Question 1

In operational amplifier applications, a voltage follower (unity-gain buffer) is configured with 100% negative feedback. Which characteristic best describes its closed-loop behavior?

Accepted Answer

Introduction: A voltage follower, also called a unity-gain buffer, is a core operational amplifier configuration used to isolate stages, present very high input impedance, and provide very low output impedance without amplifying the voltage. This question checks recognition of the defining closed-loop characteristic of a follower. Given Data / Assumptions: Ideal op-amp assumptions: infinite open-loop gain, infinite input impedance, zero output impedance, and sufficient bandwidth for the frequencies of interest. Feedback is 100% from output to inverting input; the input signal is applied to the non-inverting input. We focus on closed-loop characteristics, not open-loop parameters. Concept / Approach: With unity feedback, the closed-loop transfer drives the op-amp to make the output equal to the input. Hence the closed-loop voltage gain Av_cl equals 1 (unity). This preserves the input signal level while providing buffering between source and load. The circuit is widely used to prevent loading effects and to build active filters and instrumentation chains. Step-by-Step Solution: 1) In a follower, route Vout directly to the inverting input and apply Vin to the non-inverting input.2) The op-amp forces its differential input toward zero; thus Vout approaches Vin.3) The closed-loop gain Av_cl = Vout / Vin equals 1 for ideal conditions.4) The key benefit is buffering: high input impedance and low output impedance, minimizing source loading. Verification / Alternative check: Write the non-inverting gain formula Av_cl = 1 + (Rf / Rin). For a follower, Rin approaches infinity and Rf short equals 1. In the canonical derivation for unity feedback, the limit yields Av_cl = 1, confirming the identity. Why Other Options Are Wrong: Small open-loop voltage gain: Open-loop gain of an op-amp is typically very large, not small, and the question asks closed-loop behavior. Closed-loop bandwidth of zero: Incorrect; followers often extend bandwidth compared to higher closed-loop gains. Large closed-loop output impedance: The follower provides low output impedance, not large. Closed-loop voltage gain greater than unity: By definition the follower has a gain of 1. Common Pitfalls: Confusing open-loop and closed-loop parameters, or believing that any feedback changes voltage gain above unity. In a follower, feedback sets the gain exactly to 1 while improving drive capability and linearity. Final Answer: closed-loop voltage gain of unity

Question 2

In op-amp circuits and analog signal references, what is the key distinction between a real ground and a virtual ground?

Accepted Answer

Introduction: Virtual ground is a common term in op-amp analysis, especially for inverting configurations and split-supply emulation from a single supply. Understanding how a virtual ground differs from a true ground node is crucial for correct biasing and current return planning. Given Data / Assumptions: Ideal op-amp operation for small signals around the operating point. Virtual ground node created by feedback, not by a physical connection to ground or earth. Conventional definitions: ground is a true reference node that can source or sink return current; virtual ground is only a maintained potential. Concept / Approach: A true ground is a physical reference node at 0 V that typically serves as the return path for currents in the circuit. A virtual ground is a node held at a voltage (often 0 V) by active feedback; it behaves as a voltage reference but is not inherently a current sink or source unless the circuit around it is designed to provide that capability. Step-by-Step Explanation: 1) In an inverting op-amp stage, feedback strives to keep the inverting input at the same potential as the non-inverting input.2) If the non-inverting input is at 0 V, the inverting input node is held near 0 V by feedback, forming a virtual ground.3) That node has very high input impedance; negligible current should enter the op-amp input.4) Therefore, while the voltage is approximately 0 V, the node does not serve as a general-purpose return path for unrelated currents. Verification / Alternative check: Analyze nodal currents: the current through the input resistor flows through the feedback network, not into the op-amp input pin, demonstrating that the node is a maintained potential rather than a current return. Why Other Options Are Wrong: Current reference: A virtual ground is not guaranteed to absorb or supply return current. Power reference: Power delivery relates to voltage and current together, not just a reference node potential. Difference reference: Not a standard term for reference nodes. Earth safety return: Safety earth is a protective connection; virtual ground is unrelated to mains safety. Common Pitfalls: Using a virtual ground node as a return for high currents or unrelated subcircuits can overload the op-amp or bias network and cause large errors. Treat it as a voltage reference only unless a proper rail splitter with current capability is provided. Final Answer: voltage reference

Question 3

A differential amplifier has a common-mode gain Acm = 0.2 and a common-mode rejection ratio CMRR = 3250. If a single-ended input of 7 mV RMS is applied to one input with the other input at 0 V, what is the resulting output voltage (RMS)?

