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.
| 1. |
The depletion layer in diode is `1 mum` wide and the knee potential is `0.6 V`, then the electric field in the depletion layer will beA. zeroB. `0.6 Vm^(-1)`C. `6xx10^(4) V//m`D. `6xx10^(5)V//m` |
|
Answer» Correct Answer - D By using `E=V/d=0.6/(10^(-6))=6xx10^(5)V//m` |
|
| 2. |
The diagram shows an operational circuit. Which of the following correctely states the nature and magnitude of the amplification ? A. `{:("Nature","Magnitude of"(V_(0))/(V_(i))),("Inverting"," 0.2"):}`B. `{:("Nature","Magnitude of"(V_(0))/(V_(i))),("Inverting"," 5.0"):}`C. `{:("Nature","Magnitude of"(V_(0))/(V_(i))),("Non-inverting"," 0.2"):}`D. `{:("Nature","Magnitude of"(V_(0))/(V_(i))),("Non-inverting"," 5.0"):}` |
|
Answer» Correct Answer - B The inverting input of the operational amplifier is virtual ground, thus `(V_(1))/(R_(1))+(V_(0))/(R_(2)) =0 implies (V_(0))/(V_(i))=-(R_(2))/(R_(1)) =-(10 K)/(2K)=-5` It is thus an inverting amplifier with gain of `-5`. |
|
| 3. |
The following truth table corresponds to the logic gate `|(A,0,0,1,1),(B,0,1,0,1),(X,0,1,1,1)|`A. `NAND`B. `OR`C. `AND`D. `XOR` |
|
Answer» Correct Answer - B 0 |
|
| 4. |
Which of the following gates corresponds to the truth table given below ? `|(A,B,X),(1,1,0),(1,0,1),(0,1,1),(0,0,1)|`A. `XOR`B. `OR`C. `NAND`D. `NOR` |
|
Answer» Correct Answer - C This a truth table for `NAND` gate because `X=bar(A.B)`. |
|
| 5. |
The following truth table belongs to which one of the following four gates? `|(A,B,C),(1,1,0),(1,0,0),(0,1,0),(0,0,1)|`A. `OR`B. `NAND`C. `XOR`D. `NOR` |
|
Answer» Correct Answer - D For `NOR` gate, `bar(A+B)=Y` |
|
| 6. |
Pure sodium `(Na)` is a good conductor of electricity because the `3s` and `3p` atomic bands overlap to from a partially filled conduction band. By contrast the ionic sodium chloride `(NaCl)` crystal isA. InsulatorB. ConductorC. SemiconductorD. None of these |
|
Answer» Correct Answer - A In sodium chloride the `Na^(+)` and `Cl^(-)` ions both have noble gas electron configuration corresponding to completely filled bands. Since the bands do not overlaps, there must be a gap between the filled bands and the empty bands above them, so `NaCl` in an insulator. |
|
| 7. |
In an insulator, the forbidden energy gap between the valence band and conduction band is of the order ofA. `1 MeV`B. `0.1 MeV`C. `1 eV`D. `5 eV` |
|
Answer» Correct Answer - D In insulators, the forbidden energy gap is largest and it is of the orders of `6 eV`. |
|
| 8. |
In a good conductor the energy gap between the conduction band and the valence band isA. InfiniteB. WideC. NarrowD. Zero |
|
Answer» Correct Answer - D The conduction and valence bands in the conductors merge into each other. |
|
| 9. |
In a common emitter amplifier , when a signal of 40 mV is added to the input voltage, the base current changes by `100 muA` and emitter current changes by 2.1mA the trans-conductance isA. `1/20 Omega^(-1)`B. `1/50 Omega^(-1)`C. `50 Omega^(-1)`D. `15 Omega^(-1)` |
| Answer» Correct Answer - A | |
| 10. |
In the given circuit, calculate the ratio of currents through the battery if (1) Key `K_(1)` is pressed, `K_(2)` open and then (2) Key `K_(2)` is pressed, key `K_(1)` is opened A. `1/2`B. 4C. 2D. 1 |
| Answer» Correct Answer - D | |
| 11. |
A transistor is used in Common-emitter mode in an amplifier circuits. When a signal of 20 mV is added to the base-emitter voltage, the base current changes by `40 muA` and the collector current changes by 2mA. The load resistance is `5 kOmega` then the voltage gain isA. 500B. 200C. 400D. 1000 |
