Q: In comparator terminology, what does the phrase “zero-level detector” specifically describe?

Introduction: Comparators are often named for their reference level: a “zero-level detector” is a special case used for sign detection, zero-crossing timing, and frequency measurement. Recognizing this terminology is important when reading datasheets and schematics in instrumentation and signal conditioning. Given Data / Assumptions: Comparator compares input v_in to a fixed reference V_ref. “Zero-level” implies V_ref = 0 V (ground). Idealized output toggles between two logic or saturation levels. Concept / Approach: When V_ref = 0 V, the comparator indicates the sign of v_in. Positive portions of the waveform produce a logic “high,” negative portions produce a logic “low.” This is the basis of zero-crossing detectors used in phase-locked loops, tachometers, and waveform timing. Step-by-Step Solution: Set V_ref = 0 V at the comparator’s reference input.Feed an AC or varying signal to the other input.Whenever v_in ≥ 0, output goes to the high state; when v_in < 0, it goes low.Thus, the device detects crossings of the zero level (sign changes). Verification / Alternative check: Bench tests with a sine wave show output transitions precisely at the instants the input crosses 0 V (neglecting offsets and delays), confirming the “zero-level detector” function. Why Other Options Are Wrong: Sine-wave output: Comparators produce saturated two-level outputs. Peak detector: Captures amplitude peaks using a diode and capacitor, not sign. Limiter: Clamps amplitude but does not compare to a fixed reference in this sense. Sample-and-hold: Temporarily stores analog levels; unrelated to zero-cross detection. Common Pitfalls: Confusing zero-crossing detectors with Schmitt triggers; hysteresis adds noise immunity but shifts trip points away from exact zero. Ignoring input offset voltages that move the effective zero slightly. Final Answer: comparator with a trip point referenced to zero

Q: Capacitor current–voltage relationship: If a given capacitor experiences a voltage that changes at a constant rate with respect to time (dv/dt = constant), what happens to the current through the capacitor?

Introduction: Many timing and waveform-shaping circuits depend on the fundamental law of capacitors relating current and the time rate of change of voltage. Understanding the qualitative behavior when dv/dt is constant is crucial for recognizing why integrators produce ramps and why differentiators produce spikes. Given Data / Assumptions: Ideal capacitor with capacitance C (no leakage or series resistance). Voltage across the capacitor varies such that dv/dt is constant. Steady operating conditions (no dielectric absorption effects considered). Concept / Approach: The governing equation is i_C = C * dv/dt. If dv/dt is constant, multiplying by the fixed C yields a constant current. This is exactly what happens in an op-amp integrator receiving a constant input level—the capacitor current is constant, so the capacitor voltage changes linearly with time. Step-by-Step Solution: Start with i_C = C * dv/dt.Let dv/dt = k (constant). Then i_C = C * k (a constant).Therefore, the capacitor current neither increases nor decreases; it remains constant while dv/dt remains constant.If the sign of dv/dt changes (slope reversal), the current changes sign but its magnitude remains constant for the same |dv/dt|. Verification / Alternative check: In a triangular-wave generator, the integrator sees a constant voltage level, producing a constant capacitor current and therefore a linear (constant-slope) ramp—direct experimental confirmation of the rule. Why Other Options Are Wrong: Increase/decrease/decrease logarithmically: Would require dv/dt itself to change with time. Oscillate sinusoidally: That would imply a sinusoidal dv/dt, not a constant. Common Pitfalls: Confusing constant voltage (dv/dt = 0, giving zero current) with constant slope (nonzero current). Overlooking series resistance that slightly distorts the ideal constant-current condition in real circuits. Final Answer: be constant

Q: For a band-pass filter, the center (resonant) frequency f0 is always equal to which of the following, in terms of the lower and upper −3 dB cutoff frequencies f1 and f2?

