Question 1

Recirculating register — definition by connection Which description correctly defines a recirculating register?

Accepted Answer

Introduction / Context: A recirculating (or circulating) register feeds its serial output back to its serial input so that stored data shifts around in a loop. This is useful for repeated pattern generation and simple delay lines. Given Data / Assumptions: Register supports serial in and serial out. We are seeking the wiring characteristic that makes it recirculate. Concept / Approach: Connecting the serial output back to the serial input closes the loop. Each clock shifts the pattern by one position while preserving the sequence length. Optionally, gating or loading logic can inject new data or reset the pattern. Step-by-Step Solution: Identify serial-out node (e.g., Q of last stage).Feed it to the serial-in node (first stage input), possibly through a gate.Clock shifts circulate the bit pattern indefinitely. Verification / Alternative check: Oscilloscope observation would show the same pattern rotating through Q outputs. This is analogous to a ring counter when the pattern is one-hot. Why Other Options Are Wrong: All Qs tied together: causes contention and invalid logic levels. “A register that can be used over again” is vague and not a defining connection. Common Pitfalls: Confusing recirculating registers with Johnson or ring counters; those are special cases depending on whether feedback is inverted and what initial pattern is used. Final Answer: serial out connected to serial in

Question 2

In digital counters, a ring counter advances a single 1 (one-hot) through its flip-flops. For a modulus-12 (mod-12) ring counter, what is the minimum number of flip-flops required to realize exactly 12 unique states?

Accepted Answer

Introduction / Context: Ring counters are a special class of shift-register counters widely used in timing, control sequencing, and state machine design. They employ a one-hot code in which exactly one flip-flop is set at any time and the 1 circulates around each clock pulse. This question checks whether you understand how the modulus (the total number of distinct states) relates directly to the number of flip-flops in a ring counter. Given Data / Assumptions: Counter type: ring counter (one-hot). Desired modulus (number of states): 12. Each flip-flop stores a single bit and participates equally in the rotation of the 1. Concept / Approach: In a ring counter, the number of valid states equals the number of flip-flops when exactly one flip-flop is high and the rest are low. Therefore, a mod-N ring counter requires N flip-flops. This is unlike binary ripple or synchronous counters where the number of states is 2^n for n flip-flops. Step-by-Step Solution: 1) Define modulus requirement: we need 12 distinct circulating one-hot states.2) In a ring counter, each flip-flop position corresponds to one unique state.3) Therefore, number of flip-flops required = number of states = 12.4) Minimum flip-flop count that exactly meets mod-12 = 12 flip-flops. Verification / Alternative check: If fewer than 12 flip-flops were used (e.g., 10), only 10 distinct one-hot positions would exist, so the modulus would be 10, not 12. Conversely, using more than 12 flip-flops would create a larger modulus than required. Why Other Options Are Wrong: 10 flip-flops: gives mod-10, not mod-12.6 flip-flops: gives mod-6, not mod-12.2 flip-flops: gives mod-2, trivial and incorrect here.8 flip-flops: gives mod-8, still not 12. Common Pitfalls: Confusing ring counters with binary counters. Binary counters use 2^n states; ring counters use n states. Another pitfall is assuming Johnson (twisted-ring) logic, which yields 2n states; this problem specifies a plain ring counter, not a Johnson counter. Final Answer: 12 flip-flops

Question 3

An 8-bit serial-in/serial-out (SISO) shift register is clocked at 2 MHz. Assuming one bit shifts per clock, what total time delay (td) does the register introduce from input to output?

