Question 1

Universal shift register behavior — CLEAR inactive and mode selects HIGH: In a typical universal shift register (for example, 74xx194/74xx195), the mode control inputs S1 and S0 select the operation, while CLEAR is an asynchronous reset that is usua…

Accepted Answer

Introduction / Context: Universal shift registers such as the 74xx194/195 families support multiple modes selected by two control lines, commonly labeled S1 and S0. These modes typically include hold (no change), shift right, shift left, and parallel load. An additional CLEAR input usually provides an asynchronous reset (often active LOW). Understanding how CLEAR and the mode selects interact is essential for correct design and troubleshooting. Given Data / Assumptions: CLEAR is inactive when HIGH and active when LOW. Mode selects S1 and S0 determine operation per the standard truth table: 00 = hold, 01 = shift right, 10 = shift left, 11 = parallel load. Clocked operation: the selected action occurs on the appropriate clock transition. Concept / Approach: When CLEAR is HIGH, the device is not being forced into reset. The effective operation is then solely dictated by S1 and S0. The conventional truth table for universal shift registers assigns S1 = 1 and S0 = 1 to the parallel load function, meaning that the parallel data present at the D inputs is transferred into the internal flip-flops and then presented at the parallel outputs on the next clock event. Step-by-Step Solution: Confirm CLEAR level: CLEAR = HIGH implies no asynchronous reset is applied.Interpret mode: S1 = 1 and S0 = 1 select the parallel load mode.Apply clock: on the active edge, the word at the parallel inputs is captured into the register.Outputs: the new word appears at Q outputs as the register state. Verification / Alternative check: Consult any standard 74xx194/195 mode table: the row with S1S0 = 11 shows “Parallel Load.” A lab check by tying CLEAR HIGH, S1 = S0 = HIGH, and stepping the clock while changing the parallel inputs will show the outputs updating to the input word each clock. Why Other Options Are Wrong: Operation is inhibited: Inhibited/hold corresponds to S1S0 = 00, not 11.All stages reset: Reset requires CLEAR active (LOW), not HIGH.Outcome depends on input pattern: The function (parallel load) does not depend on the pattern; only the data value depends on the inputs. Common Pitfalls: Confusing active-LOW CLEAR with an active-HIGH reset; misreading the mode table and swapping the meaning of S1/S0; forgetting that the load happens on the clock event, not immediately upon setting S1/S0. Final Answer: Parallel inputs are loaded and appear at the parallel outputs

Question 2

Making a D flip-flop toggle like a T flip-flop: Which connection forces a clocked D flip-flop to change state (toggle) on every qualifying clock edge?

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Introduction / Context: A D (data) flip-flop copies its D input to Q on the active clock transition, subject to setup and hold times. A common technique is to turn a D flip-flop into a T (toggle) flip-flop. This is useful for building divide-by-2 counters and frequency dividers with minimal hardware. Given Data / Assumptions: The D flip-flop updates Q with the value present at D at the active clock edge. Connections may include feedback from Q or Q̅ to D. We seek a configuration that guarantees Q switches state each clock. Concept / Approach: If the D input always receives the complement of the current output, then at the next clock edge Q must change to the opposite value. The simplest way to enforce this is to connect Q̅ (the inverted output) directly to D. If Q̅ is 1, D is 1 and Q becomes 1 at the next edge; if Q̅ is 0, D is 0 and Q becomes 0 at the next edge. Because Q toggles at each qualifying edge, the flip-flop behaves like a T-FF with T implicitly at logic 1. Step-by-Step Solution: Start with Q at an arbitrary state.Connect Q̅ to D so D = not(Q).On the next active clock edge, Q takes D = not(Q), thus flipping state.Repeat each clock: the device toggles continuously, dividing the clock by 2. Verification / Alternative check: Simulate the flip-flop with D = Q̅; you will observe Q toggling on every active edge, producing a 50% duty cycle square wave at half the input clock frequency (assuming symmetric timing). Why Other Options Are Wrong: Connect Q to D: The device becomes a 1-bit memory with no toggling (Q holds state).Tie D LOW: Q asynchronously stays LOW after the first clock that captures 0; no toggling.Tie D HIGH: Q becomes HIGH after the first clock that captures 1; no toggling. Common Pitfalls: Neglecting setup/hold timing or using the wrong clock edge; forgetting that asynchronous clears/presets override the toggling behavior. Final Answer: Connect Q̅ (the complemented output) back to D

Question 3

QB behavior in a 2-bit ripple (asynchronous) counter: For a two-flip-flop ripple counter driven by a single clock into the first stage, which description best matches the QB waveform relative to the input clock?

