Operational Amplifier Fundamentals Questions and Answers

Q: The output voltage (\(V_{o}\)) of the following Op-amp circuit is :In the circuit, \(V_{2}\) = 4 V is applied through \(R_{3}\) = 1 kΩ to the inverting (-) input, with a feedback resistor \(R_{4}\) = 1 kΩ from the output \(V_{o}\) to that input; \(V_{1}\) = 2 V is applied through \(R_{1}\) = 1 kΩ to the non-inverting (+) input, which is connected to ground through \(R_{2}\) = 1 kΩ.

The output voltage is - 2 V.This is a difference amplifier. The non-inverting input sees \(V_{1}\) through the divider \(R_{1}, R_{2}\), and the gain to the output is set by \(R_{3}\) and \(R_{4}\):\[ \begin{aligned} V_{+} &= V_{1}\frac{R_{2}}{R_{1} + R_{2}} = 1\ \text{V} \\ V_{o} &= \left(1 + \frac{R_{4}}{R_{3}}\right)V_{+} - \frac{R_{4}}{R_{3}}V_{2} \\ &= 2(1) - 4 = -2\ \text{V} \end{aligned} \]With all resistors equal, the circuit simply gives \(V_{1} - V_{2}\). The option +2 reverses the subtraction, and \(\pm 4\) uses only one input.

Q: Negative feedback for an operational amplifier leads to :

Negative feedback leads to an increase in the input impedance and the bandwidth.With series (voltage) negative feedback, as in the non-inverting amplifier, the input impedance is multiplied by the factor (1 + Aβ), the output impedance is divided by the same factor, and the bandwidth is extended by it while the gain falls by it, keeping the gain-bandwidth product constant.So the output impedance decreases rather than increases, which rules out option A; option C wrongly says the bandwidth decreases; and option D ignores the effect altogether.

Q: What is the purpose of inverting (-) and non-inverting (+) input terminals on an operational amplifier ?

The two inputs provide input signals for differential amplification.An op-amp is a differential amplifier: its output is \(A(V_{+} - V_{-})\), the open-loop gain times the difference between the voltages at the non-inverting (+) and inverting (-) inputs. A signal at the + input appears at the output in phase, a signal at the - input appears inverted, and a signal common to both is rejected.The power supply is connected to separate supply pins, not the inputs. The gain and the amount of feedback are set by the external components connected around the op-amp, typically from the output to the inverting input, not by the existence of the two terminals.

Direction for [ Question No: 1 To 1 ] :

Assertion - Reason type questions : The following items consist of two statements, one labelled the 'Assertion (A)' and the other labelled the 'Reason (R)'. You are to examine these two statements and decide if the Assertion (A) and the Reason (R) are individually true and if so, whether the Reason is a correct explanation of the Assertion. Select your answers to these items using the codes given below and mark your answer sheet accordingly.

Assertion (A) : In an Op-Amp circuit when one input terminal of the Op-Amp is grounded, the other terminal becomes a virtual ground.
Reason (R) : Input impedance of the Op-Amp is high.

A.Both (A) and (R) are true and (R) is the correct explanation of (A).
B.Both (A) and (R) are true, but (R) is not the correct explanation of (A).
C.(A) is true and (R) is false.
D.(A) is false and (R) is true.

Exam Year(s)

UGC NET - 2013,

Answer: B

When a negative feedback op-amp has one terminal grounded, the high open-loop gain makes the voltage difference between the terminals almost zero, so the other terminal stays near ground potential and is called a virtual ground, so (A) is true. The input impedance of an op-amp is high, so (R) is also true. However, the virtual ground exists because of the very high gain with feedback, not because of the input impedance, so (R) is not the right explanation of (A). The answer is both are true but (R) is not the correct explanation.