Accepted Answer

Introduction / Context: Serial-in/serial-out (SISO) shift registers delay data by shifting one bit per clock. They are commonly used for timing alignment, serialization, and simple delay lines in digital systems. The total delay equals the number of stages multiplied by the clock period because a bit must traverse each flip-flop stage sequentially. Given Data / Assumptions: Register length: 8 bits (8 stages). Clock frequency fclk = 2 MHz. Each clock shifts data by exactly one stage. No additional pipeline or combinational delay is considered. Concept / Approach: The time delay through an N-bit SISO register is td = N / fclk. This follows because each shift moves the bit by one stage every clock period T = 1 / fclk, and the bit must advance through N stages to appear at the output. Step-by-Step Solution: 1) Compute the clock period: T = 1 / fclk = 1 / (2 × 10^6) seconds = 0.5 µs.2) The number of stages N = 8.3) Total delay td = N × T = 8 × 0.5 µs = 4.0 µs.4) Therefore, the correct delay is 4 µs. Verification / Alternative check: Formula check: td = N / fclk = 8 / (2 × 10^6) s = 4 × 10^−6 s = 4 µs. This matches the stage-by-stage reasoning above. Why Other Options Are Wrong: 16 µs and 8 µs: would correspond to a slower clock or more stages.2 µs and 1 µs: imply fewer stages or a faster clock than specified. Common Pitfalls: Forgetting that the delay scales linearly with the number of stages, or mixing up parallel and serial shifting behavior. Another common mistake is using frequency directly instead of converting to the period before multiplying by the number of stages. Final Answer: 4 µs

Question 4

In shift-register terminology, what is the essential difference between a shift-right register and a shift-left register in digital logic design?

Accepted Answer

Introduction / Context: Shift registers are fundamental storage and movement elements in digital systems. They appear in serializers, deserializers, counters, and digital filters. Understanding the direction of data movement—right or left—is crucial for correct interfacing and timing with serial data streams. Given Data / Assumptions: We are comparing two ideal shift-register types only in terms of bit motion. No special loading (parallel-in) or output mode (parallel-out) differences are implied. Clocking and logic families are assumed equivalent. Concept / Approach: A shift-right register moves each stored bit one position toward the least significant bit (LSB) end per clock pulse, accepting a new bit at the most significant bit (MSB) end. Conversely, a shift-left register moves bits toward the MSB, accepting new bits at the LSB end. The direction determines which end is the serial input and which bit exits at the other end. Step-by-Step Solution: 1) Identify directionality: right shift → toward LSB; left shift → toward MSB.2) Recognize interface impact: the serial-input pin location depends on shift direction.3) Realize timing impact: which bit emerges first (MSB or LSB) depends on direction.4) Conclude: the essential difference is strictly the direction of the shift. Verification / Alternative check: Referencing standard parts (e.g., 74HC164, 74HC595, 74HC194), documentation specifies shift direction and which pin is used for serial input/output. Functional block diagrams confirm that only direction changes between variants; fundamental operation remains identical otherwise. Why Other Options Are Wrong: No difference: incorrect; direction is the defining property.Asynchronous vs. synchronous: both are synchronous unless explicitly designed otherwise.Analog storage: shift registers are digital storage elements.Tri-state requirement difference: unrelated to shift direction. Common Pitfalls: Confusing the shift direction with bit significance or assuming the input/output pins are interchangeable without reconfiguring the design. Always check the data sheet for bit order on serial interfaces. Final Answer: The direction of the shift (toward LSB versus toward MSB).

Question 5

In a microprocessor system, what is the most appropriate and common use of a latch (e.g., transparent latch or D latch) when interfacing with buses and peripherals?

Accepted Answer

Introduction / Context: Latches are level-sensitive storage elements used extensively in microprocessor and microcontroller systems to hold data stable during bus transfers, timing skew, or when crossing between functional blocks. They complement edge-triggered flip-flops by offering transparent or gated behavior that is convenient for buffering and isolating buses. Given Data / Assumptions: System contains a CPU (or MCU), memory, and peripherals on shared buses. Timing is such that data may need to be held stable after address/control lines change. We are considering standard digital design practices. Concept / Approach: A latch captures data when its enable (gate) is active and holds it when the enable is inactive. In bus interfaces (address/data multiplexing, I/O ports, memory-mapped devices), latches are used to demultiplex address/data lines (e.g., holding low-order address lines), to create output ports from the CPU into external hardware, or to buffer data safely until the receiving device is ready. Step-by-Step Solution: 1) Identify a bus transaction where data is valid only for a short window.2) Assert the latch enable during the valid window so the latch captures the byte/word.3) Deassert the enable to hold the captured value, ensuring downstream logic sees a stable level even as the bus changes.4) Use the latched value to drive LEDs, control lines, or provide address hold for external memory/peripherals. Verification / Alternative check: Classic examples include using a latch to hold the multiplexed address (A0–A7) from an 8086/8051-type bus, or using octal latches (e.g., 74HC373/74HC573) to create output ports. Timing diagrams in data sheets show how enables are aligned with valid data windows. Why Other Options Are Wrong: Transportation for employees: humorous distractor, not technical.Replacing the bus with permanent connections: the opposite of what latches achieve; they provide temporary storage, not permanent wiring.Regenerating clock power from address lines: unrelated function.Measuring analog voltage: requires ADC/comparator, not a latch. Common Pitfalls: Confusing a latch with a flip-flop; flip-flops are edge-triggered, latches are level-sensitive. Also, failing to control the enable leads to unintended transparency and bus contention. Final Answer: Temporary storage (holding) of a byte/word to stabilize data between the CPU/bus and a peripheral

