1What is the input impedance of an ideal operational amplifier?
The ideal op-amp
Easy
A.Infinite
B.
C.
D.Zero
Correct Answer: Infinite
Explanation:
An ideal operational amplifier has infinite input impedance, meaning it draws no current from the input source.
Incorrect! Try again.
2What is the output impedance of an ideal op-amp?
The ideal op-amp
Easy
A.
B.Very high
C.Infinite
D.Zero
Correct Answer: Zero
Explanation:
An ideal op-amp has zero output impedance, allowing it to drive any load without a drop in voltage.
Incorrect! Try again.
3What is the open-loop voltage gain of an ideal op-amp?
The ideal op-amp
Easy
A.Unity ($1$)
B.Infinite
C.$100$
D.Zero
Correct Answer: Infinite
Explanation:
The open-loop voltage gain of an ideal op-amp is considered to be infinite.
Incorrect! Try again.
4Which of the following describes the bandwidth of an ideal op-amp?
The ideal op-amp
Easy
A.Zero
B.Dependent on the supply voltage
C.Infinite
D.Limited to
Correct Answer: Infinite
Explanation:
An ideal op-amp has infinite bandwidth, meaning it can amplify signals of any frequency with the same gain.
Incorrect! Try again.
5In the equivalent circuit of an op-amp, the output voltage is usually modeled as a dependent voltage source. What does this source depend on?
Equivalent circuit of an opamp
Easy
A.The output current
B.The input difference voltage ()
C.The common-mode voltage
D.The power supply voltage
Correct Answer: The input difference voltage ()
Explanation:
The dependent voltage source in the equivalent circuit is , where is the open-loop gain and is the difference between the two input voltages.
Incorrect! Try again.
6Where is the input impedance () placed in the equivalent circuit of an op-amp?
Equivalent circuit of an opamp
Easy
A.Between the non-inverting and inverting input terminals
B.Between the inverting input and ground
C.In series with the output
D.Between the output and ground
Correct Answer: Between the non-inverting and inverting input terminals
Explanation:
The input impedance is connected across the two input terminals (inverting and non-inverting).
Incorrect! Try again.
7In the ideal voltage transfer curve of an op-amp, what happens when the differential input voltage is positive and slightly greater than zero?
Ideal voltage transfer curve
Easy
A.Output increases linearly
B.Output saturates at
C.Output stays at zero
D.Output saturates at
Correct Answer: Output saturates at
Explanation:
Because the open-loop gain is infinite, even a tiny positive differential input causes the output to saturate at the positive supply rail .
Incorrect! Try again.
8What is the slope of the ideal voltage transfer curve in the linear region for an ideal op-amp?
Ideal voltage transfer curve
Easy
A.Negative
B.One
C.Infinite
D.Zero
Correct Answer: Infinite
Explanation:
Since an ideal op-amp has infinite open-loop gain, the transition between and is instantaneous at , giving an infinite slope.
Incorrect! Try again.
9What happens to a practical op-amp in open-loop configuration if a small DC differential input voltage is applied?
Open loop op-amp configurations
Easy
A.It is driven into saturation
B.It amplifies linearly with a gain of 10
C.It acts as a buffer
D.It oscillates at high frequency
Correct Answer: It is driven into saturation
Explanation:
In an open-loop configuration, the gain is so high that even a small input voltage will drive the op-amp output to its saturation voltage.
Incorrect! Try again.
10Which of the following applications typically uses an op-amp in an open-loop configuration?
Open loop op-amp configurations
Easy
A.Comparator
B.Inverting amplifier
C.Voltage follower
D.Integrator
Correct Answer: Comparator
Explanation:
A comparator uses an op-amp without feedback (open-loop) to compare two voltages, driving the output to saturation based on which input is higher.
Incorrect! Try again.
11What is the primary effect of introducing negative feedback in an op-amp circuit?
Op-amp with negative feedback
Easy
A.It increases the overall voltage gain
B.It increases output impedance
C.It stabilizes the overall voltage gain
D.It reduces the bandwidth
Correct Answer: It stabilizes the overall voltage gain
Explanation:
Negative feedback drastically reduces the gain to a stable, predictable value determined by external components, and improves stability.
