 10^(6))` |
|
Answer» Correct Answer - D For transistor as an oscillator, `f=(1)/(2pi sqrt(LC))` where, `f=2MHz =2xx10^(6)Hz and sqrt(LC)=?` `rArr sqrt(LC)=(1)/(2pif)=(1)/(2pi xx 2xx10^(6))=(1)/(4pi xx 10^(6))` |
|
| 626. |
The forbidden energy gap in a semiconductor is of the order ofA. 0.1 evB. 0.5 eVC. 1eVD. 10 eV |
| Answer» Correct Answer - C | |
| 627. |
The forbidden energy gap in an intrinsic semiconductor isA. very largeB. zeroC. very smallD. half of the forbidden gap in a conductor |
| Answer» Correct Answer - C | |
| 628. |
Statement 1: Conductivity of semiconductor increases with increase in temperature. Statement 2: Forbidden energy gap is highest for semiconductors.A. Both statement-1 and statement-2 are true, and statement-2 is the correct explanation of statement 1.B. Both statement-1 and statement-2 are true but statement-2 is not the correct explanation of statement-2C. Statement-1 is true but statement-2 is false.D. Statement-1 is false but statement-2 true. |
| Answer» Correct Answer - C | |
| 629. |
Statement 1: Conductivity of semiconductors decreases with increase in temperature Statement-2: More electron goes from valance band to conduction band with increase in temperature.A. Both statement-1 and statement-2 are true, and statement-2 is the correct explanation of statement 1.B. Both statement-1 and statement-2 are true but statement-2 is not the correct explanation of statement-2C. Statement-1 is true but statement-2 is false.D. Statement-1 is false but statement-2 true. |
| Answer» Correct Answer - D | |
| 630. |
Instantaneous displacement current of `1.0` A in the space between the paraller plates of `1 mu F` capacitor can be established by changing potenial difference of:A. `10^(-6)V//s`B. `10^(6)V//s`C. `10^(-8)V//s`D. `10^(8)V//s` |
| Answer» Correct Answer - B | |
| 631. |
A larger parallel plate capactior, whose plates have an area of `1m^(2)` are separated each other by `1` mm, is the plates has the dielectric constant `10`, then the displacement current at this instant is:A. `25muA`B. `11muA`C. `2.2muA`D. `1.1muA` |
| Answer» Correct Answer - C | |
| 632. |
A place electromagnetic wave `F_(s)=100 cos(6xx10^(8)t+4x) V//m` Propagates in a medium of dielectric constant. The refractive index isA. 1.5B. 2C. 2.4D. 4 |
| Answer» Correct Answer - D | |
| 633. |
The velocity of electromagnetic waves in a dielectric medium `(in_(r)=4)` isA. `3xx10^(8)m//s`B. `1.5xx10^(8)m//s`C. `6xx10^(8)m//s`D. `7.5xx10^(7)m//s` |
| Answer» Correct Answer - B | |
| 634. |
The current gain of the amplifier in the common emitter configuration is 80. What is its current gain in common base configuration ?A. 0.999B. 0.909C. 0.908D. 0.988 |
|
Answer» Correct Answer - D Given, `beta=80alpha=(beta)/(1+beta)` `rArr alpha=(80)/(1+80)=(80)/(81)=0.988` |
|
| 635. |
Applying different potential at the ends of p-n junction , current is measured for increasing potential. Which curve shows the relationship between current and potential.A. B. C. D. |
| Answer» Correct Answer - C | |
| 636. |
If the two ends of a p-n junction are joined by a wire ,A. there will not be a steady current in the circuitB. there will be a steady current from the N-side to the P-sideC. There will a steady current from the P-side to the N-sideD. There may or may not be a current depending upon the resistance of the connecting wire. |
| Answer» Correct Answer - A | |
| 637. |
Symbolic representation of photodiode isA. B. C. D. |