Question 6

Three-state (tri-state) buffers support bus sharing by offering a third output condition beyond logic HIGH and logic LOW. What are the three possible output conditions of a three-state buffer?

Accepted Answer

Introduction / Context: When multiple devices must share a common signal line (data bus), direct connection of outputs can cause contention unless inactive devices present a high-impedance state. Three-state buffers solve this by providing a third output condition that effectively disconnects the driver from the bus, allowing safe multiplexing of outputs. Given Data / Assumptions: Standard digital logic levels: logic HIGH (1) and logic LOW (0). An enable pin controls whether the buffer drives the line or goes high-impedance (Hi-Z). Terminology may use words (HIGH/LOW/float) or symbols (1/0/Hi-Z) interchangeably. Concept / Approach: The three states are: logic HIGH, logic LOW, and high-impedance (Hi-Z, sometimes called float). In Hi-Z, the buffer neither sources nor sinks current on the bus; it appears electrically disconnected, enabling another device to drive the line. Step-by-Step Solution: 1) With enable asserted and input = 1 → output drives HIGH.2) With enable asserted and input = 0 → output drives LOW.3) With enable deasserted → output goes to Hi-Z, allowing bus sharing. Verification / Alternative check: Data sheets for common bus buffers (e.g., 74HC125/126, 74LVC series) explicitly list these three states in truth tables. The Hi-Z state is indicated when OE (output enable) is inactive. Why Other Options Are Wrong: HIGH, LOW, negative voltage: three-state buffers do not generate negative rails in standard logic families.LOW, pulsed, latched: mixing behavioral descriptions with logic states; incorrect. Common Pitfalls: Confusing Hi-Z with logic LOW; Hi-Z is not a logic 0 and should not be pulled as such unless a pull-up/pull-down is provided. Also, enabling two outputs simultaneously on the same bus can still cause contention even with three-state devices; careful control of enables is essential. Final Answer: Both of the above describe the same three states

Question 7

In bused digital systems, when is it especially important to insert a three-state buffer between sources and a common line to prevent contention and ensure correct logic levels?

Accepted Answer

Introduction / Context: Three-state buffers (tri-state drivers) are key components for shared buses in microprocessor, FPGA, and memory systems. They prevent electrical contention by allowing only one device at a time to actively drive a line while others are electrically disconnected (high-impedance). Given Data / Assumptions: Multiple potential sources need to communicate with a single receiver (or a shared bus). Only one source should drive the bus at any moment. Unused sources must not load or fight the line. Concept / Approach: Whenever outputs are wired to a common line, contention can occur if two drivers assert different logic levels simultaneously. Three-state buffers add an enable-controlled Hi-Z state so nonselected devices disconnect from the bus, eliminating fights and avoiding damage or logic errors. Step-by-Step Solution: 1) Identify a shared bus scenario with N potential talkers (outputs) and one listener (input or bus).2) Assign an output-enable control so that exactly one talker is active at a time.3) Ensure all other talkers are disabled (Hi-Z) to avoid contention.4) The receiver now sees a clean logic level from the currently enabled source. Verification / Alternative check: Bus arbitration schemes (e.g., microprocessor address/data buses, I2C with open-drain, or SPI with dedicated chip selects) all hinge on the concept of only one active driver. Tri-state buffers implement this at the electrical level for push-pull buses. Why Other Options Are Wrong: All outputs normally HIGH or LOW: the problem is contention, not normal idle levels.Outputs connected to separate inputs: no shared bus, so contention is not inherent.Only one device ever drives the line: a buffer is not strictly required to prevent contention in this case. Common Pitfalls: Enabling two devices simultaneously or forgetting that high-impedance lines may need pull-ups or defined idle states to prevent floating inputs. Always coordinate enable signals via chip select or bus arbitration logic. Final Answer: When two or more outputs are connected to the same input (shared bus line)

Question 8

Digital electronics – tri-state outputs on shift registers When the output of a tri-state (three-state) shift register is disabled, what electrical state is presented at the output pins?

