Question 1

When is multiplexing required? In digital systems, multiplexing of signals is usually required under which circumstance?

Accepted Answer

Introduction / Context: Multiplexing (time, frequency, or code division) allows multiple signals to share a limited physical medium. It is widely used to reduce wiring and channel count where links are constrained. Given Data / Assumptions: Long-distance links are expensive or have limited conductors. On-chip internal data movement often has abundant parallel paths, but board-to-board or long-cable runs are scarce. Concept / Approach: Over long distances, sending many parallel wires is costly and prone to skew. Serial or multiplexed schemes combine many signals over fewer lines, then demultiplex at the destination, conserving bandwidth and conductors. Step-by-Step Solution: Identify the scenario most constrained by physical media: long distance links.Multiplexing combines multiple signals over one or a few channels.Therefore, option “moving data over long distance transmission lines” is most accurate. Verification / Alternative check: Standards like TDM in telecom, SERDES on backplanes, and network links exemplify multiplexing for distance or pin-count constraints. Why Other Options Are Wrong: Internal CPU buses often use wide parallel paths; multiplexing may occur but is not “usually required.”Memory/register transfers are typically parallel on short distances to maximize throughput.Option d restates a typical internal case, not the primary driver for multiplexing. Common Pitfalls: Assuming multiplexing is always better; for short distances with tight timing, parallel buses can be superior. Final Answer: Moving data over long distance transmission lines.

Question 2

Cascading 7485 comparators for 8-bit comparison To compare two 8-bit numbers using two 7485 (4-bit) magnitude comparators, how should the cascade connections be made?

Accepted Answer

Introduction / Context: The 7485 is a 4-bit magnitude comparator with cascade inputs (<, =, >) and outputs (P < Q, P = Q, P > Q). Cascading two devices allows comparison of wider words such as 8-bit values. Given Data / Assumptions: Two comparators: COMP A compares the lower-order 4 bits, COMP B the higher-order 4 bits. Proper use of cascade pins is required for correct multi-byte comparison. Concept / Approach: The lower nibble determines equality and less/greater status only if the higher nibbles are equal. Therefore, the outputs from the lower comparator must feed the cascade inputs of the higher comparator. The lowest-level comparator’s cascade inputs must be initialized to “equal” true and “less/greater” false to start the chain. Step-by-Step Solution: Assign COMP A to bits [3..0] and COMP B to bits [7..4].Initialize COMP A cascade inputs: = input = 1 (VCC), < input = 0 (GND), > input = 0 (GND).Connect COMP A outputs (P < Q, P = Q, P > Q) to COMP B inputs (<, =, >), respectively.Read the final comparison results from COMP B outputs. Verification / Alternative check: This is the standard cascading method in the 7485 data sheet truth table and application diagrams. Why Other Options Are Wrong: Connecting outputs to outputs (option a) is invalid and may damage parts.Option b lacks the necessary initialization of the lowest stage inputs.Option c initializes but fails to cascade the stages. Common Pitfalls: Accidentally swapping less/greater lines or forgetting to tie initial cascade inputs, resulting in indeterminate comparisons. Final Answer: Connect A’s P < Q, P = Q, P > Q outputs to B’s <, =, > inputs, and also tie A’s = input high (VCC) with < and > inputs low (ground).

Question 3

Erratic BCD-to-decimal decoder behavior on breadboard A BCD-to-decimal decoder circuit behaves erratically on a wire-wrap protoboard, although scoped waveforms look reasonable. What is the most helpful advice to correct the problem?

