Question 1

74x75/TTL knowledge check — device capacity How many storage elements (latches/flip-flops) are contained in the 7475 series IC as commonly used in digital labs?

Accepted Answer

Introduction / Context: The 7475 is a classic TTL device providing multiple level-sensitive bistable elements in a single package. Recognizing the capacity of such parts is helpful when budgeting resources for small counters, registers, or interface latches in educational and prototyping environments. Given Data / Assumptions: Device: 7475 (TTL), often described as a “4-bit bistable latch.” Each storage element holds one bit. We distinguish between level-sensitive latches (7475) and edge-triggered flip-flops (e.g., 7474). Concept / Approach: The catalog description “4-bit latch” indicates four independent storage elements in one IC. While some texts casually use “flip-flop” as a generic term, the 7475 is technically a set of transparent latches controlled by an enable signal, not edge-triggered flip-flops. Functionally, however, it provides four 1-bit storage locations. Step-by-Step Reasoning: Identify device family and function: 7475 = quad (4) latch.Count bits stored: 4 latches ⇒ 4 one-bit storage elements.Therefore, the IC contains four storage elements. Verification / Alternative check: Vendor datasheets and laboratory manuals list “Quad Latch” in the title block for 7475, confirming the count. Pinouts show repeated latch sections (Q, /Q, D, Enable) four times. Why Other Options Are Wrong: 1 or 2: underestimate the part’s capacity. 8 or 16: confuse with octal or larger register devices (e.g., 74HC373/374 are octal). Common Pitfalls: Equating 7475 with 7474 (which is a dual edge-triggered D flip-flop). Final Answer: 4

Question 2

555 timer in astable mode — adjust duty cycle to 0.5 without changing frequency Given a standard astable 555 configuration, how should R1 and R2 be adjusted to correct the duty cycle to 0.5 (50%) while keeping the output frequency unchanged?

Accepted Answer

Introduction / Context: In the classic astable 555 (without diodes), the timing is set by R1, R2, and C. The frequency is f ≈ 1.44 / ((R1 + 2R2) * C), and the duty cycle D ≈ (R1 + R2) / (R1 + 2R2). Designers often need to move D toward 0.5 while holding frequency constant. This requires coordinated changes in R1 and R2 rather than altering the capacitor or supply conditions. Given Data / Assumptions: Standard 555 astable (single timing capacitor C; charge through R1 + R2, discharge through R2). Target duty cycle D_target = 0.5. Frequency constraint: keep (R1 + 2R2)C constant. Concept / Approach: From D = (R1 + R2) / (R1 + 2R2), moving toward 0.5 requires reducing the numerator relative to the denominator. Decreasing R1 lowers the numerator more strongly than the denominator, while increasing R2 can compensate in the denominator so that (R1 + 2R2) remains unchanged to preserve frequency. Note: with the basic circuit, exact 50% ideally requires R1 → 0; in practice, we reduce R1 substantially and increase R2 appropriately (or add a steering diode for perfect 50%). Step-by-Step Solution: Hold K = (R1 + 2R2)C constant (frequency fixed).Adjust R1 ↓ to reduce D toward 0.5.Adjust R2 ↑ so R1 + 2R2 stays constant, keeping f unchanged.Result: D approaches 0.5 while f remains the same. Verification / Alternative check: Compute a sample: suppose initially R1 = 10 kΩ, R2 = 10 kΩ, C fixed. K = (10 + 20)kΩC = 30kC. To move closer to 0.5, choose R1 = 1 kΩ; then to keep K, solve 1 + 2R2 = 30 ⇒ R2 ≈ 14.5 kΩ. New D ≈ (1 + 14.5)/(1 + 29) ≈ 15.5/30 ≈ 0.517 (closer to 0.5). Why Other Options Are Wrong: Increase C: changes frequency directly. Change Vcc or RL: does not control duty cycle in the ideal 555 timing equations. Decrease both R1 and R2: frequency will change unless tuned precisely; not the directed move toward D = 0.5. Increase R1 and decrease R2: pushes D away from 0.5 for a fixed K. Common Pitfalls: Expecting exact 50% without adding a charge/discharge steering diode; the standard topology cannot achieve perfect D = 0.5 unless R1 approaches zero. Final Answer: Decrease R1 and increase R2.

Question 3

Asynchronous (ripple) counter — interpreting stage #4's first toggle Four J–K flip-flops are cascaded as a ripple divider. When the output of the fourth stage toggles for the first time, how many input clock pulses have occurred at the counter input?

