Question 1

Definition check — The ratio of differential gain (Ad) to common-mode gain (Acm) in a differential amplifier is called what?

Accepted Answer

Introduction: Differential amplifiers should amplify differences while rejecting signals common to both inputs. A precise metric quantifies this ability and appears on every op-amp datasheet, often in decibels for convenience. Given Data / Assumptions: Differential gain Ad amplifies V1 − V2 Common-mode gain Acm amplifies (V1 + V2) / 2 Linear small-signal regime Concept / Approach: The figure of merit is CMRR, defined as CMRR = Ad / Acm. In practice, it is reported in dB: CMRR_dB = 20 * log10(Ad / Acm). A large CMRR means excellent rejection of common-mode interference such as power-line hum or ground shifts. Step-by-Step Explanation: Identify desired metric: ratio of gainsWrite: CMRR = Ad / AcmOptionally convert to dB: 20 * log10(CMRR)Interpret: higher CMRR → better common-mode suppression Verification / Alternative check: Datasheets list both Ad and CMRR; dividing Ad by Acm matches published CMRR, confirming the definition. Why Other Options Are Wrong: amplitude / phase: Not ratios of the specific gains. differential-mode rejection: Nonstandard term; the amplifier should not reject differential signals. common-mode rejection: Describes the phenomenon, not explicitly the ratio unless stated as “ratio”. Common Pitfalls: Confusing CMR (concept) with CMRR (quantified ratio). Mixing power (dB) with voltage gain without the 20× factor. Final Answer: common-mode rejection ratio

Question 2

Functional identification — A circuit whose output is proportional to the algebraic difference between two input signals is best described as which type of amplifier?

Accepted Answer

Introduction: Many analog front ends and op-amp configurations are based on differential principles. Recognizing the building block that outputs a scaled version of V1 − V2 is crucial to understanding instrumentation amplifiers, comparators in linear mode, and active filters. Given Data / Assumptions: Two input signals applied to a device or network Interest in the arithmetic difference of those inputs Linear small-signal behavior Concept / Approach: A differential amplifier produces Vout = Ad * (V1 − V2). Common-mode components ideally cancel, and only the difference is amplified. An operational amplifier is a general-purpose active device; when configured appropriately, it implements a differential amplifier among many other functions. Step-by-Step Explanation: Define desired behavior: output ∝ (V1 − V2)Map behavior to circuit type: differential amplifierNote: an op-amp can be used to realize a differential amplifier but is not itself automatically that configurationPractical designs maximize CMRR to reject shared interference Verification / Alternative check: Examine instrumentation amplifier datasheets: the first stage is a precision differential amplifier where output tracks the difference between inputs. Why Other Options Are Wrong: common-mode: Refers to signals common to both inputs, not difference. darlington: A transistor pair for high current gain, not differential operation per se. operational: Describes the device class, not the configured function. transconductance: Output current proportional to input voltage, different transfer. Common Pitfalls: Assuming any op-amp circuit is inherently differential—configuration matters. Ignoring input resistor matching, which degrades CMRR. Final Answer: differential

Question 3

In analog electronics, if the input to a voltage comparator is a sinusoidal (sine) wave, what general waveform shape will appear at the comparator's output under ideal operation?

Accepted Answer

Introduction: A voltage comparator is designed to compare an input signal against a reference and rapidly switch its output high or low depending on which input is larger. This question tests understanding of how a continuous-time sinusoidal input is transformed by an ideal comparator into a two-level logic-like output, emphasizing threshold crossing behavior and slew-driven transitions rather than linear amplification. Given Data / Assumptions: Ideal comparator with very high open-loop gain. Single reference threshold (often 0 V or Vref). No hysteresis (i.e., not a Schmitt trigger), unless otherwise stated. Input is a pure sine wave of sufficient amplitude to cross the threshold twice per cycle. Concept / Approach: The comparator saturates its output to one of two rails depending on the sign of the differential input. When a sine wave crosses the threshold, the output abruptly flips between its high and low saturation levels. Because the comparator does not reproduce the shape of the input but only its sign relative to the reference, the output becomes a two-level waveform (a rectangular or square-like wave). Step-by-Step Solution: Model the comparator output as: Vout = +Vsat when Vin > Vref, else Vout = −Vsat.A sine wave Vin(t) = Vm * sin(ωt) crosses Vref twice per period.Each zero/threshold crossing triggers a fast transition of Vout to the opposite rail.Across many cycles, the output thus alternates between two levels, forming a rectangular waveform. Verification / Alternative check: Counting threshold crossings: a sine has two crossings per period, giving a 50% duty cycle when Vref = 0 and the sine is centered. With nonzero Vref or offsets, duty cycle changes but the output remains rectangular because only the sign (not magnitude) is encoded. Why Other Options Are Wrong: Ramp voltage: Requires integration, not comparison. Sine wave: Would be the input or an amplifier output, not a comparator's saturated output. Sawtooth wave: Requires a repetitive charge–discharge process, not simple thresholding. Common Pitfalls: Confusing a comparator with a linear op-amp stage; forgetting that slew and saturation dominate; assuming harmonically related wave shaping without threshold behavior; overlooking hysteresis effects of a Schmitt trigger (which affect switching points but not the rectangular nature). Final Answer: Rectangular wave.