Accepted Answer

Introduction: Differential amplifiers respond to both differential and common-mode components of their inputs. This problem assesses correct use of CMRR to find differential gain and then compute the output for a single-ended stimulus, which contains both a differential part and a common-mode part. Given Data / Assumptions: Common-mode gain Acm = 0.2 (unitless voltage gain). Common-mode rejection ratio CMRR = 3250 = Ad / Acm. Single-ended input: v1 = 7 mV RMS, v2 = 0 V. Standard definitions: v_d = v1 − v2, v_cm = (v1 + v2)/2. Concept / Approach: From CMRR, compute the differential gain Ad = CMRR * Acm. Then form the output as v_out = Ad * v_d + Acm * v_cm, noting that signs depend on which input is designated as non-inverting; we report the magnitude here, which is conventional for such problems. Step-by-Step Solution: 1) Compute Ad: Ad = 3250 * 0.2 = 650.2) Compute differential and common-mode inputs: v_d = 7 mV − 0 = 7 mV; v_cm = (7 mV + 0)/2 = 3.5 mV.3) Differential contribution: Ad * v_d = 650 * 0.007 V = 4.55 V.4) Common-mode contribution: Acm * v_cm = 0.2 * 0.0035 V = 0.0007 V (0.7 mV), which is negligible compared to 4.55 V.5) Total output magnitude ≈ 4.55 V + 0.0007 V ≈ 4.5507 V, which rounds to 4.55 V RMS. Verification / Alternative check: Since CMRR is high, the differential term dominates. Ignoring the tiny common-mode term gives 4.55 V, confirming the same answer within 0.02%. Why Other Options Are Wrong: 1.4 mV rms and 9.1 mV rms: Much too small; they ignore the large differential gain. 650 mV rms and 0.455 V rms: Off by factors of approximately 7 or 10; they misapply CMRR or confuse Ad with Acm. Common Pitfalls: Confusing CMRR as a dB quantity when a linear ratio is given, neglecting that single-ended drive creates both v_d and v_cm, or forgetting that Ad = CMRR * Acm. Final Answer: 4.55 V rms

Question 4

For a standard inverting op-amp amplifier, which adjustment will decrease the magnitude of the closed-loop gain without changing topology or supply voltages?

Accepted Answer

Introduction: Closed-loop gain control in the inverting amplifier relies on the resistor ratio. Knowing which component to adjust to reduce the magnitude of gain is fundamental for practical design and troubleshooting. Given Data / Assumptions: Inverting configuration with input resistor Rin and feedback resistor Rf. Ideal op-amp behavior for gain calculation. We are interested in adjusting component values, not the input signal level. Concept / Approach: The inverting closed-loop gain magnitude is |Av| = Rf / Rin. To decrease |Av|, either reduce Rf or increase Rin. Changes to the input signal amplitude do not alter gain; they only change output level for a given gain. Step-by-Step Solution: 1) Start with |Av| = Rf / Rin.2) To make |Av| smaller, increase the denominator (Rin) or decrease the numerator (Rf).3) Among the offered options, increasing the input resistor value directly increases Rin and reduces |Av|.4) Therefore, the correct choice is to increase the value of the input resistor. Verification / Alternative check: Example: If Rf = 10 kΩ and Rin = 1 kΩ, |Av| = 10. Doubling Rin to 2 kΩ reduces |Av| to 5, confirming the relationship. Why Other Options Are Wrong: Reducing input amplitude: Changes output level but does not change gain. Increasing the feedback resistor: Increases |Av| because the numerator grows. Removing the feedback resistor: Forces near open-loop conditions, making gain extremely large and uncontrolled. Decreasing the input resistor: Reduces Rin, which increases |Av|. Common Pitfalls: Confusing signal amplitude with gain, or overlooking that the inverting gain depends purely on resistor ratio, not on op-amp open-loop gain within its linear range. Final Answer: increasing the value of the input resistor

Question 5

In a bipolar-input operational amplifier, the input offset current I_os is defined in terms of the input bias currents into the two input transistors. Which expression matches I_os?

Accepted Answer

Introduction: Two key DC parameters of bipolar-input op-amps are input bias current and input offset current. Distinguishing these improves DC error budgeting in precision analog design. Given Data / Assumptions: Two nominally matched input transistors draw base currents IB1 and IB2. Definitions are according to standard op-amp terminology. We consider magnitudes; signs indicate which input draws slightly more current. Concept / Approach: The input bias current I_b is defined as the average of the two base currents: I_b = (IB1 + IB2) / 2. The input offset current I_os is the absolute difference: I_os = |IB1 − IB2|. These definitions separate common DC loading from mismatch-related error. Step-by-Step Solution: 1) Identify IB1 and IB2 as currents entering the op-amp inputs.2) Form the average: I_b = (IB1 + IB2)/2.3) Compute the difference: I_os = |IB1 − IB2|.4) Compare to options; the difference between the two base currents matches the definition of input offset current. Verification / Alternative check: Consult any op-amp datasheet glossary: bias current and offset current are always defined as average and difference, respectively, validating the selection. Why Other Options Are Wrong: Average of two base currents: That is the input bias current, not I_os. Collector current divided by current gain: That relates to estimating a single transistor base current, not the definition of I_os. Difference between two base-emitter voltages: Defines input offset voltage mechanisms, not currents. Sum of base currents: Not a standard DC error metric for op-amps. Common Pitfalls: Mixing up I_b and I_os, or assuming offset current is negligible. In precision circuits with high source resistance, even nanoamp differences can create millivolt-level errors. Final Answer: difference between two base currents

Question 6

In feedback amplifier terminology, what defines negative feedback in terms of the relationship between the returning feedback signal and the input signal?

