InterviewSolution
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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.
| 1. |
In the circuit below, `A` and `B` represents two inputs and `C` represents the output, the circuit represents. .A. `NOR` gateB. `AND` gatesC. `NAND`gateD. `OR`gate |
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Answer» Correct Answer - D It is diode-diode logic ckt of `OR` gate. |
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| 2. |
The Boolean expression of the output `Y` of the inputs `A` and `B` for the circuit shown in the fig. . |
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Answer» The output of `AND` gate `1` is `overline AB` The ouput of `AND` gate `2` is `A overline B` `:.` The output of `OR` gate is `Y = overline A B + A overline B`. |
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| 3. |
When `p-n` junction diode is forward biased thenA. the depletion region is reduced and barrier height is increasedB. the depletion region is widened and barrier height is reducedC. both the depletion region and barrier height are reducedD. both the depletion region and barrier height are increased |
| Answer» Correct Answer - C | |
| 4. |
There is a sudden increase in current in zener diode isA. Due to rupture of bondsB. Resistance of deplection layer becomes lessC. Due to high dopingD. Due to less doping |
| Answer» Correct Answer - A | |
| 5. |
In a `p-n` junction photo cell, the value of the photo electromotive force produced by monochromatic light is proportional toA. The barrier voltage at the `p-n` junctionB. The intensity of the light falling on the cellC. The frequency of the light falling on the cellD. The voltage applied at the `p-n` junction |
| Answer» Correct Answer - B | |
| 6. |
The following one represents logic addition isA. `1 + 1=2`B. `1 + 1 = 10`C. `1 + 1 = 1`D. `1 + 1 = 11` |
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Answer» Correct Answer - C According to Boolean algebra, `1 + 1=1`. |
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| 7. |
For a logic `0101` the waveform is.A. B. C. D. |
| Answer» Correct Answer - A | |
| 8. |
The process of converting alternating current into direct current is known asA. modulationB. amplificationC. detectionD. rectification |
| Answer» Correct Answer - D | |
| 9. |
On increasing reverse voltage in a `p-n` junction diode the value of reverse current willA. gradually increasesB. first remains constant and then suddenly increaseC. remains constantD. gradually decrease |
| Answer» Correct Answer - B | |
| 10. |
In a `p-n` junction diode, the currect `I` can expressed as `I=I_(0) exp((eV)/(2k_(B)T)-1)` where `I_(0)` is called the reverse saturation current, `V` is the voltage across the diode and is positive for forward bias and negative for reverse bias, and `I` is the current through the diode, `K_(B)` is the Boltzmann constant `(8.6 xx 10^(-5) eV//K)` and `T` is the absolute temperature. If for a given diode `I_(o) = 5 xx 10^(-12) A` and `T = 300 K`, then (a) What will be the forward current at a formward voltage of `0.6V` ? (b) What will be the increase in the current if the voltage across the diode is increased to `0.7 V` ? ( c) What is the dynamic resistance ? (d) What will be current if reverse bias voltage changes from `1 V` to `2 V` ? |
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Answer» `I_(o) = 5 xx 10^(-12) A, k = 8.6 xx 10^(-5) eVk^(-1)` `= 8.6 xx 10^(-5) xx 1.6 xx 10^(-19) Jk^(-1)` (a) `I =I_(0) (e^(.^(ev)//_(2kT))-1)`, For `V = 0.6V`, `I = 5 xx 10^(-12) ((1.6 xx 10^(-19) xx 0.6)/(e^(2 xx 8.6 xx10^(-5)xx1.6 xx10^(-19) xx300))-1)` =`5 xx 10^(-12) (e^(23.52) -1)` =`5 xx 10^(-12) (1.256 xx 10^(10) -1) = 0.0628.4` (b) For `v=0.7 v`, we have `I = 5 xx 10^(-12) ((1.6 xx 10^(-19) xx 0.7)/(e^(2 xx 8.6 xx10^(-5)xx1.6 xx10^(-19) xx300))-1)` `I = 5 xx 10^(-12) (e^(27.32) -1)` =` 5 xx 10^(-12) (6.054 xx 10^(11) -1) = 3.0271 A` `:. Delta I = 3.271 -0.0628 = 2.9643 A` ( c) `Delta I =2.9643, Delta v = 0.7 -0.6 = 0.1 V` dynamic resistance `= (Delta v)/(Delta I) = (0.1)/(2.9643) = 0.0337 Omega` (d) For change in voltage from `1` to `2 v`, the current will remain equal to `I_(0) = 5 xx 10^(-12) A`. It shows that the diode possesses practically infinite resistance in reverse biasing. |
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| 11. |
Pickout the incorrect statement regarding reverse saturation current in the `p-n` junction diode.A. this currect doubles for every `100^@C` rise of temperatureB. the current is due to minority carriersC. the curreny carriers are produced by thermal agitationD. reverse saturation current is also known as leakage current |
| Answer» Correct Answer - A | |
