Question 1

Active filter implementation: When op-amps are used to realize standard low-pass and high-pass active filter stages in audio and instrumentation (for example, Sallen–Key or similar topologies), which amplifier configuration is most commonly employed?

Accepted Answer

Introduction: Active filters use op-amps with R–C networks to realize precise frequency responses. Choosing the amplifier configuration impacts input impedance, noise gain, and ease of component selection. Given Data / Assumptions: Standard low-pass and high-pass responses at modest orders (e.g., 2nd-order). Common practical topologies such as Sallen–Key, which buffer the network. Linear operation with negative feedback (not comparator mode). Concept / Approach: The widely used Sallen–Key topology employs a noninverting op-amp buffer/amp. This configuration offers very high input impedance, making the filter less sensitive to source loading, and provides straightforward gain setting without inverting signal polarity. While inverting multiple-feedback filters exist, the most commonly taught beginner implementation for simple HP/LP sections is noninverting (Sallen–Key). Step-by-Step Solution: 1) Identify common active filter families: Sallen–Key (noninverting) and multiple-feedback (often inverting).2) Note that introductory designs prefer Sallen–Key for simplicity and buffering.3) Conclude that the noninverting configuration is the most commonly employed for basic HP/LP stages.4) Recognize that other configurations may be used for special requirements, but they are not the default choice. Verification / Alternative check: Compare input impedance: noninverting stages present high input impedance; inverting stages present input resistance equal to the input resistor, which can load the source more significantly. Why Other Options Are Wrong: Comparator: open-loop threshold device; not a linear filter element.Open-loop: an op-amp without feedback saturates; filters require controlled closed-loop gain.Inverting: used in multiple-feedback filters but is not the most common entry-level choice compared to Sallen–Key. Common Pitfalls: Assuming all active filters are inverting because of some multiple-feedback designs; many practical HP/LP stages are noninverting buffers with RC networks around them. Final Answer: noninverting

Question 2

Operational amplifier fundamentals: evaluate the statement—“An ideal inverting amplifier provides an output that is phase-shifted by 180° relative to its input (i.e., it inverts the signal polarity).”

Accepted Answer

Introduction / Context: The inverting amplifier is a canonical op-amp configuration used for scaling, summing, and filtering. A defining characteristic is the phase inversion between input and output. Recognizing this behavior is foundational for signal-chain design and phase-sensitive applications such as feedback and mixing. Given Data / Assumptions: Ideal operational amplifier (infinite gain, bandwidth sufficiently high for the signal of interest). Standard inverting configuration with input resistor R_in and feedback resistor R_f. Linear, small-signal operation (no clipping or slew-rate limiting). Concept / Approach: For an ideal inverting amplifier, the gain is V_out / V_in = −R_f / R_in. The negative sign indicates a phase inversion of 180 degrees with respect to the input. The magnitude R_f / R_in sets the amplitude scaling, independent of the inversion property. In sinusoidal steady state below the op-amp's bandwidth limits, the output is an inverted, scaled version of the input. Step-by-Step Solution: 1) Write the inverting gain relation: V_out = −(R_f / R_in) * V_in. 2) Interpret the negative sign as a 180° phase shift (polarity inversion). 3) Note that the inversion holds for any linear waveform (not only sinusoids) as a sign reversal in time domain. 4) Conclude the statement is correct for the ideal model within the usable frequency range. Verification / Alternative check: Square-wave tests show output transitions mirrored about zero; Bode plots show a constant −180° phase offset at low frequencies where the op-amp behaves ideally (before additional phase lag from finite bandwidth appears). Why Other Options Are Wrong: Incorrect: contradicts the standard inverting amplifier formula. True only for unity gain: inversion occurs for any gain magnitude R_f / R_in. True only at DC, not AC: holds across the passband where the op-amp remains linear and within bandwidth. True only for sine: inversion applies to any waveform under linear operation. Common Pitfalls: Confusing “phase inversion” with “phase shift that varies with frequency”; overlooking bandwidth and stability limits that can add extra phase lag beyond the ideal −180° at higher frequencies. Final Answer: Correct

Question 3

Op-amp polarity with inverting input drive — clarification: If an input signal is applied to the inverting (−) input of an operational amplifier while the noninverting (+) input is tied to ground (0 V reference), the closed-loop output (with standar…