| Answer» Correct Answer - A | |
| 12. |
The emitter region in a PNP-junction transistor is more heavily doped than the base region, so thatA. The flow across the base region will be only because of electronsB. The flow across the base region will be only because of holesC. Recombination will be decreased in the base regionD. Base current will be high |
| Answer» Correct Answer - C | |
| 13. |
In a `PNP` transistor working as common-base amplifier, current gain is `0.96` and current is `7.2 mA`. The base current isA. `0.4 mA`B. `0.2 mA`C. `0.29 mA`D. `0.35 mA` |
|
Answer» Correct Answer - C `alpha=(i_(c))/(i_(e)) =0.96` and `i_(e)=7.2 mA` `i_(c)=0.96xxi_(e)=0.96xx7.2=6.91mA` `:. i_(e)=i_(c)+i_(b) implies 7.2=6.91+i_(b) implies i_(b)=0.29 mA`. |
|
| 14. |
For a common base configuration of `PNP` transistor `(l_(C))/(l_(E))=0.98`, then maximum current gain in common emitter configuration will beA. `12`B. `24`C. `6`D. `5` |
|
Answer» Correct Answer - B `beta=(alpha)/(1-alpha)=0.96/(1-0.96)=24`. |
|
| 15. |
If `l_(1),l_(2),l_(3)` are the lengths of the emitter, base and collector of a transistor thenA. `l_(1)=l_(2)=l_(3)`B. `l_(3) lt l_(2) lt l_(1)`C. `l_(3) lt l_(1) lt l_(2)`D. `l_(3) gt l_(1) gt l_(2)` |
| Answer» Correct Answer - D | |
| 16. |
The emitter-base junction of a transistor is .........biased while the collector-base junctio is.......biasedA. Reverse, forwardB. Reverse,reverseC. Forward,ForwardD. Forwards, reverse |
|
Answer» Correct Answer - D The emitter base juction is forward biased while collector base juction is reversed biased. |
|
| 17. |
Electrical conductivity of a semiconductorA. Decreases with the rise in its temperatureB. Increases with the rise in its temperatureC. Does not change with the rise in its temperatureD. First increases and then decreases with the rise in its temperature |
|
Answer» Correct Answer - B Resistivity decreases with rise in temperature , conductivity increases. |
|
| 18. |
A transistor is operated in common emitter configuration at `V_(c)=2 V` such that a change in the base current from `100 muA` to `200 muA` produces a change in the collector current from `5 mA` to `10 mA`. The current gain isA. `75`B. `100`C. `150`D. `50` |
|
Answer» Correct Answer - D For a transistor `I_(E)=I_(B)+I_(C)` where `I_(E)`=emitter current `I_(B)`=base current `I_(C)`=collector current and current gain `beta =(DeltaI_(C))/(DeltaI_(B))` `DeltaI_(C)=5xx10^(-3)A` `DeltaI_(B)=100xx10^(-6)A` `beta=5/100xx1000=50` |
|
| 19. |
A transistor is operated in common emitter configuration at `V_(c)=2 V` such that a change in the base current from `100 muA` to `300 muA` produces a change in the collector current from `10 mA` to `20 mA`. The current gain isA. `75`B. `100`C. `25`D. `50` |
|
Answer» Correct Answer - D `beta=(DeltaI_(C))/(DeltaI_(B))=((20-10)mA)/((300-100)mA)=(10xx10^(-3))/(200xx10^(-6))=50` |
|
| 20. |
The number of beta particles emitter by radioactive sustance is twice the number of alpha particles emitter by it. The resulting daughter is anA. Isobar of parentB. isomer of parentC. isoton of parentD. isotope of parent |
|
Answer» Correct Answer - D Let the radioactive substance be `._(Z)^(A)X` Radioactive transition is given by `._(Z)^(A)Xoverset(-alpha)to`._(Z-2)^(A-4)X overset(-2beta)to._(Z)^(A-4)X` The atoms of element having the same atomic numbers but different mass numbers are called isotopes. So, `._(Z)^(A)X` and `._(Z)^(Z-4)X` are isotopes. |
