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Linear Integrated MCQ Set 1

1. What is Barkhausen criterion for oscillation?
a) Aß > 1
b) Aß < 1
c) Aß = 1
d) Aß ≠ 1

Answer: c [Reason:] The Barkhausen criterion for oscillation is Aß = 1. Where, A-> gain of amplifier and ß-> transfer ratio.

2. At what condition the output signal can be continuously obtained from input signal?
a) When the product of input voltage and feedback voltage is equal to 1
b) When the product of amplifier gain and transfer ratio is equal to 1
c) When the product of feedback voltage and transfer ratio is equal to 1
d) When the product of amplifier gain and input voltage is equal to 1

Answer: b [Reason:] When Aß=1, the feedback signal will be equal to the input signal. At this condition, the circuit will continue to provide output, even if the external signal is disconnected. This is because the amplifier cannot distinguish between external signal and signal from the feedback circuit. Thus, output signal is continuously obtained.

3. An oscillator is a type of
a) Feedforward amplifier
b) Feedback amplifier
c) Waveform amplifier
d) RC amplifier

Answer: b [Reason:] An oscillation is a type of feedback amplifier in which a part of output is fed back to the input via a feedback circuit.

4. Find the basic structure of feedback oscillator.
a)
b)
c)
d) None of the mentioned

Answer: c [Reason:] The above mentioned diagram is the basic structure of feedback oscillator. It consists of an amplifier, to the external input (vi) is applied and it have a feedback network from which the feedback signal (vf) is obtained.

5. What is the condition to achieve oscillations?
a) |Aß|=1
b) ∠Aß=0o
c) ∠Aß=multiples of 2π
d) All the mentioned

Answer: d [Reason:] All the conditions should be simultaneously satisfied to achieve oscillations.

6. What happens if |Aß|<1
a) Oscillation will die down
b) Oscillation will keep on increasing
c) Oscillation remains constant
d) Oscillation fluctuates

Answer: a [Reason:] If |Aß| becomes less than unity, the feedback signal goes on reducing in each feedback cycle and oscillation will die down eventually.

7. How sustained oscillation can be achieved?
a) Maintaining |Aß| slightly greater than unity
b) Maintaining |Aß| equal to unity
c) Due to non-linearity of transistor
d) Due to use of feedback network

Answer: c [Reason:] When |Aß| is kept slightly greater than unity the signal, however, cannot go on increasing and get limited due to non-linearity of the device (that is transistor enters into saturation). Thus, it is the non-linearity of the transistor because of which the sustained oscillation can be achieved.

8. Why it is difficult to maintain Barkhausen condition for oscillation?
a) Due to variation in temperature
b) Due to variation in supply voltage
c) Due to variation in components life time
d) All of the mentioned

Answer: d [Reason:] The Barkhausen condition |Aß|=1 is usually difficult to maintain in the circuit as the value of A and ß vary due to temperature variations, aging of components, change of supply voltage etc.

9. Name the type of noise signal present in the oscillation?
a) Schmitt noise
b) Schottky noise
c) Saturation noise
d) None of the mentioned

Answer: b [Reason:] Schottky noise is the noise signal always present at the input of the transistor due to variation in the carrier concentration.

10. A basic feedback oscillator is satisfying the Barkhausen criterion. If the ß value is given as 0.7072, find the gain of basic amplifier?
a) 2.1216
b) 0.7072
c) 1
d) 1.414

Answer: d [Reason:] Barkhausen criterion for oscillation is given as Aß=1 => A=1/ ß = 1/0.7072 = 1.414.

11. The feedback signal of basic sine wave oscillator is given as
a) Vf = Aß ×Vo
b) Vf = Aß ×Vi
c) Vf = Aß × (Vo/ Vi)
d) Vf = Aß × (Vi/ Vo)

Answer: b [Reason:] The feedback signal of an oscillator is given as the product of external applied signal & the loop gain of the system. => Vf= Aß ×Vi.

12. Express the requirement for oscillation in polar form
a) Aß =1∠360o
b) Aß =1∠90o
c) Aß =1∠πo
d) Aß =1∠270o

Answer: a [Reason:] There are two requirements for oscillation 1. The magnitude of Aß=1 2. The total phase shift of Aß=0o or 360o.

