Q: Asynchronous (Ripple) Counters — Key Limitation What is a major drawback of asynchronous (ripple) counters compared with synchronous counters for frequency-scaling applications?

Introduction / Context: Asynchronous (ripple) counters cascade the clock edge from one flip-flop to the next, creating sequential toggling. While simple and low-resource, this architecture accumulates propagation delays (tPD) along the chain, which can distort timing and limit usable input frequency. Given Data / Assumptions: Each flip-flop has finite clock-to-Q delay. In a ripple counter, the effective count boundary changes at slightly different times across bits. For high input frequencies, cumulative delay becomes comparable to or exceeds the period. Concept / Approach: The maximum input frequency for reliable counting is constrained by the worst-case sum of propagation delays through the toggling chain. As frequency increases, skew between bit transitions causes transient invalid states, glitching, and potential miscounts at downstream decoding logic. Step-by-Step Solution: Model a 4-bit ripple counter with tPD per stage.Cumulative delay across bits approaches 4*tPD, reducing available time before the next input edge.Above a certain input frequency, the device cannot settle to a valid count before the next transition.Result: high-frequency operation is limited by internal propagation delays. Verification / Alternative check: Compare with synchronous counters, which clock all flip-flops simultaneously; only combinational path (not cumulative toggling) limits frequency, allowing much higher speeds than ripple designs. Why Other Options Are Wrong: Low-frequency limitation: Ripple counters perform well at low frequencies; the issue is high-speed operation. No major drawbacks: Ignores well-documented ripple delay issues. No propagation delays: Physically impossible; all flip-flops have finite delay. Common Pitfalls: Using ripple counters near their frequency limits with additional decoding that is sensitive to transient states, leading to glitches or erroneous triggering. Final Answer: high-frequency applications are limited because of internal propagation delays

Q: Ripple counter timing — recycle delay through 4 cascaded flip-flops A 4-bit ripple counter uses flip-flops with 15 ns clock-to-Q propagation delay each. How long does it take to recycle from 1111 back to 0000 (worst-case total ripple time)?

Introduction / Context: Ripple (asynchronous) counters are built by daisy-chaining flip-flops so that each stage is clocked by the previous stage’s output. This simplicity comes with a cost: output bits do not change simultaneously. Instead, transitions propagate sequentially from the least significant stage to the most significant stage, producing a cumulative timing delay that matters in high-speed applications. Given Data / Assumptions: Counter width: 4 bits (four cascaded flip-flops). Clock-to-Q propagation delay per flip-flop: 15 ns. We are concerned with the recycle transition from 1111 to 0000 (the longest ripple event). Concept / Approach: In a ripple counter, each flip-flop toggles after the previous stage’s output transition has propagated. Thus, the total time for the most significant bit to settle equals the sum of the individual propagation delays through all stages involved. The worst case is typically the all-ones to all-zeros transition, requiring a toggle at each stage in sequence. Step-by-Step Solution: At t0, the LSB toggles in response to the clock; its Q settles after 15 ns.The next flip-flop sees this as its clock; it toggles and settles after an additional 15 ns.This continues stage by stage through all four flip-flops.Total worst-case recycle delay = 4 * 15 ns = 60 ns. Verification / Alternative check: Timing diagram inspection shows staggered transitions: Q0 at t0+15 ns, Q1 at t0+30 ns, Q2 at t0+45 ns, Q3 at t0+60 ns. Only after the MSB settles has the state truly become 0000. Why Other Options Are Wrong: 15 ns: counts only one stage, not the ripple through all four. 30 ns or 45 ns: account for 2 or 3 stages, still short of the full chain. Common Pitfalls: Assuming simultaneous bit changes as in synchronous counters; ignoring that ripple delay limits maximum input clock frequency and can cause decoding glitches if outputs are read during the transition window. Final Answer: 60 ns

Q: Flip-flops needed for a decade (MOD-10) counter How many flip-flops are required to construct a decade counter that counts 0 through 9 and then recycles?

Introduction / Context: A decade counter advances through ten distinct states (0–9) before recycling to 0. Determining the minimum number of flip-flops needed is a classic design question and highlights the relationship between binary capacity (2^n states) and a desired modulus that is not a power of two. The design also illustrates how decoded resets create nonbinary moduli. Given Data / Assumptions: Required modulus: 10 states (0 to 9 inclusive). Flip-flops are binary storage elements; an n-bit register can represent up to 2^n states. Counter may use synchronous or ripple architecture with decoding to reset after state 9. Concept / Approach: Find the smallest n such that 2^n ≥ 10. Since 2^3 = 8 < 10 and 2^4 = 16 ≥ 10, four flip-flops are the minimum to represent at least ten states. Additional combinational logic decodes state 10 (1010) to asynchronously/synchronously reset the counter to 0, thereby creating a MOD-10 sequence within a 16-state space. Step-by-Step Solution: Compute capacity: 2^3 = 8 (insufficient).Next power: 2^4 = 16 (sufficient to include 10 states).Add decoding: detect 1010 (decimal 10) and drive reset to return to 0000.Therefore, use 4 flip-flops with reset decoding for a decade counter. Verification / Alternative check: Simulation or truth-table enumeration confirms that states 0000 through 1001 appear in sequence, and when 1010 is reached the reset forces 0000, giving a 10-state loop. Why Other Options Are Wrong: 10 and 8 flip-flops: far exceed the minimum; 8 yields astronomically more states than required. 5 flip-flops: also works but is not minimal; the question asks how many are required at minimum. Common Pitfalls: Forgetting the need for a reset decode when the modulus is not a power of two; miscounting valid states; or assuming 3 flip-flops suffice for 0–9 (they do not—only 8 states are available). Final Answer: 4

Question 1

Counter ICs — 7490 vs. 7492 Modulus Identify the correct modulus (divide-by value) for each of these TTL counters: 7490 and 7492.