Accepted Answer

Introduction: Negative feedback is the foundation of linear amplifier stability, accuracy, and bandwidth extension. Recognizing how the feedback signal is combined with the input is critical to predict circuit behavior. Given Data / Assumptions: Standard block-diagram convention with summing junction at the input. Feedback network samples a portion of the output and returns it to the input. Signals are considered at the same reference phase for comparison. Concept / Approach: Negative feedback means the returned signal is subtracted from the input. In other words, the feedback opposes the input error. This reduces distortion, desensitizes the closed-loop gain to device parameters, and improves linearity. In contrast, positive feedback adds to the input and can cause oscillation. Step-by-Step Solution: 1) Denote input as Vin and feedback as Vf.2) In negative feedback, the error signal is Ve = Vin − Vf.3) The amplifier acts to reduce Ve toward zero, stabilizing the output.4) Therefore, the returning signal necessarily opposes the input signal. Verification / Alternative check: Write closed-loop gain: Av_cl = A / (1 + Aβ) for negative feedback. The plus sign in the denominator arises from subtraction at the summing junction, confirming opposition of Vf to Vin. Why Other Options Are Wrong: Proportional to output current: Feedback can sample voltage or current; proportionality alone does not define polarity. Proportional to differential voltage gain: Not the defining feature of negative feedback. Aids the input signal: That is positive feedback, risking instability or oscillation. Proportional to input bias current: Irrelevant to feedback sign. Common Pitfalls: Assuming any feedback is stabilizing. Only negative feedback that subtracts the appropriate proportion of output from the input yields predictable closed-loop behavior. Final Answer: opposes the input signal

Question 7

A closed-loop inverting amplifier has a gain magnitude of 3.9. If the input resistor Rin = 1.6 kΩ, what value of feedback resistor Rf is required to realize this gain?

Accepted Answer

Introduction: Inverting op-amp gain design is a straightforward resistor ratio problem. This item confirms fluency with the formula and with consistent unit handling between ohms and kilohms. Given Data / Assumptions: Desired closed-loop magnitude |Av| = 3.9. Rin = 1.6 kΩ. Ideal op-amp approximation for gain computation. Concept / Approach: For an inverting amplifier, the magnitude of the closed-loop gain is |Av| = Rf / Rin. Solve for Rf by multiplying the target gain by Rin. Step-by-Step Solution: 1) Use |Av| = Rf / Rin.2) Rearrange: Rf = |Av| * Rin.3) Substitute: Rf = 3.9 * 1.6 kΩ = 6.24 kΩ.4) Convert to ohms if desired: 6.24 kΩ = 6240 Ω. Verification / Alternative check: Check back: |Av| = 6240 Ω / 1600 Ω = 3.9, confirming the design value. Why Other Options Are Wrong: 2.4 kilohms and 0.62 kilohms: These produce gains of 1.5 and 0.39 respectively, not 3.9. 410 ohms: Yields a gain magnitude of approximately 0.256, far from the target. 6.24 megaohms: Off by 1000×; unit mistake between kΩ and MΩ. Common Pitfalls: Mixing units (kΩ vs Ω) or overlooking that the sign of gain is negative in inverting mode, while the question asks only for magnitude. Final Answer: 6240 ohms

Question 8

In op-amp networks, replacing a resistor with a capacitor changes the operation. What does the placement of that capacitor determine?

Accepted Answer

Introduction: Active integrators and differentiators are first-order building blocks for filtering, control, and analog computation. The key idea is that swapping a resistor for a capacitor in the input path or in the feedback path transforms an amplifier into a differentiator or integrator, respectively. Given Data / Assumptions: Ideal op-amp behavior in the frequency range of interest. Small-signal linear operation (no slew-rate limiting or saturation). Single capacitor substitution relative to the classic inverting topology. Concept / Approach: In the inverting configuration, placing the capacitor in the feedback path yields an integrator: Vout is proportional to the time integral of Vin. Placing the capacitor in the input path yields a differentiator: Vout is proportional to the time derivative of Vin. Proper resistor-capacitor networks are added to stabilize real circuits and limit high-frequency noise gain. Step-by-Step Explanation: 1) For an integrator, use a resistor at the input and a capacitor in feedback.2) The transfer is Vout = −(1 / (R * C)) * ∫ Vin dt, within the op-amp linear region.3) For a differentiator, use a capacitor at the input and a resistor in feedback.4) The transfer is Vout = −R * C * (dVin/dt), within the bandwidth where the ideal model is valid. Verification / Alternative check: Examine impedances: For integrator, Zf = 1 / (sC) grows at low frequency, forcing low-frequency gain to be small and high-frequency gain to be larger, consistent with integration. For differentiator, Zin = 1 / (sC) falls with frequency, so high-frequency components dominate, consistent with differentiation. Why Other Options Are Wrong: Open- or closed-loop gain: Closed-loop gain is set by network ratios; capacitor placement specifically selects integration vs differentiation behavior. Saturation or cutoff: Those are operating region terms, not topological functions determined by the capacitor location. Addition or subtraction: Summing or differencing relates to multi-input resistor networks, not capacitor placement. Slew rate or bandwidth limit directly: While capacitors affect bandwidth, the principal role in this context is to shape the transfer as an integrator or differentiator. Common Pitfalls: Building an ideal differentiator without damping resistors causes excessive high-frequency noise and potential instability. Practical designs include series and parallel resistors to bound gain. Final Answer: integration or differentiation