Linear Integrated MCQ Set 2

1. What will be the phase shift of feedback circuit in RC phase shift oscillator?
a) 360o phase shift
b) 180o phase shift
c) 90o phase shift
d) 60o phase shift

Answer: b [Reason:] The RC feedback network provide 180o phase shift and amplifier used in RC phase shift oscillator provide 180o phase shift (op-amp is used in the inverting mode) to obtain a total phase shift of 360o.

2. How many RC stages are used in the RC phase shift oscillator?
a) Six
b) Two
c) Four
d) Three

Answer: d [Reason:] The RC stage forms the feedback network of oscillator. It consist of three identical RC stages, each of 60o phase shift so as to provide a total phase shift of 180o.

3. Calculate the frequency of oscillation for RC phase shift oscillator having the value of R and C as 35Ω and 3.7µF respectively.
a) 1230 Hz
b) 204 Hz
c) 502Hz
d) 673 Hz

Answer: c [Reason:] The frequency of oscillation of RC phase shift oscillator is, fo=1/(2πRC√6) = 1/(2×3.14×√6×3.7µF×35Ω) => fo= 1/ 1.9921×10-3 = 502Hz.

4. What must be done to ensure that oscillation will not die out in RC phase shift oscillator?
a) Gain of amplifier is kept greater than 29
b) Gain of amplifier is kept greater than 1
c) Gain of amplifier is kept less than 29
d) Gain of amplifier is kept less than 1

Answer: a [Reason:] For a sustained oscillation in RC phase shift oscillator the gain of the inverting op-amp should be at least 29. Therefore, gain is kept greater than 29 to ensure the variation in circuit parameter will not make |Aß|<1, otherwise oscillation will die out.

5. Calculate the feedback voltage from phase shit network.

a) Vf = ( VoR3S3C3) / (1+ 6SRC+5S2C2R2+S3R3C3)
b) Vf = ( VoR3S3C3) / (1+ 5SRC+6S2C2R2+S3R3C3)
c) Vf = ( VoR3S3C3) / (1+ 5SRC+5S2C2R2+S3R3C3)
d) Vf = ( VoR2S3C3) / (1+ 6SRC+5S2C2R2+S3R3C3)

Answer: b [Reason:] Applying KVL equation to the circuit, we get => I1(R+(1/SC))-I2=Vo -> Equ1 => -I1R+ I2(2R+(1/SC))- I3R=0 -> Equ2 => 0- I2R+ I3(2R+(1/SC)=6 -> Equ3 WKT, Vf = I3× 2R, Solving Equ 1, 2, and 3 for I3 => I3= ( VoR2S3C3)/ (1+ 5SRC+6S2C2R2+S3R3C3) => Vf = I3× 2R =( VoR3S3C3) / (1+ 5SRC+6S2C2R2+S3R3C3).

6. Which type of op-amp is avoided for high frequencies?
a) LM318
b) Op-amp 741
c) LF 351
d) None of the mentioned

Answer: b [Reason:] Op-amp741 is generally used for low frequencies < 1 kHz.

7. Find out the constant values of α and ß in phase shift oscillator.
a) α = √6, ß = -1/29
b) α = 6, ß = -1/29
c) α = √6, ß = 1/29
d) α = 6, ß = -1/29

Answer: a [Reason:] From phase shift network, we obtain ß= 1/(1-5 α2)+j α(6- α2) -> Equ1 For Aß=1, ß should be real and the imaginary terms must be zero α(6- α2) =0 => α=√6 Now substituting α2=6 in Equ 1, we get ß=-1/29 (Negative sign indicates that the feedback network produces a phase shift of 180o).

8. A phase shift oscillator is designed to oscillate at 155Hz. Determine the value of Rf. (Take C=0.30µF)
a) 399Ω
b) 3.98MΩ
c) 13.9kΩ
d) 403kΩ

Answer: d [Reason:] R = 1/(2πC√6×fo) => R= 1/7.153×10-4= 1398=13.9kΩ.

9. The value of feedback resistor in phase shift oscillator is 180kΩ. Find its input resistance?
a) 52kΩ
b) 151kΩ
c) 209kΩ
d) 6.2kΩ

Answer: c [Reason:] To obtain sustained oscillation in phase shift oscillator. => |A|=29 or |Rf / R1|=29 => R1|= Rf / 29 = 180kΩ/29= 6.21kΩ,

10. Determine the frequency of oscillation (fo) in phase shift oscillator?
a) fo = √6/ωRC
b) fo = 0.56/ωRC
c) fo = 0.065/ωRC
d) fo = 6/ωRC

Answer: c [Reason:] The frequency of oscillation of phase shift oscillator is given as fo = 1/(2π×RC×√6) = 1/15.38×RC => fo = 0.065/RC.

