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# Multiple choice question for engineering

## Set 1

1. PIN diode is a photosensitive diode because of _______
a) large current flow in p and n region
b) depletion layer increases giving a larger surface area
c) stronger covalent bonds
d) low carrier storage

Answer: b [Reason:] An intrinsic layer that is sandwiched between p and n layers. This gives a larger surface area making it compatible for photosensitivity. Reverse bias causes an increased depleted region in a PIN diode.

2. During forward bias, the PIN diode acts as _______
a) a variable resistor
b) a variable capacitor
c) a switch
d) an LED

Answer: a [Reason:] In forward bias, the forward resistance decreases and acts as a variable resistor. The low frequency model of a PIN diode neglects the input capacitive values.

3. During reverse bias, the PIN diode acts as _______
a) Variable resistor
b) Switch
c) Variable capacitor
d) LED

Answer: c [Reason:] In reverse bias, the intrinsic layer is completely covered by depletion layer. The stored charges vanishes acting like a variable capacitor. The high frequency model of a PIN diode neglects the input resistances.

4. When the p and n regions are used for high resistivity, the depletion region at the respective places is called _________
a) Q and ϒ regions
b) ϒ and π regions
c) Q and π regions
d) π and ϒ regions

Answer: d [Reason:] When p region is used for high resistance, the depletion layer is high at p side.When n side is used the depletion layer is high at n side. They are called as π and ϒ regions respectively.

5. The applications for PIN diode are __________
a) Microwave switch
b) LED
c) Voltage regulator
d) Amplifier

Answer: a [Reason:] Being employed at 300Hz, the swept voltage is attained at π region.Then it’s used as a microwave switch. Swept voltage is nothing but, the voltage at which the complete intrinsic layer is swept out as a depleted one.

6. In high frequency model, the values of resistance ‘R’ and capacitance ‘C’ are _______
a) 0.1 to 10KΩ and 0.02 to 2pF respectively
b) 1 to 10KΩ and 0.02 to 2pF respectively
c) 10 to 100KΩ and 0.02 to 2pF respectively
d) 0.1 to 10KΩ and 2 to 20pF respectively

Answer: a [Reason:] At high frequency, the applied values for resistance and capacitance is 0.1 to 10KΩ and 0.02 to 2pF respectively. At high frequencies, it almost acts as a perfect resistor.

7. What happens in PIN diode for low frequency model?
a) reactance decreases
b) conductance increases
c) resistance increases
d) reactance increases

Answer: d [Reason:] In a low frequency model, the resistance decreases and reactance increases.Here the variable resistance is neglected. At low frequencies, the charge can be removed and the diode can be turned off.

8. Which of the following is true about a PIN diode?
a) it’s photosensitive in reverse bias
b) it offers low resistance and low capacitance
c) it has a decreased reversed breakdown voltage
d) carrier storage is low

Answer: a [Reason:] Due to increased depletion region, the covalent bonds break and increase the surface area for photosensitivity. This property is used in fields of light sensors, image scanners, artificial retina systems.

9. In the application of frequency models, the value of forward current is _____
a) IF = A(µPP + µNN)q
b) IF = A(µPN + µNP)q
c) IF = A(µPP – µNN)q
d) IF = A(µPN – µNP)q

Answer: a [Reason:] The forward current depends on mobility and carrier concentration. In frequency models, the value of forward current is IF = A*(µPP + µNN)q. Where, µP and µN are the mobility of p and n type charge carriers respectively.

10. The forward resistance for a PIN diode is given by ________
a) RF = W/σP
b) RF = W/σN
c) RF = WσP
d) RF = WσN

Answer: b [Reason:] Forward resistance for a PIN diode depends on the width, current density and positive carrier concentration of the diode. No diode is perfectly ideal. In practise, a diode offers a small resistance in forward bias which is called as forward resistance.

## Set 2

1. Diode acts as a short circuit when switched from forward to reverse bias for some time due to______
a) Accumulation of minority charge carriers when it’s in forward bias
b) Accumulation of majority charge carriers when it’s in forward bias
c) Accumulation of minority charge carriers when it’s in reverse bias
d) Accumulation of majority charge carrier when it’s in reverse bias

Answer: a [Reason:] When a diode is switched suddenly, it persists the conducting property for a short time in its reverse bias also. This leads to excess minority charge carrier settlement at potential barrier. Hence acts as a short circuit.

2. Reverse recovery time for a diode is?
a) Time taken to eliminate excess minority charge carriers
b) Sum of storage time (TS) and transition time (TT)
c) Time taken to eliminate excess majority charge carriers
d) Time elapsed to return to non conduction state

Answer: a [Reason:] The time period for which diode remains in conduction state even in reverse direction is called storage time. The time elapsed to return the non conduction state is called transition time. Their sum is called reverse recovery time.