Introduction: In narrow-band band-pass design and analysis, relating the center frequency to the cutoff points is essential for tuning, measuring, and specifying filters. The correct relationship uses a geometric mean, not an arithmetic mean, reflecting how frequency responses scale multiplicatively. Given Data / Assumptions: Lower cutoff (−3 dB) frequency: f1. Upper cutoff (−3 dB) frequency: f2. Narrow-band assumption generally applies, but the identity holds broadly for standard definitions. Concept / Approach: The center (resonant) frequency of a band-pass filter satisfies f0 = sqrt(f1 * f2). This arises from the symmetry of logarithmic frequency scaling and is consistent with constant-Q filter behavior where the passband is centered on a logarithmic axis. Step-by-Step Solution: Define bandwidth: BW = f2 − f1 and quality factor Q = f0 / BW.For many common band-pass implementations, the midband gain occurs near f0 where the reactive components balance.Using filter theory, the −3 dB points satisfy f0^2 = f1 * f2 ⇒ f0 = sqrt(f1 * f2).Therefore, the correct relationship is the geometric mean of the cutoff frequencies. Verification / Alternative check: On a Bode magnitude plot versus log frequency, f0 lies midway between f1 and f2 on the log axis, which mathematically corresponds to the geometric mean. Measured data from RLC band-pass circuits confirms the identity within approximation errors for real components. Why Other Options Are Wrong: Bandwidth or bandwidth/Q: Related parameters, but not equal to f0. −3 dB frequency: There are two such frequencies (f1 and f2); f0 is not itself a −3 dB point. Arithmetic average: Only correct for perfectly symmetric responses on a linear frequency axis, which standard band-pass filters are not. Common Pitfalls: Using arithmetic mean, which misplaces f0 for wide or even moderate bandwidths. Confusing midband gain frequency with one of the cutoff points. Final Answer: geometric average of the critical frequencies

Q: In an inverting op-amp averaging amplifier, if the feedback resistor Rf equals the value of each input resistor divided by the number of inputs (Rf = Rin/N), what is the resulting relationship between output and the individual input voltages?

Introduction: Summing and averaging amplifiers are standard op-amp applications. By choosing resistor ratios appropriately, the same circuit can output a weighted sum, a true average, or a scaled/inverted version of either. This question checks if you can map the resistor relationship onto the mathematical function realized. Given Data / Assumptions: Inverting summing topology with N inputs, each through Rin to the inverting node. Feedback resistor Rf set to Rin/N. Ideal op-amp, linear region (negative feedback), non-inverting input at ground/reference. Concept / Approach: For the inverting summer, Vout = −Rf * Σ (Vi / Rin). With Rf = Rin/N, this becomes Vout = −(1/N) * Σ Vi, which is the inverted average of the inputs. If a non-inverted average is desired, a second inverting stage (gain −1) can be added. Step-by-Step Solution: Start from KCL at virtual ground: Σ (Vi − 0)/Rin = (0 − Vout)/Rf.Rearrange: Vout = −Rf * Σ (Vi / Rin).Substitute Rf = Rin/N: Vout = −(Rin/N) * Σ (Vi / Rin) = −(1/N) * Σ Vi.Interpretation: output equals negative of the arithmetic mean of the inputs. Verification / Alternative check: Example with N = 4, inputs {1, 2, 3, 4} V gives Σ Vi = 10 V, average = 2.5 V. The circuit outputs −2.5 V, matching the formula. A unity-gain inverter afterward produces +2.5 V if a non-inverted average is required. Why Other Options Are Wrong: Average (non-inverted) / sums: sign or scaling does not match given Rf. Inverted sum: would require Rf = Rin (not divided by N). RMS: Requires squaring, averaging, and square-root operations, not a simple linear op-amp network. Common Pitfalls: Forgetting the minus sign associated with the inverting input. Using unequal Rin values unintentionally, which creates a weighted (not simple) average. Final Answer: the inverted average of the individual inputs

Q: Averaging amplifier design rule — In a multi-input op-amp averaging amplifier, how should the input resistances compare to the feedback resistance to obtain a true arithmetic average at the output?

Introduction: An averaging amplifier outputs the arithmetic mean of several input voltages. It is implemented as a special case of the inverting summing amplifier, with resistor choices that scale the sum by 1/N (N = number of inputs) so the result equals the average rather than the simple sum. Given Data / Assumptions: N input channels, each through an equal resistor Rin to the inverting node. Feedback resistor Rf from output to the inverting node. Noninverting input grounded; op-amp operates linearly. Concept / Approach: The inverting summer gives Vout = −(Rf / Rin) * Σ Vi. To obtain the average, you require Vout = −(1/N) * Σ Vi, which is achieved by choosing Rf = Rin / N. Therefore, Rin must be N times larger than Rf, meaning each input resistance is greater than the feedback resistance whenever N > 1. Step-by-Step Solution: Start from Vout = −(Rf / Rin) * Σ Vi.Set desired gain: Vout = −(1/N) * Σ Vi.Equate factors: Rf / Rin = 1 / N → Rf = Rin / N.Thus Rin = N * Rf → Rin > Rf for N ≥ 2. Verification / Alternative check: Example: N = 4, choose Rin = 40 kΩ, then Rf = 10 kΩ. Evaluate with any four inputs; the output equals the negative of their average, confirming the design rule. Why Other Options Are Wrong: equal to the feedback resistance: Produces a sum (not an average) when all Rin are equal. less than the feedback resistance: Would yield magnitude greater than an average. unequal / randomly chosen: Breaks symmetry and corrupts averaging unless precisely compensated. Common Pitfalls: Forgetting the negative sign in the inverting configuration; a noninverting averaging stage needs a different topology. Mixing Rin values; mismatch skews weighting and degrades accuracy. Final Answer: greater than the feedback resistance