11. The condition for zero phase shift in wein bridge oscillator is achieved by
a) Connecting feedback to non-inverting input terminal of op-amp
b) Balancing the bridge
c) Applying parallel combination of RC to the feedback network
d) All of the mentioned

Answer: b [Reason:] In wein bridge oscillator, the feedback signal in the circuit is connected to the non-inverting input of op-amp. So, feedback network does not provide any feedback and the condition of zero phase shift around the circuit is achieved by balancing the bridge.

Linear Integrated MCQ Set 3

1. A feedback amplifier is also called as
a) Open loop amplifier
b) Closed loop amplifier
c) Feedback network amplifier
d) Looped network amplifier

Answer: b [Reason:] A feedback amplifier is sometimes referred as a closed loop amplifier because the feedback forms a closed loop between input and the output.

2. How many types of configuration are available for feedback amplifier?
a) Six
b) Four
c) Two
d) Eight

Answer: b [Reason:] There are four type of configuration are available. They are voltage series feedback, voltage shunt feedback, Current series feedback and Current shunt feedback.

3. Which of the following is not a feedback configuration?
a) Current-series feedback
b) Voltage-shunt feedback
c) Current-Voltage feedback
c) Current-Shunt feedback

Answer: c [Reason:] In a feedback amplifier, either current or voltage can be fed back to the input, but both current and voltage cannot be feedback simultaneously.

4. When load current flows into the feedback circuit, the configuration is said to be
a) Current-shunt feedback
b) Voltage-shunt feedback
c) Voltage-series feedback
d) All of the mentioned

Answer: b [Reason:] In current-series and current-shunt feedback circuit, the load current flows into the feedback circuit.

5. Find the voltage-series feedback amplifier from the given diagram?

Answer: a [Reason:] The mentioned diagram is the voltage-series feedback amplifier because the voltage across load resistor is the input voltage to the feedback circuit.

6. On what criteria does the feedback amplifier are classified?
a) Signal fed back to input
b) Signal applied to input
c) Signal fed back to output
d) None of the mentioned

Answer: d [Reason:] The feedback amplifiers are classified according to whether the voltage or current is fed back to the input in series or in parallel.

7. The closed loop voltage gain is reciprocal of
a) Voltage gain of op-amp
b) Gain of the feedback circuit
c) Open loop voltage gain
d) None of the mentioned

Answer: b [Reason:] Comparing the equation of closed loop voltage gain (AF) and the gain of the feedback circuit (B). AF is reciprocal of B => AF = 1+( RF/ R1) ; B= R1/( R1+ RF) => B = 1+( R1/ RF) Therefore, AF = 1/B.

8. Select the specifications that implies the inverting amplifier?
a) V1 = -3v, V2 = -4v
b) V1 = -2v, V2 = 3v
c) V1 = 5v, V 2 = 15v
d) V1 = 0v, V2 = 5v

Answer: d [Reason:] In inverting amplifier, the input is applied to the inverting terminal and the non-inverting terminal is grounded. So,the input applied to inverting amplifier can be V1 = 0v, V2 = 5v.

Linear Integrated MCQ Set 4

1. Which circuit is used for obtaining desired output waveform in operational amplifier?
a) Clipper
b) Clamper
c) Peak amplifier
d) Sample and hold

Answer: a [Reason:] In an op-amp clipper circuits a rectifier diode is used to clip off certain portions of the input signal to obtain a designed output waveform.

2. The clipping level in op-amp is determined by
a) AC supply voltage
b) Control voltage
c) Reference voltage
d) Input voltage

Answer: c [Reason:] The clipping level is determined by the reference voltage which should be less than the input voltage range of an op-amp.

3. In a positive clipper, the diode conducts when
a) Vin < Vref
b) Vin = Vref
c) Vin > Vref
d) None of the mentioned

Answer: b [Reason:]In a positive clipper, the diode conducts until Vin = Vref (during the positive half cycle of the input), because when Vin < Vref, the voltage (Vref) at the negative input is higher than that at the positive input.