Accepted Answer

Introduction / Context: Tri-state (three-state) outputs are common on shift registers and bus-oriented digital ICs. They allow multiple devices to share a data bus without interfering with each other. Understanding what “disabled” means at the electrical level is essential for reliable system design and troubleshooting. Given Data / Assumptions: A shift register with tri-state outputs (for example, OE or G control). Output enable (OE) can enable the drivers or place them in the disabled condition. No external pull-ups or pull-downs are assumed unless stated. Concept / Approach: Tri-state logic has three possible output conditions: logic HIGH, logic LOW, and high-impedance (often called Hi-Z). When the output is disabled via the OE control, the internal output transistors are turned off so that the pin neither sources nor sinks current. This makes the pin effectively disconnected from the internal driver, allowing another device to drive the line, or allowing the line to float to a level set by external resistors. Step-by-Step Solution: Identify output control: If OE is inactive, internal P- and N-devices are OFF.With both devices OFF, the output driver does not drive HIGH or LOW.Therefore, the pin is in a high-impedance (Hi-Z) state, suitable for bus sharing. Verification / Alternative check: Consulting typical datasheets, the disabled output is specified as Hi-Z with only leakage current limits noted; logic thresholds are undefined because the IC is not actively driving the level. Why Other Options Are Wrong: Float state / float or high-impedance state: “Float” is colloquial and ambiguous; the formal and correct electrical term is “high-impedance state.”LOW state / pulled to VCC internally: A disabled tri-state driver does not force a defined logic level unless external biasing is provided. Common Pitfalls: Assuming a disabled output equals logic LOW; in bus systems this can cause contention if multiple devices attempt to drive simultaneously. Always use proper OE control and, if needed, external pull-ups or pull-downs to define a default level. Final Answer: high-impedance state

Question 9

Ring counter initialization – starting a one-hot sequence To start a ring shift counter so that it operates correctly, what initialization is required?

Accepted Answer

Introduction / Context: A ring counter is a shift register configured so that a single 1 circulates through a chain of flip-flops, generating non-overlapping timing phases. Correct startup is essential; otherwise the sequence may be all zeros or multiple ones, breaking the intended timing. Given Data / Assumptions: Ring counter built from N flip-flops with serial feedback. We want a one-hot pattern (only one flip-flop set at a time). No automatic power-on preset unless designed. Concept / Approach: Ring counters require an initial state with exactly one flip-flop at logic 1 and the remaining flip-flops at logic 0. On each clock, the 1 shifts to the next stage, creating a cyclic sequence of length N. If all stages were cleared, the sequence would remain zeros; if more than one stage were set, multiple ones would circulate. Step-by-Step Solution: Define desired state: 1000…0 (one-hot).Initialize by presetting one flip-flop = 1 and clearing all others = 0.Apply clock pulses to circulate the single 1 around the ring. Verification / Alternative check: Simulate a 4-stage ring: start 1000 → 0100 → 0010 → 0001 → 1000 …; any different initialization (e.g., 0000 or 1100) fails to produce the unique one-hot cycle. Why Other Options Are Wrong: Clearing all / presetting all: leads to 0000… or 1111…, neither produces a valid one-hot sequence.Clearing one and presetting all others: creates a many-hot state, not a ring counter.No initialization: most discrete designs need explicit preset/clear at startup. Common Pitfalls: Relying on random power-on states. Use synchronous or asynchronous preset/clear, or a reset controller, to guarantee the proper initial one-hot pattern. Final Answer: presetting one flip-flop and clearing all others

Question 10

Serial loading into a 5-bit shift register (right-most bit first) Bits 11001 are shifted in (LSB first) to a 5-bit parallel-output shift register whose initial state is 01110. After three clock pulses, what pattern is stored?

