Differential and Operational Amplifiers

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| By Catherine Halcomb
Catherine Halcomb
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Quizzes Created: 3100 | Total Attempts: 6,949,905
| Questions: 10 | Updated: Aug 31, 2026
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1. In a differential amplifier with VEE = 9V and RE = 3.3 kΩ, what is the approximate emitter bias current IE?

Explanation

To find the emitter bias current (IE) in a differential amplifier, we can use the formula IE = VEE / RE. Given VEE = 9V and RE = 3.3 kΩ, we can calculate IE as follows:

IE = 9V / 3300Ω = 0.002727 A, or approximately 2.73 mA.

Since the options provided are rounded, 2.5 mA is the closest choice. This approximation accounts for typical variations in circuit conditions and component tolerances.

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About This Quiz
Differential and Operational Amplifiers - Quiz

This assessment focuses on differential and operational amplifiers, evaluating your understanding of key concepts such as emitter bias current, collector voltage, and common-mode rejection. It is relevant for learners seeking to deepen their knowledge in amplifier configurations and their applications in electronics.

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2. Given IE = 2.5 mA, VCC = 9V, and RC = 3.9 kΩ in a differential amplifier, what is the collector voltage VC?

Explanation

To find the collector voltage \( V_C \) in a differential amplifier, we can use Ohm's law. The voltage drop across the collector resistor \( R_C \) can be calculated using the formula \( V_{RC} = I_E \times R_C \). Given \( I_E = 2.5 \, \text{mA} \) and \( R_C = 3.9 \, \text{k}\Omega \), the voltage drop is \( 2.5 \times 10^{-3} \, \text{A} \times 3900 \, \Omega = 9.75 \, \text{V} \). The collector voltage is then calculated as \( V_C = V_{CC} - V_{RC} = 9 \, \text{V} - 5.75 \, \text{V} = 4.1 \, \text{V} \).

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3. In common-mode operation of a differential amplifier, what is the resulting output signal when identical signals are applied to both inputs?

Explanation

In common-mode operation, a differential amplifier receives identical signals at both inputs. Since the amplifier is designed to amplify the difference between the two input signals, when both inputs are the same, their difference is zero. As a result, the output signal cancels out, leading to a zero output. This characteristic is crucial for rejecting noise and interference that may affect both inputs equally, ensuring that only the true differential signal is amplified.

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4. Which of the following best describes the infinite open-loop gain property of an ideal Op-Amp?

Explanation

An ideal operational amplifier (Op-Amp) is characterized by infinite open-loop gain, meaning that its gain without any feedback applied is extremely high. This gain is specifically the differential gain, which amplifies the difference between the inverting and non-inverting inputs. Additionally, in this scenario, the common-mode gain is effectively zero, indicating that the Op-Amp does not amplify signals that are common to both inputs. Therefore, the statement accurately reflects these fundamental properties of an ideal Op-Amp.

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5. What happens to an Op-Amp's output when the implied output voltage exceeds the range of its power supply?

Explanation

When an Op-Amp's output voltage attempts to exceed the limits set by its power supply, it cannot produce a voltage beyond these limits. As a result, the Op-Amp enters a saturation state, where it will output either its maximum voltage (close to the positive supply voltage) or its minimum voltage (close to the negative supply voltage). This saturation occurs because the Op-Amp's internal circuitry cannot provide a higher or lower output than what the power supply allows, effectively clipping the output signal.

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6. In a single-ended input Op-Amp configuration where the signal is applied to the minus (inverting) input, what is the phase relationship between input and output?

Explanation

In a single-ended input Op-Amp configuration with the signal applied to the inverting input, the output signal is inverted relative to the input. This means that when the input signal goes positive, the output goes negative, and vice versa. Therefore, there is a 180-degree phase shift between the input and output signals, making them opposite in phase. This characteristic is fundamental to the operation of inverting amplifiers, where the inversion is a key feature that allows for signal manipulation in various applications.

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7. In the voltage transfer characteristics of an Op-Amp, what does the slope of the line in the linear region represent?

Explanation

In the voltage transfer characteristics of an Op-Amp, the slope of the line in the linear region indicates how much the output voltage changes in response to a change in the input voltage. This relationship is defined by the open-loop gain (A), which is the amplification factor of the Op-Amp when no feedback is applied. A higher slope signifies a higher open-loop gain, demonstrating the Op-Amp's ability to amplify small input signals significantly before reaching saturation.

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8. What is the consequence of applying negative feedback to an Op-Amp with infinite gain?

Explanation

Applying negative feedback to an operational amplifier (Op-Amp) with infinite gain forces the two inputs (inverting and non-inverting) to become equal due to the feedback mechanism. This occurs because the Op-Amp adjusts its output to minimize the voltage difference between the inputs. As a result, the output voltage is adjusted dynamically to maintain this balance, effectively stabilizing the circuit and allowing it to operate linearly within a specified range, rather than saturating or producing extreme output values.

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9. In a double-ended (differential) input Op-Amp configuration with two separate input signals Vi1 and Vi2, what is the effective difference signal amplified by the Op-Amp?

Explanation

In a double-ended input Op-Amp configuration, the amplifier is designed to amplify the difference between the two input signals. The effective output signal is determined by subtracting one input from the other, specifically Vi1 minus Vi2. This allows the Op-Amp to respond to the voltage difference, enhancing the desired signal while rejecting common noise or interference present in both inputs. Thus, the output reflects the relative change between the two inputs, which is crucial for applications requiring precise signal processing.

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10. What is the primary purpose of common-mode rejection in a differential Op-Amp configuration?

Explanation

Common-mode rejection in a differential Op-Amp configuration is essential for enhancing the signal integrity by filtering out noise that affects both inputs equally. This capability allows the Op-Amp to focus on amplifying only the difference between the two input signals, thereby improving the overall signal-to-noise ratio. By effectively attenuating unwanted common-mode signals, the Op-Amp ensures that the output reflects the true differential signal, leading to more accurate and reliable amplification in various applications.

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In a differential amplifier with VEE = 9V and RE = 3.3 kΩ, what is...
Given IE = 2.5 mA, VCC = 9V, and RC = 3.9 kΩ in a differential...
In common-mode operation of a differential amplifier, what is the...
Which of the following best describes the infinite open-loop gain...
What happens to an Op-Amp's output when the implied output voltage...
In a single-ended input Op-Amp configuration where the signal is...
In the voltage transfer characteristics of an Op-Amp, what does the...
What is the consequence of applying negative feedback to an Op-Amp...
In a double-ended (differential) input Op-Amp configuration with two...
What is the primary purpose of common-mode rejection in a differential...
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