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# Multiple choice question for engineering

## Set 1

1. The conduction band edge in the p material is not at the same level to that of conduction band edge in the n material. Is it true or false?
a) True
b) False

Answer: a [Reason:] In a p-n junction diode, the energy levels of the p material and n material will not be at same level. They will be different. So, the conduction band edge as well as the valence band edge of the p material will not be same to that of the n material.

2. Which of the following equations represent the correct expression for the shift in the energy levels for the p-n junction?
a) Eo = Ecn – Ecp
b) Eo = Ecp – Ecn
c) Eo = Ecp + Ecn
d) Eo = -Ecp – Ecn

Answer: b [Reason:] The shift in the energy of the energy level will be the difference of the conduction band edge of the p material and conduction band edge of n material. In the energy level diagram, the conduction band edge of p material is higher than that of the n material.

3. Calculate the Eo given that Nd=1.5*1010cm-3, Na=1.5*1010cm-3 at temperature 300K?
a) 1.5*1010eV
b) 0.256eV
c) 0eV
d) 4.14*10-21eV

Answer: c [Reason:] Eo=kTln((Nd*Na)/(ni)2) Substituting k=1.38*10-23/K, T=300k and the values ofNd,Naand ni, We get Eo=0eV.

4. In a p-n junction, the valence band edge of the p material is greater than which of the following band?
a) Conduction band edge of n material
b) Valence band edge of n material
c) Conduction band edge of p material
d) Fermi level of p material

Answer: b [Reason:] When the p-n junction is formed, the energy levels of the p- material go higher than the n material. That’s why the valence band of the p material will be greater than that of the n material.

5. Which of the following equations represent the correct expression for the band diagram of the p-n junction? (E1=difference between the fermi level of material and conduction band of n material and E2=difference between the conduction band of n material and fermi level of n material)
a) Ecn – E f = (1/2)*EG – E1
b) Ecn – E f = (1/2)*EG – E2
c) Ef – Ecp = (1/2)*EG – E1
d) Ecn – Ef = (1/2)*EG + E1

Answer: a [Reason:] From the energy band diagram of the p-n junction, the option ‘a’ satisfies that band diagram.

6. Calculate the value of Eo when pno=104cm-3 and ppo=1016cm-3 at T=300K.
a) 1meV
b) 0.7meV
c) 0.7eV
d) 0.1meV

Answer: c [Reason:] Eo=kTln(ppo/pno) Substituting the values, we get Eo=0.7eV.

7. Calculate the value of Dp when µp=400cm/s and VT=25mV.
a) 1
b) 0.01
c) 0.1
d) 10

Answer: c [Reason:] Dp= µp*VT =400*10-2*25*10-3 =0.1.

8. What is the value of kT at room temperature?
a) 0.0256eV
b) 0.25eV
c) 25eV
d) 0.0025eV

Answer: a [Reason:] kT=1.38*10-23*300K =4.14*10-21/ (1.6*10-19) =0.0256eV.

9. Is Vo depends only on the equilibrium concentrations. Is it true or false?
a) True
b) False

Answer: a [Reason:] Vo is the contact potential of the junction when the junction is in equilibrium. If, the junction is not in the equilibrium, Vo can’t be calculated.

10. Calculate Vo when ppo=1016cm-3, pno=104cm-3 and Vt=25mV.
a) 69V
b) 6.9V
c) 0.69V
d) 0.069V

Answer: c [Reason:] Vo=VT ln⁡(ppo/pno ) =25*10-3*ln(1016/104) =0.69V.

## Set 2

1. Which states get filled in the conduction band when the donor-type impurity is added to a crystal?
a) Na
b) Nd
c) N
d) P

Answer: b [Reason:] When the donor-type impurity is added to a crystal, first Nd states get filled because it is of the highest energy.

2. Which of the following expression represent the correct formulae for calculating the exact position of the Fermi level for p-type material?
a) EF = EV + kTln(ND / NA )
b) EF = -EV + kTln(ND / NA )
c) EF = EV – kTln(ND / NA )
d) EF = -EV – kTln(ND / NA )

Answer: a [Reason:] The correct position of the Fermi level is found with the formula in the ‘a’ option.

3. Where will be the position of the Fermi level of the n-type material when ND=NA?
a) Ec
b) Ev
c) Ef
d) Efi

Answer: a [Reason:] When ND=NA, kTln(ND/NA )=0 So, Ef=Ec.

4. When the temperature of either n-type or p-type increases, determine the movement of the position of the Fermi energy level?
a) Towards up of energy gap
b) Towards down of energy gap
c) Towards centre of energy gap
d) Towards out of page

Answer: c [Reason:] whenever the temperature increases, the Fermi energy level tends to move at the centre of the energy gap.

5. Is it true, when the temperature rises, the electrons in the conduction band becomes greater than the donor atoms?
a) True
b) False

Answer: a [Reason:] When the temperature increases, there is an increase in the electron-hole pairs and all the donor atoms get ionized, so now the thermally generated electrons will be greater than the donor atoms.

6. If the excess carriers are created in the semiconductor, then identify the correct energy level diagram.
a) b) c) d) Answer: a [Reason:] The diagram A refers the most suitable energy level diagrams because Efp>Ef>Efi>Efp>Ev.