Accepted Answer

Introduction / Context: Operational amplifiers (op-amps) are the core of countless analog circuits. One of the first concepts learners meet is the polarity behavior of inverting versus noninverting configurations. This question asks whether driving the inverting input while grounding the noninverting input produces an output that is opposite in polarity to the input. Understanding this sign inversion is essential for signal conditioning, sensor interfaces, and control systems. Given Data / Assumptions: The inverting (−) input is driven by a source through an input network. The noninverting (+) input is at 0 V (signal ground). Standard negative feedback is present to set the closed-loop gain. Ideal op-amp assumptions: very high open-loop gain, infinite input impedance, and zero output impedance for conceptual analysis. Concept / Approach: In the classic inverting amplifier, negative feedback forces the op-amp to hold the differential input near zero (the “virtual short” concept). Since the noninverting input is at 0 V, the inverting input node sits at a “virtual ground.” The output must move in whatever direction is necessary to reduce the input difference to nearly zero. Because the input signal is applied to the inverting input, the output moves in the opposite direction to cancel the error, producing inversion. Step-by-Step Solution: 1) With negative feedback, v+ ≈ v−. Here v+ = 0 V, so v− ≈ 0 V (virtual ground).2) Apply a positive input through the input resistor to the v− node; a small positive error appears.3) The op-amp output swings negative to drive current through the feedback resistor, canceling the error and restoring v− ≈ 0 V.4) Result: v_out is proportional to −v_in (i.e., opposite polarity), with magnitude set by the resistor ratio. Verification / Alternative check: For an inverting stage, closed-loop gain is Av = −Rf/Rin. The negative sign explicitly denotes inversion. Lab measurements confirm that a sine wave at the input emerges inverted at the output when the (+) input is grounded and negative feedback is used. Why Other Options Are Wrong: Incorrect: Conflicts with the defining property of the inverting configuration. Only true for DC signals: Inversion holds for AC and DC within bandwidth limits. Only true when open-loop: Open-loop operation is not stable for linear amplification; inversion is a closed-loop attribute here. Common Pitfalls: Confusing the inverting topology with noninverting; overlooking that the inversion result depends on negative feedback, not on operating the device open-loop. Final Answer: Correct

Question 4

Differential amplifier behavior — common-mode versus differential: If the two inputs to a differential amplifier are exactly the same signal (pure common-mode), is the output equal to the input signal multiplied by 2, or does it ideally cancel to ze…

Accepted Answer

Introduction / Context: Differential amplifiers respond to the difference between their two inputs while rejecting signals that are common to both. This behavior underpins instrumentation amplifiers and noise rejection strategies. The statement under test claims that identical inputs produce an output equal to “signal × 2,” which would imply summing rather than differencing. We examine whether that is correct under ideal assumptions. Given Data / Assumptions: Two amplifier inputs receive exactly the same waveform (common-mode input). Ideal differential amplifier characteristics: infinite common-mode rejection ratio (CMRR). Linear operation and no saturation. Concept / Approach: An ideal differential amplifier multiplies the difference (v+ − v−) by the differential gain Ad. When the inputs are identical, v+ = v−, so v+ − v− = 0. Therefore the ideal output is 0 V, not twice the input. Real amplifiers have finite CMRR, so small residual output may appear, but it should be minimal compared with the differential response. Step-by-Step Solution: 1) Identify the condition: v1 = v2 → purely common-mode signal.2) Compute differential input: v_diff = v1 − v2 = 0.3) Ideal output: v_out = Ad * v_diff = 0.4) Practical detail: finite CMRR allows a small v_out = Acm * v_cm, but not “×2 of the input.” Verification / Alternative check: Instrumentation amplifier datasheets specify CMRR (often 80–120 dB). High CMRR means large rejection of common-mode signals; measured outputs under identical inputs are near zero, not doubled. Why Other Options Are Wrong: Correct: Incorrect because identical inputs should cancel in a differential stage. Only true when CMRR is infinite: Even with infinite CMRR the output is zero, not doubled. Only true for square waves: Waveform shape is irrelevant; difference is zero for any waveform. Common Pitfalls: Confusing summing amplifiers with differential amplifiers; misinterpreting common-mode behavior as additive gain rather than rejection. Final Answer: Incorrect

Question 5

Ideal op-amp output impedance — conceptual check: In the idealized operational amplifier model used for closed-loop analysis, is the output impedance taken to be 0 Ω (i.e., an ideal voltage source behavior at the output)?

Accepted Answer

Introduction / Context: Op-amp “ideal” models are intentionally simplified to make feedback analysis tractable. Among the assumptions—very high open-loop gain, infinite input impedance—is an output that behaves like an ideal voltage source with zero output impedance. This lets the amplifier maintain the commanded output voltage across a load without droop in the model. Given Data / Assumptions: Ideal op-amp model for conceptual understanding. Closed-loop operation with negative feedback. No bandwidth or slew limitations considered. Concept / Approach: The ideal op-amp is modeled with Rout = 0 Ω so that the output node is stiff and unaffected by load current changes. In reality, Rout is small but not zero and is further reduced by feedback over frequency ranges where loop gain is high. For hand analysis and intuition building, assuming 0 Ω is standard and yields accurate first-order predictions within the linear region. Step-by-Step Solution: 1) Recall ideal assumptions: Aol → ∞, Rin → ∞, Rout → 0 Ω.2) Use this to justify that v_out is set primarily by feedback, not by load current.3) Recognize that real devices approximate this via low Rout and loop-gain reduction of output impedance.4) Apply ideality for quick closed-loop gain predictions and loading robustness estimates. Verification / Alternative check: Practical output impedance can be measured via small-signal load step tests; feedback reduces the open-loop Rout substantially within bandwidth. This aligns with the idealized 0 Ω concept used in introductory design. Why Other Options Are Wrong: Incorrect: Conflicts with the standard ideal model assumption. Only true in open-loop operation: Open-loop output is not controlled; the assumption is a property of the ideal model, not of open-loop behavior. Only true above unity-gain frequency: Output impedance typically rises as loop gain falls at high frequency; the ideal 0 Ω is a simplifying assumption, not frequency-limited. Common Pitfalls: Assuming real outputs are always ideal; ignoring current limit, slew, and bandwidth that can cause droop or distortion under heavy loads. Final Answer: Correct

Question 6

Crystal oscillator quality factor (Q) — truth test: Do crystal oscillator circuits exhibit a very low Q, or are they characterized by very high Q (narrowband, low-loss resonance)?