Accepted Answer

Introduction / Context: The classic 74xx counter family contains dedicated divide-by-N counters wired internally for common moduli. Recognizing the default modulus helps when selecting parts for decade or divide-by-12 functions without external gating logic. Given Data / Assumptions: Both parts are BCD/binary ripple counters with internal resets wired for a specific modulus. 7490 is historically known as a decade counter. 7492 is a divide-by-12 counter. Concept / Approach: A MOD-N counter repeats its count every N pulses and divides the input frequency by N. The 7490 internally implements MOD-10 (decade). The 7492 is internally configured as MOD-12, saving glue logic where divide-by-12 is needed (e.g., timebases). Step-by-Step Solution: Identify 7490: decade counter → MOD-10.Identify 7492: divide-by-12 counter → MOD-12.Map to answer choice: “7490 is MOD-10, 7492 is MOD-12.” Verification / Alternative check: Standard TTL datasheets list the 7490 as “Decade (MOD-10) counter” and the 7492 as “Divide-by-12 counter,” confirming the selection. Why Other Options Are Wrong: Options A, B, C swap or misstate the moduli. Common Pitfalls: Confusing 7492 (MOD-12) with 7493 (binary MOD-16). The 7493 is the common divide-by-16 ripple counter. Final Answer: 7490 is a MOD-10, 7492 is a MOD-12

Question 2

BCD-to-7-Segment Driver 7447 — Determine Outputs for Digit 9 Given a 7447 common-anode driver with inputs RBI = 0, LT = 1, and BCD A3 A2 A1 A0 = 1 0 0 1 (decimal 9), list the states of RBO and segments a–g (active-LOW outputs).

Accepted Answer

Introduction / Context: The 7447 is a BCD-to-7-segment decoder/driver intended for common-anode displays. Its outputs are active-LOW, meaning a segment is lit when its output pin drives LOW. Understanding the mapping from BCD inputs to segment patterns, and the function of RBI/RBO and LT pins, is essential for display interfacing and ripple-blanking cascades. Given Data / Assumptions: Inputs: A3 A2 A1 A0 = 1 0 0 1 (decimal 9). RBI (Ripple Blanking Input) = 0; LT (Lamp Test) = 1 (inactive). Outputs are active-LOW: output 0 lights the segment; 1 turns it off. Concept / Approach: For the digit 9, the illuminated segments on a standard 7-segment display are a, b, c, d, f, g, while segment e is off. Since outputs are active-LOW, those lit segments must be 0, and the unlit segment e must be 1. RBO (Ripple Blanking Output) propagates blanking when displaying leading zeros; for any nonzero digit with RBI = 0, RBO = 1 (no blanking signal forwarded). Step-by-Step Solution: Map the pattern for 9: segments {a, b, c, d, f, g} ON; segment {e} OFF.Translate to active-LOW levels: a=0, b=0, c=0, d=0, f=0, g=0, e=1.Determine RBO: input is 9 (nonzero), so RBO = 1.Assemble full listing: RBO=1; a=0 b=0 c=0 d=0 e=1 f=0 g=0. Verification / Alternative check: Datasheet truth tables for the 7447 confirm the segment coding for decimal 9 and the ripple blanking operation: RBO drives LOW only when the device blanks a zero with RBI asserted; otherwise RBO remains HIGH. Why Other Options Are Wrong: Any option with d = 1: would turn segment d off, contradicting the standard 9 pattern. Any option with RBO = 0: indicates leading-zero blanking; not applicable for a nonzero digit. Common Pitfalls: Forgetting that 7447 outputs are active-LOW (0 = ON), and confusing RBI/RBO behavior for the zero digit only. Final Answer: RBO = 1, a = 0, b = 0, c = 0, d = 0, e = 1, f = 0, g = 0

Question 3

Number Systems — Convert 15,536 (Decimal) to Hexadecimal Find the hexadecimal equivalent of the decimal number 15,536, showing the place-value reasoning.

Accepted Answer

Introduction / Context: Converting from decimal to hexadecimal is routine in digital systems when addressing memory, reading registers, or interpreting dumps. Hexadecimal condenses binary into human-friendly base-16 digits using 0–9 and A–F, where A=10 through F=15. Given Data / Assumptions: Decimal input: 15,536. Hexadecimal base: 16 with weights 16^0, 16^1, 16^2, 16^3, ... No sign or fractional component. Concept / Approach: Use positional weights or repeated division by 16. For a four-digit hex result H3 H2 H1 H0, the decimal value is H316^3 + H216^2 + H116^1 + H016^0. Determine each digit by dividing and taking remainders in base 16. Step-by-Step Solution: Compute 16^3 = 4096 and 16^2 = 256.Find H3: floor(15,536 / 4096) = 3 → remainder = 15,536 − 34096 = 15,536 − 12,288 = 3,248.Find H2: floor(3,248 / 256) = 12 → 12 maps to C; remainder = 3,248 − 12256 = 3,248 − 3,072 = 176.Find H1: floor(176 / 16) = 11 → 11 maps to B; remainder = 176 − 1116 = 176 − 176 = 0.Find H0: remainder 0 → 0.Assemble digits: 3 C B 0 → 3CB0. Verification / Alternative check: Back-convert: 34096 + C256 + B16 + 0 = 12,288 + 3,072 + 176 + 0 = 15,536, confirming correctness. Why Other Options Are Wrong: 3C66, 63C0, 6300: Each changes one or more hex digits; back-conversion does not equal 15,536. Common Pitfalls: Mis-mapping letters (e.g., confusing B=11 and C=12) or forgetting to carry remainders properly during division by 16. Final Answer: 3CB0

