# 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

2. Which of the following is the Taylor’s expression?

a)

b)

c)

d)

### View Answer

3. Calculate the number of coulombs per second if the area is 4cm^{2}, 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

^{-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

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

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

7. What is the diffusion length for holes when Dp=25cm^{2}/s and τ_{p}=25s?

a) 25cm

b) 1cm

c) 0.04cm

d) 50cm

### View Answer

_{p}=√(D

_{p}*τ

_{p}) =√(25*25) =25cm.

8. Which of the following represents the continuity equation?

a) dp/dt=-(p-p0)/τp+Dp(d^{2p}/dx^{2})-µpd(ρϵ)/dx

b) dp/dt=-(p-p0)/τp-Dp(d^{2p}/dx^{2})-µpd(ρϵ)/dx

c) dp/dt=-(p-p0)/τp+Dp(d^{2p}/dx^{2})+µpd(ρϵ)/dx

d) dp/dt=(p-p0)/τp-Dp(d^{2p}/dx^{2})-µpd(ρϵ)/dx

### View Answer

9. What is the diffusion length for electrons when Dn=10cm^{2}/s and τ_{n}=40s?

a) 50cm

b) 25cm

c) 20cm

d) 15cm

### View Answer

_{n}=√(D

_{n}*τ

_{n}) =√(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

## 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

2. If for an amplifier the common mode input signal is v_{c}, the differential signal id v_{d} and A_{c} and A_{d} represent common mode gain and differential gain respectively, then the output voltage v_{0} is given by

a) v_{0} = A_{d} v_{d} – A_{c} v_{c}

b) v_{0} = – A_{d} v_{d} + A_{c} v_{c}

c) v_{0} = A_{d} v_{d} + A_{c} v_{c}

d) v_{0} = – A_{d} v_{d} – A_{c} v_{c}

### View Answer

3. If for an amplifier v_{1} and v_{2} are the input signals, v_{c} and v_{d} represent the common mode and differential signals respectively, then the expression for CMRR (Common Mode Rejection Ratio) is

a) 20 log (|A_{d}| / |A_{c}|)

b) -10 log (|A_{c}| / |Ad|)^{2}

c) 20 log (v_{2} – v_{1} / 0.5(v_{2} + v_{1}))

d) All of the mentioned

### View Answer

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

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 = R_{2}/R_{1} = R_{4}/R_{3})

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

6. For the circuit given below determine the input common mode resistance.

a) (R_{1} + R_{3}) || (R_{2}) || + (R_{4})

b) (R_{1} + R_{4}) || (R_{2} + R_{3})

c) (R_{1} + R_{2}) || (R3 + R_{4})

d) (R_{1} + R_{3}) || (R_{2} + R_{4})

### View Answer

_{1}& R

_{3}with the series combination of R

_{3}and R

_{4}is the required answer as is visible by the circuit.

7. For the circuit shown below express v_{0} as a function of v_{1} and v_{2}.

a) v_{0} = v_{1} + v_{2}

b) v_{0} = v_{2} – v_{1}

c) v_{0} = v_{1} – v_{2}

d) v_{0} = -v_{1} – v_{2}

### View Answer

8. For the difference amplifier shown below, let all the resistors be 10kΩ ± x%. The expression for the worst-case common-mode gain is

a) x / 50

b) x / 100

c) 2x / (100 – x)

d) 2x / (100 + x)

### View Answer

9. Determine A_{d} and A_{c} for the given circuit.

a) A_{c} = 0 and A_{d} = 1

b) A_{c} ≠ 0 and A_{d} = 1

c) A_{c} = 0 and A_{d} ≠ 1

d) A_{c} ≠ 0 and A_{d} ≠ 1

### View Answer

_{1}and V

_{2}. Thus obtain the value of V

_{d}and V

_{c}.

10. Determine the voltage gain for the given circuit known that R_{1} = R_{3} = 10kΩ abd R_{2} = R_{4} = 100kΩ.

a) 1

b) 10

c) 100

d) 1000

### View Answer

## Set 3

1. What is the SI unit of electron diffusion constant?

a) cm^{2}/s

b) m^{2}/s

c) m/s

d) none

### View Answer

^{2}/s.

2. Calculate the diffusion current density when the concentration of electron varies from the 1*10^{18} to 7*10^{17} cm^{-3} over a distance of 0.10 cm.D=225cm^{2}/s

a) 100 A/cm^{2}

b) 108 A/cm^{2}

c) 0.01A/cm^{2}

d) None

### View Answer

^{-19}*225*(10

^{18}-(7*10

^{17}))/0.1 =108A/cm

^{2}.

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

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

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

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

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

8. What does dn/dx represent?

a) Velocity gradient

b) Volume gradient

c) Density gradient

d) None

### View Answer

9. Calculate the diffusion constant for the holes when the mobility of the holes is 400cm^{2}/V-s and temperature is 300K?

a) 1.035m m^{2}/s

b) 0.035m m^{2}/s

c) 1.5m m^{2}/s

d) 1.9m m^{2}/s

### View Answer

_{p}=V

_{T}*μn = (1.38*10

^{-23}*300*400*10

^{-2})/ (1.6*10

^{-19}) = 1.035m m

^{2}/s.

