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. |
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. `5xx10^(6)Vm^(-1)`B. `5xx10^(-7)Vm^(-1)`C. `5xx10^(5)Vm^(-1)`D. `5xx10^(-1)Vm^(-1)` |
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Answer» Correct Answer - C In forward biasing condition, the linner electric field is given by `E=(DeltaV)/(Deltar)` `or |E|=(DeltaV)/(Deltar)=(5xx10^(-1))/(10^(-6))` `=5xx10^(5) Vm^(-1)` |
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| 2. |
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. `0.6Vm^(-1)`C. `6xx10^(4) V//m`D. `6xx10^(5) V//m` |
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Answer» Correct Answer - D By using `E=(V)/(d)=(0.6)/(10^(-6))=6xx10^(5) V//m` |
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| 3. |
The knee voltage of a p-n junction diode is 0.8 V and the width of the depletion layer is `2muA` . The electric field in the depletion layer isA. 4 MV/mB. 0.4 MV/mC. 4 KV/mD. 0.4 KV/m |
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Answer» Correct Answer - B `E=(V)/(d)=(0.8)/(2xx10^(-6))=0.4xx10^(6) V//m` |
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| 4. |
The junction diode in the following circuit requires a minimum current of `1 mA` to be above the knee point `(0.7 V)` of its `I-V` characterstic curve. The voltage across the diode is independent of current above the knee point. If `V_(B)=5 V`, then the maximum value of `R` so that the voltage is above the knee point, will be A. `4.3 k Omega`B. `860 k Omega`C. `4.3 k Omega`D. `860 Omega` |
| Answer» Correct Answer - A | |
| 5. |
In the study of transistor as an amplifier , `alpha=l_C/l_E` and `beta=I_C/I_B` where `I_C,I_E` and `I_B` are the collector, emitter and base currents respectively. The correct relation between `alpha` and `beta` is given byA. `beta=(1-alpha)/alpha`B. `beta=alpha/(1-alpha)`C. `beta=(1+alpha)/alpha`D. `beta=alpha /(1+alpha )` |
| Answer» Correct Answer - B | |
| 6. |
A silicon diode is connected to a load resistance `R_(L)` as shown in the figure. If `V_(in)=15V` and `R_(L)=10kOmega` (a) If barrier voltage, `V_(B)=0.7V` then calculate (i) output voltage across `R_(L)` (ii) current in diode and ltbRgt (iii) forward resistance. (b) If diode is assumed ideal, then what will be (i) output voltage and (ii) output current in diode? |
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Answer» (a) (i) `V_(out)=V_( i n)-V_(B)=15-0.7=14.3V` (ii) `L=(V_(out))/(R_(L))=(14.3)/(10xx10^(3))=1.43mA` (iii) `r_(f)=(V_(B))/(I)=(0.7)/(1.43xx10^(-3))=490Omega` (b) For ideal diode, `r_(f)=0,V_(B)=0` (i) `V_(out)=V_(i n)=15V` (ii) ` I=(V_(out))/(RF_(L))=(15)/(10xx10^(3))=1.5mA` |
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| 7. |
For the transistor circuit shown in figure , evaluate `V_E , R_B and R_E`. Given `I_C = 1 mA`, ` V_(CE) = 3V , V_(BE) = 0.5 V , V_(CC )= 12 V and beta = 1000` |
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Answer» Correct Answer - A::B Consider the figure given here to solve this problem ` I_C = I_E ` [As base current is very small] ` R_C = 7.8 K Omega` From the figure, ` I_C(R_C + R_E) + V_(CE) = 12` ` (R_E + R_C) xx 1 xx 10^(-3) + 3 = 12 ` ` R_E + R_C = 9 xx 10^(3) = 9 k Omega ` ` R_E = 9 -7.8 = 1.2 k Omega ` ` V_E = I_E xx R_E ` ` = 1 xx 10^(-3) xx 1.2 xx 10^(3) = 1.2 V ` Voltage , ` V_B = V_E + V_(BE) = 1.2 + 0.5 = 1.7 V ` Current, `I = V_B / (20 xx 10^(3)) = 1.7 /( 20 xx 10^(3)) ` ` = 0.085 mA ` Resistance , `R_B = (12-1.7) / (1_C / beta) + 0.085 = 10.3 / (0.01 + 0.085) ` [ Given , ` beta = 100`] ` = 108 k Omega ` |
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| 8. |
(i) calculate the value of output voltage `V_(0)` and current `I` if silicon diode and germanium diode conduct at `0.7V` and `0.3 V` respectively. Fig. (ii) If now germanium diode is cinnected to `12 V` in reverse polarity, find new values of `V_(0)` and `I`. |
