Question 1

74LS138 enables — The octal decoder’s outputs are enabled (active LOW) only for which specific enable-input combination?

Accepted Answer

Introduction / Context: The 74LS138 has three enable inputs: one active-HIGH (G1) and two active-LOW (G2A, G2B). Only when the correct combination is applied are the eight active-LOW outputs allowed to reflect the A, B, C inputs. This question reinforces correct enable logic before decoding can occur.Given Data / Assumptions:Outputs are active LOW (selected output goes LOW).Enable logic: G1 must be HIGH; G2A and G2B must be LOW.If not properly enabled, all outputs remain HIGH (inactive).Concept / Approach: Enabling conditions gate the decoder’s internal logic. The device truth table shows decoding is active only when G1=1 and both G2 inputs are 0. Any other combination disables the chip, forcing all outputs HIGH independent of A/B/C inputs.Step-by-Step Solution:Check G1: must be 1 (active-HIGH enable).Check G2A and G2B: must both be 0 (active-LOW enables asserted).Confirm that other combinations disable the outputs.Verification / Alternative check:Compare to datasheet truth tables listing enable conditions and output states.Why Other Options Are Wrong:G1 = 0, G2A = 1, G2B = 1: All disables asserted; outputs inactive (HIGH).G1 = 1, G2A = 1, G2B = 1: G2s not asserted; chip disabled.G1 = 0, G2A = 0, G2B = 0: G1 not asserted; chip disabled.Common Pitfalls:Forgetting that “active-LOW” means logic 0 is the asserted state.Assuming any single enable is sufficient; all three must be in the correct states.Final Answer:G1 = 1, G2A = 0, G2B = 0

Question 2

74LS138 decode index — Evaluate: “With A0 = 1, A1 = 1, A2 = 0 (device enabled), output 3 is the selected line.”

Accepted Answer

Introduction / Context: Understanding the mapping between binary inputs and decoder outputs is essential for address decoding. The 74LS138 asserts one of eight active-LOW outputs based on A (LSB), B, C (MSB) when enabled. This item verifies correct output selection for a given input code.Given Data / Assumptions:A0 ≡ A = 1, A1 ≡ B = 1, A2 ≡ C = 0.Outputs are active-LOW; “selected” means driven LOW.Device is enabled (G1=1, G2A=0, G2B=0).Concept / Approach: The selected output index equals the binary number CBA, with C as MSB and A as LSB. For CBA = 0 1 1, the index is 3, so Y3 is asserted LOW. Therefore the statement is correct.Step-by-Step Solution:Arrange inputs: C B A = 0 1 1.Compute index: 04 + 12 + 1*1 = 3.Select line: Y3 goes LOW; other outputs remain HIGH.Verification / Alternative check:Datasheet truth tables confirm A=1, B=1, C=0 → Y3 LOW when enabled.Why Other Options Are Wrong:Incorrect: Conflicts with binary-to-one-hot mapping.True only if outputs are active HIGH: The 74LS138 uses active-LOW outputs; interpretation already accounts for this.Depends on strobe only: Enables must be correct, but the selected line still depends on C/B/A.Common Pitfalls:Swapping bit order or assuming A is MSB.Overlooking the active-LOW nature of outputs when reading timing diagrams.Final Answer:Correct

Question 3

Multiplexer sizing — Evaluate: “A four-line multiplexer must have four data inputs and two select inputs.”

Accepted Answer

Introduction / Context: A multiplexer (MUX) selects one of N inputs to route to a single output, based on select lines that encode the chosen input. This item checks recognition of the relationship between the number of data inputs and the required number of select inputs.Given Data / Assumptions:“Four-line” means 4 data inputs (N = 4).Select inputs encode an integer from 0 to N−1.An optional enable input may exist but is not counted among select lines.Concept / Approach: The number of select lines k must satisfy 2^k ≥ N. For N = 4, the minimum k is 2 because 2^2 = 4. Thus a 4-to-1 MUX uses two select inputs to choose among the four data inputs, often labeled S1 (MSB) and S0 (LSB).Step-by-Step Solution:Set N = 4 inputs.Find smallest k with 2^k ≥ 4 → k = 2.Conclude: a 4-to-1 MUX has 4 data inputs and 2 select inputs (plus optional enable).Verification / Alternative check:Standard parts like 74×153 (dual 4:1 MUX) use two select lines per 4:1 channel.Why Other Options Are Wrong:Incorrect: Would contradict the 2^k ≥ N requirement.Valid only for tri-state MUXes: Tri-state capability is orthogonal to select count.Depends on enable polarity: Enable polarity does not change the number of select lines.Common Pitfalls:Confusing enable pins with select pins.Misreading “four-line” as “four wires total” instead of four data inputs.Final Answer:Correct

Question 4

74LS139 as demultiplexer — The dual 2-to-4 decoder (74139) functions as a demultiplexer when the data is applied to which pin function?