Question 9

In op-amp applications, which amplifier configuration produces an output that is directly proportional to the algebraic sum (addition) of two or more input voltages?

Accepted Answer

Introduction: In analog signal processing with operational amplifiers, engineers often need to combine multiple input signals into a single output that represents their algebraic sum. The configuration that performs this function is known as the summing amplifier (also called an adder). It is widely used in audio mixing, sensor fusion, DAC current-to-voltage conversion, and weighted arithmetic operations. Given Data / Assumptions: We are considering ideal op-amp behavior (very high open-loop gain, infinite input impedance, and zero output impedance) operated with negative feedback in the linear region. Multiple input sources are applied through input resistors to the inverting or non-inverting node, depending on the chosen topology. The question asks specifically for the configuration whose output is proportional to the addition of the inputs. Concept / Approach: The classic inverting summing amplifier ties several input resistors to the inverting input. Because of virtual ground at the summing node, the currents through each input resistor sum at the node and flow through the feedback resistor. The output voltage is thus proportional to the sum of the input voltages (with an overall sign and scaling factor set by resistor ratios). For equal resistors, the magnitude of the output is proportional to the simple sum of the inputs. Step-by-Step Solution: Write nodal relation at the inverting input: (V1 − 0)/R1 + (V2 − 0)/R2 + … + (Vn − 0)/Rn = (0 − Vout)/RfSolve for Vout: Vout = −Rf * (V1/R1 + V2/R2 + … + Vn/Rn)Equal-value case: if R1 = R2 = … = Rn = R, then Vout = −(Rf/R) * (V1 + V2 + … + Vn)Therefore, the output is directly proportional to the algebraic sum of inputs (with a sign defined by inversion and weights by resistor ratios). Verification / Alternative check: Consider V1 = 1 V and V2 = 2 V with R1 = R2 = R and Rf = R. Then Vout = −(1 + 2) V = −3 V, demonstrating proportional summation. A non-inverting summer can also be built by first forming a passive average and then amplifying, again yielding proportional addition. Why Other Options Are Wrong: Differentiator: Produces output proportional to the time derivative of the input, not the sum of multiple inputs. Difference / analog subtractor: Implements subtraction (V1 − V2), not general summation of many inputs. Integrator: Produces output proportional to the time integral of the input, not algebraic addition of multiple inputs. Common Pitfalls: Confusing weighted addition with simple buffering; resistor values set the weights. Ignoring op-amp saturation; large sums can exceed supply rails. Forgetting the inversion sign in the inverting summing configuration; a non-inverting adder avoids inversion but requires a different network. Final Answer: summing

Question 10

For an ideal inverting op-amp with negative feedback, the magnitude of the closed-loop voltage gain A_cl is equal to which ratio of resistances?

Accepted Answer

Introduction: The inverting amplifier is a foundational op-amp building block. Its closed-loop gain is set precisely by passive components, making the result largely independent of the very large but uncertain open-loop gain. Knowing the exact relationship between the feedback resistor and the input resistor is essential for predictable amplification in sensor interfaces, audio stages, and data acquisition. Given Data / Assumptions: Ideal op-amp operation with negative feedback (linear region). Single input applied through input resistor Rin to the inverting input; feedback resistor Rf from output to inverting input. Non-inverting input tied to ground (or a reference) for the classic inverting stage. Concept / Approach: Virtual ground holds the inverting input near 0 V. The input current through Rin must equal the feedback current through Rf. This yields a direct algebraic relation between Vout and Vin. The sign indicates inversion; the magnitude depends only on the resistor ratio. Step-by-Step Solution: Apply KCL at the inverting node: (Vin − 0)/Rin + (Vout − 0)/Rf = 0Rearrange: Vout/Rf = −Vin/RinSolve for gain: A_cl = Vout/Vin = −Rf/RinMagnitude: |A_cl| = Rf/RinThus, the magnitude equals the feedback resistance divided by the input resistance. Verification / Alternative check: If Rin = 10 kΩ and Rf = 100 kΩ, then A_cl = −100 kΩ / 10 kΩ = −10, so |A_cl| = 10. SPICE simulations and lab measurements confirm that as long as A_ol is very large and the op-amp remains unsaturated, the closed-loop gain tracks Rf/Rin closely across frequency where loop gain is high. Why Other Options Are Wrong: The ratio of input to feedback resistance: This gives 1/|A_cl|, not the magnitude. The open-loop voltage gain A_ol: Enormous and device-dependent; closed-loop gain is set by feedback, not by A_ol directly. The input resistance: A resistance value is not a dimensionless gain. The reciprocal of β: β depends on the feedback network topology; for an inverting stage, |A_cl| = Rf/Rin, not simply 1/β. Common Pitfalls: Forgetting the negative sign (inversion) when quoting A_cl. Ignoring op-amp bandwidth and slew-rate limits that can reduce effective closed-loop gain at high frequency or large signals. Using tolerances poorly; mismatch in Rin and Rf changes the exact gain and noise performance. Final Answer: the feedback resistance divided by the input resistance

