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## Linear Integrated MCQ Set 1

1. How many gates per chip are used in first generation Integrated Circuits?
a) 3-30
b) 30-300
c) 300-3000
d) More than 3000

Answer: a [Reason:] The first generation ICs belongs to small scale integration, which consists of 3-30 gates per chip (approximately).

2. Find the chip area for a Medium Scale Integration IC?
a) 8 mm3
b) 4 mm2
c) 64 mm3
d) 16 mm2

Answer: d [Reason:] The approximate length and breadth of Medium Scale Integration would be 4 mm. Therefore, its area is given as = length × breadth = 4mm × 4mm = 16mm2.

3. The number of transistors used in Very Large Scale Integration is
a) 107 transistors/chip
b) 106 – 107 transistors/chip
c) 203 – 105 transistors/chip
d) 102 – 203 transistors/chip

Answer: c [Reason:] Very Large Scale Integration (VLSI) ICs are fabricated using more than 3000 gates/chip, which is equivalent to 20,000 – 1,00,00,00 transistors/chip.

4. What type of integration is chosen to fabricate Integrated Circuits like Counters, multiplexers and Adders?
a) Small Scale Integration (SSI)
b) Medium Scale Integration (MSI)
c) Large Scale Integration (LSI)
d) Very Large Scale Integration (VLSI)

Answer: b [Reason:] Fabrication of ICs like counter, multiplexers and Adders requires 30-300 gates per chip. Therefore, Medium Scale Integration is best suitable.

5. Determine the chip area for Large Scale Integration ICs.
a) 1,00,000 mil2
b) 10,000 mil2
c) 1,60,000 mil2
d) 16,000 mil2

Answer: c [Reason:] The chip area for a Large Scale Integration IC is 1 cm2. => Area of LSI = 10mm × 10mm = 1cm × 1 cm = 1cm2. => 1,60,000mil2 (1cm=400mil).

6. Ultra Large Scale Integration are used in fabrication of
a) 8-bit microprocessors, RAM, ROM
b) 16 and 32- bit microprocessors
c) Special processors and Smart sensors
d) All of the mentioned

Answer: c [Reason:] Ultra Large Scale Integration have nearly 106 – 107 transistors/chip. Hence, it is possible to fabricate smart sensors and special processor.

7. The concept of Integrated circuits was introduced at the beginning of 1960 by
a) Texas instrument and Fairchild Semiconductor
b) Bell telephone laboratories and Fair child Semiconductor
c) Fairchild Semiconductor
d) Texas instrument and Bell telephone Laboratories

Answer: a [Reason:] The concept of Integrated circuits was introduced by Texas instrument and Fairchild Semiconductor, whereas Bell telephone laboratories developed the concept of transistors.

8. Which process is used to produce small circuits of micron range on silicon wafer?
a) Photo etching
b) Coordinatograph
c) Photolithography
d) Ion implantation

Answer: c [Reason:] It is possible to fabricate as many as 10,000 transistors on a 1cmX1cm chip, using photolithography process.

9. Mention the technique used in photolithography process
a) X-ray lithographic technique
b) Ultraviolet lithographic technique
c) Electron beam lithographic technique
d) All of the mentioned

Answer: d [Reason:] All these techniques are used to produce device dimension as small as 2µm or even down to sub micron range (<1µm).

## Linear Integrated MCQ Set 2

1. Which is not considered as a linear voltage regulator?
a) Fixed output voltage regulator
c) Switching regulator
d) Special regulator

Answer: c [Reason:] In linear regulator’s the impedance of active element may be continuously varied to supply a desired current to the load. But in the switching regulator, a switch is turned on and off.

2. What is the dropout voltage in a three terminal IC regulator?
a) |Vin| ≥ |Vo|+2v
b) |Vin| < |Vo|-2v
c) |V in| = |Vo|
d) |Vin| ≤ |Vo|

Answer: a [Reason:] The unregulated input voltage must be atleast 2V more than the regulated output voltage. For example, if Vo=5V, then Vin=7V.

