# Linear Integrated Circuit MCQ [Free PDF] – Objective Question Answer for Linear Integrated Circuit Quiz

81. The maximum current through the load in all application that uses voltage to current converter with floating is

A. 100mA
B. 75mA
C. 25mA
D. 50mA

Answer: C

The maximum current through the load cannot exceed the short circuit current of the 741c op-amp which is 25mA.

82. For voltage to current converter with grounded load, establish a relation between the non-inverting input terminals and load current

A. V1 = [Vin+Vo-(IL×R)] /2
B. V1 = [Vin-Vo-(IL×R)] /2
C. V1 = [Vin+Vo-IL+R] /2
D. V1 = [Vin+Vo+(IL×R)] /2

Answer: A

In the voltage to current converter circuit, the relationship between the voltage v1 at the non-inverting input terminal and load is given as

V1 = [Vin+Vo-(IL×R)] /2.

83. Find the gain of the voltage to the current converter with grounded load?

A. 2
B. 1
C. ∞
D. 0

Answer: A

In voltage to current converter with a grounded load all resistors must be equal in value.

∴ Gain = Vo/Vin

= [1+(RF/R1)] = 1+R/R =1+1=2.

84. Find the output voltage and the load current for the circuit given below. Assume that the op-amp is initially nulled V1 =2.5v

A. IL=0.42mA, Vo =10v
B. IL=0.42mA, Vo =3.4v
C. IL=0.42mA, Vo =6.1v
D. IL=0.42mA, Vo =5v

Answer: D

The load current IL =vin /R

=5/12kΩ =0.42mA

vo =IL×R, can be obtained when

Vo=2×V1 = 2×2.5 =5v.

85. The output current equation for MC1408 digital to analog converter would be

A. Io= -(Vref/R1)×[(D7/2)+(D6/4)+(D5/8)+(D4/16)+(D3/32)+(D2/64)+(D1/128)+(D0/256)].

B. Io= (Vref/2R1)×[(D7/2)+(D6/4)+(D5/8)+(D4/16)+(D3/32)+(D2/64)+(D1/128)+(D0/256)].

C. Io= (Vref/R1)×[(D7/2)+(D6/4)+(D5/8)+(D4/16)+(D3/32)+(D2/64)+(D1/128)+(D0/256)].

D. Io= -(Vref/2R1)×[(D7/2)+(D6/4)+(D5/8)+(D4/16)+(D3/32)+(D2/64)+(D1/128)+(D0/256)].

Answer: C

MC1408 is a combination of a DAC and current to voltage converter. If the binary signal is input to MC1408 DAC then the output current would be,

Io= (Vref/R1)×[ (D7/2)+(D6/4)+(D5/8)+(D4/16)+(D3/32)+(D2/64)+(D1/128)+(D0/256)].

86. Determine the maximum value of output current of the DAC in MC1408?

A. 0.773×(Vref/R1)
B. 0.448×(Vref/R1)
C. 0.996×(Vref/R1)
D. 0.224×(Vref/R1)

Answer: C

The output current of DAC is the maximum when all the inputs are logic 1.

Therefore, Io= (Vref/R1)×(1/2+1/4+1/8+1/16+1/32+1/64+1/128/+1/256)

=(0.996)×(Vref/R1).

87. Determine the range or the output voltage?

A. 0 – 2.51v
B. 0 – 2.22v
C. 0 – 3.74v
D. 0 – 4.93v

Answer: D

When all binary input D0 through D7 are logic 0, the current Io =0.

∴ The minimum value of Vo =0v.

When all the inputs are at logic 1, Io = (Vref/R1) ×

(1/2+1/4+1/8+1/16+1/32+1/64+1/128/+1/256)

= (3/2kΩ) × (0.996) =1.494mA.

Hence, the maximum value of output voltage is

Vo= Io×RF = 1.494×3.3kΩ =4.93v.

Thus, the output voltage range is from 0 to 4.93v.

88. Calculate the change in the output voltage if the photocell is exposed to a light of 0.61lux from a dark condition. Specification: Assume that the op-amp is initially nulled, Minimum dark resistance = 100kΩ, and resistance when illuminated (at 0.61lux) = 1.5kΩ.

A. Vo –> 23v to 50v
B. Vo –> 0v to 33.11v
C. Vo –> -1.653v to 8.987v
D. Vo –> -0.176v to -11.73v

Answer: D

The resistance RT in darkness is 100kΩ.

The minimum output voltage in darkness is

Vo min = -(Vdc×RF)/ RT

= -(3.2v×5.5kΩ)/100kΩ = -0.176v.