Accepted Answer

Introduction / Context: Decoders (such as BCD-to-decimal) are level-sensitive combinational devices. During input transitions, temporary invalid code combinations can occur and create momentary incorrect outputs called “glitches.” On protoboards, wiring capacitance and unequal path delays exacerbate this. Given Data / Assumptions: Breadboard/wire-wrap prototype with longer leads and skew. Scope may not capture brief hazards if bandwidth/triggering are not ideal. Decoder has an enable/strobe input. Concept / Approach: Glitch prevention uses a strobe (enable) to qualify decoding only after the inputs are stable. By gating outputs with a properly timed enable, momentary hazards do not propagate to the system. Step-by-Step Solution: Recognize the conditions for hazards: multiple input bits switching nearly simultaneously.Use the decoder’s enable to latch or gate outputs only when inputs have settled (e.g., after a clock edge).Optionally buffer inputs with a register to guarantee synchronous, hazard-free transitions. Verification / Alternative check: After applying a strobe synchronized to stable inputs, erratic activations disappear, confirming glitches as the root cause. Why Other Options Are Wrong: Thermal intermittence is rare and would show temperature correlation.Bad connections are possible but would show persistent faults, not momentary decoding glitches aligned with transitions.Combining both suggestions (option d) spreads focus and ignores the most likely logic-timing cause. Common Pitfalls: Relying on a slow scope timebase that misses narrow pulses, and assuming stable logic when hazards are too brief to see easily. Final Answer: Glitches are probably causing false decoding; use a strobe (enable) to time the decoder input sampling.

Question 4

Decoder fundamentals Which statement best describes the function of a digital decoder in logic systems?

Accepted Answer

Introduction / Context: Decoders translate binary input codes into one-of-N outputs (or into a coded representation) to select devices, display digits, or route signals. Examples include 2-to-4, 3-to-8 decoders and BCD-to-7-segment decoders. Given Data / Assumptions: Binary input code supplied to the decoder. Outputs correspond to specific symbols, lines, or digits. Concept / Approach: A decoder interprets the binary input and activates exactly one output line (for one-of-N) or drives a pattern (e.g., 7 segments) to represent the coded value. This is the inverse operation of an encoder. Step-by-Step Solution: Recognize “binary to specific output” mapping as the standard decoder role.Examples: 74HC138 (3-to-8 line decoder) or 74LS47 (BCD-to-7-segment).Therefore select the statement that describes converting a binary number into an output representing a character/digit. Verification / Alternative check: Datasheets and textbooks define decoders as code translators from inputs to outputs with one-hot or formatted patterns. Why Other Options Are Wrong: Decimal to binary: that is encoding or binary conversion, not decoding in logic IC terms.Preventing improper operation: too vague; decoders are functional routing/selection devices.Brand interoperability: unrelated to decoder function. Common Pitfalls: Confusing decoder with demultiplexer—though related, demultiplexers route a single data line to multiple outputs under binary select inputs. Final Answer: A decoder converts a binary number into a specific output representing a particular character or digit.

Question 5

Multiplexer select sequencing for parallel-to-serial conversion A 16-input multiplexer (16:1 MUX) is used to perform parallel-to-serial data conversion. Which counter modulus is required to drive the MUX's data-select inputs so that all 16 inputs ar…

Accepted Answer

Introduction / Context: This question checks your understanding of how a multiplexer performs parallel-to-serial conversion and how many select states are required to visit each input exactly once per frame. Correctly choosing the driving counter ensures that every parallel bit is sampled in order. Given Data / Assumptions: A single 16:1 multiplexer is used. The goal is to serialize 16 parallel lines by time-multiplexing. Select inputs are binary encoded (S3 S2 S1 S0). Concept / Approach: A 16:1 MUX requires 4 select lines to address inputs 0 through 15. To sweep all inputs cyclically, the select lines must iterate through 16 unique states. Therefore, a 4-bit counter that counts 0..15 (a modulus-16 counter) is required. Step-by-Step Solution: Number of inputs N = 16 → required select states = 16.Binary states needed = 2^k ≥ 16 → k = 4 select lines.A counter that produces 16 distinct states is a MOD-16 counter. Verification / Alternative check: Any lesser modulus (for example, 8 or 4) would not reach all inputs; any greater modulus would repeat or produce unused states. Thus MOD-16 is both necessary and sufficient. Why Other Options Are Wrong: MOD 8 / MOD 4 / MOD 2: These provide 8, 4, or 2 states respectively, which are insufficient to address 16 inputs. Common Pitfalls: Confusing number of select lines (4) with modulus. Both are linked: 4 lines imply a counter cycling through 2^4 = 16 states. Final Answer: MOD 16

Question 6

Troubleshooting a DMM's 7-segment display that occasionally skips digits A digital multimeter's seven-segment display sometimes fails to show certain numbers. What is a safe and effective test procedure to identify a defective segment?