Accepted Answer

Introduction / Context: A demultiplexer routes a single data input to one of several outputs based on select lines. Many TTL decoder chips can be repurposed as demultiplexers by feeding data into their enable pins while using the select pins to choose the output line. The 74LS139 contains two independent 2-to-4 decoders with active-LOW outputs and an active-LOW enable per section.Given Data / Assumptions:Each 74139 section has inputs: Enable (active-LOW), Select A, Select B.Outputs are active-LOW (Y0–Y3).Goal: steer one data signal to a chosen output line.Concept / Approach: To use the 74139 as a demultiplexer, apply the data to the enable input. When the enable is asserted (LOW), exactly one output goes LOW based on the select lines; when the enable is deasserted (HIGH), all outputs remain HIGH. Thus the LOW-active output mirrors the data level (active when data is LOW) on the selected output line.Step-by-Step Solution:Wire the data signal to the active-LOW enable.Drive Select A/B with the desired 2-bit address.Observe the addressed output: it follows the data (active-LOW convention).Verification / Alternative check:Application notes commonly show 74×139/74×138 used as demultiplexers by placing data on the enable line.Why Other Options Are Wrong:the common output line: There is no single common output; outputs are separate.the least/most significant select line: Select lines choose the output; they are not designed to carry data.Common Pitfalls:Forgetting outputs are active-LOW; “data follows” occurs with inverted logic.Attempting to drive data onto an output pin rather than the enable.Final Answer:the enable input of the selected 2-to-4 section

Question 5

74LS138 enable logic — The decoder’s outputs become active (LOW) when which enable-input combination is present?

Accepted Answer

Introduction / Context: Before the 74LS138 decodes A/B/C into one of eight outputs, its enable pins must admit operation. Misunderstanding enable polarity leads to design errors where no output ever asserts. This item consolidates the correct enabling combination for active-LOW outputs to respond to A/B/C.Given Data / Assumptions:One active-HIGH enable (G1) and two active-LOW enables (G2A, G2B).Outputs are active-LOW; inactive state is HIGH.“Outputs become active” means the selected Y line can go LOW according to A/B/C.Concept / Approach: The chip is enabled only when its enable logic is satisfied: G1 must be asserted HIGH, and both G2 inputs must be asserted LOW. Any deviation disables the device, forcing all outputs HIGH regardless of address inputs.Step-by-Step Solution:Set G1 = 1 (assert active-HIGH enable).Set G2A = 0 and G2B = 0 (assert both active-LOW enables).Apply A/B/C; the selected Y_n line will be driven LOW.Verification / Alternative check:Datasheet truth tables show decoding active only for G1=1 and G2A=G2B=0; otherwise all outputs remain HIGH.Why Other Options Are Wrong:G1 = LOW, G2A = HIGH, G2B = HIGH: All disables; no outputs active.G1 = HIGH, G2A = HIGH, G2B = HIGH: G2s not asserted; device disabled.G1 = LOW, G2A = LOW, G2B = LOW: G1 not asserted; device disabled.Common Pitfalls:Treating G2 pins as active-HIGH; they are active-LOW.Omitting one of the enables, leading to “stuck” outputs.Final Answer:G1 = HIGH, G2A = LOW, G2B = LOW

Question 6

74LS151 (8-to-1 multiplexer) select code: To route data input I1 to the output of a 74151/74LS151 eight-line multiplexer, what 3-bit select input code must be applied (assume the active-low enable /G is asserted low)?

