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# Multiple choice question for engineering

## Set 1

(Q.1-Q.4) For the given circuits and input waveform determine the output waveform

1.

Answer: d [Reason:] Diode is off when Vi < 5 and for Vi > 5, Vo = 5V.

2.

Answer: c [Reason:] Diode is off when Vi + 2 > 0.

3.

Answer: d [Reason:] For V1 < 4 the diode is on otherwise its off.

4.

Answer: c [Reason:] During +ve half cycle when Vi < 8 both the diodes are off. For V1 > 8, D1 is on. Similarly for the negative half cycle.

5. The maximum load current that can be drawn is

a) 1.4 mA
b) 2.3 mA
c) 1.8 mA
d) 2.5 mA

Answer: a [Reason:] At regulated poer supply, Is = 30-9/15k or 1.4 mA. The load current will remain less than this current.

6. For the circuit shown diode cutting voltage is Vin = 0. The ripple voltage is to be no more than vrip = 4 V. The minimum load resistance, that can be connected to the output is (in kilo ohm)

a) 6.25
b) 12.5
c) 25
d) 30

7. The circuit inside the box in fig. P3.1.31. contains only resistor and diodes. The terminal voltage vo is
connected to some point in the circuit inside the box. The largest and smallest possible value of vo most nearly to is respectively

a) 15 V, 6 V
b) 24 V, 0 V
c) 24 V, 6 V
d) 15 V, 9 V

Answer: d [Reason:] The output voltage cannot exceed the positive power supply voltage and cannot be lower than the negative power supply voltage.

8. The Q-point of the zener diode in the circuit shown below is

a) (0.34 mA, 4 V)
b) (0.34 mA, 4.93 V)
c) (0.94 mA, 4 V)
d) (0.94 mA, 4.93 V)

## Set 2

1. The static resistance R of the diode is given by __________
a) V/I
b) V*I
c) V+I
d) V-I

Answer: a [Reason:] According to Ohms law the electric current in the circuit is directly proportion to voltage and inversely proportional to resistance so, R=V/I.

2. In the volt ampere characteristics of the diode, the slope of the line joining the operating point to the origin at any point is equal to reciprocal of the _________
a) resistance
b) conductance
c) voltage
d) current

Answer: a [Reason:] In the diode’s volt ampere characteristics, the line joining the operating point and the origin, at any point of the line is equal to the conductance so, it is reciprocal of the resistance.

3. At room temperature (VT = 26) what will be the approximate value of r when n=1 and I=100mA?
a) 26 ohms
b) 2.6 ohms
c) 260 ohms
d) 2600 ohms

Answer: c [Reason:] We know that R= (n*VT) /I, by substituting the value of n, VT, I we get R= 260 ohms, (1*26)/100*10-3 = 260 ohms.

4. In the diode volt ampere characteristics what will be the resistance if a slope is drawn between the voltages 50 to 100 and corresponding current 5 to 10?
a) 5 ohms
b) 10 ohms
c) 50 ohms
d) 100 ohms

Answer: b [Reason:] We know that, in volt ampere characteristics the resistance is equal to the reciprocal of the line joining the origin and operating point, R = dV/dI, by substituting the value of dV and dI we get R= 10ohms.

5. In piecewise linear characteristics what will be the RF value if the slope is 0.5?
a) 25 m ohms
b) 50 m ohms
c) 2 ohms
d) 10 ohms

Answer: c [Reason:] In piecewise linear characteristics the forward resistance will be equal to reciprocal of the slope so, RF = 1/slope, RF = 1/0.5 which is equal to 2 ohms.

6. A diode will behave as an open circuit if the voltage in the circuit is less than __________
a) cut off voltage
b) saturation voltage
c) leakage voltage
d) threshold voltage

Answer: d [Reason:] The diode made up of semiconductor has a certain threshold voltage only after which it behave as closed circuit in the sense it performs some operation if the threshold voltage is greater than the voltage in circuit.

7. What will be the approximate value of thermal voltage of diode?
a) 25mV at 300K
b) 30mV at 180K
c) 25mV at 180K
d) 30mV at 300K

Answer: a [Reason:] We know that the thermal voltage of diode is approximately equal to room temperature which is 300K then for all practical purpose the thermal voltage of diode is taken as 25mV so it will be 25mV at 300K.

