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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.
| 251. |
In common emitter transistor amplifier circuit, the input signal voltage and the output collector voltage are ..............phase. |
| Answer» Correct Answer - `180^(@)` out of | |
| 252. |
In common emitter transistor amplifier circuit, the input signal voltage and the output collector voltage are ..............phase.A. `135^(@)`B. `180^(@)`C. `45^(@)`D. `90^(@)` |
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Answer» Correct Answer - B In common emitter amplifier circuit, the input and output are in opposite phase i.e., the phase difference between input and output voltages is `180^(@)` |
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| 253. |
For a common emiter configuration if `a` and `beta` have their usualy meaning , the incorrrect relationship between `a and beta` is :A. `1/alpha=1/beta+1`B. `alpha=beta/(1-beta)`C. `alpha=beta/(1+beta)`D. `alpha=beta^(2)/(1+beta^(2))` |
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Answer» Correct Answer - B::D `alpha=I_(c)/I_(e)=I_(c)/(I_(c)+I_(B))=(I_(c)//I_(B))/((I_(c)//I_(B))+1)=beta/(beta+1)` or `1/alpha=(beta+1)/beta=1+1/beta` |
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| 254. |
How many NAND gates are required to realise (i) OR gates and (ii) AND gate. |
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Answer» (i) OR gate can be realised from three NAND gates. (ii) AND gate can be realised from two NAND gate. |
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| 255. |
In a `n`-type semiconductor, which of the following statement is true?A. Electrons are majority carriers and trivalent atoms are the dopantsB. Electrons are minority carriers and pentavalent atoms are the dopantsC. Holes are minority carriers and pentavalent atoms are the dopantsD. Holes are majority carriers and trivalent atoms are the dopants |
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Answer» Correct Answer - C n-type is obtained by doping the Ge or Si with pentavalent atoms. In n-type semiconductor, electrons are majority carriers and holes are minority carriers, hence |
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| 256. |
Write the truth table for the circuit shown in Fig. Name the gate of which this circuit is performing. |
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Answer» Here, `y_(1)=bar(A.B)=barA+barB` `y_(2)=A+B` `y=y_(1).y_(2)` `=(bar(A.B)).(A+B)` `=(barA+barB).(A+B)` `=bar.(A+B)+bar.(A+B)` `=bar.A+bar.B+bar.A+bar.B` `0+bar.B+bar.A+0` `=A.bar+B.bar` It is a Boolean expression of XOR gate. Thus XOR gate is produced. Truth table of this gate is given below. `|{:(A,B,y_(1)=bar(A.B),y_(2)=A+B,y=(bar(A.B)).(A+B)),(0,0,1,0,0),(0,1,1,1,1),(1,0,1,1,1),(1,1,0,1,0):}|` |
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| 257. |
Two car garages have a common gate which needs to open automatically when a car enters either of the garages or cars enter both. Devise a circuit that resembles this situation using diodes for the situation . |
| Answer» The illustration resembles with OR gate as shown in Fig.The truth table of OR gate is given in Fig., which illustrates the given problem. | |
| 258. |
The energy gap between conductionband and valence band is of the order of 0.07 eV. It is a/anA. insulatorB. conductorC. semiconductorD. alloy |
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Answer» Correct Answer - B The energy gap between conduction band and valency band is 0.07 eV, which is very small. Therefore, the electrons can go very easily from valence band to conduction band after gaining a little energy. Hence the material is a conductor. |
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| 259. |
The currect voltage relation of a diode is given by `1 = (e^(van v//T) -1 )mA` where the applied volied `V` is in volts and the tempetature `T` is in degree kelvin if a student make an error meassurting `+- 01 V` while measuring the current of `5 mA at 300 K` what be the error in the value of current in `mA`A. 0.5 mAB. 0.05 mAC. 0.2 mAD. 0.02 mA |
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Answer» Correct Answer - C Given, `I=(e^(1000V//T)-1) mA` When `I=5mA, then 5=e^(1000V//T)-1` or `e^(1000V//T)=6mA` Now `dI=(e^(1000V//T))xx1000/TdV` `=(6 mA)xx1000/300xx(0.01)=0.2 mA` |
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| 260. |
A `Ge` specimen is dopped with `Al`. The concentration of acceptor atoms is `~10^(21) at oms//m^(3)`. Given that the intrinsic concentration of electron hole pairs is `~10^(19)//m^(3)`, the concentration of electron in the speciman isA. ` 10^(17)//m^(3)`B. `10^(15)//m^(3)`C. `10^(4)//m^(3)`D. `10^(2)//m^(3)` |
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Answer» Correct Answer - A Here `n_(i)=10^(19)m^(3), n_(h)=10^(21) "atoms"//m^(3)` The intrinsic concentration of electron-hole pair is given by `n_(i)^(2)=n_(e)n_(h)` `:. n_(e)=n_(i)^(2)/n_(h)=((10^(19))^(2))/10^(21)=10^(38)/10^(21)=10^(17)//m^(3)` |
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| 261. |
In a forward biased `PN`- junction diode, the potential barrier in the depletion region is of the from...A. B. C. D. |