Accepted Answer

Introduction / Context: Shift registers are fundamental sequential circuits. Problems often specify the order of serial data (for example, “right-most bit first”) and the direction of shifting. Careful tracking of each clock edge avoids off-by-one errors. Given Data / Assumptions: 5-bit register initial contents: 0 1 1 1 0 (MSB→LSB). Incoming serial pattern: 11001, right-most bit first → input sequence = 1, 0, 0, 1, 1. On each clock the register shifts to the right; the serial input enters the MSB. Concept / Approach: Apply one clock at a time, moving each bit right by one position. The value entering the MSB is the next serial input bit. After the specified number of clocks, read out the five stored bits. Step-by-Step Solution: Initial: 0 1 1 1 0.Clock 1 (in=1): 1 0 1 1 1.Clock 2 (in=0): 0 1 0 1 1.Clock 3 (in=0): 0 0 1 0 1.Stored pattern after 3 clocks (MSB→LSB): 0 0 1 0 1 → 00101. Verification / Alternative check: Repeat the sequence with a quick table of states per clock; both methods give 00101 after the third pulse. Why Other Options Are Wrong: 01110: original state, ignores shifting.00001 / 00110: off-by-one or wrong input ordering.11101: does not match the right-shift with given inputs. Common Pitfalls: Confusing “right-most bit first” with least-significant bit last, or assuming the serial input enters the LSB instead of the MSB for a right-shifting register. Final Answer: 00101

Question 11

8-bit shift register initially clear – serial entry (right-most bit first) The sequence 10011100 is shifted in (LSB first) to an 8-bit parallel-out shift register initially 00000000. After four clock pulses, what are the Q outputs?

Accepted Answer

Introduction / Context: Serial loading order and shift direction must be interpreted correctly to predict the intermediate register contents. With an initially cleared register, early states reveal exactly how data propagates inward from the serial input. Given Data / Assumptions: Initial contents: 00000000 (MSB→LSB). Input pattern: 10011100, right-most bit first → serial input sequence = 0, 0, 1, 1, 1, 0, 0, 1. Register shifts right each clock; serial input enters the MSB. Concept / Approach: Track the first four input bits and states. Because the register starts at all zeros, the first two inputs are zero and do not change the visible pattern. Once ones arrive, they collect at the upper bits. Step-by-Step Solution: Clock 1 (in=0): 00000000.Clock 2 (in=0): 00000000.Clock 3 (in=1): 10000000.Clock 4 (in=1): 11000000.Therefore, after four pulses, Q = 11000000. Verification / Alternative check: Simulate with a quick script or hand table; both agree on 11000000 after four clocks with this loading order. Why Other Options Are Wrong: 10011100: would require the full eight clocks to load all bits.00001100 / 00000011: imply LSB-entry or left-shift direction; not the stated operation.11110000: corresponds to a different initial state or more clocks. Common Pitfalls: Assuming the serial input enters the LSB, or forgetting that the first two zeros leave the register unchanged from its cleared state. Final Answer: 11000000

Question 12

SISO register timing – loading nibble 1100 (LSB first) A 4-bit serial-in/serial-out shift register is initially 0000. You wish to store 1100 with right-most bit first. What 4-bit pattern is present after the second clock pulse?

Accepted Answer

Introduction / Context: With a cleared SISO register, early input bits may not yet be visible in the upper positions. Recognizing the serial order and the number of elapsed clocks is crucial to predict intermediate states. Given Data / Assumptions: Register width: 4 bits, initial = 0000. Target nibble: 1100, right-most bit first → input sequence = 0, 0, 1, 1. Right shift; serial input enters MSB each clock. Concept / Approach: Feed the first two bits (both zeros) and observe that shifting zeros into a cleared register keeps it at zero for these first two steps. Only when ones start entering will the MSB become 1 on later clocks. Step-by-Step Solution: Clock 1 (in=0): 0000 → 0000.Clock 2 (in=0): 0000 → 0000.Therefore after two clocks, content remains 0000. Verification / Alternative check: Continue two more clocks to see the progression: Clock 3 (in=1): 1000; Clock 4 (in=1): 1100. This confirms the intermediate 0000 at the second clock. Why Other Options Are Wrong: 1100 / 0011 / 1111 / 1000: These require more than two clocks or a different shift direction/serial entry point. Common Pitfalls: Assuming the desired final nibble appears immediately; under-counting clock pulses; mixing left-shift with right-shift behavior. Final Answer: 0000

Question 13

Johnson counter sequencing (4-bit) On the third clock pulse, a 4-bit Johnson sequence is Q0 = 1, Q1 = 1, Q2 = 1, Q3 = 0. What are Q0..Q3 on the fourth clock pulse?