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Answer» As Ge diode conducts at `0.3V`, both Ge and Si are in parallel, so `p.d.` across each is `0.3V`. At `0.3V`, Si diode offers infinite resistance and it does not conduct. `p.d.` across `R=5kOmega` is `12-0.3=11.7V` `V_(0)11.7V` `i=(V_(0))/(R )=(11.7)/(5xx10^(3))=2.34mA` When the connection of Ge diode are reversed, it will be in reverse biased and for conduction, the `p.d.` across the silicon diode will be `0.7V` `p.d.` across `R=5kOmega` is `12-0.7=11.3V` `V_(0)=11.3V` `i=(V_(0))/(R )=(11.3)/(5xx10^(3))=2.26mA` |
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| 9. |
The emitter of transistor is doped the heaviest because itA. receives the inputB. is supplier of charge carriersC. dissipates minimum powerD. should have low resistance |
| Answer» Correct Answer - B | |
| 10. |
A device whose one end is connected to `-ve` terminal and other end connected to `+ve` terminal. If both ends are interchanged with suppy then current is not flowing then device will be-A. P-N junctionB. TransistorC. Zener diodeD. Triode |
| Answer» Correct Answer - A | |
| 11. |
In a `p-n` junction diode, the barrier potential opposes diffusion ofA. minority carriers in both regions onlyB. Majority carriersC. Electrons in N-regionD. Holes in P-Region |
| Answer» Correct Answer - B | |
| 12. |
The barrier potential in a p-n junction diode is 0.3 volts . The current required is 6 mA. If a resistance of `200 Omega` is connected in series with the junction diode then the e.m.f. of the cell required for use in the circuit isA. 0.3 VB. 1.2 VC. 0.9 VD. 1.5 V |
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Answer» Correct Answer - D `V_(b)=0.3 V, I=6 mA , R= 200 Omega , E=?` `I=(E-V_(b))/(R)` `E=IR+V_(b)` `=6xx10^(-3) xx200+0.3` `=1.2+0.3` `=1.5 V` |
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| 13. |
In an intrinsic semiconductor, the fermi energy level isA. in the middle of forbidden gapB. Below the middle of forbidden gapC. Above the middle of forbidden gapD. outside the forbidden gap |
| Answer» Correct Answer - A | |
| 14. |
The energy of a photon of sodium light `(lambda=589 nm)`equal the band gap of a semiconducting material.(a)Find the minimum energy E requried to create a hole-electron pair.(b)Find the value of `E//kT`at a temperature of 300K. |
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Answer» (a) `E=(hc)/(lambda)=(1242)/(589)=2.1eV` (b) `(E)/(kT)=(2.1xx1.6xx10^(-19))/(1.38xx10^(-23)xx300)=81` It is difficult for the thermal energy to create the hole-electron pair but a photon of light can do it easily. |
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| 15. |
The energy of a photon of sodium light wavelength `5890 Å` equals the energy gap of a semiconducting material. Find the minimum energy E required to create a hole-electron combination. the value of `E//kT` at a temperature of `27^(@)C`, where `k=8.62xx10^(-5) eV//K, h=6.63xx10^(-34)Js`.A. 0.026 eVB. 0.31 eVC. 2.1 eVD. 6.4 eV |
| Answer» Correct Answer - C | |
| 16. |
If n-type semiconductor is heated thenA. the number of electrons increases and the number of holes decreasesB. the number of holes increases and the number of electrons decreasesC. at the number of electrons and holes both remains equalD. the number of both electrons and holes increases. |
| Answer» Correct Answer - D | |
| 17. |
When impurity atoms like trivalent or pentavalent are added insuitable amounts, to a semiconductor thenA. its resistance increasesB. its conductivity decreasesC. resistances has no effectD. conductivity increase. |
| Answer» Correct Answer - D | |
| 18. |