Accepted Answer

Introduction / Context: Ripple (asynchronous) counters are built by cascading flip-flops so that the output of one stage clocks the next. In a 2-bit ripple counter, the first stage (QA) toggles at half the input clock frequency, while the second stage (QB) toggles at half the QA frequency. Understanding these frequency divisions and phase relationships is essential for timing and decoding. Given Data / Assumptions: Two T-like stages (or JK/D configured to toggle) are cascaded. Clock drives the first stage; QA output feeds the clock of the second stage. Idealized behavior is considered; propagation delays exist but are ignored for the frequency description. Concept / Approach: The first flip-flop divides the clock by 2, producing QA at f_clk/2. The second flip-flop uses QA as its clock and divides again by 2, yielding QB = f_clk/4. The duty cycle of each ideal toggling stage is 50%. Because the second stage clocks on transitions of QA, QB toggles when QA changes state, establishing a fixed phase relation behind QA. Step-by-Step Solution: Let the input clock be f_clk; QA toggles at f_clk/2.QB toggles whenever QA transitions, resulting in QB at f_clk/4.Ideally, each stage produces a 50% duty cycle square wave.Therefore, QB is a clean square wave at one-quarter the input frequency, lagging QA by one toggle event. Verification / Alternative check: A timing diagram or quick simulation confirms QA transitions at every other clock, and QB at every other QA transition. Counting sequence 00 → 01 → 10 → 11 → 00 reinforces the divide-by-4 behavior at QB. Why Other Options Are Wrong: f_clk/2 aligned to the input clock: That describes QA, not QB.Leads QA by one toggle: In a ripple chain, later stages lag earlier stages.Random glitches: Properly designed counters produce deterministic waveforms; hazards are minimized by using flip-flops for state holding. Common Pitfalls: Misinterpreting asynchronous propagation as frequency error; ignoring that brief ripple delays can affect combinational decoding but not the fundamental divide-by-N behavior at Q outputs. Final Answer: A clean square wave at f_clk/4 with 50% duty cycle, toggling on QA transitions

Question 4

Retaining values in synthesizable VHDL: To remember a value across clock cycles in hardware, where must the value ultimately be stored?

Accepted Answer

Introduction / Context: VHDL can describe both combinational and sequential logic. Remembering a value across time requires state, which in hardware corresponds to storage elements such as flip-flops or latches. The common way to infer such storage in synthesizable VHDL is to update a signal inside a clocked process. Given Data / Assumptions: We are targeting synthesis, not just simulation. State retention across clock cycles implies flip-flops or equivalent. Clocked processes infer registers when signals are assigned conditionally on a clock edge. Concept / Approach: Although variables exist in VHDL, a variable by itself does not create hardware storage unless used within a clocked process in a way that infers a register, and even then the physical storage is ultimately realized by flip-flops on signals. The safe and explicit method is to assign to a signal inside a rising_edge or falling_edge process, which synthesis maps to a register. Step-by-Step Solution: Declare a signal representing the state (for example, count).Create a clocked process: if rising_edge(clk) then … end if.Assign next-state to the signal within the clocked process.Synthesis produces a flip-flop that holds the value across cycles. Verification / Alternative check: Examine synthesis reports or technology schematics: a clocked process with a signal assignment produces a D flip-flop. Simulation confirms that the value persists across edges, not just within delta cycles. Why Other Options Are Wrong: Pure function: Functions are combinational by nature and hold no hardware state.Type declaration: Types define data representations, not storage.Combinational variable: Without a clock, no sequential element is inferred; the value does not persist across cycles in hardware. Common Pitfalls: Assuming variables inherently create memory; forgetting reset logic for deterministic startup; unintentionally inferring latches by incomplete combinational assignments. Final Answer: In a registered signal updated inside a clocked process

Question 5

Identifying a Johnson counter topology: Which name correctly describes a shift-register counter formed by feeding the inverted output of the last stage back to the first stage?