4. What happens if the potentiometer Rp is connected to negative supply?

a) Output waveform below -Vref will be clipped off
b) Output waveform above +Vref will be clipped off
c) Output waveform above -Vref will be clipped off
d) Output waveform below +Vref will be clipped off

Answer: c [Reason:] If the potentiometer Rp is connected to negative supply -VEE instead of +VCC, the reference voltage Vref will be negative. As a result the entire output waveform above -Vref to be clipped off.

5. Find the output waveform for when Vin < Vref

Answer: c [Reason:] The negative portion of the output voltage below -Vref is clipped off because, diode will be in off condition when Vin < Vref.

6. What happens if the input voltage is higher than reference voltage in a positive clipper?
a) Output voltage = Reference voltage
b) Output voltage = DC Positive voltage
c) Output voltage = Input voltage
d) All of the mentioned

Answer: a [Reason:] When input voltage is higher than reference voltage, the op-amp operates in open loop and diode become reverse biased. Thus, the output voltage will be equal to reference voltage.

7. A positive small signal halfwave rectifier can
a) Rectify signals with peak value only
b) Rectify signals with value of few millivolts only
c) Rectify signals with both peak value and down to few millivolts
d) None of the mentioned

Answer: c [Reason:] A positive small signal halfwave rectifier can rectify signals with peak values down to few millivolts, because the high open loop gain of the op-amp automatically adjusts the voltage drive to the diode, so that the rectified output peak is the same as the input.

8. Determine the output waveform of negative small signal half wave rectifier.

Answer: d [Reason:] During the positive alteration of Vin, D1 is reverse biased. Therefore, Vo =0v. On the other hand, during the negative alteration, D1 is forward biased and hence Vo follows Vin.

9. Diode in small signal positive halfwave rectifier circuit acts as
a) Ideal diode
b) Clipper diode
c) Clamper diode
d) Rectifier diode

Answer: a [Reason:] The diode acts as an ideal diode, since the voltage across the ON diode is divided by the open loop gain of the op-amp. As the input voltage starts increasing in the positive direction, the output of the op-amp also increases positively till the diode become forward biased.

10. How to minimize the response time and increase the operating frequency range of the op-amp?
a) Positive halfwave rectifier with two diodes
b) Positive halfwave rectifier with one diode
c) Negative halfwave rectifier with two diodes
d) Negative halfwave rectifier with one diode

Answer: c [Reason:] Negative halfwave rectifier circuit with two diodes are used so that the output of the op-amp does not saturate. Thus, minimizes the response time and increases the operating frequency range.

11. Why a voltage follower stage is connected at the output of the negative small signal half wave rectifier?
a) Due to Non-uniform input resistance
b) Due to Non-uniform output resistance
c) Due to Uniform output voltage
d) None of the mentioned

Answer: b [Reason:] The output resistance of the circuit is non-uniform as it depends on the state of diode. That is, the output impedance is low when diode is on and high when diode is off.

12. A circuit with a predetermined dc level is added to the output voltage of the op-amp is called
a) Clamper
b) Positive clipper
c) Halfwave rectifier
d) None of the mentioned

Answer: a [Reason:] A clamper clamps the output to a desired dc level.

13. Determine the output waveform for a peak amplifier with input =4Vpsinewave and Vref=1V.

Answer: a [Reason:] In a peak amplifier the input waveform peak is clamped at Vref. The output voltage Vo=2Vp+Vref=(2×4v)+1v = 9v.

14. An op-amp clamper circuit is also referred as
a) DC cutter
b) DC inserter
c) DC lifter
d) DC leveller

Answer: b [Reason:] In an op-amp clamper circuit, a pre-determined dc level is deliberately inserted at the output voltage. For this reason, the clamper is sometimes called as DC inserter.

15. At what values of Ci and Rd a precision clamping can obtained in peak clamper when the time period of the input waveform is 0.4s?
a) Ci=0.1µF and Rd=10kΩ
b) Ci=0.47µF and Rd=10kΩ
c) Ci=33µF and Rd=10kΩ
d) Ci=2.5µF and Rd=10kΩ

Answer: a [Reason:] For precision clamping, Ci and Rd << T/2. So, (0.1µF×10 kΩ) << (0.4/2) = 1×10-3 << 0.2. Therefore, Ci=0.1µF and Rd=10kΩ.

Linear Integrated MCQ Set 5

1. Open loop configuration is not preferred in op-amps because
a) First break frequency is too large
b) First break frequency is very small
c) Second break frequency is too large
d) All of the mentioned

Answer: b [Reason:] The bandwidth of the op-amp is simply the first break frequency. As the value of bandwidth is very small, the open loop configuration is of little use or not preferred in op-amp.