Incorrect! Try again.
12Which concept states that in a closed-loop op-amp circuit, the voltage difference between the inverting and non-inverting terminals is almost zero?
Op-amp with negative feedback
Easy
A.Virtual ground / Virtual short
B.Common-mode rejection
C.Barkhausen criterion
D.Miller effect
Correct Answer: Virtual ground / Virtual short
Explanation:
The virtual short concept implies that negative feedback forces the voltage at the inverting terminal to be equal to the voltage at the non-inverting terminal.
Incorrect! Try again.
13In the block diagram of a feedback amplifier, what does the 'mixer' network do?
Block diagram representation of feedback configurations
Easy
A.It filters high frequencies
B.It provides the main amplification
C.It combines the input signal and the feedback signal
D.It samples the output signal
Correct Answer: It combines the input signal and the feedback signal
Explanation:
The mixer network at the input combines the external input signal with the feedback signal coming from the output.
Incorrect! Try again.
14In a feedback block diagram, the parameter typically represents:
Block diagram representation of feedback configurations
Easy
A.Open-loop gain
B.Output voltage
C.Input impedance
D.Feedback factor
Correct Answer: Feedback factor
Explanation:
The symbol represents the feedback factor, which is the fraction of the output signal that is fed back to the input.
Incorrect! Try again.
15Which basic op-amp circuit is a classic example of a voltage series feedback amplifier?
Voltage series feedback amplifier
Easy
A.Differentiator
B.Inverting amplifier
C.Integrator
D.Non-inverting amplifier
Correct Answer: Non-inverting amplifier
Explanation:
A non-inverting amplifier samples the output voltage and feeds it back in series with the input, making it a voltage series feedback amplifier.
Incorrect! Try again.
16How does voltage series feedback affect the input and output impedances of an op-amp?
Voltage series feedback amplifier
Easy
A.Decreases both
B.Decreases input, increases output
C.Increases both
D.Increases input, decreases output
Correct Answer: Increases input, decreases output
Explanation:
Voltage series feedback increases the input impedance (due to series connection at input) and decreases output impedance (due to voltage sampling at output).
Incorrect! Try again.
17Which op-amp circuit represents a voltage shunt feedback amplifier?
Voltage shunt feedback amplifier
Easy
A.Differential amplifier
B.Non-inverting amplifier
C.Voltage follower
D.Inverting amplifier
Correct Answer: Inverting amplifier
Explanation:
The inverting amplifier uses voltage shunt feedback, where the output voltage is sampled and fed back in parallel (shunt) with the input signal.
Incorrect! Try again.
18What is the effect of voltage shunt feedback on the input impedance?
Voltage shunt feedback amplifier
Easy
A.It decreases it
B.It increases it
C.It has no effect
D.It makes it infinite
Correct Answer: It decreases it
Explanation:
Shunt mixing at the input lowers the overall input impedance of the amplifier circuit.
Incorrect! Try again.
19What is the main function of a differential amplifier?
Differential amplifiers
Easy
A.To amplify the difference between two signals
B.To amplify the sum of two signals
C.To generate a sine wave
D.To act as a low pass filter
Correct Answer: To amplify the difference between two signals
Explanation:
A differential amplifier amplifies the voltage difference between its two input terminals.
Incorrect! Try again.
20What happens to the output of an ideal differential amplifier if the exact same voltage (common-mode voltage) is applied to both inputs?
Differential amplifiers
Easy
A.The output is maximized
B.The output saturates
C.The output is half the input voltage
D.The output is zero
Correct Answer: The output is zero
Explanation:
An ideal differential amplifier completely rejects common-mode signals, so if both inputs are identical, the difference is zero, and the output is zero.
Incorrect! Try again.
21An ideal op-amp is connected in a circuit with a feedback resistor. What happens to the input impedance seen by the source connected to the non-inverting terminal?