Question 11

In a free-running (astable) multivibrator, what initiates the oscillations at power-up and sustains continuous switching without any external trigger?

Accepted Answer

Introduction: An astable multivibrator is a classic oscillator that generates a continuous stream of pulses or square waves with no stable operating point. It is used in clock generation, flashing indicators, and timing circuits. The question probes understanding of what starts and sustains such oscillation in the absence of external triggers. Given Data / Assumptions: A standard astable multivibrator using positive feedback and RC timing (implemented with op-amps, comparators, or cross-coupled transistors). Power is applied to the circuit, and no external trigger input is provided. Active devices and passive components are subject to unavoidable small variations and noise. Concept / Approach: Because the astable has no stable equilibrium, any infinitesimal disturbance pushes the output away from its metastable point. Thermal noise, device offset voltages, and component mismatches provide the initial nudge immediately after power-up. Positive feedback then drives the output toward one state while an RC network times the transition to the opposite state, creating ongoing oscillation without any external assistance. Step-by-Step Solution: At power-up, offsets and noise create a small voltage difference at the comparator or amplifier input.Positive feedback amplifies this difference rapidly, forcing the output to a rail.The timing capacitor charges or discharges through a resistor toward a reference threshold.When the threshold is crossed, the feedback polarity flips and the output switches to the opposite rail.This cycle repeats indefinitely, establishing a free-running oscillation without any trigger. Verification / Alternative check: In practice, even perfectly symmetrical designs begin oscillating due to microscopic asymmetries or noise. Simulation with a tiny initial condition bias or noise source reproduces the same spontaneous startup behavior. Why Other Options Are Wrong: A trigger / an input signal / an external circuit: These are not required for astable operation; they are used in monostable or externally clocked multivibrators. A one-shot pulse: Describes monostable startup, not an astable that runs continuously. Common Pitfalls: Confusing astable (free-running) with monostable (one-shot) or bistable (flip-flop). Assuming a perfect balance would stop oscillation; in reality, offsets and noise always exist. Overlooking power-supply ramp behavior, which itself can provide the initial disturbance. Final Answer: nothing

Question 12

An operational amplifier has differential gain A_d = 50,000 and common-mode gain A_cm = 2. What is the common-mode rejection ratio (CMRR) expressed in decibels (dB)?

Accepted Answer

Introduction: Common-mode rejection ratio (CMRR) quantifies how well a differential amplifier suppresses signals that are common to both inputs relative to signals that are different between them. A high CMRR is critical for rejecting power-line hum, reference shifts, and electromagnetic interference in sensor front-ends and instrumentation systems. Given Data / Assumptions: Differential gain A_d = 50,000. Common-mode gain A_cm = 2. Standard definition of CMRR, and conversion to decibels using 20 * log10(ratio). Concept / Approach: CMRR (ratio) is defined as A_d / A_cm. To express it in decibels, use CMRR_dB = 20 * log10(A_d / A_cm). A larger ratio means the amplifier strongly prefers differential signals and rejects common disturbances. Step-by-Step Solution: Compute ratio: CMRR = A_d / A_cm = 50,000 / 2 = 25,000Convert to dB: CMRR_dB = 20 * log10(25,000)Evaluate log10(25,000) ≈ 4.39794CMRR_dB ≈ 20 * 4.39794 ≈ 87.9588 dBRounded to one decimal place: 87.9 dB Verification / Alternative check: As a quick estimate, 20 * log10(10,000) = 80 dB and 20 * log10(2.5) ≈ 8 dB; total ≈ 88 dB, consistent with the precise calculation. The resulting value is typical for general-purpose op-amps and below that of precision instrumentation amplifiers. Why Other Options Are Wrong: −87.9 dB / −43.9 dB: CMRR is a suppression ratio and is conventionally reported as a positive dB magnitude. 43.9 dB: Would correspond to a ratio of roughly 155, far below the computed 25,000. 25,000 (unitless ratio): This is the raw ratio, not the requested dB value. Common Pitfalls: Using 10 * log10 instead of 20 * log10 for voltage or gain ratios. Confusing CMRR with PSRR (power-supply rejection ratio). Reporting the ratio without specifying units or scale, which can mislead design decisions. Final Answer: 87.9 dB

Question 13

In legacy op-amp packaging, how many leads are commonly available for the TO-5 style metal can version of an operational amplifier?