Accepted Answer

Introduction / Context: A Johnson (twisted-ring) counter feeds back the inverted last stage into the first stage. It produces a 2N-state sequence using N flip-flops and is widely used for timing and divide-by counters. Given Data / Assumptions: 4-bit Johnson counter with feedback = NOT(Q3) into the first stage. Third pulse state: Q0 Q1 Q2 Q3 = 1 1 1 0. Positive-edge shifting to the right (conventional tabulations). Concept / Approach: On each clock, every Q shifts toward the MSB, and the inverted last stage enters the first stage. Given Q3 = 0, NOT(Q3) = 1 becomes the new incoming bit, so the next state turns on the remaining stage. Step-by-Step Solution: Present state: 1 1 1 0.Feedback bit = NOT(Q3) = 1.Next state after clock: 1 1 1 1. Verification / Alternative check: Enumerate a typical 4-bit Johnson sequence: 0000 → 1000 → 1100 → 1110 → 1111 → 0111 → 0011 → 0001 → 0000 … The transition 1110 → 1111 matches the given step. Why Other Options Are Wrong: Other states do not follow from feedback = 1 at this point or belong to earlier/later steps in the cycle. Common Pitfalls: Confusing ring vs. Johnson counters, or forgetting to invert the last stage when computing the incoming bit. Final Answer: Q0 = 1, Q1 = 1, Q2 = 1, Q3 = 1

Question 14

Serial load into an 8-bit parallel-out register (LSB first) Bits 10110111 are shifted in right-most bit first to an 8-bit parallel-out shift register with initial state 11110000. What is the register state after two clock pulses?

Accepted Answer

Introduction / Context: Intermediate states in serial loading depend on the initial contents, the input bit order, and the shift direction. Recognizing where the serial input enters helps avoid common off-by-one mistakes. Given Data / Assumptions: Initial register: 11110000 (MSB→LSB). Input pattern: 10110111, right-most bit first → input sequence = 1, 1, 1, 0, 1, 1, 0, 1. On each clock the register shifts right; the serial input enters the MSB. Concept / Approach: At each clock, the current MSB receives the next serial bit, and all other bits shift one place toward the LSB. With the first two inputs both equal to 1, the upper two bits become 1 while the lower bits are previous contents shifted down. Step-by-Step Solution: Initial: 1 1 1 1 0 0 0 0.Clock 1 (in=1): 1 1 1 1 1 0 0 0.Clock 2 (in=1): 1 1 1 1 1 1 0 0 → 11111100. Verification / Alternative check: Quick simulation or a state table confirms that two leading 1s populate the top two positions while lower positions are the earlier upper bits shifting down. Why Other Options Are Wrong: 11111000: corresponds to a different intermediate count of clocks.Other options either show no change, the final fully loaded pattern, or a wrong shift direction. Common Pitfalls: Misinterpreting where the serial input enters or counting the number of applied clocks incorrectly. Final Answer: 11111100

Question 15

Definition – parallel load in a shift register What is meant by a “parallel load” operation for a shift register?

Accepted Answer

Introduction / Context: Shift registers support two fundamental ways to enter data: serial and parallel. Parallel load is used whenever an entire word must be captured at once, such as sampling a bus or synchronizing external inputs. Given Data / Assumptions: A generic N-bit shift register with individual flip-flops. Control inputs select between shift mode and load mode. In load mode, external lines D[N−1:0] present a data word. Concept / Approach: Parallel load means that on a single strobe (often a clock edge while a load control is active), every flip-flop captures its corresponding input bit simultaneously. This contrasts with serial loading, which requires N clock pulses to enter N bits. Step-by-Step Solution: Assert the parallel load control (e.g., /PL = 0).Present the data word on the parallel inputs.On the qualifying clock or strobe, all flip-flops latch their input bits at the same time, completing the load in one operation. Verification / Alternative check: Datasheets for universal shift registers (e.g., 74xx194/195) show mode tables where the “load” mode captures all D inputs concurrently on a clock edge. Why Other Options Are Wrong: One-at-a-time or MSB/LSB-only actions describe serial loading or specific preset/clear functions, not parallel load.“Shifted from MSB” is, by definition, serial shifting, not parallel loading. Common Pitfalls: Confusing “preset/clear” with “parallel load,” and assuming that a parallel load requires multiple clocks. It is a one-cycle snapshot of the entire word. Final Answer: All flip-flops are loaded simultaneously with a data word