When a p-n junction is reverse-biased, the current becomes almost constant at 25 μA. When it is forward-biased at 200 mV, a current of 75 μA is obtained. Find the magnitude of diffusion current when the diode is(a) unbiased, (b) reverse-biased at 200 mV and (c) forward-biased at 200 mV. |
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Answer» i1 = 25 μA, V = 200 mV, i2 = 75 μA a) When in unbiased condition drift current = diffusion current Diffusion current = 25 μA. b) On reverse biasing the diffusion current becomes ‘O’. c) On forward biasing the actual current be x. x – Drift current = Forward biasing current => x – 25 μA = 75 μA => x = (75 + 25) μA = 100 μA. |
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| 19. |
How many `NAND` gate are used to from `AND` gate?A. 4B. 3C. 2D. 1 |
| Answer» Correct Answer - C | |
| 20. |
Which logic gate is represented by the following truth table? A. ANDB. ORC. NANDD. NOR |
| Answer» Correct Answer - B | |
| 21. |
Which one of the following gates can be served as a building block for any digital circuit ?A. ORB. ANDC. NOTD. NAND |
| Answer» Correct Answer - D | |
| 22. |
Which of the following truth table represents an AND gate? A. 4B. 3C. 2D. 1 |
| Answer» Correct Answer - C | |
| 23. |
Which logic gate is represented by the following truth table ? A. OR gateB. AND gateC. NAND gateD. NOT gate |
| Answer» Correct Answer - B | |
| 24. |
Th truth table corresponds toA. NAND gateB. NOR gateC. AND gateD. NOT gate |
| Answer» Correct Answer - B | |
| 25. |
Which gate is represeneted by the symbolic diagram given here A. AND gateB. OR gateC. NOT gateD. NAND gate |
| Answer» Correct Answer - A | |
| 26. |
The truth table of a logic gate is as follows : It corresponds toA. OR gateB. NOR gateC. AND gateD. NAND gate |
| Answer» Correct Answer - D | |
| 27. |
Which of the following gate corresponds to the truth table given below : `{:(A,B,Y),(0,0,1),(0,1,1),(1,0,1),(1,1,0):}`A. NANDB. ANDC. XORD. OR |
| Answer» Correct Answer - A | |
| 28. |
Which gate is represeneted by the symbolic diagram given here A. AND gateB. NAND gateC. OR gateD. NOR gate |
| Answer» Correct Answer - D | |
| 29. |
To which logic gate does the truth table given below correspond ? A. ANDB. ORC. NANDD. XOR |
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Answer» Correct Answer - D Here, X=0, if A=0 and B=0 and X=0, if A=1 and B=1 Hence this is the output of XOR gate. |
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| 30. |
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. NANDB. ORC. ANDD. XOR |
| Answer» Correct Answer - B | |
| 31. |
To which logic gate does the truth table given below correspond ? A. OR gateB. AND gateC. NAND gateD. NOR gate |
| Answer» Correct Answer - C | |
| 32. |
To which logic gate does the truth table given below correspond ? A. OR gateB. AND gateC. NAND gateD. NOR gate |
| Answer» Correct Answer - D | |
| 33. |
The truth table of a logic gate is a tableA. giving only the true numbersB. rejecting only the wrong numbersC. giving the relation between the input and output variables of a logic gateD. which gives all the possible input logic levels and the corresponding resultant logic levels in the output |
| Answer» Correct Answer - D | |
| 34. |
To which logic gate does the truth table given below correspond ? A. OR gateB. AND gateC. NAND gateD. NOR gate |
| Answer» Correct Answer - B | |
| 35. |
The truth table of the logic circuit shown-A. `{:(A,B,Y),(0,0,0),(1,0,1),(1,0,1),(1,1,1):}`B. `{:(A,B,Y),(0,0,0),(0,1,1),(1,0,1),(1,1,0):}`C. `{:(A,B,Y),(0,0,1),(0,1,0),(1,0,0),(1,1,0):}`D. `{:(A,B,Y),(0,0,0),(1,0,0),(0,1,0),(1,1,1):}` |
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Answer» Correct Answer - B It is also called XOR gate the boolean expression for XOR gate is `Y=A.overline(B)+overline(A)+B` |
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| 36. |