7. If excess charge carriers are created in the semiconductor then the new Fermi level is known as Quasi-Fermi level. Is it true?
a) True
b) False

Answer: a [Reason:] Quasi-fermi level is defined as the change in the level of the Fermi level when the excess chare carriers are added to the semiconductor.

8. Ef lies in the middle of the energy level indicates the unequal concentration of the holes and the electrons?
a) True
b) False

Answer: b [Reason:] When the Ef is in the middle of the energy level, it indicates the equal concentration of the holes and electrons.

9. Consider a bar of silicon having carrier concentration n0=1015 cm-3 and ni=1010cm-3. Assume the excess carrier concentrations to be n=1013cm-3, calculate the quasi-fermi energy level at T=300K?
a) 0.2982 eV
b) 0.2984 eV
c) 0.5971 eV
d) 1Ev

Answer: b [Reason:] =1.38*10-23*300*ln(1013+1015/1013) =0.2984 eV.

10. From the above equation, assuming the same values for the for ni, n= p and T. Given that p0=105cm-3. Calculate the quasi-fermi energy level in eV?
a) 0.1985
b) 0.15
c) 0.1792
d) 0.1

Answer: c [Reason:] Using the same equation, Substituting the respective values, EFi – EFp=0.1792 eV.

## Set 3

(Q.1-Q.3) A parallel resonant circuit has a resistance of 2k ohm and half power frequencies of 86 kHz and 90 kHz.
1. The value of capacitor is
a) 6 µF
b) 20 nF
c) 2 nF
d) 60 µF

Answer: b [Reason:] BW = 2p(90-86)k = 1/RC or C = 19.89 nF.

2. The value of inductor is
a) 4.3 mH
b) 43 mH
c) 0.16 mH
d) 1.6 mH

Answer: c [Reason:] w = (86 + 90)k/2 = 88 = (1/LC)(0.5) or L = 0.16 mH.

3. The quality factor is
a) 22
b) 100
c) 48
d) 200

Answer: a [Reason:] Q = w/BW = 176pK/8pk = 22.

(Q.4-Q.5) A parallel resonant circuit has a midband admittance of 25 X 10(-3) S, quality factor of 80 and a resonant frequency of 200 krad s.

4. The value of R (in ohm) is
a) 40
b) 56.57
c) 80
d) 28.28

Answer: a [Reason:] At mid-band frequency Y = 1/R or R = 1000/25 or 40 ohm.

5. The value of C is
a) 2 µF
b) 28.1 µF
c) 10 µF
d) 14.14 µF

Answer: c [Reason:] Q = wRC or C = 80/(200 x 1000 x 40) or 10 µF.

6. A parallel RLC circuit has R 1 k and C 1 F. The quality factor at resonance is 200. The value of inductor is
a) 35.4 H
b) 25 H
c) 17.7 H
d) 50 H

Answer: b [Reason:] Use Q = R (L/C)0.5.

7. A parallel circuit has R = 1k ohm , C = 50 µF and L = 10mH. The quality factor at resonance is
a) 100
b) 90.86
c) 70.7
d) None of the above

Answer: c [Reason:] Use Q = R (L/C)0.5.

8. A series resonant circuit has L = 1 mH and C = 10 F. The required R (in ohm) for the BW = 15 9 . Hz is
a) 0.1
b) 0.2
c) 0.0159
d) 500

Answer: a [Reason:] Use BW = R/L.

9. For the RLC parallel resonant circuit when R = 8k, L = 40 mH and C = 0.25 F, the quality factor
Q is
a) 40
b) 20
c) 30
d) 10

Answer: b [Reason:] use Q = R (C/L)0.5.

10. A series resonant circuit has an inductor L = 10 mH. The resonant frequency w = 10^6 rad/s and bandwidth is BW = 103 rad/s. The value of R and C will be
a) 100 F, 10 ohm
b) 100 pF, 10 ohm
c) 100 pF, 10 Mega-ohm
d) 100 µF, 10 Meg-ohm

Answer: b [Reason:] Use wxw = 1/LC and BW = R/L.

## Set 4

1. Consider a voltage amplifier having a frequency response of the low-pass STC type with a dc gain of 60 dB and a 3-dB frequency of 1000 Hz. Then the gain db at
a) f = 10 Hz is 55 db
b) f = 10 kHz is 45 db
c) f = 100 kHz is 25 db
d) f = 1Mhz is 0 db

Answer: d [Reason:] Use standard formulas for frequency response and voltage gain.

2. STC networks can be classified into two categories: low-pass (LP) and high-pass (HP). Then which of the following is true?
a) HP network passes dc and low frequencies and attenuate high frequency and opposite for LP network
b) LP network passes dc and low frequencies and attenuate high frequency and opposite for HP network
c) HP network passes dc and high frequencies and attenuate low frequency and opposite for LP network
d) LP network passes low frequencies only and attenuate high frequency and opposite for HP network

Answer: b [Reason:] By definition a LP network allows dc current (or low frequency current) and an LP network does the opposite, that is, allows high frequency ac current.