Question 16

Serial loading time vs. clock rate At a 200 kHz clock frequency, how long does it take to serially load eight bits into a shift register?

Accepted Answer

Introduction / Context: Serial interfaces and shift registers are clocked devices, so throughput and latency are directly tied to the clock frequency. Converting between frequency, period, and total transfer time is a routine skill in digital systems. Given Data / Assumptions: Clock frequency f = 200 kHz. Word length = 8 bits (8 clock pulses to load). One bit is entered per clock pulse. Concept / Approach: The time per clock is the period T = 1 / f. The total time to clock N bits serially is N * T. With f in kilohertz, convert to seconds or microseconds carefully to avoid powers-of-ten mistakes. Step-by-Step Solution: Compute period: T = 1 / 200,000 s = 5 μs.Total time for 8 bits: 8 * 5 μs = 40 μs. Verification / Alternative check: Use frequency scaling: at 100 kHz, 10 μs/bit; doubling to 200 kHz halves the per-bit time to 5 μs, and 8 bits then take 40 μs. The result is consistent. Why Other Options Are Wrong: 4 μs / 0.4 μs: off by factors of 10 or 100.400 μs / 40 ms: orders of magnitude too slow for 200 kHz operation. Common Pitfalls: Confusing kHz with kbits/s or forgetting that 1 kHz = 1000 Hz (not 1024) when doing rough timing estimates. Final Answer: 40 μs

Question 17

74×395A family behavior – role of Output Enable (OE) On the 74395A shift-register device, what function is controlled by the Output Enable pin?

Accepted Answer

Introduction / Context: Many 74×× shift registers include tri-state output buffers so their parallel outputs can share a bus. The Output Enable (OE) pin controls whether the device drives the bus or leaves it undriven (Hi-Z). Given Data / Assumptions: Device: 74395A (representative tri-state parallel-out shift register). OE is a dedicated control input. Mode selection and loading are handled by separate pins. Concept / Approach: OE does not change the stored data; it gates the output drivers. When OE is asserted (device-specific polarity), the buffer connects the internal register to the output pins. When deasserted, the outputs go to high-impedance, allowing bus sharing without contention. Step-by-Step Solution: OE active → outputs enabled → pins reflect register contents.OE inactive → outputs disabled → pins are Hi-Z regardless of internal data. Verification / Alternative check: Datasheets show separate timing for shift/load and OE gating; OE changes the electrical drive state, not the data path inside the register. Why Other Options Are Wrong: “Forces HIGH/LOW” would be a direct output override, not a tri-state control.“Determines when data can be loaded” or “selects serial/parallel” are functions of mode or load inputs, not OE. Common Pitfalls: Confusing OE with asynchronous clear/load controls. OE only affects whether the device is driving the bus. Final Answer: It activates the three-state (tri-state) output buffer

Question 18

Digital electronics – Ring counter state evolution An 8-bit ring counter has the initial state 10111110 (bit 7 to bit 0). After four clock pulses (circular shifting each clock), what will be the new 8-bit state?

