All questions of Operational Amplifier for Electrical Engineering (EE) Exam

The given problem involves designing a non-inverting amplifier using an operational amplifier (op amp) with specific characteristics. The op amp has a gain of 106 dB at dc (direct current) and a single-pole frequency response with a transition frequency (ft) of 2 MHz. We are required to determine the 3-dB frequency of the closed-loop gain for the amplifier.

Understanding the Problem:
To solve this problem, we need to consider the characteristics of the op amp and the properties of the non-inverting amplifier configuration. The closed-loop gain of the non-inverting amplifier is determined by the feedback resistor (Rf) and the input resistor (Rin).

Non-Inverting Amplifier Configuration:
In a non-inverting amplifier, the input signal is applied to the non-inverting terminal of the op amp, and the feedback resistor is connected between the output and the inverting terminal. The input resistor is connected between the non-inverting terminal and the input signal source.

Determining the Closed-Loop Gain:
The closed-loop gain (Av) of a non-inverting amplifier is given by the formula:
Av = 1 + (Rf/Rin)

In this problem, the desired nominal dc gain is 100, so we can write:
100 = 1 + (Rf/Rin)

Determining the Gain at the 3-dB Frequency:
The transition frequency (ft) of the op amp is the frequency at which the gain starts to roll off. In a single-pole response, the gain decreases at a rate of -20 dB per decade after the transition frequency.

The 3-dB frequency (f3dB) is the frequency at which the gain drops by 3 dB (half-power point). We can determine the 3-dB frequency using the formula:
f3dB = ft / Av

Substituting the given values:
f3dB = 2 MHz / 100

Solving the Equation:
Now, we can solve the equation to find the value of f3dB:
f3dB = 20 kHz

Therefore, the 3-dB frequency of the closed-loop gain is 20 kHz, which corresponds to option B.

Advantages of Weighted Summer Operational Amplifier:
Advantages:
- Summing Capability: A weighted summer operational amplifier is capable of summing various input voltages together. This is especially useful in applications where multiple signals need to be combined or added together.
- Independent Adjustment: Each input signal may be independently adjusted by adjusting the corresponding input resistance. This level of control allows for precise tuning of each input signal, ensuring accurate summation.
- Sign of Voltage Signal: If one needs both the positive and negative aspects of a voltage signal, two operational amplifiers are typically needed. However, with a weighted summer operational amplifier, this can be achieved in a single component, making it more efficient and cost-effective.
- All of the mentioned: The correct answer is all of the mentioned advantages. A weighted summer operational amplifier offers the ability to sum input voltages, independently adjust each input signal, and handle both positive and negative voltage signals in a single component. This makes it a versatile and valuable tool in various electrical engineering applications.

Understanding the Requirements for Closed Loop Amplifiers
To design a closed-loop amplifier with a specific DC gain and bandwidth, we need to consider the gain-bandwidth product (GBP) of the operational amplifier (op-amp).
Key Concepts
- DC Gain: Given as -2 V/V, which indicates an inverting amplifier configuration.
- 3dB Bandwidth: Required bandwidth is 10 MHz.
Calculating the Required Gain-Bandwidth Product
The gain-bandwidth product (GBP) is a constant for a given op-amp. It can be expressed as:
GBP = Gain × Bandwidth
In this case, the gain is -2 (absolute value is considered):
- Gain (|A|) = 2 V/V
- Bandwidth (BW) = 10 MHz
Thus, the required GBP will be:
GBP = 2 V/V × 10 MHz = 20 MHz
Choosing the Appropriate Op-Amp
For an op-amp to successfully implement the closed-loop amplifier, its GBP must be equal to or greater than the calculated value of 20 MHz.
The required frequency for internally compensated op-amps typically needs to be higher than the desired bandwidth to account for the phase margin and stability.
Final Calculation for Required Frequency
To ensure stability and accommodate the closed-loop gain, we can estimate a safe operating frequency:
- A common rule of thumb is to use a frequency that is 1.5 to 2 times the GBP.
Therefore,
Required Frequency = 2 × 10 MHz = 20 MHz
Considering additional safety margins, a frequency of about 30 MHz is recommended for stability.
Conclusion
Based on these calculations, the correct option for the required frequency is:
Option 'D': 30 MHz

Differential input and single-ended output type amplifier:
Operational amplifiers (op-amps) are electronic devices that amplify the difference between two input voltages. They have various configurations depending on their input and output types. Option 'A' refers to a differential input and single-ended output type amplifier.

Differential Input:
A differential input means that the amplifier has two input terminals, commonly labeled as the non-inverting input (+) and the inverting input (-). The voltage difference between these two input terminals determines the amplifier's output.

Single-Ended Output:
A single-ended output means that the amplifier has only one output terminal, which produces an amplified version of the voltage difference between the input terminals.

Working of Differential Input and Single-Ended Output Amplifier:
1. The differential input stage: The amplifier's input stage consists of a differential pair of transistors. The non-inverting input (+) is connected to the base of one transistor, and the inverting input (-) is connected to the base of the other transistor.
2. Differential amplification: The differential pair amplifies the voltage difference between the two input terminals. The amplified signal appears at the collectors of the transistors.
3. Single-ended output stage: The output stage of the amplifier is designed to convert the differential voltage at the collectors of the transistors into a single-ended output voltage.
4. Voltage amplification: The output stage amplifies the differential voltage and provides a single-ended output voltage that is proportional to the difference between the input voltages.
5. Output coupling: A coupling capacitor is often used to remove any DC component and pass only the AC component of the output voltage.
6. Load resistance: The output voltage is connected to a load resistance, which determines the current flowing through it and the voltage level seen by the load.