Question 1

In a practical triangular-wave generator built with op-amps, which block must follow an op-amp comparator so that the comparator’s two-level output is converted into a linear ramp (triangular) waveform?

Accepted Answer

Introduction: A common way to generate a triangular wave in analog electronics is to use a fast switching stage that produces a two-level output, and then a block that converts those levels into a linear ramp. This question checks whether you know the classic comparator-plus-integrator topology used in many waveform generators and function generators. Given Data / Assumptions: The first stage is an op-amp comparator (often with hysteresis like a Schmitt trigger) producing a symmetrical high/low output. The second stage should transform a constant high or low level into a steadily changing voltage. Idealized op-amp behavior for conceptual understanding. Concept / Approach: An integrator outputs the time integral of its input. When the comparator output is at a fixed positive level, the integrator output changes linearly with negative or positive slope depending on polarity. When the comparator flips to the opposite level, the slope reverses. Cascading them yields a continuous up-down ramp, i.e., a triangular wave. Step-by-Step Solution: Use a comparator to produce a two-state (±V) signal as the control input.Feed that signal into an op-amp integrator with a resistor at the input and a capacitor in the feedback path.For a constant input ±V, the integrator output changes at a constant rate dv_o/dt = ±V/(R*C).The output ramps until a threshold in the comparator is reached; the comparator toggles and the slope reverses.The repeating reversal generates a triangular waveform at the integrator output. Verification / Alternative check: From i = C * dv/dt and i = (V_in − V_node)/R for an inverting integrator, a constant V_in produces a constant dv/dt, ensuring linear ramps. Practical circuits add hysteresis in the comparator to set the amplitude and frequency via R, C, and thresholds. Why Other Options Are Wrong: Differentiator: Produces spikes from edges, not ramps from levels. Amplifier: Only scales amplitude; does not create time integration. Multivibrator: A general oscillator class; in this chain, the necessary function is integration. Schmitt trigger: Often part of the comparator stage, not the ramp-forming stage. Common Pitfalls: Confusing differentiators and integrators because both use RC networks with op-amps. Omitting hysteresis in the comparator, which can cause jitter or non-oscillation. Ignoring slew-rate and saturation limits that distort the triangular waveform at high frequencies. Final Answer: integrator

Question 2

For a two-pole (second-order) active high-pass filter stage, what is the magnitude of its asymptotic roll-off rate in the stopband?

Accepted Answer

Introduction: Roll-off rate describes how quickly a filter attenuates frequencies in the stopband. Recognizing roll-off versus filter order is fundamental when selecting between first- and second-order stages to meet attenuation targets at a given frequency offset from the cutoff. Given Data / Assumptions: Two-pole (second-order) active high-pass filter. Asymptotic stopband slope is requested (far from cutoff). Standard definitions: first order → 20 dB/decade; each additional pole adds another 20 dB/decade. Concept / Approach: A filter’s order N gives the stopband slope magnitude ≈ 20 * N dB/decade. Thus, a second-order stage yields 40 dB/decade. For a high-pass, the stopband lies below the cutoff; the magnitude slope is the same as for a low-pass above cutoff, differing only in frequency region and sign convention. Step-by-Step Solution: Identify order: N = 2 (two poles).Compute slope magnitude: 20 * N = 40 dB/decade.Interpretation: every 10× decrease in frequency within the stopband yields ≈ 40 dB more attenuation (asymptotically).Near cutoff, the slope is gentler; the 40 dB/dec holds well in the far stopband. Verification / Alternative check: Compare with a first-order RC high-pass which gives ~20 dB/decade in the stopband; cascading two such sections (or using a biquad) doubles the slope to ~40 dB/decade, matching second-order theory. Why Other Options Are Wrong: −40 dB/decade: Same magnitude but sign; many texts quote magnitude only. The question asks for the magnitude. 20/−20 dB/decade: First-order behavior, not second order. 60 dB/decade: Would require a third-order (three-pole) filter. Common Pitfalls: Confusing sign convention (increase vs decrease) with magnitude; design typically uses magnitude. Assuming the exact slope at cutoff equals the asymptotic slope; it does not. Final Answer: 40 dB/decade

Question 3

For reliable start-up of a Wien-bridge sine-wave oscillator, what should the initial closed-loop gain A_cl of the amplifier be relative to the ideal value of 3 required by the feedback network?

