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Multiple choice question for engineering

Set 1

1. What does p/τ represent?
a) holes
b) time
c) holes per second lost
d) p per unit time

View Answer

Answer: c [Reason:] Option (c) represents the holes per second lost by recombination per unit volume.

2. Which of the following is the Taylor’s expression?
a)electronic-devices-circuits-questions-answers-continuity-equation-q2
b) electronic-devices-circuits-questions-answers-continuity-equation-q2a
c) electronic-devices-circuits-questions-answers-continuity-equation-q2b
d) electronic-devices-circuits-questions-answers-continuity-equation-q2c

View Answer

Answer: a [Reason:] Option a represents the correct formula for the Taylor’s expression.

3. Calculate the number of coulombs per second if the area is 4cm2, recombination rate of hole is 1000 cm-3/s and the differential length is 2mm.
a) 1.28*10-23
b) 1.28*10-22
c) 1.28*10-21
d) 1.28*10-20

View Answer

Answer: b [Reason:] number of coulombs per second= eAdxp/τ =1.6*10-19*4*10-4*2*10-3*1000 =1.28*10-22.

4. The current entering the volume at x is I and leaving is I+Δi , the number of coulombs per second will be equal to δI. Is it true or false?
a) True
b) False

View Answer

Answer: a [Reason:] Coulombs per second is known as the current. The differential current will be the current through the semiconductor.

5. The change in the carrier density is due to
a) Flow of incoming flux
b) Flow of outgoing flux
c) Difference of flow between incoming and outgoing flux
d) Difference of flow between incoming and outgoing flux plus generation and minus recombination

View Answer

Answer: d [Reason:] The change in the carrier density describes the continuity equation which is equal to the difference between the incoming and outgoing flux plus generation and minus recombination.

6. What of the following conditions satisfies when the number of holes which are thermally generated is equal to the holes lost by recombination?
a) I≠0
b) dp/dt≠0
c) g=p/τ
d) g≠p/τ

View Answer

Answer: c [Reason:] Under the equilibrium conditions, I will be zero and then the dp/dt will aso be equal to zero in the continuity equation. Then, g= p/τ is left which is option c.

7. What is the diffusion length for holes when Dp=25cm2/s and τp=25s?
a) 25cm
b) 1cm
c) 0.04cm
d) 50cm

View Answer

Answer: a [Reason:] Lp=√(Dpp) =√(25*25) =25cm.

8. Which of the following represents the continuity equation?
a) dp/dt=-(p-p0)/τp+Dp(d2p/dx2)-µpd(ρϵ)/dx
b) dp/dt=-(p-p0)/τp-Dp(d2p/dx2)-µpd(ρϵ)/dx
c) dp/dt=-(p-p0)/τp+Dp(d2p/dx2)+µpd(ρϵ)/dx
d) dp/dt=(p-p0)/τp-Dp(d2p/dx2)-µpd(ρϵ)/dx

View Answer

Answer: a [Reason:] Option a represents the correct equation of the continuity equation for holes.

9. What is the diffusion length for electrons when Dn=10cm2/s and τn=40s?
a) 50cm
b) 25cm
c) 20cm
d) 15cm

View Answer

Answer: c [Reason:] Ln=√(Dnn) =√(10*40) =20cm.

10. Which of the following represents the best definition for the diffusion length for holes?
a) Average distance which an electron is injected travels before recombining with an electron
b) Average distance which a hole is injected travels before recombining with an electron
c) Average distance which a hole is injected travels before recombining with a hole
d) Average distance which an electron is injected before recombining with a hole

View Answer

Answer: b [Reason:] Diffusion length for holes is represented as the average distance which a hole is injected travels before recombining with an electron. It is the distance into the semiconductor at which the injected concentration falls to 1/ϵ of its value at x=0.

Set 2

1. For the difference amplifier which of the following is true?
a) It responds to the difference between the two signals and rejects the signal that are common to both the signal
b) It responds to the signal that are common to the two inputs only
c) It has a low value of input resistance
d) The efficacy of the amplifier is measured by the degree of its differential signal to the preference of the common mode signal

View Answer

Answer: a [Reason:] All the statements are not true except for the fact that it responds only when there is difference between two signals only.