Accepted Answer

Introduction / Context: Counter topologies based on shift registers include the ring counter and the Johnson (twisted-ring) counter. Recognizing their feedback patterns helps predict sequence length, duty cycles, and decoding methods used in timing and control applications. Given Data / Assumptions: The last stage output is inverted before being fed back to the first stage input. All stages clock simultaneously (synchronous shifting). Initialization places the register in a valid starting pattern. Concept / Approach: In a simple ring counter, the last stage feeds back directly (non-inverted) to the first stage, causing a single 1 (or 0) to circulate and producing a modulus equal to the number of stages. In a Johnson counter, the inverted last-stage output is fed back to the first stage. This yields a sequence that is twice the number of stages in length (MOD = 2N), with patterns of consecutive 1s transitioning to consecutive 0s as the sequence progresses. Step-by-Step Solution: Inspect feedback: presence of inversion indicates Johnson (twisted-ring) type.Predict modulus: with N stages, a Johnson counter generates 2N unique states.Understand decoding: non-overlapping windowed outputs can be derived from specific state patterns.Conclude the correct identification is “Johnson counter.” Verification / Alternative check: Draw a 4-stage example and simulate the shift: you will observe 8 distinct states before repeating, confirming the Johnson behavior and differentiating it from a 4-state simple ring counter. Why Other Options Are Wrong: Ring counter: Uses non-inverted feedback and has MOD = N.Decade counter / BCD counter: These are modulus-10 counters based on binary/BCD logic, not defined by twisted-ring feedback. Common Pitfalls: Confusing Johnson and ring counters due to similar schematics; overlooking the inversion bubble in feedback; assuming both have the same modulus. Final Answer: Johnson counter

Question 6

Meaning of counter modulus (MOD): In counter terminology, “modulus” refers to which property of the counting sequence?

Accepted Answer

Introduction / Context: Counters are frequently described by their modulus, often written as MOD-N. Understanding what “modulus” means is fundamental for designing dividers, timing chains, and display drivers. Given Data / Assumptions: Counter progresses through a defined sequence of states. After a fixed number of unique states, it returns to the starting state. MOD denotes the count of unique states in one full cycle. Concept / Approach: The modulus is the sequence length. A binary counter with N flip-flops has up to 2^N states, so its modulus is 2^N if all states are used. Custom modulo counters reset early to achieve a smaller modulus (for example, MOD-10 for decade/BCD counting). This number also represents the division ratio when using a counter as a frequency divider. Step-by-Step Solution: Define MOD-N: the sequence repeats after N unique states.Example: 4-bit binary counter → 16 states → MOD-16.Example: BCD decade counter → 10 states → MOD-10.Therefore, modulus equals the maximum number of states per counting cycle. Verification / Alternative check: Analyze a counter truth table or simulation: counting from 0 up to N-1 and then resetting confirms the modulus as N. Datasheets specify MOD-N to indicate sequence length and division factor. Why Other Options Are Wrong: Fabrication method / elasticity: These are unrelated to digital counting.UP/DOWN control: This selects direction but does not define the number of states. Common Pitfalls: Confusing modulus with maximum binary value only; overlooking that unused or skipped states also define the effective modulus. Final Answer: The maximum number of states in the counter sequence (MOD-N)

Question 7

Largest decimal value in a 4-bit binary counter: What is the decimal equivalent of the maximum binary number representable with 4 bits?

Accepted Answer

Introduction / Context: Binary counters use a certain number of flip-flops to represent states. With 4 bits, the representable unsigned values range from 0 to 2^4 - 1. Knowing the maximum value helps determine range, modulus, and overflow behavior in digital designs. Given Data / Assumptions: We consider an unsigned 4-bit binary number. Binary range formula: 0 through 2^N - 1 for N bits. No sign bit or special encoding is assumed. Concept / Approach: For N bits, the highest unsigned value is 2^N - 1. Substituting N = 4 gives 2^4 - 1 = 16 - 1 = 15. In binary, this is 1111. Many counters therefore have a modulus of 16 when all 4-bit states are used. Step-by-Step Solution: Compute 2^4 = 16.Subtract 1 to get 15.Confirm representation: 1111₂ = 8 + 4 + 2 + 1 = 15.Thus, the largest decimal value is 15. Verification / Alternative check: List the 4-bit codes from 0000 to 1111 and sum the weights of 1111 to verify 15. Simulation or a counter datasheet (MOD-16) corroborates this. Why Other Options Are Wrong: 8: This is 1000₂, not the maximum 4-bit value.16: Requires 5 bits to represent 10000₂.32: Requires 6 bits to represent 100000₂. Common Pitfalls: Off-by-one errors when applying 2^N; confusing the count of states (16) with the largest numeric value (15). Final Answer: 15

Question 8

Maintaining reset on a J-K flip-flop through a clock event: A J-K flip-flop is currently reset and must remain reset after the next active clock edge. Which input condition guarantees that Q stays LOW?