Question 4

Cascaded Decade Counters — Overall Divide Ratio If three decade (MOD-10) counters are cascaded, by what factor is the input frequency divided at the final output?

Accepted Answer

Introduction / Context: Counters are ubiquitous for frequency division, timebase generation, and event counting. When multiple counters are cascaded (the output of one drives the clock of the next), the overall divide ratio equals the product of individual moduli. Decade counters, by definition, divide by 10 each cycle. Given Data / Assumptions: Three identical decade counters (each MOD-10). Cascaded ripple or synchronous configuration; modulus multiplication still applies. Only one clock input applied to the first stage. Concept / Approach: Each stage reduces the frequency by its modulus. Therefore, the net division is the product of the moduli: 10 * 10 * 10. The final output toggles at one-thousandth of the input frequency, a common arrangement in digital clocks and frequency meters. Step-by-Step Solution: Stage 1: divide by 10 → f1 = fin / 10.Stage 2: divide by 10 again → f2 = f1 / 10 = fin / 100.Stage 3: divide by 10 again → f3 = f2 / 10 = fin / 1000.Hence, overall division = 1,000. Verification / Alternative check: Consider a 1 MHz input. After three cascaded decades: 1 MHz → 100 kHz → 10 kHz → 1 kHz. That is a 1,000:1 reduction. Why Other Options Are Wrong: 10 or 100: Too small; ignore cumulative effect of three stages. 20: Not a power-of-ten divisor for decade chains. Common Pitfalls: Forgetting that cascaded counters multiply moduli, or confusing decade (base-10) with binary (base-2) counters when estimating total division. Final Answer: 1,000

Question 5

Binary Counter — Terminal Count in DOWN Mode (3-Bit) For a 3-bit binary counter operating in DOWN mode, what is the terminal count (the state at which it rolls under to the maximum state)?

Accepted Answer

Introduction / Context: Terminal count (TC) denotes the count at which a counter rolls over (or under) and optionally generates a carry/borrow pulse for cascading. For a DOWN counter, the sequence decrements each clock and wraps when it reaches the minimum representable state. Given Data / Assumptions: Width: 3 bits (states 000 through 111). DOWN mode: sequence proceeds 111 → 110 → … → 000 → wrap to 111. Binary, no gray or BCD coding involved. Concept / Approach: In 3-bit binary, the smallest state is 000, and the largest is 111. A DOWN counter decrements until it reaches 000. The next decrement causes wraparound to 111. Therefore, the terminal count for DOWN operation is the minimum state 000 (often the condition that asserts a “borrow” or terminal-count flag). Step-by-Step Solution: Enumerate downward: 111, 110, 101, 100, 011, 010, 001, 000.Identify wrap point: after 000, the sequence returns to 111.Thus, terminal count (before wrap) = 000. Verification / Alternative check: Examine a simple synchronous 3-bit subtract-one design; comparators flag the 000 state to force reload/borrow and wrap back to 111 on the next tick. Why Other Options Are Wrong: 111: That is the terminal count for UP mode (wrap from max to 000). 101, 010: Intermediate states, not wrap points. Common Pitfalls: Confusing UP vs DOWN sequences or misinterpreting “terminal count” as any max/min rather than the wrap state specific to the chosen count direction. Final Answer: 000

Question 6

One-Shot in HDL — Make the Circuit Fire Once per Rising Edge In a hardware description language (HDL), which technique ensures a circuit responds once to each positive transition (rising edge) on a trigger input, implementing a one-shot behavior?

Accepted Answer

Introduction / Context: A one-shot (monostable) responds once per trigger edge, generating a single pulse of defined width. In HDL, this behavior is implemented by detecting transitions and gating logic so that internal processes trigger only on the specified edge, regardless of input level duration. Given Data / Assumptions: Trigger input may remain HIGH after an edge. We require exactly one response per positive edge. Pulse width may be fixed (one clock) or extended via counters after detection. Concept / Approach: Edge trapping (edge detection) captures the moment when a signal changes from 0 to 1 by comparing the current sample with a one-cycle delayed sample. The expression rising_edge = (curr = 1) and (prev = 0) produces a single-cycle strobe that can start a monostable or increment a counter once per event. Step-by-Step Solution: Register the trigger: prev <= trigger at each clock.Form the edge detect: pulse <= trigger and (not prev).Use pulse as an enable/start for the one-shot timing (e.g., load a down-counter or set a state).Ensure re-arm logic so the next rising edge is seen only after pulse completes. Verification / Alternative check: Simulate with long HIGH levels and bouncy edges; the edge-detect strobe fires exactly once per 0→1 transition, regardless of how long the trigger stays HIGH, proving one-shot behavior. Why Other Options Are Wrong: By using a counter: A counter can shape pulse width, but without edge detection it may retrigger or miss edges. By using an active clock: Clocks are required but do not by themselves create one-shot behavior. By using an immediate reload: Reload is a mechanism after detection; not the detection method. Common Pitfalls: Neglecting to register the previous state (causing level-sensitive behavior) and forgetting to synchronize asynchronous triggers to avoid metastability. Final Answer: By using edge trapping

Question 7

Asynchronous (Ripple) Counters — Key Limitation What is a major drawback of asynchronous (ripple) counters compared with synchronous counters for frequency-scaling applications?

