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

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 A_OL). 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.

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.

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:

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