Accepted Answer

Introduction / Context: The J-K flip-flop generalizes the S-R and T flip-flops. Its truth table shows how inputs J and K control setting, resetting, holding, and toggling on each clock edge. Ensuring that a reset condition persists through a clock requires choosing inputs that force Q to remain LOW. Given Data / Assumptions: Edge-triggered J-K flip-flop with synchronous control via J and K. Initial state: Q = 0 (reset). We want Q to remain 0 after the next clock event. Concept / Approach: The canonical J-K table indicates: J = 0, K = 0 → hold; J = 0, K = 1 → reset; J = 1, K = 0 → set; J = 1, K = 1 → toggle. Starting from Q = 0, both hold and reset rows preserve LOW. However, to guarantee staying reset even if transiently disturbed or to enforce reset from an unknown state, the unambiguous forcing condition is J = 0 and K = 1, which actively drives Q to 0 on the clock. Step-by-Step Solution: Start with Q = 0 (reset).Apply J = 0, K = 1.On the next clock edge, the flip-flop takes the reset action: Q → 0.Thus Q remains LOW as required. Verification / Alternative check: Compare to the hold case (J = 0, K = 0): it preserves the current state but does not force reset if noise or an earlier error had set Q. The reset case is more definitive for ensuring LOW output. Why Other Options Are Wrong: J = 0, K = 0: Holds the present state; it does not actively reset if Q was uncertain.J is don’t care, K = 0: Includes possibilities that would set Q when J = 1.J = 0 and K is don’t care: Allows K = 0 (hold) and is not a guaranteed forced reset. Common Pitfalls: Confusing the hold and reset actions; applying J = K = 1 which toggles and can break the required condition. Final Answer: J = 0 and K = 1

Question 9

States in a BCD (decade) counter: How many unique states are present in a Binary-Coded Decimal (BCD) decade counter sequence?

Accepted Answer

Introduction / Context: BCD counters are used to represent decimal digits 0 through 9 using binary patterns. Understanding their number of states is essential for designing display drivers, timers, and digital instruments that need decimal outputs. Given Data / Assumptions: BCD encodes decimal digits 0..9 with four-bit binary codes. After reaching 9, the counter resets to 0. Illegal codes (1010..1111) are skipped or corrected. Concept / Approach: A BCD decade counter cycles through exactly ten valid states, corresponding to digits 0 to 9: 0000 to 1001. It then resets or recycles, giving a MOD-10 sequence. This behavior is different from a straight 4-bit binary counter, which would have 16 states (MOD-16). Step-by-Step Solution: List valid codes: 0000 (0) through 1001 (9).Count them: there are 10 valid states.Recognize that codes 1010..1111 are excluded by design.Therefore, the number of states in a BCD counter is 10. Verification / Alternative check: Datasheets for BCD counters (e.g., decade counters) label them MOD-10 and show reset logic to skip the six unused codes. Why Other Options Are Wrong: 8 or 9: Too few; BCD must represent ten decimal digits.11: Too many; only ten valid BCD states are used. Common Pitfalls: Assuming a 4-bit register always cycles through 16 states; forgetting that BCD purposely avoids 1010..1111. Final Answer: 10

Question 10

Duty cycle of the MSB in a 4-bit BCD (0–9) counter: Over one full 0 to 9 counting cycle, what is the duty cycle of the most significant bit (the 8’s place) of the BCD output?

