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## Linear Integrated MCQ Set 1

1. To increase the input resistance, the differential amplifier replaces transistor by
a) Current mirror
b) Current repeater
c) Darlington pair
d) All of the mentioned

Answer: c [Reason:] Higher value of input resistance can be obtained by using Darlington pair in place of transistor.

2. In Darlington pair differential amplifier the current gain is given as 100. Where IB1=5µA and IC1=0.35mA. Determine IC2
a) 0.5mA
b) 1.5mA
c) 2mA
d) 0.15mA

Answer: d [Reason:] The current gain in Darlington pair differential amplifier is given as β=( IC1+IC2)/IB1. Substituting the values in the equation, we get IC2=(β×IB1)-IC1 =(100×5µA)-0.35mA =0.15mA.

3. In the circuit shown, find the overall current gain?

a) 456218
b) 444878
c) 444210
d) 455734

Answer: b [Reason:] From the circuit given, IB= IB1 = 5.6µA. IE1= IB1+ IC1 = 1.43mA + 5.6µA = 1.435mA. IE1= IB2 = 1.435mA. The individual current gain values, β1=IC1/ IB1 => β1 = 1.43mA/5.6µA= 255.36. Similarly,β2=IC2/ IB2 => β2 = 2.5A / 1.435mA =1742.16 Therefore, the overall current gain, β = β1 × β2 = 255.36 × 1742.16 = 444878.

4. Introducing FET differential amplifier pair at the input stage of differential amplifier produces
a) High output resistance
b) High input resistance
c) Low input impedance
d) All of the mentioned

Answer: b [Reason:] Input resistance of the order 1012 Ω is possible with JFET at the input stage of differential amplifier.

5. Why active load is used in amplifier to obtain large gain in intermediate stage of amplifier?
a) To obtain a very large voltage gain
b) To get High input resistance
c) To reduce the noises
d) To increase current gain

Answer: a [Reason:] To increase gain usually large collector resistance value as gain is proportional to load resistor. However, due to limitation of maximum value load resistor, active loads are used in amplifier to obtain large gain in intermediate stages of amplifier.

6. Which circuit is used as active load for an amplifier
a) Wildar Current source
b) Darlington pair
c) Current Mirror
d) All of the mentioned

Answer: c [Reason:] Current mirror has DC resistance (order of few kΩ), as quiescent voltage across it is a fraction of supply voltage and current in milliampere.

7. What is the equation of load current for a differential amplifier with an active load?
a) IL = gm×vd
b) IL = Iq /2
c) IL = β×Iq×( Vin1 – Vin2)
d) IL = 2×gm/( Vin1 – Vin2)

Answer: a [Reason:] The load current is given as product of difference between input & output voltage and transconductance. Therefore, the equation of load current is , IL = gm×vd.

8. The input voltage of a difference amplifier are 2.5v and 4.9v. If the transconductance is 0.065Ω-1, determine the load current entering the next stage
a) 0.156A
b) 1.56A
c) 0.156mA
d) 15.6µA

Answer: a [Reason:] Load current entering the next stages of amplifier is the sum of individual load current, which is given by IL = IL1+ IL2 (Since only two input voltages are given). IL = gm×Vin1 + gm×Vin2 = gm×( Vin1 – Vin2) = 0.065 Ω-1×(4.9v-2.5v) = 0.156A.

9. Calculate the VI – VO for the level shifter shown in the figure (Assume identical silicon transistor and very large value of β). Transistor QA and QB form current mirror.

a) 5.56V
b) 6.00v
c) 7.98v
d) 6.65v

Answer: d [Reason:] Since the transistor QA and QB form current mirror, ICA= ICB = I. => I = (VCC – VBE) / R0 = (15v-0.7)/12k Ω (for β>>1, output current =input current) => I= 1.19mA. The shift in level is given as VI – VO = VBE + I×R1 =0.07v+1.19mA×5kΩ =6.65v.

10. Load resistors (Re) is neglected for maximizing the voltage gain in amplifier because,
a) Requires large chip are
b) Requires large power supply
c) Quiescent drop across Re increases
d) All of the mentioned

Answer: d [Reason:] As gain is proportional to load resistor, large resistance value is required. Due to limitation mentioned, it is neglected.

