Question 1

Wien-bridge oscillator condition: For a Wien-bridge oscillator to produce a sustained sine-wave output, must the net phase shift around the positive-feedback loop be greater than 0° (i.e., nonzero)?

Accepted Answer

Introduction / Context: The Wien-bridge oscillator is a classic RC sine-wave source built around an operational amplifier with a frequency-selective feedback network. Its steady, low-distortion oscillation hinges on satisfying the Barkhausen criteria: a loop gain of unity at the oscillation frequency and a net loop phase shift of 0° (or an integer multiple of 360°). This item probes whether the required loop phase must be greater than 0°, which would contradict the fundamental condition for sustained oscillation. Given Data / Assumptions: Single-op-amp Wien-bridge topology with an RC lead-lag network in the positive feedback path. Linear, small-signal analysis at the intended oscillation frequency f0. Amplitude control (lamp/JFET/AGC) ensures loop gain settles to unity but does not alter the phase criterion. Concept / Approach: The Barkhausen phase condition requires the total phase shift around the closed loop to be 0° (mod 360°) at f0 so that the fed-back signal reinforces the input. In the Wien network, the RC bridge contributes 0° phase shift exactly at f0, while at other frequencies it leads or lags. The op-amp stage is arranged to provide non-inverting amplification so the overall phase is 0° at f0. Any net phase greater than 0° (i.e., a non-zero residual phase at f0) would reduce constructive reinforcement and prevent a steady, sustained sinusoid. Step-by-Step Solution: State Barkhausen: loop phase = 0° (mod 360°), loop gain magnitude = 1 at f0.At f0, the Wien RC lead-lag network has zero net phase shift.The amplifier path is configured non-inverting (0°) so loop phase remains 0°.Therefore, the claim “phase shift must be greater than 0°” is wrong; it must be 0°. Verification / Alternative check: Frequency response of the Wien network shows a flat phase of 0° at its center frequency, with positive/negative phase away from f0. Practical oscillators add an amplitude-control element (e.g., incandescent lamp or JFET) to trim loop gain to unity while preserving the 0° phase at f0. Why Other Options Are Wrong: “Only during start-up” misattributes phase; start-up requires loop gain > 1, not phase > 0°. LC tanks are not part of the Wien bridge. Transistor type is irrelevant to the phase condition at f0. Common Pitfalls: Confusing sign convention: “greater than 0°” would imply a nonzero phase error, violating Barkhausen. Also mixing amplitude criterion (gain control) with the independent phase criterion. Final Answer: Incorrect

Question 2

Active filter order vs roll-off: A two-pole (second-order) active filter exhibits a stopband roll-off of −20 dB/decade. Is this statement accurate?

Accepted Answer

Introduction / Context: Filter order determines asymptotic attenuation slope beyond the cutoff region. Each reactive pole contributes an additional −20 dB/decade (or −6 dB/octave) in the stopband. Understanding this principle is core to predicting selectivity and choosing between first- and second-order topologies in active RC designs such as Sallen–Key or multiple-feedback filters. Given Data / Assumptions: “Two-pole” means a second-order low-pass, high-pass, band-pass, or band-stop realized actively. Asymptotic stopband behavior is considered (well beyond cutoff). Standard linear, time-invariant response without special equalization. Concept / Approach: For a first-order (one-pole) filter, the far-stopband slope is −20 dB/decade. For a second-order (two-pole) filter, slopes add to −40 dB/decade. This is independent of whether the active form is Sallen–Key, multiple-feedback, or biquad. While passband ripple/peaking depends on Q or damping ratio, the asymptotic slope is set by order alone. Step-by-Step Solution: Identify order: two poles ⇒ second order.Each pole contributes −20 dB/decade in the stopband.Total asymptotic slope = −40 dB/decade.Therefore, the statement claiming −20 dB/decade is incorrect. Verification / Alternative check: Bode magnitude plots of standard 2nd-order Butterworth, Bessel, or Chebyshev sections all show a −40 dB/decade asymptote. Cascading two first-order stages demonstrates the same cumulative slope. Why Other Options Are Wrong: “True for unity-gain Sallen–Key” is false; gain does not change the asymptotic order. “Only at cutoff” confuses the local slope near the corner with the far-stopband asymptote. High-pass and low-pass of the same order share the same magnitude slope magnitude in their respective stopbands. Common Pitfalls: Equating peaking or damping with slope; mistaking the −3 dB point behavior for the asymptotic roll-off far from cutoff. Final Answer: Incorrect

Question 3

Shunt regulator architecture: In a basic zener shunt voltage regulator, is the zener diode the control element that actively modulates current to regulate the output?