Accepted Answer

Introduction / Context: Asynchronous (ripple) counters cascade the clock edge from one flip-flop to the next, creating sequential toggling. While simple and low-resource, this architecture accumulates propagation delays (tPD) along the chain, which can distort timing and limit usable input frequency. Given Data / Assumptions: Each flip-flop has finite clock-to-Q delay. In a ripple counter, the effective count boundary changes at slightly different times across bits. For high input frequencies, cumulative delay becomes comparable to or exceeds the period. Concept / Approach: The maximum input frequency for reliable counting is constrained by the worst-case sum of propagation delays through the toggling chain. As frequency increases, skew between bit transitions causes transient invalid states, glitching, and potential miscounts at downstream decoding logic. Step-by-Step Solution: Model a 4-bit ripple counter with tPD per stage.Cumulative delay across bits approaches 4*tPD, reducing available time before the next input edge.Above a certain input frequency, the device cannot settle to a valid count before the next transition.Result: high-frequency operation is limited by internal propagation delays. Verification / Alternative check: Compare with synchronous counters, which clock all flip-flops simultaneously; only combinational path (not cumulative toggling) limits frequency, allowing much higher speeds than ripple designs. Why Other Options Are Wrong: Low-frequency limitation: Ripple counters perform well at low frequencies; the issue is high-speed operation. No major drawbacks: Ignores well-documented ripple delay issues. No propagation delays: Physically impossible; all flip-flops have finite delay. Common Pitfalls: Using ripple counters near their frequency limits with additional decoding that is sensitive to transient states, leading to glitches or erroneous triggering. Final Answer: high-frequency applications are limited because of internal propagation delays

Question 8

Cascading counters — overall modulus calculation When two digital counters are cascaded (one drives the next), the overall MOD (modulus) of the combined counter is equal to the ________ of their individual MOD numbers.

Accepted Answer

Introduction / Context: Counters are sequential circuits that step through a fixed number of states, known as the MOD or modulus. In practical designs we often cascade counters to obtain large divide ratios (for example, frequency dividers in timing circuits). Understanding how the overall modulus combines when counters are connected in series is a foundational skill in digital electronics and clock-generation design. Given Data / Assumptions: Two counters are cascaded so that the terminal count (or ripple output) of the first counter drives the clock/enable of the second. Each counter has a well-defined modulus: MOD_A and MOD_B. No skipped states or special decoding are introduced other than standard cascading. Concept / Approach: When counters are cascaded, the second counter advances once for every complete cycle of the first. Therefore, the total number of unique states before the overall system repeats is the number of states in the first counter multiplied by the number of states in the second. This is analogous to a base-N odometer where one wheel increments only after the lower wheel completes a full revolution. Step-by-Step Solution: Let the first counter have MOD_A states and the second have MOD_B states.The first counter cycles through MOD_A input pulses to produce a single increment for the second counter.The second counter then requires MOD_B increments to return to its initial state.Total input pulses for a complete overall cycle = MOD_A * MOD_B, hence overall modulus = product. Verification / Alternative check: Example: cascade a MOD-10 (decade) with a MOD-6 counter to form a minutes counter (seconds → minutes). The total modulus is 10 * 6 = 60, matching the 00–59 range. This simple cross-check demonstrates the multiplicative rule. Why Other Options Are Wrong: Sum: would imply the system repeats after MOD_A + MOD_B pulses, which contradicts the odometer behavior. Log: bears no relation to state counting. Reciprocal: has no meaning in this discrete-state context. Common Pitfalls: Confusing synchronous gating with asynchronous ripple; forgetting that decode/reset logic can change the effective modulus (but the basic cascade without extra decoding multiplies). Also, ensure the cascade uses appropriate enables or ripple outputs to avoid missed counts. Final Answer: product

Question 9

Ripple counter timing — recycle delay through 4 cascaded flip-flops A 4-bit ripple counter uses flip-flops with 15 ns clock-to-Q propagation delay each. How long does it take to recycle from 1111 back to 0000 (worst-case total ripple time)?