Accepted Answer

Introduction / Context: Duty cycle of a counter output describes the fraction of time a particular bit remains HIGH during a complete counting period. For BCD counters that count 0–9, the bit patterns differ from straight binary because six states (10–15) do not occur. This affects the duty cycle of each bit. Given Data / Assumptions: We have a 4-bit BCD output representing digits 0..9. MSB is the 8’s bit (weight 8). One complete cycle includes ten counts (0 to 9). Concept / Approach: Determine for which BCD digits the 8’s bit is HIGH. In BCD, the 8’s bit is 1 for values 8 (1000) and 9 (1001) only. That is two counts out of the ten total in the cycle. Step-by-Step Solution: List BCD codes 0..9: 0000, 0001, 0010, 0011, 0100, 0101, 0110, 0111, 1000, 1001.Identify when MSB (8’s bit) is 1: digits 8 and 9 only.Compute duty cycle: HIGH counts = 2, total counts = 10.Duty cycle = (2 / 10) * 100% = 20%. Verification / Alternative check: Plot the MSB across the sequence or simulate a BCD counter. You will see the MSB LOW for counts 0..7, then HIGH for 8 and 9, then back LOW at 0, confirming a 20% HIGH fraction. Why Other Options Are Wrong: 10%: Would imply only one count out of ten is HIGH, which is incorrect.50%: True for the MSB of a 4-bit straight binary counter over 16 counts, not for BCD over 10 counts.80%: Inverts the correct ratio; only two out of ten are HIGH. Common Pitfalls: Confusing BCD with straight binary; assuming each bit has roughly 50% duty without checking the actual sequence length and valid codes. Final Answer: 20%

Question 11

Identify the function and resulting output level A small digital system (commonly built with a PISO shift register and a DATA pin) is used to convert multiple parallel input lines into a single serial stream for transmission. For such a parallel-to-…

Accepted Answer

Introduction / Context: Parallel-to-serial interfaces are widely used to minimize wiring by sending several bits over a single line. A common building block is a PISO (parallel-in/serial-out) shift register whose DATA (or QH) pin outputs one bit per clock. This item asks you to first recognize the function and then state the logic level at the DATA output at a labeled instant in a typical demonstration timing diagram. Given Data / Assumptions: The block is a shift-register style interface with parallel inputs latched, then shifted out bit-by-bit. DATA is an active-high logical output that carries the current serial bit. Standard operation: load on a load/shift control signal; shift on clock edges, MSB- or LSB-first depending on device. Concept / Approach: The distinguishing features of a PISO are a bank of parallel input latches and a single serial DATA output. Recognizing the presence of a shift clock and a load (or shift enable) identifies parallel-to-serial conversion. The DATA level at any instant equals the bit presently at the output stage after the most recent shift. Step-by-Step Solution: Identify shift-register presence from control and single DATA line.Infer function: parallel bits are serialized across time.At the indicated instant, DATA reflects the active-high logic of the current shifted bit, stated here as HIGH. Verification / Alternative check: Compare with a 74HC165/74HC595 timing chart: after a load, sequential clocks present each bit on DATA as HIGH or LOW. A labeled frame showing a 1 at the output corresponds to HIGH. Why Other Options Are Wrong: “multiplexing, 1” confuses time-division selection with shift-register serialization. “demultiplexing, 0” reverses the function. “parallel-to-serial conversion, 0” asserts the opposite level from the indicated HIGH in the referenced moment. Common Pitfalls: Mixing up multiplexers (select 1-of-N inputs) with PISO shift registers (serializing N inputs). Also, forgetting that DATA shows the bit after the most recent shift edge. Final Answer: parallel-to-serial conversion, HIGH

Question 12

Validating the clear (CLR) input on a binary counter How should you correctly test the CLR function so that all flip-flops return to the defined reset state when the CLR pin is driven to its specified active level?

Accepted Answer

Introduction / Context: Most counter ICs provide an asynchronous clear (CLR) pin that forces all internal flip-flops to a known reset state regardless of the clock. A correct bench test must exercise the CLR pin at its specified active level and confirm the outputs return to the documented reset code. Given Data / Assumptions: CLR is asynchronous and overrides counting. Active level (HIGH or LOW) depends on the device datasheet and pin marking (e.g., CLR, CLR̅). Reset state is defined; for binary up-counters it is typically all zeros. Concept / Approach: To test CLR, assert the pin at its active level and directly observe outputs. Since the function is asynchronous, clocking is unnecessary during the assertion. Release CLR and verify normal counting resumes. Step-by-Step Solution: Identify the CLR polarity from the datasheet (active-HIGH or active-LOW).Drive CLR to its active level.Observe all Q outputs; expect the defined reset pattern (commonly all LOW).Release CLR; apply clock pulses and verify counting continues from the reset state. Verification / Alternative check: Use a logic analyzer or LEDs on Q pins to visually confirm synchronous behavior resumes post-clear starting at the reset code. Why Other Options Are Wrong: Grounding or tying to Vcc without regard to active level may be incorrect for devices with opposite polarity. Expecting “toggling” during an asserted CLR is wrong because clear forces a static state. Common Pitfalls: Misreading an overbar on CLR̅ and driving the wrong polarity; failing to allow for propagation time before reading Q pins after assertion. Final Answer: apply the active level to the CLR input and check all of the Q outputs to see if they are all in their reset state