8. What will be the thermal voltage of the diode if the temperature is 300K?
a) 25.8 mV
b) 50 mV
c) 50V
d) 19.627 mV

Answer: a [Reason:] The thermal voltage of the diode is given by, VT = KT/q, by substituting the values of T, K which is Boltzmann constant and q which is the charge of the electron we get VT = (300*1.38*10-23)/ (1.602*10-19), VT= 25.8mV.

9. What will be the diode resistance if the current in the circuit is zero?
a) 0 ohms
b) 0.7 ohms
c) 0.3 ohms
d) 1 ohms

Answer: a [Reason:] When the current in the circuit is zero there will be no flow of charges to resist hence the diode resistance will be zero.

10. Which of these following is not a characteristic of an ideal diode?
a) Perfect conductor when forward bias
b) Zero voltage across it when forward bias
c) Perfect insulator when reverse bias
d) Zero current through it when forward bias

Answer: d [Reason:] The diode acts as an ideal diode when it is a perfect conductor and has zero voltage across it during forward bias, a perfect insulator and zero current through it during reverse bias.

## Set 3

1. At what temperature the donor states are completely ionized?
a) 0 K
b) ROOM
c) 300K
d) 900K

Answer: b [Reason:] At room temperature, the donors have donated their electrons to the conduction band.

2. The opposite of ionization takes place at which temperature?
a) 0 K
b) ROOM
c) 300 K
d) 900K

Answer: a [Reason:] AT 0 K, all the donors and acceptors are in their lowest energy levels.

3. What do you mean by the tem ‘FREEZE-OUT’?
a) All the electrons are frozen at room temperature
b) None of the electrons are thermally elevated to the conduction band
c) All the electrons are in the conduction band
d) All the holes are in the valence band

Answer: b [Reason:] Freeze out means none of the electrons are transmitted to the conduction band.

4. Which of the following expressions represent the correct formula for the density of electrons occupying the donor level?
a) nd=Nd-Nd+
b) nd=Nd-Nd
c) nd=Nd+Nd+
d) nd=Nd+Nd

Answer: a [Reason:] The density of the electrons is equal to the electrons present in the substrate minus the number of donors present.

5. Which of the following band is just above the intrinsic Fermi level for n-type semiconductor?
a) Donor band
b) Valence band
c) Acceptor band
d) Conduction band

Answer: a [Reason:] For n-type semiconductors, the donor band is just above the intrinsic Fermi level.

6. At absolute zero temperature, which level is above the Fermi energy level in the case of donors?
a) Donor energy level
b) Acceptor energy level
c) Conduction Band
d) Valence Band

Answer: c [Reason:] At T=0 K, the tem exp(-∞)=0 in the expression of Thus, EF>ED So, only conduction band lies above the Fermi energy level.

7. At T=0 K, the location of Fermi level with respect to the Ec and Ed for the n type material is?
a) Above than conduction band
b) Midway
c) Lower than Ed
d) Greater than Ed

Answer: b [Reason:] At T=0 K, the Fermi level of n band lies between the midway of Ec and Ed as intrinsic Fermi level always lies between the Ec and Ev.

8. At absolute zero temperature, which level is below the Fermi energy level in the case of acceptors?
a) Donor energy level
b) Valence Band
c) Conduction band
d) Acceptor energy level

Answer: b [Reason:] At T=0 K, the tem exp(-∞)=0 in the expression of So, only valence band lies below the Fermi energy level of the acceptors.

9.

For the above n-type semiconductor, what is B knows as?
a) Valence Band
b) Conduction Band
c) Donor Energy level
d) Acceptor energy level

Answer: c [Reason:] For n-type semiconductors, the donor energy level is always greater than the Fermi level energy.

10.

For the above given figure, identify the correct option for satisfying the above semiconductor figure?
a) P type, A-Conduction band, B-donor energy band, C- Valence band
b) P type, A-Conduction band, B-acceptor energy band, C- Valence band
c) n type, A-Conduction band, B-donor energy band, C- Valence band
d) n type, A-Conduction band, B-acceptor energy band, C- Valence band

Answer: c [Reason:] The given figure has B band below the intrinsic Fermi level, so that would be acceptor energy band and will be a p-type semiconductor.