2. If for an amplifier the common mode input signal is vc, the differential signal id vd and Ac and Ad represent common mode gain and differential gain respectively, then the output voltage v0 is given by
a) v0 = Ad vd – Ac vc
b) v0 = – Ad vd + Ac vc
c) v0 = Ad vd + Ac vc
d) v0 = – Ad vd – Ac vc

View Answer

Answer: c [Reason:] It is a standard mathematical expression.

3. If for an amplifier v1 and v2 are the input signals, vc and vd represent the common mode and differential signals respectively, then the expression for CMRR (Common Mode Rejection Ratio) is
a) 20 log (|Ad| / |Ac|)
b) -10 log (|Ac| / |Ad|)2
c) 20 log (v2 – v1 / 0.5(v2 + v1))
d) All of the mentioned

View Answer

Answer: d [Reason:] Note that all the expressions are identical.

4. The problem with the single operational difference amplifier is its
a) High input resistance
b) Low input resistance
c) Low output resistance
d) None of the mentioned

View Answer

Answer: b [Reason:] Due to low input resistance a large part of the signal is lost to the source’s internal resistance.

5. For the difference amplifier as shown in the figure show that if each resistor has a tolerance of ±100 ε % (i.e., for, say, a 5% resistor, ε = 0.05) then the worst-case CMRR is given approximately by (given K = R2/R1 = R4/R3)
electronic-devices-circuits-questions-answers-difference-amplifiers-q5
a) 20 log [K+1/4ε].
b) 20 log [K+1/2ε].
c) 20 log [K+1/ε].
d) 20 log [2K+2/ε].

View Answer

Answer: a

6. For the circuit given below determine the input common mode resistance.
electronic-devices-circuits-questions-answers-difference-amplifiers-q5
a) (R1 + R3) || (R2) || + (R4)
b) (R1 + R4) || (R2 + R3)
c) (R1 + R2) || (R3 + R4)
d) (R1 + R3) || (R2 + R4)

View Answer

Answer: c [Reason:] Parallel combination of series combination of R1 & R3 with the series combination of R3 and R4 is the required answer as is visible by the circuit.

7. For the circuit shown below express v0 as a function of v1 and v2.
electronic-devices-circuits-questions-answers-difference-amplifiers-q7
a) v0 = v1 + v2
b) v0 = v2 – v1
c) v0 = v1 – v2
d) v0 = -v1 – v2

View Answer

Answer: b [Reason:] Considering the fact that the potential at the input terminals are identical and proceeding we obtain the given result.

8. For the difference amplifier shown below, let all the resistors be 10kΩ ± x%. The expression for the worst-case common-mode gain is
electronic-devices-circuits-questions-answers-difference-amplifiers-q7
a) x / 50
b) x / 100
c) 2x / (100 – x)
d) 2x / (100 + x)

View Answer

Answer: d

9. Determine Ad and Ac for the given circuit.
electronic-devices-circuits-questions-answers-difference-amplifiers-q9
a) Ac = 0 and Ad = 1
b) Ac ≠ 0 and Ad = 1
c) Ac = 0 and Ad ≠ 1
d) Ac ≠ 0 and Ad ≠ 1

View Answer

Answer: a [Reason:] Consider the fact that the potential at the input terminals are identical and obtain the values of V1 and V2. Thus obtain the value of Vd and Vc.

10. Determine the voltage gain for the given circuit known that R1 = R3 = 10kΩ abd R2 = R4 = 100kΩ.
electronic-devices-circuits-questions-answers-difference-amplifiers-q5
a) 1
b) 10
c) 100
d) 1000

View Answer

Answer: b [Reason:] Voltage gain is 100/10. 

Set 3

1. What is the SI unit of electron diffusion constant?
a) cm2/s
b) m2/s
c) m/s
d) none

View Answer

Answer: b [Reason:] J=eDdn/dx So D=q*m/ (q*s*(1/m)) =m2/s.

2. Calculate the diffusion current density when the concentration of electron varies from the 1*1018 to 7*1017 cm-3 over a distance of 0.10 cm.D=225cm2/s
a) 100 A/cm2
b) 108 A/cm2
c) 0.01A/cm2
d) None

View Answer

Answer: b [Reason:] J=eDdn/dx J=1.6*10-19*225*(1018-(7*1017))/0.1 =108A/cm2.