Accepted Answer

Introduction / Context: Intermittent missing numerals on a seven-segment readout usually point to a single weak or open segment. A controlled bench test that individually drives segments helps you separate a defective LED segment from a driver or interconnect problem. Given Data / Assumptions: The DMM display is a seven-segment package (likely multiplexed). Segments can be driven individually for test with proper polarity and current limiting. We want a procedure that isolates the display from the rest of the circuit. Concept / Approach: By applying a known test voltage to each segment (through an appropriate series resistor to limit current), you directly verify whether the LED segment illuminates. If a segment fails to light when correctly driven, the display module is suspect. If all segments light reliably, the fault likely lies in the driver IC or multiplexing circuitry. Step-by-Step Solution: Identify display type: common anode or common cathode.Use a bench supply (for example, 5 V) with a ~330–1 kΩ series resistor to limit current to a safe LED value.Drive each segment pin with correct polarity while holding the common pin at the proper rail.Record any segment that does not illuminate or flickers abnormally. Verification / Alternative check: If all segments pass, reconnect the display and probe the driver outputs during normal operation. A logic probe or oscilloscope will reveal missing drive pulses due to a faulty driver or bad solder joints. Why Other Options Are Wrong: Remove and diode-check (A): Impractical and may mislead due to internal multiplexing; also risks damage. Tie all segments to Vcc (C): Forces “8” but does not isolate a single bad segment versus driver faults. Slowly increase DMM supply (D): Varying entire supply is not a targeted segment test and may stress the instrument. Common Pitfalls: Forgetting the series resistor, which can overcurrent and damage the LED segment. Applying reversed polarity on common-anode versus common-cathode modules. Final Answer: Apply 5 V directly to each segment (with proper current limiting) and replace any segment that does not light.

Question 7

7447 BCD-to-seven-segment decoder: which segments light for the digit “1”? For a 7447 code converter driving a common-anode display, which segments are active to display the numeral 1?

Accepted Answer

Introduction / Context: Understanding standard seven-segment coding is essential for designing front-panel displays, debugging instrument readouts, and writing HDL driving logic. The TTL 7447 is a classic BCD-to-seven-segment decoder for common-anode LEDs. Given Data / Assumptions: IC: 7447 BCD-to-seven-segment decoder/driver. Target digit: decimal 1. Common-anode display (7447 outputs are active-LOW). Concept / Approach: A seven-segment display uses segments a through g. For the digit “1,” only segments b and c should illuminate. With the 7447 and a common-anode LED, the decoder pulls b and c LOW (active-LOW), while other outputs remain HIGH (off). Step-by-Step Solution: Recall the standard encoding: “1” → segments b and c on; a, d, e, f, g off.Map to the 7447: outputs for b and c go LOW; others stay HIGH.Therefore, the correct description is “segments b and c are active.” Verification / Alternative check: Check a segment truth table from a datasheet or memorize common patterns: 0 (a,b,c,d,e,f), 1 (b,c), 2 (a,b,d,e,g), 3 (a,b,c,d,g), etc. Why Other Options Are Wrong: a only / b only / a and b: None match the standard “1” pattern, which always requires b and c together. Common Pitfalls: Mistaking common-anode for common-cathode: the 7447 uses active-LOW outputs, but the lit segments for a given digit are the same logical set (polarity just inverts the drive). Final Answer: Segments b and c are active.

Question 8

HDL design focus for multiplexers and demultiplexers In hardware description languages such as VHDL or Verilog, what is the key issue when designing MUX and DEMUX blocks?

Accepted Answer

Introduction / Context: Writing synthesizable HDL for data routing primitives like multiplexers (MUX) and demultiplexers (DEMUX) relies on correctly expressing conditional data movement. This ensures that tools infer efficient logic consistent with the intended selection behavior. Given Data / Assumptions: Target: generic MUX/DEMUX in HDL. We care about the central design concern that yields correct inference in synthesis. Concept / Approach: Both MUX and DEMUX are fundamentally conditional assignments. A MUX selects one of many inputs to drive a single output based on select lines; a DEMUX routes one input to one of many outputs. In HDL, this is captured by conditional signal assignments (if/else or case) that map select values to the proper data path with defined defaults to avoid latches. Step-by-Step Solution: Identify that selection logic = conditions on select lines.Express behavior with concurrent selected signal assignment or sequential if/case in a clockless process.Ensure all branches are covered to prevent unintended storage elements. Verification / Alternative check: Synthesis reports should show inferred multiplexers (e.g., LUT-based mux trees) or demultiplexing gates without latches. Simulation waveforms must match truth tables. Why Other Options Are Wrong: Library availability: Helpful but not the core design issue. Using case specifically: One way, but if/else works equally well; the key is the conditional assignment itself. Describing the functions: Too vague; the actionable requirement is conditional assignments tied to select inputs. Common Pitfalls: Leaving out a default branch, causing latch inference. Mixing blocking/non-blocking assignments incorrectly in Verilog when clocked logic is added later. Final Answer: Assigning signals under certain conditions.