Accepted Answer

Introduction: The Wien-bridge oscillator is prized for low-distortion sine waves. It relies on a frequency-selective feedback network that requires a loop gain slightly above unity at start-up to initiate oscillation, then near-exactly unity to sustain a constant amplitude. Understanding how A_cl relates to the required loop gain is key to practical design. Given Data / Assumptions: Wien bridge RC network demands an amplifier gain of about 3 for unity loop gain at the oscillation frequency. Start-up must overcome losses and component tolerances. Amplitude stabilization (e.g., lamp, JFET, diodes) will later reduce gain toward the ideal value. Concept / Approach: To start oscillation, the Barkhausen magnitude condition must be slightly exceeded (loop gain > 1). In a Wien-bridge, that equates to setting A_cl a bit greater than 3 initially. A stabilization element then automatically lowers A_cl toward 3 as the output grows, maintaining steady amplitude with low distortion. Step-by-Step Solution: Set initial A_cl slightly > 3 to ensure loop gain > 1.As the output amplitude rises, a nonlinear element (lamp resistance increase, JFET resistance change, diode limiter) reduces effective gain.The loop gain converges to 1 at steady state (A_cl ≈ 3).Result: stable sinusoidal oscillation at the design frequency. Verification / Alternative check: Classic Hewlett-Packard designs used a small incandescent lamp in the feedback path; its resistance rises with temperature, automatically trimming the gain from >3 at start-up to ≈3 during steady operation—experimental proof of the principle. Why Other Options Are Wrong: A_cl < 3 or ≅ 1: Loop gain below unity prevents start-up. 0: No amplification, no oscillation. 3/2: Far too low to start. Common Pitfalls: Setting A_cl exactly 3 without any start-up margin; unit tolerances can suppress oscillation. Excessive initial gain causing heavy clipping and distortion if stabilization is too slow. Final Answer: A_cl > 3

Question 4

If a pure sine-wave signal is applied to the input of an ideal voltage comparator (with no hysteresis), what general waveform shape appears at the comparator’s output?

Accepted Answer

Introduction: A comparator is a decision-making device: it compares an input voltage to a reference and switches its output to one of two saturation levels depending on the sign of the difference. This question checks whether you recognize that “thresholding” a sine produces a two-level output (a rectangular/square wave). Given Data / Assumptions: Ideal comparator, very high gain, rail-to-rail saturation. Reference at 0 V (or any fixed V_ref); input is a clean sine wave that crosses the reference each cycle. No hysteresis unless stated; edges are idealized. Concept / Approach: Whenever the input exceeds the reference, the output saturates “high”; when the input falls below the reference, it saturates “low.” As a result, the output toggles between two constant levels, forming a rectangular waveform whose duty cycle depends on the reference level and input symmetry. Step-by-Step Solution: Apply sine wave v_in(t) to the comparator with threshold V_ref.For v_in > V_ref, output → +V_sat; for v_in < V_ref, output → −V_sat.Crossings occur twice per cycle; the output is high for the portion of the cycle above the threshold and low otherwise.Thus, the waveform is rectangular (often called square if 50% duty cycle). Verification / Alternative check: Oscilloscope tests in lab confirm that thresholding a sine produces fast transitions to fixed levels. Adding hysteresis (Schmitt trigger) improves noise immunity but retains the rectangular nature. Why Other Options Are Wrong: Ramp/sawtooth/triangle: Produced by integrators or ramp generators, not by simple thresholding. Sine wave: That is the input, not the comparator’s two-level output. Common Pitfalls: Expecting a “perfect square”; real comparators have finite slew and propagation delay. Ignoring the effect of V_ref offset, which skews duty cycle away from 50%. Final Answer: rectangular wave

Question 5

In comparator terminology, what does the phrase “zero-level detector” specifically describe?