Question 13

Decade counter range clarification A decade (mod-10) counter cycles through how many decimal counts before recycling? Choose the best completion to reflect its 0–9 sequence.

Accepted Answer

Introduction / Context: The term “decade counter” means a modulo-10 counter. It is common in digital systems for generating decimal digits or timing sequences. This item asks how many decimal counts the device produces before rolling over. Given Data / Assumptions: Modulus is the number of distinct states in the counting sequence. A typical decade counter cycles through 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 and then returns to 0. Some decades are BCD-encoded; others can be decoded one-of-ten, but both have modulus 10. Concept / Approach: The key idea is that a decade counter encompasses ten states. While the highest displayed value is 9, the total number of distinct counts is 10. Therefore, the correct completion emphasizes the modulus rather than the printed maximum value. Step-by-Step Solution: Define modulus: number of unique states before repeating.List states: 0 through 9.Count the states: 10 total.Select the option that states 10. Verification / Alternative check: Inspect a 74xx90/74xx160/74xx4017 datasheet; the decade sections clearly indicate ten counts per cycle. Why Other Options Are Wrong: “9” confuses the highest numeral with count total. “15” corresponds to a 4-bit binary full modulus. “0” is not a count quantity. Common Pitfalls: Equating the maximum displayed digit (9) with the modulus. Always count the number of states, not the largest state label. Final Answer: 10

Question 14

Altera/Intel FPGA megafunction identification Which counter megafunction from the Altera library supports up/down counting, synchronous parallel load, and asynchronous cascading controls?

Accepted Answer

Introduction / Context: When building counters in FPGAs, designers often use parameterizable megafunctions from the vendor library rather than hand-coding every feature. In the Altera (now Intel) ecosystem, the Library of Parameterized Modules (LPM) provides such reusable blocks. Given Data / Assumptions: The question references a counter capable of up/down operation. It also supports synchronous parallel load (loading a preset value on a clock) and asynchronous features for cascading. Altera’s LPM suite includes LPM_COUNTER, configurable for width, direction, load, clear, and carry/borrow for cascading. Concept / Approach: Match the feature set to the vendor library term. “LPM” refers to a catalog (e.g., LPM_COUNTER). “74134” is a fixed-function TTL part number. “synchronous” and “AHDL” are not names of specific megafunctions; AHDL is a hardware description language, not a counter block name. Step-by-Step Solution: Identify the need: up/down control, synchronous load, cascade.Recall Altera’s LPM megafunctions: LPM_COUNTER offers these options.Eliminate distractors that are not megafunction identifiers.Select “LPM.” Verification / Alternative check: Vendor documentation for LPM_COUNTER lists generics/parameters enabling direction, synchronous load, asynchronous clear, and carry-chains for cascading. Why Other Options Are Wrong: “74134” is unrelated TTL. “synchronous” describes clocking style, not a library element. “AHDL” is a language, not a ready-made counter block. Common Pitfalls: Confusing language/tool names with library component names; looking for a specific part number instead of the generic FPGA megafunction category. Final Answer: LPM

Question 15

Identify the monostable type A one-shot that begins an output pulse on a trigger edge and restarts (extends) the internal pulse timing if another trigger arrives before completion is called ________.