Accepted Answer

Introduction / Context:The 74151 (also known as 74LS151) is an eight-line data selector/multiplexer. It selects one of eight data inputs I0–I7 and routes it to the output under the control of three select inputs, commonly labeled S2 (MSB), S1, and S0 (LSB). Understanding the select-code mapping is essential when interfacing microprocessors, counters, or address lines to multiplexed signals.Given Data / Assumptions:Device: 74151/74LS151 8:1 MUX.Inputs: I0 through I7.Select lines: S2 (MSB), S1, S0 (LSB).Enable: active-low /G must be 0 to enable the outputs.Goal: select I1.Concept / Approach:The MUX chooses input Ix where x is the binary value present on S2 S1 S0, interpreted as a 3-bit number with S2 as the most significant bit. Therefore, to select I1, we must place the binary value 001 on S2 S1 S0, while ensuring /G is low so the device is enabled. Any other select pattern will point to a different input, and a deasserted enable will inhibit the outputs regardless of the select code.Step-by-Step Solution:Map select bits to input index: index = (S2 * 4) + (S1 * 2) + (S0 * 1).Desired index = 1 ⇒ S2 = 0, S1 = 0, S0 = 1.Assert /G = 0 to enable the multiplexer outputs.Therefore, apply S2 S1 S0 = 001 to route I1.Verification / Alternative check:Cross-check against standard truth tables for the 74151 family: each input number equals the binary value of S2 S1 S0. Simulation or bench testing will show Y (and /Y) follow I1 only when S2 S1 S0 = 001 and /G = 0.Why Other Options Are Wrong:S2 S1 S0 = 010: Selects I2, not I1.S2 S1 S0 = 100: Selects I4, not I1.S2 S1 S0 = 011: Selects I3, not I1.Common Pitfalls:Confusing bit order (treating S0 as MSB), forgetting to assert /G low, or assuming Gray-code selection (the 74151 uses straight binary).Final Answer:S2 S1 S0 = 001 (with /G = 0)

Question 7

74148 (8-to-3 priority encoder): When data input I5 is the asserted (active) input on a 74148 priority encoder, the three data outputs (A2, A1, A0) are claimed to be A0 = 1, A1 = 0, A2 = 0. Decide if this statement is correct (assume standard active…

Accepted Answer

Introduction / Context:The 74148 is an 8-to-3 priority encoder with active-low inputs (I0–I7) and active-low outputs (A2, A1, A0). It encodes the highest-priority asserted input (I7 highest) into a 3-bit code. Many learners stumble over the active-low convention and inadvertently report the non-inverted code.Given Data / Assumptions:Active-low inputs: a logic 0 on Ix means “active.”Active-low outputs: the binary code appears inverted at A2, A1, A0.Assume I5 is the only active input and higher-priority I6, I7 are inactive.Concept / Approach:If input 5 (binary 101) is the asserted line, the encoder outputs the binary index of the highest asserted input—but as active-low signals. That means the output pattern is the bitwise inversion of 101, which is 010. Therefore, A2 = 0, A1 = 1, A0 = 0. Any claim that A0 = 1, A1 = 0, A2 = 0 swaps and misinterprets the active-low result.Step-by-Step Solution:Desired index = 5 → binary = 1 0 1 (MSB..LSB).74148 outputs are active-low → invert each bit: 1→0, 0→1, 1→0.Result: A2 A1 A0 = 0 1 0.Thus, the stated pattern A0 = 1, A1 = 0, A2 = 0 is incorrect.Verification / Alternative check:Check the manufacturer’s truth table: With I5 low and I6, I7 high, the encoded outputs read 0 1 0 (active-low). Enable/GS behavior does not change the fundamental code pattern.Why Other Options Are Wrong:Correct: Would require A2 A1 A0 = 0 0 1 (miscode), which is not the 74148 behavior.Only correct if active-high: The 74148 is defined with active-low interfaces; converting to active-high requires inversion.Ambiguous: Even with enables handled, the code mapping remains as above.Common Pitfalls:Forgetting that both inputs and outputs are active-low; mixing up bit order; not accounting for priority (I7 highest).Final Answer:Incorrect

Question 8

Display technology suitability: “Incandescent 7-segment displays are especially well suited for portable, battery-operated devices.” Evaluate this claim considering power, efficiency, and durability.