|
| 21. |
In a `p-n` junction diode, change in temperature due to heatingA. affects only reverse resistanceB. affects only forward resistanceC. does not affect resistance of `p-n` junctionD. affects the overall `V-I` characterstics of `p-n` junction |
|
Answer» Correct Answer - D Affects the overall `V-I` characteristics of `p-n` junction |
|
| 22. |
If a small amount of antimony is added to germanium crystalA. the antimony becomes an acceptor atomB. there will be more free electrons than holes in the semiconductorC. its resistance is increasedD. it becomes a `p`-type semiconductor |
|
Answer» Correct Answer - B When a small amount of antimony is added to germanium crystal, the crystal becomes `n`-type semiconductor because antimony is a pentavelent substrate. It excess free electrons. |
|
| 23. |
In the circuit given below, `V(t)` is the sinusiodal voltage source, voltage drop `V_(AB)(t)` across the resistance `R` is A. Is half wave rectifiedB. Is full wave rectifiedC. Has the same peak value in the positive and negative half cyclesD. Has different peak values during positive and negative half cycle |
|
Answer» Correct Answer - D In positive half cycle one diode is in forward biasing and other is in reverse biasing while in negative half cycle thier polarity reverses, and direction of current is opposite through `R` for positive and negative half cycles so output is not rectified. |
|
| 24. |
In the circuit given below, `V(t)` is the sinusiodal voltage source, voltage drop `V_(AB)(t)` across the resistance `R` is A. Is half wave rectifiedB. Is full wave rectifiedC. Has the same peak value in the positive and negative half cyclesD. Has different peak values during positive and negative half cycle |
|
Answer» Correct Answer - D In positive half cycle one diode is in forward biasing and other is in reverse biasing while in negative half cycle thier polarity reverses, and direction of current is opposite through `R` for positive and negative half cycles so output is not rectified. Since `R_(1)` and `R_(2)` are different hence the peaks during positive half and negative half of the input signal will be different. |
|
| 25. |
In a common emitter amplifier, using output reisistance of `5000` ohm and input resistance fo `2000`ohm, if the peak value of input signal voltage is `10 m V` and `beta=50`, then the peak value of output voltage isA. `5xx10^(-6)` voltB. `2.5xx10^(-4)`C. `1.25` voltD. `125` volt |
|
Answer» Correct Answer - B In common emitter mode, the transistor is current amplifier `DeltaI_(B)=(10xx10^(-3))/2000=5xx10^(-6) A` Again, `beta=(DeltaI_(C))/(DeltaI_(B))` or `DeltaI_(C)=betaxxDeltaI_(B)` `=50xx5xx10^(-6)A =250xx10^(-6)A` `=2.5xx10^(-4) A` Peak value of output voltage `=2.5xx10^(-4)xx5000` volt=`1.25` volt |
|
| 26. |
A sinusiodal voltage of `rms` value `200` volt is connected to the diode and capacitor `C` in the shown so that half wave reactification occurs. The final potential difference in volt across `C` is A. `500`B. `200`C. `283`D. `141` |
|
Answer» Correct Answer - C Junction diode conductors during alternate half-cycles of conduction, the capacitor will change itself to peak value of supply voltage. `:.` Voltage across capacitor `=E_(rms)sqrt(2)=200xxsqrt(2)V=282.8 V~~283V` |
|
| 27. |
In a common emitter amplifier, using output reisistance of `5000` ohm and input resistance fo `2000`ohm, if the peak value of input signal voltage is `10 m V` and `beta=50`. The power gain isA. `125xx50`B. `125/90`C. `1.25xx50`D. `2.5xx10^(4)` |