| 12. |
A `p-n` junction `(D)` shown in the figure can act as a rectifier. An alternating current source `(V)` is connected in the circuit. .B. C. D. |
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Answer» Correct Answer - C Given figure is halfwave rectifier. |
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| 13. |
Truth table for system of four `NAND` gates as shown in figure is : .A. `{:(A,B,Y),(0,0,0),(0,1,1),(1,0,1),(1,1,0):}`B. `{:(A,B,Y),(0,0,0),(0,1,0),(1,0,1),(1,1,1):}`C. `{:(A,B,Y),(0,0,1),(0,1,1),(1,0,0),(1,1,0):}`D. `{:(A,B,Y),(0,0,1),(0,1,0),(1,0,0),(1,1,1):}` |
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Answer» Correct Answer - A `Y=overline(overline((A. overline(A.B))).(overline(overline(A.B).B)))= (overline(overline(A. overline(A.B))))+(overline(overline(overline(A.B)B)))` `Y=A.(overline(A)+ overline(B))+(overline(A)+overline(B)).B` `=A.overline(A)+A. overline(B)+ overline(A).B+ B. overline(B)` `Y=(A.overline(B))+(overline(A).B)`. |
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| 14. |
Among the following is not the function of `NOT` gate isA. stop a signalB. invert an input signalC. complement a signalD. change the logic in a digital circuit |
| Answer» Correct Answer - A | |
| 15. |
Decimal number `15` is equivalent to the binary number :A. `110001`B. `000101`C. `101101`D. `001111` |
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Answer» Correct Answer - D `2|__ul(15)` `2|__ul(7-1)`. `:. 15_((10))=1111_((2))`. |
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| 16. |
In the Binary number system the number `100` represents :A. oneB. threeC. fourD. hundred |
| Answer» Correct Answer - C | |
| 17. |
The equivalent of `110` in the decimal number isA. 2B. 4C. 8D. 6 |
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Answer» Correct Answer - D `{:(2,1,1),(1,1,0):}_((2))=1 xx 2^(2) + 1 xx 2^(1) + 0 xx 2^@ = 6`. |
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| 18. |
If the resistivity of copper is `1.7 xx 10^(-6) Omega cm`, then the mobility of electrons in copper, if each atom of copper contributes one free electron for conduction, is [The amomic weight of copper is `63.54` and its density is `8.96 g//c c]` :A. `23.36 cm^(2)//Vs`B. `503.03cm^(2)//Vs`C. `43.25 cm^(2)//Vs`D. `88.0 cm^(2)//Vs` |
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Answer» Correct Answer - C Mobility of electron `mu =(sigma)/(n e)-(1)` also respectivity `rho = (1)/(sigma)` From Eqs. (i) and (ii), we get `mu = (1)/(n e rho)` Here, `n` = number of free electrons per unit volume `n=(N_(0) xxd)/("atomicweight")` `n=(6.023 xx10^(23) xx8.96)/(63.54) =8.5 xx 10^(22)` From Eqs. (iii) and (iv), we get `mu=(1)/(8.5 xx 10^(22) xx 1.6 xx 10^(-19) xx 1.7 xx 10^(-6))` `:. mu=43.25 cm^(2)//Vs`. |
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| 19. |
The main cause of avalence breakdown isA. collision by ionisationB. high dopingC. recombination of electrons and holesD. low doping |
| Answer» Correct Answer - A | |
| 20. |
When `p-n` junction is forward biased, the current across the junction is mainly due toA. diffusion of chargesB. drifting of chargesC. both diffusion and drifting of chargesD. holes only |
| Answer» Correct Answer - A | |
| 21. |
Considering `a.p-n` junction as a capacitor, forward with `p` and `n` material acting as thin metal electrodes and depletion layer width acting as seperation between them. Basing on this assume that a `n-p-n` transistor is working as an amplifier in `CE` configuration. If `C_(1)` and `C_(2)` are base-emiter and collector emitter junction capacitances, then :A. `C_(1) gt C_(2)`B. `C_(1) lt C_(2)`C. `C_(1) = C_(2)`D. `C_(1) = C_(2) = 0` |
| Answer» Correct Answer - A | |
| 22. |
The potential in the depletion layer due to.A. ElectronsB. HolesC. IonsD. Forbidden band |
| Answer» Correct Answer - C | |
| 23. |
`p-n` junction in reverse bias behaves likeA. an inductorB. a condenserC. amplifierD. an off switch |
| Answer» Correct Answer - D | |
| 24. |
In forward bias the depletion layer behaves likeA. an insulatorB. a conductorC. a semiconductorD. capacitor |
| Answer» Correct Answer - B | |
| 25. |
`p-n` junction diode acts asA. ohmic resistanceB. non-ohmic resistanceC. both `1` and `2`D. amplifier |
| Answer» Correct Answer - B | |
| 26. |
The current through any `p-n` junction is due to (a) drift of charge carriers (b) diffusion of charge carriers ( c) different concentrations of same type of charge carriers in different regions. (d) Same concentrations of same type of charge carriers in different regionsA. `a,b` and `c`B. `a` and `b` onlyC. only `d`D. `a,b,c,d` |
| Answer» Correct Answer - A | |
| 27. |