Accepted Answer

Introduction / Context: Asynchronous (ripple) counters cascade toggling stages, each dividing the frequency of the previous stage by 2. Understanding the relationship between stage number and division factor enables quick reasoning about count values based on observed toggles at intermediate stages. Given Data / Assumptions: Four J–K flip-flops configured to toggle (J = K = 1) in a ripple chain. Stage numbering: divider #1 sees the input clock; divider #4 is the fourth stage in the chain. Each stage divides by 2 relative to its input. Concept / Approach: Stage n in a ripple counter has output frequency f_in / 2^n and period 2^n input clock periods. A single transition (LOW→HIGH or HIGH→LOW) occurs every half-period of that stage, i.e., every 2^(n−1) input pulses. Therefore, the first toggle at stage #4 happens after 2^(4−1) = 8 input pulses. Step-by-Step Solution: Divider #1: toggles every 1 input pulse (2^0).Divider #2: toggles every 2 pulses (2^1).Divider #3: toggles every 4 pulses (2^2).Divider #4: toggles every 8 pulses (2^3) ⇒ first change at 8. Verification / Alternative check: Think in binary count: after 8 input clocks, the 4-bit count is 1000₂, which corresponds to the MSB (stage #4) making its first transition from 0 to 1, matching the reasoning above. Why Other Options Are Wrong: 16: that would be a full cycle of stage #4 (two toggles), not the first toggle. 4 or 2: correspond to stages #3 and #2, respectively. 1: is the toggle interval of stage #1 only. Common Pitfalls: Confusing “toggle” (half-period) with “full period.” Stage #4’s period is 16 input pulses; its first toggle appears at 8. Final Answer: 8

Question 4

555 monostable timing — compute the output pulse width For a one-shot (monostable) 555 timer with R1 = 3.3 kΩ and C1 = 0.02 µF (20 nF), determine the expected output pulse width using the standard formula t = 1.1 * R * C.

Accepted Answer

Introduction / Context: The 555 timer configured as a monostable generates a single output pulse of width t when triggered. The pulse width is set by the external timing resistor and capacitor according to the widely used relation t = 1.1 * R * C (for the basic one-shot connection). Correct unit handling is vital when choosing component values or interpreting lab measurements. Given Data / Assumptions: R1 = 3.3 kΩ = 3300 Ω. C1 = 0.02 µF = 20 nF = 20 × 10^-9 F. Formula: t = 1.1 * R * C. Concept / Approach: Substitute the given component values into t = 1.1 * R * C and convert units carefully. Convert kilo-ohms to ohms and microfarads to farads before multiplying. The resulting time will be in seconds; convert to microseconds (µs) for a convenient expression. Step-by-Step Solution: Convert R: 3.3 kΩ = 3300 Ω.Convert C: 0.02 µF = 20 nF = 20 × 10^-9 F.Compute RC: 3300 * 20 × 10^-9 = 66 × 10^-6 s = 66 µs.Apply 1.1 factor: t = 1.1 * 66 µs = 72.6 µs ≈ 73 µs. Verification / Alternative check: Cross-check with a calculator or scope: expected pulse near 73 µs. Small deviations may occur due to component tolerances (e.g., ±5% resistors, ±10% capacitors) and the timer’s internal thresholds, but the nominal value matches 73 µs. Why Other Options Are Wrong: 7.3 µs and 0.73 µs: off by factors of 10 or 100 due to unit mistakes. 7.3 ms and 73 ms: millisecond values are three orders of magnitude too large for the given RC. Common Pitfalls: Forgetting the 1.1 multiplier or mixing microfarads and nanofarads. Final Answer: 73 µs

Question 5

Latch behavior — why cross-coupled gates hold state NOR- and NAND-based latches tend to remain latched. Which configuration feature is responsible for this bistable memory action?

Accepted Answer

Introduction / Context: Set–reset latches made from basic NOR or NAND gates are the simplest memory elements in digital electronics. Their ability to “remember” a previous input depends on mutual feedback between the gates, establishing two stable operating points. This question probes your understanding of the structural reason behind latch behavior. Given Data / Assumptions: Two-gate latch constructed from NOR or NAND devices. Inputs briefly drive the latch to a desired state. The latch must maintain that state after the inputs are released. Concept / Approach: Cross coupling means the output of each gate feeds an input of the other, creating positive feedback. When one output is HIGH and the other LOW, the feedback reinforces this condition. A short set or reset pulse flips the pair into the alternative stable state, which is again self-reinforced after the pulse ends. Step-by-Step Reasoning: Apply a brief SET on one input: one gate output goes HIGH, forcing the other LOW.Feedback maintains the complementary outputs once the input returns inactive.Later, a RESET input can swap the states, which again persist via feedback. Verification / Alternative check: Plot simple DC transfer characteristics or simulate the pair; you will observe two stable equilibrium points and one unstable transition point, illustrating classic bistability due to positive feedback through cross coupling. Why Other Options Are Wrong: Gate impedance / low input voltages: not the fundamental cause of memory; latching results from topology, not absolute impedances. Asynchronous operation: describes timing discipline, not the physical mechanism of storage. Schmitt triggering: introduces hysteresis for clean switching but is not required for basic latching. Common Pitfalls: Confusing the presence of feedback (cross coupling) with edge-triggering; edge-triggered flip-flops are built from latches but include additional gating. Final Answer: cross coupling

Question 6

JK flip-flop truth table — identify the “no change” input condition For a JK flip-flop, which input combination leaves the current state unchanged on the active clock edge?