The ideal op-amp
Medium
A.It is equal to the source resistance
B.It depends on the feedback resistor value
C.It becomes zero
D.It approaches infinity
Correct Answer: It approaches infinity
Explanation:
For an ideal op-amp, the input impedance is infinite regardless of the external feedback network. Therefore, the source connected to the non-inverting terminal sees an infinite impedance.
Incorrect! Try again.
22In the equivalent circuit of a practical op-amp, how is the output voltage modeled?
equivalent circuit of an opamp
Medium
A.As a voltage source in series with an output resistance
B.As a voltage source parallel with an output resistance
C.As an independent voltage source
D.As a current source controlled by input voltage
Correct Answer: As a voltage source in series with an output resistance
Explanation:
The equivalent circuit of an op-amp models the output as a voltage-controlled voltage source given by in series with the internal output resistance .
Incorrect! Try again.
23According to the ideal voltage transfer curve of an op-amp, if the differential input voltage is positive and extremely small, what is the output voltage?
ideal voltage transfer curve
Medium
A.Equal to
B.
C.Zero
D.
Correct Answer:
Explanation:
An ideal op-amp has infinite open-loop gain. Thus, any infinitesimally small positive differential input voltage will drive the output to the positive saturation voltage .
Incorrect! Try again.
24In an open-loop inverting amplifier configuration, a 1 V sine wave is applied to the inverting input while the non-inverting input is grounded. If the open-loop gain is , what is the output?
open loop op-amp configurations
Medium
A.A 0.1 V sine wave, 180 out of phase with the input
B.A 0.1 V sine wave, in phase with the input
C.A saturated DC voltage
D.Zero volts
Correct Answer: A 0.1 V sine wave, 180 out of phase with the input
Explanation:
The output is . The negative sign indicates a 180 phase shift.
Incorrect! Try again.
25Which of the following is a direct consequence of introducing negative feedback in an op-amp circuit?
op-amp with negative feedback
Medium
A.Increase in distortion
B.Decrease in bandwidth
C.Reduction in output impedance
D.Increase in overall voltage gain
Correct Answer: Reduction in output impedance
Explanation:
Negative feedback in op-amps typically reduces the output impedance (especially in voltage-series and voltage-shunt configurations), increases bandwidth, and reduces distortion, at the cost of reduced overall gain.
Incorrect! Try again.
26In the block diagram of a feedback amplifier, if the feedback network samples the output voltage and mixes it in series with the input voltage, what is the feedback topology?
block diagram representation of feedback configurations
Medium
A.Voltage-shunt
B.Current-series
C.Current-shunt
D.Voltage-series
Correct Answer: Voltage-series
Explanation:
Sampling the output voltage means 'voltage' sampling, and mixing it in series with the input means 'series' mixing. Hence, it is a voltage-series feedback topology.
Incorrect! Try again.
27Calculate the closed-loop gain of a non-inverting op-amp (voltage-series feedback) if the feedback resistor and the input resistor .
voltage series feedback amplifier
Medium
A.-6
B.5
C.6
D.-5
Correct Answer: 6
Explanation:
The closed-loop gain for a non-inverting amplifier is .
Incorrect! Try again.
28In an inverting amplifier (voltage-shunt feedback), what is the effect of the virtual ground on the input impedance of the circuit as seen by the signal source?
voltage shunt feedback amplifier
Medium
A.The input impedance is equal to the feedback resistor
B.The input impedance becomes infinite
C.The input impedance is exactly equal to the input resistor
D.The input impedance is zero
Correct Answer: The input impedance is exactly equal to the input resistor
Explanation:
Because of the virtual ground at the inverting terminal, the input signal source sees the input resistor directly connected to ground. Thus, .
Incorrect! Try again.
29A differential amplifier has a differential gain of 100 and a Common-Mode Rejection Ratio (CMRR) of 60 dB. What is its common-mode gain ()?
differential amplifiers
Medium
A.0.1
B.1
C.100
D.10
Correct Answer: 0.1
Explanation:
CMRR in dB is . Given CMRR = 60 dB, . Since , .
Incorrect! Try again.