Question 9

Who “controls” a demultiplexer's outputs? In a standard digital demultiplexer (1-to-N), which signals directly determine which output line is driven?

Accepted Answer

Introduction / Context: Demultiplexers route one input signal to one of several outputs. Knowing which signals govern the routing is crucial for timing analysis, HDL design, and system debugging. Given Data / Assumptions: Device: 1-to-N DEMUX with select lines and a single data input. Enable input may also exist but is not central to the question. Concept / Approach: The select lines (also called address or control inputs) encode the index of the output line that will receive the data. Internally, logic gates decode the select pattern to enable exactly one output path; however, the external, user-controlled signals that choose the path are the select lines. Step-by-Step Solution: Recognize that data alone cannot choose its destination; it is merely replicated/steered.Select lines provide the binary address of the chosen output.Therefore, the outputs are under the direct control of the select lines. Verification / Alternative check: Truth tables show that for each select code, exactly one output equals the data input (or its enabled form), confirming select-line control. Why Other Options Are Wrong: Internal gates (AND/OR): Implementation detail, not user-controlled signals. Input data line: Provides the value to be routed, but does not choose the destination. Common Pitfalls: Confusing data enables with address selection; enable controls whether routing occurs, not which line is chosen. Final Answer: input data select lines.

Question 10

Purpose of control inputs in HDL magnitude comparators Why do HDL or IC magnitude comparators (for example, 7485-style designs) include control inputs?

Accepted Answer

Introduction / Context: Magnitude comparators determine whether A < B, A = B, or A > B. For wide words (more than the base bit-width), multiple comparator slices must be chained. Control inputs enable clean multi-slice operation. Given Data / Assumptions: Comparator slices operate on limited width (for example, 4 bits). We must compare multi-nibble or multi-byte values by cascading stages. Concept / Approach: Control inputs such as AB_in feed the results from less-significant stages into more-significant ones. The most-significant slice outputs the final comparison after considering these chained inputs, thereby enabling scalable word sizes. Step-by-Step Solution: Identify the need: compare words wider than a single comparator's width.Use control/cascade inputs to import lower-order comparison results.Produce final AB at the top stage. Verification / Alternative check: Datasheets for devices like the 7485 show pins named “cascade inputs” and “cascade outputs” for exactly this purpose. Why Other Options Are Wrong: Generic phrases (B, C, D): Too vague and do not capture the specific purpose of cascading multi-chip comparisons. Common Pitfalls: Driving cascade inputs incorrectly (for example, not setting the initial least-significant comparator inputs), which yields erroneous results. Final Answer: For cascading the chips

Question 11

Parity error output always HIGH although DATA IN equals DATA OUT In the shown data transmission system, a logic analyzer confirms that DATA IN (left) matches DATA OUT (right), yet the parity error flag remains asserted HIGH. What is the most reasona…