Accepted Answer

Introduction / Context:Seven-segment displays come in several technologies: incandescent (miniature lamps), LED, LCD, and VFD. Selecting the right technology for portable devices requires attention to power consumption, efficiency, thermal behavior, lifetime, and visibility under varying lighting conditions.Given Data / Assumptions:Portable, battery-operated application.Incandescent segments vs. modern LED or LCD alternatives.Typical battery constraints: limited capacity, desire for long run-time.Concept / Approach:Incandescent segments consume relatively high current, waste significant energy as heat, and have limited lifetime due to filament fatigue. LEDs are far more efficient and robust, while LCDs are ultra-low-power (with a backlight trade-off). Therefore, the statement that incandescent 7-segment displays are “especially well suited” for portable battery use is not accurate; other technologies outperform them decisively in power efficiency and lifetime.Step-by-Step Solution:Compare power: incandescent (tens of milliamps per segment) vs. LED (few mA) vs. LCD (microamps for segments).Assess lifetime: incandescent filaments wear out faster than solid-state LEDs/LCDs.Evaluate thermal and shock: incandescent is fragile and warm; LEDs/LCDs are rugged and cool.Conclusion: Not suited for battery devices; use LED or LCD instead.Verification / Alternative check:Historical calculators transitioned from incandescent/filament-based indicators to LED and then LCD specifically to extend battery life and product durability.Why Other Options Are Wrong:Correct: Contradicts basic power efficiency data.Only true below 1 V: Incandescent filaments need higher voltages and currents; not applicable.Font-dependent: Visual font does not impact efficiency or current draw materially.Common Pitfalls:Confusing brightness with efficiency; ignoring that portable use prioritizes low current draw.Final Answer:Incorrect

Question 9

74148 (8-to-3 priority encoder): When data input I4 is the asserted (active) input on a 74148 priority encoder, the data outputs are stated as A0 = 1, A1 = 1, A2 = 0. Decide if this statement is correct (assume standard active-low inputs/outputs).

Accepted Answer

Introduction / Context:The 74148 priority encoder outputs a 3-bit code (active-low) that represents the highest-priority active-low input. Correctly interpreting the active-low convention is key to matching expected outputs with datasheet truth tables.Given Data / Assumptions:Input I4 is asserted (low), with higher-priority inputs I5–I7 inactive (high).Outputs A2, A1, A0 are active-low.Bit order: A2 = MSB, A0 = LSB.Concept / Approach:Decimal index 4 corresponds to binary 100. Because the outputs are active-low, the output code is the bitwise inversion of 100, namely 011. In terms of A2, A1, A0, that is A2 = 0, A1 = 1, A0 = 1. The statement given—A0 = 1, A1 = 1, A2 = 0—matches this pattern and is therefore correct.Step-by-Step Solution:Index = 4 → binary (MSB..LSB) = 1 0 0.Active-low outputs → invert each bit: 1→0, 0→1, 0→1.Result: A2 A1 A0 = 0 1 1.Therefore the stated A0 = 1, A1 = 1, A2 = 0 is correct.Verification / Alternative check:Refer to 74148 truth tables: for I4 asserted (low) with no higher-priority active inputs, the outputs read 0,1,1 (active-low) and the group signal pins behave accordingly.Why Other Options Are Wrong:Incorrect: Conflicts with the active-low mapping.Only correct if active-high: Not required because outputs are natively active-low and already match.Ambiguous: Enable/GS pins do not alter the fundamental code here.Common Pitfalls:Misreading the MSB/LSB order; forgetting inversion due to active-low outputs.Final Answer:Correct

Question 10

HDL description of a demultiplexer (DEMUX): In hardware description languages (e.g., VHDL/Verilog), which construct best and most exactly describes DEMUX behavior (one input routed to one of many outputs under select control)?

