Select Page
Generic selectors
Exact matches only
Search in title
Search in content
Search in posts
Search in pages
Filter by Categories
nmims post
Objective Type Set
Online MCQ Assignment
Question Solution
Solved Question
Uncategorized

Linear Integrated MCQ Set 1

1. How many sets of electrical specification are there for 741C op-amp?
a) Five
b) Eight
c) Two
d) Ten

Answer: c [Reason:] There are two sets of electrical specifications. One set of specification applies at room temperature of 25oC , whereas the other set applies to the commercial temperature range from 0 to 70oC.

2. Calculate the input offset current from the circuit shown below:

a) +1.55mA
b) ±1.55mA
c) -1.55mA
d) None of the mentioned

Answer: a [Reason:] The algebraic difference between the current into the inverting and non-inverting terminal is referred to as input offset current. Iio = |IB1-IB2| =|3.2 – 4.75|=|-1.55| => Iio = +1.55mA.

3. What is the value of current drawn from power supply 741c op-amp?
a) Is = 1.5mA
b) Is = 3.2mA
c) Is = 4.0mA
d) Is = 2.8mA

Answer: a [Reason:] Supply current is the current drawn by the op-amp from the power supply; this parameter is not given on most op-amp data. For 741c op-amp, the supply current Is=2.8mA response.

4. Which is a time varying response?
b) Transient response
c) Both Steady state and transient response
d) None of the mentioned

Answer: b [Reason:] The rise time and percentage of overshoot are the characteristic of transient response.

5. How is the equivalent input noise voltage and current?
a) Noise voltage = V2/Hz and Noise current = A2/Hz
b) Noise voltage = V3/Hz and Noise current = A2/Hz
c) Noise voltage = V2/Hz and Noise current = A3/Hz
d) Noise voltage = V3/Hz and Noise current = A3/Hz

Answer: a [Reason:] As the electrical noise is random in nature, it is expressed as a root mean square value. The equivalent input noise voltage is expressed as square voltage (V2/Hz) and the equivalent input noise current as square noise current (A2/Hz).

6. The physical closeness of dual and quad package op-amp results in
a) Op-amp coupling
b) Amplifier to amplifier coupling
c) Channel coupling
d) Signal to noise coupling

Answer: b [Reason:] When a signal is applied to the input of only one op-amp, some signal will appear at the output of other op-amp due to the physical closeness of op-amps in dual and quad packages, this causes amplifier to amplifier coupling.

7. A parameter which is applicable only to dual and quad op-amp is
a) Channel separation
b) Gain bandwidth product
c) Long term input offset voltage stability
d) Equivalent input noise voltage and current

Answer: a [Reason:] Channel separation is specified in the dual and quad op-amp data sheet because it is a measure of the amount of electrical coupling between op-amps that are integrated on the same chip.

8. Select the parameter that is not included in the evaluation for ac applications.
a) Gain-bandwidth product
b) Channel separation
c) Slew rate
d) All of the mentioned

Answer: c [Reason:] Slew rate is one of the important factors in selecting the op-amp for ac application particularly at relatively high frequencies application in oscillator, comparators and filters.

9. Which information is not included in a typical data sheet of op-amp?
a) Brief description of the basic type of the device
b) Pin configuration, package type & order information
c) Internal schematic diagram
d) Characteristics analysis of amplifier applications

Answer: d [Reason:] The characteristics of amplifier is included in the datasheet whereas, the application’s characteristics will not be included. Rather a collection of amplifier applications with circuit diagram can be found in the datasheets.

Linear Integrated MCQ Set 2

1. 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 [Reason:] 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.

2. Find the circuit that is used to compress the dynamic range of a signal?

Answer: a [Reason:] Log amps are used to compress the dynamic range of a signal. The fundamental log amp circuit consists of a grounded base transistor in the feedback path.

3. Find the output voltage of the log-amplifier
a) VO = -(kT)×ln(Vi/Vref)
b) VO = -(kT/q)×ln(Vi/Vref)
c) VO = -(kT/q)×ln(Vref/Vi)
d) VO = (kT/q)×ln(Vi/Vref)

Answer: b [Reason:] the output voltage is proportional to the logarithm of input voltage. VO =-(kT/q)×ln(Vi / Vref).