2. Calculate the value of open loop frequency response curve at any point beyond break frequency in 741C op-amp?
a) 1000000Hz
b) 1000Hz
d) 10000000Hz
d) 100000Hz

Answer: a [Reason:] For a 741C op-amp the gain=200000 and the break frequency = 5Hz. Therefore, the unity gain bandwidth or the product of the coordinates (gain and frequency) of any point beyond the break frequency = 200000×5Hz =1MHz =1000000Hz.

3.

Frequency response of µA709C for various closed loop gain is shown. Choose the curve that has wider bandwidth and high gain?
a) Curve 4
b) Curve 3
c) Curve 2
d) Curve 1

Answer: d [Reason:] To get a high gain and relatively wider bandwidth, the compensating components for curve 1 should be used. Curve 1 has the gain equal to 60dB and bandwidth nearly equivalent to 10MHz.

4. When does a system said to be stable?
a) Output reaches a minimum value at finite time
b) Output reaches a maximum value at any time
c) Output reaches a fixed value at finite time
d) Output reaches a fixed value at any time

Answer: c [Reason:] A circuit or a group of circuits connected together as a system is said to be stable, if its output reaches a fixed value in a finite time.

5. Why unstable systems are considered to be impractical?
a) None of the mentioned
b) Output decreases with time
c) Output reaches fixed value
d) Output increases with time

Answer: d [Reason:] A circuit is said to be unstable if its output increases with time instead of achieving fixed value. In fact the output keeps on increasing until the system breaks down. Therefore, unstable systems are impractical and need to be made stable.

6. How is the criterion for the system determined?
a) Graphical method
b) Theoretical method
c) Analytical method
d) All of the mentioned

Answer: d [Reason:] The criterion for stability of system is tested practically using one of the above mentioned methods.

7. A standard block diagram of closed loop system composes of
a) Two blocks
b) Single block
c) Three blocks
d) None of the mentioned

Answer: a [Reason:] In closed loop system (control system) the standard form of representing a system is composed of two blocks. (i) Forward block and (ii) Feedback block.

8. “Transfer function” in control system refers to
a) Feedback block
b) Content of each block
c) Forward block
d) Input and output blocks

Answer: b [Reason:] The transfer function in control system refers the content of each block and it depends on the complexity of a system.

9. Which method is considered to be a graphical method in testing the system?
a) Bode plot
b) Routh-Hurwitz criteria
c) Circuit testing
d) None of the mentioned

Answer: a [Reason:] A graphical method used in determining the stability / testing the system is the bode plots. It is composed of magnitude versus frequency and phase angle versus frequency plots.

10. Measure taken to increase the bandwidth of an op-amp.
a) Increase the frequency for the configuration
b) Reduce the gain of the configuration
c) Closed loop configuration is used
d) Open loop configuration is used

Answer: c [Reason:] Closed loop configuration is preferred to increase the bandwidth of an op-amp.

11. Name the block connected in the feedback path
a) Feedback block
b) Forward block
c) Output block
d) Summation of feedback, forward and output block

Answer: a [Reason:] The block in the feedback path is referred to as the feedback blocks, which connects the block between the output signal and the feedback signal.

12. When the magnitude of (AoL )×( ß ) = 0dB, state the condition at which system become stable?
a) Phase angle is > -180o
b) Phase angle is < -180o
c) Phase angle is > +180o
d) Phase angle is < +180o

Answer: a [Reason:] Closed loop voltage gain AF = AoL/ (1+AoL×ß) At lower frequency the contribution of AoL is zero. So, AoL×ß >0 and obviously AF < AoL and system is stable.

13. Mention the phase condition that leads to sustained oscillation
a) ∠-Aß =0
b) All of the mentioned
c) ∠-Aß = multiple of 2π
d) ∠Aß = odd multiple of π

Answer: b [Reason:] The circuit leads to oscillation if the characteristic equation (1+AoL×ß )=0 => Loop gain, AoL×ß =1 Since AoL*ß is a complex quantity , the magnitude is |AoL×ß |=1 and the phase condition are ∠-Aß=0 (or multiple of 2π) or ∠Aß= π (or odd multiple of π).

14. Determine the state of the system

a) Stable system
b) Unstable system
c) Casual system
d) Bounded system