30Which parameter allows an ideal op-amp to amplify DC signals without attenuation?
The ideal op-amp
Medium
A.Zero output impedance
B.Infinite slew rate
C.Infinite bandwidth
D.Zero common-mode gain
Correct Answer: Infinite bandwidth
Explanation:
An ideal op-amp has infinite bandwidth, meaning it can amplify all frequencies from DC (0 Hz) up to infinity with the same gain.
Incorrect! Try again.
31In the equivalent circuit of an op-amp, if the input resistance is very large but finite, what is the primary effect on the circuit when driven by a high-impedance source?
equivalent circuit of an opamp
Medium
A.Output voltage saturation
B.Loading effect on the source
C.Infinite common-mode rejection
D.Increased slew rate
Correct Answer: Loading effect on the source
Explanation:
A finite input resistance will draw some current from the signal source. If the source has a high internal impedance, this will cause a voltage drop across the source's internal resistance, leading to a loading effect.
Incorrect! Try again.
32Why is an open-loop op-amp generally not used as a linear amplifier?
open loop op-amp configurations
Medium
A.Its bandwidth is too wide
B.It produces too much noise
C.It has low input impedance
D.Its voltage gain is highly unstable and leads to saturation
Correct Answer: Its voltage gain is highly unstable and leads to saturation
Explanation:
The open-loop gain of an op-amp is extremely high and varies with temperature and manufacturing. Even tiny noise or offset voltages will drive the output into saturation, making it unsuitable for linear amplification.
Incorrect! Try again.
33What is the primary reason the concept of 'virtual ground' is valid in an inverting amplifier with negative feedback?
op-amp with negative feedback
Medium
A.The output is shorted to the input
B.The input bias currents are zero
C.The feedback resistor is extremely large
D.The non-inverting terminal is grounded, and the infinite open-loop gain drives the differential input voltage to zero
Correct Answer: The non-inverting terminal is grounded, and the infinite open-loop gain drives the differential input voltage to zero
Explanation:
With negative feedback and infinite open-loop gain, the op-amp adjusts its output to keep the voltage difference between its input terminals at zero (). Since the non-inverting terminal is grounded, the inverting terminal is at 'virtual ground'.
Incorrect! Try again.
34If an op-amp feedback topology decreases both the input and output impedances, which topology is it?
block diagram representation of feedback configurations
Medium
A.Current-shunt
B.Voltage-series
C.Voltage-shunt
D.Current-series
Correct Answer: Voltage-shunt
Explanation:
Voltage sampling (output) decreases output impedance, and shunt mixing (input) decreases input impedance. Therefore, the voltage-shunt topology decreases both.
Incorrect! Try again.
35In a voltage follower (a special case of voltage-series feedback), what are the values of and ?
voltage series feedback amplifier
Medium
A.,
B.,
C.
D.,
Correct Answer: ,
Explanation:
A voltage follower has a direct short from output to the inverting input () and an open circuit to ground (), yielding a closed-loop gain of 1.
Incorrect! Try again.
36Determine the output voltage of an inverting amplifier with , , and an input voltage .
voltage shunt feedback amplifier
Medium
A.-10 V
B.12 V
C.10 V
D.-12 V
Correct Answer: 10 V
Explanation:
The gain is . The output voltage is .
Incorrect! Try again.
37A basic op-amp differential amplifier uses four resistors: and at the inverting input, and and at the non-inverting input. To ensure optimal common-mode rejection, what condition must the resistor values satisfy?
differential amplifiers
Medium
A.
B.
C.
D.
Correct Answer:
Explanation:
To balance the bridge and reject common-mode signals completely in a basic one-op-amp differential amplifier, the ratio of the feedback to input resistor on the inverting side must equal the ratio on the non-inverting side: .
Incorrect! Try again.
38On the ideal voltage transfer curve, the region between and is a vertical line. What does this vertical line represent mathematically?
ideal voltage transfer curve
Medium
A.Unity gain
B.Zero gain
C.Infinite input impedance
D.Infinite open-loop voltage gain
Correct Answer: Infinite open-loop voltage gain
Explanation:
The slope of the transfer curve represents the open-loop gain (). A vertical line indicates that an infinitesimally small input causes a full output swing, implying .
