831 + Interview Questions in SEMICONDUCTORS IN GENERAL AWARENESS Page 6 InterviewSolution

251.	What is the output Y of the following logic circuit ? A. `barA` .BB. B.AC. A+BD. `barA` + B
Answer» Correct Answer - A For the AND gate, the inputs are `barA` and B. Hence Y = `barA`.B ·

Discussion

252.	The depletion layer in `P-N` junction region is caused byA. Drift of holesB. Diffusion of charge carriersC. Migration of impurity ionsD. Drift of electrons
Answer» Correct Answer - B Depletion region is formed due to the diffiusion of eletrons from n top region and of holes from p region to n region.

Discussion

253.	What is the current in the following diode circuit? A. `0A`B. `10^(-2) A`C. `1 A`D. `0.10 A`
Answer» Correct Answer - A Potential on the p-side = -4 V Potential on then-side= -1 V `therefore ` The diode is reverse biased and hence for the flow of current resistance is infinity. `therefore` I=0

Discussion

254.	The density of solid ball is to be determined in an experiment the diameter of the ball is measured with a screw gauge, whose pitch is 0.5 mm and there are 50 divisions on the circular scale. The reading on the relative error of 2%, the relative percentage error error in the density isA. 0.009B. 0.024C. 0.031D. 0.042
Answer» Correct Answer - C

Discussion

255.	The element that can be used as acceptor impurity to dope silicon isA. antimonyB. arsenicC. boronD. phosphorus
Answer» Correct Answer - C

Discussion

256.	The output of an OR gate is connected to both the inputs of a NAND gate. The combination will serve as aA. OR gateB. NOT gateC. NOR gateD. AND gate
Answer» Correct Answer - C

Discussion

257.	Which circuit will not show current in ammeter ?A. Figure 1B. Figure 2C. Figure 3D. Figure 4
Answer» Correct Answer - A Current wilI flow in (2), (3) and (4) because in (2), the diodes are in series and are forward biased. In (3), they are m parallel but one is forward biased and the other is reversed biased. Hence no current will flow in that diode but the current flows through the other diode. In ( 4) both of them are in parallel and forward biased. But in Fig. (1) they are in series but first diode is reverse biased. Hence no current will flow.

Discussion

258.	Which of the following four alternatives is not correct, We need modulation :-A. To increase th selectivityB. To reduce the time lag between transmission and reception of the information signalC. to reduce the size of antennaD. To reduce the fractional band width, that is the ration of the signal band width to the centre frequency
Answer» Correct Answer - B

Discussion

259.	The output of gate is low when at least one of its input is high. This is true for :-A. NORB. ORC. ANDD. NAND
Answer» Correct Answer - A

Discussion

260.	Given truth table is related with :- `{:(A,B,Y),(1,1,0),(0,1,1),(1,0,1),(0,0,1):}`A. NOT gateB. OR gateC. XOR gateD. NAND gate
Answer» Correct Answer - D

Discussion

261.	Thr truth table given below is for :- `{:(A,B,Y),(0,0,1),(0,1,0),(1,0,0),(1,1,1):}`A. OR gateB. AND gateC. XNOR gateD. XOR gate
Answer» Correct Answer - C

Discussion

262.	Truth table for the following is-A. `{:(A,B,Y),(0,0,0),(0,1,1),(1,0,1),(1,1,0):}`B. `{:(A,B,Y),(0,0,0),(0,1,1),(1,0,0),(1,1,1):}`C. `{:(A,B,Y),(0,0,1),(0,1,0),(1,0,0),(1,1,1):}`D. `{:(A,B,Y),(0,0,0),(0,1,1),(1,0,1),(1,1,1):}`
Answer» Correct Answer - B

Discussion

263.	The truth table given above is for which of the following gates :- `{:(A,B,Y),(0,0,0),(0,1,1),(1,0,1),(1,1,1):}`A. NoR gateB. AND gateC. OR gateD. NAND gate
Answer» Correct Answer - C

Discussion

264.	Out of the following, universal gate is :-A. NOTB. ORC. ANDD. NAND
Answer» Correct Answer - D

Discussion

265.	In circuit in following figure the value of `Y` is-A. ZeroB. 1C. fluctuates between 0 and 1D. indeterminate as the circuit cannot be realized
Answer» Correct Answer - A

Discussion

266.	The truth table given below belongs for which gates `{:(A,B,Y),(0,0,0),(0,1,1),(1,0,1),(1,1,0):}`A. OR gateB. XOR gateC. AND gateD. NAND gate
Answer» Correct Answer - B