Question 4

In feedback amplifier design, a portion of the output is fed back for circuit stabilization. What is this type of feedback called?

Accepted Answer

Introduction: Stability and predictable gain are central goals in analog amplifier design. Feeding a fraction of the output back to the input can either stabilize or destabilize the system depending on polarity and magnitude. This question assesses knowledge of how feedback polarity influences stability, bandwidth, and linearity in practical circuits. Given Data / Assumptions: An amplifier stage with large open-loop gain. A feedback network that returns a portion of the output to the input. Linear operation around a bias point; small-signal assumptions. Concept / Approach: Negative feedback subtracts a portion of the output from the input drive, thereby reducing overall gain sensitivity to parameter variations and extending bandwidth. It also lowers distortion and noise referred to the input. Positive feedback reinforces deviations and is used to create oscillation or hysteresis (e.g., Schmitt triggers), not stabilization. Open-loop operation means no feedback is applied, which maximizes gain but sacrifices stability and accuracy. Step-by-Step Solution: Identify the feedback polarity: stabilized, linear, wider-band response → negative feedback.Relate closed-loop gain: Acl ≈ Aol / (1 + β * Aol), where β is feedback factor.Observe benefits: reduced distortion, predictable gain set by the feedback network, improved input/output impedances.Conclude that stabilization is the hallmark of negative feedback, not positive feedback. Verification / Alternative check: Closed-loop sensitivity S ≈ 1 / (1 + β * Aol) demonstrates desensitization to device parameter drift when negative feedback is applied. Empirically, output distortion and temperature drift diminish with appropriate β and phase margin. Why Other Options Are Wrong: Distortion: A performance defect, not a feedback topology. Open-loop: Lacks stabilization; gain and offset vary widely. Positive feedback: Encourages oscillation or bistability; used in comparators and oscillators. Common Pitfalls: Assuming any feedback stabilizes; ignoring phase shift that can turn negative feedback effectively positive at high frequency; forgetting compensation for adequate phase margin to avert oscillation. Final Answer: Negative feedback.

Question 5

In multivibrator terminology, an astable multivibrator (with no stable states) is commonly referred to as what kind of multivibrator?

Accepted Answer

Introduction: Multivibrators are fundamental timing and waveform-generation circuits. Understanding the difference between astable, monostable, and bistable configurations is key to selecting the right building block for clocks, pulse generators, and timing logic. This question focuses on the common alias for an astable multivibrator. Given Data / Assumptions: Astable multivibrator has no stable state. It alternates between two quasi-stable states continuously. Implemented with transistors, op-amps, or timers (e.g., 555) with RC timing. Concept / Approach: An astable multivibrator oscillates autonomously, requiring no external trigger once powered. Consequently, it is called a free-running multivibrator. In contrast, a monostable (one-shot) has one stable state and produces a single pulse when triggered; a bistable (flip-flop) has two stable states and toggles between them upon triggering. Step-by-Step Solution: Recognize that “no stable states” implies continuous oscillation.Continuous oscillation without trigger equals “free-running.”Therefore, the correct common name is free-running multivibrator. Verification / Alternative check: Examining timing equations (e.g., for a 555 astable: T = 0.693 * (RA + 2RB) * C) shows output repeats indefinitely; no external triggers are needed, confirming free-running behavior. Why Other Options Are Wrong: One-shot (monostable): Needs a trigger; not free-running. Bistable: Two stable states; not oscillatory by itself. Monostable: Same as one-shot; not continuously oscillatory. Common Pitfalls: Confusing the names because all use RC timing; assuming any RC network oscillates without the required feedback polarity; overlooking that astable implies no rest state. Final Answer: Free-running multivibrator.