A transistor is connected in a common emitter configuration. The collector supply is 8 V and the voltage drop across a resistor of `800 Omega ` in the collector circuit is 0.5 V. If the current gain factor `(alpha) ` is 0.96 , Find the base current. |
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Answer» Correct Answer - B ` beta = alpha / (1 - alpha) = 0.96 / (1 - 0.96) = 24` The collector current is , ` i_c = "voltage drop across collector resistor" / "resistance" = 0.5 / 800 A = 0.625 xx 10^(-3) A ` From the definition of ` beta = i_c / i_b ` the base current ` i_b = i_c / beta = (0.625 xx 10^(-3)) / 24 A ` `= 26 mu A ` . |
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| 37. |
The collector supply in a common emitter amplifier is 8 V and the voltage drop across the load of 800 `Omega` is 0.4 V. If the current gain for common base be `alpha`=0.96 , then the base currentA. `15 muA`B. `21 muA`C. `25 muA`D. `30 muA` |
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Answer» Correct Answer - B `V_("cc")=8V, R_(l)=800 Omega, V_(0)=0.4 V, alpha=0.96, I_(b)=? ` `V_(0)=I_(c)R_(l)` `therefore I_(c)=(V_(0))/(R_(l))=(0.4)/(800)=5xx10^(-4)A` `I_(c)=500 muA` Now `alpha=(I_(c))/(I_(e))=(I_(c))/(I_(c)+I_(b))` `therefore I_(c)=alphaI_(c)+ alpha I_(b)` `I_(b)=((1-alpha)/(alpha))I_(c)` `=((1-0.96)/(0.96))500` `=21 muA` |
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| 38. |
In `NPN` transistor the collector current is `10mA`. If `90%` of electrons emitted reach the collector, theA. emitter current will be `9mA`B. emitter current will be `11.1mA`C. base current will be `0.1mA`D. base current will be `0.01mA` |
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Answer» Correct Answer - B `I_(c)=10mA` `I_(E)=(10)/(0.9)=11.1mA` |
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| 39. |
Variation of current passing through a conductor as the voltage supplied across its ends as varied is shown in the adjoining diagram. If the resistance (R) is determined at the points A, B, C and D we will find that-A. `R_(C)=R_(D)`B. `R_(B) gt R_(A)`C. `R_(C) gt R_(B)`D. `R_(A) gt R_(C)` |
| Answer» Correct Answer - D | |
| 40. |
Which one of the following is not necessary in a feedback oscillator ?A. AmplifierB. Feedback circuitC. External input signalD. Frequency |
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Answer» Correct Answer - C The feedback oscillator generates a.c. output signals without using external input signals |
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| 41. |
In `NPN` transistor, if doping in base region is increased then collector currentA. DecreasesB. IncreasesC. Remain sameD. None of these |
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Answer» Correct Answer - A For a transistor, `i_e=i_b+i_c therefore i_c=i_e - I_b` If the doping in the base region is increased, then the base current (`I_b`) also increases. Hence `l_c` will decrease. |
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| 42. |
Zener diode is a specially desined p-n junction diode, in which both p-side and n-side of p-n junction are heavily doped. The zener diode is designed especially to operate in the reverse break down voltage region continuosly without being damaged? Zener diode is used to remove the fluctuations from the given voltage and thereby provides a voltage of constant magnitude (i.e., Zener diode is used as voltage regulator). Read the above pragraph and answer the following question: (i) What is the most important use of Zener diode? (ii) What are the essential conditions for proper working of Zener diode? (iii) What do you learn from the above atudy?A. constant voltage across applied loadB. any desired current at constant voltageC. a p-n junction working under constant regulated voltage conditionsD. a p-n junction to operate at high voltages. |