## Set 4

1. Which of the following expressions represents the correct distribution of the electrons in the conduction band? (gc(E)=density of quantum states, fF(E)=Fermi dirac probability
a) n(E)=gc(E)*fF(E)
b) n(E)=gc(-E)*fF(E)
c) n(E)=gc(E)*fF(-E)
d) n(E)= gc(-E)*fF(-E)

Answer: a [Reason:] The distribution of the electrons in the conduction band is given by the product of the density into Fermi-dirac distribution.

2. What is the value of the effective density of states function in the conduction band at 300k?
a) 3*1019 cm-3
b) 0.4*10-19 cm-3
c) 2.5*1019 cm-3
d) 2.5*10-19 cm-3

Answer : c [Reason:] : Substituting the values of mn=m0 ,h=6.626*10-34J/s ,k=1.38*10-23 and T=300K, we get Nc=2.5*1019 cm-3.

3. In a semiconductor which of the following carries can contribute to the current?
a) Electrons
b) Holes
c) Both
d) None

Answer : c [Reason:] : In a semiconductor, two types of charges are there by which the flow of the current takes place. So, both the holes and electrons take part in the flow of the current.

4. Which of the following expressions represent the Fermi probability function?
a) fF(E)=exp(-[E-EF]/KT)
b) fF(E)=exp(-[EF-E]/KT)
c) fF(E)=exp([E-EF]/KT)
d) fF(E)=exp(-[EF-E]/KT)

Answer : b [Reason:] : It is the correct formula for the Fermi probability function.

5. Electrons from valence band rises to conduction band when the temperature is greater than 0 k. Is it True or False?
a) True
b) False

Answer : a [Reason:] : As the temperature rises above 0 k, the electrons gain energy and rises to the conduction band from the valence band.

6. What is the intrinsic electrons concentration at T=300K in Silicon?
a) 1.5*1010cm-3
b) 1.5*10-10cm-3
c) 2.5*1019cm-3
d) 2.5*10-19cm-3

Answer : a [Reason:] : Using the formula, We get, ni=1.5*1010cm-3.

7. The intrinsic Fermi level of a semiconductor depends on which of the following things?
a) Emidgap
b) mp*
c) mn*
d) All of the mentioned

Answer : d [Reason:] : From the above formula, Efidepends on all of the options given.

8. What is the difference between the practical value and theoretical value of ni?
a) Factor of 1
b) Factor of 2
c) Factor of 3
d) Factor of 4

Answer : b [Reason:] : This is practically proved.

9. The thermal equilibrium concentration of the electrons in the conduction band and the holes in the valence band depends upon?
a) Effective density of states
b) Fermi energy level
c) Both A and B
d) Neither A nor B

Answer : c [Reason:] : The electrons and holes depends upon the effective density of the states and the Fermi energy level given by the formula, .

10. In which of the following semiconductor, the concentration of the holes and electrons is equal?
a) Intrinsic
b) Extrinsic
c) Compound
d) Elemental

Answer : a [Reason:] : In the intrinsic semiconductor, ni=pi that is the number of the electrons is equal to the number of the holes. Whereas in the extrinsic conductor ni is not equal to pi.

## Set 5

1. The percentage voltage regulation (VL) is given by_________
a) (VNL-VL)/VNL*100
b) (VNL+VL)/VNL*100
c) (VNL-VL)/VL*100
d) (VNL+VL)/VL*100

Answer: a [Reason:] The change in the output voltage from no load to full load condition is called as voltage regulation, where VNL is the voltage at no load condition. It is used to maintain a nearly constant output voltage. If the regulation is high, the output voltage is stable.

2. The limiting value of the current resistor used in a Zener diode (when used as a regulator)
a) (R)min=[(Vin)max + VZ/R
b) (R)min=[(Vin)max-VZ]/R
c) (R)min=[(Vin)max-VZ]R
d) (R)min=[(Vin)max+ VZ]R

Answer: b [Reason:] When the input voltage is maximum, the load current is minimum, the Zener current should not increase the maximum rated value. Therefore there should be a minimum value of resistor.

3. When the regulation by a Zener diode is with a varying input voltage, what happens to the voltage drop across the resistance?
a) Decreases
b) Has no effect on voltage
c) Increases
d) The variations depend on temperature

Answer: c [Reason:] When the input voltage varies, the input current also varies. This makes more current to flow in the diode. This increase in the current should balance a change in the load current. Hence the voltage drop increases across the resistor.