Accepted Answer

Introduction / Context: The JK flip-flop generalizes the SR flip-flop by resolving the forbidden input while providing useful functions—set, reset, toggle, and hold—via different input combinations. Recognizing the “no change” condition is essential when designing counters and state machines that must preserve state under certain inputs. Given Data / Assumptions: Synchronous edge-triggered JK flip-flop behavior is assumed. Inputs J and K are sampled on the active clock edge. Asynchronous preset/clear (if present) are inactive. Concept / Approach: The canonical JK truth table is: J=0,K=0 → no change (hold); J=0,K=1 → reset (Q←0); J=1,K=0 → set (Q←1); J=1,K=1 → toggle (Q←/Q). Thus the specific “no change” case is J=0 and K=0. Step-by-Step Solution: List the four input cases and their outcomes.Identify the case that preserves Q: J=0, K=0.Conclude that this is the required “hold” condition. Verification / Alternative check: Examine a vendor timing diagram; with J=K=0 at the clock edge, Q follows its previous value with no transition, confirming the hold behavior. HDL simulation with a JK primitive (or a behavioral model) will show the same. Why Other Options Are Wrong: J=1,K=1: toggles, not hold. J=1,K=0: sets Q=1. J=0,K=1: resets Q=0. J=X,K=X: not a defined, guaranteed hold condition; behavior depends on specific X values. Common Pitfalls: Assuming J=K=1 is a hold; it is the toggle case for JK. Final Answer: J = 0, K = 0

Question 7

SR latch polarity (NAND version) In a cross-coupled NAND flip-flop (latch), the set and reset inputs are normally active-LOW. How can we modify the circuit so that it instead has active-HIGH S and R control inputs (i.e., asserting a HIGH performs se…

Accepted Answer

Introduction / Context: NAND- and NOR-based cross-coupled latches implement the same memory function but with different input polarities. A standard NAND latch uses active-LOW S and R (bringing an input LOW asserts that function). Many systems, however, specify active-HIGH control signals. This question asks how to reconcile that by simple logic-level adaptation. Given Data / Assumptions: Device: cross-coupled NAND SR latch. Default polarity: S and R are active-LOW. Goal: make S and R active-HIGH without changing the fundamental latch core. Concept / Approach: Polarity conversion is achieved by inversion. If a function is active-LOW and we require active-HIGH behavior, we place an inverter in series with each control line. A HIGH at the external interface becomes a LOW into the NAND input, thus producing the original active-LOW action inside the latch while presenting an active-HIGH interface externally. Step-by-Step Solution: Identify the present polarity: S_activeLOW and R_activeLOW into a NAND latch.Insert one inverter on S and one on R (or use NAND gates wired as inverters).Now external S_high → inverter → internal S_low (asserts set). External R_high → inverter → internal R_low (asserts reset).Outputs Q and Q̄ are unchanged; only control polarity is adapted. Verification / Alternative check: Truth-check one case: External S=1, R=0 should set Q. After inversion, internal S=0, R=1 into the original NAND latch, which indeed sets Q=1. All other cases map consistently. Why Other Options Are Wrong: It can't be done: Incorrect—simple inversion is standard practice. Invert the Q outputs: Changes output polarity, not input control polarity. Insert RC networks: Affects timing/edges, not logic-level polarity. Common Pitfalls: Confusing active-level conventions, or attempting to swap NAND for NOR (which changes internal behavior rather than simply inverting inputs). Final Answer: Invert the S-R inputs.

Question 8

Master–slave flip-flop timing In a master–slave flip-flop built from level-sensitive latches, during which clock (gate) level is the master latch enabled (transparent)?