Accepted Answer

Introduction / Context: Simple shunt regulators are widely taught because they illustrate the concept of clamping a node voltage by diverting excess current to ground. However, terminology matters: which component is the “control element”? Distinguishing the reference function from the current-modulating element prevents confusion when students progress to series regulators and op-amp-based regulators. Given Data / Assumptions: Classic zener shunt regulator: source → series resistor → output node; a zener diode from output to ground. Load connected in parallel with the zener at the regulated node. Zener operates in breakdown region providing a near-constant reference voltage Vz. Concept / Approach: In a shunt regulator, the element that modulates current to maintain the node voltage is the shunt path itself. In the simplest circuit, the zener both provides the reference voltage and sinks current changes, but it is not called the “control element” in the same sense as the series pass element in a series regulator. When a transistor is added (zener referenced, transistor shunt), the transistor clearly becomes the control element while the zener remains the reference. Even in the bare-bones circuit, the more precise language is: the zener acts as a voltage reference and a current sink; it does not “control” the series current— the series resistor sets that—and regulation happens by shunting excess current. Step-by-Step Solution: Model the zener as Vz with small dynamic resistance rz.Total current through the series resistor splits between load and shunt path.The zener clamps the node near Vz by sinking surplus current; it is a reference/sink, not a pass-element controller.Thus the statement that it is “the control element” is inaccurate in standard terminology. Verification / Alternative check: Data sheets specify zener “dynamic impedance” and knee current for proper regulation; series regulators, by contrast, specify pass-transistor control characteristics, highlighting the different roles. Why Other Options Are Wrong: Qualifiers about load level or knee current affect regulation quality, not the definition of which device is the control element. Adding a pass transistor changes topology; the original statement referred to a “basic shunt regulator.” Common Pitfalls: Calling the zener a “controller” rather than a reference/sink; ignoring the role of the series resistor that sets current available to be shunted. Final Answer: Incorrect

Question 4

Wien-bridge oscillator feedback type: Which type of feedback is fundamentally used to satisfy the Barkhausen criterion for oscillation in a Wien-bridge oscillator?

Accepted Answer

Introduction / Context: Oscillators are feedback amplifiers engineered so that, at one frequency, the fed-back signal reinforces the input and sustains a waveform without an external drive. The Wien-bridge oscillator uses an RC lead-lag network to feed a selected frequency back to the noninverting input of an amplifier. Understanding which kind of feedback creates the oscillation mechanism is crucial for correct design and troubleshooting. Given Data / Assumptions: Standard Wien-bridge RC network between output and noninverting input. Amplifier configured noninverting to avoid 180° phase inversion. Optional amplitude control (lamp/JFET) may add negative feedback for stabilization. Concept / Approach: To meet Barkhausen at the oscillation frequency f0, the loop must deliver a net phase of 0° and loop gain of 1. The role of the Wien network is to select the frequency and provide positive feedback of that component to the amplifier input. Additional circuitry often introduces slow, nonlinear negative feedback to regulate amplitude (so distortion remains low), but this auxiliary path does not create the oscillation—positive feedback does. Step-by-Step Solution: Identify the frequency-selective path: RC bridge feeds output back to + input.At f0 the bridge phase = 0°, enabling constructive reinforcement (positive feedback).Set loop gain ≈ 1 at f0 for sustained oscillation; higher at start-up for build-up.Amplitude control then reduces effective gain via negative feedback to hold distortion low. Verification / Alternative check: Practical designs (e.g., lamp-stabilized Hewlett–Packard 200 series) show noninverting gain ~3 with bridge attenuation 1/3, yielding loop gain ≈ 1 and stable sinusoidal output—an explicit positive-feedback arrangement. Why Other Options Are Wrong: “Negative feedback” alone would not sustain oscillations. “Both with equal roles” misstates the primary role: positive feedback sets oscillation; negative feedback only stabilizes amplitude. LC tank networks are unnecessary; Wien oscillators are RC based. Parametric feedback is not used here. Common Pitfalls: Confusing amplitude stabilization (negative feedback) with the oscillation criterion (positive feedback). Final Answer: Positive feedback

Question 5

Barkhausen amplitude criterion in practice: To sustain oscillations in a feedback oscillator (e.g., Wien-bridge), how should the loop gain be managed so that the closed-loop condition settles?

