Op-Amp and Its Applications MCQ Quiz - Objective Question with Answer for Op-Amp and Its Applications - Download Free PDF

Last updated on Jul 9, 2025

Latest Op-Amp and Its Applications MCQ Objective Questions

Op-Amp and Its Applications Question 1:

In the OP AMP circuit shown below, what is the input voltage if output voltage Vout = -0.62 V?

qImage684c41285c5f79025153292a

  1. -20 mV
  2. +20 mV
  3. -30 mV
  4. +30 mV

Answer (Detailed Solution Below)

Option 1 : -20 mV

Op-Amp and Its Applications Question 1 Detailed Solution

Concept:

An inverting amplifier has the configuration shown below. The voltage gain of an inverting OP-AMP is given by: \( A_v = \frac{V_{out}}{V_{in}} = -\frac{R_f}{R_{in}} \)

Given:

Feedback resistor, \(R_f = 1.2~M\Omega\), Input resistor, \(R_{in} = 40~k\Omega\), Output voltage, \(V_{out} = -0.62~V\)

Calculation:

Gain of the amplifier: \( A_v = -\frac{1.2 \times 10^6}{40 \times 10^3} = -30 \)

Using the gain formula: \( V_{in} = \frac{V_{out}}{A_v} = \frac{-0.62}{-30} = 0.0207~V = 20.7~mV \)

Since it is an inverting amplifier, input voltage must be negative: \( V_{in} \approx -20~mV \)

Answer:

Option 1) -20 mV

Op-Amp and Its Applications Question 2:

The differential voltage gain and common mode voltage gain of a differential amplifier are 50 dB and 3 dB respectively, then common mode rejection ratio is

  1. 48 dB
  2. 47 dB
  3. 52 dB
  4. 53 dB

Answer (Detailed Solution Below)

Option 2 : 47 dB

Op-Amp and Its Applications Question 2 Detailed Solution

Concept:

The output voltage of op-amp is given by:

V0 = Ad Vd + Ac Vc

Where, Vd = V1 – V2

\(V_C=\frac{V_1+V_2}{2}\)

Ad = differential voltage gain

Ac = common mode voltage gain

\(CMRR=20~log (\frac{A_d}{A_c})\)

Calculation:

Given, 

Differential voltage gain = 50 dB

Common mode voltage gain = 3 dB

\(CMRR = 20~log(\frac{A_d}{A_c})\)

CMRR = 20 log(Ad) - 20 log(Ac)

= 50 - 3 = 47 dBVc=V1+V22" id="MathJax-Element-34-Frame" role="presentation" style="display: inline; position: relative;" tabindex="0">Vc=V1+V" id="MathJax-Element-7-Frame" role="presentation" style="position: relative;" tabindex="0">Vc=V1+V Given that Differential voltage gain = 47 dBCommon mode voltage gain = 2 

Op-Amp and Its Applications Question 3:

An Opamp has a bandwidth of 5 MHz when operating at a gain of 20. Its unity gain Bandwidth is

  1. 0.25 MHz
  2. 100 MHz
  3. 4 MHz
  4. 25 MHz

Answer (Detailed Solution Below)

Option 2 : 100 MHz

Op-Amp and Its Applications Question 3 Detailed Solution

Concept:

The gain-bandwidth product (GBW) of an op-amp is a constant and is defined as:

\(\text{GBW} = \text{Gain} \times \text{Bandwidth}\)

For unity gain (gain = 1), the bandwidth is equal to the GBW.

Given:

Bandwidth = 5 MHz at Gain = 20

Calculation:

\(GBW= 20 \times 5~\text{MHz} = 100~\text{MHz}\)

Therefore, the unity gain bandwidth = 100 MHz

Hence, the correct option is 2

Op-Amp and Its Applications Question 4:

Which of the following options is correct for an open-loop configuration of an op-amp?

  1. The output voltage is linearly proportional to the input voltage
  2. It has a feedback connection between output and input. 
  3. It has zero output voltage
  4. It has a very high open-loop voltage gain.

Answer (Detailed Solution Below)

Option 4 : It has a very high open-loop voltage gain.