Accepted Answer

Introduction / Context: A master–slave flip-flop is formed by cascading two level latches with opposite clock polarity so that an overall edge-triggered behavior results. Understanding which level enables the master versus the slave is crucial to predict sampling and update times. Given Data / Assumptions: Topology: master latch followed by slave latch. Clocking: complementary enabling—when master is open, slave is closed, and vice versa. Conventional positive-edge design: master is enabled when CLK = HIGH; slave is enabled when CLK = LOW. Concept / Approach: With a positive-edge master–slave, the master captures input while the clock is HIGH. At the transition to LOW, the master closes, the slave opens, and the captured state transfers to the output, producing an effective capture on the rising edge (HIGH-to-LOW handoff). Step-by-Step Solution: Clock HIGH → master transparent; input D can propagate into master storage.Clock transitions LOW → master latches its value; slave becomes transparent and copies master to Q.Next rising edge will again open the master, repeating the sampling cycle.Edge-triggered appearance arises from this alternating transparency. Verification / Alternative check: Check a timing diagram: input changes during CLK=HIGH can affect master but do not reach Q until the subsequent LOW phase enables the slave, aligning observable update to the edge boundary. Why Other Options Are Wrong: Gate LOW: That enables the slave in the standard positive-edge arrangement, not the master. Both / neither: Contradict the complementary nature of the two latches. Common Pitfalls: Assuming both latches enable on the same level (which would form a transparent path) or confusing positive- and negative-edge configurations. Final Answer: when the gate is HIGH

Question 9

Synchronous SR inputs in a gated flip-flop Why are the S (set) and R (reset) inputs of a gated SR flip-flop referred to as synchronous inputs?

Accepted Answer

Introduction / Context: Gated SR and JK flip-flops often include a level-sensitive gate (or clock enable) that qualifies when data inputs can affect the internal state. Inputs qualified by the clock or gate are called synchronous, while separate preset/clear pins are asynchronous. Given Data / Assumptions: Device: gated SR flip-flop (or latch) with gate/clock input. Inputs: S and R affect the state only when the gate is asserted. Asynchronous inputs (if present) are distinct lines such as PRE and CLR (usually active-LOW). Concept / Approach: Synchronous means “timed with” the clock/gate. Therefore, S and R are effective only during the interval when the gating signal is active. Outside that window the storage element holds its previous state regardless of S and R levels (barring asynchronous controls). Step-by-Step Solution: Gate LOW (for a positive-level gate) → internal path is closed → S and R changes do not alter the stored state.Gate HIGH → internal path opens → S and R can set or reset the latch according to truth table.On gate falling edge (closing), the current state is stored and retained.Asynchronous lines (if used) act immediately regardless of gate—hence they are not synchronous. Verification / Alternative check: Examine datasheets: timing diagrams show input setup/hold relative to the gate (or clock), further underscoring synchronous behavior for S and R. Why Other Options Are Wrong: Independent of the gate: That would define asynchronous, not synchronous. Asynchronous and always override: Describes PRE/CLR, not S and R on a gated input. Power-on only: Not related to normal synchronous operation. Common Pitfalls: Mixing up latch gating with asynchronous preset/clear, or assuming S/R act at any time without regard to the enabling level. Final Answer: They must occur with the gate.

Question 10

Diagnosing a JK flip-flop with asynchronous controls A circuit fails to function. Logic probe readings show CLK, J1, J2, J3, K1, K2, and K3 are pulsing. Q is HIGH. PRE is LOW. (CLR is not asserted.) Which condition best explains the fault?

Accepted Answer

Introduction / Context: JK flip-flops commonly include asynchronous preset (PRE) and clear (CLR) inputs, often active-LOW. When asserted, these override synchronous J, K, and CLK activity. Correctly interpreting probe indications is essential for debugging nonresponsive sequential circuits. Given Data / Assumptions: J and K inputs are pulsing (valid activity). CLK is pulsing (clock present). Q output is stuck HIGH. PRE line reads LOW; asynchronous controls are active-LOW on most TTL/CMOS JK devices. Concept / Approach: If PRE is active-LOW and held LOW, the flip-flop is asynchronously forced to Q=1 regardless of J, K, or clock. As long as PRE remains asserted, no synchronous toggling can change the state, which explains a constant HIGH Q amidst active inputs and clock. Step-by-Step Solution: Interpret PRE=LOW as an active asynchronous preset condition.While PRE=LOW, the internal forcing path dominates, setting Q=1 and Q̄=0.J/K/CLK pulses cannot propagate state changes due to the override.Therefore, the likely fault is a stuck-LOW PRE signal or short to ground. Verification / Alternative check: Temporarily release PRE (drive it HIGH via pull-up). If the device then responds to clock and J/K, the diagnosis is confirmed. Checking CLR should show it inactive (HIGH) to avoid forcing reset. Why Other Options Are Wrong: No problem: Contradicted by lack of state changes at Q with active J/K/CLK. Clock should be held HIGH: Holding CLK HIGH would disable edge/level sampling, not fix a stuck state. CLR stuck HIGH: On active-LOW CLR, HIGH is the inactive state, so this would not force Q HIGH. Common Pitfalls: Forgetting that PRE/CLR are usually active-LOW, or assuming synchronous inputs can override asynchronous controls (they cannot). Final Answer: The PRE is stuck LOW.