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Answer» Correct Answer - D When p-n junction is forward biased, the applied voltage opposes the barrier potential. Due to it, the width of depletion layer decreases and effective barrier potential becomes less. The option `` and `` show the potential barrier in reverse bias, whereas option `` and `` show potential barrier on forward bias. Since the width of depletion layerin option `` is less than shown in option ``. Hence, the potential barrier in the deplection region will be the form as shown in option ``. |
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| 262. |
The barrier potential of a p-n junction depends on : (i) type of semiconductor material (ii) amount of doping (iii) temperature. Which is one of the following is correct?A. (i) and (ii) onlyB. (ii) onlyC. (ii) and (iii) onlyD. (i), (ii) and (iii) |
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Answer» Correct Answer - D Potential barrier of p-n junction depends on all the given condition. |
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| 263. |
What is the location of acceptor energy levels in the energy band diagram of p-type semiconductor. |
| Answer» In the energy band digram of p-type silicon semiconductor, the acceptor energy levels lie in energy gap slightly above the valence band. The highest acceptor energy levels lies at 0.045 eV above the top of valence band. | |
| 264. |
A p-type semiconductor has acceptor energy level 53meV above the valence band. Find the maximum wavelength of light that can creat a hole. Use `h=6.63xx10^(-34)Js, c=3xx10^(8)ms^(-1)`. |
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Answer» Here, `E=53 meV=53xx10^(-3) eV` `=53xx10^(-3)xx1.6xx10^(-19)J` `=53xx1.6xx10^(-22)J` Max. wavelength, `lambda_(max)=(hc)/E` `=((6.63xx10^(-34))xx(3xx10^(8)))/(53xx1.6xx10^(-22))=2.35xx10^(-5)m` |
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| 265. |
Find the maximum wavelength of electromagnatic radiation which can creat a hole-electron pair in silicon. The band is silicon is `1.1 eV`. [Use `h = 6.6xx10^(-34) Js , c=3xx10^(8)ms^(-1), e=1.6xx10^(-19)C.` |
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Answer» Here, `E_(g)=0.72 eV` `=0.72xx(1.6xx10^(-19))J`. If `lambda` is the maximum wavelength of electromagnetic radation, which can creat a hole-electron pair in germanium, then `E^(g)=(hc)/lambda` or `lambda=(hc)/E^(g)=((6.62xx10^(-34))xx(3xx10^(8)))/(0.72xx1.6xx10^(-19))` `=1.724xx10^(-6)m` |
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| 266. |
Find the maximum wave- length of electromagnatic radiation which can creat a hole-electron pair in silicon. The band is silicon is `1.1 eV`. [Use `h = 6.6xx10^(-34) Js , c=3xx10^(8)ms^(-1), e=1.6xx10^(-19)C]` |
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Answer» Here, `E_(g) =1.1 eV` `=1.1xx1.6xx10^(-19)J` Let `lambda` be the wavelength of electromagnetic radiation which can create a hole-electron pair in silicon. Then `E_(g)=(hc)/lambda` or `lambda=(hc)/(E_(g))=((6.6xx10^(-34))xx(3xx10^(8)))/(1.1xx1.6xx10^(-19))` `=11.25xx10^(-7)m` |
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| 267. |
The electrical conductivity of a semiconductor increases when electromagnatic radiation of wavelength shorter than 2480 nm is incident on it. The band gap (in eV) for the semiconductor is `[hc=1242 eV nm]`A. `0.9`B. `0.7`C. `0.5`D. `1.1` |
| Answer» Correct Answer - C | |
| 268. |
Suppose the energy liberated in the recombination of a hole-electron pair is converted into electromagnatic radiation. If the maximum wavelength emitted is 660nm, what is the band width? (Use `h=6.6xx10^(-34) J-s)`A. 0.1875 eVB. 1.875 eVC. 0.938 eVD. 0.625 eV |
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Answer» Correct Answer - B Here `lambda=660 nm=660xx10^(-9)m` Band gap `E_(g)=(hc)/(lambda)=((6.6xx10^(-34))xx(3xx10^(8)))/(660xx10^(-9))` `(6.6xx10^(-34)xx3xx10^(8))/(660xx10^(-9)xx1.6xx10^(-19))ev` `=1.875 eV` |
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| 269. |
A semiconductor `X` is made by dopping a germanium crystal with arsenic `(Z=33)`. A scond semiconductor `Y` is made by dopping germanium with indium `(Z=49)`. The two are joined end to end and connected to a battery as shown. Which of the following statements is correct? A. X is P-type, Y is N-type and the junction is forward biasedB. X is N-type, Y is P-type and the junction is forward biasedC. X is P-type, Y is N-type and the junction is reverse biasedD. X is N-type, Y is P-type and the junction is reverse biased |
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Answer» Correct Answer - D Arsenic has valence five and indium has valeance three. So X is N-type and Y is p-type. As positive terminal of battery is connected to N-type, and negative terminal to p-type of P-N junction, hence the junction is reverse biased. |
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| 270. |
Justify the output wavefrom (y) of the OR gate for input and as gives in Fig. |
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Answer» From the given wavefrom, we note the followings: `{:("Time interval",Input A, Input B, Output y),("For time"t lt t_(1),0,0,0),("For time"t_(1) "to" t_(2),1,0,1),("For time" t_(2) "to" t_(3),1,1,1),("For time" t_(3) "to" t_(4),0,1,1),("For time" t_(4)"to" t_(5),0,0,0),("For time" t_(5) "to" t_(6),1,0,0),("For time" t gt t_(6),0,1,1):}` Since the output is 1 in level when if one or more input assume 1 in level), so the Boolean expression A+B=y is satisfied, which is for OR gate. |