3. Which is the correct formula for the Jp?
a) Jp=qDdp/dx
b) Jp=pDdn/dx
c) Jp=-qDdp/dx
d) None

View Answer

Answer: c [Reason:] Jp is negative for the p type of semiconductors.

4. What is the direction of the electron diffusion current density relative to the electron flux?
a) Same direction
b) Opposite to each other
c) Perpendicular to each other
d) At 270 degrees to each other

View Answer

Answer: b [Reason:] From the graph between electron concentration and the distance, we can see that the direction of the electron diffusion current density is opposite to the electron flux.

5. In diffusion, the particles flow from a region of _______ to region of ___________
a) High, low
b) Low , high
c) High , medium
d) Low, medium

View Answer

Answer: a [Reason:] Diffusion is the process of flow of particles form the region of the high concentration to a region of low concentration.

6. Which of the following parameter describes the best movement of the electrons inside a semiconductor?
a) Velocity gradient
b) Diffusion
c) Mobility
d) Density gradient

View Answer

Answer: c [Reason:] Mobility is defined as the movement of the electrons inside a semiconductor. On the other hand, velocity gradient is the ratio of velocity to distance.

7. Which of the following term isn’t a part of the total current density in a semiconductor?
a) Temperature
b) µ
c) e
d) E

View Answer

Answer: a [Reason:] J=enµE+epµE+eDdn/dx-eDdp/dx So, temperature isn’t a part of the equation.

8. What does dn/dx represent?
a) Velocity gradient
b) Volume gradient
c) Density gradient
d) None

View Answer

Answer: c [Reason:] dn/dx represent velocity gradient.

9. Calculate the diffusion constant for the holes when the mobility of the holes is 400cm2/V-s and temperature is 300K?
a) 1.035m m2/s
b) 0.035m m2/s
c) 1.5m m2/s
d) 1.9m m2/s

View Answer

Answer: a [Reason:] Dp=VT*μn = (1.38*10-23*300*400*10-2)/ (1.6*10-19) = 1.035m m2/s.

10. Calculate the diffusion constant for the electrons when the mobility of the electrons is 325cm2/V-s and temperature is 300K?
a) 0.85 m2/s
b) 0.084 m2/s
c) 0.58 m2/s
d) 0.95 m2/s

View Answer

Answer: c [Reason:] Dn=VT*μn = (1.38*10-23*300*325*10-2)/ (1.6*10-19) = 0.084 m2/s.

Set 4

1. Assuming that the signal is quantized to satisfy the condition of previous question and assuming the approximate bandwidth of the signal is W. The minimum required bandwidth for transmission of a binary PCM signal based on this quantization scheme will be
a) 5 W
b) 10 W
c) 20 W
d) None of the mentioned

View Answer

Answer: b [Reason:] The minimum bandwidth requirement for transmission of a binary PCM signal is BW= vW. Since v 10, we have BW = 10 W.

2. In PCM system, if the quantization levels are increased form 2 to 8, the relative bandwidth requirement will
a) Remain same
b) Be doubled
c) Be tripled
d) Become four times

View Answer

Answer: c [Reason:] If L = 2, then 2 = 2n or n = 1 ND. If L = 8, then 8 = 2n or n = 3. So relative bandwidth will be tripled.

3. A speech signal occupying the bandwidth of 300 Hz to 3 kHz is converted into PCM format for use in digital communication. If the sampling frequency is8 kHz and each sample is quantized into 256 levels, then the output bit the rate will be
a) 3 kb/s
b) 8 kb/s
c) 64 kb/s
d) 256 kb/s

View Answer

Answer: c [Reason:] fs = 8 kHz, 2n = 256 or n = 8. Bit Rate = 8 x 8k = 64 kb/s.

4. Analog data having highest harmonic at 30 kHz generated by a sensor has been digitized using 6 level PCM. What will be the rate of digital signal generated?
a) 120 kbps
b) 200 kbps
c) 240 kbps
d) 180 kbps

View Answer

Answer: d [Reason:] Nyquist Rate = 2 x 30k = 60 kHz 2n should be greater than or equal to 6. Thus n 3, Bit Rate = 60×3 = 18 kHz.