Question 11

Monostable 74122 (retriggerable) timing An RC network for a 74122 retriggerable one-shot uses REXT = 100 kΩ and CEXT = 0.005 µF. What is the output pulse width (approximate)?

Accepted Answer

Introduction / Context: The TTL 74122 retriggerable one-shot generates a pulse whose width depends on an external resistor and capacitor. While exact constants vary slightly by logic family and conditions, datasheets provide a proportionality factor so designers can estimate the output pulse width from REXT and CEXT values. Given Data / Assumptions: Device: 74122 retriggerable monostable multivibrator. REXT = 100 kΩ. CEXT = 0.005 µF = 5 nF. We use the typical timing relation tW ≈ K * REXT * CEXT with family-typical constant K. Concept / Approach: For the 74122 family, a commonly used approximation is tW ≈ 0.14 * REXT * CEXT under nominal conditions. This constant captures the internal threshold ratios and discharge currents used by the monostable timing network. Calculate RC first, then apply the constant to estimate tW. Step-by-Step Solution: Compute RC: REXT * CEXT = 100,000 Ω * 5 × 10^-9 F = 5 × 10^-4 s = 500 µs.Apply constant: tW ≈ 0.14 * 500 µs = 70 µs.Thus the expected pulse width is about 70 microseconds.Because the 74122 is retriggerable, additional triggers within this interval will extend tW. Verification / Alternative check: Cross-check by consulting device timing graphs or using a bench setup: measured tW should fall near 70 µs, with tolerance due to component spread and supply/temperature variations. Why Other Options Are Wrong: 16 µs / 32 µs: Too short compared to the RC and constant typical for 74122. 160 µs: Exceeds the estimated value; would require a larger constant or component values. Common Pitfalls: Using the constant from a different device (for example, 74121) or mis-converting microfarads and ohms, leading to 10× errors. Final Answer: 70 µs

Question 12

VHDL component instantiation syntax In VHDL, each instance of a component is addressed (labeled) using what syntactic form?

Accepted Answer

Introduction / Context: Structural VHDL connects components by instantiating them with labels. The label uniquely identifies each instance in the design hierarchy for simulation, debugging, and synthesis reports. Given Data / Assumptions: VHDL syntax relies on a label (instance name) and a component/entity name. The general form is followed by a port map association list. Tools reference instances by their labels in messages and waveforms. Concept / Approach: The canonical instantiation form is: label : entity work.entity_name port map (...); or, when using components: label : component_name port map (...); The crucial part is the label followed by a colon, then the referenced entity/component identifier. Step-by-Step Solution: Choose an instance label, e.g., U1.Write: U1 : my_component port map ( ... signal associations ... );Optionally qualify with library/entity if not using component binding: U1 : entity work.my_entity(arch) port map (...);Simulate or synthesize; reports will list the instance by label U1. Verification / Alternative check: Any VHDL reference manual shows examples using the label-colon-component form. Vendor templates in IDEs auto-generate this structure. Why Other Options Are Wrong: Semicolon after the name: Semicolons terminate statements, not label association. Following with the library name or a library number: Libraries are referenced in the entity path; they do not replace the label:component form. Common Pitfalls: Confusing component instantiation (component/label) with direct entity instantiation (entity work.X), or omitting the label, which reduces readability and hierarchical traceability. Final Answer: A name followed by a colon and the name of the library primitive

Question 13

Hold condition of an SR/JK flip-flop Under which input condition does a flip-flop hold (retain) its current state without changing?

Accepted Answer

Introduction / Context: All basic latches and flip-flops provide a hold condition in which the stored bit is preserved. Recognizing the hold input combination is essential to understanding timing diagrams and writing correct state-transition tables. Given Data / Assumptions: Device: SR or JK flip-flop (synchronously gated). Active inputs change state only when the clock/gate permits. We consider the logical activation of set and reset (polarity depends on the specific device). Concept / Approach: The hold condition is the case where neither set nor reset is asserted. For an SR latch, that corresponds to S=0 and R=0 if inputs are active-HIGH (or S=1 and R=1 if inputs are active-LOW). With no command to set or reset, the device maintains its previous Q and Q̄ values across the clock interval. Step-by-Step Solution: Identify which polarity the device uses (active-HIGH or active-LOW).Select the input combination that means “no request” (both deasserted).During this combination and with the appropriate clock level, the internal feedback preserves the stored bit.On the next active input assertion, the state may change according to the truth table. Verification / Alternative check: Consult SR truth tables: the line corresponding to “no active S or R” is labeled “no change” or “hold.” The same concept applies to JK when J=K=0 (no change). Why Other Options Are Wrong: Both S and R active: For SR, this is invalid or forbidden (or invokes an asynchronous override if present). Only S active: Sets the flip-flop. Only R active: Resets the flip-flop. Common Pitfalls: Forgetting the active level of inputs and misreading the “forbidden” SR condition as a hold; in JK, J=K=0 is the explicit hold state. Final Answer: no active S or R input