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Direction for [ Question No: 2 To 4 ] :

Read the following paragraph and answer the questions :

An operational amplifier is a direct coupled high gain amplifier with feedback. It is employed to perform a wide variety of functions such as summer, integrator, differentiator, logarithmic amplifier, oscillators, filters, comparators voltage to frequency converter, amplitude modulators, sample and hold circuit and so on. An ideal Op-Amp has infinite input resistance, zero output resistance, infinite voltage gain and bandwidth, zero offset and no drift. An ideal Op-Amp is a perfectly balanced circuit such that output voltage is zero for two equal input signals. However a real Op-Amp exhibits imbalance due to mismatch at the input differential stage. It gives rise to offsets. In practical applications offsets are removed using balancing networks. In general an Op-Amp consists of the following four stages :
(1) Differential amplifier first stage
(2) Differential amplifier second stage
(3) Emitter follower
(4) Level translator and output stage
It is most widely used analog device.

Which is a non-linear application ?

A.I-V converter
B.Sample-and-hold
C.Voltage follower
D.Active filter

Exam Year(s)

UGC NET - 2012,

Answer: B

The sample-and-hold circuit is the non-linear application, option (B). Its output follows the input only during the sampling interval and then holds a fixed value, so the transfer is not a straight line proportional to the input, and it involves a switch. The I-V converter, option (A), the voltage follower, option (C), and the active filter, option (D), are linear circuits, since they operate in the linear region of the amplifier, and their outputs vary linearly with the inputs. Logarithmic amplifiers, comparators and rectifiers are other non-linear applications named in the passage.

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Analog multiplication is carried out with

A.Scalar
B.Precision rectifier
C.Sample-and-hold circuit
D.Logarithmic amplifier

Exam Year(s)

UGC NET - 2012,

Answer: D

Analog multiplication is carried out with a logarithmic amplifier, option (D). The method takes the logarithms of the two inputs, adds them in a summing amplifier, and then takes the antilogarithm of the sum with an exponential amplifier, since \(\log(xy) = \log x + \log y\). The passage names the logarithmic amplifier among the functions of an op-amp. A precision rectifier, option (B), only rectifies small signals. A sample-and-hold circuit, option (C), stores a voltage. Option (A), scalar, is not a circuit used for multiplication of two signals.

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What will be the feedback voltage ?

An op-amp on plus and minus 12 V supplies. Its output Vo feeds a divider of 95 k and 5 k to ground; the divider midpoint returns to the non-inverting input; the inverting input is a free terminal.

A.2.4 V
B.1.2 V
C.0.6 V
D.0.3 V

Exam Year(s)

UGC NET - 2012,

Answer: C

The feedback voltage is 0.6 V, option (C). The output of the op-amp is fed through the divider of 95 k and 5 k to the non-inverting input, so the feedback is positive and the circuit works as a comparator with the output at one of its supply limits, +12 V or -12 V. The feedback fraction is

\[ \beta = \frac{5}{95+5} = 0.05 \]

\[ V_{f} = \beta\,V_{sat} = 0.05\times12 = 0.6\ \text{V} \]

The magnitude is 0.6 V, with the sign of the output. Option (A), 2.4 V, would need a fraction of 0.2. Option (B), 1.2 V, would need 0.1. Option (D), 0.3 V, would need 0.025.

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The output voltage (\(V_{o}\)) of the following Op-amp circuit is :

In the circuit, \(V_{2}\) = 4 V is applied through \(R_{3}\) = 1 kΩ to the inverting (-) input, with a feedback resistor \(R_{4}\) = 1 kΩ from the output \(V_{o}\) to that input; \(V_{1}\) = 2 V is applied through \(R_{1}\) = 1 kΩ to the non-inverting (+) input, which is connected to ground through \(R_{2}\) = 1 kΩ.

A.+ 2
B.- 2
C.+ 4
D.- 4

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: B

The output voltage is - 2 V.

This is a difference amplifier. The non-inverting input sees \(V_{1}\) through the divider \(R_{1}, R_{2}\), and the gain to the output is set by \(R_{3}\) and \(R_{4}\):

\[ \begin{aligned} V_{+} &= V_{1}\frac{R_{2}}{R_{1} + R_{2}} = 1\ \text{V} \\ V_{o} &= \left(1 + \frac{R_{4}}{R_{3}}\right)V_{+} - \frac{R_{4}}{R_{3}}V_{2} \\ &= 2(1) - 4 = -2\ \text{V} \end{aligned} \]

With all resistors equal, the circuit simply gives \(V_{1} - V_{2}\). The option +2 reverses the subtraction, and \(\pm 4\) uses only one input.