Accepted Answer

Introduction: Comparators are often named for their reference level: a “zero-level detector” is a special case used for sign detection, zero-crossing timing, and frequency measurement. Recognizing this terminology is important when reading datasheets and schematics in instrumentation and signal conditioning. Given Data / Assumptions: Comparator compares input v_in to a fixed reference V_ref. “Zero-level” implies V_ref = 0 V (ground). Idealized output toggles between two logic or saturation levels. Concept / Approach: When V_ref = 0 V, the comparator indicates the sign of v_in. Positive portions of the waveform produce a logic “high,” negative portions produce a logic “low.” This is the basis of zero-crossing detectors used in phase-locked loops, tachometers, and waveform timing. Step-by-Step Solution: Set V_ref = 0 V at the comparator’s reference input.Feed an AC or varying signal to the other input.Whenever v_in ≥ 0, output goes to the high state; when v_in < 0, it goes low.Thus, the device detects crossings of the zero level (sign changes). Verification / Alternative check: Bench tests with a sine wave show output transitions precisely at the instants the input crosses 0 V (neglecting offsets and delays), confirming the “zero-level detector” function. Why Other Options Are Wrong: Sine-wave output: Comparators produce saturated two-level outputs. Peak detector: Captures amplitude peaks using a diode and capacitor, not sign. Limiter: Clamps amplitude but does not compare to a fixed reference in this sense. Sample-and-hold: Temporarily stores analog levels; unrelated to zero-cross detection. Common Pitfalls: Confusing zero-crossing detectors with Schmitt triggers; hysteresis adds noise immunity but shifts trip points away from exact zero. Ignoring input offset voltages that move the effective zero slightly. Final Answer: comparator with a trip point referenced to zero

Question 6

Capacitor current–voltage relationship: If a given capacitor experiences a voltage that changes at a constant rate with respect to time (dv/dt = constant), what happens to the current through the capacitor?

Accepted Answer

Introduction: Many timing and waveform-shaping circuits depend on the fundamental law of capacitors relating current and the time rate of change of voltage. Understanding the qualitative behavior when dv/dt is constant is crucial for recognizing why integrators produce ramps and why differentiators produce spikes. Given Data / Assumptions: Ideal capacitor with capacitance C (no leakage or series resistance). Voltage across the capacitor varies such that dv/dt is constant. Steady operating conditions (no dielectric absorption effects considered). Concept / Approach: The governing equation is i_C = C * dv/dt. If dv/dt is constant, multiplying by the fixed C yields a constant current. This is exactly what happens in an op-amp integrator receiving a constant input level—the capacitor current is constant, so the capacitor voltage changes linearly with time. Step-by-Step Solution: Start with i_C = C * dv/dt.Let dv/dt = k (constant). Then i_C = C * k (a constant).Therefore, the capacitor current neither increases nor decreases; it remains constant while dv/dt remains constant.If the sign of dv/dt changes (slope reversal), the current changes sign but its magnitude remains constant for the same |dv/dt|. Verification / Alternative check: In a triangular-wave generator, the integrator sees a constant voltage level, producing a constant capacitor current and therefore a linear (constant-slope) ramp—direct experimental confirmation of the rule. Why Other Options Are Wrong: Increase/decrease/decrease logarithmically: Would require dv/dt itself to change with time. Oscillate sinusoidally: That would imply a sinusoidal dv/dt, not a constant. Common Pitfalls: Confusing constant voltage (dv/dt = 0, giving zero current) with constant slope (nonzero current). Overlooking series resistance that slightly distorts the ideal constant-current condition in real circuits. Final Answer: be constant

Question 7

For a band-pass filter, the center (resonant) frequency f0 is always equal to which of the following, in terms of the lower and upper −3 dB cutoff frequencies f1 and f2?

Accepted Answer

Introduction: In narrow-band band-pass design and analysis, relating the center frequency to the cutoff points is essential for tuning, measuring, and specifying filters. The correct relationship uses a geometric mean, not an arithmetic mean, reflecting how frequency responses scale multiplicatively. Given Data / Assumptions: Lower cutoff (−3 dB) frequency: f1. Upper cutoff (−3 dB) frequency: f2. Narrow-band assumption generally applies, but the identity holds broadly for standard definitions. Concept / Approach: The center (resonant) frequency of a band-pass filter satisfies f0 = sqrt(f1 * f2). This arises from the symmetry of logarithmic frequency scaling and is consistent with constant-Q filter behavior where the passband is centered on a logarithmic axis. Step-by-Step Solution: Define bandwidth: BW = f2 − f1 and quality factor Q = f0 / BW.For many common band-pass implementations, the midband gain occurs near f0 where the reactive components balance.Using filter theory, the −3 dB points satisfy f0^2 = f1 * f2 ⇒ f0 = sqrt(f1 * f2).Therefore, the correct relationship is the geometric mean of the cutoff frequencies. Verification / Alternative check: On a Bode magnitude plot versus log frequency, f0 lies midway between f1 and f2 on the log axis, which mathematically corresponds to the geometric mean. Measured data from RLC band-pass circuits confirms the identity within approximation errors for real components. Why Other Options Are Wrong: Bandwidth or bandwidth/Q: Related parameters, but not equal to f0. −3 dB frequency: There are two such frequencies (f1 and f2); f0 is not itself a −3 dB point. Arithmetic average: Only correct for perfectly symmetric responses on a linear frequency axis, which standard band-pass filters are not. Common Pitfalls: Using arithmetic mean, which misplaces f0 for wide or even moderate bandwidths. Confusing midband gain frequency with one of the cutoff points. Final Answer: geometric average of the critical frequencies

