Question 1

7476 vs. 74LS76 — family and triggering style What is the correct relationship between the classic 7476 and the 74LS76 J–K flip-flop devices with preset and clear?

Accepted Answer

Introduction / Context: The 74xx TTL family and the 74LSxx low-power Schottky TTL family include many functionally equivalent parts with different speed and power characteristics. The 7476 and 74LS76 are both dual J–K flip-flops with asynchronous preset and clear. It is easy to confuse these with strictly edge-triggered variants such as the 74LS109, so knowing which are master–slave helps you read datasheets correctly and design reliable counters and registers. Given Data / Assumptions: Both parts implement J–K flip-flops with preset and clear. Part numbering: 7476 (standard TTL), 74LS76 (low-power Schottky). We are distinguishing master–slave vs. edge-triggered operation. Concept / Approach: Master–slave J–K flip-flops effectively sample on one phase and transfer on the opposite phase of the clock, achieving edge-like behavior but using two level-sensitive stages. Datasheet summaries and logic diagrams show that both 7476 and 74LS76 are master–slave devices; truly edge-triggered J–K devices carry different part numbers (e.g., 74LS109, 74LS112 families). Step-by-Step Solution: 1) Identify function blocks: both include master and slave latches.2) Confirm asynchronous inputs (preset, clear) exist on both.3) Recognize that the LS suffix denotes process improvement, not a change to edge-triggered architecture.4) Conclude both are master–slave J–K flip-flops. Verification / Alternative check: Cross-reference catalog listings: 7476/74LS76 are master–slave; 74LS109 is explicitly positive edge-triggered, highlighting the architectural distinction. Why Other Options Are Wrong: Both edge-triggered: confuses them with 74LS109-type devices. Mixed claims (edge vs. master–slave): contradict canonical datasheets. Common Pitfalls: Equating LS variants with functional changes; LS affects speed/power, not the fundamental logic function. Final Answer: the 7476 is master-slave, the 74LS76 is master-slave

Question 2

Transparent D latch — enable-controlled behavior For a level-sensitive D latch, which statement correctly describes the relationship between input D and output Q as a function of the enable input EN?

Accepted Answer

Introduction / Context: A D latch is a level-sensitive storage element. When enabled, it is transparent and passes input changes to the output; when disabled, it holds the last captured value. Recognizing this behavior is critical when building registers, pipelines, and two-phase latch-based timing systems. Given Data / Assumptions: Device: D (data) latch with enable input EN. Polarity: EN is active-HIGH in the typical configuration. Outputs: Q (and often Q' as complement). Concept / Approach: Transparency means that whenever EN is asserted HIGH, the internal gating path opens and Q immediately follows D (limited by propagation delay). When EN goes LOW, the path closes and the latch stores the last D value at that moment, maintaining Q until EN is asserted again. Step-by-Step Solution: 1) Consider EN = HIGH: data path enabled → Q reflects D.2) Consider EN = LOW: data path disabled → Q holds previous state.3) Therefore, the correct summary is: Q follows D when EN is HIGH.4) Complement output, if present, is simply Q' = NOT(Q). Verification / Alternative check: Examine timing diagrams: with EN HIGH, toggling D produces corresponding toggles on Q after a small delay; with EN LOW, Q remains flat regardless of D activity. Why Other Options Are Wrong: Following at EN LOW or always HIGH output contradict latch operation. Opposite at EN LOW implies inversion behavior that is not part of a standard D latch. Common Pitfalls: Confusing a D latch with an edge-triggered D flip-flop (which samples only at an edge), or misinterpreting EN polarity in specific vendor symbols. Final Answer: the Q output follows the D input when EN is HIGH

Question 3

Pulse-triggered flip-flops — functional description Which statement best describes the action of a pulse-triggered flip-flop in a synchronous digital system?