5. Four voice signals. each limited to 4 kHz and sampled at Nyquist rate are converted into binary PCM signal using 256 quantization levels. The bit transmission rate for the time-division multiplexed signal will be
a) 8 kbps
b) 64 kbps
c) 256 kbps
d) 512 kbps

View Answer

Answer: c [Reason:] Nyquist Rate 2 x 4k = 8 kHz Total sample 4 x 8 = 32 k sample/sec 256 = 28, so that 8 bits are required Bit Rate 32k x 8 = 256 kbps.

6. A TDM link has 20 signal channels and each channel is sampled 8000 times/sec. Each sample is represented by seven binary bits and contains an additional bit for synchronization. The total bit rate for the TDM link is
a) 1180 K bits/sec
b) 1280 K bits/sec
c) 1180 M bits/sec
d) 1280 M bits/sec

View Answer

Answer: b [Reason:] Total sample 8000 x 20 = 160 k sample/sec Bit for each sample 7 + 1 = 8 Bit Rate = 160k x 8 = 1280 kilobits/sec.

7. Four signals each band limited to 5 kHz are sampled at twice the Nyquist rate. The resulting PAM samples are transmitted over a single channel after time division multiplexing. The theoretical minimum transmissions bandwidth of the channel should be equal to
a) 5 kHz
b) 20 kHz
c) 40 kHz
d) 80 kHz

View Answer

Answer: d [Reason:] fm = 5 kHz, Nyquist Rate 2 x 5 = 10 kHz Since signal are sampled at twice the Nyquist rate so sampling rate 2 x 10 = 20 kHz. Total transmission bandwidth 4 x 20 = 80 kHz.

8. A sinusoidal massage signal m(t) is transmitted by binary PCM without compression. If the signal to-quantization-noise ratio is required to be at least 48 dB, the minimum number of bits per sample will be
a) 8
b) 10
c) 12
d) 14

View Answer

Answer: a [Reason:] 3(L2)/2 = 48 db or L = 205.09. Since L is power of 2, so we select L = 256 Hence 256 = 28, So 8 bits per sample is required.

9. A speech signal has a total duration of 20 sec. It is sampled at the rate of 8 kHz and then PCM encoded. The signal-to-quantization noise ratio is required to be 40 dB. The minimum storage capacity needed to accommodate this signal is
a) 1.12 KBytes
b) 140 KBytes
c) 168 KBytes
d) None of the mentioned

View Answer

Answer: b [Reason:] (SNR)q = 1.76 + 6.02(n) = 40 dB, n = 6.35 We take the n = 7. Capacity = 20 x 8k x 7 = 1.12 Mbits = 140 Kbytes.

10. A linear delta modulator is designed to operate on speech signals limited to 3.4 kHz. The sampling rate is 10 time the Nyquist rate of the speech signal. The step size is 100 m V. The modulator is tested with a this test signal required to avoid slope overload is
a) 2.04 V
b) 1.08 V
c) 4.08 V
d) 2.16 V

View Answer

Answer: b [Reason:] Amax = (0.1 x 68k)/(2000p) or 1.08V.

Set 5

1. Compared to a PN junction with NA=1014/CM3, which one of the following is true for NA=ND= 1020/CM3?
a) depletion capacitance decreases
b) depletion capacitance increases
c) depletion capacitance remains same
d) depletion capacitance can’t be predicted

View Answer

Answer: b [Reason:] We know, CT=Aε/W and W ∝ (1/NA+1/ND) 1/2. So, CT ∝ (1/NA+1/ND)-1/2 So when NA and ND increases, depletion capacitance CT increases.

2. If CT is the transition capacitance, which of the following are true?
1) in forward bias, CT dominates
2) in reverse bias, CT dominates
3) in forward bias, diffusion capacitance dominates
4) in reverse bias, diffusion capacitance dominates
a) 1 only
b) 2only
c) 2 and 3
d) 3 only

View Answer

Answer: c [Reason:] In reverse bias condition, depletion region increases and acts as an insulator or dielectric medium. So, the transition capacitance increases. In forward bias condition, due to stored charge of minority carriers, diffusion capacitance increases.