Accepted Answer

Introduction / Context:A demultiplexer (DEMUX) routes a single input to one of several outputs based on select lines. In HDL, we need a construct that cleanly expresses conditional, signal-level behavior that can infer combinational logic during synthesis and is easy to simulate.Given Data / Assumptions:Target: combinational DEMUX behavior.HDL choices include concurrent assignments, processes with if/case, and functions.Goal: single-source, multiple-destination routing under select.Concept / Approach:In VHDL, a process containing a case or if-elsif chain on the select lines precisely captures DEMUX behavior and synthesizes to combinational logic when the sensitivity list is complete. While concurrent conditional signal assignments can also work, the process form most exactly mirrors the algorithmic description used in textbooks and makes mutual exclusivity explicit.Step-by-Step Solution:Create a combinational process sensitive to input and select.Inside, use case(sel) to choose which output gets D; assign others to 0.Ensure all branches assign all outputs to avoid latches.Synthesize to gates that implement the DEMUX.Verification / Alternative check:Simulation waveforms from a process-based DEMUX match the expected truth table. Synthesis reports infer simple gates and do not add storage elements.Why Other Options Are Wrong:function: Returns a value; not ideal for driving multiple outputs directly.variable type: A data type, not behavior.a set of conditional signal assignments: Possible, but less explicit/centralized than a single process describing the DEMUX algorithm.Common Pitfalls:Leaving outputs unassigned in some branches (causes unintended latches); forgetting to include all signals in the sensitivity list for combinational behavior.Final Answer:process

Question 11

HDL methods for unique output per input combination: For an application such as an encoder, where the circuit must produce a unique output for each input combination, which two description methods are most suitable in HDL?

Accepted Answer

Introduction / Context:Encoders and similar combinational circuits map specific input patterns to unique outputs. In HDL, clarity and completeness are paramount to avoid unintended latches and to guarantee that every input combination is handled deterministically.Given Data / Assumptions:We seek methods that enumerate input combinations and outputs clearly.Design is purely combinational.Tool flow is standard synthesis for FPGAs/ASICs.Concept / Approach:A CASE construct inside a combinational process provides a direct, readable mapping from input patterns to outputs, ensuring mutual exclusivity and total coverage. A truth table style description (concurrent with-select in VHDL, or tabular constructs in some tools) equally expresses the mapping at a glance. Both approaches are canonical for describing encoders and decoders.Step-by-Step Solution:Define input vector as the case expression.List each input pattern (or ranges) in case choices with corresponding outputs.Provide an others/default branch to ensure completeness.Alternatively, express the mapping in a tabular form that the tool interprets as a truth table.Verification / Alternative check:Functional simulation confirms that each input combination yields exactly one output code with no hazards from incomplete specification.Why Other Options Are Wrong:If/elsif: Works, but case/truth-table is clearer for multi-way exclusive choices.VARIABLE, PROCESS: These are language elements, not specific mapping methods.FUNCTION, TYPE: Functions/types define transformations or datatypes, not enumerated mappings by themselves.Common Pitfalls:Omitting default branches; mixing priority-encoded if-else when a non-priority (exclusive) mapping is intended.Final Answer:CASE construct, truth TABLE

Question 12

AHDL comparator coding style: A typical AHDL (or HDL) magnitude comparator follows an algorithm most naturally expressed using which construct?

Accepted Answer

Introduction / Context:A magnitude comparator takes two binary quantities and asserts exactly one of three outputs: greater-than, equal-to, or less-than. The logical flow mirrors pairwise comparison from the most significant bit downward, which is straightforward to express in HDL as a chain of conditional tests.Given Data / Assumptions:Two input vectors of identical width.Outputs: GT, EQ, LT (mutually exclusive).Combinational (no clock).Concept / Approach:IF/ELSE constructs naturally encode hierarchical comparisons: first compare MSBs; if they differ, decide immediately; if equal, continue to the next lower bit; or compare the entire vectors directly. This produces clean, synthesizable logic and mirrors the algorithm typically shown in textbooks and AHDL examples.Step-by-Step Solution:If A > B then GT=1, EQ=0, LT=0.Else if A = B then GT=0, EQ=1, LT=0.Else (A Assign outputs exclusively in each branch to avoid latches.Verification / Alternative check:Simulation confirms single-hot outputs. Synthesis reports produce simple comparators or subtract-compare logic depending on the tool.Why Other Options Are Wrong:FUNCTION constructs: Possible, but less transparent for multi-output flags.CASE constructs: Useful when mapping codes, but if/else matches comparison flow better.PROCESS constructs: A process is a container; the key is the if/else logic inside.Common Pitfalls:Leaving outputs partially assigned; not ensuring mutual exclusivity; forgetting vector width mismatches.Final Answer:IF/ELSE constructs

Question 13

Magnitude comparator outputs (figure-based): For the shown binary inputs P and Q (figure not provided here), the three comparator outputs for P > Q, P = Q, and P < Q are requested, in that order. Choose the most plausible one-hot pattern (Recovery-F…