Accepted Answer

Introduction / Context: Parity systems append a single bit so that the total number of 1s is either even or odd. If a parity error flag is stuck HIGH despite correct end-to-end data, the issue is likely within the parity generation/checking logic or in the way the flag is formed and latched. Given Data / Assumptions: DATA IN equals DATA OUT bit-for-bit (verified on a logic analyzer). Parity error output remains HIGH continuously. We do not have evidence that parity bits themselves are correct or that the error flag logic is properly reset. Concept / Approach: If data integrity appears intact, persistent error indication points to faults in the parity generation/checker path or the error flag logic (gate or latch). Possible culprits include a failed parity generator/checker IC, a stuck flag-forming gate, or a defective storage element that holds the error bit after a transient. Step-by-Step Solution: Confirm the parity convention (even/odd) and verify the parity bit line at both ends.Probe the parity checker outputs and the error gate input to see if they agree.Check the error flag latch or storage circuit and its reset/clear control.If any of these blocks misbehaves, replace or rework accordingly. Verification / Alternative check: Inject known patterns with both correct and intentionally wrong parity. If the error flag never clears on correct parity patterns, the checker or flag logic is faulty. If it never asserts on wrong parity, the generator/checker path is at fault. Why Other Options Are Wrong: No single component is guaranteed to be the only failure mode; any of the listed blocks could hold the flag HIGH even with good data lines. Common Pitfalls: Assuming good data implies correct parity; the parity bit could be corrupted or the checker miswired. Final Answer: Any of the above.

Question 12

7485 magnitude comparator outputs for A = 1001 and B = 1010 For a 7485-style 4-bit magnitude comparator, determine the outputs A<B, A=B, and A>B when inputs are A = 1001 (decimal 9) and B = 1010 (decimal 10).

Accepted Answer

Introduction / Context: Magnitude comparators output three mutually exclusive flags indicating less-than, equal, or greater-than relationships between binary words. Practicing with specific inputs develops fluency useful for timing and control logic. Given Data / Assumptions: A = 1001 (binary) = 9 (decimal). B = 1010 (binary) = 10 (decimal). No cascading inputs are asserted; we consider only local comparison. Concept / Approach: Compare bit by bit from MSB to LSB. The first position where A and B differ determines the outcome. Alternatively convert to decimal and compare numerically. Step-by-Step Solution: A = 1001, B = 1010 → MSB bits are 1 for both; continue.Next bits: 0 (A) vs 0 (B) → equal; continue.Next: 0 (A) vs 1 (B) → A < B at this position; we can conclude A < B overall.Therefore outputs: AB = 0. Verification / Alternative check: Decimal comparison: 9 < 10, so A Why Other Options Are Wrong: Equal or A>B asserted: These contradict the bitwise or decimal comparison. Multiple flags true: In a proper comparator only one of AB can be 1 at a time. Common Pitfalls: Comparing from LSB upward; significant bits must be examined first because they carry higher weight. Final Answer: A < B is 1, A = B is 0, A > B is 0.

Question 13

Digital logic decoders — what do the address and enable inputs actually do? In combinational logic, a decoder activates exactly one output line based on its binary address inputs, often gated by enable pins. What is the primary purpose of a decoder'…

Accepted Answer

Introduction / Context: Decoders are fundamental combinational logic blocks that convert an n-bit binary code at their inputs into one-of-2^n active outputs. They are used for address decoding in memory systems, chip selects, multiplexed display driving, and many other digital applications. Understanding the role of the address inputs versus any enable pins is essential for correct design and troubleshooting. Given Data / Assumptions: A standard n-to-2^n decoder has n address inputs (often labeled A, B, C, …). Many families include one or more enable inputs (active-HIGH or active-LOW) that globally gate the outputs. Only one output should be active at a time when the device is enabled. Concept / Approach: The decoder's address inputs select which single output line becomes active. If the device has enable inputs, they merely permit or inhibit that normal decoding action. When disabled, all outputs take the inactive state regardless of the address inputs. Thus, the primary purpose of the decoder's inputs (the address lines) is to cause the correct output gate to assert according to the input code. Step-by-Step Solution: Identify the role of address inputs: select exactly one output line.Identify the role of enable pins: gate the entire device on or off.Match the options: the statement that focuses on selecting the correct output gate via inputs is most accurate.Therefore choose option (a). Verification / Alternative check: Look at common parts such as 74HC138 (3-to-8) or 74LS154 (4-to-16). Their data sheets show truth tables where the outputs follow the address inputs only when enable conditions are satisfied; otherwise, all outputs are inactive. This confirms the interpretation above. Why Other Options Are Wrong: (b) Enable pins can disable outputs, but that is not the primary purpose of the address inputs. (c) No standard decoder activates all outputs simultaneously; that would defeat decoding. (d) Mixes two different roles and is misleading about “activate all outputs.” Common Pitfalls: Confusing address inputs with enable inputs; they serve different functions. Forgetting the active-LOW polarity of some enables, leading to miswiring. Final Answer: To allow the decoder to respond to the inputs to activate the correct output gate.