Op-Amp and Its Applications Question 4 Detailed Solution

Explanation:

  • An open-loop configuration of an operational amplifier (op-amp) refers to a setup in which there is no feedback connection between the output and the input of the op-amp. In this configuration, the op-amp operates with its inherent gain, which is extremely high, typically in the range of tens of thousands to millions.
  • In an open-loop configuration, the op-amp amplifies the voltage difference between its inverting input (-) and non-inverting input (+). The gain of the op-amp in this configuration is called the open-loop gain (denoted as AOL). Since the open-loop gain is extremely high, even a small input voltage difference results in a very large output voltage, typically driving the op-amp to saturation (either the positive or negative supply voltage).

Characteristics of Open-Loop Configuration:

  • Very High Gain: The open-loop voltage gain of an op-amp is extremely high, typically ranging from 104 to 107.
  • No Feedback: There is no feedback loop connecting the output to the input. This lack of feedback makes the configuration highly sensitive to input voltage changes.
  • Non-Linear Operation: Due to the high gain, the output voltage often saturates at either the positive or negative supply voltage, depending on the polarity of the input voltage difference.

Op-Amp and Its Applications Question 5:

If all the resistances of an inverting op-amp adder are equal, then the output voltage is _______.

  1. difference of input voltages
  2. sum of input voltages
  3. negative sum of the input voltages
  4. zero as all the resistances are equal 

Answer (Detailed Solution Below)

Option 3 : negative sum of the input voltages

Op-Amp and Its Applications Question 5 Detailed Solution

Concept:

An inverting op-amp adder sums multiple input voltages applied through equal resistors into the inverting terminal. Due to the inverting configuration, the output voltage is proportional to the negative of the sum of input voltages.

The output voltage is given by: \( V_{out} = - (V_1 + V_2 + V_3 + \dots) \)

where all input resistors and feedback resistor are equal.

Calculation:

Given:

All resistances of the inverting adder are equal.

Output voltage: \( V_{out} = - (V_1 + V_2 + V_3 + \dots) \)

Answer:

The output voltage of an inverting op-amp adder when all resistances are equal is the negative sum of the input voltages.

Top Op-Amp and Its Applications MCQ Objective Questions

The given figure represents? 

EC Analog Circuit subject test1 final Images-Q29

  1. Integrator 
  2. Precision rectifier  
  3. Differentiator 
  4. Summing amplifier 

Answer (Detailed Solution Below)

Option 2 : Precision rectifier  

Op-Amp and Its Applications Question 6 Detailed Solution

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The given figure represents the precision rectifier.

Precision rectifier

EC Analog Circuit subject test1 final Images-Q29

Integrator

F2 U.B D.K 28.09.2019 D 2

Differentiator

 RRB JE LIC PART 1 66 14 Q Hindi - Final images deepak Q14

Summing amplifier

 correction diagram shubham

Voltage follower :
 EC Analog Circuit subject test1 final Images-Q29.2

Positive peak detector :
 EC Analog Circuit subject test1 final Images-Q29.3

 

Half wave rectifier :
 EC Analog Circuit subject test1 final Images-Q29.1

In differential amplifier differential gain is 20000 and CMRR is 80 dB. Find common-mode gain? (take log 10000 = 4)

  1. 2
  2. 1
  3. 0.5
  4. 0

Answer (Detailed Solution Below)

Option 1 : 2

Op-Amp and Its Applications Question 7 Detailed Solution

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Concept:

CMRR (Common mode rejection ratio) is defined as the ratio of differential-mode voltage gain (Ad) and the common-mode voltage gain (Ac).

Mathematically, in dB this is expressed as:

\(CMRR = 20\log \left| {\frac{{{A_d}}}{{{A_{cm}}}}} \right| \)    ----(1)

Ad = Differential gain.

Acm = Common mode gain.

Calculation:

Given:

Ad = 20000

CMRR = 80 dB

From equation 1;

\(80= 20\log \frac{{{20000}}}{{{A_{cm}}}}\)

\(\log \frac{{{20000}}}{{{A_{cm}}}}=4\)

On solving we'll get,

Acm = 2

What is the typical value of open-loop voltage gain, AVOL, for a 741 op-amp?