11. What is the need for level shifter in operational amplifier?
a) Level the quiescent voltage
b) Remove distortion at output
c) Limits the output voltage
d) Increase the quiescent voltage

Answer: c [Reason:] Because of direct couple, Dc level rises stages to stage and tends to shift operating point. This limits output swing (Voltage).

12. Limitation of an output stage amplifier, if it emitter follower with complementary transistor
a) Cross-over distortion
b) Low impedance output
c) Shift in level

Answer: a [Reason:]The limitation in the amplifier is that , the output voltage remains zero until the input voltage exceeds cut in voltage VBE= 0.5v, which is known as cross-over distortion.

13. An output stage amplifier can produce output signal, when the input signal is
a) 0.48v
b) 0.9v
c) 1.2v
d) 0.5v

Answer: c [Reason:] In an Output stage amplifier, due to cross-over distortion output voltage produces input voltage is greater than two times of cut-in voltage which is equal to 1v. Since, VBE= 0.5v => 2×VBE= 1v.

14. Find the disadvantage in the following circuit diagram:

a) Voltage get attenuated by R1
b) Voltage get attenuated by R2
c) Voltage get attenuated by R1 and R2
d) Voltage shift get increased by the drop across R1 and R2

Answer: b [Reason:] The output taken at the junction of R1 and R2 increases the voltage shift. However, the disadvantage is that, the signal voltage gets attenuated by R2. => R2/(R1 + R2).

## Linear Integrated MCQ Set 2

1. Find out the op-amp which does not have same specifications and behaviour as that of N5741?
a) MC1741
b) CA3741
c) SN52741
d) None of the mentioned

Answer: d [Reason:] All these op-amp have the same they specifications and behave the same because the last three digits in each manufacturer designation is 741. For example Fairchild’s original µA741 is also manufactured by various other manufacturers under their own designation.

2. The op-amps 741C and 741E are identical to op-amps
a) 741s and 741A
b) 741 and 741A
c) 741A and 741Sc
d) 741 and 741S

Answer: b [Reason:] The 741C and 741E are identical to 741 and 741A except that the former have their performance guaranteed over a temperature range 0o to 70o or 75oc.

3. Which of the following is a military grade op-amp?
a) 741
b) 741C
c) 741S
d) 741SC

Answer: c [Reason:] The 741S is a military grade op-amp with a higher slew rate (rate of change of output voltage per unit of time) than the 741 op-amp.

4. Which of the following is in correct? “First generation op-amp ”
a) Consists of hundred transistors/chip
b) Requires an external frequency compensating network for stable operation
c) Have short circuit protection
d) Has latch-up problem

Answer: c [Reason:] First generation op-amp has no short circuit protection. The op-amp is susceptible to burnout if output accidentally shorted to ground.

5. What is the disadvantage of integrated circuit?
a) Parameter within the IC cannot be modified
b) Low power requirement
c) ICs are considered to use minimum number of external connections
d) None of the mentioned

Answer: a [Reason:] The disadvantage of IC is that, a lack of flexibility an IC. It is generally not possible to modify the parameters within which an integrated circuit will operate.

6. Whenever an IC is designed manufacturers use a
a) Specific code and manufacturer’s name
b) Specific code and specific type number
c) Specific code and specific value
d) Specific type number

Answer: b [Reason:] Each manufacturer uses a specific code and assign a specific type number to the IC’s produced. That is, each manufacturer uses their own identifying initial followed by IC type number.

7. General purpose op-amp cannot be used for the application
a) Integrator
b) Audio power amplifier
c) Differentiation
d) Summing amplifier

Answer: b [Reason:] Audio power amplifier is a special purpose op-amp and is used only for the specific application they are designed for.

8. Low volume production methods are best suited to hybrid IC technology because
a) It require stipulated temperature to fabricate a circuit
b) It require several steps to fabricate a circuit
c) It require large components to fabricate a circuit
d) It require various designers to fabricate a circuit

Answer: b [Reason:] In hybrid IC, first the individual components are made, wired or metallic interconnection is done and finally they are diffused to form a single circuit.

9. Which is the different version of IC 741C?
a) 741A
b) 741E
c) 741S
d) 741SC

Answer: d [Reason:] 741SC is the different version of 741C IC, which is a commercial grade op-amp with higher slew rate and operating in the same temperature range.