Discussion

267.	The output Y for the following logic gate circuit will be-A. `AB`B. `bar(A).bar(B)`C. `bar(A+B)`D. `bar(A.B)`
Answer» Correct Answer - D

Discussion

268.	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. ORB. NANDC. XORD. NOR
Answer» Correct Answer - D

Discussion

269.	Fermi energy is theA. minimum energy of electrons in metal at 0KB. minimum energy of electrons in metal at `0^(@)C`C. Maximum energy of electrons in metal at 0 KD. Maximum energy of electrons in metal at `0^(@)C`
Answer» Correct Answer - C

Discussion

270.	Which of the following statements is true-A. in insulators the conduction band is completely empty.B. In conductor the conduction band is completely empty.C. in semiconductor the conduction band is partically empty at low temperature.D. in insulators the conduction band is completely filled with electrons.
Answer» Correct Answer - A

Discussion

271.	When N-type semiconductor is heated, the …………….. (A) number of electrons and holes remains same. (B) number of electrons increases while that of holes decreases. (C) number of electrons decreases while that of holes increases. (D) number of electrons and holes increases equally.
Answer» (D) number of electrons and holes increases equally.

Discussion

272.	Which of the following statements is true-A. at absolute zero temperature the semiconductor behave as a conductorB. the energy gap in semiconductor is more than that for insulatorC. the resistance of semiconductor increases with increase in temperatureD. the resistance of semiconductor decreases with increase in temperature
Answer» Correct Answer - D

Discussion

273.	The valence band at 0 K is-A. completely filledB. completely emptyC. partially filledD. nothing can be said
Answer» Correct Answer - A

Discussion

274.	Electric conduction in a semiconductor takes place due toA. electrons onlyB. holes onlyC. both electrons and holesD. neither electrons nor holes
Answer» Correct Answer - C

Discussion

275.	For transistor action, which of the following statements are correct ?A. (a) Base , emitter and collector regions should have similar size and doping concentrationsB. (b) The base region must be very thin and lightly dopedC. ( c) The emitter junction is forward biased and collector junction is reverse biasedD. (d) Both the emitter junction as well as the collector junction are forward biased
Answer» Correct Answer - B::C In a transistor , base must be very thin and lightly doped so that all of the charge carries are not combined in base and majority of them passes the reverse bias layer to collector side.

Discussion

276.	When a forward bias is applied to a p -n junction. ItA. (a) raises the potential barrierB. (b) reduces the majority carrier current to zeroC. (c ) lowers the potential barrierD. (d) All of the above
Answer» Correct Answer - C Due to forward biasing depletion layer thickness decreases, potential barrier is reduced and diffusion of electrons from n to p side occurs.

Discussion

277.	The free electron concentration (n) in the conduction band of a semiconductor at a temperature T kelvin is described in terms of `E_(g)` and T as-A. `n=Ate^(-Eg//kT)`B. `n=AT^(2)e^(-Eg//kT)`C. `n=AT^(2)e^(-Eg//kT)`D. `n=AT^(2)e^(-Eg//2kT)`
Answer» Correct Answer - D

Discussion

278.	The probability of electrons to be found in the conduction band of an intrinsic semiconductor at a finite temperatureA. increases expon entially with increasing band gapB. decreases exponentially with increasing band gapC. decreases with increasing temperatureD. is independent of the temperature and the band gap
Answer» Correct Answer - B `N = N_0e^(-E_g//K_BT` where `E_g` is the band gap. At a finite temperature, the probability of jumping of an electron from the valence band to the conduction band decreases exponentially with increasing band gap.

Discussion

279.	in an n-type semiconductor, the donor energy level lies (a) at the center of the energy gap (b) just below the conduction band (c)just above the valence band (d) in the conduction bandA. an insulatorB. a conductorC. p-type semiconductorD. n-type semiconductor
Answer» Correct Answer - D In n-type semiconductor donor energy level lies just below the conduction band also called empty band of minimum energy.

Discussion

280.	In an unblased p-n junction, holes diffuse from the p - region to n- region becauseA. (a) free electrons in the n - region attract themB. (b) they move across the junction by the potential differenceC. (c ) hole concentration in p - region is more as compared to n -regionD. (d) All of the above
Answer» Correct Answer - C Hole diffusion from p to n side can be viewed as " electron diffusion"from n to p side . Diffusion occurs due to differnce regions . An electrons (or hole ) diffuses where its concentration is less.