Question 6

Operational amplifiers: What does the term open-loop voltage gain (Aol) specifically refer to in the context of an op-amp?

Accepted Answer

Introduction: Op-amp specifications distinguish between open-loop and closed-loop operation. Open-loop voltage gain (Aol) is a key intrinsic parameter describing how the device amplifies a very small differential input voltage without feedback. This question checks the correct interpretation of Aol and how it differs from application-level closed-loop gain (Acl). Given Data / Assumptions: Linear operation near the input common-mode range. No external feedback network (open loop). Small-signal differential excitation. Concept / Approach: Aol is the op-amp's internal gain between its differential input and output when no feedback is applied. In practice Aol is very large (often 10^5 to 10^6 or higher at low frequency) but varies with frequency and temperature. Closed-loop gain (Acl) is set primarily by the external feedback network and is much more predictable and stable than Aol. Step-by-Step Solution: Define Aol = Vout / Vdiff with feedback disabled.Recognize Acl ≈ 1/β when Aol ≫ 1/β in negative feedback applications.Note Aol is internal and not “controlled” precisely; it rolls off with frequency (dominant pole). Verification / Alternative check: Data sheets show Aol vs. frequency plots; Aol collapses as frequency rises, whereas Acl remains nearly constant within bandwidth, confirming the internal nature of Aol versus externally set Acl. Why Other Options Are Wrong: External voltage gain: That is closed-loop behavior defined by external components. Most controlled parameter: Aol varies widely; closed-loop gain is the controlled one. Same as Acl: They differ; Acl depends on feedback factor β and Aol. Common Pitfalls: Assuming Aol equals the gain you design for; neglecting that without feedback the op-amp saturates even for microvolts of input difference; ignoring frequency dependence of Aol. Final Answer: Internal voltage gain the device is capable of.

Question 7

Phase-locked loop (PLL) fundamentals: For a PLL to initially acquire lock with an input signal, the signal frequency must satisfy which condition?

Accepted Answer

Introduction: PLLs synchronize a voltage-controlled oscillator (VCO) to the phase and frequency of an input signal. Two important ranges govern behavior: capture range (for acquisition) and lock range (for tracking once locked). This question distinguishes those concepts and identifies the condition for initial lock acquisition. Given Data / Assumptions: Standard analog PLL with phase detector, loop filter, and VCO. Input is a near-sinusoidal signal with slowly varying frequency. Loop filter is linear and stable with adequate phase margin. Concept / Approach: The capture range is the frequency span within which the loop can pull the VCO into phase lock starting from an unlocked condition. The lock range is typically wider and represents how far the VCO can track the input once locked. Therefore, initial acquisition requires the input frequency to lie within the capture range, not merely the lock range. Step-by-Step Solution: Define capture range: frequency offsets where the loop can acquire lock.Define lock range: wider tracking range after lock.Conclude: for initial lock, input must be within capture range. Verification / Alternative check: Loop simulations show that starting outside capture range results in beats at the detector output with no net control voltage convergence; starting inside capture range yields control voltage settling to align VCO and input phases. Why Other Options Are Wrong: Come within the lock range: Lock range applies after acquisition; not sufficient for initial capture. Be less/greater than the capture frequency: Capture is a range, not a single frequency inequality. Common Pitfalls: Using lock range as a proxy for capture; ignoring loop filter bandwidth and nonlinearity that make capture narrower than lock; forgetting frequency steps can momentarily break lock even within lock range if the step exceeds capture dynamics. Final Answer: Come within the capture range.

Question 8

Op-amp characteristics: How does common-mode voltage gain (Ac) typically compare with differential voltage gain (Ad) in a high-quality differential amplifier stage?