Question 8

In an inverting op-amp averaging amplifier, if the feedback resistor Rf equals the value of each input resistor divided by the number of inputs (Rf = Rin/N), what is the resulting relationship between output and the individual input voltages?

Accepted Answer

Introduction: Summing and averaging amplifiers are standard op-amp applications. By choosing resistor ratios appropriately, the same circuit can output a weighted sum, a true average, or a scaled/inverted version of either. This question checks if you can map the resistor relationship onto the mathematical function realized. Given Data / Assumptions: Inverting summing topology with N inputs, each through Rin to the inverting node. Feedback resistor Rf set to Rin/N. Ideal op-amp, linear region (negative feedback), non-inverting input at ground/reference. Concept / Approach: For the inverting summer, Vout = −Rf * Σ (Vi / Rin). With Rf = Rin/N, this becomes Vout = −(1/N) * Σ Vi, which is the inverted average of the inputs. If a non-inverted average is desired, a second inverting stage (gain −1) can be added. Step-by-Step Solution: Start from KCL at virtual ground: Σ (Vi − 0)/Rin = (0 − Vout)/Rf.Rearrange: Vout = −Rf * Σ (Vi / Rin).Substitute Rf = Rin/N: Vout = −(Rin/N) * Σ (Vi / Rin) = −(1/N) * Σ Vi.Interpretation: output equals negative of the arithmetic mean of the inputs. Verification / Alternative check: Example with N = 4, inputs {1, 2, 3, 4} V gives Σ Vi = 10 V, average = 2.5 V. The circuit outputs −2.5 V, matching the formula. A unity-gain inverter afterward produces +2.5 V if a non-inverted average is required. Why Other Options Are Wrong: Average (non-inverted) / sums: sign or scaling does not match given Rf. Inverted sum: would require Rf = Rin (not divided by N). RMS: Requires squaring, averaging, and square-root operations, not a simple linear op-amp network. Common Pitfalls: Forgetting the minus sign associated with the inverting input. Using unequal Rin values unintentionally, which creates a weighted (not simple) average. Final Answer: the inverted average of the individual inputs

Question 9

From a circuit-classification standpoint, a voltage comparator (which switches between two saturation levels based on input polarity) is best described as which type of circuit?

Accepted Answer

Introduction: Comparators are ubiquitous building blocks used to make binary decisions from analog signals. Because their output snaps between two levels depending on input sign relative to a threshold, their behavior is inherently nonlinear, unlike linear amplifiers that preserve proportionality and superposition over a range. Given Data / Assumptions: Ideal comparator with very high open-loop gain. Operates in saturation (two-level output), not in the op-amp’s linear region with feedback. Reference threshold fixed; input is arbitrary analog signal. Concept / Approach: Linearity requires superposition: if input x1 produces y1 and x2 produces y2, then a * x1 + b * x2 should produce a * y1 + b * y2. A comparator violates this because small input changes around the threshold cause large discontinuous output changes; the transfer characteristic is a hard nonlinearity (sign function with saturation limits). Step-by-Step Solution: Examine transfer: y = +V_sat for v_in > V_ref; y = −V_sat for v_in < V_ref.This mapping is not proportional (gain depends on input region) and is discontinuous at v_in = V_ref.Therefore, the device is nonlinear by definition.Linear behavior would require y = k * x over an operating range with constant k, which a comparator does not provide. Verification / Alternative check: Plotting the comparator’s characteristic shows a steep vertical transition at the threshold with flat saturated regions—classic nonlinearity. In contrast, an op-amp with negative feedback configured as an amplifier has a near-linear region with limited gain and bandwidth. Why Other Options Are Wrong: Active filter / resonant network: Frequency-selective linear networks, not decision devices. Current source: A source element, not a comparator function. Linear circuit: Contradicts the discontinuous transfer and lack of superposition. Common Pitfalls: Assuming all op-amp-based circuits are linear; comparators deliberately avoid linear operation. Confusing Schmitt triggers (with hysteresis) for linear amplifiers; they remain nonlinear. Final Answer: nonlinear circuit

Question 10

Op-amp applications — A digital-to-analog converter (DAC) can be implemented using which classic operational-amplifier building block, where weighted digital inputs are summed to produce a proportional analog output voltage?