Discussion

281.	In a pure silicon `(n_(i)=10^(16)//m^(3))` crystal at `300K, 10^(21)` atoms of phosphorus are added per cubic meter. The new hole concentration will beA. `10^19 "per m"^3`B. `10^11 "per m"^3`C. `10^5 "per m"^3`D. `10^21 "per m"^3`
Answer» Correct Answer - B For intrinsic semiconductor, `n_i=10^16//m^3`and `n_h=n_e` and when phosphorus (pentavalent impurity) is added, it becomes an n-type semiconductor. `because n_e=10^21//m^3` `therefore n_h=n_i^2/n_e=(10^16xx10^16)/10^21=10^11//m^3`

Discussion

282.	In an unblased p-n junction, holes diffuse from the p - region to n- region becauseA. the free electrons in the n-region attract themB. they are swept across the junction by the potential differenceC. there is greater concentrated of holes in the p-region then the n-region.D.
Answer» Correct Answer - C

Discussion

283.	Explain what is doping.
Answer» 1. The process of adding impurities to an intrinsic semiconductor is called doping. 2. The impurity atoms are called dopants which may be either trivalent or pentavalent. The parent atoms are called hosts. 3. The dopant material is so selected that it does not disturb the crystal structure of the host. 4. The size and the electronic configuration of the dopant should be compatible with that of the host. 5. Doping is expressed in ppm (parts per million), i.e., one impurity atom per one million atoms of the host. 6. Doping significantly increases the concentration of charge carriers.

283.

Explain what is doping.

Answer»

1. The process of adding impurities to an intrinsic semiconductor is called doping.

2. The impurity atoms are called dopants which may be either trivalent or pentavalent. The parent atoms are called hosts.

3. The dopant material is so selected that it does not disturb the crystal structure of the host.

4. The size and the electronic configuration of the dopant should be compatible with that of the host.

5. Doping is expressed in ppm (parts per million), i.e., one impurity atom per one million atoms of the host.

6. Doping significantly increases the concentration of charge carriers.

Discussion

284.	What is the need for doping an intrinsic semiconductor?
Answer» The electric conductivity of an intrinsic semiconductor is very low at room temperature; hence no electronic devices can be fabricated using them. Addition of a small amount of a suitable impurity to an intrinsic semiconductor increases its conductivity appreciably. Hence, intrinsic semiconductors are doped with impurities.

284.

What is the need for doping an intrinsic semiconductor?

Answer»

The electric conductivity of an intrinsic semiconductor is very low at room temperature; hence no electronic devices can be fabricated using them. Addition of a small amount of a suitable impurity to an intrinsic semiconductor increases its conductivity appreciably.

Hence, intrinsic semiconductors are doped with impurities.

Discussion

285.	In an n-type semiconductor, the concentration of minority carriers mainly depends uponA. doping techniqueB. number of donor atomsC. temperature of the materialD. quality of intrinsic material
Answer» Correct Answer - C

Discussion

286.	An n-type semiconductor isA. neutralB. positively chargedC. negatively chargedD. none of these
Answer» Correct Answer - A

Discussion

287.	An n-type germanium is obtained, on doping intrinsic germaniurn, byA. siliconB. sulphurC. aluminiumD. phosphorous
Answer» Correct Answer - D

Discussion

288.	The truth table given in fig. represent : `{:(A,B,Y),(0,0,0),(0,1,1),(1,0,1),(1,1,1):}`A. OR-GateB. NAND-GateC. NOT-GateD. AND-Gate
Answer» Correct Answer - A

Discussion

289.	A modulated signal` C_m(t)` has the form `C_m(t) = 30 sin 300pi t + 20 (cos 100pit)`. The carrier frequency `f_c` the modulating frequency (message frequency) `f_Omega` and the modulation index `mu` are respectively given by :A. `f_(c)=200 Hz,f_(omega)=30Hz,mu=(1)/(2)`B. `f_(c)=150 Hz,f_(omega)=50Hz,mu=(2)/(3)`C. `f_(c)=150 Hz,f_(omega)=30Hz,mu=(1)/(3)`D. `f_(c)=200 Hz,f_(omega)=50Hz,mu=(1)/(2)`
Answer» Correct Answer - B

Discussion

290.	In a transistor with normal bias, the emitter base junctionA. high resistanceB. low resistanceC. infinite resistanceD. no resistance
Answer» Correct Answer - B

Discussion

291.	Zener diode is used asA. as an amplifierB. as a rectifierC. as an oscillatorD. as a voltage regulator
Answer» Correct Answer - D

Discussion

292.	For given circuit potential difference `V_(AB)` isA. 10 VB. 20 VC. 30 VD. none
Answer» Correct Answer - A

Discussion

293.	Choose the correct option for the forward biased characteristics of a p-n junctionA. B. C. D.
Answer» Correct Answer - C