Accepted Answer

Introduction / Context: Ripple (asynchronous) counters are built by daisy-chaining flip-flops so that each stage is clocked by the previous stage’s output. This simplicity comes with a cost: output bits do not change simultaneously. Instead, transitions propagate sequentially from the least significant stage to the most significant stage, producing a cumulative timing delay that matters in high-speed applications. Given Data / Assumptions: Counter width: 4 bits (four cascaded flip-flops). Clock-to-Q propagation delay per flip-flop: 15 ns. We are concerned with the recycle transition from 1111 to 0000 (the longest ripple event). Concept / Approach: In a ripple counter, each flip-flop toggles after the previous stage’s output transition has propagated. Thus, the total time for the most significant bit to settle equals the sum of the individual propagation delays through all stages involved. The worst case is typically the all-ones to all-zeros transition, requiring a toggle at each stage in sequence. Step-by-Step Solution: At t0, the LSB toggles in response to the clock; its Q settles after 15 ns.The next flip-flop sees this as its clock; it toggles and settles after an additional 15 ns.This continues stage by stage through all four flip-flops.Total worst-case recycle delay = 4 * 15 ns = 60 ns. Verification / Alternative check: Timing diagram inspection shows staggered transitions: Q0 at t0+15 ns, Q1 at t0+30 ns, Q2 at t0+45 ns, Q3 at t0+60 ns. Only after the MSB settles has the state truly become 0000. Why Other Options Are Wrong: 15 ns: counts only one stage, not the ripple through all four. 30 ns or 45 ns: account for 2 or 3 stages, still short of the full chain. Common Pitfalls: Assuming simultaneous bit changes as in synchronous counters; ignoring that ripple delay limits maximum input clock frequency and can cause decoding glitches if outputs are read during the transition window. Final Answer: 60 ns

Question 10

Flip-flops needed for a decade (MOD-10) counter How many flip-flops are required to construct a decade counter that counts 0 through 9 and then recycles?

Accepted Answer

Introduction / Context: A decade counter advances through ten distinct states (0–9) before recycling to 0. Determining the minimum number of flip-flops needed is a classic design question and highlights the relationship between binary capacity (2^n states) and a desired modulus that is not a power of two. The design also illustrates how decoded resets create nonbinary moduli. Given Data / Assumptions: Required modulus: 10 states (0 to 9 inclusive). Flip-flops are binary storage elements; an n-bit register can represent up to 2^n states. Counter may use synchronous or ripple architecture with decoding to reset after state 9. Concept / Approach: Find the smallest n such that 2^n ≥ 10. Since 2^3 = 8 < 10 and 2^4 = 16 ≥ 10, four flip-flops are the minimum to represent at least ten states. Additional combinational logic decodes state 10 (1010) to asynchronously/synchronously reset the counter to 0, thereby creating a MOD-10 sequence within a 16-state space. Step-by-Step Solution: Compute capacity: 2^3 = 8 (insufficient).Next power: 2^4 = 16 (sufficient to include 10 states).Add decoding: detect 1010 (decimal 10) and drive reset to return to 0000.Therefore, use 4 flip-flops with reset decoding for a decade counter. Verification / Alternative check: Simulation or truth-table enumeration confirms that states 0000 through 1001 appear in sequence, and when 1010 is reached the reset forces 0000, giving a 10-state loop. Why Other Options Are Wrong: 10 and 8 flip-flops: far exceed the minimum; 8 yields astronomically more states than required. 5 flip-flops: also works but is not minimal; the question asks how many are required at minimum. Common Pitfalls: Forgetting the need for a reset decode when the modulus is not a power of two; miscounting valid states; or assuming 3 flip-flops suffice for 0–9 (they do not—only 8 states are available). Final Answer: 4

Question 11

System fails to produce data output — flip-flops test OK individually A multistage sequential circuit produces no data output. Each flip-flop checks OK with a logic probe and pulser when tested individually. Which fault could still explain the syste…

Accepted Answer

Introduction / Context: Troubleshooting sequential logic requires distinguishing device-level health from system-level connectivity. A logic probe and pulser can verify that each flip-flop toggles when stimulated directly, yet the overall chain may still fail due to faults in shared resources (like the clock) or broken links between stages. Recognizing symptoms that point to a global timing path problem is a key diagnostic skill. Given Data / Assumptions: No overall data output is observed at the system output node. Each flip-flop responds correctly when locally pulsed at its input/clock during isolated tests. Interconnections and the common system clock distribute signals between stages under normal operation. Concept / Approach: If a clock input line is open (broken trace, bad via, cracked solder joint), stages will not receive legitimate clock transitions, so they will never advance during system operation. Local testing with a hand pulser can bypass the broken clock path, making each flip-flop appear healthy in isolation and masking the real fault. This discrepancy between local OK tests and global failure is a hallmark of a missing common stimulus. Step-by-Step Solution: Note symptom: system output stuck or inactive despite all ICs passing individual tests.Infer that a shared signal (clock or enable) might be absent along the chain.Hypothesize an open clock input line to one stage; that stage never toggles under real operation, blocking data propagation.Confirm by tracing the clock distribution with a scope and continuity tester to locate the open segment. Verification / Alternative check: Probe the clock at each stage; the missing edge will reveal where the line is open. Wiggling or thermal stressing the board may intermittently restore contact, further indicating a cracked joint or via. Why Other Options Are Wrong: Data output line grounded: would show a hard LOW even during local pulsing at the last stage; the prompt suggests all FFs appear to work when pulsed. Interconnect line shorted to ground: a direct short on a data link would also defeat local pulsing across that link; such a fault typically manifests during the isolated test. Solder bridge to Vcc on a flip-flop input: this would usually prevent correct local toggling on that device during probe/pulser tests. Common Pitfalls: Testing only at device pins without verifying the shared clock/enable network; overlooking hairline cracks that behave intermittently; forgetting to check connectors or jumpers that distribute the master clock. Final Answer: One of the clock input lines may be open.