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| 271. |
The Fig shown input waveforms A and B to a logic gate. Draw the output waveform for an OR gate. Write the truth table for this logic gate and draw its logic symbol. |
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Answer» The Boolean expression for OR gate is `y=A+B`. It means, the output of an OR gate will be maximum if one or more inputs will be maximum. Thus the output waveform will be as showing in Fig.by curve y. The truth table of OR gates is shown below: `|{:(,A,B,y=A+B,),("For" t_(1)"to"t_(2),0,0,0,),("For" t_(2)"to" t_(3),1,0,1,),("For" t_(3)"to"t_(4),1,1,1,),("For" t_(4) "to" t_(5),0,1,1,),("For"t_(5) "to"t_(6),0,0,0,),("For" t_(6)"to"t_(7),1,0,0,),("For" gtt_(7),1,0,1,):}|` |
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| 272. |
You are given two circuit as shown in Fig. and . Show that circuit acts as OR gate while the circuit acts as AND gate |
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Answer» In Fig . , the output of NOR gate is made input for NOT gate. Then Boolean expression is `y=bar(bar(A+B))=A+B` Hence OR gate is formed. In Fig. , the output of two NOT gate is made input for NOR gate. The Boolean expression is `y=bar(barA+barB)=bar(barA).bar(barB)=A.B` Hence AND gate is formed. |
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| 273. |
Write the truth table for a NAND gate as given in Fig.9. Hence identify the exact logic operation carried out by this circuit. |
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Answer» In Fig. we have NAND gate, with one input A and one output y. The Boolean expression is `y=bar(A.A)=barA+barA=barA` i.e. NOT gate is produced. The truth table is as shown in table, Fig. |
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| 274. |
Define a hole. State its characteristics. |
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Answer» Hole is a seat of positive charge which is produced when an electron breaks away from a covalent bond in a semiconductor. Characteristics of a hole 1. Hole carries a unit positive charge. 2. It has the same magnitude of charge as that of electron. 3. Energy of a hole is high as compared to that of electron. 4. The mobility of hole is smaller than that of electron. 5. In external electric field holes move in a direction opposite to that of electron. |
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| 275. |
Which semiconductor has more mobility : p-type or n-type ? Explain. |
| Answer» The mobility of electron n-type semiconductor is more than the mobility of holes in p-type semiconductor. Since the n-type semiconductor has electrons as majority carriers and holes as minority carriers, whereas the p-type semiconductor has holes as majority carriers and electrons as minority carriers, therefore mobility of n-type is more than that of p-type. | |
| 276. |
Distinguish between n-type and p-type semiconductors on the basis of energy-band diagram. |
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Answer» The energy band diagram of n-type semiconductor has been shown in Fig. and of p-type semiconductor in Fig. From these energy band diagrams, we note that, in n-type semiconductor the forbidden energy gap is reduce by donore energy state and electrons are majority carriers. In p-type semiconductor, the forbidden energy gap is reduced by acceptor energy state and holes are majority carriers. n-type semiconductor is obtained by doping a pure germanium or silicon semiconductor with a suitable impurity atoms of valence five like phosphosous, antimony and bismuth etc. p-type semiconductor is obtained by doping a pure germanium or silicon semiconductor with a suitable impurity atoms of valence three like aluminium, boron, indium, etc. |
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| 277. |
The energy gap of silicon is 1.14 eV. Find the maximum wavelenth at which silicon starts energy absorption. |
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Answer» Here, `E=1.14eV=1.14xx1.6xx10^(-19)J`. `lambda=(hc)/E=(6.62xx10^(-34)xx3xx10^(8))/(1.6xx10^(-19)xx1.14)=10,888A^(@)`. |
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| 278. |
Distinguish between intrinsic and extrinsic semi-conductors on the basis of energy band diagram. |
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Answer» The energy band diagram of instrinsic semiconductor and extrinsic semiconductor have been shown in Fig. and or . From these energy band diagrams we note that (i) The energy gap is of 1.1 eV for intrinsic semiconductor and less than 1.1 eV (due to donor energy state or acceptor energy state) for extrinsic semiconductor. (ii) The conductivity of extrinsic semiconductor is more than intrinsic semiconductor because of smaller energy gap for extrinsic semiconductor. (iii) The resistivity of intrinsic semiconductor is more than extrinsic semi-conductor as resistivity is reciprocal of conductivity. |
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