Question 14

Validity of BCD codes — evaluate 10110110 as a binary-coded decimal value Binary-coded decimal (BCD) encodes each decimal digit separately using four bits (0000 to 1001 only). Which is the correct interpretation of the 8-bit BCD sequence 10110110?

Accepted Answer

Introduction / Context: Binary-coded decimal (BCD) is a weighted code where each decimal digit 0–9 is represented by a 4-bit nibble. Because only 0000 through 1001 are valid for a single digit, any nibble from 1010 to 1111 indicates an invalid BCD encoding. This question checks your ability to validate BCD patterns and avoid accidental interpretation as straight binary. Given Data / Assumptions: Given 8-bit pattern: 1011 0110. BCD validity requires each nibble to be in the range 0000–1001. Do not treat the entire 8 bits as a single binary number; treat as two BCD digits. Concept / Approach: Split the 8-bit code into high nibble and low nibble. Validate each nibble individually against the allowed BCD range. If either nibble falls outside 0000–1001, the overall 2-digit BCD value is invalid, regardless of what the straight-binary value would be. Step-by-Step Solution: Write the code as nibbles: 1011 0110.High nibble 1011 equals decimal 11, which is outside 0–9 and therefore invalid for BCD.Low nibble 0110 equals decimal 6, which is valid, but the presence of an invalid high nibble makes the entire BCD code invalid.Conclusion: the pattern is not a valid BCD number. Verification / Alternative check: Cross-check against a BCD table: only 0000–1001 represent digits 0–9. Codes 1010–1111 are forbidden in pure BCD. The leading nibble 1011 falls in the forbidden range, so the code is invalid. Why Other Options Are Wrong: (a) 182: This would require a three-digit BCD; also, the high nibble cannot be 10 or 11. (b) 36: Would require nibbles 0011 0110; the given high nibble is 1011. (c) 116: Would require three BCD digits or an entirely different encoding. Common Pitfalls: Interpreting the whole 8-bit pattern as straight binary. In straight binary, 10110110 is 182, but that is not how BCD works. Forgetting to validate each nibble independently. Final Answer: 10110110 is not a valid BCD number.

Question 15

VHDL design fundamentals — key attribute of the conditional signal assignment In VHDL, a conditional signal assignment (using when ... else) is a concurrent statement that tests conditions in order. Which property is the most important attribute of …

Accepted Answer

Introduction / Context: VHDL offers multiple ways to describe combinational logic: processes with if/elsif, selected or conditional signal assignments, and with-select constructs. The conditional signal assignment (CSA) is a concise concurrent form that models priority logic. Understanding how its conditions are evaluated prevents unintended latches or priority bugs. Given Data / Assumptions: CSA syntax: target <= expr1 when cond1 else expr2 when cond2 else expr_default; It is a concurrent statement, but its conditions are checked in a defined order. The first true condition determines the assigned expression. Concept / Approach: Although the CSA is concurrent (evaluated whenever any RHS signal changes), the list of conditions is prioritized: evaluation proceeds top-to-bottom until one condition is true. This gives the statement a sequential or priority flavor, similar to an if/elsif chain inside a process. Recognizing this ordering is crucial when multiple conditions might be true simultaneously. Step-by-Step Solution: Identify the nature of the construct: concurrent but ordered.Note that only the first true condition drives the output; later conditions are ignored.Map to hardware: synthesizers infer a priority multiplexer network.Therefore, the standout attribute is sequential (prioritized) evaluation of the listed conditions. Verification / Alternative check: Compare with the with-select form, which implies mutually exclusive choices without priority (one-hot style). If overlapping choices exist, a conditional statement enforces priority; with-select typically requires unique choices to avoid conflicts. Why Other Options Are Wrong: (a) Tristate outputs: CSA does not inherently imply tristate behavior. (c) Library components: Unrelated to the construct’s semantics. (d) Fast activation times: Speed depends on resulting logic, not on the statement form. Common Pitfalls: Writing overlapping conditions without intending priority, which can cause unexpected synthesis results. Omitting a final else, inadvertently inferring storage elements. Final Answer: Its sequential evaluation

Question 16

Decoder notation in IEEE/ANSI symbols A logic symbol shows the internal designation “bcd/dec”. In digital electronics, what does this indicate about the decoder’s function and its input–output size?