Accepted Answer

Introduction / Context: Pulse-triggered flip-flops are designed to respond to a narrow clock pulse rather than a sustained logic level. They bridge the conceptual gap between level-sensitive latches and ideal edge-triggered flip-flops by reducing the effective sampling window to a short interval. Understanding their behavior helps you reason about timing windows, metastability risk, and clock conditioning in real hardware. Given Data / Assumptions: The device expects a brief clock pulse (generated internally or externally). Data inputs are considered synchronous and must meet setup/hold relative to the pulse. We are describing functional behavior, not specific pin polarities. Concept / Approach: When a properly timed clock pulse arrives, the flip-flop samples the input and updates the output accordingly. Outside of that short pulse interval, input changes are ignored, thereby preventing level tracking. This emulates edge-triggered behavior sufficiently for many applications, assuming pulse width is controlled to avoid multiple toggles. Step-by-Step Solution: 1) Recognize that the device is inert until a clock pulse occurs.2) During the pulse, the data path is enabled and the input state is captured.3) After the pulse ends, the output holds the captured value.4) Proper setup/hold timing around the pulse prevents metastability. Verification / Alternative check: Inspect timing diagrams: the output transitions only during the short pulse; making the pulse too wide effectively reintroduces level sensitivity. Why Other Options Are Wrong: Pulsing S–R or synchronous inputs specifically is not required; the clock pulse is the trigger. Leading-edge enter and trailing-edge transfer describes master–slave operation, not pulse-triggered directly. Common Pitfalls: Using excessively long clock pulses that allow unintended multiple transitions; failing to meet setup/hold around the pulse edges. Final Answer: A pulse on the clock transfers data from input to output.

Question 4

J–K flip-flop hazard — naming the issue while clock is HIGH The phenomenon where unwanted signals on J and K are interpreted (causing spurious toggles) while the clock input Cp is HIGH is called what?

Accepted Answer

Introduction / Context: Level-sensitive J–K flip-flops can exhibit problematic behavior if their inputs change while the clock is at the active level. Understanding the specific terminology for these hazards helps in recognizing, explaining, and fixing erratic toggling in counters or control logic during lab diagnostics. Given Data / Assumptions: Device: J–K flip-flop with level-sensitive gating when Cp is HIGH. Inputs J and K might experience glitches or asynchronous changes. We focus on the undesired toggling that occurs during the active clock level. Concept / Approach: When Cp is HIGH and the internal gating is transparent, a stray 1 present on J or K may be captured, producing an unintended state change. This specific behavior is colloquially termed ones catching, reflecting that a momentary logic-1 on an input is caught and affects the output when it should not. Step-by-Step Solution: 1) Recognize the vulnerable window: Cp = HIGH.2) Identify the cause: transient or asynchronous 1s on J or K.3) Observe the effect: unintended toggle or set/reset action.4) Name the phenomenon: ones catching. Verification / Alternative check: Scope measurements often show brief spikes on J/K coincident with Cp HIGH leading to unexpected Q transitions; adding input synchronizers or converting to edge-triggered devices mitigates this. Why Other Options Are Wrong: Parity error checking: unrelated to flip-flop internal timing; pertains to data integrity methods. Digital discrimination/digital filtering: generic signal-processing terms, not the classic flip-flop hazard name. Common Pitfalls: Failing to synchronize asynchronous inputs, letting combinational hazards feed J/K directly, or using excessive clock HIGH width with level-sensitive designs. Final Answer: ones catching

Question 5

Classic S–R flip-flop — key limitation What is one commonly cited disadvantage of the basic S–R (Set–Reset) flip-flop implementation?

Accepted Answer

Introduction / Context: The S–R flip-flop is the simplest bistable built from cross-coupled gates (NOR or NAND). While educationally valuable, it has a notable limitation that makes it less desirable as a universal storage element in synchronous designs. Recognizing this limitation helps you choose D, J–K, or T flip-flops where deterministic behavior is required. Given Data / Assumptions: Standard S–R behavior using either NOR (active-HIGH S,R) or NAND (active-LOW S,R) implementations. Two outputs: Q and Q' (complement). No explicit clock in the simplest configuration; synchronous versions exist but inherit the same limitation. Concept / Approach: For the NOR-based S–R latch, the input combination S = 1 and R = 1 is forbidden because both outputs attempt to be driven LOW simultaneously, violating the intended complementary relationship. For the NAND version (active-LOW inputs), the forbidden combination is S = 0 and R = 0. This undefined/invalid state leads to unpredictable results when the inputs return to the non-asserted condition, including possible metastability or output ambiguity. Step-by-Step Solution: 1) Identify the invalid input combination for the chosen gate style (NOR or NAND).2) Understand that the outputs do not have a unique, stable resolution under that condition.3) Conclude that the device is unsuitable when both set and reset can be asserted simultaneously.4) Prefer D or J–K flip-flops to avoid undefined input states in synchronous systems. Verification / Alternative check: Simulate the latch: drive the forbidden inputs, then release them; observe output dependency on gate delays (race), illustrating the unpredictability. Why Other Options Are Wrong: No enable input: not a fundamental disadvantage; enables can be added externally. No clock input: true for the simple latch, but clocked S–R flip-flops exist. Single output: incorrect; S–R latches provide Q and Q'. Common Pitfalls: Assuming the invalid state is harmless, or forgetting which input polarity (NOR vs. NAND) defines the forbidden combination. Final Answer: It has an invalid state.