4. How to provide saturation current and temperature compensation in log-amp?
a) Applying reference voltage alone to two different log-amps
b) Applying input and reference voltage to same log-amps
c) Applying input and reference voltage to separate log-amps
d) None of the mentioned

Answer: c [Reason:] The emitter saturation current varies from transistor to transistor with temperature. Therefore, the input and reference voltage are applied to separate log-amps and two transistors are integrated close together in the same silicon wafer. This provides a close match of the saturation currents and ensures good thermal tracking.

5. The input voltage, 6v and reference voltage, 4 v are applied to a log-amp with saturation current and temperature compensation. Find the output voltage of the log-amp?
a) 6.314(kT/q)v
b) 0.597(kT/q)v
c) 0.405(kT/q)v
d) 1.214(kT/q)v

Answer: c [Reason:] The output voltage of saturation current and temperature compensation log-amp, VO = (kT/q)×ln(Vi / Vref) =(kT/q)×ln(6v/4v) =(kT/q)×ln(1.5) VO = 0.405(kT/q)v.

6. Find the circuit used for compensating dependency of temperature in the output voltage?

Answer: c [Reason:] The temperature dependence on the output voltage is compensated by connecting an op-amp which provide a non-inverting gain of [1+ (R2/ RTC)] at the output of the log-amp with saturation current compensation. Now the output voltage becomes,VO = [1+ (R2/ RTC)]×[(kT/q)×ln(Vi / Vref)] Where, RTC –> temperature sensitive resistance with a positive co-efficient of temperature.

7. Determine the output voltage for the given circuit

a) VO = Vref/(10-k’vi)
b) VO = Vref+(10-k’vi)
c) VO = Vref×(10-k’vi)
d) VO = Vref-(10-k’vi)

Answer: c [Reason:] The output voltage of an antilog amp is given as, VO = Vref (10-k’vi) Where k’ = 0.4343 (q/kt)×[(RTC/ (R2 +RTC)].

8. Calculate the base voltage of Q2 transistor in the log-amp using two op-amps?

a) 8.7v
b) 5.3v
c) 3.3v
d) 6.2v

Answer: c [Reason:] The base voltage of Q2 transistor, VB = [RTC / (R2 +RTC)]×(Vi) = [10kΩ/(5kΩ+10kΩ)]×5v =3.33v.

Linear Integrated MCQ Set 3

1. What is the conversion ratio of the phase detector in 565 PLL?
a) 0.14
b) 0.35
c) 0.4458
d) 0.7

Answer: c [Reason:] The conversion ratio of the phase detector of 565 PLL (Monolithic PLL) Kφ = 1.4/π = 0.4458.

2. Given fo = 1.2kHz and V = 13v, find the lock-in range of monolithic Phase-Locked Loop.
a) ±575Hz
b) ±720Hz
c) ±150Hz
d) ±1kHz

Answer: b [Reason:] The lock-in range of monolithic PLL, △fL = ±(7.8×fo)/V = ±(7.8×1.2kHz)/13 = ±720Hz.

3. Find out the incorrect statement.
Monolithic phase detector is preferred for critical applications as it is:
1. Independent of variation in amplitude
2. Independent of variation in duty cycle of the input waveform
3. Independent of variation in response time
a) 1 & 2
b) 1 & 3
c) 2 & 3
d) 1, 2 & 3

Answer: a [Reason:] Monolithic phase detectors are not sensitive to harmonics of the input signal and change in duty cycle of input and output frequency.

4. Determine the capture range of IC PLL 565 for a lock-in range of ± 1kHz.

a) △fc = ±31.453Hz
b) △fc = ±66.505Hz
c) △fc = ±87.653Hz
d) None of the mentioned

Answer: b [Reason:] The capture range is △fc = ±[△fL/ (2π×3.6×103×C]0.5 = ±[1kHz/(2π×3.6×kΩ×10µF)]0.5 = ±[1kHz/226.08×-6]0.5 = [4423]0.5 = ±66.505Hz.

5. Find the lock-in range of monolithic Phase-Locked Loop from the given diagram.

a) -fo-△fL to fo-△fL
b) -fo-△fL to -fo-△fC
c) fo-△fL to fo-△fC
d) -fo-△fC to fo-△fC

Answer: a [Reason:] Lock-in range of monolithic PLL is from -fo-△fL to fo-△fL.