Accepted Answer

Introduction: Operational amplifiers have been produced in many package styles. Before plastic dual in-line packages and modern surface-mount options dominated, metal can packages were common in laboratory, military, and high-reliability gear. Knowing typical lead counts helps when reading historical schematics or selecting compatible replacements. Given Data / Assumptions: The focus is on the TO-5/TO-99 style round metal can versions used for op-amps. We consider the range of commercially common lead counts that supported various pinouts (standard op-amp pins, offset null pins, and extra connections). Question asks for how many leads are commonly available, not just a single exact count for one specific device. Concept / Approach: Classic metal can op-amp families appeared with multiple header configurations to accommodate different functions. While the most familiar versions provide 8 leads for basic op-amp pins, other variants with 10 or 12 leads exist to bring out offset null, balance, or additional pins. Therefore, the common range encompasses 8, 10, or 12 leads for the metal can style. Step-by-Step Solution: Identify baseline: 8-lead cans are widely used for standard op-amps (power, inputs, output, and optional null pins combined).Account for expanded pinouts: 10- and 12-lead cans support extra functions like offset adjustment and special connections.Therefore, the commonly encountered lead counts are 8, 10, or 12. Verification / Alternative check: Surveys of historical datasheets for popular can-packaged op-amps show 8-lead as the norm and 10/12-lead variants for special versions, aligning with the stated range and typical parts catalogs from the era. Why Other Options Are Wrong: 6, 8, or 10: Six leads are uncommon for general-purpose op-amps; insufficient pins for standard functions. 8 or 14 / 8 or 16: 14 and 16 are standard DIP or SOIC counts, not typical for round metal cans. 4 or 6: Far too few for a full-feature op-amp with dual supplies and offset functions. Common Pitfalls: Confusing metal can lead counts with modern DIP/SOIC counts. Assuming all cans are 8-lead only; extended versions exist for specialized functions. Overlooking the difference between device families (general-purpose versus instrumentation types). Final Answer: 8, 10, or 12

Question 14

A circuit that combines an active device (such as an op-amp) with resistors and capacitors to realize a frequency-selective response is referred to as what type of filter?

Accepted Answer

Introduction: Filters are used to shape spectra by passing desired frequency bands and attenuating others. When a filter uses an active element such as an operational amplifier together with passive resistors and capacitors, it is called an active filter. These are ubiquitous in audio, instrumentation, and communications front-ends due to their predictable response and ability to provide gain and buffering. Given Data / Assumptions: Presence of an amplifier (op-amp, BJT, FET) in combination with RC networks. Goal is to realize a frequency-selective transfer function (low-pass, high-pass, band-pass, or band-stop). No inductors are required; the active device can synthesize inductor-like behavior when needed. Concept / Approach: Active filters use negative feedback around an amplifier to implement precise poles and zeros. By choosing component values, one can realize standard responses (Butterworth, Bessel, Chebyshev, multiple feedback, Sallen–Key) with controllable cutoff frequency, quality factor, and gain. The amplifier also isolates stages, presenting high input impedance and low output impedance. Step-by-Step Solution: Identify components: op-amp + R and C form the building block.Select topology: e.g., Sallen–Key for low-pass/high-pass or multiple-feedback for band-pass.Set design targets: choose fc, Q, and passband gain by calculating resistor and capacitor values.Implement and verify: simulate or measure magnitude and phase to confirm the intended frequency response. Verification / Alternative check: Active filters provide gain (unlike purely passive RC filters), maintain load isolation, and allow precise tuning. Replacing the amplifier with a passive network would remove these advantages and change the classification to a passive filter. Why Other Options Are Wrong: Relaxation or harmonic oscillator: These generate periodic signals rather than selective filtering of an input. Signal generator: Broad term for sources, not selective networks. Differential amplifier: A building block that amplifies differences; not itself a filter unless combined with RC networks in a feedback configuration. Common Pitfalls: Assuming an op-amp alone makes a circuit an active filter; the RC network and feedback define the filtering. Ignoring op-amp bandwidth and slew-rate limits that can distort high-frequency responses. Loading effects between cascaded stages when buffering is omitted. Final Answer: active filter

Question 15

Operational amplifier basics — Inverting amplifier input impedance (Zin): In a standard inverting op-amp configuration using an ideal op-amp, with input resistor Ri feeding the inverting (−) input and a feedback resistor Rf from output to the invert…