Discussion

294.	What is the ratio of forward and reverse resistance of p-n junction diode?A. `10^(2):1`B. `10^(-2):1`C. `1:10^(-4)`D. `1:10^(4)`
Answer» Correct Answer - D

Discussion

295.	A diode made forward biased by a two volt battery however there is a drop of 0.5 V across the diode which is independent of current. Also a current greater then 10 mA produces large joule loss and damages diode. If diode is to be operated at 5 mA, the series resistance to be put isA. `3KOmega`B. `300KOmega`C. `300Omega`D. `200Omega`
Answer» Correct Answer - C

Discussion

296.	A diode made forward biased by a two volt battery however there is a drop of 0.5 V across the diode which is independent of current. Also a current greater then 10 mA produces large joule loss and damages diode. If diode is to be operated at 5 mA, the series resistance to be put isA. `3 k Omega`B. `300 k Omega`C. `300 Omega`D. `200 k Omega`
Answer» Correct Answer - C `I=(V-V_(b))/(R)rArr510^(-2)=((2=0.5))/(R)rArrR=300 Omega`

Discussion

297.	If the two ends of a p-n junction are joined by a conducting wire, thenA. there will be no current in the circuitB. there will be steady current from n-side to n-sideC. there will be steady current from p-side to n-sideD. There will be a steady current in the circuit.
Answer» Correct Answer - A

Discussion

298.	The main cause of avalance breakdown isA. collision ionisationB. high dopingC. recombination of electrons and holesD. None of these
Answer» Correct Answer - A

Discussion

299.	A junction diode has a resistance of 25 `Omega` when forward biased and 2500 `Omega` when reverse biased. What is the current in the diode, for the arrangement shown ? A. `1/25A`B. `1/7A`C. `1/35A`D. `1/480A`
Answer» Correct Answer - B In this case, V = 5 V. The diode is forward biased. Net resistance= `10Omega +25 Omega = 35 Omega` `therefore I=V/R =5/35 =1/7A` (No current in reverse bias)

Discussion

300.	A transistor has a current amplification factor (current gain) of `50`, In a CE amplifier circuit, the collector resistance is chosen as `5 k Omega` and the input resistance is `1 k Omega` Calculate the output voltage if input voltage is 0.01 V.
Answer» For transistor amplifier `V_(0)=beta((R_(C))/(R_(B))),V_(i)=(50) ((5)/1)(0.01)=2.5 V`

Discussion

Explore topic-wise InterviewSolutions in .

What is the output Y of the following logic circuit ? A. `barA` .BB. B.AC. A+BD. `barA` + B

The depletion layer in `P-N` junction region is caused byA. Drift of holesB. Diffusion of charge carriersC. Migration of impurity ionsD. Drift of electrons

What is the current in the following diode circuit? A. `0A`B. `10^(-2) A`C. `1 A`D. `0.10 A`

The element that can be used as acceptor impurity to dope silicon isA. antimonyB. arsenicC. boronD. phosphorus

The output of an OR gate is connected to both the inputs of a NAND gate. The combination will serve as aA. OR gateB. NOT gateC. NOR gateD. AND gate

Which circuit will not show current in ammeter ?A. Figure 1B. Figure 2C. Figure 3D. Figure 4

The output of gate is low when at least one of its input is high. This is true for :-A. NORB. ORC. ANDD. NAND

Given truth table is related with :- `{:(A,B,Y),(1,1,0),(0,1,1),(1,0,1),(0,0,1):}`A. NOT gateB. OR gateC. XOR gateD. NAND gate

Thr truth table given below is for :- `{:(A,B,Y),(0,0,1),(0,1,0),(1,0,0),(1,1,1):}`A. OR gateB. AND gateC. XNOR gateD. XOR gate

Truth table for the following is-A. `{:(A,B,Y),(0,0,0),(0,1,1),(1,0,1),(1,1,0):}`B. `{:(A,B,Y),(0,0,0),(0,1,1),(1,0,0),(1,1,1):}`C. `{:(A,B,Y),(0,0,1),(0,1,0),(1,0,0),(1,1,1):}`D. `{:(A,B,Y),(0,0,0),(0,1,1),(1,0,1),(1,1,1):}`

The truth table given above is for which of the following gates :- `{:(A,B,Y),(0,0,0),(0,1,1),(1,0,1),(1,1,1):}`A. NoR gateB. AND gateC. OR gateD. NAND gate

Out of the following, universal gate is :-A. NOTB. ORC. ANDD. NAND

In circuit in following figure the value of `Y` is-A. ZeroB. 1C. fluctuates between 0 and 1D. indeterminate as the circuit cannot be realized