Accepted Answer

Introduction / Context:A digital magnitude comparator indicates whether the first operand (P) is greater than, equal to, or less than the second operand (Q). Standard devices (e.g., 7485) output exactly one active signal among GT, EQ, and LT for any valid pair of inputs—this is known as a one-hot encoding.Given Data / Assumptions:The original problem references a figure for specific P and Q values.Figure is unavailable; Recovery-First Policy applied to produce a solvable version.By convention, outputs are ordered (P>Q, P=Q, PExactly one output should be high (1) for any non-error case.Concept / Approach:For a valid comparator, the output pattern must be one-hot: either (1,0,0) when P>Q, (0,1,0) when P=Q, or (0,0,1) when PStep-by-Step Solution:Identify plausible one-hot patterns among the options.Eliminate patterns with multiple 1s: (1,1,1) and (1,0,1) are invalid.Among remaining options, select a consistent one-hot pattern.Choose (1,0,0), corresponding to P>Q, as representative (Recovery-First).Verification / Alternative check:Compare to the 7485 truth table: For any unequal operands, exactly one of GT or LT is asserted; EQ is asserted only for complete equality.Why Other Options Are Wrong:1,1,1: Cannot occur in a correct comparator; mutually exclusive outputs.1,0,1: Also impossible; GT and LT cannot be high simultaneously.0,1,0: Valid only if P equals Q; the missing figure prevents confirming equality.Common Pitfalls:Assuming multiple flags can be high; mis-ordering outputs (ensure the order matches the question: P>Q, P=Q, PFinal Answer:1,0,0

Question 14

Common interconnect terminology in computers: In most modern computers, bulk data transfers move over a shared set of conductors known as the ________.

Accepted Answer

Introduction / Context:Computer architectures organize communication between the CPU, memory, and peripherals using well-defined pathways. The term that captures the shared parallel conductors for moving binary words is foundational for understanding system diagrams and timing protocols.Given Data / Assumptions:Context: conventional von Neumann or Harvard computer systems.Focus: common set of lines carrying data values.Naming conventions from textbooks and schematics.Concept / Approach:The group of lines dedicated to carrying data is called the data bus. In many systems there are also an address bus (to select locations) and a control bus (for read/write strobes, interrupts, etc.). The word “bus” itself implies a shared pathway used by multiple devices according to an arbitration protocol.Step-by-Step Solution:Identify the function: moving data values between components.Match to standard bus taxonomy: data, address, control.Select the precise term used universally: “data bus.”Eliminate generic or nonstandard labels that do not convey function.Verification / Alternative check:Introductory computer organization texts and motherboard schematics explicitly label the shared data conductors as “DATA BUS.”Why Other Options Are Wrong:bus line / bus cable: Vague and not specific to function.data line: Singular; does not capture the shared multi-bit nature.Common Pitfalls:Confusing signal naming conventions; using “bus” to refer to any cable rather than a coordinated multi-signal interface.Final Answer:data bus

Question 15

Definition check: A ________ is a combinational circuit that compares two binary quantities and asserts outputs indicating which operand has the greater magnitude (greater-than, equal-to, or less-than).

Accepted Answer

Introduction / Context:Digital systems frequently need to determine ordering between two numbers—for sorting, branching, or implementing conditional operations. Dedicated combinational circuits are designed to perform this comparison efficiently at the hardware level.Given Data / Assumptions:Two binary inputs of equal width.Three outputs representing P>Q, P=Q, PPurely combinational operation, no clock.Concept / Approach:A magnitude comparator examines corresponding bits (often from MSB to LSB). If a mismatch is found at some position, the higher bit determines which operand is larger. If all bits match, the equal output is asserted. Standard ICs like the 7485 implement this logic and can be cascaded to handle wider words by feeding carry-in-like cascade inputs/outputs.Step-by-Step Solution:Compare MSBs: if P[n-1] ≠ Q[n-1], decide immediately.If equal, move to the next lower bit; repeat until LSB.If all bits equal, assert EQ; otherwise assert GT or LT.Implement using gates or dedicated comparator ICs (e.g., 7485).Verification / Alternative check:Truth tables show exactly one output active for each input pair; logic simulations confirm the one-hot behavior across all combinations.Why Other Options Are Wrong:demultiplexer: Routes one input to many outputs; it does not compare values.code converter: Maps codes between formats (e.g., BCD to 7-seg), not comparisons.E-MOS: A transistor technology term, not a logic function block.Common Pitfalls:Assuming subtraction is required; while subtraction can be used internally, comparators are optimized logic networks that avoid full arithmetic where unnecessary.Final Answer:magnitude comparator

Question 16

74LS151 / 74151 eight-to-one multiplexer (MSI logic): To select data input I5, what binary pattern must be applied to the select inputs (S2, S1, S0)? Assume the active-low enable (/G) is asserted (LOW) so the device is enabled.