Accepted Answer

Introduction: A digital-to-analog converter (DAC) transforms an N-bit digital word into a proportional analog quantity. One of the most common analog circuit realizations for a voltage-output DAC is based on the op-amp summing amplifier, often called a scaling adder, where each digital bit drives a weighted branch whose contribution is summed at the op-amp input. Given Data / Assumptions: Goal: convert a digital word into an analog voltage. Available op-amp topologies: summing (scaling) adder, noninverting amplifier, voltage-to-current converter, etc. Binary weighting can be implemented by resistor networks (R-2R ladder or binary-weighted resistors). Concept / Approach: A scaling adder sums currents from multiple input resistors into the op-amp’s inverting node (virtual ground). Each input branch is weighted so that its contribution is proportional to the digital bit’s significance. With an R-2R ladder or binary-weighted resistors, the summed current becomes an output voltage via the op-amp’s feedback resistor, creating a precise, linear mapping from code to voltage. Step-by-Step Solution: Represent each digital bit (b_k ∈ {0,1}) as a switch feeding a weighted resistor.Currents add at the inverting node: Itotal = Σ (Vref * b_k / Rk).Op-amp converts current to voltage: Vout = −Rf * Itotal.Choose Rk so each bit contributes 2^k weighting (or use R-2R network); linear DAC achieved. Verification / Alternative check: R-2R ladders are industry-standard DAC cores. Their front end is exactly a current summing (scaling) network terminated at an op-amp that performs I-to-V conversion, i.e., a scaling adder behaviorally. Why Other Options Are Wrong: voltage-to-current converter: Often part of DACs but alone does not implement code-weighted summation. noninverting amplifier: Provides gain to a single analog input, not code-weighted summation. adjustable bandwidth circuit: A filter concept, not a DAC core. instrumentation amplifier: Precision differential gain, unrelated to code summation. Common Pitfalls: Confusing the op-amp I-to-V stage with the entire DAC function; the essential feature is weighted summation. Ignoring resistor matching and reference stability, which set DAC linearity and accuracy. Final Answer: scaling adder

Question 11

Operational-amplifier integrator behavior — The ramp (output) voltage of an ideal op-amp integrator under a constant input signal will:

Accepted Answer

Introduction: An op-amp integrator converts an input voltage into the time integral at its output. With a constant (DC) input, the output of an ideal integrator is a linearly changing ramp. Recognizing this linear relationship is essential when designing ramps, function generators, and control systems that rely on integrator dynamics. Given Data / Assumptions: Ideal op-amp with a feedback capacitor Cf and input resistor Rin. Constant input voltage (or piecewise constant segments). Linear region of operation (no saturation of the output rails). Concept / Approach: For an inverting integrator, the fundamental relation is Vout(t) = −(1 / (Rin * Cf)) * ∫ Vin dt + Vout(t0). When Vin is constant, the integral is Vin * t, producing a linear ramp with slope proportional to Vin and inversely proportional to Rin * Cf. The sign is inverted for the classic inverting topology. Step-by-Step Solution: Write integrator law: d(Vout)/dt = −Vin / (Rin * Cf).For Vin = V0 (constant), slope m = −V0 / (Rin * Cf) → linear in time.If V0 > 0, output ramps negative; if V0 < 0, output ramps positive.Changing Vin stepwise changes the slope piecewise, still linear between changes. Verification / Alternative check: Bench observation with a square-wave input shows triangular (piecewise linear) output—direct evidence of linear rate change. Simulation (SPICE) reproduces the same slope proportionality to Vin and 1/(Rin * Cf). Why Other Options Are Wrong: exponentially: Exponential arises in RC passive charging, not ideal op-amp integration under DC input. always increasing: Sign depends on Vin; negative inputs produce rising ramps. constant: Only when Vin = 0, otherwise a slope exists. oscillate sinusoidally: Requires additional dynamics; not a property of a simple integrator. Common Pitfalls: Forgetting saturation limits; real outputs clip when rails are reached. Ignoring bias currents and offsets that can cause unintended drift (pseudo-ramp) at Vin ≈ 0. Final Answer: increase or decrease at a linear rate

Question 12

Averaging amplifier design rule — In a multi-input op-amp averaging amplifier, how should the input resistances compare to the feedback resistance to obtain a true arithmetic average at the output?