  1. More than 50,000
  2. More than 2,00,000
  3. More than 10,000
  4. More than 1,00,000

Answer (Detailed Solution Below)

Option 2 : More than 2,00,000

Op-Amp and Its Applications Question 8 Detailed Solution

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The following are the basic specifications of IC 741:

1) Power Supply: Requires a Minimum voltage of 5 V and can withstand up to 18 V. 

2) Input Impedance: About 2 MΩ

3) Output impedance: About 75 Ω

4) Voltage Gain: 200,000 for low frequencies. 

5) Input Offset: Ranges between 2 mV and 6 mV

6) Slew Rate: 0.5 V/µS. 

Which type of BUS structure is used to connect the I/O devices?

  1. Multiple BUS structure
  2. Single BUS structure
  3. Star BUS structure
  4. Node to Node BUS structure

Answer (Detailed Solution Below)

Option 2 : Single BUS structure

Op-Amp and Its Applications Question 9 Detailed Solution

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The correct answer is option 2.

Concept:

A single BUS structure is used to connect the I/O devices. One common bus is utilized to communicate between peripherals and the CPU in a single bus configuration. It has drawbacks owing to the usage of a single common bus.​

  • All units are connected to a single bus, so it provides the sole means of interconnection. A single bus structure has the advantages of simplicity and low cost.
  • Because only two units may participate in data transmission at the same time, single bus structures have the drawback of the restricted speed.
  • This necessitates the use of an arbitration mechanism, as well as the forced waiting of units.
  • An example is a communication between the processor and printer.

F1 Savita Engineering 07-5-22-D5

Hence the correct answer is Single BUS structure.

Additional Information

  •  MULTIBUS II is an open system bus architecture for designing general-purpose 8-, 16-, and 32-bit microcomputer systems.

The given figure represents.

F2 U.B D.K 28.09.2019 D 2

  1. Summing amplifier
  2. Precision rectifier
  3. Integrator
  4. Differentiator 

Answer (Detailed Solution Below)

Option 3 : Integrator

Op-Amp and Its Applications Question 10 Detailed Solution

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Integrator:

  • The Op-amp Integrator is an operational amplifier circuit that performs the mathematical operation of Integration.
  • We can cause the output to respond to changes in the input voltage over time as the op-amp integrator produces an output voltage which is proportional to the integral of the input voltage.

           F2 U.B D.K 28.09.2019 D 2

 

Important:

Summing amplifier:

  • We can design an op-amp circuit to combine a number of input signals and to produce single output as a weighted sum of input signals.
  • The summing amplifier is basically an op-amp circuit that can combine numbers of input signals to a single output that is the weighted sum of the applied inputs.

          F1 R.K. Nita 23.10.2019 D 7

 

Precision rectifier:

  • The precision rectifier, also known as a super diode, is a configuration obtained with an operational amplifier in order to have a circuit behave like an ideal diode and rectifier.
  • It is very useful for high-precision signal processing. With the help of a precision rectifier, the high-precision signal processing can be done very easily.

          F1 R.K. Nita 23.10.2019 D 8

Differentiator:

  • The basic operational amplifier differentiator circuit produces an output signal which is the first derivative of the input signal.
  • The input signal to the differentiator is applied to the capacitor. The capacitor blocks any DC content so there is no current flow to the amplifier summing point, X resulting in zero output voltage.
  • The capacitor only allows AC type input voltage changes to pass through and whose frequency is dependent on the rate of change of the input signal.

          F1 R.K. Nita 23.10.2019 D 9

Which of the following Op-Amp (operational amplifier) system is non-linear?

  1. Voltage to current converter
  2. Voltage follower
  3. Active filter
  4. Sample and hold circuit

Answer (Detailed Solution Below)

Option 4 : Sample and hold circuit

Op-Amp and Its Applications Question 11 Detailed Solution

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A circuit is said to be linear if there exists a linear relationship between its input and the output. Similarly, a circuit is said to be non-linear if there exists a non-linear relationship between its input and output.