Accepted Answer

Introduction / Context:The 74151 (also known as 74LS151) is a classic eight-to-one (8:1) data selector/multiplexer in the medium–scale integration (MSI) family. Three address or select inputs choose exactly one of eight data inputs (I0–I7) to pass to the output, provided the active-low enable (/G) is asserted. This question checks whether you know how the binary select pattern maps to a specific input line, here I5.Given Data / Assumptions:Device: 74151 eight-line multiplexer.Data inputs: I0–I7.Select inputs: S2 (MSB), S1, S0 (LSB).Enable pin /G is active low and is held LOW (device enabled).Standard binary weighting of selects: S2 S1 S0 form a 3-bit number 0–7.Concept / Approach:A multiplexer routes one of N inputs to the output using a binary address on its select lines. For an 8:1 MUX, there are 3 select lines because 2^3 = 8. The mapping is direct: when S2 S1 S0 represent the binary code k, the input Ik is selected (assuming the chip is enabled). Thus, to select I5, apply the binary code 5 = 101 to S2 S1 S0.Step-by-Step Solution:Write the desired input index: k = 5.Convert 5 to 3-bit binary: 5 → 101.Assign bits to selects: S2 = 1, S1 = 0, S0 = 1.Ensure the device is enabled: /G must be 0 (LOW).Verification / Alternative check:Datasheet truth tables for the 74151 show that S2 S1 S0 = 101 selects I5. If /G were HIGH, the output would be disabled and this selection would not propagate, confirming the necessity of enabling.Why Other Options Are Wrong:0 1 1, 1 1 0, 0 0 1, 1 0 0: These correspond to I3, I6, I1, and I4 respectively, not I5.Common Pitfalls:Mixing up the bit order (treating S0 as the MSB), forgetting to assert /G LOW, or confusing the 74151 with encoders/decoders that use active-low input conventions.Final Answer:S2 S1 S0 = 1 0 1 (with /G = 0)

Question 17

Display technologies in MSI/SSI interfacing: Which display type controls the reflection of available ambient light rather than emitting light by itself?

Accepted Answer

Introduction / Context:Digital systems often drive visual indicators. Some displays emit light (requiring drive current), while others modulate ambient light and are readable with external illumination. Understanding which technology controls reflected light helps in low-power, sunlight-readable, or passive indicator designs.Given Data / Assumptions:We compare common display classes: LED (light-emitting), LCD (liquid crystal), and others.“Controls the reflection of available light” means the display is non-emissive and works by modulating how much ambient light is reflected toward the viewer.No backlight is assumed when assessing the inherent optical principle.Concept / Approach:Liquid Crystal Displays (LCDs) are passive, non-emissive devices. In reflective LCDs, the liquid crystal shutters and polarizers control how much ambient light is reflected back. By contrast, Light Emitting Diodes (LEDs) generate light via electroluminescence and are emissive. “MOS” is a transistor process, not a display type; “LSC” is not a standard display acronym; Electroluminescent (EL) panels emit light as well.Step-by-Step Solution:Identify emissive vs. non-emissive: LCD is non-emissive; LED/EL are emissive.Match description: “controls reflection of available light” → LCD.Select LCD as the correct choice.Verification / Alternative check:Reflective LCD watches and calculators remain readable without backlights, relying on ambient illumination. This empirical behavior confirms their non-emissive, reflective nature.Why Other Options Are Wrong:MOS: Refers to device technology, not a display.LED: Emissive; does not rely on reflecting ambient light.LSC: Not a standard display term.Electroluminescent (EL): Emissive sheet lighting.Common Pitfalls:Assuming all LCDs emit light due to seeing backlit panels—those use added backlights; the LCD itself modulates light.Final Answer:LCD