3. Switching speed of P+ junction depends on.
a) Mobility of minority carriers in P junction
b) Life time of minority carriers in P junction
c) Mobility of majority carriers in N junction
d) Life time of minority carriers in N junction

Answer: d [Reason:] Switching leads to move holes in P region to N region as minority carriers. Removal of this accumulation determines switching speed. P+ regards to a diode in which the p type is doped excessively.

4. Time taken for a diode to reach 90% of its final value when switched from steady state is______
a) 2.3*time constant
b) 2.2*time constant
c) 1.5*time constant
d) equals the time constant

Answer: b [Reason:] Time constant = RC. To reach 90% of the final value, time taken is 2.2 of RC. Time constant is the time required to discharge the capacitor, through the resistor, by 36.8%.

5. Which of the following are true?
1) In reverse bias, the diode undergoes stages of storage and transition times
2) Minority charge carriers accumulation makes the diode as a short circuit
3) Storage time is the sum of recovery and transition times
a) 1 only
b) 2 and 3
c) 3 only
d) 1 and 2 only

Answer: d [Reason:] When a diode is switched from forward to reverse bias, storage and transition times takes place. The accumulation time or the life time of minority carriers makes it a short circuit. The conduction property is holds for a short period of time in reverse bias also.

6. In a circuit below, the switch is at position 1 at t<0 and at position 2 when t=0. Assume diode has zero voltage drop and storage time. For 0<t<ts, the VR at 1k ohm resistor is given by_____ a) 5V
b) -5V
c) 0v
d) 10V

Answer: b [Reason:] At position ‘1’ when connected to +5V, the diode is forward biased and acts as a short circuit. So, VR is 5V. For 0<t<ts VR is -5V as the diode is in reverse bias. But it holds the conductive property within the storage time period. So, V is -5V.

7. The switch is at position shown in the figure initially and steady state is from t=0 to t=to. The switch suddenly is thrown to the other position. The current flowing through the 10K resistor from t=0 is? a) 1mA
b) 2mA
c) -2mA
d) -1mA

Answer: c [Reason:] Initially, the diode is in forward bias. When suddenly switched to reverse bias, upto a storage time limit, it conducts during storage time period. We know that, current I=V/R=-20/10K=-2mA.

8. A PN junction diode with 100Ω resistor is forward biased such that 100A current flows. If voltage across this combination is instantaneously reversed to 10V at t=0, the reverse current that flows through diode at t=0 is? a) 10mA
b) 100mA
c) -100mA
d) -10mA

Answer: b [Reason:] At t=0, V=-10V. During storage time, current still flows. We know that, current I=V/R=10/100Ω=100mA from N to P region.

9. The delay in switching between the ON and OFF states is due to _________
a) The time required to change amount of excess minority carriers stored in quasi-neutral regions
b) The time required to change amount of excess majority carriers stored in quasi-neutral regions
c) The conduction between storage time and recovery time
d) The exponential increase in carriers in N region

Answer: a [Reason:] When switched instantaneously it stays in a quasi state i.e.., temporary state which stores charges. The delay is produced due to this charge settlement. The diode needs to discharge these excess carriers in order to return the non conduction stage.

10. The delay time can be reduced by?
a) decreasing lifetime and increasing ratio of reverse to forward current
b) increasing lifetime and decreasing ratio of reverse to forward current
c) increasing lifetime and increasing ratio of reverse to forward current
d) decreasing lifetime and decreasing ratio of reverse to forward current

Answer: a [Reason:] When the current increases the depletion layer decreases and the storage and transition time decreases. A decreased depletion layer can easily discharge the excess carrier and thereby lessens the delay time.

## Set 3

1. How many junction/s do a diode consist?
a) 0
b) 1
c) 2
d) 3

Answer: b [Reason:] Diode is a one junction semiconductor device which has one cathode and anode. The junction is of p-n type.

2. If the positive terminal of the battery is connected to the anode of the diode, then it is known as
a) Forward biased
b) Reverse biased
c) Equilibrium
d) Schottky barrier

Answer: a [Reason:] When a positive terminal is connected to the anode, the diode is forward biased which lets the flow of the current in the circuit.

3. During reverse bias, a small current develops known as
a) Forward current
b) Reverse current
c) Reverse saturation current
d) Active current

Answer: c [Reason:] When the diode is reverse biased, a small current flows between the p-n junction which is of the order of the Pico ampere. This current is known as reverse saturation current.