Incorrect! Try again.
39How does negative feedback affect the gain-bandwidth product (GBW) of an op-amp circuit?
op-amp with negative feedback
Medium
A.It increases the GBW
B.It decreases the GBW
C.It makes the GBW infinite
D.It leaves the GBW approximately constant
Correct Answer: It leaves the GBW approximately constant
Explanation:
For a typical internally compensated op-amp, the gain-bandwidth product is a constant. Negative feedback decreases the gain and increases the bandwidth by the same factor, keeping the product constant.
Incorrect! Try again.
40In a feedback amplifier block diagram, the term represents the feedback factor. For a non-inverting amplifier with resistors and , what is the expression for ?
block diagram representation of feedback configurations
Medium
A.
B.
C.
D.
Correct Answer:
Explanation:
The feedback network acts as a voltage divider sampling the output. The voltage fed back to the inverting terminal is . Thus, .
Incorrect! Try again.
41An ideal op-amp is connected in a circuit with infinite input impedance and zero output impedance. If a non-ideal parasitic capacitance is introduced at the inverting input node in a high-gain closed-loop configuration, what is the most significant consequence on the system dynamics?
The ideal op-amp
Hard
A.It increases the closed-loop bandwidth significantly.
B.It perfectly cancels the dominant pole of the op-amp.
C.It effectively increases the input bias current to infinity at DC.
D.It introduces a pole in the feedback loop, degrading phase margin and potentially causing instability.
Correct Answer: It introduces a pole in the feedback loop, degrading phase margin and potentially causing instability.
Explanation:
A parasitic capacitance at the inverting input interacts with the feedback resistors to create a pole in the feedback path ( network). This adds phase lag to the loop gain, degrading the phase margin and leading to peaking or instability.
Incorrect! Try again.
42In the equivalent circuit of a non-ideal op-amp, the finite open-loop gain is represented as a frequency-dependent voltage-controlled voltage source, . If this op-amp is used in a voltage series feedback topology with feedback factor , what is the exact expression for the closed-loop pole frequency ?
equivalent circuit of an opamp
Hard
A.
B.
C.
D.
Correct Answer:
Explanation:
The closed-loop gain is . Substituting , the denominator becomes . Thus, the new pole frequency is .
Incorrect! Try again.
43Consider an op-amp with an ideal voltage transfer curve bounded by . If a sinusoidal input is applied to an open-loop non-inverting configuration, and a small DC offset is present at the inverting terminal, what is the duty cycle of the resulting square wave output?
ideal voltage transfer curve
Hard
A.
B.
C.
D.
Correct Answer:
Explanation:
The output switches when . The switching angle is . The high time corresponds to radians out of . Thus, the duty cycle is .
Incorrect! Try again.
44An op-amp operating in an open-loop configuration has a finite CMRR. If the differential gain is and the common-mode gain is , which of the following represents the total output voltage given inputs and at the non-inverting and inverting terminals respectively, considering edge case input combinations?
open loop op-amp configurations
Hard
A.
B.
C.
D.
Correct Answer:
Explanation:
The total output is . Since , we can write . With and , the expression follows directly.
Incorrect! Try again.
45A negative feedback op-amp circuit exhibits a loop gain . To ensure a phase margin of exactly , a compensation capacitor is added that shifts the dominant pole. What must be the new dominant pole frequency approximately, assuming the second pole remains at rad/s?
op-amp with negative feedback
Hard
A.$1$ rad/s
B.$100$ rad/s
C.$10$ rad/s
D. rad/s
Correct Answer: $10$ rad/s
Explanation:
For a phase margin, the unity-gain frequency of the loop gain must equal the second pole frequency, so rad/s. Assuming a single dominant pole roll-off of dB/decade, the gain crosses $0$ dB at . Thus, , yielding rad/s.
Incorrect! Try again.