Accepted Answer

Introduction: Differential amplifiers and op-amps are designed to amplify differences between inputs while rejecting signals common to both inputs (common-mode). This question checks understanding of the relative magnitudes of differential gain (Ad) and common-mode gain (Ac) and their relationship to common-mode rejection ratio (CMRR). Given Data / Assumptions: Balanced differential input pair with proper biasing. Common-mode signals are ideally rejected. High-performance integrated op-amp topology. Concept / Approach: Ideal differential amplifiers have Ac = 0, meaning perfect rejection of common-mode input. Real amplifiers have small but nonzero Ac. Since Ad is intentionally large, the ratio CMRR = Ad / Ac is also large (often specified in dB). Thus, Ac is far smaller than Ad in a good design. Step-by-Step Solution: Recognize design goal: maximize Ad, minimize Ac.State relation: CMRR = Ad / Ac.High CMRR implies Ac ≪ Ad, hence Ac is smaller than differential gain. Verification / Alternative check: Data sheets typically show CMRR values of 80–120 dB, implying Ac is orders of magnitude lower than Ad. Bench tests injecting identical signals into both inputs yield minimal output variation compared to differential stimulation. Why Other Options Are Wrong: Equal/greater than differential gain: Would yield poor common-mode rejection and is contrary to op-amp purpose. None of the above: Unnecessary; the correct qualitative relationship is clear. Common Pitfalls: Assuming infinite CMRR; forgetting that resistor mismatch and finite transistor Early effects raise Ac; ignoring frequency dependence—CMRR typically degrades at higher frequency. Final Answer: Smaller than differential voltage gain.

Question 9

If one input of an operational amplifier is grounded while a signal is applied to the other input, what mode or configuration is the op-amp said to be operating in?

Accepted Answer

Introduction: Op-amps can be driven in different ways depending on whether one or both inputs carry a signal. Understanding single-ended versus differential (double-ended) stimulation is key to interpreting gain, noise, and common-mode behavior. This question identifies the correct descriptor when one input is grounded and the other receives the signal. Given Data / Assumptions: Two-input op-amp (inverting and noninverting terminals). One input explicitly connected to ground (0 V reference). Other input receives the signal, possibly through a network. Concept / Approach: When only one input is driven and the other is at ground, the amplifier is being driven single-ended. If both inputs receive equal and opposite signals, it is double-ended (fully differential). “Common-mode” refers to identical signals at both inputs, not one grounded and one driven. “Noninverting mode” is a particular single-ended topology but requires clarifying which terminal is driven and feedback arrangement; the question asks more generally about the drive type, not the specific feedback configuration. Step-by-Step Solution: Identify that only one input has a signal → single-ended drive.Grounding the other input eliminates common-mode signal on that input.Terminology focuses on how the amplifier is excited, independent of exact feedback network. Verification / Alternative check: In instrumentation amplifiers, double-ended (fully differential) drive improves CMRR; by contrast, the described setup is clearly single-ended, confirming the terminology. Why Other Options Are Wrong: Common-mode: Requires identical signals on both inputs. Double-ended: Needs both inputs driven with signals. Noninverting mode: Possible only if the + input is driven and feedback is configured appropriately; the question does not assert that specific topology. Common Pitfalls: Equating “one input grounded” with a specific polarity (inverting vs noninverting); forgetting that “mode” can refer to drive type (single vs double-ended) or feedback topology (inverting/noninverting) depending on context. Final Answer: Single-ended.

Question 10

Unity-gain amplifier terminology: What is another common name for a unity-gain buffer stage built with an op-amp?

Accepted Answer

Introduction: Unity-gain stages are used to buffer signals without changing their amplitude while providing high input impedance and low output impedance. Recognizing the common names for such stages helps in reading schematics and selecting the right building block for impedance matching and isolation. Given Data / Assumptions: Op-amp configured with 100% negative feedback from output to inverting input. Input is applied to the noninverting terminal (+). Ideal op-amp assumptions for conceptual clarity. Concept / Approach: The voltage follower (also called a unity-gain buffer) forces Vout to follow Vin because the feedback drives the differential input toward zero. This provides a gain of approximately 1 while transforming impedance. It is not a comparator (which saturates between rails) and not a difference amplifier (which amplifies V+ − V− with settable gain using a resistor network). Step-by-Step Solution: Connect output directly to inverting input (feedback factor β = 1).Apply input to noninverting input.Closed-loop gain Acl ≈ 1; output tracks input. Verification / Alternative check: Impedance transformation: input impedance is very high (limited by input bias network), output impedance is low (≈ ohms with feedback). Bench measurements show negligible gain error at low frequency when Aol is large. Why Other Options Are Wrong: Difference amplifier: Requires specific resistor ratios; not unity by default. Comparator: Operates open-loop, saturating rather than following. Single ended: Describes drive type, not a unity-gain buffer topology. Common Pitfalls: Confusing buffer with follower-plus-offset circuits; assuming a comparator can be “used as an op-amp” (its output stage and input structures differ). Final Answer: Voltage follower.