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Negative feedback for an operational amplifier leads to :

A.Increase the input and output impedance
B.Increase the input impedance and the bandwidth
C.Decrease the output impedance and bandwidth
D.Does not affect impedance or bandwidth

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: B

Negative feedback leads to an increase in the input impedance and the bandwidth.

With series (voltage) negative feedback, as in the non-inverting amplifier, the input impedance is multiplied by the factor (1 + Aβ), the output impedance is divided by the same factor, and the bandwidth is extended by it while the gain falls by it, keeping the gain-bandwidth product constant.

So the output impedance decreases rather than increases, which rules out option A; option C wrongly says the bandwidth decreases; and option D ignores the effect altogether.

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What is the purpose of inverting (-) and non-inverting (+) input terminals on an operational amplifier ?

A.Determine Op-Amp's power supply
B.Control gain of Op-Amp
C.Define the amount of feedback
D.Provide input signals for differential amplification

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: D

The two inputs provide input signals for differential amplification.

An op-amp is a differential amplifier: its output is \(A(V_{+} - V_{-})\), the open-loop gain times the difference between the voltages at the non-inverting (+) and inverting (-) inputs. A signal at the + input appears at the output in phase, a signal at the - input appears inverted, and a signal common to both is rejected.

The power supply is connected to separate supply pins, not the inputs. The gain and the amount of feedback are set by the external components connected around the op-amp, typically from the output to the inverting input, not by the existence of the two terminals.

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List - I	List - II
(a) Voltage regulated IC	(i) Figure of merit
(b) Bistable multivibrator	(ii) CMRR
(c) OPAMP	(iii) Capture range and locking range
(d) Phase Locked Loop	(iv) flip-flop

Codes :
(a) (b) (c) (d)

A.(i) (iii) (iv) (ii)
B.(i) (iv) (ii) (iii)
C.(iv) (iii) (i) (ii)
D.(iii) (ii) (iv) (i)

Exam Year(s)

UGC NET - 2013,

Answer: B

The matching is (a)-(i), (b)-(iv), (c)-(ii), (d)-(iii), which is option (B). A voltage regulator IC is rated by its figure of merit, the regulation quality. A bistable multivibrator is a flip-flop. An op-amp is characterised by its common-mode rejection ratio (CMRR), and a phase locked loop by its capture range and locking range. Options (A), (C) and (D) pair the op-amp or the PLL with the flip-flop or the figure of merit, so each is wrong on at least two pairs.

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An ideal op-amp has

A.infinite common mode gain as well as differential gain.
B.infinite common mode gain and zero differential gain.
C.infinite differential gain and zero common mode gain.
D.zero differential gain and infinite common mode gain.

Exam Year(s)

UGC NET - 2013,

Answer: C

An ideal op-amp has infinite differential gain, so that a tiny difference between the inputs gives a large output, and zero common mode gain, so that infinite CMRR follows. The first and second options give it infinite common mode gain, which would make it useless as a differential amplifier because the output would respond to noise common to both inputs. The fourth option has zero differential gain, which would mean that it amplifies nothing of the wanted signal. The answer is infinite differential gain and zero common mode gain.

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10.

Match the following lists :

List - I	List - II
a. Virtual ground	i. Time constant
b. \(\frac{L}{R}\)	ii. Q-factor
c. \(\frac{WL}{R}\)	iii. Op-Amp
d. Intrinsic stand-off ratio	iv. UJT

A.ii i iii iv
B.iii iv i ii
C.iii i ii iv
D.iv i ii iii

Exam Year(s)

UGC NET - 2013,

Answer: C

The correct match is virtual ground with Op-Amp (iii), L/R with time constant (i), WL/R with Q-factor (ii) and intrinsic stand-off ratio with UJT (iv), which is the code iii i ii iv, option (C).

The virtual ground belongs to an inverting op-amp, L/R is the time constant of an RL circuit, \(\omega L/R\) (printed WL/R) is the quality factor of a coil, and the intrinsic stand-off ratio is a parameter of the unijunction transistor. Options A, B and D mismatch at least one pair.

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Operational Amplifier Fundamentals Questions and Answers

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