46In the general block diagram of a feedback amplifier, if the forward network has a strong reverse transmission and the feedback network has a forward transmission , what is the exact generalized expression for the closed-loop gain using Blackman's theorem or signal flow graph analysis?
block diagram representation of feedback configurations
Hard
A.
B.
C.
D.
Correct Answer:
Explanation:
Using signal flow graph analysis, the forward paths from input to output are and . The loops are (forward through , reverse through ) and (forward through , reverse through ). Applying Mason's Gain Formula gives the exact closed-loop gain.
Incorrect! Try again.
47A voltage series feedback amplifier is constructed using an op-amp with finite open-loop gain , input resistance , and output resistance . The feedback network consists of and (where ). What is the exact exact closed-loop output resistance without assuming ?
voltage series feedback amplifier
Hard
A.
B.
C.
D.
Correct Answer:
Explanation:
To find , we apply a test voltage at the output and measure the current. The feedback network loads the output, changing the open-loop resistance to . The effective loop gain is modified by the voltage division at the input due to the finite , leading to the exact expression.
Incorrect! Try again.
48In a voltage shunt feedback topology (transimpedance amplifier) with feedback resistor , an op-amp has a finite gain-bandwidth product (GBW) and an input capacitance . To maximally flat the closed-loop frequency response (Butterworth response), a feedback capacitor is placed across . What is the required value of ?
voltage shunt feedback amplifier
Hard
A.
B.
C.
D.
Correct Answer:
Explanation:
For a maximally flat response in a transimpedance amplifier, the phase margin should be , which corresponds to a damping factor of . Solving the characteristic equation of the second-order system yields the optimum feedback capacitance .
Incorrect! Try again.
49A practical differential amplifier is built using a single op-amp with four resistors: and on the inverting side, and and on the non-inverting side. If resistor tolerances cause a worst-case mismatch of (where ), what is the approximate worst-case Common Mode Rejection Ratio (CMRR) of the circuit, assuming an ideal op-amp?
differential amplifiers
Hard
A.
B.
C.
D.
Correct Answer:
Explanation:
With a worst-case mismatch in the 4 resistors, the common-mode gain becomes approximately . Since , for small .
Incorrect! Try again.
50If an ideal op-amp is configured as a positive feedback Schmitt trigger with a reference voltage applied to the inverting terminal, and feedback resistors and at the non-inverting terminal, what are the exact upper () and lower () threshold voltages given saturation limits ?
The ideal op-amp
Hard
A.
B.
C.
D.
Correct Answer:
Explanation:
Assuming is connected between the non-inverting input and ground/input, and is the feedback resistor. Superposition at the non-inverting node gives . Equating to and solving for yields the thresholds depending on .
Incorrect! Try again.
51In the AC small-signal equivalent circuit of an op-amp, the input stage generates a noise voltage spectral density and noise current spectral density . In a non-inverting amplifier with source resistance and feedback resistors (to ground) and (feedback), what is the total equivalent input noise voltage spectral density ?
equivalent circuit of an opamp
Hard
A.
B.
C.
D.
Correct Answer:
Explanation:
The total input-referred noise includes the op-amp's voltage noise , the current noise flowing through the source resistance , the current noise flowing through the equivalent feedback resistance , and the Johnson noise of the resistors referred to the input.
Incorrect! Try again.
52An op-amp with an open-loop gain of and output saturation voltages of operates under open-loop conditions. A noisy DC signal is applied differentially. What is the maximum peak-to-peak amplitude of that will keep the op-amp strictly in the linear region of its voltage transfer curve?
ideal voltage transfer curve
Hard
A.
B.
C.
D.
Correct Answer:
Explanation:
The linear region bounds are . The DC offset is . The distance to the positive limit is . The distance to the negative limit is . The maximum symmetric noise swing without clipping is , which is a peak-to-peak of .
Incorrect! Try again.
53A real op-amp with a slew rate of is driven by a step input in an open-loop configuration. Due to internal architecture, the input differential pair exhibits a maximum tail current and the compensation capacitor is . Which fundamental relationship correctly bounds the open-loop large-signal transient behavior?
open loop op-amp configurations
Hard
A.