3. To get a maximum output current, IC regulation are provided with
b) Heat sink
c) Peak detector
d) None of the mentioned

Answer: b [Reason:] The load current may vary from 0 to rated maximum output current. To maintain this condition, the IC regulator is usually provided with a heat sink; otherwise it may not provide the rated maximum output current.

4. For the given circuit, let VEB(ON)=1v, ß= 15 and IO=2mA. Calculate the load current

a) IL = 23.45A
b) IL = 46.32A
c) IL = 56.87A
d) IL = 30.75A

Answer: d [Reason:] The equation for load current, IL = [(ß+1)IO]-[ß×(VEB(ON)/R1)]=[(15+1)×2]–[15×(1v/12 Ω)] =32-1.25 =30.75A.

5. Which type of regulator is considered more efficient?
a) All of the mentioned
b) Special regulator
c) Fixed output regulator
d) Switching regulator

Answer: d [Reason:] The switching element dissipates negligible power in either on or off state. Therefore, the switching regulator is more efficient than the linear regulators.

6. State the reason for thermal shutdown of IC regulator?
a) Spikes in temperature
b) Decrease in temperature
c) Fluctuation in temperature
d) Increase in temperature

Answer: d [Reason:] The IC regulator has a temperature sensor (built-in) which turn off the IC, when it becomes too hot (usually 125oC-150oC). The output current will drop and remains there until the IC has cooled significantly.

7. Find the difference between output current having a load of 100Ω and 120Ω for 7805 IC regulator. Consider the following specification: Voltage across the load = 5v; Voltage across the internal resistor= 350mv.
a) 8.4mA
b) 7mA
c) 9mA
d) 3.4mA

Answer: a [Reason:] Given the voltage across the internal resistor to be 350mv, which is less than 0.7v. Hence the transistor in 7805 is off. When load = 100Ω, IL= IO= Ii= 5v/100 Ω = 50mA When load=120Ω, IO= 5v/120 Ω = 41.6mA. So, the difference between the output voltage = 50-41.6mA = 8.4mA.

8. The change in output voltage for the corresponding change in load current in a 7805 IC regulator is defined as
a) All of the mentioned
b) Line regulation
d) Input regulation

Answer: c [Reason:] Load regulation is defined as the change in output voltage for a change in load current and is also expressed in millivolts or as a percentage of output voltage.

9. An IC 7840 regulator has an output current =180mA and internal resistor =10Ω. Find the collector current in the output using the transistor specification: ß=15 and VEB(ON) =1.5v.
a) 270mA
b) 450mA
c) 100mA
d) 50mA

Answer: b [Reason:] The collector current from transistor, IC= ßIB Where, IB= IO-(VEB(ON)/R1) = 180mA-(1.5v-10Ω) = 0.03A. Therefore, IC= 15×0.03 = 0.45A = 450mA.

10. How the average temperature coefficient of output voltage expressed in fixed voltage regulator?
a) miilivolts/oC
b) miilivoltsoC
c) None of the mentioned
d) oC/ miilivolts

Answer: a [Reason:] The temperature stability or average temperature coefficient of output voltage, is the change in the output voltage per unit change in temperature and expressed in miilivolts/oC.

11. In the circuit given below, let VEB(ON)=0.8v and ß=16. Calculate the output current coming from 7805 IC and collector current coming from transistor Q1 for a load of 5Ω.

a) IO =111mA, IC= 808mA
b) IO =111mA, IC= 829mA
c) IO =111mA, IC= 881mA
d) IO =111mA, IC= 889mA

Answer: d [Reason:] When load = 5Ω, IL= 5v/5Ω =1A. The voltage across R1 is 7Ω × 1A=7v. Since, IL is more than 100mA, the transistor Q1 turns on and supplies the extra current required. Therefore, IL =(ß+1)IO-[ß×(VEB(ON)/R1) IO = [IL/(ß+1)]+ [ß×(VEB(ON)/R1) = [1/(16+1)]+[16×(0.8/2Ω)] ≅111mA. => IC=IL-IO=1A-111mA =889mA.