| Answer» Correct Answer - C | |
| 638. |
Symbol of zener diode-A. B. C. D. |
| Answer» Correct Answer - A | |
| 639. |
In a chemical process, alarm systems are to be activated whenever either the pressure or the temperature in the reaction chamber exceeds certain limits. This is done by using a logic gate whose inputs will be the voltages corresponding to the high temperature or high pressure in the reaction chamber. Which logic gate shouJd be used to activate the alarms?A. AND gateB. NAND gateC. NOR gateD. OR gate |
| Answer» Correct Answer - D | |
| 640. |
Name the logic gate realised using p-n junction diode in the given Fig.Give its logic symbol. A. NAND gateB. OR gateC. AND gateD. NOR gate |
|
Answer» Correct Answer - B It represents an OR gate because we get the output by using either diode `D_1` or diode `D_2` or both `D_1` and `D_2`. |
|
| 641. |
A researcher wants an alarm to sound when the temperature of air in his controlled research chamber rises above `40^@C` or falls below `20^@C`.The alarm can be triggered by the output ofA. an AND gateB. a NAND gateC. a NOT gateD. an OR gate |
|
Answer» Correct Answer - D The alarm can be triggered by the output of an OR gate. For both the temperatures above `40^@C` and below `20^@C`. |
|
| 642. |
Statement-I : When base region has larger width, the collector current decreases. Statement-II : In transistor, sum of base current and collector current is equal to emitter current.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 | |
| 643. |
A gate in which all the inputs must be low to get a high output is calledA. A NAND gateB. An inverterC. A NOR gateD. An AND gate |
| Answer» Correct Answer - C | |
| 644. |
The materials resistance of which decreases with increases in temperature (i.e. the temperature coefficient of resistance is negative) are calledA. conductorsB. insulatorsC. semiconductorsD. all of the above |
| Answer» Correct Answer - B::C | |
| 645. |
In the circuit below, `A` and `B` represents two inputs and `C` represents the output, the circuit represents. .A. AND gateB. NAND gateC. OR gateD. NOR gate |
|
Answer» Correct Answer - C The circuit represents an OR gate. Both diodes are forward biased. `therefore` For A= 0, B = 1, C = 1 or A= 1, B = 0, C = 1 or A = 1, B = 1, C = 1 or A= 0, B = 0, C = 0 |
|
| 646. |
If A and B are two inputs in OR gate , then OR gate has an output of 0 when the value of A and B areA. A=0, B=0B. A=1,B=1C. A=1,B=0D. A=0,B=1 |
| Answer» Correct Answer - A | |
| 647. |
In the circuit below, `A` and `B` represents two inputs and `C` represents the output, the circuit represents. .A. NoR gateB. AND gateC. NAND gateD. OR gate |
| Answer» Correct Answer - D | |
| 648. |
Which of the following represents correctly the truth table of configuration of gates shown here A. `{:(A,B,Y),(0,0,0),(0,1,1),(1,0,1),(1,1,1):}`B. `{:(A,B,Y),(0,0,1),(0,1,0),(1,0,0),(1,1,1):}`C. `{:(A,B,Y),(0,0,0),(0,1,1),(1,0,1),(1,1,0):}`D. `{:(A,B,Y),(0,0,1),(0,1,1),(1,0,1),(1,1,0):}` |
| Answer» Correct Answer - C | |
| 649. |
Pure Si at 300 K has equal electron `(n_(e))` and hole `(n_(h))` concentrastions of `1.5xx10^(16)m^(-3)` doping by indium increases `n_(h)` to `4.5xx10^(22)m^(-3)`. Caculate `n_(theta)` in the doped Si- |
|
Answer» `n_(e)n_(h)=n_(i)^(2)` `n_(h)=4.5xx10^(22)m^(-3)` so, `n_(e)=5.0xx10^(9)m^(-3)` |
|
| 650. |
The majority charge carriers in `P`-type semiconductors areA. electronsB. holes (holes)C. neutronsD. protons |
| Answer» Correct Answer - B | |