Question 14

Monostable 74121 (nonretriggerable) timing A 74121 one-shot uses REXT = 49 kΩ and CEXT = 0.2 µF. Approximately what is the output pulse width tW?

Accepted Answer

Introduction / Context: The nonretriggerable 74121 monostable produces a fixed-width pulse determined by an external resistor and capacitor. Designers use a simple proportional formula with a device-specific constant to estimate pulse duration from component values. Given Data / Assumptions: Device: 74121 (nonretriggerable) monostable. REXT = 49 kΩ. CEXT = 0.2 µF. Approximate timing relation: tW ≈ 0.7 * REXT * CEXT (typical for 74121). Concept / Approach: First compute the RC time constant, then multiply by the typical coefficient. Ensure unit conversions are correct (kΩ to Ω, µF to F) to avoid order-of-magnitude mistakes. Step-by-Step Solution: Compute RC: 49,000 Ω * 0.2 × 10^-6 F = 9.8 × 10^-3 s = 9.8 ms.Apply coefficient: tW ≈ 0.7 * 9.8 ms ≈ 6.86 ms.Round to one decimal place: 6.9 ms.Therefore, the expected pulse width is about 6.9 ms. Verification / Alternative check: A bench check with nominal parts will produce a pulse near this value, noting tolerance on R and C (often ±5% to ±20%) and supply/temperature effects; the actual width will vary accordingly. Why Other Options Are Wrong: 6.9 µs: Off by a factor of 1000 due to unit error (µF vs nF). 69 ms / 690 ms: Too large; would require larger C or R. Common Pitfalls: Mixing microfarads and nanofarads or forgetting that kΩ * µF yields milliseconds when scaled by a sub-unity constant. Final Answer: 6.9 ms

Question 15

555 timer threshold and trigger levels In a standard 555 timer, the internal three 5 kΩ resistors form a divider that sets the trigger and threshold levels to what fractions of VCC?

Accepted Answer

Introduction / Context: The ubiquitous 555 timer uses an internal resistor ladder (three equal resistors) to establish two key comparator thresholds that govern charging/discharging of the timing capacitor in astable and monostable modes. Given Data / Assumptions: Three equal resistors from VCC to ground create nodes at 2/3 VCC and 1/3 VCC. Lower comparator (trigger) compares the capacitor voltage to 1/3 VCC. Upper comparator (threshold) compares the capacitor voltage to 2/3 VCC. Concept / Approach: Equal-value series resistors divide VCC into three equal drops. The midpoints are at 2/3 VCC and 1/3 VCC. The 555 uses these as reference levels: falling below 1/3 VCC triggers the flip-flop to start charging; rising above 2/3 VCC sets the threshold comparator to end the timing interval. Step-by-Step Solution: Compute divider outputs: with three equal resistors, node voltages are VCC*(1/3) and VCC*(2/3).Associate to 555 comparators: trigger at 1/3 VCC, threshold at 2/3 VCC.These fixed fractions hold across supply voltages so long as the resistors track.Design equations reference these ratios to set timing and hysteresis. Verification / Alternative check: Refer to any 555 internal schematic; the two comparator references clearly show 1/3 VCC and 2/3 VCC nodes from the 5 kΩ ladder. Why Other Options Are Wrong: 1/4 and 1/2, 3/4: Not produced by three equal resistors. Other pairings mix correct and incorrect values, which do not match 555 operation. Common Pitfalls: Confusing trigger with threshold or assuming absolute voltages rather than fractions of VCC, which makes 555 behavior supply-scaled. Final Answer: 1/3 VCC and a threshold level 2/3 VCC

Question 16

Cross-coupled NOR latch polarity For a cross-coupled NOR flip-flop (latch), are the set and reset control inputs active-HIGH or active-LOW?