10. An example of second generation IC used in greatest percentage of application
a) µA748
b) MC1558
c) µA741
d) LM101

Answer: c [Reason:] All the IC belongs to second generation op-amp. In that general purpose op-amp’741’ is used widely in greatest percentage of application.

11. What is the advantage of Hybrid Integrated Circuit?
a) Miniaturized circuits are made of individual components
b) Insulate components by protection
c) Circuit designer can choose the component value
d) All of the mentioned

Answer: d [Reason:] In hybrid integrated circuit miniaturization can be achieved and allow circuit designer a complete freedom in choosing the resistor values, whereas monolithic IC cannot use some important components on construction.

12. Find the pin configuration of µA741operational amplifier?

Answer: a [Reason:] The metal can configuration of µA741 op-amp has eight pins with pin number 8 identified by a tab. The other pins are numbered counter clockwise from pin 8, beginning with pin1.

## Linear Integrated MCQ Set 3

1. Which of these statements is false?
a) The open loop gain AOL(f) dB is approximately constant from 0Hz to fo
b) When input signal frequency and f is equal to break frequency fo, the gain frequency is called -3dB frequency
c) The open loop gain AOL(f) dB is approximately constant upto break frequency fo, but there after it increase 20dB each time there is a tenfold increase in frequency.
d) At unity gain crossover frequency, the open loop gain AOL(f) dB is zero

Answer: c [Reason:] When AOL(f) is approximately constant up to break frequency, there will be 20dB decrease each time there is a tenfold increase in frequency. Therefore it may be considered that the gain roll of at the rate of 20dB/decade.

2. What is the maximum phase shift that can occur in an op-amp with single capacitor?
a) 180o
b) 60o
c) 270o
d) 90o

Answer: d [Reason:] At corner frequency, the phase angle is -45o (lagging) and at infinite frequency the phase angle is -90o. Therefore, a maximum of 90 o phase change can occur in an op-amp with a single capacitor.

3. How can the gain roll off represented in dB/octave?
a) 12 dB/octave
b) 6 dB/octave
c) 10 dB/octave
d) 8 dB/octave

Answer: b [Reason:] Octave represents a two fold increase in frequency. Therefore, 20 gain roll off at the rate 20 dB/decade is equivalent to 6 dB/octave.

4. Select the correct magnitude and phase for the frequency range.

 List-I List-II 1. f<1 i. Gain is 3dB down from the value of AOL in dB 2. f=f1 ii. Gain roll off at the rate of 20 dB/decade 3. f>>f1 iii. Magnitude of the gain is 20logxAOL in dB

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

Answer: a [Reason:] Properties of magnitude and phase angle characteristics equations.

5. The specific frequency at which AOL (dB) is called
a) Gain bandwidth product
b) Closed loop bandwidth
c) Small signal bandwidth
d) All of the mentioned

Answer: d [Reason:] AOL (dB) is zero at some specific value of input signal frequency called unity gain band width. The mentioned terms are the equivalent terms for unity gain band width.

6. Find out the expression for open loop gain magnitude with three break frequency?
a) A/[1+ j(f/fo1)]x[1+ j(f/fo2)]x[1+ j(f/fo3)].
b) A/[1+ j(f/fo)]3
c) A/A/[1+ j(f/fo1)]+ [1+ j(f/fo2)]+ [1+ j(f/fo3)].
d) All of the mentioned

Answer: c [Reason:] The gain equation for op-amp is AOL(f) = A/[1+ j(f/fo1)]x[1+ j(f/fo2)]x[1+ j(f/fo3)].

7. Find out the frequencies that are avoided in frequency response plot?
a) None of the mentioned
b) Single break frequency
c) Upper break frequency
d) Lower break frequency

Answer: c [Reason:] Op-amps have more than one break frequency because only a few capacitors are present. Often upper break frequencies are well above the unity gain bandwidth. So, they are avoided in frequency response plot.

8. In a phase response curve of MC1556, the phase shift is -162.5o about 4MHz. If the approximate value of first break frequency is 5.5Hz. Determine the approximate value of the second break frequency?
a) 945.89MHz
b) 945.89kHz
c) 945.89GHz
d) 945.89Hz

Answer: b [Reason:] The phase angle equation is φ(f) = – tan-1 (f/fo1)- tan-1(f/fo2) => -162.5o = – tan-1(3MHz/5.5) – tan-1(3MHz/fo2) => tan-1(3MHz/fo2) = +162.5o-89.99o => 3MHz/fo2= tan(72.50o) => fo2 = 3MHz/tan(72.50o)=945.89kHz.