Question 11

For an op-amp output to swing above and below a 0 V reference (i.e., provide positive and negative output voltages), what supply arrangement is required for the op-amp circuit?

Accepted Answer

Introduction: Op-amp output swing is fundamentally limited by its supply rails. Generating signals that cross 0 V (true bipolar swing) requires appropriate supply configuration. This question targets the practical requirement for achieving symmetrical positive and negative output excursions around ground. Given Data / Assumptions: Conventional op-amp (not necessarily rail-to-rail) driving a load referenced to ground. Desired output swings both above and below 0 V. Moderate load currents; linear operation with feedback. Concept / Approach: To produce negative output voltages relative to ground, the op-amp must be powered with a negative rail below 0 V (e.g., −V) and a positive rail above 0 V (e.g., +V). A single-supply configuration (0 to +V) typically requires bias shifting (virtual ground) to keep the signal within 0…+V, not around 0 V. Feedback networks or bandwidth do not by themselves create negative swing; they only shape response within supply limits. Step-by-Step Solution: Set supplies to ±V (e.g., −12 V and +12 V).Bias inputs around 0 V so the closed-loop output can cross zero.Confirm required headroom given the device’s output swing limits relative to rails. Verification / Alternative check: Data sheets list output swing vs. supply and load; only with a negative rail does the device produce negative output without level-shifting. Lab tests with single-supply show clipping at 0 V unless a mid-supply reference is used (which shifts “ground” for the signal). Why Other Options Are Wrong: Resistive feedback: Controls gain, not swing polarity. Zero offset: Reduces DC error but does not create negative swing. Wide bandwidth: A speed metric; unrelated to crossing 0 V. Common Pitfalls: Assuming rail-to-rail input/output guarantees negative swing on a single supply; forgetting to reference the load and measurement ground correctly; overlooking output headroom limits near rails. Final Answer: A negative and positive supply.

Question 12

Phase-locked loop architecture: A classical analog PLL is composed of which three fundamental subcircuits?

Accepted Answer

Introduction: A phase-locked loop (PLL) synchronizes an internal oscillator to an external reference in phase and frequency. Recognizing its core blocks is crucial for understanding acquisition, tracking, and noise performance in communication and control systems. Given Data / Assumptions: Classical, linear PLL model. Phase detector (phase comparator), loop filter, and voltage-controlled oscillator (VCO). Continuous-time implementation with standard low-pass loop filter. Concept / Approach: The phase comparator multiplies or otherwise compares phases to produce an error proportional to phase difference. The low-pass loop filter removes high-frequency components (sum terms) and shapes loop dynamics. The VCO converts the filtered error voltage into a frequency/phase adjustment, closing the feedback loop. Step-by-Step Solution: Identify detector: phase comparator produces error proportional to phase difference.Filter the error with a low-pass network to set stability and bandwidth.Drive the VCO so its frequency/phase aligns to minimize the error. Verification / Alternative check: Linearized small-signal loop models show a type-1 or type-2 control loop where the low-pass filter contributes poles/zeros for desired phase margin. Simulations confirm that replacing the low-pass with a bandpass prevents DC error correction, breaking lock behavior. Why Other Options Are Wrong: Phase comparator, comparator, and VCO: Redundant and missing the essential loop filter. Phase comparator with bandpass: A bandpass would reject DC error needed for correction. Demodulator: Not a core PLL block (though PLLs can be used for FM demodulation). Common Pitfalls: Confusing frequency discriminators with phase detectors; assuming any filter topology suffices—PLL needs low-pass characteristics for DC error propagation and stability; overlooking VCO gain (Kvco) and detector gain (Kd) impact on loop design. Final Answer: Phase comparator, low-pass filter, and VCO.

Question 13

In closed-loop analog design, a noninverting operational amplifier (op-amp) connects the input signal to the op-amp's noninverting (+) terminal and uses a feedback divider from the output to the inverting (−) terminal. In typical textbook implementa…