12. Calculate the output voltage for LM314 regulator. The current IADJ is very small in the order of 100µA. (Assume VREF=1.25v)

a) 17.17v
b) 34.25v
c) 89.34v
d) 23.12v

Answer: a [Reason:] The output voltage, VO =VREF[1=(R2/R1)]+(IADJ×R2)=1.25Vin× [1+(3kΩ/240Ω)] +( 100µA×3kΩ )= 16.875 +0.3. => VO=17.17v.

13. Compute the input voltage of 7805c voltage regulator with a current source that will deliver a 0.725A current to 65Ω, 10w load. (Assume reference voltage =5v)
a) Vin = 84v
b) Vin = 34v
c) Vin = 54v
d) Vin = 64v

Answer: c [Reason:] VO=VREF+VL =VREF+(IL×RL) = 5v+(0.725A×65Ω) = 52.125v => Input voltage, Vin = VO + dropout voltage = 52.125v+2v. => Vin ≅54v.

14. Which of the following is not a characteristic of adjustable voltage regulators?
a) Non-versatile
b) Better performance
c) Increased reliability
d) None of the mentioned

Answer: a [Reason:] Adjustable voltage regulators are versatile; it has improved over-load protection allowing greater output current over operating temperature range.

## Linear Integrated MCQ Set 3

1. Input bias current is defined as
a) Average of two input bias current
b) Summing of two input bias current
c) Difference of two input bias current
d) Product of two input bias current

Answer: a [Reason:] Input bias current is the average of two input bias current flowing into the non-inverting and inverting input of an op-amp.

2. Although the value of input bias current is very small, it causes
a) Output voltage
b) Input offset voltage
c) Output offset voltage
d) All of the mentioned

Answer: c [Reason:] Even a very small value of input bias current can cause a significant output offset voltage in circuits using relatively large feedback resistors.

3. The formula for output offset voltage of an op-amp due to input bias current
a) VOIB= RF*IB
b) VOIB= (RF+R1)/IB
c) VOIB= (1+RF)*IB
d) VOIB= [1+(RF/R1)]*IB

Answer: a [Reason:] The output offset voltage due to input bias current is VOIB = RF*IB.

4. Find the input bias current for the circuit given below

a) 10mA
b) 2mA
c) 5mA
d) None of the mentioned

Answer: c [Reason:] Input bias current, IB=(IB1+ IB2)/2 => IB =(4mA+6mA)/2 = 5mA.

5. Mention a step to reduce the output offset voltage caused due to input bias current?
a) Use small feedback resistor and resistance at the input terminal
b) Use small feedback resistors
c) Reduce the value of load resistors
d) None of the mentioned

Answer: b [Reason:] Since the output offset voltage is proportional to feedback resistor and input bias current. The amount of VOIB can be reduced by reducing the value of feedback resistor.

6. Given below is a differential amplifier in which V1=V2. What happens to VOIB at this condition?

a) VOIB= 0
b) VOIB= VOIB×10-10
c) VOIB= VOIB/2
d) VOIB= -1

Answer: a [Reason:] The voltage V1 and V2 are caused by the current IB1 and IB2. Although this bias current are very small, if they are made equal, then there will be no output voltage VOIB.

7. Name the resistor that is connected in the non-inverting terminal of op-amp which is in parallel combination of resistor connected in inverting terminal and feedback resistor.
a) Random minimizing resistor
b) Offset minimizing resistor
c) Offset reducing resistors
d) Output minimizing resistors

Answer: b [Reason:] The voltage is product of resistors and input bias current. Therefore, the value of the resistors are adjusted such that the resistors are connected at the inverting input terminal is made equal to resistor connected in non-inverting input terminal. The use of this resistors minimize the amount of output offset voltage and therefore, they are referred to as offset minimizing resistors.

8. Calculate ROM, if the value of IB1 = IB2 in the given circuit.

a) 1173.11Ω
b) 171.31Ω
c) 1171.43Ω
d) 1071.43Ω

Answer: d [Reason:] Offset minimizing resistor, ROM =(R1* RF)/( R1+RF). => ROM = (1.2kΩ*10kΩ)/(1.2kΩ+10Ω) = 1071.43Ω.