6. At what range the PLL can maintain the lock in the circuit?
a) Lock in range
b) Input range
c) Feedback loop range
d) None of the mentioned

Answer: a [Reason:] The change in frequency of the incoming signal can be tracked when the PLL is locked. So, the range of frequencies over which PLL maintains the lock with the incoming signal is called as the lock in range.

7. At which state the phase-locked loop tracks any change in input frequency?
a) Free running state
b) Capture state
c) Phase locked state
d) All of the mentioned

Answer: c [Reason:] In the phase-locked, the output frequency is exactly same as the input signal frequency. So the circuit tracks any change in the input frequency through its repetitive action.

Linear Integrated MCQ Set 4

1. Determine output voltage of analog multiplier provided with two input signal Vx and Vy.
a) Vo = (Vx ×Vx) / Vy
b) Vo = (Vx ×Vy / Vref
c) Vo = (Vy ×Vy) / Vx
d) Vo = (Vx ×Vy) / Vref2

Answer: b [Reason:] The output is the product of two inputs divided by a reference voltage in analog multiplier. Thus, the output voltage is a scaled version of x and y inputs. => Vo =Vx ×Vy / Vref.

2. Match the list-I with list-II

 List-I List-II 1. One quadrant multiplier i. Input 1- Positive, Input 2- Either positive or negative 2. Two quadrant multiplier ii. Input 1- Positive, Input 2 – Positive 3. Four quadrant multiplier iii. Input 1- Either positive or negative, Input 2- Either positive or negative

a) 1-ii, 2-i, 3-iii
b) 1-ii, 2-ii, 3-ii
c) 1-iii, 2-I, 3-ii
d) 1-I, 2-iii, 3-i

Answer: a [Reason:] If both inputs are positive, the IC is said to be a one quadrant multiplier. A two quadrant multiplier function properly, if one input is held positive and the other is allowed to swing. Similarly, for a four quadrant multiplier both the inputs are allowed to swing.

3. What is the disadvantage of log-antilog multiplier?
a) Provides four quadrant multiplication only
b) Provides one quadrant multiplication only
c) Provides two and four quadrant multiplication only
d) Provides one, two and four quadrant multiplication only

Answer: b [Reason:] Log amplifier requires the input and reference voltage to be of the same polarity. This restricts log-antilog multiplier to one quadrant operation.

4. An input of Vsinωt is applied to an ideal frequency doubler. Compute its output voltage?
a) Vo = [(Vx×Vy) /Vref2] × [1-cos2ωt/2].
b) Vo = [(Vx2×Vy2) /Vref] × [1-cos2ωt/2].
c) Vo = [(Vx×Vy)2 /Vref] × [1-cos2ωt/2].
d) Vo = [(Vx×Vy) /( Vref] × [1-cos2ωt/2].

Answer: d [Reason:] In an ideal frequency doubler, same frequency is applied to both inputs. ∴ Vx = Vxsinωt and Vy = Vysinωt => Vo = (Vx×Vy × sin2ωt) / Vref = [(Vx×Vy) / Vref] × [1-cos2ωt/2].

5. Find the output voltage for the squarer circuit given below, choose input frequency as 10kHz and Vref =10v

a) Vo = 5.0-(5.0×cos4π×104t)
b) Vo = 2.75-(2.75×cos4π×104t)
c) Vo = 1.25-(1.25×cos4π×104t)
d) None of the mentioned

Answer: c [Reason:] Output voltage of frequency Vo =Vi2 / Vref => Vi = 5sinωt = 5sin2π×104t Vo = [5×(sin2π×104t)2 ]/10 = 2.5×[1/2-(1/2cos2π ×2×104t)] = 1.25-(1.25×cos4π×104t).

6. Calculate the phase difference between two input signals applied to a multiplier, if the input signals are Vx= 2sinωt and Vy= 4sin(ωt+θ). (Take Vref= 12v).
a) θ = 1.019
b) θ = 30.626
c) θ = 13.87
d) θ = 45.667

Answer: a [Reason:] Vo= [Vmx×Vmy /(2×Vref)] ×cosθ => (Vo×2×Vref)/ (Vmx × Vmy) = cosθ => cosθ = (10×2×12)/(2×4) = 30. => θ = cos-130 =1.019.