Question 6

Positive edge-triggered D flip-flop — operational description Which statement correctly describes how a positive edge-triggered D flip-flop captures and presents data at its output?

Accepted Answer

Introduction / Context: The positive edge-triggered D (data) flip-flop is the workhorse memory element of synchronous logic. Its behavior is simple: sample D at the rising edge of the clock and hold that value until the next rising edge. Understanding this sampling action is essential for designing pipelines, registers, and safe clock-domain crossings (with additional synchronizers). Given Data / Assumptions: Device: positive edge-triggered D flip-flop. Inputs: D and CLK (plus optional asynchronous set/clear not discussed here). Output: Q (and often Q'). Concept / Approach: At each LOW-to-HIGH clock transition, the flip-flop captures the current logic level present at D, provided setup and hold times are satisfied. Immediately after the edge (allowing for propagation delay), Q updates to that sampled value and remains stable for the rest of the clock period, regardless of subsequent changes on D. Step-by-Step Solution: 1) Identify the sampling instant: the leading (rising) edge.2) Apply the sampling rule: Q(next) = D(at the edge).3) Between edges: Q holds its previous state even if D toggles.4) Repeat at the next rising edge to update Q again. Verification / Alternative check: Review a timing diagram: align D transitions to meet setup/hold around the rising edge; Q transitions occur shortly after each rising edge only. Why Other Options Are Wrong: Toggle behavior with both inputs HIGH describes a J–K flip-flop, not a D FF. Invalid state with both LOW applies to S–R latches, not D FFs. Leading-edge capture and trailing-edge output transfer describes master–slave staging, not a single edge-triggered D FF. Common Pitfalls: Ignoring setup/hold requirements leading to metastability, or assuming Q follows D continuously during the HIGH level of the clock. Final Answer: The output will follow the input on the leading edge of the clock.

Question 7

Flip-Flop Concepts — Master–Slave J–K Behavior A master–slave J–K flip-flop is sometimes described as a pulse-triggered or level-triggered device. The statement claims: “input data is read during the entire time the clock pulse is at a LOW level.” S…

Accepted Answer

Introduction / Context: A classic master–slave J–K flip-flop consists of two latches clocked on opposite phases. The “master” is enabled during one clock level, while the “slave” is enabled during the opposite level. Overall behavior appears edge-triggered because the slave transfers the master state only when the clock transitions. Understanding which clock level is sampled avoids timing mistakes. Given Data / Assumptions: Device: master–slave J–K flip-flop built from two gated latches. Conventional design: if the device is labeled as positive-edge triggered, the master is transparent while the clock is HIGH, not LOW. The statement asserts that sampling happens the entire time the clock is LOW. Concept / Approach: In a master–slave arrangement, one latch is transparent while the clock is at its active level and the other is opaque, then roles swap on the opposite level. Hence, for a positive-edge equivalent device, the master reads inputs during the HIGH level. The slave updates the output when the clock goes LOW, producing an effective edge-triggered transfer. Step-by-Step Solution: Identify the active sampling window: for positive-edge behavior, inputs affect the master while CLK is HIGH.Determine when outputs update: the slave becomes transparent when CLK goes LOW, passing the master state to the output.Compare with the claim: the statement says inputs are read during the entire LOW level, which is not the general case.Conclude the statement is incorrect. Verification / Alternative check: Timing diagrams show J and K gating the master while CLK is HIGH. When CLK falls, the slave latches the master state, yielding behavior indistinguishable from edge triggering. Why Other Options Are Wrong: True: Does not match typical master–slave timing. Only true for asynchronous inputs: Asynchronous inputs (preset/clear) bypass the clock and are unrelated. True only for T mode: T toggling derives from J = K = 1 but does not change which level is sampled. Common Pitfalls: Assuming the slave sampling window equals the input sampling window; confusing level-sensitive internal latches with the overall edge-triggered behavior. Final Answer: False

Question 8

Shift Registers — Load and Unload Cycles For a serial-in/serial-out 8-bit shift register, how many clock pulses are required to completely load the 8-bit word and then completely unload it from the register output?