Accepted Answer

Introduction / Context: Oscillator stability, phase noise, and spectral purity are tightly linked to a resonator’s quality factor, Q. Quartz crystals are famous for exceptionally high Q compared with LC or RC networks. This item tests recognition that crystal oscillators are high-Q devices, not low-Q. Given Data / Assumptions: Quartz crystal used as the resonant element. Linear small-signal resonator model (motional arm and shunt capacitance). Focus on qualitative Q property, not exact numerical values. Concept / Approach: The quality factor Q measures energy stored versus energy lost per cycle. Crystals have extremely low loss, leading to very high Q (often 10^4–10^6 range depending on cut and frequency). High Q yields a narrow resonance bandwidth and superior frequency stability, which is why crystals dominate timekeeping and reference applications. Therefore the statement “crystal oscillator circuits have a very low Q” is false. Step-by-Step Solution: 1) Define Q qualitatively: higher Q → narrower bandwidth, lower loss.2) Recognize quartz crystals as low-loss resonators with extremely high Q.3) Infer that crystal oscillators inherit this high Q, giving excellent selectivity and stability.4) Conclude the statement about “very low Q” is incorrect. Verification / Alternative check: Compare with RC phase-shift or Wien-bridge oscillators: their Q is modest; LC tanks are higher; crystals are highest among common electrical resonators, which aligns with their ultra-stable performance in clocks and radios. Why Other Options Are Wrong: Correct: Not true; crystals are high-Q. Only correct for RC oscillators: RC types are lower-Q than crystals, but the statement concerns crystals. Only correct at low temperatures: Temperature affects frequency slightly but does not convert high-Q into low-Q. Common Pitfalls: Generalizing from RC oscillators to crystals; confusing narrow bandwidth (high Q) with low Q due to terminology. Final Answer: Incorrect

Question 7

Open-loop vs. closed-loop operation — feedback usage: Is feedback used in an “open-loop” operational amplifier circuit, or is open-loop explicitly the no-feedback condition used primarily for comparators and conceptual analysis?

Accepted Answer

Introduction / Context: Op-amp applications are commonly divided into open-loop and closed-loop. Clear terminology prevents design errors. “Open-loop” means the output is not fed back to the input for linear gain-setting; the device runs at its enormous open-loop gain and quickly saturates, which is useful for comparator behavior but not linear amplification. This question asks whether feedback is used in open-loop circuits. Given Data / Assumptions: Standard op-amp definitions (open-loop vs. closed-loop). Idealized large open-loop gain. No stability or slew considerations required for the concept. Concept / Approach: Open-loop operation by definition means no feedback path is applied to control the gain. Closed-loop operation introduces negative feedback to set a predictable, finite gain determined by external components. Therefore the statement “Feedback is used in an open-loop op-amp circuit” is incorrect because it contradicts the definition of open-loop. Step-by-Step Solution: 1) Recall: Open-loop → no feedback path; closed-loop → feedback present.2) In open-loop, small input differences drive the output to saturation.3) For linear amplifiers, negative feedback is introduced, creating closed-loop operation.4) Hence, feedback is not “used” in open-loop circuits. Verification / Alternative check: Comparator circuits are open-loop uses of op-amps: outputs rail high or low based on input polarity; no feedback sets a linear gain. Conversely, inverting/noninverting amplifiers are closed-loop examples with resistive feedback networks. Why Other Options Are Wrong: Correct: Conflicts with the definition of open-loop. Only true for unity-gain buffers: Buffers require feedback (closed-loop) to achieve unity gain. Only true when offset is zero: Offset does not change the loop classification. Common Pitfalls: Using “open-loop” loosely to mean “no external resistor values given”; equating comparator behavior with linear amplification. Final Answer: Incorrect

Question 8

Astable multivibrator — number of stable states: Does an astable multivibrator have two stable states, or does it have no stable states and continuously oscillate between two levels?