When the photocell is illuminated, its resistance RT =1.5kΩ.

Therefore, the maximum output voltage is

Vo max = -(Vdc×RF)/ RT

= -(3.2v×5.5kΩ)/1.5kΩ =-11.73v.

Thus, Vo varies from -0.176v to -11.73 as the photocell is exposed to light from a dark condition.

89. Which cell can be used instead of a photocell to obtain an active transducer in photosensitive devices?

A. Photovoltaic cell
B. Photodiode
C. Photosensor
D. All of the mentioned

Answer: A

A photovoltaic cell is a semiconductor junction device that converts radiation energy into electrical energy and hence it does not require external voltage.

90. If the input applied to DAC using the current to voltage converter is 10110100, determine the reference voltage (Assume Io= 2mA and R1=1.2kΩ)

A. 53.1v
B. 3.41v
C. 9.21v
D. 67.34v

Answer: B

Io=Vref/R1×[(D7/2)+(D6/4)+(D5/8)+(D4/16)+(D4/32)+(D4/64)]

Vref =(Io×R1)/ (1/2+0+1/8+1/16+0+1/64+0+0)

=(2mA×1.2kΩ)/0.703 .

=> Vref = 3.41v.

91. The current to voltage converter photosensitive device can be used as

A. Light intensity meter
B. Light radiating meter
C. Light deposition meter
D. None of the mentioned

Answer: A

The photosensitive device can be used as a light intensity meter by connecting a meter at the output that is calibrated for light intensity.

92. For a full-wave rectification, in a low voltage ac voltmeter, the meter current can be expressed as

A. Io = (1.9×Vin)/R1
B. Io = (3.9×Vin)/R1
C. Io = (0.9×Vin)/R1
D. Io = (2.9×Vin)/R1

Answer: C

For full-wave rectification, meter current is expressed as

Io = 0.9 ×Vin/R1.

93. Which of the following circuit has the highest input resistance?

A. Voltage follower
B. Inverting amplifier
C. Differential amplifier
D. None of the mentioned

Answer: A

The voltage follower has the highest positive input resistance of an op-amp circuit. For this reason, it is used to reduce voltage error caused by source loading.

94. Find the bias current from the given circuit

A. 30mA
B. 3mA
C. 0.30mA
D. 0.03mA

Answer: C

The bias current is given as

Iin =Vin/Rin = 3v/10kΩ.

Where, Iin= Ib =0.3mA.

95. How to choose an op-amp when working with high input source resistance?

A. Op-amp with low bias current
B. Op-amp with higher slew rate
C. Buffer or voltage follower
D. All of the mentioned

Answer: D

When the op-amp is driven by a high input source resistance, the output and input voltage will not be equal due to an error at the input. A remedy to this problem is an op-amp with a low input bias current and a high slew rate should be chosen as a voltage follower.

96. What must be done to block the ac input voltage riding on a dc level?

A. Use RC network
B. Use coupling capacitor
C. Use resistive transducer
D. None of the mentioned

Answer: B

To block the dc level a coupling capacitor must be used in series with the input of the voltage follower.

97. To get higher input resistance in AC coupled voltage follower,

A. The output resistance is bootstrapped
B. The input resistance is bootstrapped
C. The bias resistance is bootstrapped
D. The feedback resistance is bootstrapped

Answer: C

Bias resistor connected to the ground to provide a path in an AC coupled voltage follower, drastically reducing the input resistance of the circuit. Therefore, to get high input resistance, the bias resistance is bootstrapped.

98. Voltage follower circuits are used in

A. Active filter
B. All of the mentioned
C. Sample and hold circuit
D. Bridge circuit with transducer

Answer: B

Voltage followers are useful for all the above-mentioned applications because they involve working with high-input source resistance.

99. Compute the reference voltage for a fundamental log-amp, if its internal resistance=5MΩ.

A. 0.5µv
B. 0.05µv
C. 5µv
D. None of the mentioned

Answer: A

The reference voltage, Vref = R1 × Is

Where, Is~10-13A (for an emitter saturation current).

∴ Vref = 10-13 × 5MΩ

= 5×10-7 = 0.5µv.

100. Which of the following functions does the antilog computation required to perform continuously with log-amps?

A. In(x)
B. log(x)
C. Sinh(x)
D. All of the mentioned

Answer: D

Log-amp can easily perform function such as In(x), Log(x), Sinh(x) to have direct dB display on digital voltmeter and spectrum analyser.

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