Question 12

Designing synchronous nonbinary counters The process for creating a synchronous counter that counts in a nonbinary sequence (custom MOD) is primarily based on:

Accepted Answer

Introduction / Context: Synchronous counters change state on a common clock edge, enabling precise custom sequences that are not simple powers of two. Building a nonbinary (arbitrary MOD) synchronous counter relies on steering the flip-flop inputs so that only the desired state transitions occur, while undesired states are skipped or reset predictably. Given Data / Assumptions: Target is a nonbinary state sequence (e.g., MOD-10, MOD-12, Gray order, or custom codes). Flip-flops are clocked synchronously. Design freedom includes using combinational logic to drive J, K, T, or D inputs as functions of present state. Concept / Approach: The canonical method is to derive next-state equations from a state table or Karnaugh maps. External combinational logic decodes the present state (and possibly inputs) to generate the required control levels for the flip-flop inputs (e.g., J and K). This logic ensures that on each clock the counter advances to exactly the next desired state, creating the specified sequence and modulus. Step-by-Step Solution: Define desired sequence and assign binary codes to states.Write next-state equations Q_next = f(Q_present, inputs).Map equations onto flip-flop excitation tables to get J, K (or D/T) expressions.Implement the expressions with gates that decode the current state and drive the flip-flops. Verification / Alternative check: Simulate the counter or step through states manually to verify that every present state transitions to the correct next state and that unintended states self-correct to the sequence (if needed via reset gating). Why Other Options Are Wrong: Modifying BCD or asynchronous counters to “every second pulse” does not generalize to arbitrary sequences and misses the synchronous design principle. Eliminating counter stages is impossible; the flip-flops are what store state. Combinational logic alone cannot “count.” Common Pitfalls: Forgetting to analyze unused states (self-start or self-correcting behavior), neglecting hazards in the decode logic, or not meeting setup/hold timing on the synchronous inputs. Final Answer: external logic circuits that decode the various states of the counter to apply the correct logic levels to the J-K inputs

Question 13

Retriggerable edge-triggered one-shot in VHDL — impossible condition at the clock edge In a retriggerable, edge-triggered one-shot described in VHDL using a counter, which of the following conditions will not exist at a clock edge?

Accepted Answer

Introduction / Context: Retriggerable one-shots (monostables) generate a pulse of programmable width that restarts if another trigger arrives before timeout. A common synthesizable VHDL model uses a counter and edge detection on the trigger input. At each system clock edge, the process updates the counter: load on a trigger, decrement while active, or hold at zero. Recognizing which state descriptors are relevant avoids confusion with unrelated storage structures. Given Data / Assumptions: Implementation uses a synchronous process with a counter representing the remaining pulse duration. Edge-trigger detection logic identifies a new trigger on the current clock edge. No shift registers are used in the specified one-shot architecture. Concept / Approach: At any given clock edge, exactly one of these actions applies: load the counter to the preset timeout if a trigger edge is detected (retriggerable behavior), decrement the counter by one when it is nonzero (pulse ongoing), or keep it at zero if it already timed out. A description involving a “shift register reset” does not belong to a counter-based monostable and thus cannot be the operative condition for this design. Step-by-Step Solution: If trigger_edge = 1 → counter := preset_value (retrigger).Else if counter > 0 → counter := counter − 1.Else → counter remains 0 (no pulse).Output is typically 1 while counter > 0 and 0 otherwise. Verification / Alternative check: Examine typical VHDL templates for retriggerable monostables; they use counters/timers and optional synchronizers, not shift registers. Simulation waveforms show the three mutually exclusive branches above executing at successive clock edges. Why Other Options Are Wrong: Loading on a trigger: valid and required for retriggerability. Keeping counter at zero: valid when idle. Counting down by one: valid while the pulse is active. Common Pitfalls: Mixing unrelated design templates (e.g., serial shift register timing) into counter-based monostables; forgetting to synchronize the trigger to the clock domain (which could otherwise cause missed or double triggers). Final Answer: The shift register is reset.

Question 14

Correct start-up for a ring counter To ensure correct operation at power-up, starting a ring counter requires:

Accepted Answer

Introduction / Context: A ring counter circulates a single 1 (or single 0) through a chain of flip-flops. For deterministic behavior, exactly one stage must start in the opposite state to the others. If the counter powers up with all zeros (or multiple ones), it may stick or behave unpredictably. Proper initialization ensures the intended cyclic sequence begins immediately. Given Data / Assumptions: Ring counter built from n flip-flops with feedback from last stage to first. Active HIGH “token” design (one 1 surrounded by zeros). Asynchronous preset/clear inputs are available for initialization. Concept / Approach: To create the single circulating 1, one flip-flop must be preset to 1 while all others are cleared to 0. On each clock, the 1 shifts to the next stage, producing a one-hot sequence of period n. Without this initial condition, the ring can land in the all-zero state (no token to move) or in a multi-one state (generating erroneous patterns). Step-by-Step Solution: Select one stage (often the first) to receive PRESET = 1 at initialization.Assert CLEAR = 1 on all other stages to force their Q = 0.Release asynchronous controls; the ring now contains exactly one 1.Clock the chain; the 1 advances around the ring each cycle. Verification / Alternative check: Scope the outputs Q0..Qn−1: a single HIGH should appear and step through each stage in turn. If the ring stalls, verify that precisely one stage was preset and the rest cleared. Why Other Options Are Wrong: Clearing all stages leaves no 1 to circulate (dead state). Presetting all stages yields many 1s, not a one-hot pattern. “Clearing one and presetting the others” inverts the required token convention and would not match a 1-hot ring unless the logic were inverted everywhere. Common Pitfalls: Relying on random power-up states; neglecting reset synchronization; or mis-wiring preset/clear polarities so the token is not created reliably. Final Answer: presetting one flip-flop and clearing all the others.