Accepted Answer

Introduction / Context: Multivibrators—astable, monostable, and bistable—are foundational timing circuits. Each type is defined by how many stable states it possesses. This question asks whether an astable multivibrator has two stable states or none, which determines whether it oscillates on its own or requires triggering. Given Data / Assumptions: Standard definitions of multivibrator types. Positive feedback timing network that charges and discharges to switch states. Idealized analysis without component tolerances considered. Concept / Approach: An astable multivibrator has no stable equilibrium. It repeatedly switches between two levels due to charging/discharging dynamics in the timing network, creating a free-running square wave. A bistable (flip-flop) has two stable states and requires a trigger to change state. A monostable has one stable state and one quasi-stable state, returning to stable after a pulse. Therefore, saying “an astable has two states” is incorrect because those two states are not stable; the circuit cannot rest in either. Step-by-Step Solution: 1) Identify type: astable → free-running oscillator with no stable resting point.2) Recognize that it toggles between two levels but neither is stable.3) Conclude: claim of “two stable states” describes a bistable, not an astable.4) Thus, the statement is false. Verification / Alternative check: 555 timer in astable mode auto-oscillates; in bistable mode it latches; in monostable mode one-shot pulses occur—textbook confirmations of the state definitions. Why Other Options Are Wrong: Correct: Not correct; two stable states belong to bistable devices. True only with positive feedback removed: Removing positive feedback will not produce two stable states characteristic of a bistable; it typically halts oscillation without forming a latch. True only when supply ripple is zero: Supply ripple does not change the state-count definition. Common Pitfalls: Equating “two output levels” with “two stable states”; overlooking the role of triggering in monostable and bistable circuits. Final Answer: Incorrect

Question 9

Role of the feedback resistor — does it “add to” op-amp gain? In closed-loop op-amp amplifiers, does the feedback resistor simply add to the intrinsic gain of the op-amp, or does it set the closed-loop gain via a ratio with the input network?

Accepted Answer

Introduction / Context: Students often conflate the op-amp’s huge open-loop gain with the closed-loop gain obtained using external resistors. The feedback resistor Rf does not “add to” the op-amp’s gain; instead, Rf, together with the input resistor(s), sets the closed-loop gain by determining how much output is fed back to the input node. Given Data / Assumptions: Negative feedback topology (inverting or noninverting). Ideal op-amp model for intuition: Aol → ∞. Linear operation within output swing and bandwidth limits. Concept / Approach: Closed-loop gain is a function of the external network. For an inverting amplifier, Av = −Rf/Rin. For a noninverting amplifier, Av = 1 + Rf/Rg. In neither case is the open-loop gain “added” to by Rf. Rather, feedback forces the effective gain to the value implied by resistor ratios, largely independent of the enormous Aol so long as loop gain is high over the signal band. Step-by-Step Solution: 1) Write inverting case: Av = −Rf/Rin (ratio, not sum).2) Write noninverting case: Av = 1 + Rf/Rg (again, ratio-based).3) Recognize that Rf shapes feedback fraction β, which with Aol determines closed-loop gain ≈ 1/β.4) Conclude the phrase “adds to the gain” is inaccurate; Rf defines gain via ratios. Verification / Alternative check: Circuit simulators and lab benches show that changing Rf while holding Rin (or Rg) fixed scales the closed-loop gain in the predicted ratio; the open-loop gain is not “added to.” Why Other Options Are Wrong: Correct: Misstates the role of Rf. Only correct for open-loop comparator use: Open-loop comparators do not use Rf to set linear gain. Only correct when Rf = Rin: Even with Rf = Rin, gain is −1 (inverting) or 2 (noninverting with proper Rg), still ratio-defined. Common Pitfalls: Thinking of gain as additive; ignoring the feedback fraction and loop-gain concepts. Final Answer: Incorrect

Question 10

Slew rate calculation — check the claim: It takes 4 µs for an op-amp output to transition from −14 V to +14 V (a 28 V step). Is the slew rate 3.5 V/µs as stated, or a different value?

Accepted Answer

Introduction / Context: Slew rate (SR) is the maximum rate of change of an op-amp’s output voltage and sets a hard limit on how fast the output can swing. It is crucial for large-signal performance, preventing distortion in fast waveforms and step responses. This problem checks a numeric SR claim using simple arithmetic. Given Data / Assumptions: Voltage change ΔV = +14 − (−14) = 28 V. Transition time Δt = 4 µs. Definition: SR = ΔV / Δt (in V/µs). Concept / Approach: Compute SR = 28 V / 4 µs = 7 V/µs. The statement “3.5 V/µs” is exactly half of the correct value and therefore incorrect. No RMS versus peak distinction applies here; slew rate is a time-derivative limit on the instantaneous output voltage, determined by internal current limiting and compensation capacitances. Step-by-Step Solution: 1) Calculate ΔV = 28 V.2) Use SR = ΔV / Δt = 28 / 4 = 7 V/µs.3) Compare with the claim 3.5 V/µs → mismatch.4) Conclude the claim is false; correct SR is 7 V/µs. Verification / Alternative check: If the amplifier had a 3.5 V/µs limit, the same 28 V step would require 8 µs, not 4 µs. This cross-check confirms that 4 µs implies 7 V/µs. Why Other Options Are Wrong: Correct: Not correct, as the computed value is 7 V/µs. Only correct for peak values, not RMS: Slew rate is not an RMS concept. Only correct with capacitive loading: External load may influence behavior but does not alter the basic ΔV/Δt calculation given. Common Pitfalls: Confusing full-swing step with half-swing; mixing bandwidth limits (small-signal) with slew-rate (large-signal) limitations. Final Answer: Incorrect

Question 11

Common-mode input definition — op-amp inputs tied to the same source: When both inputs of an op-amp are connected to the same signal source (i.e., the same waveform appears at v+ and v−), is this a common-mode input condition?