Accepted Answer

Introduction / Context: Decoder symbols in the IEEE/ANSI standard often include short internal designations that summarize the input code and the output code. Understanding these tags is essential when reading schematics quickly and selecting the correct integrated circuit for tasks such as driving numeric displays or selecting one of many lines. Given Data / Assumptions: The internal tag on the logic symbol is “bcd/dec”. We are dealing with common combinational decoders used in digital systems. Binary Coded Decimal (BCD) uses four input bits to represent decimal digits 0–9. Concept / Approach: Notation of the form “input/output” indicates the coding at the input and the coding at the output. Thus “bcd/dec” literally means the device accepts BCD and outputs decimal. A BCD-to-decimal decoder takes a 4-bit BCD input (A3..A0) and asserts exactly one of ten outputs (Y0..Y9) corresponding to the decimal digit, which is ideal for driving a 10-line interface such as a Nixie/indicator or for one-of-ten selection. Step-by-Step Solution: Interpret the tag: “bcd/dec” → BCD in, decimal out.Determine sizes: BCD requires 4 input lines (to represent 0–9).Decimal out implies 10 distinct outputs (one per digit).Therefore: 4 inputs and 10 outputs, BCD-to-decimal. Verification / Alternative check: Classic parts such as 7442/7445 are BCD-to-decimal decoders. Their datasheets show 4 inputs and 10 mutually exclusive outputs, matching the interpretation of “bcd/dec”. Why Other Options Are Wrong: Decimal-to-BCD (options a and c) reverses the direction; that would be an encoder, not a decoder. “Ten inputs and four outputs” (options a and b) is the wrong sizing; decoders expand lines, not compress them. Common Pitfalls: Confusing decoders with encoders; decoders generally expand N inputs to up to 2^N (or 10) outputs. Overlooking that BCD uses only ten of the sixteen 4-bit combinations; the remaining codes are invalid. Final Answer: BCD-to-decimal decoder with four inputs and ten outputs.

Question 17

Modeling a priority encoder in VHDL Which VHDL description style is appropriate for a priority encoder, so that higher-priority inputs dominate the output encoding?

Accepted Answer

Introduction / Context: A priority encoder outputs a binary code for the highest-priority asserted input. In hardware description languages such as VHDL, capturing the “priority” behavior requires a coding style that enforces ordered checks and drives outputs deterministically. Choosing the right technique ensures synthesizable, readable logic that matches the hardware intent. Given Data / Assumptions: The device is a priority encoder (e.g., 8-to-3 or 16-to-4). VHDL will be used for synthesis into standard logic. Outputs of a module are typically declared as signals (ports) in VHDL. Concept / Approach: A common and synthesizable approach is to use sequential statements inside a clockless process and assign to signal outputs with ordered if–elsif–else clauses that check the highest-priority input first. This realizes “priority coding.” Although variables can be used locally within a process, the entity/module outputs are signals; assigning to signals with explicit priority ensures correct hardware behavior after synthesis. Step-by-Step Solution: Declare outputs as signals (entity ports).Write a combinational process with a full sensitivity list.Inside the process, use ordered if–elsif checks from highest to lowest priority input.Assign the encoded value to signal outputs; provide a default (e.g., “others” or a zero code). Verification / Alternative check: Simulation will show that when multiple inputs are HIGH, the code corresponds to the highest-priority line. Synthesis reports map the style into multiplexers and gates implementing priority logic, confirming correctness. Why Other Options Are Wrong: Integer outputs: VHDL outputs are signals; integers may be used internally, but not as direct port types in most synthesized styles. Tristate outputs: Unrelated to priority encoding; tristate pertains to bus driving. Variables and priority coding: Variables can be used internally, but the essential concept is that outputs are signals. The option is misleading. Common Pitfalls: Forgetting a default assignment, leading to inferred latches. Using unordered concurrent equations that do not enforce priority. Final Answer: Signal outputs and priority coding

Question 18

AHDL BCD→binary conversion detail In an AHDL design of a BCD-to-binary converter, how is multiplication by 10 typically implemented efficiently at the logic level?