7. Express the output voltage equation of divider circuit
a) Vo= -(Vref/2)×(Vz/Vx)
b) Vo= -(2×Vref)×(Vz/Vx)
c) Vo= -(Vref)×(Vz/Vx)
d) Vo= -Vref2×(Vz/Vx)

Answer: c [Reason:] The output voltage of the divider, Vo= -Vref×(Vz/Vx). Where Vz –> dividend and Vx –> divisor.

8. Find the divider circuit configuration given below

Answer: a [Reason:] Division is the complement of multiplication. So, the divider can be accomplished by placing the multiplier circuit element in the op-amp feedback loop.

9. Find the input current for the circuit given below.

a) IZ = 0.5372mA
b) IZ = 1.581mA
c) IZ = 2.436mA
d) IZ =9.347mA

Answer: b [Reason:] Input current, IZ = -(Vx×Vo)/(Vref×R) = -(4.79v×16.5v)/(10×5kΩ) = 1.581mA.

10. Find the condition at which the output will not saturate?

a) Vx > 10v ; Vy > 10v
b) Vx < 10v ; Vy > 10v
c) Vx < 10v ; Vy < 10v
d) Vx > 10v ; Vy < 10v

Answer: c [Reason:] In an analog multiplier, Vref is internally set to 10v. As long as Vx < Vref and Vy < Vref, the output of multiplier will not saturate.

Linear Integrated MCQ Set 5

1. The major source of interference with the desired signal in electronic system is called
a) Error signal
b) Interference signal
c) Noise signal
d) Faulty signal

Answer: c [Reason:] Any unwanted signal associated with the desired signal is noise.

2. Identify the external noise sources
a) Switching of rotating machinery
b) Control circuits
c) Ignition system
d) All of the mentioned

Answer: b [Reason:] In any electronics system, noise can come from many external sources as a result of circuitry itself.

3. Which among the following is not a source of internal noise?
a) AC random voltages generated within conductors
b) Lightning
c) Switching of circuits
d) None of the mentioned

Answer: b [Reason:] Except lightning the remaining are self induced or internal noises. Lightning is a natural phenomenon of external noise source.

4. The type of noise phenomena that increases with increase in temperature is
s) None of the mentioned
b) 1/f noise
c) Schottky noise
d) Thermal noise

Answer: d [Reason:] The thermal noise increases with an increase in temperature.

5. The factors that determine the amount of noise induced is
a) Type of coupling between two circuits
b) Rate of change of current per unit time
c) Speed of operation of circuit
d) All of the mentioned

Answer: a [Reason:] All the factors determine the amount of noise induced in a given circuit.

6. Which type of noise is produced when the op-amp has wider bandwidth?
a) 1/f noise
b) Schottky noise
c) Thermal noise

Answer: b [Reason:] When the op-amp has wide bandwidth, schottky noise is produced. The amount of schottky noise is greater with wider bandwidth and large resistance.

7. Mention a scheme to reduce the effect of electrical noise on ICs?
a) Isolating ICs on different boards
b) Add a filter near ICs
c) Physical shielding of ICs
d) Use of compensating networks near ICs

Answer: c [Reason:] Physical shielding of the ICs and associated wiring helps to prevent external electromagnetic radiation from inducing noise into the internal circuitry.

8. Internal noise generation can be reduced by keeping
a) None of the mentioned
c) Both input and output lead length short

Answer: c [Reason:] The common way of reducing the internal noise generation is by keeping input and output lead length as short as practically possible.

9. How to choose an IC for a high electrical noise environment?
a) Low degree of noise immunity
b) High degree of noise immunity
c) Low degree of noise reduction
d) High degree of noise reduction

Answer: b [Reason:] In a high electrical noise environment an IC with a high degree of noise immunity will minimize the amount of special care needed for proper circuit operation.

10. Which among the following has best immunity to induced noise?
a) Non-inverting amplifier
b) Inverting amplifier
c) Differential amplifier
d) Voltage follower

Answer: c [Reason:] The differential amplifier offers best immunity to induced noise because, the ratio of the output noise voltage to the input noise voltage in practice will be much smaller that unity.

11. Find the output voltage for differential amplifier, Where Vin->Induced input noise.

a) VO= -[(RF/R1)×Vd]+Vno
b) VO= -(RF/R1)×(Vd/Vno)
c) VO= -[(RF/R1)×Vd]-Vno
d) VO= -[(RF/R1)×Vd]×Vno