Applications:
- Differential input and single-ended output amplifiers are commonly used in instrumentation systems, audio amplifiers, and data acquisition systems.
- They are suitable for applications where the input signal is differential, but the output signal can be single-ended.
- They can be used for amplifying small differential signals and rejecting common-mode noise.
- Single-ended output makes it easier to interface with other circuits or devices that accept single-ended signals.

In conclusion, a differential input and single-ended output type amplifier is a common configuration of operational amplifiers. It amplifies the voltage difference between two input terminals and provides a single-ended output voltage. This configuration is widely used in various applications in electrical engineering.

Explanation:

High Resistance Source to Low Resistance Load:
- In the non-inverting closed loop configuration, the feedback resistor is in parallel with the input resistor, resulting in a high input impedance.
- When connecting a high resistance source to a low resistance load, the voltage division can cause a significant drop in voltage across the load due to the mismatch in resistance.
- To prevent this voltage drop and maintain signal integrity, a unity gain follower or buffer amplifier with a high input impedance is used.
- The buffer amplifier presents a high input impedance to the high resistance source, preventing any significant loading effect.
Therefore, the correct answer is option 'D', which states that a unity gain follower or buffer amplifier is used to connect a high resistance source to a low resistance load.

Single-pole model is also known as 'Dominant pole'.

A single-pole model is a simplified representation of a system or circuit that consists of a single dominant pole. A pole is a point in the complex frequency domain where the transfer function of a system becomes infinite or the response of a system becomes undefined. The dominant pole is the pole that has the greatest influence on the system's behavior.

Importance of Single-pole Model:

The single-pole model is commonly used in control system analysis and design because it provides a good approximation for systems with dominant poles. It allows engineers to simplify complex systems and focus on the most significant dynamics. The dominant pole determines the overall response characteristics of the system, such as stability, transient response, and frequency response.

Characteristics of Dominant Pole:

1. Stability: The presence of a dominant pole ensures system stability. A stable system means that the output remains bounded for any bounded input.

2. Transient Response: The dominant pole determines the speed and damping of the system's transient response. A system with a dominant pole close to the imaginary axis will have a slower response, while a pole farther away will result in a faster response.

3. Frequency Response: The location of the dominant pole influences the frequency response of the system. It determines the bandwidth and gain of the system at different frequencies.

Advantages of Single-pole Model:

1. Simplicity: The single-pole model simplifies complex systems into a single dominant pole, making it easier to analyze and design control systems.

2. Insightful Analysis: The single-pole model allows engineers to gain insight into the system's behavior without dealing with the complexities of higher-order models.

3. Design Flexibility: By focusing on the dominant pole, engineers can make design decisions to achieve desired system characteristics, such as stability, response speed, and frequency response.

4. Model Validation: The single-pole model can be used to validate the behavior of a more complex system by comparing the responses obtained from both models.

In conclusion, the single-pole model, also known as the dominant pole, is an important concept in control system analysis and design. It simplifies complex systems and allows engineers to focus on the most significant dynamics that determine stability, transient response, and frequency response.

Given data:
Symmetrical square wave of 20-V peak-to-peak, 0 average, and 2-ms period applied to a Miller integrator.
We need to find the value of the time constant CR such that the triangular waveform at the output has a 20-V peak-to-peak amplitude.

Miller Integrator:
Miller integrator is a type of operational amplifier integrator circuit with a capacitor connected between the output and the input terminals of the amplifier. The capacitor C is used to integrate the input signal and produces an output signal that is proportional to the time integral of the input signal.

Solution:
The triangular waveform at the output of a Miller integrator is given by the equation:

Vout = - (Vin/CR) * t + V0

Where, Vin is the input voltage, V0 is the initial output voltage, CR is the time constant of the circuit and t is the time.

For a symmetrical square wave with a 2-ms period, the frequency of the waveform is 1/2ms = 500 Hz. The peak-to-peak voltage of the square wave is 20V. The average value of the square wave is zero.

The triangular waveform at the output of the Miller integrator will have a maximum amplitude when the input signal changes polarity. At this point, the output voltage should be equal to the peak-to-peak voltage of the triangular waveform, which is 20V.

The time taken for the input voltage to change from 0V to 10V is half the period of the waveform, which is 1 ms.

Substituting the values in the equation for the output voltage, we get:

Vout = - (10/CR) * 0.001 + V0

When the input voltage changes from 0V to 10V, the output voltage changes from V0 to V0 - (10/CR) * 0.001.

Similarly, when the input voltage changes from 10V to 0V, the output voltage changes from V0 - (10/CR) * 0.001 to V0 + (10/CR) * 0.001.

The peak-to-peak voltage of the output triangular waveform is given by:

Peak-to-peak voltage = (V0 - (10/CR) * 0.001) - (V0 + (10/CR) * 0.001)

Peak-to-peak voltage = - (20/CR) * 0.001

For the peak-to-peak voltage of the output waveform to be 20V, we have:

(20/CR) * 0.001 = 20

CR = 0.5 ms

Therefore, the value of the time constant CR such that the triangular waveform at the output has a 20-V peak-to-peak amplitude is 0.5 ms.