Linear application of op-amp:

  • Inverting amplifiers
  • Voltage to current converter
  • Voltage follower
  • Integrator Differentiation
  • Instrumentation amplifier
  • Log and antilog amplifiers


Non-linear applications of op-amp:

  • Schmitt trigger
  • Sample and hold circuits
  • Zero crossing detector
  • Nonzero crossing detector
  • Multivibrators: astable, monostable, bistable
  • Precision rectifier or super diode with the connection of op-amp

A special case of non-inverting amplifier in which all of the output voltage is fed back to the inverting input of the op-amp is called:

  1. differentiator
  2. integrator
  3. logarithmic amplifier
  4. voltage follower

Answer (Detailed Solution Below)

Option 4 : voltage follower

Op-Amp and Its Applications Question 12 Detailed Solution

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Voltage Follower:

  • A voltage follower is an op-amp circuit whose output voltage straight away follows the input voltage. i.e. output voltage equivalent to the input voltage.
  • The Op-amp circuit does not provide any amplification thus, its voltage gain is unity.
  • The voltage follower is used as a buffer amplifier, isolation amplifier, unity gain amplifier as the output follows the input.
  • The voltage follower provides no alternation or no amplification but only buffering.

F2 U.B  Madhu 17.06.20 D 10

Characteristics:

  • High input impedance
  • Low output impedance
  • Current Gain & power gain high
  • Voltage gain unity

 

Derivation:

The Voltage follower circuit is a Non-Inverting Amplifier that has negative feedback.

The gain of Non-Inverting Amplifier is given by:

\(A = 1 + \frac{{{R_f}}}{{{R_1}}}\)

For voltage follower, Rf is 0

Hence gain (A) = 1

An amplifier circuit has a voltage gain of 500. If the output voltage is 5 V, then the input voltage is:

  1. 1 / 2500 V
  2. 2500 V
  3. 100 mV
  4. 10 mV

Answer (Detailed Solution Below)

Option 4 : 10 mV

Op-Amp and Its Applications Question 13 Detailed Solution

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Concept:

The voltage gain (Av) for an amplifier is defined as the ratio of the output voltage to the input voltage, i.e.

\(A_v=\frac{V_0}{V_{in}}\)

V0 = Output voltage

Vin = Input voltage

Calculation:

Given: Av = 500, and V0 = 5 V

\(500=\frac{5}{V_{in}}\)

Vin = 1/100 V

Vin = 10 mV

In the circuit shown below, the output at B is

F1 Shubham 16-9-2020 Swati D1

If A is given as:

F1 Shubham 16-9-2020 Swati D13

  1. F1 Shubham 16-9-2020 Swati D12
  2. F1 Shubham 16-9-2020 Swati D13
  3. F1 Shubham 16-9-2020 Swati D14
  4. F1 Shubham 16-9-2020 Swati D15

Answer (Detailed Solution Below)

Option 1 : F1 Shubham 16-9-2020 Swati D12

Op-Amp and Its Applications Question 14 Detailed Solution

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Inverting amplifier:

The configuration of the inverting amplifier is shown here:

F1 Shubham 16-9-2020 Swati D1

The gain is defined as:

\(\frac{{{V_0}}}{{{V_{in}}}} = - \frac{{{R_f}}}{R}\)

If the impedances Rf and R are equal in magnitude and phase, then the closed-loop voltage gain is -1, and the input signal will undergo a 180° phase shift at the output. Hence, such a circuit is also called a phase inverter.

Analysis:

If the input is given as:

F1 Shubham 16-9-2020 Swati D13

Then the output is a 180o shifted version of the input.

F1 Shubham 16-9-2020 Swati D12

Hence, option 1 is correct.

26 June 1

Non-inverting amplifier:

The configuration of the non-inverting amplifier is shown here.

F2 Shubham Madhu 21.08.20 D6

The gain is defined as:

\(\frac{{{V_0}}}{{{V_{in}}}} = 1 + \frac{{{R_f}}}{R}\)

In an RC coupled amplifier, frequency response is improved with

  1. Lower R1
  2. Higher Cc
  3. Less gain
  4. More bias

Answer (Detailed Solution Below)

Option 2 : Higher Cc

Op-Amp and Its Applications Question 15 Detailed Solution

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In an RC coupled amplifier, frequency response is improved with higher Cc. The capacitor CC is the coupling capacitor that connects two stages and prevents DC interference between the stages and controls the shift of operating point.
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