3. For an abrupt PN junction diode, small signal capacitance is 1nF/cm2 at zero bias condition.If the built in voltage, Vbi is 1V, the capacitance at reverse bias of 99V is?
a) 0.1nF/cm2
b) 1nF/cm2
c) 1.5nF/cm2
d) 2nF/cm2

View Answer

Answer: a [Reason:] Cjo is the capacitance at zero bias, that is VR=0V, Cjo=Cj for VR=0V. We know, Cj = Cjo/(1+(VR/Vbi))m , m=1/2 for abrupt. So, putting Cj=0.1nF/cm2 where, VR=99V and Vbi=1V we get, Cjo= 0.1(1+99)1/2 = 0.1nF/cm2.

4. The built in capacitance V0 for a step graded PN junction is 0.75V. Junction capacitance Cj at reverse bias when VR=1.25V is 5pF. The value of Cj when VR=7.25V is?
a) 0.1pF
b) 1.7pF
c) 1pF
d) 2.5Pf

View Answer

Answer: d [Reason:] We know, Cj1/ Cj2=[(V0+VR2)/(V0+VR2)]1/2 So, Cj2=Cj1/ {(0.75+7.25)/(0.75+1.25)}1/2 we get Cj2=Cj1 /2 =5/2=2.5Pf.

5. Consider an abrupt PN junction. Let V0 be the built in potential of this junction and VR be the reverse bias voltage applied. If the junction capacitance Cj is 1pF for V0+VR =1V, then for V0+VR =4V what will be the value of Cj?
a) 0.1pF
b) 1.7pF
c) 1pF
d) 0.5Pf

View Answer

Answer: d [Reason:] We know, Cj1/ Cj2=[(V0+VR2)/(V0+VR1)]1/2 Cj2=Cj1(1/4)1/2=1/2 . We get Cj2=1/2=0.5pF.

6. A silicon PN junction diode under revers bias has depletion width of 10µm, relative permittivity is 11.7 and permittivity, ε0 =8.85×10-12F/m. Then depletion capacitance /m2 =?
a) 0.1µF/m2
b) 1.7µF/m2
c) 10µF/m2
d) 0.5µF/m2

View Answer

Answer: c [Reason:] We know, CT =Aε0εr /W CT/A= (8.85×10-12)(11.7)/10 =10 By putting the values we get 10µF/m2.

7. The transition capacitance, CT of a PN junction having uniform doping in both sides, varies with junction voltage as ________
a) (VB )1/2
b) (VB )-1/2
c) (VB )1/4
d) (VB )-1/4

View Answer

Answer: b [Reason:] CT = K/(V0+VB)1/2 As it’s having uniform doping on both sides, the voltage V0 will be zero. So, CT=K/(VB)1/2. The variation of transition capacitance with built in capacitance is (VB )-1/2.

8. The CT for an abrupt PN junction diode is ________
a) CT = K/(V0+VB)1/2
b) CT = K/(V0+VB)-1/2
c) CT = K/(V0+VB)1/3
d) CT = K/(V0+VB)-1/3

View Answer

Answer: a [Reason:] For an abrupt PN junction diode, CT = K/(V0+VB)n. Here, n=1/2 for abrupt PN junction diode and 1/3 for linear PN junction diode. When the doping concentration of a diode varies within a small scale of area, then the diode is called as an abrupt diode.

9. The diffusion capacitance of a PN junction _______
a) decreases with increasing current and increasing temperature
b) decreases with decreasing current and increasing temperature
c) increasing with increasing current and increasing temperature
d) doesnot depend on current and temperature

View Answer

Answer: b [Reason:] CD =τ I /n0 VT Where, I is the current and VT is temperature factor. The diffusion capacitance is directly proportional to current and indirectly proportional to the temperature.

10. Transition capacitance is also called as _______
a) diffusion capacitance
b) depletion capacitance
c) conductance capacitance
d) resistive capacitance

View Answer

Answer: b [Reason:] Transition capacitance occurs in reverse bias. We obtain a depletion layer in that case. Hence it’s also called as depletion capacitance. The diffusion capacitance occurs in forward bias.