| Answer» Correct Answer - A | |
| 43. |
A full wave rectifier uses two diode, the internal resistance if each diode is `20ohm`. The transformer rms secondary voltage from center tap to each end of secondary is 50 V and load resistance is `980Ohm`. Find (i) the mean load current (ii) the rms value of load current.A. 0.05 AB. 45 mAC. 0.25 AD. 25 mA |
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Answer» Correct Answer - B Maximum load current, `I_(m)=(V_(m))/(r_(f)+R_(L))` `=(50sqrt(2)V)/((20+980)Omega)=7.07mA` `therefore ` Mean load current, `I_(DC) =(2I_(m))/(pi)=(2xx70.7)/(50xx70.7)=45mA` |
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| 44. |
A student is given a transister. He is asked to find ot the teminals of p-n-p transistor as emitter, base and collector. He is told that the terminal marked with red dot is emitter. He touches red probe with known terminal as emitter and marks other two lead wires as A and B. He measures resistance between emitter and lead A. Then measured resistance between emitter and lead B and finds that resistance increases. This shows-A. A is base and B is collectorB. A is collector and B is baseC. eithher can be collector or baseD. multimeter cannot be used to test the terminals |
| Answer» Correct Answer - A | |
| 45. |
Semiconductor is damaged by the strong current due toA. lack of free electronB. excess of electronC. excess of protonD. None of the above |
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Answer» Correct Answer - D Breakdown of semiconductor device occurs due to the flow of large current. |
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| 46. |
In a `p-n` junction photo cell, the value of the photo electromotive force produced by monochromatic light is proportional toA. the voltage applied at p-n junctionB. the barrier voltage at p-n junctionC. the intensity of light falling on cellD. the frequency of light falling on cell |
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Answer» Correct Answer - C When a light (wavelength sufficient to break the covalent bond) falls on the junction. New hole electron pairs are created. Number of produced electron hole pair depend upon number of photons. So, photoemf or current proportional to intensity of light. |
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| 47. |
The specimen of an intrinsic semiconductor contains `1.2xx10^15` holes/`m^3`. If it is doped by phosphorous atoms in a small proportion, then the number of holes/`m^3` in the doped semiconductor willA. slighty increaseB. slightly decreaseC. remain constant at `1.2xx10^15` holes/`m^3`D. be doubled |
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Answer» Correct Answer - C Phosphorous increases the number of electrons. |
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| 48. |
The characteristic curve for a diode is shown in the figure for forward bias mode. The cut-off voltage for this diode is approximately-A. 0.5 VB. 0.8 VC. 1 VD. gt1V |
| Answer» Correct Answer - A | |
| 49. |
If `V_(1) gt V_(2), r` is resistance offered by diode in forward bias then current through the diode is. . |
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Answer» Correct Answer - C `I=V/R=(V_(1)-V_(2))/(R+r)` |
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| 50. |
If an ideal diode is used in the given circuit, find the current through each resistance A. `I_(1)=(9)/(6)A, I_(2)=3A`B. `I_(1)=3A, I_(2)=3/2 A`C. `I_(1)=2/3 A, I_(2)=1/3 A`D. `I_(1)=1/3 A, I_(2)=2/3 A` |
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Answer» Correct Answer - A `I=(E)/(R+r)=(9)/(2+r)=9/2 A` Now `I_(1)=((R_(2))/(R_(1)+R_(2)))I=((3)/(6+3))(9)/(2)=3/2=9/4A` and `I_(2)=((R_(1))/(R_(1)+R_(2)))I=((6)/(6+3))9/2 =3A` |
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