Accepted Answer

Introduction / Context: Common-mode and differential-mode inputs are fundamental in understanding what an op-amp amplifies versus rejects. Noise immunity, instrumentation amplifier design, and sensor interfacing depend heavily on these definitions. This item asks you to recognize the common-mode condition: the same signal applied to both inputs simultaneously. Given Data / Assumptions: Both op-amp inputs see the same time-varying signal. Linear operation assumed; no saturation. Focus on definition, not on numeric rejection performance. Concept / Approach: Common-mode input is the component of the input that is identical at both terminals. Differential input is the difference between them. An ideal differential amplifier responds only to the differential component, rejecting common-mode signals (output ideally zero for pure common-mode). Real devices have finite CMRR, so some common-mode leaks into the output, but the definition of common-mode remains the same. Step-by-Step Solution: 1) Define v_cm = (v+ + v−) / 2; define v_diff = v+ − v−.2) If v+ = v−, then v_diff = 0 and the entire input is common-mode.3) Therefore the condition of tying both inputs to one source is common-mode.4) Ideal output is zero; practical output depends on CMRR. Verification / Alternative check: Instrumentation amplifiers specify high CMRR because they must measure small differences riding on large common-mode voltages, precisely the case of identical signals at both inputs plus a tiny difference. Why Other Options Are Wrong: Incorrect: Contradicts the definition of common-mode. Only true for differential pairs without feedback: Feedback does not alter the definition of common-mode. Only true for DC offsets: AC or DC can be common-mode if identical at both inputs. Common Pitfalls: Confusing “what is common-mode” with “how much is rejected”; the former is a definition, the latter is device performance. Final Answer: Correct

Question 12

Magnitude of open-loop gain — op-amp fundamentals: Do operational amplifiers have very low open-loop gain, or is the open-loop gain typically very high (e.g., 10^5 or more at low frequency) before feedback sets the usable closed-loop gain?

Accepted Answer

Introduction / Context: The op-amp s power comes from its enormous open-loop gain (Aol), which allows negative feedback to precisely set a smaller, predictable closed-loop gain. Misstating the open-loop gain as “very low” inverts this fundamental idea and can lead to incorrect expectations about precision and linearity without feedback. Given Data / Assumptions: Typical general-purpose op-amps exhibit Aol on the order of 10^5 to 10^6 at low frequency. Frequency compensation rolls off Aol with frequency, but DC/low-frequency Aol is still very high. Closed-loop gain is set by external networks under feedback. Concept / Approach: High open-loop gain ensures that, under negative feedback, even tiny differential input errors produce the needed output to satisfy the feedback condition. This is why linear closed-loop gain can be accurate and largely resistor-ratio controlled. Therefore, the statement “an op-amp has very low open-loop gain” is false; the opposite is true. Step-by-Step Solution: 1) Recognize typical Aol values (e.g., 100 dB ≈ 10^5).2) Note that closed-loop gain Av_cl ≈ 1/β, provided Aol * β ≫ 1.3) The high Aol makes the input differential voltage extremely small in linear operation.4) Conclude that describing Aol as “very low” is incorrect. Verification / Alternative check: Datasheets for classic op-amps (e.g., 741, 358, 5532) list typical DC open-loop gains from tens to hundreds of thousands, confirming the high Aol premise. Why Other Options Are Wrong: Correct: Not correct; Aol is very high, not low. Only correct for unity-gain stable op-amps: Stability compensation does not reduce DC Aol to “very low.” Only correct above 1 MHz: Aol decreases with frequency, but not to “very low” at DC; the claim remains false. Common Pitfalls: Assuming the specified closed-loop gain (e.g., 10 or 100) equals the device’s open-loop gain; overlooking how feedback leverages large Aol for accuracy. Final Answer: Incorrect

Question 13

Op-amp topology identification (repaired for solvability): Without the actual schematic or component values, can we conclude that “this circuit is a differentiator” (i.e., an op-amp with a series input capacitor and a feedback resistor that outputs …