9. Calculate the output voltage for the given circuit using the specification: R1 = 820Ω; ROM=811.882Ω; Vin=10mVpp; VOIB≅0.

a) 1.025Vpp
b) 1.8Vpp
c) 1Vpp
d) 2Vpp

Answer: c [Reason:] Offset minimizing resistor, ROM = (R1*RF)/(R1+ RF) => RF = (ROM* R1)/( R1– ROM) = (812Ω*811.882Ω)/(820Ω-811.882Ω) = 82kΩ. ∴ Vo = -(RF/ R1)* Vin = -(82kΩ/820Ω)*10mVpp = 1Vpp.

10. Analyse the given circuit and determine the correct option

a) Voo ≥ VIOB
b) Voo = VIOB
c) Voo >> VIOB
d) Voo << VIOB

Answer: c [Reason:] 741op-amp has Vio = 6mvdc and IB =500nA. The output offset voltage due to input offset voltage is given as Voo =[1+(RF/R1)]*Vio = [1+(4.7kΩ/47Ω)]*6mv = 0.606v. The output offset voltage due to input bias current is given as VIOB = RF*IB =4.7kΩ*500nA = 2.35mv. =>∴ Voo >> VIOB.

11. The specification for LM101A op-amp is given as IB =75nA. Determine the value of VIOB– V1.

a) 0.112v
b) 0.750v
c) 0.374v
d) 0.634v

Answer: a [Reason:] The voltage at non-inverting terminal is given as V1 = ROM*IB1 = 148Ω*7.5nA = 1.11µv. => ∵ ROM = (R1*RF)/(R1+ RF) = (15kΩ*150Ω)/(15kΩ+150Ω) =148Ω The output offset voltage is given as VIOB = RF*IB => VIOB = 15kΩ*7.5nA = 112.5mv => ∴ VIOB– V1 = 0.112v.

## Linear Integrated MCQ Set 4

1. The maximum amount by which the two input bias current may differ is known as
a) Input null current
b) Average input bias current
c) Input offset current
d) None of the mentioned

Answer: c [Reason:] The input offset current is defined as the algebraic difference between two input bias current.

2. A 741 type op-amp has a maximum input offset current of 200nA dc. What conclusion can be derived from this statement?
{ IB1 – Input bias current at inverting input terminal and IB2 – Input bias current at non-inverting input terminal}
a) IB1 may be larger than IB2 by 200nA
b) IB2 may be larger than IB1 by 200nA
c) Iio and IB2 may be equal to 200nA
d) All of the mentioned

Answer: d [Reason:] In a 741 op-amp, Iio = 200nA dc, which means that the maximum difference between IB1 and IB2 can be as large as 200nA.

3. The maximum magnitude of the output offset voltage is
a) VOIio = RF*Iio
b) VOIio = RF*( IB1+IB2)
c) VOIio = RF*IB1
d) VOIio = RF*( IB1-IB2)

Answer: a [Reason:] The output offset voltage due to input offset current is given as VOIio = RF*Iio.

4. Find the output offset voltage of an 741 op-amp; If the gain of the non-inverting amplifier is 8.5 and feedback resistor = 15kΩ? (IB=200nA for 741 op-amp)
a) 1µv
b) 4 µv
c) 3 µv
d) 2 µv

Answer: c [Reason:] The circuit diagram of inverting amplifier is given below Gain, A=1+(RF/R1) => R1 = RF/(A-1) = 15kΩ/(8.5-1) = 2kΩ. => ROM =(R1* RF)/( R1+RF) = 1.76kΩ. The output offset voltage, VOIB1= VOIB= RF*IB => VOIB1= 1.76kΩ*200nA*8.5 = 2.9×10-6 ≅ 3 µv.

5. Find out the input offset current from the circuit

a) Iio = |IBA*IBC|
b) Iio = |IBA+ IBC|
c) Iio = |IBA/ IBC|
d) Iio = |IBA– IBC|

Answer: d [Reason:] the input offset current is Iio = |IBA– IBC|.