Accepted Answer

Introduction / Context: Shift registers move data one bit per clock edge. Knowing how many clock pulses are required to fill and then drain a serial register is fundamental for interface timing, buffering, and serial communication chores. Given Data / Assumptions: Register length: 8 bits. Serial-in/serial-out operation (one bit shifted per clock). Definition: “Load” means shifting an entire 8-bit word into the register; “Unload” means shifting that 8-bit word out to the serial output. Concept / Approach: Each clock shifts one bit. To load N bits serially, N pulses are required. After loading, the word occupies the register stages; to shift the same N bits out, another N pulses are needed unless there is parallel readback. Step-by-Step Solution: Loading: 8 bits → 8 clock pulses.Unloading: the stored 8 bits must be shifted to the output → 8 additional pulses.Total pulses = 8 + 8 = 16.Therefore, 16 clock pulses are required. Verification / Alternative check: Simulate with an 8-stage shift register: after the first 8 pulses, the word is fully inside; after an additional 8, it has fully exited at the serial output. Why Other Options Are Wrong: 2 or 4: Far too few for serial shifting of 8 bits. 8: Enough to load or unload, but not both. Common Pitfalls: Confusing serial and parallel load/unload modes; assuming data appears at the output immediately without counting the full shift length. Final Answer: 16

Question 9

555 Timer (Monostable) — Find R1 for Given Pulse Width A 555 timer operates as a monostable multivibrator with C1 = 100 µF. Determine the resistor R1 needed to obtain a pulse width of 500 ms using the standard relation t = 1.1 * R1 * C1.

Accepted Answer

Introduction / Context: The 555 timer in monostable mode produces a single output pulse whose width depends on an external resistor R1 and capacitor C1. Correct dimensioning of R1 for a target pulse width ensures timing accuracy in applications like debouncing and pulse stretching. Given Data / Assumptions: Pulsed width target t = 500 ms = 0.5 s. Capacitance C1 = 100 µF = 100 × 10^-6 F. Monostable formula: t = 1.1 * R1 * C1. Concept / Approach: Rearrange the monostable formula to solve for R1: R1 = t / (1.1 * C1). Substitute the given values and compute. Choose the closest standard value among the options. Step-by-Step Solution: Compute denominator: 1.1 * C1 = 1.1 * 100×10^-6 = 110×10^-6 = 0.00011.Compute R1: R1 = 0.5 / 0.00011 ≈ 4545.45 Ω.Express in kΩ: ≈ 4.545 kΩ, which rounds to 4.5 kΩ.Select the nearest offered value: 4.5 kΩ. Verification / Alternative check: Back-calculate t with R1 = 4.5 kΩ: t ≈ 1.1 * 4500 * 100×10^-6 ≈ 0.495 s, which is within 1% of 0.5 s and typical component tolerances dominate anyway. Why Other Options Are Wrong: 45 Ω or 455 Ω: Yield pulse widths near 5 ms and 50 ms, far too short. 455 kΩ: Produces t near 50 s, far too long. Common Pitfalls: Confusing µF with nF, or forgetting the factor 1.1; either error produces pulse widths off by orders of magnitude. Final Answer: 4.5 kΩ

Question 10

Flip-Flop Trigger Integrity — Clock Edge Rates For stable flip-flop triggering in synchronous digital circuits, what is the general rule regarding the rise and fall times of the clock pulse?

Accepted Answer

Introduction / Context: Clock quality strongly influences the reliability of edge-triggered flip-flops. Slow clock transitions can cause multiple internal crossings of threshold regions, leading to runt pulses, extra sampling windows, and metastability. Given Data / Assumptions: Flip-flops are edge-triggered devices with specified setup and hold constraints. Clock rise/fall time affects how cleanly the edge is detected. Noise and ringing are more problematic with sluggish edges. Concept / Approach: Fast (short) rise and fall times ensure the clock quickly crosses threshold regions, producing a single, unambiguous triggering event. Designers therefore strive for steep edges within the device specifications. Step-by-Step Solution: Identify the desired behavior: one clean trigger per cycle.Relate edge rate to threshold crossing: faster edges reduce time in the ambiguous region.Conclude that very short rise/fall times are preferred for stable triggering. Verification / Alternative check: Datasheets specify maximum allowable rise/fall times on the clock pin. Staying below these limits avoids double-clocking and skewed timing. Why Other Options Are Wrong: Very long: Increases risk of false triggers. Maximum value to stabilize inputs: Stabilization is handled by setup/hold windows, not by slowing the clock edges. Of no consequence: Rise/fall times are explicitly constrained in specifications. Common Pitfalls: Confusing clock frequency with edge rate; even low-frequency clocks should have fast edges. Final Answer: very short.