Question 15

Shift-register counter types — identification Which of the following is a type of shift-register-based counter?

Accepted Answer

Introduction / Context: Shift registers can be used not only for data movement but also to implement counters by circulating specific patterns. Recognizing these counter types helps in selecting the right topology for timing, sequencing, or state machine applications where one-hot patterns or specific code sequences are desirable. Given Data / Assumptions: A “shift-register counter” refers to a counter realized by shifting bits rather than by incrementing binary numbers. Examples include ring counters and Johnson (twisted ring) counters. Options include other familiar counter families for contrast. Concept / Approach: A ring counter is a shift register configured with feedback so that a single HIGH circulates around the stages, producing a one-hot sequence. This is the quintessential shift-register counter. Johnson counters similarly use feedback of the inverted output to generate a 2n-state sequence. By contrast, “binary,” “BCD,” or “decade” refer to modulus or code families, usually implemented with standard binary add-one logic rather than pure shifting of a one-hot token. Step-by-Step Solution: Identify which choices are defined by shifting a pattern: ring fits exactly.Recognize that “binary” and “BCD/decade” denote arithmetic counting schemes, not shift-pattern circulation.Therefore, select “Ring” as the shift-register counter type. Verification / Alternative check: Inspect example circuits: a 4-stage ring counter produces 1000→0100→0010→0001→repeat, purely by shifting. Why Other Options Are Wrong: Decade and BCD describe 10-state sequences typically formed with binary counters plus reset decoding, not shift-register circulation. Binary refers to a general counting scheme based on adders/toggles, not a shift-register structure. Common Pitfalls: Equating “decade/BCD” with any implementation technology; although one could create a decade sequence with a shift register, the canonical “type of shift register counter” is the ring (and Johnson), not the code family items. Final Answer: Ring

Question 16

Master Reset (MR) on 4-bit binary counters — behavior and usage Select the statement that best describes how the Master Reset inputs (MR1 and MR2) work on typical 4-bit binary counters (e.g., TTL/CMOS families).

Accepted Answer

Introduction / Context: The Master Reset (MR) inputs on standard 4-bit binary counters (such as 74xx90/93 asynchronous types and 74xx16x synchronous types) provide an immediate way to force the counter outputs to a known state. Knowing the polarity and effect of MR is essential for designing reliable power-on initialization and for clearing counters during operation. Given Data / Assumptions: Two reset inputs labeled MR1 and MR2 are present. Typical counters require both MR inputs to be asserted simultaneously. Unless otherwise specified, MR is asynchronous and active-HIGH. Concept / Approach: In many TTL/CMOS counter families, Master Reset is implemented as an asynchronous clear that overrides counting. The device datasheets commonly specify that MR1 and MR2 must both be HIGH to activate reset. When asserted, all Q outputs go to 0 and the internal counting chain is cleared immediately, independent of the clock. For normal counting, both MR inputs must be held LOW (inactive). Step-by-Step Solution: Identify polarity: MR is active-HIGH on common parts.Activation condition: both MR1 = HIGH and MR2 = HIGH are required.Effect: all Q outputs are forced to 0 (reset state).Count mode: MR1 and MR2 LOW keep reset inactive so the device can count. Verification / Alternative check: Check a standard counter truth table. You will find entries indicating that when MR inputs are HIGH simultaneously, Q outputs immediately become 0000 and counting halts. When MR returns LOW, counting resumes from 0000 on subsequent clock pulses. Why Other Options Are Wrong: Reset to one: typical MR clears to 0000, not 1111. Synchronous reset: most MR implementations are asynchronous; synchronous clears use different pins/control logic. “To enable count mode, MR must be LOW”: partially true as a condition, but it does not describe the use; the key behavior is that both HIGH reset to zero. Preset via MR: presetting uses dedicated load inputs, not MR. Common Pitfalls: Confusing asynchronous MR with synchronous clear or parallel load control lines. Forgetting that some families may use active-LOW clears; always confirm datasheets. Final Answer: When MR1 and MR2 are both HIGH, all Qs will be reset to zero.

Question 17

Ripple (asynchronous) MOD-16 counter — output frequencies per stage A 22 MHz clock drives a MOD-16 binary ripple counter (4 cascaded divide-by-2 stages). What are the output frequencies Q1 through Q4?

Accepted Answer

Introduction / Context: A ripple (asynchronous) binary counter is a cascade of divide-by-2 stages. Each flip-flop output frequency is half that of its driving input. Quickly determining the output frequencies is vital when designing frequency dividers, timing chains, and prescalers in digital and mixed-signal systems. Given Data / Assumptions: Input clock fin = 22 MHz. MOD-16 counter implemented by 4 toggle stages (divide-by-2 each). Outputs labeled Q1 (first stage) through Q4 (fourth stage). Concept / Approach: Each stage divides by 2: fQn = fin / 2^n where n is the stage number (1..4). The total modulus is 2^4 = 16, hence MOD-16. Compute successively to avoid mistakes in powers of two and units. Step-by-Step Solution: Q1 = 22 MHz / 2 = 11 MHz.Q2 = 22 MHz / 4 = 5.5 MHz.Q3 = 22 MHz / 8 = 2.75 MHz.Q4 = 22 MHz / 16 = 1.375 MHz. Verification / Alternative check: Confirm modulus: Q4 period equals 16 input cycles. 22 MHz / 16 = 1.375 MHz, matching the final value. A quick binary check: every stage halves the frequency—no exceptions in a clean ripple chain. Why Other Options Are Wrong: Options with unchanged frequencies (all 22 or all 11 MHz) ignore the divide-by-2 nature. Any set not following exact halving at each stage is incorrect. The 0.6875 MHz value corresponds to a fifth division (not present here). Common Pitfalls: Mixing up stage numbering or forgetting that Q1 is the first divider output. Arithmetic slips when halving odd MHz values; keep one extra decimal place if needed. Final Answer: Q1 = 11 MHz, Q2 = 5.5 MHz, Q3 = 2.75 MHz, Q4 = 1.375 MHz