3. Single-time-constant (STC) networks are those networks that are composed of, or can be reduced to
a) One reactive component (L or C) and a resistance (R)
b) Only capacitive component (C) and resistance (R)
c) Only inductive component (L) and resistance (R)
d) Reactive components (L, C or both L and C) and resistance (R)

Answer: a [Reason:] STC has only one reactive component and one resistive component.

4. The signal whose waveform is not effected by a linear circuit is
a) Triangular Waveform signal
b) Rectangular waveform signal
c) Sine/Cosine wave signal
d) Sawtooth waveform signal

Answer: c [Reason:] Only sine/cosine wave are not affected by a linear circuit while all other waveforms are affected by a linear circuit.

5. Which of the following is not a classification of amplifiers on the basis of their frequency response?
a) Capacitively coupled amplifier
b) Direct coupled amplifier
c) Bandpass amplifier
d) None of the mentioned

Answer: d [Reason:] None of the options provided are correct.

6. General representation of the frequency response curve is called
a) Bode Plot
b) Miller Plot
c) Thevenin Plot
d) Bandwidth Plot

Answer: a [Reason:] General representation of frequency response curves are called Bode plot. Bode plots are also called semi logarithmic plots since they have logarithmic values values on one of the axes.

7. Under what condition can the circuit shown be called a compensated attenuator. a) C1R1 = C2R2
b) C1R2 = C2R1
c) C1C2 = R1R2
d) R1 = 0

Answer: a [Reason:] Standard condition of a compensated attenuator. Here is the derivation for the same.  8. When a circuit is called compensated attenuator?
a) Transfer function is directly proportional to the frequency
b) Transfer function is inversely proportional to the frequency
c) Transfer function is independent of the frequency
d) Natural log of the transfer function is proportional to the frequency

Answer: c [Reason:] Transfer function does not has frequency in its mathematical formula.

9. Which of the following is true?
a) Coupling capacitors causes the gain to fall off at high frequencies
b) Internal capacitor of a device causes the gain to fall off at low frequencies
c) All of the mentioned
d) None of the mentioned

Answer: d [Reason:] Both the statements are false.

10. Which of the following is true?
a) Monolithic IC amplifiers are directly coupled or dc amplifiers
b) Televisions and radios use tuned amplifiers
c) Audio amplifiers have coupling capacitor amplifier
d) All of the mentioned

Answer: d [Reason:] These all are practical applications of different types of amplifiers.

## Set 5

1. What is the range of the carrier lifetime?
a) Nanoseconds to microseconds
b) Nanoseconds to hundreds of microseconds
c) Nanoseconds to tens of microseconds
d) Nanoseconds to milliseconds

Answer: b [Reason:] Carrier lifetime is defined as the existence of any carrier for τ seconds. Carrier lifetime ranges from nanoseconds to hundreds of microseconds.

2. What is the process number of Schokley-Read-Hall Theory processes?
Process-‘ The capture of an electron from the conduction band by an initially neutral empty trap’
a) Process1
b) Process2
c) Process3
d) Process4

Answer: a [Reason:] This is the first process of Schokley-Read-Hall theory of Recombination.

3. Calculate the recombination rate if the excess carrier concentration is 1014cm-3 and the carrier lifetime is 1µseconds.
a) 108
b) 1010
c) 1020
d) 1014

Answer: c [Reason:] The recombination rate, R=δn/τ. So, R=1014 / 10-6 R=1020.

4. Calculate the capture rate where Cn=10, Nt=1010cm-3, n=1020 and fF (Et)=0.4.
a) 6*1030
b) 5*1030
c) 36*1030
d) 1.66*1029

Answer: a [Reason:] Rcn=Cn*N_t*(1-fF (Et))*n Substituting the values, Rcn=6*1030.

5. Calculate the emission rate where En=2.5, Nt=1010cm-3 and fF (Et)=0.6 .
a) 15*1010
b) 1.5*1010
c) 15*1011
d) 1.5*1011

Answer: b [Reason:] Ren=En*Nt*(fF (Et)) Substituting the values, Ren=1.5*1010.

6. At what condition, the rate of electron capture from the conduction band and the rate of the electron emission back into the conduction band must be equal?
a) Thermal equilibrium
b) At room temperature
c) T=250K
d) At boiling temperature

Answer: a [Reason:] At thermal equilibrium, the electron capture rate and the emission rate will be same in the conduction band.

7. Calculate the carrier lifetime when Cp=5 and Nt=1010cm-3.
a) 2*1011
b) 2*10-11
c) 20*10-11
d) 20*1011

Answer: b [Reason:] τp=1/(Cp*Nt ) =1/(5*1010) =2*10-11.

8. The number of majority carriers that are available for recombining with excess minority carriers decreases as the excess semiconductor becomes intrinsic. Is it true?
a) True
b) False

Answer: a [Reason:] With the increase in the number of the majority carriers, the carriers for the recombination will be decreasing with the excess minority carriers and will finally become intrinsic as the concentrations will be same.

9. Which of the following is used as the recombination agent by semiconductor device manufactures?
a) Silver
b) Gold
c) Platinum
d) Aluminium