Accepted Answer

Introduction / Context: Oscillation start-up and steady-state behavior differ. Designers initially allow loop gain slightly above unity to grow a sinusoid from noise, then a control mechanism trims the gain to exactly unity at the oscillation frequency to prevent runaway or decay. This question clarifies that steady oscillation requires settling the effective closed-loop condition to unity loop gain. Given Data / Assumptions: Linear analysis around the oscillation frequency f0. Positive feedback provides the frequency-selective path. Automatic amplitude control (lamp/JFET/compressor) is available. Concept / Approach: The Barkhausen magnitude criterion states |Aβ| = 1 at f0 for sustained oscillation. Practical circuits start with |Aβ| > 1 to guarantee build-up, then introduce negative feedback or gain compression so that the effective loop gain is reduced to 1. If it remains > 1, the waveform clips; if < 1, oscillations die out. Therefore, “reduced, equals one” describes steady operation best. Step-by-Step Solution: Permit start-up with loop gain slightly > 1 to amplify noise.As amplitude rises, engage control element to lower amplifier gain.Settle at |Aβ| = 1 for a stable, low-distortion sine wave.Maintain 0° loop phase at f0 to satisfy the phase criterion. Verification / Alternative check: Classic lamp-stabilized Wien oscillators use the lamp’s resistance rise with temperature to decrease gain as amplitude increases, naturally converging on unity loop gain with low distortion. Similar behavior is achieved with JFET-based AGC or diode limiters with careful biasing. Why Other Options Are Wrong: “Reduced, less than one” kills oscillation; “increased, more/ much more than one” leads to runaway and clipped output; leaving gain uncontrolled invites severe distortion or saturation. Common Pitfalls: Thinking Barkhausen requires > 1 at all times; in fact, > 1 is only for start-up. Long-term, unity magnitude and 0° phase are required. Final Answer: Reduced, equals one

Question 6

Summing amplifier principle: Which property of the op-amp input node enables the standard inverting summing amplifier to add multiple input voltages linearly?

Accepted Answer

Introduction / Context: The inverting summing amplifier is a workhorse op-amp configuration that produces an output proportional to a weighted sum of several input voltages. The key enabler is the behavior of the inverting input node under negative feedback, often described by the “virtual ground” concept. Grasping this concept clarifies why currents from multiple inputs add algebraically at a single node without interaction. Given Data / Assumptions: Ideal op-amp with infinite open-loop gain, infinite input impedance, and zero output impedance. Negative feedback via a feedback resistor from output to the inverting input. Noninverting input tied to ground (reference 0 V). Concept / Approach: With large open-loop gain and negative feedback, the op-amp drives its output so that the differential input voltage approaches zero. If the noninverting input is at 0 V, the inverting input node is held at approximately 0 V as well, even though it is not physically connected to ground. This is the “virtual ground.” Because the inverting input draws negligible current, currents from each source resistor flow into the summing node and then through the feedback resistor, adding linearly by Kirchhoff’s Current Law. Step-by-Step Solution: Tie the noninverting input to 0 V reference.Negative feedback forces v− ≈ v+ = 0 V ⇒ virtual ground at the inverting node.Each input Vi through Ri contributes current Ii = Vi / Ri into the node.KCL: sum(Ii) = If = −Vo / Rf ⇒ Vo = −Rf * Σ(Vi / Ri). Verification / Alternative check: SPICE simulations show the inverting node clamped near 0 V while the output adjusts to satisfy KCL. Practical op-amps with finite gain still maintain the node within microvolts to millivolts of ground in linear operation. Why Other Options Are Wrong: Virtual ground at the noninverting input does not create the summing node. Low/high-voltage ratings are irrelevant to the summing mechanism. Positive feedback would break linear summation and risk oscillation. Common Pitfalls: Assuming significant current enters the op-amp input; forgetting that resistor values set weights; ignoring output swing limits that can cause clipping. Final Answer: Virtual ground at the inverting input

Question 7

Rate of change terminology in math & circuits: What is the mathematical operation that returns the instantaneous rate of change (slope) of a function or waveform?

Accepted Answer

Introduction / Context: Signal processing and analog computing frequently invoke basic calculus. Differentiation measures how quickly a quantity changes with respect to an independent variable (often time). In op-amp circuits, differentiators approximate this operation across a bandwidth, converting a ramp to a constant level and a constant to zero (ideally). Recognizing the correct terminology prevents confusion with integration, which performs the inverse accumulation process. Given Data / Assumptions: Standard definitions from calculus applied to continuous signals. Idealized operations; real circuits have finite bandwidths and noise limits. Independent variable commonly time, but can be any continuous variable. Concept / Approach: Differentiation computes the derivative dy/dx, representing the slope of y with respect to x at a point. In time domain signals, v_out ∝ dv_in/dt for an ideal differentiator. Practical op-amp differentiators place a series capacitor at the input and a resistor in feedback, often with additional components to tame high-frequency noise and low-frequency drift. This operation is used for edge detection, motion sensing, and shaping in control systems. Step-by-Step Solution: Define derivative: slope = limit of Δy/Δx as Δx → 0.Map to circuits: v_out = −RC * dv_in/dt (ideal inverting differentiator).Interpret examples: sine in → cosine out (phase lead of 90° in ideal band).Note constraints: add R and C limits to prevent noise amplification. Verification / Alternative check: Fourier view: differentiation multiplies spectra by jω, increasing magnitude with frequency; lab Bode plots show +20 dB/decade slope within the designed band of an active differentiator. Why Other Options Are Wrong: Integration is the inverse (area under curve). Averaging smooths noise but does not yield instantaneous slope. Linear regression fits a global trend, not pointwise derivative. Fourier synthesis builds waveforms from sinusoids and is unrelated to instantaneous slope. Common Pitfalls: Expecting ideal differentiation at all frequencies; in reality, noise and saturation limit usable bandwidth. Final Answer: Differentiation

Question 8

Area under a curve terminology: Which mathematical operation finds the accumulated area under the curve of a function over an interval?