6. Determine the maximum output offset voltage caused by input offset current

a) 5.4mv
b) 7.3mv
c) 6.9mv
d) 8.1mv

Answer: a [Reason:] For a 741 op-amp, Iio = 200nA(Maximum). => Therefore, VOIio = RF*Iio = 27kΩ*200nA =5.4mv.

7. Find the total output offset voltage of non-inverting op-amp?
a) VooT = {[1+(RF/R1)]*Vio} + RF*Iio
b) VooT = -[(RF/R1)*Vio] + RF*Iio
c) VooT = -[(RF/R1)*Vio] + Iio
d) None of the mentioned

Answer: b [Reason:] The gain of the inverting amplifier A=-(RF/R1). Therefore, the total output offset voltage is given as VooT = -[(RF/R1)*Vio] + RF*Iio.

8. Analyse the circuit and determine the value of total output offset voltage?

The op-amp is the MC1536 with the following specifications:
Vio = 7.5mv, Iio = 5nA, IB =250nA.
a) 0.12v
b) 1.4v
c) 0.76v
d) 8.5v

Answer:c [Reason:] The total output offset voltage due to input offset voltage, VooT = -[(RF/R1)*Vio] + RF*Iio => VooT = [1+(47kΩ/470Ω)]*7.5mv+(47kΩ*50nA) = 0.7598 ≅ 0.76v.

## Linear Integrated MCQ Set 5

1. What makes the output voltage equals to zero in practical op-amp?
a) Input offset voltage
b) Output offset voltage
c) Offset minimizing voltage
d) Error voltage

Answer: a [Reason:] Input offset voltage is the differential input voltage that exists between two input terminals of an op-amp without any external input and force the output voltage to zero.

2. What happens due to mismatch between two input terminals in an op-amp?
a) Input offset voltage
b) Output offset voltage
c) Bothe the input and output offset voltage
d) None of the mentioned

Answer: b [Reason:] The input offset voltage in op-amp force the output voltage to zero due to the mismatch between two input terminal, there will be voltage produced at the output and this voltage is called output offset voltage.

3. Define polarity of the output offset voltage in a practical op-amp?
a) Positive polarity
b) Negative polarity
c) Positive or negative polarity
d) None of the mentioned

Answer: c [Reason:] The output offset voltage is a DC voltage, it may be positive or negative in polarity depending on whether the potential difference between two input terminal is positive or negative.

4. The input offset voltage of 741 op-amp has an absolute maximum value of 6mv, which means
a) Minimum difference between input terminals in 741 op-amp can be large as 6mv DC
b) Minimum difference between input terminals in 741 op-amp can be large as 6mv AC
c) Maximum difference between input terminals in 741 op-amp can be large as 6mv DC
d) Maximum difference between input terminals in 741 op-amp can be large as 6mv AC

Answer: c [Reason:] Given, the absolute maximum value for a 741 is Vio= 6mv. Therefore, voltage at the non-inverting input terminal may differ from that at the inverting input terminal by as much as 6mv dc. Also the output offset voltage is a DC voltage and it cannot be AC voltage.

5. If three different 741 op-amps are taken and the corresponding output offset voltage for each of them is measured. The output voltage in these three op-amps have
a) Same amplitude and polarity
b) Different amplitude and polarity
c) Same amplitude and different polarity
d) Different amplitude and same polarity

Answer: b [Reason:] Even though the op-amps are of the same type, the output voltage in these three op-amps are not of the same amplitude and polarity, because of mass production.

6. To reduce the output offset voltage VooT to zero
a) Input offset voltage compensating network is added at the inverting input terminal
b) Input offset voltage compensating network is added at the non-inverting input terminal
c) Input offset voltage compensating network is added at the output terminal
d) None of the mentioned

Answer: d [Reason:] To reduce the VooT to zero, the external circuit is added at the input terminal of the op-amp that will give the flexibility of obtaining input offset voltage of proper amplitude and polarity. The input terminal can be inverting or non-inverting.

7. Which of the following op-amp does not need compensating network?
a) 777
b) 741
c) 748
d) All of the mentioned

Answer: d [Reason:] The compensating network is not needed for these op-amps because, they have offset null pins.

8. Find out the voltage offset null circuit for the 741 op-amp?