4. If the voltage of the potential barrier is V0. A voltage V is applied to the input, at what moment will the barrier disappear?
a) V< V0
b) V= V0
c) V> V0
d) V<< V0

Answer: b [Reason:] When the voltage will be same that of the potential barrier, the potential barrier disappears resulting in flow of current.

5. During the reverse biased of the diode, the back resistance decrease with the increase of the temperature. Is it true or false?
a) True
b) False

Answer: a [Reason:] Due to the increase in the reverse saturation current due to the increase in the temperature, the back resistance decrease with the increasing temperature.

6. What is the maximum electric field when Vbi=2V , VR=5V and width of the semiconductor is 7cm?
a) -100V/m
b) -200V/m
c) 100V/m
d) 200V/m

Answer: b [Reason:] Emax=-2(Vbi+VR)/W =-2(2+5)/ (7*10-2) =-200V/m.

7. When the diode is reverse biased with a voltage of 6V and Vbi=0.63V. Calculate the total potential.
a) 6V
b) 6.63V
c) 5.27V
d) 0.63V

Answer: b [Reason:] Vt=Vbi+VR =0.63+6 =6V.

8. It is possible to measure the voltage across the potential barrier through a voltmeter?
a) True
b) False

Answer: b [Reason:] The contacts of the voltmeter have some resistance which will not accurately measure the voltage across the potential barrier. Thus, it is not possible to measure the voltage across the potential barrier.

9. What will be the output of the following circuit? (Assume 0.7V drop across the diode) a) 12V
b) 12.7V
c) 11.3V
d) 0V

Answer: c [Reason:] V=12-0.7 =11.3V.

10. Which of the following formula represents the correct formula for width of the depletion region?
a) b) c) d) Answer: a [Reason:] Option a is the correct formula.

## Set 4

1. The materials that are used in the construction of point contact diode are _________
a) Silicon
b) SnTe or Bi2Te3
c) GaS or CdS
d) HgI

Answer: b [Reason:] The diode base of SnTe or Bi2Te3 is highly detection sensitive. They are mechanically stable over long periods of use either as harmonic generators or mixers. They are emphasized in the 2-200THz region.

2. Which of the following are true?
1) point contact diode has a metal whisker to make pressure contact during its operation
2) it has high voltage rating
3) its V-I characteristics are constant
4) it has low breakdown voltage
a) 1 and 2
b) 3 only
c) 1 and 4
d) 2 only

Answer: c [Reason:] The diode contains a die of a semiconductor material on which an epitaxial layer is deposited. It uses the metal whisker to make pressure contact against that layer. It has a low breakdown voltage.

3. In the forward bias condition, the resistance of point contact diode is_________
a) less than that of a general PN diode
b) greater than that of a general PN diode
c) equal to that of a general PN diode
d) varies exponentially than that of a general PN diode

Answer: a [Reason:] The current flow of the point contact diode is not independent of voltage applied to the crystal unlikely to a general PN diode. This characteristic of contact diode makes its capacitance high at high frequency. A small capacitive current flows in the circuit.

4. The barrier layer capacitance of a point contact diode is_________
a) 0.1pF to 1pF
b) 5pF to 50pF
c) 0.2pF to 2pF
d) 0.008µF to 20µF

Answer: a [Reason:] The barrier capacitance at the point is very low about 0.1pF to 1pF. The capacitance between the cat whisker and crystal is less compared to junction diode capacitance between both sides of the diode. For a general PN diode is 0.008µF to 20µF.

5. The cat whisker wire present in the contact diode is used for_________
a) for heat dissipation
b) for charge transfer between sections
c) maintaining the pressure between sections
d) preventing current flow

Answer: c [Reason:] The operation of a contact diode depends on the pressure of contact between semiconductor crystal and point. The cat whisker wire presses against the crystal to form a section and the section allows the current flow. This is similar to the behaviour of PN diode.

6. The semiconductor junctions those are present in a contact diode_________
a) beryllium-copper and bronze-phosphor
b) beryllium-phosphor and bronze-copper
c) mercury-iodine
d) tin-tungsten

Answer: a [Reason:] The diode contains two sections having a small rectangular crystal of N type silicon and a fine beryllium-copper and bronze-phosphor. It has a tungsten wire which is called as a cat whisker wire. This helps in pressing one section to other.

7. The application of a contact diode is_________
a) Clampers and clippers
b) Voltage multipliers
c) Rectifiers
d) AM detectors

Answer: d [Reason:] The point contact diodes are the oldest microwave semiconductor devices. They were developed during world war 2. They have excessive applications in microwave fields and used as receivers and detectors.