The depletion region isA. region of opposite chargesB. neutral regionC. region of infinite energyD. regon of free current carriers |
| Answer» Correct Answer - D | |
| 28. |
A semiconductor has an electron concentration of `0.45 xx 10^(12) m^(-3)` and a hole concentration of `5.0 xx 10^(20) m^(-3)`. Calculate its conductivity. Given electron mobility `= 0.135 m^(2) V^(-1) s^(-1)`, hole mobility `= 0.048 m^(2) V^(-1) s^(-1)`, |
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Answer» The conductivity of a semicon ductor is the sum of the conductivities due to electrons and holes and is given by `sigma = sigma_(e) + sigma_(h) =n_(e)e mu_(e) +n_(h) e mu_(h) =e(n_(e)mu_(e)+n_(h)mu_(h))` As per given date, `n_(e)` is negligible as compared to `n_(h)`, so that we can write `sigma = en_(h) mu_(h)` =`(1.6 xx 10^(-19) C) (5.0 xx 10^(20) m^(-3))(0.048 m^(2) V^(-1) s^(-1))` =`3.84 Omega^(-1) m^(-1) =3.84 m^(-1)`. |
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| 29. |
The thickness of depletion layer is approximatelyA. `1 mu m`B. `1 mm`C. `1 cm`D. `1 m` |
| Answer» Correct Answer - A | |
| 30. |
In positive logic, the logic state `1` corresponds toA. positive voltageB. zero voltageC. lower voltage levelD. higher voltage level. |
| Answer» Correct Answer - D | |
| 31. |
In the Boolean algebra : `A + B =`A. `overline(A)+overline(B)`B. `A.B`C. `overline(overline(A))+overline(overline(B))`D. `overline(overline(A))+overline(B)` |
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Answer» Correct Answer - C `overline(overline(A))A and overline(overline(B)) =B`. |
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| 32. |
Consider the following statement `A` and `B` and identify the correct choice of the given answers `A:` The width of the depletion layer in a `P-N` junction diode increases in forwards biase `B:` In an intrinsic semiconductor the fermi energy level is exactely in the middle of the forbidden gapA. `A` is true and `B` is falseB. Both `A` and `B` are falseC. `A` is false and `B` is trueD. Both `A` and `B` are true |
| Answer» Correct Answer - C | |
| 33. |
The main cause of Zener breakdown is.A. the base semiconductor being germaniumB. production of electron - hole pairs due to thermal exitationC. low dopingD. high doping |
| Answer» Correct Answer - D | |
| 34. |
In a given circuit as shown the two input wave forms `A` and `B` are applied simultaneously. The resultant wave format at `Y` is.A. B. C. D. |
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Answer» Correct Answer - A Logic of `A` is `1010`, Logic of `B` is `1001` Logic of `Y` is `0111`. |
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| 35. |
A `PN` junction diode cannot be usedA. as rectifierB. for converting light energy to electric energyC. for gettting light radiationD. both majority and minority carriers on either side of junction |
| Answer» Correct Answer - D | |
| 36. |
The potential barrier at `PN` junction is due toA. fixed acceptor and donor ions on either side of the junctionB. minority carriers on either side of the junctionC. majority carriers on either side of the junctionD. both majority and minority carriers on either side of junction |
| Answer» Correct Answer - A | |
| 37. |
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 is forward bias of junctionB. `3` corresponds to forward bias of junction and `1` corresponds to reverse bias junctionC. `1` corresponds to forward bias and `3` corresponds to reverse bias junctionD. `3` and `1` both correspond to reverse bias and `3` corresponds to revers bias of junction |
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Answer» Correct Answer - B When `p-n` junction is forward biased, it opposes the potential barrier junction, when `p-n` junction is reverse biased, it supports the potential barrier junction resulting increases in potential barrier across the junction. |
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| 38. |
In the diagram, the input is across the terminals `A` and `C` and the output is across the terminals `B` and `D`, then the outputs is A. ZeroB. Same as the inputC. Full wave rectifiedD. Half-wave rectified |
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Answer» Correct Answer - C The given question is based on Bridge Rectifier. This is the most widely used full-wave rectifier. It makes use of four diodes `D_(1),D_(2),D_(3),D_(4)` connected in the four arms of a bridge. Bridge rectifier does not require a centre-tapped transformer. |
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| 39. |