Question 11

Propagation Delay Definitions — tPLH Measurement For a synchronous flip-flop, propagation delay time tPLH is measured from which reference to which output transition?

Accepted Answer

Introduction / Context: Datasheets specify separate propagation delays for LOW-to-HIGH and HIGH-to-LOW output transitions. Correct interpretation of tPLH and tPHL ensures accurate timing analysis and setup of constraints in synchronous designs. Given Data / Assumptions: tPLH refers to the output transition from LOW to HIGH. Trigger reference is the active clock edge for edge-triggered devices. Asynchronous preset/clear are measured separately and are not part of tPLH unless explicitly stated. Concept / Approach: By definition, tPLH is the time from the specified input transition (usually the triggering clock edge) to the resulting LOW-to-HIGH transition at the output pin under given load and supply conditions. Step-by-Step Solution: Identify the input event: the triggering clock edge.Identify the output event: Q transitions from LOW to HIGH.Measure the time difference under specified conditions: that interval is tPLH. Verification / Alternative check: Open any logic family datasheet: tPLH and tPHL are listed separately with distinct typical and maximum values. Why Other Options Are Wrong: Clock to HIGH-to-LOW transition: That is tPHL. Preset or clear references: Those are asynchronous timing specs, not the synchronous tPLH definition. Common Pitfalls: Swapping tPLH and tPHL; mixing asynchronous and synchronous timing definitions. Final Answer: triggering edge of the clock pulse to the LOW-to-HIGH transition of the output

Question 12

555 Timer (Monostable) — Compute Output Pulse Width A 555 monostable circuit uses R1 = 4.7 kΩ and C1 = 47 µF. Using t = 1.1 * R1 * C1, determine the output pulse width and choose the closest option.

Accepted Answer

Introduction / Context: The 555 timer in monostable mode generates a single pulse each time it is triggered. Pulse width depends on the external timing resistor and capacitor. Accurate calculation is essential for timing, control, and pulse-shaping tasks. Given Data / Assumptions: R1 = 4.7 kΩ = 4700 Ω. C1 = 47 µF = 47 × 10^-6 F. Formula: t = 1.1 * R1 * C1. Concept / Approach: Apply the standard monostable timing relation directly and express the result in milliseconds for readability. Step-by-Step Solution: Compute t = 1.1 * 4700 * 47×10^-6.First multiply R1 * C1 = 4700 * 47×10^-6 ≈ 0.2209.Multiply by 1.1: 0.2209 * 1.1 ≈ 0.243 s.Convert to milliseconds: 0.243 s ≈ 243 ms. Verification / Alternative check: Dimensional check: Ω * F yields seconds; numerical value aligns with typical 555 timing ranges for kΩ and tens of µF. Why Other Options Are Wrong: 24 s: Off by three orders of magnitude. 24 ms: Off by approximately a factor of 10. 243 µs: Off by a factor of 1000. Common Pitfalls: Misplacing metric prefixes (µF vs nF) or forgetting the 1.1 factor. Final Answer: 243 ms

Question 13

Clock Input Symbol — Meaning of the Triangle on a J–K Flip-Flop In logic symbols, a small triangle drawn at the clock input pin indicates which clocking characteristic for a J–K flip-flop?

Accepted Answer

Introduction / Context: Schematic symbols communicate device behavior at a glance. The clock input styling on flip-flops tells the designer whether the device responds to levels or edges, and sometimes which edge is active. Given Data / Assumptions: The symbol in question shows a triangle at the clock pin. No bubble is mentioned; a bubble would indicate inversion (negative edge). Standard digital schematic notation is assumed. Concept / Approach: A triangle at the clock input denotes edge triggering. If a bubble accompanies the triangle, it indicates triggering on the falling edge; without a bubble it typically indicates rising-edge triggering. A plain input without a triangle usually indicates level sensitivity (gated latches). Step-by-Step Solution: Identify symbol element: triangle at CLK.Map meaning: triangle → edge-triggered behavior.Infer edge polarity: absence or presence of a bubble refines rising vs falling edge, but the triangle itself means edge-triggered. Verification / Alternative check: Datasheet symbols and IEEE standards use the triangle to denote edge-sensitive clocking; a small circle indicates inversion. Why Other Options Are Wrong: Level enabled: That would be a gate without the triangle, as in a transparent latch. Asynchronous operation: Asynchronous set/clear pins have their own labels and are not implied by the triangle. Schmitt-trigger input: Usually shown by a hysteresis symbol, not a triangle at CLK. Common Pitfalls: Assuming the triangle specifies rising edge only; polarity is indicated by an additional bubble, not by the triangle alone. Final Answer: edge-triggered

Question 14

One-Shot Terminology — Alternate Name In pulse generation and timing circuits, a “one-shot” is commonly referred to by which formal multivibrator classification?