Accepted Answer

Introduction: A noninverting closed-loop op-amp is foundational in analog electronics. This question tests whether you know the characteristic gain behavior of the noninverting topology, which is widely used for signal buffering and amplification with high input impedance and predictable closed-loop gain. Given Data / Assumptions: Standard noninverting op-amp configuration using a resistive feedback divider. Ideal op-amp assumptions for clarity: infinite open-loop gain, infinite input impedance, zero output impedance. No special gain-reducing elements like attenuators at the input. Concept / Approach: The closed-loop gain of the canonical noninverting amplifier is Av = 1 + (Rf / Rin), where Rf is the feedback resistor (output to − input) and Rin is the resistor from the − input to ground (or reference). Because Rf / Rin ≥ 0 for passive resistors, Av is always ≥ 1, equaling 1 only when Rf = 0 (voltage follower) and strictly greater than 1 when Rf > 0. Step-by-Step Solution: 1) Write the noninverting gain: Av = 1 + (Rf / Rin).2) Note that passive resistors imply Rf ≥ 0 and Rin > 0 in typical designs.3) Therefore, Rf / Rin ≥ 0, so Av ≥ 1.4) For amplification (not just buffering), choose Rf > 0 to obtain Av > 1. Verification / Alternative check: Consider the limiting case Rf = 0. Then Av = 1 (a unity-gain buffer). Any positive Rf raises gain above 1, confirming the general rule that noninverting closed-loop gain is typically greater than one. Why Other Options Are Wrong: Less than one: contradicts Av = 1 + (Rf / Rin) for passive components.Of zero: physically meaningless for closed-loop voltage gain.Equal to one: only for the buffer case; generally the configuration is designed for > 1. Common Pitfalls: Confusing noninverting with inverting gain formula or forgetting the added +1 term specific to the noninverting topology. Final Answer: greater than one

Question 14

Digital interfacing and signal conditioning: A Schmitt trigger is a two-state circuit with hysteresis used at the front end of digital systems to clean up real-world signals. What application best captures its primary purpose?

Accepted Answer

Introduction: Real sensors and switches often produce slow, noisy, or bouncing waveforms. A Schmitt trigger adds hysteresis to create clean, fast transitions between logic states. This question checks if you recognize its main role in conditioning noisy inputs before they enter digital logic. Given Data / Assumptions: Two-state device with distinct upper and lower switching thresholds. Used at a digital interface where noisy or slow-rising signals are common. Objective is robust logic-level detection. Concept / Approach: Schmitt triggers employ hysteresis: once the input crosses a high threshold, the output switches and will not switch back until the input drops below a lower threshold. This dead-band resists chatter from noise near the threshold and debounces slow or mechanical inputs, producing clean logic-level square edges. Step-by-Step Solution: 1) Identify problem: noisy, slow, or bouncing inputs cause multiple unintended transitions.2) Apply hysteresis: two distinct thresholds eliminate dithering around a single boundary.3) Result: crisp, stable outputs suitable for digital circuits.4) Outcome: improved noise immunity and reliable state recognition. Verification / Alternative check: Compare a simple comparator (single threshold) to a Schmitt trigger (two thresholds). Only the Schmitt trigger prevents rapid toggling when noise rides on a slowly varying input, proving superior input noise rejection. Why Other Options Are Wrong: Pulse shaping: a side effect, but the core benefit is resisting noise-induced chatter.Peak detection: requires a rectifier/hold circuit, not a Schmitt trigger.Filtering: uses RC/L networks to attenuate frequencies; a Schmitt trigger is a thresholding device. Common Pitfalls: Thinking hysteresis is a filter; it does not remove noise energy, it prevents noise from causing state flips near the threshold. Final Answer: input noise rejection

Question 15

Operational amplifier fundamentals: Considering the standard ideal-op-amp model used for preliminary circuit analysis and hand calculations, which of the following properties correctly belongs to an ideal op-amp?

Accepted Answer

Introduction: The ideal op-amp is a theoretical construct that simplifies design intuition and first-pass analysis. Knowing which traits belong to the ideal model helps distinguish practical limitations from analytical assumptions. Given Data / Assumptions: Ideal op-amp used as a simplifying model. Focus on classic assumptions: gain, input/output impedances, bandwidth, and offset. Single-corner model without slew-rate or noise limits. Concept / Approach: Common ideal-op-amp assumptions: infinite open-loop gain, infinite input impedance, zero output impedance, zero input offset voltage/currents, and infinite bandwidth (constant gain over all frequencies). These assumptions support rules like input nodes having equal voltages under negative feedback and zero input currents. Step-by-Step Solution: 1) Check output impedance: ideal is zero, not infinite.2) Check input impedance: ideal is infinite, not zero.3) Check bandwidth: ideal is infinite, implying constant gain with frequency.4) Therefore, among the options, only infinite bandwidth matches the ideal model. Verification / Alternative check: Textbook nodal analyses rely on zero input current (infinite input impedance) and a perfect source behavior at the output (zero output impedance); these are inconsistent with options a and b, confirming that c is the only valid statement. Why Other Options Are Wrong: Infinite output impedance: would prevent the op-amp from driving loads; contradicts ideal source behavior.Zero input impedance: would load the source heavily; contradicts the infinite input impedance ideal.All of the above: includes incorrect statements, so it cannot be true. Common Pitfalls: Assuming practical op-amp limits (finite gain/bandwidth) apply to the ideal model; keep the two concepts separate to avoid analysis errors. Final Answer: infinite bandwidth