9. Consider a practical op-amp having two break frequency due to a number of RC pole pairs. Determine the gain equation using the given specifications:
A ≅ 102.92dB; fo1 =6Hz ; fo2 =2.34MHz .
a) 140000/[1+ j(f/6)]x[1+ j(f/2.34)].
b) 140000/[1+ j(f/6)]x[1+ j(f/234)].
c) 140000/[1+ j(f/6)]x[1+ j(f/23400)].
d) 140000/[1+ j(f/6)]x[1+ j(f/2340000)].

Answer: d [Reason:] Gain of op-amp at 0Hz, A = 10102.92/20 => A = 105.146 =139958.73 ≅ 140000. The gain equation with two break frequency value is given as AOL(f) = A/[1+ j(f/fo1)]x[1+ j(f/fo2)] =140000/[1+ j(f/6)]x[1+ j(f/2.34)].

10. What is the voltage transfer function of op-amp in s-domain.
a) S×A/(s+ωo)
b) A/(s+ωo)
c) (A×ωo)/(s+ωo)
d) None of the mentioned

Answer: c [Reason:] For the voltage transfer function in s-domain is expressed as AOL(f) = A/[1+ j(f/fo)] =A/[1+j(ω/ωo)]= A×ωo/(jω/jωo) = A×ωo / (s+ωo).

11. Calculate the unity gain bandwidth of an op-amp. Given, A≅ 0.2×106 & fo =5Hz?
a) 106
b) 108
c) 10-5
d) 10-10

Answer: a [Reason:] The break frequency, fo = UGB/A => =fo×A = 0.2×106 × 5 = 106.

## Linear Integrated MCQ Set 4

1. What happens if R1>>RF in the circuit

a) Some amount of output offset voltage is present
b) Some amount of input offset voltage is present
c) Some amount of gain voltage is present
d) All of the mentioned

Answer: a [Reason:] If R1 >>RF, the gain Aoo≅1, which makes Voo ≅ Vio. Thus, all op-amp circuit has some output offset voltage.

2. Determine the voltage gain for the circuit.

a) 1.1
b) 1.6
c) 1.2
d) 2.2

Answer: d [Reason:] The voltage gain, AF={1+[RF/( R1+ Rc)]} = 1+[15kΩ/(2.5kΩ+10kΩ)] = 2.2.

3. Where does the compensating network connected in an inverting amplifier.
a) Non-inverting input terminal
b) Inverting input terminal
c) Between non-inverting and output terminal
d) Between inverting and output terminal

Answer: a [Reason:] The offset voltage compensating network is connected in the non-inverting terminal for the inverting amplifier and vice versa.

4. Why closed loop differential amplifiers are difficult to null?
a) Due to compensating network
b) Due to feedback loop
c) Due to input offset voltage
d) None of the mentioned

Answer: a [Reason:] The closed loop differential amplifiers are more difficult to null because the use of compensating network can change the common mode rejection mode.

5. How to achieve maximum CMRR in the given circuit?

a) R1 = RF
b) RF = R3|| RC+ RB and R1= R2
c) R1= R2 and RF= R3+ RC
d) None of the mentioned

Answer: c [Reason:] To achieve maximum CMRR in the circuit the value of R1= R2 and RF= R3+ RC.

6. What is the advantage of compensated differential amplifier?
a) All of the mentioned
b) Slightly complex circuit
c) Does not affect CMRR
d) Balanced op-amp

Answer: d [Reason:] Since the compensated differential amplifier uses the op-amp with offset voltage null pins. The offset null circuit does not affect the CMRR.

7. The offset voltage in the voltage follower is balanced using
a) Voltage drop across the load resistor
b) Voltage drop across feedback resistor
c) Compensating network connected to inverting input terminal
d) Compensating network connected to non- inverting input terminal

Answer: b [Reason:] Voltage drop across the feedback resistor connect to inverting input terminal is used to cancel the offset voltage in voltage follower.