|
Answer» Correct Answer - A Voltage gain `=(Output Vol tag e)/(Input vol tag e)=1.25/(10xx10^(-3))=125` Power gain =voltage gainxcurrent gain `125xx50` |
|
| 28. |
In a `CE` transistor amplifier, the audio signal voltage across the collector resistance of `2 k Omega` is `2 V`. If the base resistance is `1 k Omega` and the current amplification of the transistor is `100`, the input signal voltage isA. `10 mV`B. `0.1 V`C. `1.0 V`D. `1mV` |
|
Answer» Correct Answer - A Current amplification factor `beta=(DeltaI_(C))/(DeltaI_(B))` Collector resistance `DeltaI_(C)=(2V)/(2xx10^(3) Omega)=1xx10^(-3) A` Base current `Delta I_(B)=(V_(B))/(R_(B))=(V_(B))/(R_(B))=(V_(B))/(1xx10^(3))=V_(B)xx10^(-3)` Given, `beta=100` Now, `100=(10^(-3))/(V_(B)xx10^(-3))` `V_(B)=1/100 V=10 mV` |
|
| 29. |
For `CE` transistor amlifier, the audio signal voltage across the collector resistance of `2 k Omega` is `4V`. If the currents amplification factor of the transistor is `100` and the base resistance is `1 k Omega`, then the input signal voltage isA. `30 mV`B. `15 mV`C. `10 mV`D. `20mV` |
|
Answer» Correct Answer - D `beta=100, V_(0)=4V, R_(i)=10^(3) Omega`, `R_(0)=2xx10^(3) Omega, V_(i)=?` `(V_(0))/(V_(i))=beta(R_(0))/(R_(i)) implies 4/(V_(i))=100xx(2xx10^(3))/(10^(3))` `implies V_(i)=20 mV` |
|
| 30. |
A semiconductor is known to have an electron concentration of `6xx10^(12)` per cubic centimeter and a hole concentration of `8xx10^(13)` per cubic centimeter. Is this semiconductor N-type or P-type ? |
| Answer» Correct Answer - p-type | |
| 31. |
In the given circuit the voltage drop across the diode is 0.8 V, if the diode can withstand current upto maximum of 30 mA, then find the maximum voltage of the battery |
|
Answer» Let V is the maximum forward biased voltage then , `V=lR+V_(D)` `V=30xx10^(-3) xx500+0.8` `V=15+0.8` V=15.8V |
|
| 32. |
C, Si and Ge have same lattice structure. Why is C insulator, while Si and Ge intrinsic semiconductors ? |
| Answer» Energy required to take out outer orbit electrons is less for Ge and Si but very large for C. so conduction electron is not easily available is not easily available for C and hence it is insulator . | |
| 33. |
Assertion: LED is used in display units because it emits light when current passes through it. Reason: In LED, electrons comes from conduction band to valence band when it emits energy. |
| Answer» Correct Answer - B | |
| 34. |
Intrinsic semiconductor at absolute zero temperature is aA. Good conductorB. Good semiconductorC. Perfect insulatorD. Perfect conductor |
| Answer» Correct Answer - C | |
| 35. |
In a p-n junction diode the value of drift current through depletion regionA. Decreases in forward biasingB. Decreases in reverse biasingC. Remains unchanged during forward or reverse biasingD. Increases during reverse biasing |
| Answer» Correct Answer - D | |
| 36. |
The probability of electrons to be found in the conduction band of an intrinsic semiconductor at a finite temperatureA. increases exponentially with increasing band gapB. decreases exponentially with increasing band gapC. decreases with increasing temperature.D. is independent of the temperature and band gap. |
| Answer» Correct Answer - B | |
| 37. |
Assertion:To make p type semiconductor , pentavalent impurity like phosphorus is mixed with Si. Reason: Pentavalent impurity produces free electrons. |
| Answer» Correct Answer - D | |
| 38. |