Accepted Answer

Introduction: In the classic inverting amplifier, understanding input impedance helps you size the driving source and limit noise and loading. With an ideal operational amplifier, the virtual short between inputs and the very high input resistance at the op-amp terminals simplify the small-signal input impedance seen by the source. Given Data / Assumptions: Ideal op-amp (infinite open-loop gain, infinite input resistance, zero input bias currents) Inverting topology: source → Ri → inverting node; Rf from output to inverting node Noninverting input tied to ground (0 V) Linear (unsaturated) operation with negative feedback Concept / Approach: In linear operation, the inverting node is held near 0 V (virtual ground). The source therefore “sees” a resistor to virtual ground equal to Ri. Because the op-amp input draws negligible current, essentially all source current flows through Ri, not into the op-amp pin. Step-by-Step Solution: Assume virtual ground: V− ≈ 0 V because V+ = 0 V and Aol ≫ 1Compute input current: Iin ≈ Vin / Ri (no current into op-amp input)Small-signal input impedance: Zin ≈ Vin / Iin = RiFeedback resistor Rf does not appear in Zin for the source at the inverting input Verification / Alternative check: More exact analysis with finite open-loop gain still yields Zin ≈ Ri for high gains; deviations are small for typical op-amps in closed-loop use. Why Other Options Are Wrong: Rf + Ri or Rf − Ri: Rf is in the feedback path, not in series with the source at small-signal input. ∞ (infinite): That describes the op-amp input pin, not the network seen by the source. Depends primarily on op-amp open-loop gain: In practice Zin is dominated by Ri in closed loop. Common Pitfalls: Confusing op-amp input pin resistance with network input impedance. Forgetting the virtual-ground assumption only holds in linear, unsaturated operation. Final Answer: Ri

Question 16

Op-amp virtual ground concept — If ground is applied to the noninverting (+) terminal of an inverting amplifier, what is the voltage condition at the inverting (−) terminal under linear operation?

Accepted Answer

Introduction: The idea of “virtual ground” is central to analyzing inverting op-amp circuits. It enables simple current and gain calculations and explains why the inverting topology presents a predictable input impedance and precise closed-loop gain. Given Data / Assumptions: Ideal or high-gain op-amp Negative feedback from output to inverting input Noninverting input tied to ground (0 V) Linear (unsaturated) operation with sufficient supply rails Concept / Approach: With negative feedback, the op-amp drives its output so that the differential input voltage (V+ − V−) is approximately zero. If V+ is fixed at 0 V (ground), then V− is held very near 0 V. The point is called a “virtual ground” because it is at (approximately) ground potential without being physically connected to ground. Step-by-Step Solution: Set V+ = 0 V (physical ground)With Aol ≫ 1 and feedback present, enforce V+ − V− ≈ 0 → V− ≈ 0 VNo significant current flows into op-amp input; currents go through external resistorsThis allows simple current summation at the inverting node for gain derivations Verification / Alternative check: Write node equation at V− with ideal op-amp input current ≈ 0 and solve; solution converges to V− ≈ V+ = 0 V under closed-loop operation. Why Other Options Are Wrong: not need an input resistor: The input resistor sets gain and input impedance; it is required. have high reverse current: Ideal input current is ~0; reverse conduction is not the concept. not invert the signal: Inverting stage by definition inverts. float at an undefined potential: Feedback fixes the node near 0 V. Common Pitfalls: Assuming virtual ground equals a physical short to ground—there is no direct connection. Applying virtual ground logic when the op-amp is saturated or feedback is broken. Final Answer: be virtual ground

Question 17

Noninverting amplifier node voltage — A noninverting op-amp has RIN = 1.0 kΩ (from inverting node to ground) and RFB = 2.5 kΩ (from output to inverting node). If 1.42 mV is applied to the noninverting (+) input, what is the voltage at the RIN node (…

Accepted Answer

Introduction: In the noninverting amplifier, the inverting node is forced to follow the noninverting input via negative feedback. Recognizing this “virtual short” (V− ≈ V+) lets you immediately find node voltages and then, if needed, the closed-loop gain and output voltage. Given Data / Assumptions: RIN = 1.0 kΩ from inverting node to ground RFB = 2.5 kΩ from output to inverting node Vin at noninverting (+) = 1.42 mV Ideal/high-gain op-amp with negative feedback and linear operation Concept / Approach: For a noninverting configuration, the closed-loop action makes V− ≈ V+. Therefore, the inverting node sits at approximately the same voltage as the applied noninverting input. This is independent of the actual RIN and RFB values so long as the loop is closed and not saturated. Step-by-Step Solution: Virtual short condition: V− ≈ V+Given V+ = 1.42 mV → V− ≈ 1.42 mVTherefore, the node at the top of RIN (the inverting input node) is ≈ 1.42 mV above ground(Optional) Output voltage: Vout = (1 + RFB/RIN) * Vin = (1 + 2.5/1.0) * 1.42 mV ≈ 4.97 mV Verification / Alternative check: Solving the resistor divider with V− constrained to equal V+ yields the same node voltage; the feedback forces the equality to minimize differential input. Why Other Options Are Wrong: 3.5 mV / 0.56 mV: Do not follow from the virtual short condition. ground: The node is at the input value, not 0 V. 4.97 mV: That is the approximate output voltage, not the inverting-node voltage. Common Pitfalls: Confusing the inverting node voltage with the output voltage. Assuming RIN and RFB form a fixed divider to ground regardless of feedback (they do not; V− is constrained by the loop). Final Answer: 1.42 mV

Question 18

Differential vs. common-mode inputs — What is the key difference between common-mode and differential-mode input signals for a differential amplifier (such as an op-amp front end)?