8. Find the maximum possible output offset voltage, which is caused by the input offset voltage Vio=15mv?

a) 0.075v
b) 0.75v
c) 0.75v
d) 7.5v

Answer: a [Reason:] Aoo=[1+(RF/R1)] =1+(10kΩ/2.5kΩ) = 5. Voo=5*15mv = 75mv.

9. Compute the output voltage for voltage follower with offset voltage compensating network?

a) 3.6v
b) 10.8v
c) 26v
d) 33v

Answer: b [Reason:] The output voltage is given as Vo= {1+[ RC/( Rb+ (Rmax/4))]}*Vin. Rmax=Ra/4 = 20kΩ/4 = 5kΩ. Vo=[1+(39kΩ/(10kΩ+5kΩ))]*3v = 10.8v.

## Linear Integrated MCQ Set 5

1. A Biphasic oscillator produces output signal that have
a) Cosine wave only
b) Both sine and cosine wave
c) Two identical sine wave
d) Two identical cosine wave

Answer: c [Reason:] A biphasic oscillator is a sine wave oscillator. Therefore, the output signals will be two identical sine waves that are 180o out of phase with each other.

2. If the output of the sinewave oscillator is Vsinωot, then determine the output of the inverter in the Biphasic oscillator.
a) Vsin(ωot+(3π/2))
b) Vsin(ωot+2π)
c) Vsin(ωot+π)
d) Vsin(ωot+(π/2))

Answer: b [Reason:] Since the output of Biphasic oscillator are 180o out of phase with each other. The sine wave passing through the inverting amplifier (with gain of one) outputs the signal with 180o phase shift.

3. A function generator can produce
a) Many identical waves
b) Square and sine waves only
c) Different types of waves simultaneously
d) None of the mentioned

Answer: Oscillator circuits generate only one type of waveform like square, triangular or sine wave separately, whereas a function generator can produce all three types of waves simultaneously.

4. Which of the following have distorted sinewave?
a) Function generator
b) Biphasic oscillator
c) RC phase shit oscillator
d) Wein bridge oscillator

Answer: a [Reason:] The sine shaper in function generator produces a sine wave by rounding off the tips of the triangular wave. The distortion of the sine wave thus produced is ver high compared to the sine waves generated by other oscillator.

5. Usually circuit producing sine waves are called as
a) Oscillators
b) Generators
c) Multivibrators
d) All of the mentioned

Answer: a [Reason:] The generators and the oscillators are equivalent terms, they can be interchangeable. Therefore, the circuits producing sine waves are called oscillator, while those generating a square wave or triangular wave are generators.

6. An IC function generator can allow the signals for
a) Amplitude Modulation (AM)
b) Frequency Modulation (FM)
c) Frequency Shift Keying (FSK)
d) All of the mentioned

Answer: d [Reason:] IC function generator allows FM, AM and FSK of signals in addition to producing different waveform with variable duty cycle pulse.

7. The current in the XR-2206 Function generator is determined by
a) External resistors
b) External capacitors
c) External capacitors or external resistors
d) Both external capacitor and external resistor

Answer: c [Reason:] The constant current is determined by the external resistors connected to current switches or the external capacitor connected to current controlled oscillator.

8. What will be the output, if a 180Ω resistor which was connected between 13th and 14th pin terminal is removed.

a) A triangular wave
b) A sine wave
c) A square wave
d) A rectangular wave

Answer: a [Reason:] When pin 13 and 14 are open, the gain of the sine shaper become linear and the output from the sine shaper will be a triangular wave.

9. Find the difference of potential that can be applied to XR-2206 function generator.
a) 26v
b) 5v
c) 19v
d) 30v

Answer: c [Reason:] If the power is applied to both V+ and V inputs. Then, the difference of potential (i.e. V+ – V) should be at least +10v but less than 26v.

10. The value of current and frequency of the output waveform are 5A and 13.33kHz. Find the capacitance value in function generator?
a) 250µF
b) 120µF
c) 850µF
d) 370µF

Answer: b [Reason:] The frequency of the output waveform f=(0.32×I)/C => C=(0.32×I)/f = (0.32×5A)/13.33kHz = 120µF.

11. How many set of pins are required to control the frequency in XR-2206 function generator?
a) Two
b) Three
c) Four
d) None of the mentioned