Accepted Answer

Introduction / Context: Multivibrators are fundamental timing circuits classified by the number of stable states. The term “one-shot” is widely used in practical electronics to describe a circuit that produces a single pulse of fixed duration in response to a trigger. Given Data / Assumptions: The device generates one pulse per trigger and then returns to its resting state. It is not free-running; it requires an external trigger. Classic implementations include 555 timers and monostable logic configurations. Concept / Approach: A monostable multivibrator has one stable state and one quasi-stable state. When triggered, it temporarily switches to the quasi-stable state for a duration set by R and C, then returns to the stable state, hence “one-shot.” Step-by-Step Solution: Define stable states: monostable → one stable, bistable → two stable, astable → none stable (free-running).Match behavior: a single output pulse per trigger corresponds to a monostable.Conclude that “one-shot” equals “monostable.” Verification / Alternative check: 555 monostable application notes describe the same behavior and explicitly use “one-shot” and “monostable” interchangeably. Why Other Options Are Wrong: Multivibrator: Generic class name; not specific. Bistable: Two stable states (flip-flop), not a one-shot pulse. Astable: Free-running oscillator with no trigger required. Common Pitfalls: Confusing monostable with astable due to similar RC timing formulas; remembering that astable oscillates continuously while monostable needs a trigger. Final Answer: Monostable

Question 15

Active-HIGH S–R latch truth conditions In an active-HIGH input S–R latch (set–reset latch using positive logic), under which input condition does an INVALID state occur?

Accepted Answer

Introduction / Context: The set–reset (S–R) latch is the most fundamental memory element in digital electronics. With active-HIGH inputs, asserting S requests Q = 1 (set), and asserting R requests Q = 0 (reset). Understanding which input combinations are allowed and which are forbidden is crucial for reliable design, hazard avoidance, and debug of sequential circuits. Given Data / Assumptions: Latch uses active-HIGH inputs S and R (positive logic). Outputs are Q and Q_bar, ideally complementary. We seek the condition that produces an invalid or indeterminate state. Concept / Approach: In an active-HIGH S–R latch, the truth behavior is: S = 1, R = 0 → set (Q = 1); S = 0, R = 1 → reset (Q = 0); S = 0, R = 0 → hold last state; S = 1, R = 1 → invalid/forbidden because both cross-coupled gates are driven to assert their outputs simultaneously, breaking the complement relationship and risking metastability when released. Step-by-Step Solution: List valid cases: (S,R) = (1,0) → set; (0,1) → reset; (0,0) → no change.Identify forbidden case: (S,R) = (1,1) drives both storage nodes toward the same value.Conclude: simultaneous HIGH on S and R is the invalid state. Verification / Alternative check: Analyze cross-coupled NOR implementation (common for active-HIGH): when both inputs = 1, both NOR outputs tend to 0. Releasing to 0,0 can leave the latch in an unpredictable state depending on tiny timing differences, demonstrating the forbidden nature. Why Other Options Are Wrong: Both LOW: this is the normal memory/hold condition. S = 0, R = 1: valid reset request. S = 1, R = 0: valid set request. Common Pitfalls: Confusing active-HIGH with active-LOW conventions; forgetting that the invalid combination is the simultaneous assertion of both control inputs; assuming the latch will always recover to a known state after (1,1) is released—it may not. Final Answer: HIGHs are applied simultaneously to both inputs S and R

Question 16

555 timer in astable (free-running) mode — timing network contents In a standard 555 astable configuration, which external components form the timing network that sets the output frequency and duty cycle?