Accepted Answer

Introduction / Context: Cross-coupled latches can be realized with either NOR or NAND gates. Their input polarities differ: understanding which version uses active-HIGH versus active-LOW controls is essential for interfacing with the rest of a digital system. Given Data / Assumptions: Device: cross-coupled NOR latch. Inputs: S (set) and R (reset) applied to NOR gate inputs. Outputs: Q and Q̄ taken from the gates’ outputs in the standard configuration. Concept / Approach: A NOR gate outputs HIGH only when all inputs are LOW. Therefore, applying a HIGH on one input forces that NOR output LOW, which—through cross-coupling—drives the opposite output HIGH. Thus, asserting a HIGH on S sets Q=1; asserting a HIGH on R resets Q=0. Hence the NOR latch uses active-HIGH set and reset inputs. Step-by-Step Solution: Apply S=1, R=0: Left NOR output goes 0; right NOR sees 0 on its feedback input and holds 1 → Q=1 (set).Apply R=1, S=0: Right NOR output goes 0; left NOR rises to 1 → Q=0 (reset).With S=0 and R=0: Both NORs are governed by feedback, retaining the last state (hold).With S=1 and R=1: Both outputs forced LOW (invalid/forbidden for the SR latch), leading to metastability upon release. Verification / Alternative check: Compare with the NAND latch: it uses active-LOW controls (asserting a LOW performs the action). The two are duals with inverted input polarities. Why Other Options Are Wrong: active-LOW: That describes the NAND-based latch, not NOR-based. Both / neither: Input polarity is well-defined; the latch is level-sensitive to explicit HIGH assertions for control. Common Pitfalls: Mixing up NOR and NAND latch conventions, or assuming output inversion changes input active level (it does not). Final Answer: active-HIGH

Question 17

Positive edge-triggered S–R flip-flop — timing of state sampling On a positive edge-triggered S–R flip-flop, when do the Q and Q' outputs actually reflect (sample and latch) the S and R input condition?

Accepted Answer

Introduction / Context: Edge-triggered flip-flops are fundamental memory elements in synchronous digital systems. Unlike level-sensitive latches, they sample their inputs only at a fleeting instant defined by the clock transition. Understanding precisely when a positive edge-triggered S–R (Set–Reset) flip-flop latches its inputs prevents race conditions, timing violations, and misinterpretation of waveforms during debugging. Given Data / Assumptions: Device type: positive edge-triggered S–R flip-flop. Inputs: S (set) and R (reset), assumed synchronous and valid around the sampling instant. Outputs: Q and Q' (complement). Clock polarity: responds to the LOW-to-HIGH transition. Concept / Approach: An edge-triggered device uses a pulse transition detector internally so that input sampling occurs only during an infinitesimal window at the clock's specified edge. For a positive edge-triggered device, that window is the rising edge (LOW→HIGH). Provided setup and hold constraints are met, the state after the edge remains stable until the next active clock edge, regardless of subsequent input chatter. Step-by-Step Solution: 1) Identify the triggering characteristic: positive edge-triggered.2) Map this to a timing event: the active instant is the clock's LOW-to-HIGH transition.3) Determine behavior: outputs reflect S/R evaluated at that instant and then hold.4) Confirm that during HIGH or LOW levels (between edges), an ideal edge-triggered FF ignores input changes (subject to setup/hold timing). Verification / Alternative check: Examine a timing diagram: vary S and R safely before the rising edge (meeting setup time), keep them stable briefly after (meeting hold time), and observe Q updates exactly at the rising edge. Why Other Options Are Wrong: Clock LOW or HIGH levels: those describe latch-like behavior, not edge-triggered sampling. HIGH-to-LOW transition: corresponds to a negative edge-triggered device, not a positive one. Common Pitfalls: Violating setup/hold around the edge, assuming outputs track inputs while the clock is HIGH, or confusing edge-triggered flip-flops with level-sensitive latches. Final Answer: the clock pulse transitions from LOW to HIGH

Question 18

555 timer in monostable mode — compute R1 for a target pulse width A 555 timer is configured as a monostable multivibrator with timing capacitor C1 = 0.01 µF. Determine the resistor R1 required to obtain a pulse width of 2 ms (use T = 1.1 * R1 * C1).