In doped semiconductor one dopent atom is kept typically for how many silicon atoms ?A. `10^(7)`B. `10^(4)`C. `10^(3)`D. `10^(9)` |
| Answer» Correct Answer - A | |
| 39. |
Carbon , silicon and germanium have four valence elcectrons each . These are characterised by valence and conduction bands separated by energy band - gap respectively equal to ` (E_g)_(c) (E_g)_(si) ` and ` (E_g)_(Ge) `. Which of the following statements ture ?A. `(E_g)_(Si) lt (E_g)_(Ge) lt (E_g)_C`B. `(E_g)_(C) lt (E_g)_(Ge) lt (E_g)_(Si)`C. `(E_g)_(C) gt (E_g)_(Si) gt (E_g)_(Ge)`D. `(E_g)_(C) = (E_g)_(Si) =(E_g)_(Ge)` |
| Answer» Correct Answer - C | |
| 40. |
A transistor has `alpha=0.90`, the emitter current is 15mA, what is (a) The collector current ? (b) The base current ? (c) Gain `beta` ? |
|
Answer» Given `alpha=0.95 ` and `I_(E)=15mA` (a) as `alpha=(I_(C))/(-I_(E))` `rArr I_(C)=alphaI_(E)` `rArr I_(C)=0.95xx15` `rArr I_(C)=14.25 mA` (b) As `I_(E)=I_(B)+I_(C)` `rArr I_(B)=I_(E)-I_(C)` `rArr I_(B)=15-14.25` `rArr I_(B)=0.75 mA` (c) `beta=(alpha)/(1-alpha)` `rArr beta=0.90/(1-0.90)` `rArr =0.90/0.10` `rArr beta=9` |
|
| 41. |
Serious draw back of the semiconductor device isA. Are costlyB. Do not last for long timeC. Pollute the environmentD. Cannot be used with high voltage |
| Answer» Correct Answer - D | |
| 42. |
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 ? |
| Answer» Correct Answer - 50 | |
| 43. |
Function of rectifier isA. To convert ac into dcB. To convert dc into acC. Both (1) and (2)D. None of these |
| Answer» Correct Answer - A | |
| 44. |
Zener breakdown takes place ifA. Doped impurity is lowB. Doped impurity is highC. Less impurity in N-partD. Less impurity in p-type |
| Answer» Correct Answer - B | |
| 45. |
What is the output form of full-wave rectifier ? |
| Answer» A pulsating unidirectional voltage | |
| 46. |
In a zener-regulated power supply a zener diode with `V_(Z)=6 ` volt is used for regulation. The load current is to be 4mA and the unregulated input is 10 volt. What should be the value of series resistor `R_(S)`, if the current through diode is five times the load current ? |
| Answer» Correct Answer - `R_s=167 Omega` | |
| 47. |
For a `CE` transistor amplifier, the audio signal voltage across the collector resistance of `2k Omega` is `2V`. Suppose the current amplification factor of the transistor is `100`. The value of `R_(B)` in series with `V_(BB)` supply of `2V`, if the `DC` base current has to be `10` times the signal current is. |
|
Answer» `R_B=14 kOmega` dc drop =20 volt |
|
| 48. |
The V-I characteristic of a silicon diode is shown in the figure. Calculated the resistance of the diode at `(a) I_(D)=15mA` `(b) V_(D)=-10 V` |
|
Answer» `(a)r=(DeltaV)/(Deltal)=((0.8-0.7))/((20-10)mA)=10 Omega` (b)`r="10V"/(1 mu A)=10^7 Omega` |
|
| 49. |
In a P-N junction , for a 0.2 V change in the forward bias voltage, the corresponding change in current is 20 mA. The dynamic resistance of the P-N junction isA. `30 Omega`B. `10 Omega`C. `40 Omega`D. `20 Omega` |
| Answer» Correct Answer - B | |
| 50. |
In case of `NPN`-transistor the collector current is always less than the emitter current becauseA. Collector side is reverse biased and emitter side is forward biasedB. After electrons are lost in the base and only remaining ones reach the collectorC. Collector is positive and emitter is at theD. Collector being reverse biased attracts less electrons |
|
Answer» Correct Answer - B `i_(e)=i_(b)+i_(c) implies i_(c)=i_(e)-i_(b)` |
|