Accepted Answer

Introduction / Context: Ring counters are sequential circuits formed by connecting flip-flops in a loop so that the bit pattern circulates with each clock pulse. This question assesses understanding of circular shifting and careful bit tracking over multiple clock edges. Given Data / Assumptions: Counter width: 8 bits (bit7 … bit0). Initial state: 10111110. On each clock, the register performs a circular shift (ring) so the pattern rotates one position. No resets or additional injections of 1s or 0s occur. Concept / Approach: In a ring counter, the least significant bit feeds back to the most significant bit (or vice versa, depending on the implementation). A circular shift preserves the number of 1s and 0s, merely moving their positions. We track the pattern after each clock pulse until the fourth pulse is reached. Step-by-Step Solution: Start: 10111110After 1st pulse (circular shift): 01011111After 2nd pulse: 10101111After 3rd pulse: 11010111After 4th pulse: 11101011 Verification / Alternative check: Another way is to note that a 4-position circular shift of an 8-bit word equals a rotate by 4. Performing either four right-rotates or four left-rotates on 10111110 yields 11101011, confirming the result. Why Other Options Are Wrong: 00010111: Not reachable in exactly four pure circular shifts from the given pattern. 11110000: Changes the count of 1s and 0s (violates preservation under rotation), so it cannot be a rotation of the original pattern. 00000000: Would require clearing bits; rotation does not change bit counts. Common Pitfalls: Confusing a ring counter (rotate) with a simple shift register (shifts in zeros or ones). Forgetting that rotations preserve the number of 1s and 0s. Shifting the wrong direction; here, four rotates left or right both yield the same final state due to symmetry. Final Answer: 11101011

Question 19

Digital systems – Definition of a transceiver What is a transceiver circuit in the context of shared digital data buses?

Accepted Answer

Introduction / Context: Transceivers are widely used on shared data buses to allow bidirectional data movement while preventing contention. Understanding their role is essential for bus-oriented microprocessor and FPGA designs. Given Data / Assumptions: A single signal path may need to carry data in either direction at different times. Control signals enable or disable driving the bus to avoid conflicts. Concept / Approach: Transceiver = transmitter + receiver in one device. It typically has direction control (DIR) and output enable (OE). When DIR selects A→B, the device drives from side A to side B; when toggled, it drives from B→A. OE places outputs in high-impedance when disabled, isolating the device from the bus. Step-by-Step Solution: Recognize the term: 'transceiver' implies bidirectional capability.Identify the option that states operation in both directions.Select the answer that explicitly mentions two-way buffering. Verification / Alternative check: Common TTL/CMOS parts such as 74xx245 are marketed as 'octal bus transceivers' and include DIR and OE pins; their datasheets confirm bidirectional buffering as the primary function. Why Other Options Are Wrong: a buffer that transfers data from input to output: One-direction only. a buffer that transfers data from output to input: Also one-direction only (reversed). blank option: Not applicable. Common Pitfalls: Confusing a simple buffer/line driver (unidirectional) with a transceiver (bidirectional). Forgetting that high-impedance control is vital to prevent bus contention. Final Answer: a buffer that can operate in both directions

Question 20

Parallel in / Parallel out (PIPO) shift register – effect of clocking after a parallel load A 4-bit PIPO register is loaded with D3 D2 D1 D0 = 0 1 1 1 (i.e., D0 = 1, D1 = 1, D2 = 1, D3 = 0). After three left-shifts on the subsequent clock pulses (ze…

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Introduction / Context: Many parallel-access registers support both a parallel load and shift operations. This question checks whether you can track the bit pattern through several left shifts after an initial parallel load. Given Data / Assumptions: Initial parallel load: Q3 Q2 Q1 Q0 = 0 1 1 1 (from D3=0, D2=1, D1=1, D0=1). Three left shifts are applied on subsequent clock pulses. On left shift, zero is shifted into Q0 (zero fill). No additional loads during the three clocks. Concept / Approach: A left shift moves each bit toward the MSB. Q3 is discarded each shift, Q0 receives a zero. Track the pattern step by step for three clocks. Step-by-Step Solution: Initial after load: 0 1 1 1After 1st left shift: 1 1 1 0After 2nd left shift: 1 1 0 0After 3rd left shift: 1 0 0 0 Verification / Alternative check: Count the number of ones: initially three ones. After three left shifts with zero fill in a 4-bit word, only the most significant position remains 1 and the rest become 0, giving 1000 as obtained. Why Other Options Are Wrong: 1110: This is the result after a single left shift, not after three shifts. 0001: Would correspond to three right shifts with zero fill from 0111, not left shifts. 1100: This is the state after two left shifts, not three. Common Pitfalls: Confusing left shift with right shift direction. Assuming circular rotation; here, it is a shift with zero fill, not a rotate. Forgetting that the parallel load sets the starting Q3..Q0. Final Answer: 1000

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