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

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

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">CM\(V_C=\frac{V_1 ~+~ V_2}{2}\)Vc=V1+V22Vc=V1+V" id="MathJax-Element-7-Frame" role="presentation" style="position: relative;" tabindex="0">Vc=V1+VVc=V1+V Vc=V1+V Given that Differential voltage gain = 47 dBCommon mode voltage gain = 2 

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

Concept:

The given circuit is an op-amp differentiator. For an ideal differentiator with input \( v_i(t) \), the output is given by:

\( v_o(t) = -RC \cdot \frac{d}{dt}v_i(t) \)

Given that \( RC = 1 \), the equation simplifies to:

\( v_o(t) = -\frac{d}{dt}v_i(t) \)

Given:

The input signal \( v_i(t) \) is a triangular waveform.

• During the rising edge (slope = +1): \( \frac{d}{dt}v_i(t) = 1 \Rightarrow v_o(t) = -1 \)
• During the falling edge (slope = –1): \( \frac{d}{dt}v_i(t) = -1 \Rightarrow v_o(t) = +1 \)

Thus, the output waveform will be a square wave alternating between +1 and –1, switching at every point where the slope of \( v_i(t) \) changes.

Answer:

Option 2: Square wave output alternating between +1 and –1 (in response to triangular wave input)

Explanation:

Step-by-Step Solution:

Step 1: Analyze the Sine Wave Input

The sine wave is given by the equation:

V(t) = A × sin(ωt)

Where:

• A: Amplitude of the sine wave = 2V
• ω: Angular frequency of the sine wave

The sine wave oscillates between -2V and +2V, crossing the reference voltage (1V) twice per cycle.

Step 2: Determine When the Sine Wave Exceeds the Reference Voltage

To find the time intervals during which the sine wave is greater than the reference voltage (1V), we solve the equation:

A × sin(ωt) > V_ref

Substituting the values:

2 × sin(ωt) > 1

Dividing through by 2:

sin(ωt) > 0.5

From trigonometric principles, sin(ωt) > 0.5 occurs between two angles:

ωt = π/6 and ωt = 5π/6

Therefore, the sine wave exceeds the reference voltage for the duration:

Δt = (5π/6 - π/6)/ω = (4π/6)/ω = (2π/3)/ω

Step 3: Calculate the Duty Cycle

The total time period of the sine wave is:

T = 2π/ω

The duty cycle is the ratio of the time the output is high (Δt) to the total time period (T):

Duty Cycle = Δt / T

Substituting the values:

Duty Cycle = [(2π/3)/ω] / [2π/ω]

Simplifying:

Duty Cycle = (2π/3) / (2π) = 1/3

Expressing this as a decimal:

Duty Cycle = 0.3333

Final Answer: The duty cycle of the output signal is 0.3333.

Hence the correct answer is 33.33 %

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:

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