Question 16

Feedback basics in amplifier design: When negative feedback is applied around an amplifier stage, what is the nature of the feedback signal with respect to the input signal at the summing node?

Accepted Answer

Introduction: Negative feedback underpins accurate, stable, and linear amplifier behavior. Recognizing how the returned signal combines with the input clarifies why gain becomes predictable and distortion is reduced. Given Data / Assumptions: Classical negative feedback loop with a portion of the output sampled and returned. Summing junction compares input to the feedback signal. Standard linear small-signal operation. Concept / Approach: In negative feedback, the feedback signal is subtracted from the input at the summing node. If the output momentarily increases, the feedback increases in a way that reduces the effective error signal, pushing the output back toward the desired value. This opposition stabilizes gain and reduces sensitivity to parameter drift. Step-by-Step Solution: 1) Define error: Verr = Vin − βVout, where β is the feedback factor.2) If Vout rises, βVout rises, decreasing Verr.3) The amplifier drives its output to force Verr toward zero, creating a self-correcting loop.4) Net effect: the feedback signal always opposes the input change at the summing node. Verification / Alternative check: Block-diagram algebra shows closed-loop gain Acl = A / (1 + Aβ). The positive denominator term (1 + Aβ) indicates a negative feedback sign convention leading to opposition at the summing node. Why Other Options Are Wrong: Aids the input signal: describes positive feedback, not negative.Proportional to output current: feedback may sample voltage or current; the key attribute here is opposition, not proportionality to current.Proportional to differential voltage gain: gain is a parameter, not the signal itself. Common Pitfalls: Confusing feedback polarity or assuming feedback always samples voltage; regardless of sample type, negative feedback opposes the input at the summing node. Final Answer: opposes the input signal

Question 17

Ideal amplifier expectation: Considering offset errors in precision measurement chains, which characteristic should an ideal amplifier exhibit to prevent unwanted output deviation when the input is zero?

Accepted Answer

Introduction: Offset errors corrupt accuracy in precision amplifiers, sensor interfaces, and ADC front ends. An ideal amplifier model removes these errors to simplify analysis and target best-case performance. Given Data / Assumptions: Idealized amplifier model for conceptual understanding. Zero input signal condition considered. Focus on DC accuracy and bias behavior. Concept / Approach: Input offset voltage and bias currents create nonzero output even when the input is zero. The ideal amplifier specifies zero offset (and, by extension, negligible bias effects), ensuring the output is exactly at the intended reference when there is no input stimulus. Step-by-Step Solution: 1) Define offset: a small equivalent input voltage that, when amplified, produces unwanted output shift.2) Ideal requirement: offset = 0 to keep output centered with zero input.3) Practical implication: minimize offset via device selection, chopping/auto-zero techniques, or calibration.4) Result: improved measurement accuracy and long-term stability. Verification / Alternative check: Compare outputs of two amplifiers at Vin = 0 with identical gains; the one with lower offset yields an output closer to zero, validating the ideal of zero offset. Why Other Options Are Wrong: High input current: undesirable; it loads the source and increases error via source impedance.High output impedance: prevents effective load drive; ideal is low output impedance.Moderate gain: ideal open-loop gain is extremely high; closed-loop gain is set by feedback, not by being merely moderate. Common Pitfalls: Conflating offset with drift; drift is temperature/time variation, whereas offset is the baseline error present even at constant conditions. Final Answer: zero offset

Question 18

Power regulation taxonomy: In a series dissipative voltage regulator, a pass element (such as a BJT or MOSFET) operates in its linear region to drop excess voltage and maintain a steady output. To which regulator class does this topology belong?