10. Calculate the diffusion constant for the electrons when the mobility of the electrons is 325cm^{2}/V-s and temperature is 300K?

a) 0.85 m^{2}/s

b) 0.084 m^{2}/s

c) 0.58 m^{2}/s

d) 0.95 m^{2}/s

### View Answer

_{n}=V

_{T}*μn = (1.38*10

^{-23}*300*325*10

^{-2})/ (1.6*10

^{-19}) = 0.084 m

^{2}/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

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

^{n}or n = 1 ND. If L = 8, then 8 = 2

^{n}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

^{n}= 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

^{n}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

^{8}, 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

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

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

^{2})/2 = 48 db or L = 205.09. Since L is power of 2, so we select L = 256 Hence 256 = 2

^{8}, 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

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

## Set 5

1. Compared to a PN junction with N_{A}=10^{14}/CM^{3}, which one of the following is true for N_{A}=N_{D}= 10^{20}/CM^{3}?

a) depletion capacitance decreases

b) depletion capacitance increases

c) depletion capacitance remains same

d) depletion capacitance can’t be predicted

### View Answer

_{T}=Aε/W and W ∝ (1/N

_{A}+1/N

_{D})

^{ 1/2}. So, C

_{T}∝ (1/N

_{A}+1/N

_{D})

^{-1/2}So when N

_{A}and N

_{D}increases, depletion capacitance C

_{T}increases.

2. If C_{T} is the transition capacitance, which of the following are true?

1) in forward bias, C_{T} dominates

2) in reverse bias, C_{T} 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

3. For an abrupt PN junction diode, small signal capacitance is 1nF/cm^{2} at zero bias condition.If the built in voltage, V_{bi} is 1V, the capacitance at reverse bias of 99V is?

a) 0.1nF/cm^{2}

b) 1nF/cm^{2}

c) 1.5nF/cm^{2}

d) 2nF/cm^{2}

### View Answer

_{jo}is the capacitance at zero bias, that is V

_{R}=0V, C

_{jo}=C

_{j}for V

_{R}=0V. We know, C

_{j}= C

_{jo}/(1+(V

_{R}/V

_{bi}))m , m=1/2 for abrupt. So, putting C

_{j}=0.1nF/cm

^{2}where, V

_{R}=99V and V

_{bi}=1V we get, C

_{jo}= 0.1(1+99)

^{1/2}= 0.1nF/cm

^{2}.

4. The built in capacitance V_{0} for a step graded PN junction is 0.75V. Junction capacitance C_{j} at reverse bias when V_{R}=1.25V is 5pF. The value of C_{j} when V_{R}=7.25V is?

a) 0.1pF

b) 1.7pF

c) 1pF

d) 2.5Pf

### View Answer

_{j1}/ C

_{j2}=[(V

_{0}+V

_{R2})/(V

_{0}+V

_{R2})]

^{1/2}So, C

_{j2}=C

_{j1}/ {(0.75+7.25)/(0.75+1.25)}

^{1/2}we get C

_{j2}=C

_{j1}/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 C_{j} is 1pF for V_{0}+V_{R} =1V, then for V_{0}+V_{R} =4V what will be the value of C_{j}?

a) 0.1pF

b) 1.7pF

c) 1pF

d) 0.5Pf

### View Answer

_{j1}/ C

_{j2}=[(V

_{0}+V

_{R2})/(V

_{0}+V

_{R1})]

^{1/2}C

_{j2}=C

_{j1}(1/4)

^{1/2}=1/2 . We get C

_{j2}=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^{-12}F/m. Then depletion capacitance /m^{2} =?

a) 0.1µF/m^{2}

b) 1.7µF/m^{2}

c) 10µF/m^{2}

d) 0.5µF/m^{2}

### View Answer

_{T}=Aε

_{0}ε

_{r}/W C

_{T}/A= (8.85×10

^{-12})(11.7)/10 =10 By putting the values we get 10µF/m

^{2}.

7. The transition capacitance, C_{T} 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

_{T}= K/(V

_{0}+V

_{B})

^{1/2}As it’s having uniform doping on both sides, the voltage V

_{0}will be zero. So, C

_{T}=K/(VB)

^{1/2}. The variation of transition capacitance with built in capacitance is (V

_{B})

^{-1/2}.

8. The C_{T} for an abrupt PN junction diode is ________

a) C_{T} = K/(V_{0}+V_{B})^{1/2}

b) C_{T} = K/(V_{0}+V_{B})^{-1/2}

c) C_{T} = K/(V_{0}+V_{B})^{1/3}

d) C_{T} = K/(V_{0}+V_{B})^{-1/3}

### View Answer

_{T}= K/(V

_{0}+V

_{B})

^{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

_{D}=τ I /n

_{0}V

_{T}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