B.
C.
D.
Correct Answer:
Explanation:
In most standard op-amp architectures, the slew rate is limited by the rate at which the input stage can charge and discharge the internal compensation capacitor . The maximum current available is the tail current , so the maximum rate of change of voltage is .
Incorrect! Try again.
54In a negative feedback configuration, the sensitivity of the closed-loop gain to changes in the open-loop gain is given by . If , what must be the minimum feedback factor to guarantee that varies by no more than ?
op-amp with negative feedback
Hard
A.
B.
C.
D.
Correct Answer:
Explanation:
We require . Thus, , meaning . With minimum , , but with nominal , . The closest strictly correct bound under nominal is $0.0999$.
Incorrect! Try again.
55When modeling an inverting op-amp configuration using the standard feedback block diagram, the forward gain and feedback factor are defined to match two-port network parameters. For an op-amp with gain and feedback resistor and input resistor , what is the correct block diagram parameter (feedback ratio) if ?
block diagram representation of feedback configurations
Hard
A.
B.
C.
D.
Correct Answer:
Explanation:
In an inverting amplifier, the feedback network acts as a voltage divider between the output and the inverting input terminal. The feedback factor (or ) is the fraction of output voltage fed back to the input, which is determined by the voltage division .
Incorrect! Try again.
56A voltage series feedback circuit uses an op-amp with an open-loop gain that has two poles: rad/s and rad/s. If the low-frequency loop gain is , what is the maximum value of to prevent the system from exhibiting an underdamped step response (i.e., to maintain a damping factor )?
voltage series feedback amplifier
Hard
A.
B.
C.
D.
Correct Answer:
Explanation:
The closed-loop characteristic equation is . This expands to . For critical damping, . Since , .
Incorrect! Try again.
57In a voltage shunt feedback topology acting as a transimpedance amplifier, the source is a photodiode modeled as a current source in parallel with a large capacitance . The op-amp has an input resistance , open loop gain , and feedback resistor . What is the input impedance seen by the photodiode?
voltage shunt feedback amplifier
Hard
A.
B.
C.
D.
Correct Answer:
Explanation:
Using Miller's theorem or direct nodal analysis, the feedback resistor is reflected to the input as . This is in parallel with the op-amp's internal input resistance . The source capacitance is external and drives this impedance.
Incorrect! Try again.
58An instrumentation amplifier is composed of three op-amps. The first stage consists of two op-amps with a gain resistor and feedback resistors . The second stage is a standard differential amplifier with resistors and . If and have a high common-mode voltage , what is the common-mode voltage present at the outputs of the first stage op-amps?
differential amplifiers
Hard
A.
B.
C.
D.$0$
Correct Answer:
Explanation:
In an instrumentation amplifier, the first stage amplifies the differential signal by . However, the common-mode signal does not create a potential difference across , so no common-mode current flows. Consequently, the common-mode voltage is passed through with unity gain, leaving at the outputs of the first stage.
Incorrect! Try again.
59An ideal op-amp is configured as an integrator with a resistor and capacitor . If the input is (a Dirac delta impulse) and the initial capacitor voltage is zero, what is the output for ?
The ideal op-amp
Hard
A.
B.
C.
D.
Correct Answer:
Explanation:
The output of an ideal inverting integrator is given by . Integrating the Dirac delta function yields the unit step function . Thus, .
Incorrect! Try again.
60A non-inverting amplifier is designed for a closed-loop gain of $100$. The op-amp has a DC open-loop gain of , but exhibits a dominant pole at $10$ Hz and a second non-dominant pole at $1$ MHz. What is the approximate phase margin of the closed-loop amplifier?
op-amp with negative feedback
Hard
A.
B.
C.
D.
Correct Answer:
Explanation:
The GBW product is Hz. The closed-loop bandwidth is Hz. The loop gain crosses 0 dB at Hz. The phase contribution from the dominant pole at $10$ Hz is . The second pole is at Hz, which contributes a phase lag of . Total phase shift is . Phase margin = .