Accepted Answer

Introduction / Context: Monostable multivibrators (one-shots) generate a single output pulse in response to a trigger. Their behavior upon receiving additional triggers during the active pulse is the key distinction between retriggerable and non-retriggerable types. Given Data / Assumptions: A trigger edge starts the timing cycle. Another trigger may occur before the pulse ends. We must classify whether the device ignores or extends based on that second trigger. Concept / Approach: A retriggerable one-shot restarts or extends the pulse width each time a valid trigger is applied during the current pulse, effectively lengthening the total output duration. A non-retriggerable one-shot ignores additional triggers until it times out. Step-by-Step Solution: Define retriggerable behavior: new trigger restarts timing.Define non-retriggerable behavior: extra triggers are ignored.Match the described behavior (restart/extend) to “retriggerable.”Select the correct term. Verification / Alternative check: Review timing diagrams for datasheets like 74HC123 (retriggerable) versus 74HC121 (non-retriggerable). Why Other Options Are Wrong: “non-retriggerable” does not extend the pulse. “high-level triggered” and “edge-triggered” describe trigger sensitivity, not the retrigger behavior. Common Pitfalls: Confusing input triggering mode with retrigger capability; failing to account for minimum re-trigger interval and input conditioning. Final Answer: retriggerable

Question 16

Fundamental property of ring counters For an n-stage ring counter implemented with flip-flops and a circulating single 1, the ________ equals the number of flip-flops in the register.

Accepted Answer

Introduction / Context: Ring counters are simple sequential circuits formed by connecting a shift register in a loop with a single circulating 1 (or 0). They are used for sequencing and time-slot generation. Understanding their modulus is key to predicting cycle length. Given Data / Assumptions: There are n flip-flops connected in a ring. A single 1 circulates; at any time, exactly one flip-flop is HIGH. No fault states (e.g., multiple ones) are present after initialization. Concept / Approach: Each clock shifts the 1 to the next position. After n shifts, the 1 returns to its original position, so the number of unique states visited equals n. Therefore, the modulus (the number of states before repetition) equals the number of flip-flops. Step-by-Step Solution: Let the initial state be 1000…0 (n bits).Clock the register once: the 1 moves to the next bit position.After k clocks, the 1 is at position k (mod n).After n clocks, the state repeats, so modulus = n. Verification / Alternative check: Simulate a 4-bit ring (states: 1000, 0100, 0010, 0001 → repeat). Four states confirm modulus 4. Why Other Options Are Wrong: “number of invalid states” depends on initialization, not fixed at n. “number of CASE statements” is a coding artifact. “other states are” is not a proper property statement. Common Pitfalls: Confusing ring counters with Johnson counters (twisted ring) whose modulus is 2n; assuming a ring naturally recovers from illegal states without additional logic. Final Answer: modulus is

Question 17

Converting an asynchronous up-counter to a down-counter For a ripple (asynchronous) binary counter built from leading edge-triggered flip-flops, which change produces decrementing behavior?

Accepted Answer

Introduction / Context: Ripple counters propagate a derived clock from one flip-flop to the next. Whether the counter counts up or down depends on which output (Q or Q̅) is used to clock subsequent stages and on the triggering edge. Given Data / Assumptions: Each stage is edge-triggered on the same edge type. The first stage is clocked by the external clock source. Subsequent stages are clocked by outputs of the preceding stage. Concept / Approach: In a binary ripple up-counter, each next stage is typically clocked from the Q output of the previous stage. To reverse the count direction, clock each next stage from the complementary output Q̅ (the “other side”), which inverts the sequencing, producing a down-count. Simply changing the sampling edge or taking the final readout from Q̅ does not universally implement a proper down-count in the typical configuration. Step-by-Step Solution: Start with the standard up-counter chaining: stage k+1 clocked by Q_k.Swap the chain drive to Q̅_k for stages k≥0.Verify sequence by writing first few states; observe decrementing pattern.Confirm that changing trigger polarity alone is insufficient and affects all stages timing. Verification / Alternative check: Simulate a 3-bit counter and compare Q2 Q1 Q0 sequences for the two chaining methods; the Q̅-chained design walks the binary states in reverse order. Why Other Options Are Wrong: Taking the output from the other side for display (option a) doesn’t change the internal chaining. Changing to trailing-edge devices (option c) without re-chaining does not guarantee reverse counting. “All of the above” is therefore incorrect. Common Pitfalls: Forgetting that ripple counters are sensitive to propagation delays; re-chaining must still satisfy timing and fanout limits. Final Answer: clocking of each succeeding flip-flop from the other side ( instead of Q)

Question 18

Sequential circuit design objective When you deliberately design the next-state logic so that the machine follows a non-natural order of states (e.g., 0 → 5 → 2 → 7 …), the sequential circuit design is used to ________.