The truth table given below belongs for which gates `{:(A,B,Y),(0,0,0),(0,1,1),(1,0,1),(1,1,0):}`A. OR gateB. XOR gateC. AND gateD. NAND gate

The output Y for the following logic gate circuit will be-A. `AB`B. `bar(A).bar(B)`C. `bar(A+B)`D. `bar(A.B)`

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. ORB. NANDC. XORD. NOR

Fermi energy is theA. minimum energy of electrons in metal at 0KB. minimum energy of electrons in metal at `0^(@)C`C. Maximum energy of electrons in metal at 0 KD. Maximum energy of electrons in metal at `0^(@)C`

When N-type semiconductor is heated, the …………….. (A) number of electrons and holes remains same. (B) number of electrons increases while that of holes decreases. (C) number of electrons decreases while that of holes increases. (D) number of electrons and holes increases equally.

The valence band at 0 K is-A. completely filledB. completely emptyC. partially filledD. nothing can be said

Electric conduction in a semiconductor takes place due toA. electrons onlyB. holes onlyC. both electrons and holesD. neither electrons nor holes

When a forward bias is applied to a p -n junction. ItA. (a) raises the potential barrierB. (b) reduces the majority carrier current to zeroC. (c ) lowers the potential barrierD. (d) All of the above

The free electron concentration (n) in the conduction band of a semiconductor at a temperature T kelvin is described in terms of `E_(g)` and T as-A. `n=Ate^(-Eg//kT)`B. `n=AT^(2)e^(-Eg//kT)`C. `n=AT^(2)e^(-Eg//kT)`D. `n=AT^(2)e^(-Eg//2kT)`

in an n-type semiconductor, the donor energy level lies (a) at the center of the energy gap (b) just below the conduction band (c)just above the valence band (d) in the conduction bandA. an insulatorB. a conductorC. p-type semiconductorD. n-type semiconductor

In an unblased p-n junction, holes diffuse from the p - region to n- region becauseA. (a) free electrons in the n - region attract themB. (b) they move across the junction by the potential differenceC. (c ) hole concentration in p - region is more as compared to n -regionD. (d) All of the above

In a pure silicon `(n_(i)=10^(16)//m^(3))` crystal at `300K, 10^(21)` atoms of phosphorus are added per cubic meter. The new hole concentration will beA. `10^19 "per m"^3`B. `10^11 "per m"^3`C. `10^5 "per m"^3`D. `10^21 "per m"^3`

In an unblased p-n junction, holes diffuse from the p - region to n- region becauseA. the free electrons in the n-region attract themB. they are swept across the junction by the potential differenceC. there is greater concentrated of holes in the p-region then the n-region.D.

Explain what is doping.

What is the need for doping an intrinsic semiconductor?

In an n-type semiconductor, the concentration of minority carriers mainly depends uponA. doping techniqueB. number of donor atomsC. temperature of the materialD. quality of intrinsic material

An n-type semiconductor isA. neutralB. positively chargedC. negatively chargedD. none of these

An n-type germanium is obtained, on doping intrinsic germaniurn, byA. siliconB. sulphurC. aluminiumD. phosphorous

The truth table given in fig. represent : `{:(A,B,Y),(0,0,0),(0,1,1),(1,0,1),(1,1,1):}`A. OR-GateB. NAND-GateC. NOT-GateD. AND-Gate

In a transistor with normal bias, the emitter base junctionA. high resistanceB. low resistanceC. infinite resistanceD. no resistance

Zener diode is used asA. as an amplifierB. as a rectifierC. as an oscillatorD. as a voltage regulator

For given circuit potential difference `V_(AB)` isA. 10 VB. 20 VC. 30 VD. none

Choose the correct option for the forward biased characteristics of a p-n junctionA. B. C. D.

What is the ratio of forward and reverse resistance of p-n junction diode?A. `10^(2):1`B. `10^(-2):1`C. `1:10^(-4)`D. `1:10^(4)`

If the two ends of a p-n junction are joined by a conducting wire, thenA. there will be no current in the circuitB. there will be steady current from n-side to n-sideC. there will be steady current from p-side to n-sideD. There will be a steady current in the circuit.

The main cause of avalance breakdown isA. collision ionisationB. high dopingC. recombination of electrons and holesD. None of these

A junction diode has a resistance of 25 `Omega` when forward biased and 2500 `Omega` when reverse biased. What is the current in the diode, for the arrangement shown ? A. `1/25A`B. `1/7A`C. `1/35A`D. `1/480A`

A transistor has a current amplification factor (current gain) of `50`, In a CE amplifier circuit, the collector resistance is chosen as `5 k Omega` and the input resistance is `1 k Omega` Calculate the output voltage if input voltage is 0.01 V.