Accepted Answer

Introduction / Context: Integration is fundamental in physics and engineering because many quantities are cumulative: charge is the time integral of current, energy is the integral of power, and distance is the integral of velocity. In analog circuitry, op-amp integrators implement this operation over a designed bandwidth, enabling functions such as ramp generation, active filters, and control loop compensation. Given Data / Assumptions: Continuous signals and standard calculus conventions. Ideal integrator model: output proportional to the time integral of input. Awareness that practical integrators include stabilizing resistors and finite limits. Concept / Approach: Integration computes the accumulated area under a function: for a signal v_in(t), an ideal inverting integrator produces v_out(t) = −(1/(R*C)) * ∫ v_in(t) dt over time. The area interpretation connects directly to measurement (e.g., integrating a sensor’s output to estimate position from acceleration). In active filters, integration shapes frequency response with a −20 dB/decade slope within its designed passband, complementing differentiation’s +20 dB/decade behavior. Step-by-Step Solution: Define integral as accumulated area between the function and axis over limits.Map to circuits: inverting integrator uses input resistor and feedback capacitor.Example behaviors: constant input → ramp output; sine input → negative cosine output (90° lag within band).Add practical elements: parallel resistor across Cf to set DC gain and prevent saturation. Verification / Alternative check: Frequency domain: integration multiplies by 1/(jω), reducing magnitude with frequency; measured Bode plots show −20 dB/decade slope. Scope traces of a square-wave input show a triangular/ramp output, visually demonstrating area accumulation. Why Other Options Are Wrong: Differentiation gives instantaneous slope, not area. Averaging provides mean value, not cumulative area. Linear regression estimates best-fit lines; correlation measures similarity between signals. Common Pitfalls: Expecting ideal behavior down to DC (practical integrators need leakage paths); ignoring op-amp saturation and input bias currents that cause drift. Final Answer: Integration

Question 9

Low-pass filter behavior above cutoff (fc): in a standard active or passive low-pass filter used in basic op-amp circuits, what happens to the output voltage gain as frequency increases beyond the critical (cutoff) frequency?

Accepted Answer

Introduction / Context: Low-pass filters are fundamental in analog signal processing. They pass low-frequency components while attenuating higher frequencies. The “critical” or cutoff frequency (often noted as fc) is commonly defined at the point where the magnitude response falls to 0.707 of the passband value (−3 dB for first order). This question asks what happens to the output gain beyond that point. Given Data / Assumptions: Idealized low-pass filter behavior (active or passive). Standard definition of cutoff at the −3 dB corner for first-order networks. No special peaking or resonance unless otherwise noted. Concept / Approach: A first-order low-pass has a magnitude |H(jω)| = 1 / sqrt(1 + (ω/ωc)^2). For ω > ωc (i.e., f > fc), the denominator grows, so magnitude decreases with frequency. In Bode terms, the slope is approximately −20 dB/decade per pole. Multi-pole low-pass networks attenuate even more steeply (e.g., −40 dB/decade for two poles). Thus, the gain does not increase or remain constant above cutoff; it decreases progressively. Step-by-Step Solution: 1) Identify region: frequencies above cutoff (f > fc).2) Recall magnitude trend: each pole contributes roughly −20 dB/decade beyond fc.3) Conclude the gain falls (decreases) with further increase in frequency. Verification / Alternative check: You can confirm by evaluating |H| at f = 10 * fc for a single-pole filter: |H| ≈ 1/√(1 + 10^2) ≈ 1/√101 ≈ 0.099 (about −20 dB), which is clearly lower than at fc and much lower than in the passband. Why Other Options Are Wrong: Increases: contradicts the standard roll-off. Remains unchanged: only true well below fc. Doubles per 1 kHz: arbitrary and incorrect. Becomes strictly zero above fc: ideal filters asymptotically approach zero; they do not become exactly zero. Common Pitfalls: Confusing low-pass with high-pass behavior; assuming “brick-wall” attenuation where output is exactly zero above fc; overlooking multi-pole slopes. Final Answer: Decreases

Basic Op-Amp Circuits Questions