4. In the given limiter circuit, an input voltage Vi=10sin100πt is applied. Assume that the diode drop is 0.7V when it’s forward biased. The zener breakdown voltage is 6.8V.The maximum and minimum values of outputs voltage are _______

a) 6.1V,-0.7V
b) 0.7V,-7.5V
c) 7.5V,-0.7V
d) 7.5V,-7.5V

Answer: c [Reason:] With VI= 10V when maximum, D1 is forward biased, D2 is reverse biased. Zener is in breakdown region. VOMAX=sum of breakdown voltage and diode drop=6.8+0.7=7.5V. VOMIN=negative of voltage drop=-0.7V. There will be no breakdown voltage here.

5. Determine the maximum and minimum values of load current for which the Zener diode shunt regulator will maintain regulation when VIN=24V and R=500Ω. The Zener diode has a VZ=12V and (IZ)MAX=90mA.
a) 40mA, 0mA respectively
b) 36mA, 5mA respectively
c) 10mA, 6mA respectively
d) 21mA, 0mA respectively

Answer: d [Reason:] The current through the resistance R is given by, I=(VIN-VZ)/R= (24-12)/500=24mA. (IL)MAX=I-(IZ)MIN=24-3=21mA .This current is less than (IZ)MAX. So, we assume that all the input current flows through the Zener diode. Under this condition, (IL)MIN is 0mA.

6. Determine the minimum value of load resistance that can be used in the circuit with (IZ)Min=3mA. The input voltage is 10V and the resistance R is 500Ω. The Zener diode has a VZ=6V 0and (IZ)MAX=90mA.
a) 1KΩ
b) 2.4KΩ
c) 1.2KΩ
d) 3.6KΩ

Answer: c [Reason:] The I=(VIN-VZ)/R=(10-6)/500=8mA. (IL)MAX=I-(IZ)MIN=8-3=5mA. (RL)MIN=VZ/(IL)MAX=6/5m=1.2KΩ.

7. A Zener regulator has to handle supply voltage variation from 19.5V to 22.5V. Find the magnitude of regulating resistance, if the load resistance is 6KΩ. The Zener diode has the following specifications: breakdown voltage =18V, (IZ)Min=2µA, maximum power dissipation=60mW and Zener resistance =20Ω.
a) 0 < R < 500Ω
b) 77.8 < R < 500Ω
c) 77.8 < R < 100Ω
d) 18 < R < 500Ω

Answer: b [Reason:] (PZ)MAX/rZ=(IZ)MAX2 . So, (IZ)MAX =60m/20=54.8µA. IL=VO/RL=18/6000=3mA. RMAX=(VMin-VZ)/[( IZ)Min+( IL)MAX]=(19.5-18)/(2µ+3m)=500Ω. RMin=(VMAX-VZ)/[( IZ)MAX+( IL)Min]=(22.5-18)/(54.8m+3m)=77.8Ω.

8. A transistor series regulator has the following specifications: VIN=15V, VZ=8.3V, β=100, R=1.8KΩ, RL=2KΩ. What will be the Zener current in the regulator circuit?
a) 4.56mA
b) 3.26mA
c) 4.56mA
d) 3.68mA

Answer: d [Reason:] We know, VO=VZ-VBE=8.3-0.7=7.6V. VCE=VIN-V0=15-7.6=7.4V. So, IR=(VIN-VZ)/R=(15-8.3)/1.8m=3.72mA. IL=VO/RL=7.6/2000=3.8mA. IB=IL/ β=3.8mA/100=0.038mA. Finally, IZ=IR-IB=3.72-0.038=3.682mA.

9. When is a regulator used?
a) when there are small variations in load current and input voltage
b) when there are large variations in load current and input voltage
c) when there are no variations in load current and input voltage
d) when there are small variations in load current and large variations in input voltage

Answer: a [Reason:] The regulator has following limitations: 1.It has low efficiency for heavy load currents 2. The output voltage changes slightly due to Zener impedance. Hence, it is used when there are small variations in load current and input voltage.

10. A transistor in a series voltage regulator acts like a variable resistor. The value of its resistance is determined by _______
a) emitter current
b) base current
c) collector current
d) it is not controlled by the transistor terminals