Question 18

Modulus understanding — output repetition rate For a modulus-8 (MOD-8) counter, the final output pattern repeats how often relative to the input clock pulses?

Accepted Answer

Introduction / Context: The modulus (MOD) of a counter indicates how many unique states it cycles through before repeating. Understanding this lets you predict output repetition and design timing divisions without simulating or breadboarding every case. Given Data / Assumptions: A binary counter with modulus 8. States advance by one on each active clock edge. We consider the full cycle of the counter outputs. Concept / Approach: A MOD-N counter has N distinct states. Thus, one full cycle of the count requires N clock pulses. When the Nth pulse occurs, the counter returns to the initial state and the output sequence repeats. Step-by-Step Solution: MOD-8 ⇒ 8 unique states.One state is produced per clock pulse.Therefore, repetition (final output occurs) once every 8 clock pulses. Verification / Alternative check: Enumerate the binary sequence 000 → 001 → 010 → 011 → 100 → 101 → 110 → 111 → back to 000. Count the transitions: eight pulses are consumed to return to the start. Why Other Options Are Wrong: 16/24/32 pulses: correspond to larger moduli that do not apply. 4 pulses: would be MOD-4 behavior, not MOD-8. Common Pitfalls: Confusing edge counts with state counts; each new state needs one clock. Final Answer: 8 clock pulses

Question 19

Complete state decoding for a MOD-64 counter — gate count and inputs How many AND gates are required to decode every state of a MOD-64 counter, and how many inputs must each AND gate have?

Accepted Answer

Introduction / Context: Decoding a counter state means generating a unique output that is HIGH for exactly one binary pattern and LOW otherwise. This is common in address decoding, time-slot generation, and sequence control. For an n-bit binary counter, each state can be decoded using an n-input AND (or NAND with polarity adjustments) that matches the required combination of bits and complements. Given Data / Assumptions: MOD-64 counter ⇒ 64 states ⇒ requires 6 bits (since 2^6 = 64). Full decoding implies one line per state (one-hot outputs). Assume positive-logic AND decoding with complemented inputs where needed. Concept / Approach: Each distinct 6-bit value (b5…b0) must be recognized. A single 6-input AND gate, fed by the appropriate direct or inverted bit lines, asserts HIGH only when its 6 inputs match the target pattern. Repeating this for all 64 patterns yields a full decoder. Step-by-Step Solution: Number of states = 64 ⇒ number of decode lines = 64.Each state requires examining all 6 bits ⇒ 6 inputs per gate.Therefore: 64 gates, each with 6 inputs. Verification / Alternative check: Compare to standard parts: a 6-to-64 decoder (e.g., built from smaller 3-to-8 blocks) provides 64 one-hot outputs; internally, the logic effectively realizes 64 minterms over 6 variables—conceptually equivalent to 64 AND terms of 6 inputs each. Why Other Options Are Wrong: 128 gates choices: would duplicate lines unnecessarily; only 64 unique states exist. 5 inputs per gate: cannot uniquely identify all 6-bit patterns. 32 gates: insufficient to cover 64 states. Common Pitfalls: Forgetting to account for complemented inputs; practical designs often use NAND/NOR plus inverters. Final Answer: 64 gates, 6 inputs to each gate

Question 20

4-bit counter capacity — maximum modulus A binary counter built from 4 flip-flops has what maximum modulus (number of unique states before repeating)?

Accepted Answer

Introduction / Context: The number of flip-flops determines the maximum count range of a binary counter. Each flip-flop adds one binary digit, doubling the number of available states and thereby increasing the modulus by a factor of two. This is a fundamental design relationship in digital electronics. Given Data / Assumptions: Counter uses 4 flip-flops. Binary counting (no skipped states), starting from 0000. Concept / Approach: An n-bit binary counter can represent 2^n distinct states. For n = 4, the number of states is 2^4 = 16, typically ranging from 0 to 15 in unsigned representation. Unless additional logic truncates the sequence, the natural modulus equals 16. Step-by-Step Solution: n = 4 flip-flops ⇒ states = 2^4.Compute 2^4 = 16.Therefore, maximum modulus = 16. Verification / Alternative check: Enumerate outputs from 0000 to 1111. Counting the number of unique 4-bit patterns yields 16, after which the counter repeats at 0000. Why Other Options Are Wrong: 3, 6, 8: correspond to other moduli (e.g., MOD-8 uses 3 flip-flops). 32: requires 5 flip-flops (2^5). Common Pitfalls: Confusing modulus with maximum count value; modulus is the number of states (0–15 inclusive gives 16 states). Final Answer: 16

Counters Questions