Accepted Answer

Introduction / Context: BCD-to-binary converters often need to multiply intermediate values by 10 when building positional weights (tens, hundreds). In hardware description languages like AHDL, designers prefer structures that map to fast, small hardware rather than generic arithmetic that may infer larger multipliers. Given Data / Assumptions: Target operation is multiply by 10. Implementation is in AHDL for synthesis to logic. Hardware efficiency is desired (few LUTs/gates, short delay). Concept / Approach: Multiplying by 10 can be decomposed into shifts and adds: 10 * X = (8 * X) + (2 * X) = (X << 3) + (X << 1). Shift operations are wiring (bit-reindexing) with essentially zero logic cost, and addition uses small carry chains. This approach avoids inferring a general-purpose multiplier, which would be overkill for constant-factor multiplies used in BCD conversion pipelines. Step-by-Step Solution: Express 10 * X as (8 * X) + (2 * X).Implement 8 * X via a left shift by 3 bits (X<<3).Implement 2 * X via a left shift by 1 bit (X<<1).Sum the shifted vectors with an adder → compact hardware. Verification / Alternative check: Resource and timing reports from synthesis show fewer logic elements compared to a generic “* 10” operator (if that operator infers a multiplier). Simulation waveforms confirm functional equivalence for all input cases. Why Other Options Are Wrong: Library multiplication function or explicit “*” operator: works functionally but may infer larger multiplier blocks, wasting area. Integer types: a data type choice, not a hardware technique; does not perform the multiplication itself. Common Pitfalls: Forgetting to handle BCD digit carries properly when aggregating tens/hundreds. Neglecting bit growth after shifts; ensure adequate vector width. Final Answer: By using the shifting of bits

Question 19

Diagnosing a blank MSD in a display multiplexer In a system using 74157 multiplexers, a 7449 decoder/driver, and a 74139 decoder, the most significant digit (MSD) display is blank while the least significant digit (LSD) works. Inputs and outputs on …

Accepted Answer

Introduction / Context: Troubleshooting multiplexed display systems requires understanding enable logic and active-LOW outputs. The 74139 is a dual 2-to-4 decoder with active-LOW outputs and active-LOW enables. If its outputs do not switch as the select inputs change, the selected display may blank or remain stuck. Given Data / Assumptions: 74139: active-LOW enable (EN = 0 enables) and active-LOW outputs (Yn = 0 when selected). Measured: A input toggles; B = 0 and EN = 0 (so the channel is enabled, selecting between Y0/Y1 as A toggles). Observed: Y0 stuck LOW, Y1 stuck HIGH regardless of A. Concept / Approach: With EN = 0 and B = 0, Y0 and Y1 should alternate: when A = 0, Y0 = 0 and Y1 = 1; when A = 1, Y1 = 0 and Y0 = 1. If Y0 is always 0 and Y1 always 1, either A is not truly reaching the chip (bad connection) or the internal logic is defective. Because the scope already shows A toggling at the pin and EN is reported LOW, a chip fault is likely; however, intermittent or lifted EN connections can mimic the symptom too. Step-by-Step Solution: Confirm at the 74139 pins: EN = 0, B = 0, A toggling.Expect Y0/Y1 to alternate LOW/HIGH.Since they do not, suspect a defective 74139 or an enable pin connection fault.Action: Reseat/verify EN pin continuity; if good, replace 74139. Verification / Alternative check: Temporarily swap the two sections of the 74139 (if practical) or substitute a known-good 74139; if the fault follows the chip, it confirms device failure. Continuity testing on EN validates the alternate hypothesis. Why Other Options Are Wrong: Definite “replace only” (a) ignores possible EN connection fault. “EN bad connection only” (b) ignores the strong chance of device failure. “Display bad” (d) is unlikely because the select logic itself is not switching. Common Pitfalls: Forgetting that 74139 outputs are active-LOW; stuck LOW looks like a constant selection. Probing upstream nodes only and missing a broken enable line. Final Answer: The 74139 is defective and must be replaced; or the EN pin on the 74139 could have a bad connection, which should be checked.

Question 20

The device that is an application of SOP logic is a multiplexer.

Accepted Answer

True

MSI Logic Circuits Questions