Accepted Answer

Introduction / Context: The ubiquitous 555 timer can operate in astable mode to generate a continuous square or rectangular waveform. The oscillation frequency and duty cycle depend on a simple timing network connected around the internal comparators and discharge transistor. Recognizing the exact component set used in the astable circuit is a foundational skill for electronics students and technicians. Given Data / Assumptions: Mode: 555 astable (free-running oscillator). We are considering the classic single-capacitor, two-resistor timing network. No exotic variants (such as CMOS 555 tweaks) are assumed. Concept / Approach: In the standard astable configuration, two resistors (often named RA and RB) and one timing capacitor (C) determine the charge and discharge intervals between thresholds of approximately 1/3 Vcc and 2/3 Vcc. The discharge transistor periodically grounds the node between RB and C to create the low interval. Frequency and duty cycle are derived from RA, RB, and C with well-known formulas. Step-by-Step Solution: Identify components: RA (from Vcc to timing node), RB (from timing node to discharge pin), C (from timing node to ground).Charging path: Vcc → RA → RB → C until voltage reaches 2/3 Vcc.Discharging path: C → RB → discharge transistor to ground until 1/3 Vcc.Thus, exactly two resistors and one capacitor set f and duty cycle. Verification / Alternative check: Re-derive frequency: f ≈ 1 / (0.693 * (RA + 2RB) * C). Duty cycle ≈ (RA + RB) / (RA + 2RB). Both expressions explicitly reference two resistors and one capacitor. Why Other Options Are Wrong: Three resistors: unnecessary for the basic astable. One resistor, two capacitors: not the standard 555 astable; some duty-cycle-symmetry tricks exist but are not the canonical circuit. No external R/C: impossible; the 555 requires an external timing network to oscillate. Common Pitfalls: Mixing astable with monostable component counts; forgetting that duty cycle can be made 50% using a diode across RB (still the same basic parts plus a diode). Always remember: 2 resistors + 1 capacitor define the timing. Final Answer: two external resistors and an external capacitor are used

Question 17

Active-HIGH S–R latch behavior after removing a reset An active-HIGH S–R latch has S = 0 and R = 1 (reset asserted). When R transitions to 0 while S remains 0, what is the resulting latch state?

Accepted Answer

Introduction / Context: Latches store one bit of information by feeding outputs back to inputs. With an active-HIGH S–R latch, S requests Q = 1 and R requests Q = 0. Understanding what happens when a momentary assertion is removed is essential for predicting memory behavior and designing glitch-free control signals. Given Data / Assumptions: Initial inputs: S = 0, R = 1 (reset command active). Transition: R → 0 while S remains 0. Latch is level-sensitive; after inputs return to 0, the latch holds its last state. Concept / Approach: Input combinations for active-HIGH S–R latch are: (1,0) set, (0,1) reset, (0,0) hold, (1,1) invalid. The moment R = 1 forces Q = 0 (reset). When R goes back to 0 and S is 0, inputs become (0,0), which is the hold condition. Therefore, the latch remains in the previously established reset state. Step-by-Step Solution: Apply reset: with R = 1 and S = 0, Q is driven to 0.Remove reset: now S = 0 and R = 0.Hold condition: (0,0) stores the last value, which is Q = 0.Conclude: the latch remains RESET. Verification / Alternative check: Cross-coupled NOR realization confirms the above: asserting R drives its NOR output low and the complementary feedback high. Returning to (0,0) leaves the feedback loop stable at the reset condition. Why Other Options Are Wrong: SET: would require S = 1 at some point, which did not occur. clear: ambiguous wording; in common usage it means reset, but the explicit correct term here is RESET. invalid: only when S = 1 and R = 1 simultaneously. Common Pitfalls: Believing that releasing a control input automatically returns the latch to a default; in fact, latches hold the last commanded state when both controls are inactive. Final Answer: RESET

Question 18

7475 (transparent latch) vs. 7474 (edge-triggered D flip-flop) What is the key difference between the 7475's enable input and the 7474's clock input?

Accepted Answer

Introduction / Context: Different sequential devices capture data under different timing rules. The 7475 is a pair of level-sensitive (transparent) latches, while the 7474 is a pair of edge-triggered D flip-flops. Designers must understand whether data passes through continuously while an enable is asserted or whether it is sampled only on a specific clock edge to avoid hazards and race conditions. Given Data / Assumptions: Part 7475: transparent latch with an enable (often active HIGH). Part 7474: D flip-flop with a clock (typically edge-triggered, positive edge in the standard part). We compare sensitivity to level vs. edge. Concept / Approach: A transparent latch (7475) passes D to Q while the enable level is active; when the enable is inactive, the output holds the last value. An edge-triggered flip-flop (7474) samples D only at a transition (rising edge for positive-edge-triggered) of the clock, holding the sampled value thereafter, regardless of D changes until the next edge. Step-by-Step Solution: Identify behavior of 7475: level-sensitive enable → transparent during enable = 1.Identify behavior of 7474: edge-triggered clock → captures data at a clock edge only.State the key difference succinctly: 7474 is edge-triggered; 7475 is level-sensitive. Verification / Alternative check: Examine timing diagrams: with a changing D while enable is HIGH, 7475 Q follows; for 7474, Q changes only at the clock edge and ignores mid-cycle D variation. Why Other Options Are Wrong: 7475 edge-triggered: incorrect—this device is level-sensitive. Enable asynchronous vs clock synchronous: misleading; both participate in synchronous design, but their sensitivities differ (level vs edge). Both level-sensitive: false for the 7474. Common Pitfalls: Using a transparent latch where an edge-triggered FF is needed can create races; conversely, assuming an FF will pass data during the high level of the clock is also wrong. Final Answer: The 7474 is edge-triggered.