Accepted Answer

Introduction: Differential amplifiers respond to differences between two input nodes while rejecting signals that are common to both. Distinguishing between common-mode and differential-mode signals is crucial for instrumentation, noise rejection, and accurate sensor interfacing. Given Data / Assumptions: Two input terminals: noninverting (+) and inverting (−) Linear operation of the differential front end Focus on signal composition rather than absolute amplitudes Concept / Approach: Common-mode (CM) signals are those that are substantially the same on both inputs (same magnitude and phase). Differential-mode (DM) signals are equal in magnitude but opposite in polarity (180° phase difference) or simply unequal, creating a nonzero difference. Thus, the defining distinction is their phase/polarity relationship across the two inputs. Step-by-Step Explanation: Represent inputs: V1 and V2Common-mode component: Vcm = (V1 + V2) / 2 (same phase on both inputs)Differential component: Vd = V1 − V2 (opposite polarity increases |Vd|)Differential amplifier ideally amplifies Vd and rejects Vcm Verification / Alternative check: Examine waveforms: identical waveforms in phase on both inputs are CM; two equal-amplitude, opposite-phase waveforms are DM. This shows phase relationship is the salient differentiator. Why Other Options Are Wrong: voltage/current/apparent power/impedance: These may vary, but they do not fundamentally define CM vs DM; the phase/polarity relationship does. Common Pitfalls: Assuming any equal amplitudes imply CM even when phases differ. Ignoring that practical amplifiers have finite CMRR, so some CM leaks into the output. Final Answer: phase relationship

Question 19

Closed-loop gain from resistor ratio — A feedback amplifier has a feedback/input resistor ratio RFB/RIN = 4.6 and is driven at the noninverting (+) input with 1.7 V. Assuming a standard noninverting configuration in linear operation, what is the out…

Accepted Answer

Introduction: For a noninverting amplifier, the closed-loop gain depends on the feedback network ratio. Correctly mapping the stated “feedback/input resistor ratio” to the noninverting gain expression ensures accurate output predictions and avoids misinterpreting saturation or cutoff conditions when rails are not specified. Given Data / Assumptions: Noninverting configuration RFB/RIN = 4.6 Vin (+ input) = 1.7 V Ideal op-amp, linear range (supply rails wide enough) Concept / Approach: The closed-loop gain for the noninverting amplifier is Av = 1 + (RFB / RIN). Multiply this by the input voltage to obtain the output voltage, provided the result is within supply rails and the amplifier remains unsaturated. Step-by-Step Solution: Compute gain: Av = 1 + 4.6 = 5.6Compute output: Vout = Av * Vin = 5.6 * 1.7 VVout = 9.52 VAssume rails permit 9.52 V so the amplifier stays linear Verification / Alternative check: Rearrange to check: Vin = Vout / Av = 9.52 / 5.6 = 1.7 V, confirming consistency. Why Other Options Are Wrong: 7.82 V: Uses an incorrect gain factor. saturation / cutoff: No evidence of rail limits or biasing to force nonlinearity. 4.6 V: Misreads the ratio as the gain itself and ignores the “+1” term. Common Pitfalls: Forgetting the “+1” in the noninverting gain formula. Confusing RFB/RIN with RIN/RFB. Final Answer: 9.52 V

Question 20

Open-loop polarity test — In an op-amp with open loop (no negative feedback), if the inverting (−) input is at a lower potential than the noninverting (+) input (i.e., V− < V+), what direction will the output swing?

Accepted Answer

Introduction: Understanding open-loop polarity is foundational for troubleshooting op-amp circuits. It also explains why negative feedback stabilizes the output while positive feedback can lead to latching or oscillation. The sign of the differential input dictates the extreme to which the output saturates in open loop. Given Data / Assumptions: No feedback (open-loop operation) V+ and V− compared directly by very large gain Aol Supply rails available to allow saturation Concept / Approach: The op-amp output is approximately Vout ≈ Aol * (V+ − V−). If V+ > V−, the quantity (V+ − V−) is positive, so the output rushes toward the positive rail. Conversely, if V− > V+, the output rushes negative. Step-by-Step Reasoning: Given V− < V+ → (V+ − V−) > 0Large Aol multiplies this small differenceTherefore Vout drives toward + saturationSign determines direction; magnitude set by Aol and rails Verification / Alternative check: Introduce a tiny positive offset at V+ and observe simulation or bench result: the output saturates positive in the absence of feedback, matching the sign rule. Why Other Options Are Wrong: swing negative: Would require V− > V+. close the loop / be balanced: These are not behaviors of an open-loop configuration. oscillate about zero: Not inherently; open-loop DC conditions saturate to a rail. Common Pitfalls: Assuming the output stays near zero without feedback—it does not. Confusing the labels of + and − inputs, which flips the outcome. Final Answer: swing positive

Operational Amplifiers Questions