Accepted Answer

Introduction / Context: The 555 timer in monostable mode generates a single pulse whose duration depends on an external resistor and capacitor. The standard timing relation is T = 1.1 * R * C for the high output interval. This question tests your ability to rearrange the formula and select a practical resistor value for a specified pulse width and known capacitance, a frequent task in embedded timing and pulse-generation circuits. Given Data / Assumptions: Monostable mode (one-shot) using a 555 timer IC. Target pulse width T = 2 ms. Timing capacitor C1 = 0.01 µF = 10 nF = 10 * 10^-9 F. Timing relation: T = 1.1 * R1 * C1. Concept / Approach: Rearrange the timing equation to solve for R1. Ensure unit consistency (seconds, farads, ohms). After computing R1, compare the result to the offered options, rounding to a standard resistor value where appropriate. Step-by-Step Solution: 1) Start with T = 1.1 * R1 * C1.2) Solve for R1: R1 = T / (1.1 * C1).3) Substitute values: R1 = 0.002 s / (1.1 * 10e-9 F).4) Compute denominator: 1.1 * 10e-9 = 11e-9.5) R1 = 0.002 / 11e-9 = (2e-3) / (11e-9) ≈ 181,818 Ω ≈ 182 kΩ. Verification / Alternative check: Back-calculate the pulse: T_est = 1.1 * 182,000 * 10e-9 ≈ 1.1 * 0.00182 ≈ 0.002002 s ≈ 2.00 ms, matching the requirement. Why Other Options Are Wrong: 200 kΩ: gives T ≈ 2.2 ms (too long). 91 kΩ: gives T ≈ 1.0 ms (half the target). 182 Ω: off by three orders of magnitude; pulse would be microseconds. Common Pitfalls: Confusing microfarads with nanofarads, omitting the 1.1 constant, or rounding to a non-standard value that misses timing by a wide margin. Final Answer: 182 kΩ

Question 19

Cascaded J–K flip-flops as a binary divider — output frequency Four J–K flip-flops are cascaded (J and K of each tied HIGH so they toggle). If the input frequency to the first stage is fin = 32 kHz, what is the output frequency after the fourth stag…

Accepted Answer

Introduction / Context: Cascading toggle flip-flops creates a binary frequency divider or ripple counter. Each stage divides the incoming frequency by 2, so chaining N stages yields division by 2^N. Recognizing this behavior is crucial for clock generation, baud-rate derivation, and timing state machines in digital design. Given Data / Assumptions: Four J–K flip-flops in series, each with J = K = 1 (toggle mode). Input frequency fin = 32 kHz at the first stage. We observe the output of the fourth stage. Concept / Approach: For toggle flip-flops, each positive edge (or designated clock event) causes the Q output to toggle, effectively halving the frequency. Thus, fout = fin / 2^N, where N is the number of cascaded stages. Step-by-Step Solution: 1) Determine division factor: 2^N with N = 4 → 2^4 = 16.2) Apply to fin: fout = 32 kHz / 16.3) Compute: 32 / 16 = 2 → fout = 2 kHz.4) Report the result in kHz as requested. Verification / Alternative check: Track stage-by-stage: 32 kHz → 16 kHz → 8 kHz → 4 kHz → 2 kHz. The fourth output is 2 kHz, confirming the calculation. Why Other Options Are Wrong: 1 kHz or 4 kHz: correspond to division by 32 or 8 (wrong stage count). 16 kHz: only a single division (first stage) rather than four. Common Pitfalls: Forgetting that each stage divides by 2 or miscounting how many stages the signal has passed through. Final Answer: 2 kHz

Question 20

Edge-triggered flip-flops — necessary internal mechanism Which design requirement must an edge-triggered flip-flop include to ensure it responds only at the clock transition and not during the entire clock level?

Accepted Answer

Introduction / Context: Edge-triggered flip-flops are engineered to change state only at a specific instant: the transition (edge) of the clock. This protects synchronous systems from hazards that occur when inputs vary during the clock level. The key to this behavior lies in how the flip-flop detects and gates that tiny moment in time. Given Data / Assumptions: We are comparing architectural requirements, not external pin polarities. The device is intended to respond to clock edges only. Standard synchronous timing (setup/hold) is assumed. Concept / Approach: An internal pulse transition detector creates a narrow sampling pulse from the clock edge (rising or falling). That pulse briefly enables the internal storage path, causing the flip-flop to sample inputs only during that short interval. Outside the pulse, the data path is blocked, preventing level-sensitive tracking associated with latches. Step-by-Step Solution: 1) Identify the desired behavior: edge-only sensitivity.2) Provide mechanism: generate a very narrow pulse coincident with the edge.3) Gate storage with this pulse so the device samples once per edge.4) Maintain state until the next sampling pulse. Verification / Alternative check: Examine typical master–slave vs. edge-triggered schematics; the latter commonly synthesize a brief pulse internally (via differentiating networks or logic) to emulate instantaneous sampling. Why Other Options Are Wrong: Very fast response: helpful but not sufficient to enforce edge-only sensitivity. At least two inputs for rising/falling: not required; many devices trigger on one edge only. Active-LOW inputs/complemented outputs: mere signal conventions, unrelated to edge detection. Common Pitfalls: Confusing latches with edge-triggered flip-flops or assuming polarity conventions determine edge sensitivity. Final Answer: a pulse transition detector.

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