Accepted Answer

Introduction / Context: Finite-state machines and counters can be engineered to step through states in virtually any desired sequence. Beyond simple up/down patterns, designers often implement custom (pseudo-random or application-specific) sequences to meet timing or control requirements. Given Data / Assumptions: “Random order” here means an arbitrary, pre-defined sequence, not truly random noise. Next-state logic can be derived to realize any permutation over the valid state set, with unused states redirected safely. Clocking is synchronous; transitions are governed by combinational logic feeding the state register. Concept / Approach: The question asks for the broad purpose of sequential design: shaping the next-state map to traverse a chosen sequence. Classical examples include Johnson counters, Gray counters, or application-specific “random-looking” sequences used in scanners, LED chasers, or test pattern generators. Step-by-Step Solution: Define the set of desired states.Specify the transition relation (who follows whom).Synthesize next-state equations using Karnaugh maps or HDL.Verify that the circuit visits states in the intended non-natural order. Verification / Alternative check: Simulate the state machine for several cycles; confirm it follows the custom sequence and handles illegal states with resets or self-correction. Why Other Options Are Wrong: “count up” and “count down” are special cases of sequential design but not the general aim stated. “decode an end count” is a separate task performed by decoders or comparators. Common Pitfalls: Neglecting unused states can cause lock-up. Always define recovery for illegal states and ensure synchronous reset behavior. Final Answer: count in a random order

Question 19

Compute the divided clock frequency A digital counter chain divides a 4 MHz input clock by an overall factor of 5000 before presenting the labeled output. Determine the resulting output frequency.

Accepted Answer

Introduction / Context: Many timing applications derive a lower-frequency clock from a high-frequency source by cascading divide-by-N counters. The output frequency is computed by dividing the input by the total modulus. Given Data / Assumptions: Input clock frequency f_in = 4 MHz. Total division factor N = 5000. Output frequency f_out = f_in / N. Concept / Approach: Use the simple frequency division relationship. Ensure units are consistent (hertz). Avoid common mistakes such as multiplying or misreading the modulus. Step-by-Step Solution: Convert 4 MHz to Hz: 4 MHz = 4 * 10^6 Hz.Apply formula: f_out = 4 * 10^6 / 5000.Compute the quotient: 4,000,000 / 5,000 = 800.Express the result with units: f_out = 800 Hz. Verification / Alternative check: Break the factor into 100 * 50: 4 MHz / 100 = 40 kHz; 40 kHz / 50 = 800 Hz, confirming the same result. Why Other Options Are Wrong: “4 MHz” implies no division. “20 kHz” would correspond to a 200 division. “210.5 kHz” is unrelated to an integer divide of 5000. Common Pitfalls: Mixing kHz and Hz or dropping zeros; always track powers of ten carefully in frequency division problems. Final Answer: 800 Hz

Question 20

Structured troubleshooting of digital counters One effective method is to actively drive (exercise) the circuit with a known stimulus or test pattern and then observe the output sequence to verify proper bit patterns. Complete the statement.

Accepted Answer

Introduction / Context: Digital troubleshooting is most productive when you control inputs and observe outputs under known conditions. Applying a defined stimulus or test pattern allows you to compare actual behavior to the expected sequence, quickly isolating faults in counters and state machines. Given Data / Assumptions: The circuit under test is a counter or sequential device producing bit patterns. Test equipment can inject signals and capture outputs. Expected patterns are known from datasheets or simulations. Concept / Approach: “Exercising” a circuit means feeding it with deliberate inputs: clocks, resets, enables, and known data. A “stimulus” or “test pattern” is a predefined input sequence used to validate functionality. Observing outputs with logic probes, analyzers, or scopes and comparing to the expected sequence confirms correctness or reveals specific stage failures. Step-by-Step Solution: Define the expected count or state sequence.Apply a controlled clock and known control signals (reset, load).Record output patterns across Q pins for several cycles.Compare observed patterns with the expected; note deviations to locate faults. Verification / Alternative check: Use a pattern generator to automate stimulus; capture with a logic analyzer for clear timing correlation and state decoding. Why Other Options Are Wrong: “voltmeter, ohmmeter” provide static measurements, not dynamic sequence validation. “scope, logic analyzer” are observation tools, but the missing action is providing a known stimulus. “smashing, hammer, axe” is obviously not a troubleshooting method. Common Pitfalls: Failing to reset the device to a known state before testing; using uncontrolled, noisy clocks that complicate analysis. Final Answer: exercising, stimulus, test pattern

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