Accepted Answer

Introduction / Context: In analog electronics, an op-amp differentiator is a specific closed-loop configuration that produces an output proportional to the time-derivative of the input voltage. The classic inverting differentiator uses a capacitor in series at the input and a resistor in the feedback path. Without seeing the actual schematic, a caption like “this circuit is a differentiator” is not verifiable. This repaired question checks whether you can recognize when data is insufficient, while recalling what makes a differentiator distinct from other op-amp stages such as integrators or amplifiers with standard RC shaping networks. Given Data / Assumptions: No circuit diagram, node labeling, or component values are provided. We do not know if the input element is a capacitor (Cin) and if the feedback element is a resistor (Rf). We do not know biasing, compensation, or any frequency-limiting components that may alter the ideal differentiator response. Concept / Approach: The canonical inverting differentiator satisfies Vout = −Rf * Cin * d(Vin)/dt under small-signal, linear conditions. In practice, differentiators include added series/parallel resistors or capacitors to stabilize noise gain and limit high-frequency amplification. Determining whether a given schematic realizes this function requires inspection of the input/feedback elements and the intended bandwidth constraints. Step-by-Step Solution: 1) Define differentiator signature: input path contains a capacitor; feedback path contains a resistor.2) Confirm polarity and closed-loop path via the inverting input node.3) Check for stabilizing networks that still preserve the differentiating behavior over a target band.4) Without the actual diagram, none of these verifications can be performed. Verification / Alternative check: If a schematic were available, write the transfer function using impedance algebra: Z_in = 1/(s*Cin), Z_f = Rf, so Vout/Vin = −Z_f/Z_in = −Rf * s * Cin, which is proportional to s (d/dt). Absent the diagram, the transfer cannot be derived. Why Other Options Are Wrong: Always true / always false: Ignore topology; functionality depends on the exact RC placement.True only at very low frequencies or only for sine waves: A differentiator is defined by topology and transfer function, not by a single test signal. Common Pitfalls: Assuming any op-amp plus an RC is a differentiator; confusing integrator (input resistor, feedback capacitor) with differentiator (input capacitor, feedback resistor). Final Answer: Cannot be determined from the information provided

Question 14

Timer IC application — a 555 timer can be configured as an astable multivibrator (free-running oscillator) that produces a continuous square(ish) wave without an external trigger. Is this statement appropriate?

Accepted Answer

Introduction / Context: The ubiquitous 555 timer IC is used in three classic modes: monostable (one-shot), astable (free-running), and bistable (flip-flop). In astable mode, the device continuously charges and discharges a timing capacitor between two threshold levels using external resistors, producing a periodic output. The question asks whether the 555 can indeed be configured as an astable multivibrator without external triggering. Given Data / Assumptions: Standard bipolar or CMOS 555 device. External components RA, RB, and C are available for timing. Supply within device ratings; load does not excessively clamp the output. Concept / Approach: In astable mode, the 555 uses its internal comparators and SR latch to toggle when the capacitor voltage crosses 2/3 Vcc (threshold) and 1/3 Vcc (trigger). With RA and RB arranged so the capacitor charges through RA + RB and discharges through RB, the output oscillates continuously. The approximate formulas are: f ≈ 1 / (0.693 * (RA + 2RB) * C) and duty ≈ (RA + RB) / (RA + 2RB). A 50% duty cycle can be approached with additional diodes or alternative topologies but is not mandatory for oscillation. Step-by-Step Solution: 1) Tie threshold and trigger to the timing capacitor node.2) Choose RA, RB, C for the desired frequency and duty cycle.3) The discharge transistor alternately sinks the capacitor via RB, creating periodic charging and discharging.4) The output toggles high/low in sync with the comparator transitions, forming a free-running waveform. Verification / Alternative check: Prototyping on a breadboard with RA=10 kΩ, RB=10 kΩ, C=0.01 µF yields a frequency near 1/(0.693*(10k+20k)*0.01µF) ≈ 4.8 kHz, confirming astable operation. Why Other Options Are Wrong: Incorrect: contradicts standard application circuits.Only with dual supplies: the 555 operates from a single supply.Only in monostable mode: monostable is a different configuration.Only if 50% duty: duty can be asymmetrical; oscillation does not require exactly 50%. Common Pitfalls: Expecting an exact 50% duty without additional components; miswiring threshold/trigger nodes; using timing parts outside safe value ranges. Final Answer: Correct

Question 15

Op-amp internals — do operational amplifiers use internal capacitive coupling between stages, or are they primarily DC-coupled with internal frequency compensation capacitors?

Accepted Answer

Introduction / Context: Modern integrated op-amps are designed to amplify signals down to DC (0 Hz). To achieve this, their internal stages are DC-coupled, enabling the device to maintain bias conditions and linearity for very low frequencies. Confusion arises because many op-amps contain internal capacitors; those are not for stage-to-stage AC coupling but for frequency compensation and stability. Given Data / Assumptions: Monolithic bipolar or CMOS op-amps with differential input, high-gain intermediate stage(s), and output buffer. Target behavior includes amplification at DC (offset and bias considerations apply). Internal capacitors exist, commonly forming Miller compensation networks. Concept / Approach: Capacitive coupling between stages would deliberately block DC, which would prevent true DC amplification. Instead, op-amps use DC-coupled transistor stages (differential pairs, current mirrors, active loads) so that the device supports DC and low-frequency operation. Internal capacitors are used to roll off the open-loop gain (dominant pole), improving phase margin and closed-loop stability, not to AC-couple the signal chain. Step-by-Step Solution: 1) Identify the op-amp's intended function: amplify from DC to high frequency within limits.2) Recognize that DC coupling among stages is necessary to pass DC.3) Note that the internal capacitor implements compensation (e.g., Miller capacitor), shaping A_OL(jω) for stability.4) Conclude that “internal capacitive coupling between stages” is a mischaracterization. Verification / Alternative check: Datasheets and application notes show open-loop gain falling at the dominant pole set by compensation, while small-signal DC gain remains extremely high (thus permitting closed-loop DC amplification). Why Other Options Are Wrong: Correct/AC-coupled claim: would preclude DC operation and contradicts standard op-amp design.Only vacuum-tube or only rail-to-rail DC-coupled: DC coupling is common across technologies.Supply voltage dependence: coupling choice is architectural, not set by Vcc alone. Common Pitfalls: Confusing “compensation capacitor” with “coupling capacitor”; overlooking offset and bias networks that support DC accuracy. Final Answer: Incorrect — op-amp stages are DC-coupled; internal capacitors are for compensation, not AC coupling.