In figure , assuming the diodes to be ideal , A. `D_(1)` is forward biased and `D_(2)` is reverse biased and hence current flows from `A` to `B`B. `D_(2)` is forward biased and `D_(1)` is reverse biased and hence no current flows from `B` to `A` and vice versaC. `D_(1)` and `D_(2)` are both forward biased and hence current flows form `A` to `B`D. `D_(1)` and `D_(2)` are both revrese biased and hence no current flows from `A` and `B` and vica-versa. |
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Answer» Correct Answer - B In the given circuit p-side of `p-n` function `D_(1)` is connected to lower voltage and n-side of `D_(1)` to higher voltage. Thus `D` is reverse biased. The p-side of `p-n` junction `D_(2)` is lower potential. Therefore `D_(2)` is forward biased. Hence, current flows through the junction `B` to `A`. |
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| 40. |
In the circuit shown. The potential drop across each capacitor is (assuming the two diodes are ideal). .A. 12 V,12 VB. 16 V,8 VC. zero, 24 VD. 8 V, zero |
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Answer» Correct Answer - B The diode connected parallel to the battery is reverse bias. So current will not pass through it. So total emf divided among `C_(1)` and `C_(2)` and in the inverse ratio of their capacities `V=(q)/(C),(V_(1))/(V_(2)) =(C_(2))/(C_(1))` `V_(1)=(VC_(2))/(C_(1)+C_(2)),V_(2)=(VC_(1))/(C_(1)+C_(2))`. |
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| 41. |
The correct curve between potential `(V)` and distance `(d)` near `p-n` junction is.A. B. C. D. |
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Answer» Correct Answer - A In depletion layer barrier potential `V=E.d` and beyond depletion layer potential is constant. |
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| 42. |
Find the maximum voltage across `AB` in the circuit shown in Fig. Assume that diode is ideal. . |
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Answer» As the diode is treated ideal, its forward resistance `R_(f) = zero`. It acts as short circuit. So `10 k Omega` is in parallel with `15 k Omega` and the effective resistance across `AB` is `R_(AB) =(10 xx 15)/(10+15)=(10 xx15)/(25) =6k Omega` `6 k Omega` is in series with `5 kOmega` `:.` total resistance `= R_(T) = 6 k Omega + 5 k Omega = 11 k Omega`, `V=30 V`. Current drawn from the battery is `I=(V)/(R_(T))=(30V)/(11 k Omega) =2.72 mA` `V_(AB) =IR_(AB)=2.72 mA xx 6k Omega = 16.32 V`. |
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| 43. |
A voltage amplifier operated from a `12 `volt battery has a collector load `6 k Omega`. Calculate the maximum collector current in the circuit.A. `0.5 mA`B. `1 mA`C. `3 mA`D. `2mA` |
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Answer» Correct Answer - D `I_(c)=(V_(c c))/(R_(L))`, |
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| 44. |
A `Si` and a `Ge` diode has identical physical dimensions. The band gap in `Si` is larger than that in `Ge`. An indentical reverse bias is applied across the diodes.A. The reverse current in `Ge` is larger than that in `Si`B. The reverse current in `Si` is larger than that in `Ge`C. The reverse current is identical in the two diodesD. The relative magnitude of the reverse currents cannot be determined from the given data only. |
| Answer» Correct Answer - C | |
| 45. |
`Ge` transistor can be operated at a temperatureA. upto `90^@ C`B. upto `40^@ C`C. upto `100^@ C`D. upto `20^@ C` |
| Answer» Correct Answer - A | |
| 46. |
An `n-p-n` transistor power amplifier in `C-E` configuration gives.A. Voltage amplification onlyB. Current amplification onlyC. Both current and voltage amplificationD. Only power gain of unity |
| Answer» Correct Answer - C | |
| 47. |
A `n-p-n` transistor conducts whenA. both collector and emitter are positive when respect to the baseB. collector is positive and emitter is negative with respect to the baseC. collector is positive and emitter is at same potential as the baseD. both collector and emitter are negative with respect to the base |
| Answer» Correct Answer - B | |
| 48. |
In a transistor.A. both emitter and the collector are equally dopedB. base is more heavily doped than collectorC. collector is more heavily doped than the emitterD. the base is made very thin and is lightly doped |
| Answer» Correct Answer - D | |
| 49. |
When N-P-N transistor is used as an amplifier-A. electrons move from collector to baseB. holes move from collector to baseC. electrons move from base to emitterD. holes move from base to emitter |
| Answer» Correct Answer - D | |
| 50. |
When N-P-N transistor is used as an amplifier-A. electrons move from base to collectorB. holes moves from emitter to baseC. holes move from collector to baseD. holes move from base to emitter |
| Answer» Correct Answer - A | |