Accepted Answer

Introduction: An averaging amplifier outputs the arithmetic mean of several input voltages. It is implemented as a special case of the inverting summing amplifier, with resistor choices that scale the sum by 1/N (N = number of inputs) so the result equals the average rather than the simple sum. Given Data / Assumptions: N input channels, each through an equal resistor Rin to the inverting node. Feedback resistor Rf from output to the inverting node. Noninverting input grounded; op-amp operates linearly. Concept / Approach: The inverting summer gives Vout = −(Rf / Rin) * Σ Vi. To obtain the average, you require Vout = −(1/N) * Σ Vi, which is achieved by choosing Rf = Rin / N. Therefore, Rin must be N times larger than Rf, meaning each input resistance is greater than the feedback resistance whenever N > 1. Step-by-Step Solution: Start from Vout = −(Rf / Rin) * Σ Vi.Set desired gain: Vout = −(1/N) * Σ Vi.Equate factors: Rf / Rin = 1 / N → Rf = Rin / N.Thus Rin = N * Rf → Rin > Rf for N ≥ 2. Verification / Alternative check: Example: N = 4, choose Rin = 40 kΩ, then Rf = 10 kΩ. Evaluate with any four inputs; the output equals the negative of their average, confirming the design rule. Why Other Options Are Wrong: equal to the feedback resistance: Produces a sum (not an average) when all Rin are equal. less than the feedback resistance: Would yield magnitude greater than an average. unequal / randomly chosen: Breaks symmetry and corrupts averaging unless precisely compensated. Common Pitfalls: Forgetting the negative sign in the inverting configuration; a noninverting averaging stage needs a different topology. Mixing Rin values; mismatch skews weighting and degrades accuracy. Final Answer: greater than the feedback resistance

Question 13

Series voltage regulator essentials — A basic series (linear) regulator must include which internal functional blocks to maintain a stable output?

Accepted Answer

Introduction: Series (linear) regulators keep the output voltage constant by comparing a scaled sample of the output to a stable reference and adjusting a pass element to correct any error. Identifying the mandatory functional blocks clarifies how regulation is achieved and what distinguishes a regulator from a simple follower or resistor dropper. Given Data / Assumptions: Regulator topology: series pass element between source and load. Feedback loop present. Goal: hold Vout steady against line and load changes. Concept / Approach: A precision reference (e.g., bandgap) provides a stable yardstick. An error detector (usually an op-amp/comparator) compares the sampled output to this reference and drives the pass device to minimize the error. The load is external and not a functional block of the regulator itself. Step-by-Step Solution: Sense: Sample Vout via divider → Vsense.Compare: Error = Vsense − Vref generated by the error amplifier.Control: Adjust pass transistor conduction to drive Error → 0.Stability: Compensation ensures loop stability over operating range. Verification / Alternative check: Block diagrams of common linear regulators (e.g., 78xx, LM317) show a reference and an error amplifier controlling the pass device—both are fundamental elements. Why Other Options Are Wrong: an error detector: Needed but incomplete without a reference. a load: External to the regulator; not an internal block. a reference voltage: Needed but alone cannot regulate without error detection and control. only a pass transistor: Lacks feedback; cannot regulate. Common Pitfalls: Confusing line regulation (input changes) with load regulation (output current changes) and thinking different blocks are needed; both are handled by the same loop. Ignoring loop compensation, which affects stability but is separate from the core functional requirement. Final Answer: both an error detector and a reference voltage

Question 14

When the capacitance in an active high-pass filter increases, the critical frequency decreases.

Accepted Answer

True

Question 15

The feedback component in an op-amp integrator is a capacitor.

Accepted Answer

True

Question 16

A zero-level detector is a type of comparator circuit.

Accepted Answer

True

Question 17

When used as a comparator, an op-amp is wired in a closed-loop configuration.

Accepted Answer

False

Question 18

To operate properly, an oscillator requires an external ac input signal.

Accepted Answer

False

Question 19

The output voltage of a summing amplifier is proportional to the product of the input voltages.

Accepted Answer

False

Question 20

The effectiveness of a filter in rejecting signals beyond cutoff frequency is a function of the filter roll-off rating.

Accepted Answer

True

Basic Op-Amp Circuits Questions