Question 16

Package/lead-count sanity check (repaired): Can we assert, without a datasheet or drawing, that a TO-5 metal-can operational amplifier package “can have 5 pins” as a standard op-amp configuration?

Accepted Answer

Introduction / Context: Lead counts for op-amp packages vary with function and era. Classic metal cans (TO-99, TO-5 variants) commonly expose 8 leads for single op-amps to accommodate inputs, supply rails, output, and optional trim pins. Some specialized or simplified devices can function with fewer external pins if features are omitted. This repaired question evaluates whether a blanket statement about “5 pins” can be accepted without a specific device reference. Given Data / Assumptions: No part number, pinout, or mechanical drawing is given. Different JEDEC outlines (TO-5 headers with varying lead counts) have existed. Functional requirements for op-amps include +IN, −IN, V+, V−, and OUT at minimum. Concept / Approach: An absolute claim about pin count cannot be validated without a datasheet. While many classic op-amps in metal cans used 8 leads, in principle a minimal single-supply op-amp could expose 5 leads (two inputs, output, positive rail, ground). Conversely, many precision op-amps need extra pins for offset null, compensation, or balance, making 5 pins insufficient. Therefore, the correct stance—given no device context—is that the claim cannot be confirmed or denied definitively. Step-by-Step Solution: 1) List minimum required pins for a basic op-amp: +IN, −IN, OUT, supply terminals.2) Recognize optional pins (offset null, compensation) change lead count.3) Observe that TO-5 headers exist with different lead configurations historically.4) Conclude the assertion needs a part-specific reference. Verification / Alternative check: By consulting an actual datasheet (e.g., a classic 741 in metal can), you would see typical 8-lead pinouts; other families might differ, proving that a blanket “5-pin” statement is not universally valid. Why Other Options Are Wrong: Always true / always false: overgeneralizations that ignore part-to-part variation.Vacuum-tube hybrids / offset-null-only caveats: speculative without specific part data. Common Pitfalls: Assuming a single historical package defines all op-amps; forgetting optional/legacy pins alter counts. Final Answer: Cannot be determined from the information provided

Question 17

Feedback mode identification (repaired): Without the actual schematic or signal routing, can we conclude that “this circuit is operating in closed-loop mode” (i.e., output fed back to an input to set defined gain/response)?

Accepted Answer

Introduction / Context: An op-amp operates in “closed-loop mode” when some form of feedback (usually negative) connects output to input to define gain, bandwidth, and linearity. In “open-loop mode,” there is no external feedback path, and the very high open-loop gain makes the amplifier saturate with tiny differential inputs. This repaired question tests your ability to recognize when insufficient information prevents classification of operating mode. Given Data / Assumptions: No schematic, feedback network, or node labeling is provided. We do not know if any resistor, capacitor, or network returns a fraction of the output to the inverting or non-inverting input. We assume a generic op-amp and generic source. Concept / Approach: Determining closed-loop operation requires identifying a feedback path and a reference node. Typical examples include inverting/non-inverting amplifiers, integrators, differentiators, and filters, all of which show a clear path from output to an input. Without the schematic, one cannot verify the presence or polarity of feedback or whether the device is saturated in open-loop. Step-by-Step Solution: 1) Inspect the diagram (if available) for a conductor from output to an input node through a network.2) Determine feedback polarity (negative vs positive) and the intended closed-loop gain.3) Confirm biasing so the op-amp remains in its linear region.4) With no diagram, none of the above can be verified, so no conclusion is possible. Verification / Alternative check: Given a schematic, derive the transfer function by writing node equations around the feedback network. Absence of a feedback term in the transfer indicates open-loop operation. Why Other Options Are Wrong: Always true/false: Mode depends on actual wiring.“Only if saturated” or “only unity gain”: Closed-loop gain can be any set value and saturation is not a requirement. Common Pitfalls: Assuming all op-amp circuits are closed-loop; overlooking intentional open-loop comparators that use op-amps. Final Answer: Cannot be determined from the information provided

Question 18

Voltage follower property — a unity-gain buffer (voltage follower) has a fixed closed-loop voltage gain approximately equal to 1, not 10. Is the statement “gain ≈ 10” appropriate for a voltage follower?