8. The operating frequencies of the point contact diode is_________
a) 30KHz or above
b) 10GHz or above
c) 30GHz or above
d) 10KHz or above

Answer: b [Reason:] It’s used in high frequency conversions and circuits in the order of 10KHz or above. The reactance due to capacitance is high and at high frequency a very small capacitive current flows.

9. During the manufacture of point contact diode, why is a relatively large current passed from cat whisker to silicon crystal?
a) to control the amount of current flow
b) to form small region of p type material
c) to allow mechanical support for the sections
d) to form anode and cathode regions

Answer: b [Reason:] The behaviour of a contact diode is similar to that of a PN diode. It has a tungsten wire which is called as a cat whisker wire. This helps in pressing one section to other.

10. What is the capacitive reactance across the point contact diode when compared to normal PN junction diode
a) lower
b) higher
c) equal
d) cannot be determined

Answer: a [Reason:] The current flow of the point contact diode is not independent of voltage applied to the crystal unlikely to a general PN diode. A small capacitive current flows in the circuit.

## Set 5

1. Consider an inverting amplifier with a nominal gain of 1000 constructed from an op amp with an input offset voltage of 3 mV and with output saturation levels of ±10 V. What is (approximately) the peak sine-wave input signal that can be applied without output clipping?
a) 7 mV
b) 10 mV
c) 13 mV
d) 9mV

Answer: a [Reason:] The maximum that can be sent without clipping is 10V – 1000 X 3mV or 7V.

(Q2 & Q.3) Consider an inverting amplifier with a nominal gain of 1000 constructed from an op amp with an input offset voltage of 3 mV and with output saturation levels of ±10 V. If the effect of VOs(input offset voltage) is nulled at room temperature (250C), how large an input can one now apply if:

2. The circuit is to operate at a constant temperature?
a) 8.5 mV
b) 9 mV
c) 9.5 mV
d) 10 mV

Answer: d [Reason:] Explanation: Maximum signal that will not be clipped is 10mV because 10mV X 1000 = 10V.

3. The circuit is to operate at a temperature in the range 0°C to 75°C and the temperature coefficient of VOS is 10 μV/°C?
a) 8.5 mV
b) 9 mV
c) 9.5 mV
d) 10 mV

Answer: c [Reason:] Since the effect is nullified at 25oC, the peak that can be sent now is given by 10 – (75-25) X 0.1 mV.

4. One of the DC imperfections of the amplifiers are dc offset voltage which is
a) Existence of output signal even when the common mode signal is zero
b) Existence of common mode signal causing zero output signal
c) Existence of output signal even when the differential signal is zero
d) Existence of differential signal causing zero output signal

Answer: c [Reason:] DC offset voltage is existence of output signal even when the differential signal is zero.

5. For the amplifier shown determine the value of the bias current (Ib) and input offset current (Io) respectively. a) Ib = IB1 + IB2 Io = IB1 – IB2
b) Ib = IB1 + IB2 Io = | IB1 – IB2 |
c) Ib = 0.5(IB1 + IB2) Io = | IB1 – IB2 |
d) Ib = 0.5(IB1 + IB2) Io = IB1 – IB2

Answer: c [Reason:] Standard mathematical expressions are used with the given variables.

6. Consider the circuit shown below which reduces the impact of the input bias current. If IB1 = IB2 = Input bias current, then determine the value of R3 so that the output voltage (v0) is not impacted by the input bias current. a) (R1 R2)/(R1+R2)
b) (R1 R2)/(R1-R2)
c) R1-(R1 R2)/(R1+R2)
d) R2- (R1 R2)/(R1+R2)

Answer: a [Reason:] This will be possible when R3 has the same value as the net effect of R1 and R2.

7. Consider an inverting amplifier circuit designed using an op amp and two resistors, R1 = 10 kΩ and R2 = 1 MΩ. If the op amp is specified to have an input bias current of 100 nA and an input offset current of 10 nA, find the output dc offset voltage resulting.
a) 0.1 mV
b) 1 mV
c) 10 mV
d) 100 mV

Answer: d [Reason:] Use the mathematical definition of bias current and offset current.

(Q.8-Q.10) Consider a Miller integrator with a time constant of 1ms and an input resistance of 10 kΩ. Let the op amp have VOS (offset voltage) = 2 mV and output saturation voltages of ±12 V.

8. Assuming that when the power supply is turned on the capacitor voltage is zero, how long does it take for the amplifier to saturate?
a) 3s
b) 6s
c) 9s
d) 12s

Answer: b [Reason:] Use vO = VOS (VOS/CR)t.

9. Select the largest possible value for a feedback resistor RF so that at least ±10 V of output signal swing remains available.
a) 10 kΩ
b) 100 kΩ
c) 1 MΩ
d) 10 MΩ