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

An operational amplifier is an integrated circuit that can amplify weak electric signals. An operational amplifier has two input pins and one output pin. Its basic role is to amplify and output the voltage difference between the two input pins.

OP-AMP Integrator:

The expression for the output voltage of OP-AMP integrator as:

\({V_0} = - \frac{1}{{{R_1}{C_f}}}\mathop \smallint \limits_0^t {V_{in}}dt\)

Where,

Vin = Input voltage

Vo  = Output voltage

Cf  = Feedback capacitor

R1 = Input resistance

VA, VB  = Virtual ground potentials

I = Input current

 Additional Information

OP-AMP differentiator:

The expression for the output voltage of OP-AMP differentiator as:

\({V_0} = - {R_f}C\frac{{d{V_{in}}}}{{dt}}\)

Vin = Input voltage

Vo = Output voltage

Rf  = Feedback resistance

C = Input capacitance

VA, VB = Virtual ground points

Important Points

By changing positions of Resistance and Capacitor OP-AMP Integrator converts to Differentiator and vice-versa.

The correct option is 1

Concept:

The various stages of the op-amp and configuration of Transistors is shown below:

From the figure, it is evident that the second stage which is an intermediate stage is made of Dual Input Balanced Output Differential Amplifier.

Concept:

Op-Amp Voltage Gain Calculation

• In a non-inverting op-amp configuration, the voltage gain is determined by the ratio of the resistances in the feedback network.
• The voltage gain (A_V) of a non-inverting op-amp is given by the formula:
• A_V = 1 + (R_f / R_in), where R_f is the feedback resistance and R_in is the resistance between the input and the inverting terminal.

Given:

• Feedback resistance, R_f = 20 kΩ
• Resistance between input and inverting terminal, R_in = 4 kΩ

Using the formula for the voltage gain of a non-inverting op-amp:

A_V = 1 + (R_f / R_in)

Substitute the given values:

A_V = 1 + (20 kΩ / 4 kΩ)

A_V = 1 + 5

A_V = 6

Therefore, the value of the voltage gain is 6. Thus, the correct option is 6

Explanation:

Open-Loop Differential Amplifier

Definition: An open-loop differential amplifier is an electronic amplifier that amplifies the difference between two input voltages without any feedback loop. It is commonly used in various analog circuits to amplify small differential signals. The term "open-loop" indicates that there is no feedback path from the output back to the input of the amplifier. 

Analysis of Other Options:

Option 1: "Input is applied at the inverting terminal only." This statement is incorrect because an open-loop differential amplifier requires inputs to be applied to both the inverting and non-inverting terminals to operate correctly. Applying input to only one terminal would not allow the amplifier to perform differential amplification.

Option 2: "Input is applied at non-inverting terminal only." Similar to Option 1, this statement is incorrect. An open-loop differential amplifier needs inputs at both the inverting and non-inverting terminals to function as intended.

Option 4: "Output is always zero irrespective of the input applied at terminals." This statement is incorrect because, in an open-loop differential amplifier, the output is highly dependent on the difference between the input voltages. Given the high gain, even a small input difference can produce a substantial output voltage.

Explanation:

The open-loop gain of an operational amplifier (op-amp) refers to the amplification factor it provides without any external feedback applied to the circuit. This gain is typically very high for op-amps, which allows them to amplify very small input signals to a much larger output signal. In the given problem, the open-loop gain (A_OL) of the op-amp is 10⁵ (100,000), and the input voltage (V_in) is 100 millivolts (0.1 volts).

The formula to calculate the output voltage (V_out) in an open-loop configuration is given by:

V_out = A_OL × V_in

Substituting the given values:

V_out = 10⁵ × 0.1 V

V_out = 100,000 × 0.1 V

V_out = 10,000 V

However, it is essential to consider the power supply voltage of the op-amp. In this case, the op-amp has a power supply of ±15 volts. This means that the maximum output voltage the op-amp can provide is limited to +15 volts and the minimum output voltage is limited to -15 volts, regardless of the calculated value.

Since the calculated output voltage (10,000 V) far exceeds the maximum supply voltage of the op-amp, the actual output voltage will be limited to the maximum supply voltage, which is +15 volts.

Therefore, the correct value of the output voltage is 15 V.

Important Information:

Analyzing the other options:

• Option 1 (10⁴ V): This is the result of the calculation without considering the power supply limitation. This value is impractical because it exceeds the op-amp's supply voltage.
• Option 3 (0 V): This value would be correct if the input voltage was 0 V or if the op-amp was not powered. However, neither condition applies here.
• Option 4 (100 V): This value does not consider the power supply limitation and is also much higher than the maximum output voltage the op-amp can provide.

The given figure represents the precision rectifier.

Precision rectifier

Integrator

Differentiator

Summing amplifier

Voltage follower :
 

Positive peak detector :
 

Half wave rectifier :
 

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.

A_cm = 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,

A_cm = 2

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. 

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.

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.

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.

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.

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.

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.

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

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.

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

Concept:

The voltage gain (A_v) 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

Inverting amplifier:

The configuration of the inverting amplifier is shown here:

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:

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

Hence, option 1 is correct.

Non-inverting amplifier:

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

The gain is defined as:

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