Accepted Answer

Introduction: Voltage regulators are categorized by how the pass device controls energy flow. Recognizing whether a topology is linear or switching is essential for predicting efficiency, noise, and thermal behavior. Given Data / Assumptions: Series pass device drops voltage continuously. Operation is in the linear (ohmic) region, not as a saturated switch. Goal is a stable DC output with minimal ripple. Concept / Approach: A series dissipative regulator regulates by varying the pass element's resistance so that Vout remains constant despite Vin or load changes. Energy is burned as heat across the pass device, a hallmark of linear regulation. In contrast, switching regulators modulate a switch and use inductors/capacitors to transfer energy at high efficiency. Step-by-Step Solution: 1) Identify pass element behavior: continuously variable conduction, not on/off switching.2) Recognize energy path: excess voltage * load current is dissipated as heat.3) Classify accordingly: that is the definition of a linear regulator.4) Distinguish from switching types that pulse energy and filter it. Verification / Alternative check: Efficiency trend confirms classification: linear regulators typically show efficiency ≈ Vout/Vin, while switching regulators can exceed that in many cases, reinforcing that a dissipative series pass design is linear. Why Other Options Are Wrong: Switching regulator: relies on high-frequency switching; not linear operation.Shunt regulator: uses a variable shunt path to ground; not a series pass element.DC-to-DC converter: umbrella term that often implies a switching topology; the specific class here is linear. Common Pitfalls: Assuming all regulators with a transistor are switching; operating region (linear vs switching) is the key differentiator. Final Answer: linear regulator

Question 19

Op-amp input connection styles: Textbooks describe several basic ways to drive the two op-amp inputs. Which of the following is NOT considered a standard mode of intended operation?

Accepted Answer

Introduction: Op-amps can be excited in different ways depending on whether one or both inputs carry signal. Understanding intended modes helps in analyzing gain and common-mode rejection properly. Given Data / Assumptions: Standard linear applications of op-amps. Idealized behavior with negative feedback. Typical textbook terminology. Concept / Approach: Common intended modes: inverting (signal into − input, + at reference), noninverting (signal into + input), and differential/double-ended (signals applied to both inputs with opposite polarity). Common-mode refers to the same signal simultaneously applied to both inputs; it is generally an undesired stimulus used to quantify rejection (CMRR), not a purposeful operating mode for gain. Step-by-Step Solution: 1) List intended modes: inverting, noninverting, and differential (double-ended).2) Define common-mode: identical signal on both inputs; ideally produces zero output.3) Since common-mode is measured to characterize rejection, it is not a standard gain mode.4) Therefore, the exception is common-mode. Verification / Alternative check: CMRR metric explicitly quantifies how little output should result from common-mode input, reinforcing that common-mode is not an intended amplification configuration. Why Other Options Are Wrong: Inverting mode: fundamental textbook topology.Double-ended: another way to say differential input, a standard mode.Single-ended: describes inverting or noninverting cases with one driven input and the other at reference. Common Pitfalls: Confusing differential mode (desired) with common-mode (undesired) due to similar wording; the sign and symmetry differ. Final Answer: common-mode

Question 20

Sequential logic basics: The S–R (Set–Reset) flip-flop is a bistable element used for storing 1 bit of information. Counting only its stable memory conditions, how many logic states does an S–R flip-flop have?

Accepted Answer

Introduction: Flip-flops are the building blocks of registers and state machines. The S–R flip-flop is the simplest latch-type circuit, demonstrating how digital systems store a single binary value. Given Data / Assumptions: Bistable device, meaning two stable outputs. Inputs labeled S (set) and R (reset). Invalid or metastable conditions are not counted as normal stable states. Concept / Approach: A bistable circuit has exactly two stable operating points. For the S–R flip-flop, those are Q = 1 (set) and Q = 0 (reset). When inputs are driven to certain combinations (e.g., S = 1, R = 1 in the basic NOR implementation), the state becomes invalid or leads to uncertainty, but that does not create an additional stable memory condition. Step-by-Step Solution: 1) Identify stable states: stored 1 and stored 0.2) Recognize that state persists when inputs return to the hold condition.3) Exclude invalid/metastable scenarios from the count of stable states.4) Conclude that the device has two stable logic states. Verification / Alternative check: Truth tables for NOR- or NAND-based S–R latches show two valid stable outputs and one forbidden input combination, reinforcing the count of two stable states. Why Other Options Are Wrong: 3: would imply an additional stable or tri-stable state; not present in S–R latches.4 and 5: exceed the storage capacity of one flip-flop, which holds exactly one bit. Common Pitfalls: Misinterpreting the forbidden input condition as a third state; it is not a valid stable memory condition. Final Answer: 2

Operational Amplifiers Questions