Accepted Answer

Introduction / Context: A voltage follower (also called a unity-gain buffer) connects the op-amp output directly to the inverting input and applies the signal to the non-inverting input. The closed-loop gain is approximately +1, providing high input impedance, low output impedance, and isolation without amplifying the voltage magnitude. The prompt claims a fixed gain near 10, which describes a different non-inverting amplifier configuration with an external resistor divider, not a follower. Given Data / Assumptions: Standard op-amp operated within linear region and bandwidth limits. Unity-feedback topology for a voltage follower. Neglect small gain error due to finite open-loop gain and frequency response. Concept / Approach: For a non-inverting amplifier with feedback network, the closed-loop gain is Av = 1 + (Rf/Rg). A follower is the special case Rf → ∞ and Rg → ∞ replaced by a direct short from output to inverting input, yielding Av ≈ 1. Any gain near 10 would require explicit resistor values (e.g., Rf/Rg ≈ 9), which is not a unity follower. Step-by-Step Solution: 1) Recognize unity-feedback topology of a follower.2) Write closed-loop gain for non-inverting form: Av = 1 + (Rf/Rg).3) For a follower (Rf = 0 from output to − input; Rg open), Av ≈ 1.4) Conclude that “gain ≈ 10” is incorrect for a voltage follower. Verification / Alternative check: Measure Vout/Vin of an actual follower over frequency within the device’s bandwidth; it will track ~1 with minimal phase/amp error until roll-off. Why Other Options Are Wrong: Claiming gain ≈ 10 conflates follower with a designed non-inverting amplifier of gain 10.Dual supplies or high frequency do not convert a follower into gain 10 without adding a divider.“Cannot be determined”: topology alone establishes the expected gain. Common Pitfalls: Using the term “buffer” for any non-inverting stage regardless of gain; ignoring finite bandwidth and stability considerations which slightly alter unity gain at extremes. Final Answer: Incorrect — a voltage follower has gain ≈ 1

Question 19

Astable timing sanity check — for an oscillator frequency of 54.86 kHz, is the period 1.82 s or something else? Use T = 1 / f to evaluate.

Accepted Answer

Introduction / Context: Whether using a 555 astable or any periodic oscillator, the fundamental relationship between frequency and period is T = 1 / f. An assertion that a 54.86 kHz signal has a period of 1.82 s is off by many orders of magnitude. This question reinforces quick order-of-magnitude checks that prevent unit mistakes when reading instruments or solving design problems. Given Data / Assumptions: Frequency f = 54.86 kHz = 54,860 Hz. Ideal periodic signal for which T = 1/f applies. We are checking numerical consistency only. Concept / Approach: Use T = 1 / f with careful unit handling. Since kilohertz means 10^3 Hz, a period for tens of kilohertz must be in tens of microseconds, not seconds. This mental check helps catch common calculator and unit-entry mistakes. Step-by-Step Solution: 1) Convert the frequency: f = 54,860 Hz.2) Compute T = 1 / 54,860 ≈ 0.00001823 s.3) Convert to microseconds: 0.00001823 s ≈ 18.23 µs.4) Compare to the claimed 1.82 s: the claim is wrong by a factor of ~100,000. Verification / Alternative check: Quick magnitude check: 50 kHz roughly corresponds to 20 µs (since 1/50,000 = 20 × 10^−6 s). Why Other Options Are Wrong: “Correct — 1.82 s”: inconsistent with T = 1/f.“Cannot be determined”: frequency alone is sufficient to compute period.Device type (CMOS vs bipolar) or duty cycle does not change T = 1/f for a given measured f. Common Pitfalls: Confusing kHz with Hz; misplacing decimal points; using minutes or milliseconds accidentally; relying on calculator without sanity checks. Final Answer: Incorrect — the period is approximately 18.2 µs

Question 20

Topology identification (repaired): Without the actual schematic, can we decide that “this circuit is a voltage follower” (unity-gain buffer with output tied to the inverting input and input applied to the non-inverting node)?

Accepted Answer

Introduction / Context: A voltage follower (unity buffer) is a cornerstone op-amp configuration used for impedance buffering and isolation. Identifying it requires seeing a direct feedback from output to the inverting input and the signal applied to the non-inverting input. Without a diagram, any assertion that a pictured circuit “is a follower” cannot be confirmed. This repaired item emphasizes correct identification criteria rather than guessing. Given Data / Assumptions: No schematic image or node labeling is available. We do not know whether the output is shorted to the inverting input. We do not know biasing or supply details. Concept / Approach: To verify a follower, confirm: (1) output returns to the inverting input with a direct connection, and (2) the input signal is applied to the non-inverting input. If a resistor divider exists instead, the circuit is a non-inverting amplifier with gain > 1; if the capacitor and resistor roles are swapped, it may be an integrator or differentiator. No such verification is possible without the schematic. Step-by-Step Solution: 1) Look for the unity feedback path (output → − input short).2) Confirm the signal enters at the + input.3) Ensure the op-amp is biased to operate linearly (not saturated).4) Since the drawing is absent, the classification cannot be made. Verification / Alternative check: Once a schematic is provided, measure Vout/Vin; a follower yields ≈1 over the usable bandwidth, with high input and low output impedance. Why Other Options Are Wrong: Always true/false: topology matters and cannot be assumed.Rail-to-rail/Gain 10/Dual supplies caveats are unrelated to identifying the follower topology. Common Pitfalls: Assuming any non-inverting stage is a follower; ignoring the effect of resistor ratios on gain. Final Answer: Cannot be determined from the information provided

Operational Amplifiers Questions