Question 19

Origin of the J and K input names on the J–K flip-flop What is the commonly accepted significance of the letters J and K in the J–K flip-flop input designations?

Accepted Answer

Introduction / Context: Flip-flops are standardized sequential elements, and the J–K flip-flop generalizes the S–R with additional behavior when both inputs are HIGH (toggle). Many learners wonder why the inputs are labeled J and K rather than S and R or other letters. Understanding the naming origin prevents propagation of myths and focuses attention on function rather than folklore. Given Data / Assumptions: Device: J–K flip-flop with inputs J and K. We are evaluating the significance of the letters themselves. We are not questioning the functional behavior of the device. Concept / Approach: Historically, textbooks and manufacturer literature do not attribute a definitive meaning to J and K as acronyms. Various anecdotes circulate (e.g., J for jump, named after Jack Kilby), but these lack authoritative corroboration and are considered apocryphal. The safest and most widely accepted stance in academic and exam contexts is that there is no special significance to J and K beyond being distinct labels for the two control inputs. Step-by-Step Solution: Survey common explanations: 'jump,' 'kill,' or homages—none are standard or consistently referenced in datasheets.Compare to the consistent, formal functional definition: when J = 1 and K = 0 → set; J = 0 and K = 1 → reset; J = K = 0 → hold; J = K = 1 → toggle.Conclude that function, not name origin, is what matters; the letters have no official significance. Verification / Alternative check: Examine manufacturer datasheets and reputable textbooks; none assign a semantic expansion to J and K, reinforcing that they are simply identifiers. Why Other Options Are Wrong: 'Jump' explanation: not a formal or universal definition. 'In honor of Jack Kilby': a common myth; not supported by documentation. 'All other letters used': humorous but not factual. Common Pitfalls: Relying on folklore in technical justifications; always prefer documented functional behavior over anecdotal naming stories. Final Answer: There is no known significance in their designations.

Question 20

Clocked S–R flip-flop response with S = 0 and R = 0 If both synchronous inputs of an S–R flip-flop are LOW, what happens at the active clock edge?

Accepted Answer

Introduction / Context: Clocked (synchronous) S–R flip-flops sample inputs only at a clock edge, unlike level-sensitive latches. Knowing the action associated with each input combination at the sampling instant is essential for designing state machines and avoiding forbidden operations. Given Data / Assumptions: Synchronous S and R inputs are both LOW (S = 0, R = 0). A valid clock edge occurs (positive or negative depending on device). Asynchronous inputs (if any) are inactive. Concept / Approach: For a clocked S–R flip-flop, the behavior at the active edge mirrors the basic latch logic: S = 1, R = 0 → set; S = 0, R = 1 → reset; S = 0, R = 0 → no change; S = 1, R = 1 → invalid. With both inputs LOW, the flip-flop simply holds its prior state through the clock event. Step-by-Step Solution: Identify condition at sample time: S = 0 and R = 0.Apply synchronous truth behavior: (0,0) maps to 'hold'.Therefore, the outputs remain unchanged at the clock edge. Verification / Alternative check: Consult a standard S–R FF characteristic table: the entry for S = R = 0 at the active clock indicates Q_next = Q_previous (no change). Why Other Options Are Wrong: Invalid: occurs only when S and R are both asserted (1,1) at the sampling moment. Toggle: that is characteristic of a J–K FF with J = K = 1, not of an S–R FF. Reset: would require R = 1 at the sampling instant. Common Pitfalls: Confusing the clocked S–R FF with a level latch; assuming that a clock edge forces a change even when inputs request hold; mixing S–R behavior with J–K toggling rules. Final Answer: No change will occur in the output.

Flip-Flops Questions