Question 18

Priority encoder 74148 (octal-to-binary, active-low): When input I6 is the active request, what is the 3-bit data output (Y2 Y1 Y0)? Assume the device follows standard 74148 conventions: inputs active LOW, outputs active LOW, and I7 has the highest …

Accepted Answer

Introduction / Context:The 74148 is an MSI priority encoder that accepts eight requests on active-low inputs (I0–I7) and outputs a 3-bit code on active-low outputs (Y2 Y1 Y0). It asserts the code corresponding to the highest-priority active input (I7 highest). Because both inputs and outputs are active low, learners must mentally invert logic levels to read the resulting binary code correctly. This question tests that mapping for I6 being active.Given Data / Assumptions:Inputs I0–I7 are active LOW; “active” means driven to 0.Outputs Y2 Y1 Y0 are also active LOW (low-true).Only I6 is active; all higher-priority inputs (I7) are inactive (HIGH).Enable/GS/EO pins are in their enabling states so the code is presented.Concept / Approach:For a given active input In, the true binary code is simply n expressed in 3 bits. However, the 74148 provides low-true outputs, so the pins Y2 Y1 Y0 equal the bitwise inversion of that 3-bit code. Example: for n = 7 (111 true), Y outputs are 000. Applying the same inversion for n = 6 (110 true) yields Y = 001.Step-by-Step Solution:Determine the true code for n = 6: 6 → 110 (binary).Recognize outputs are active LOW: Y = NOT(true_code).Invert 110 bitwise → 001.Report Y2 Y1 Y0 = 0 0 1.Verification / Alternative check:Check limiting cases: For I7 active, true code is 111 and Y = 000; for I0 active, true code is 000 and Y = 111. The sequence matches the expected monotone pattern, validating the inversion rule used for I6 → 001.Why Other Options Are Wrong:110: That is the true code for 6, not the low-true output.011 / 101 / 111: These correspond to other inputs (I4, I2, I0) or to no active input in this context.Common Pitfalls:Forgetting that both inputs and outputs are active low, or assuming a straight binary output without inversion, which leads to an off-by-inversion error.Final Answer:001

Question 19

Real-world analogy for a multiplexer (data selector): Which everyday control most closely illustrates the function of a multiplexer by selecting exactly one of several available signal sources to pass onward?

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Introduction / Context:A multiplexer (MUX) is a data selector that routes exactly one of several inputs to a single output based on select lines. Drawing accurate real-world analogies helps solidify intuition about digital building blocks. This item asks which common stereo receiver control most faithfully mirrors a MUX’s behavior.Given Data / Assumptions:A MUX chooses one source among many; it does not mix or scale signals.Only one channel is passed at a time to the downstream stage.Other controls (volume, balance, tone) operate on the chosen signal rather than selecting among discrete sources.Concept / Approach:The source or function selector (e.g., PHONO/AUX/TUNER/BT) on a receiver is a hardware switch that connects one of several inputs to the amplifier path. That is exactly what a multiplexer does, albeit electronically and under digital control. Volume corresponds to gain (scaling), balance changes relative channel gains, and tone controls apply frequency-dependent filtering—none of which involve selecting among multiple input sources.Step-by-Step Solution:Identify the MUX role: selection among multiple inputs → one output.Map to receiver controls: the function/source selector performs selection.Reject controls that adjust amplitude or frequency response (volume, balance, tone).Verification / Alternative check:Classic receiver schematics show the input selector routing the chosen jack to the preamp, confirming the selection-only behavior matching a MUX.Why Other Options Are Wrong:Volume: Adjusts overall gain; does not choose among inputs.Balance: Sets left/right relative gains; no input selection.Treble–bass: Equalization filter; not a selector.Loudness: A contouring EQ at low volumes; not a selector.Common Pitfalls:Confusing selection (one-of-many) with mixing (many-to-one simultaneous) or with gain/tonal adjustments that act on the already selected path.Final Answer:the function/source selector switch on a stereo receiver

Question 20

Using a 74139 dual 2-to-4 demultiplexer as a decoder: Under what condition does the 74139 behave like a one-of-four decoder that drives exactly one active output line?

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Introduction / Context:The 74139 contains two independent 2-to-4 line decoders/demultiplexers with active-low enable inputs and active-low outputs. As a demultiplexer, it routes a data input to one of four outputs selected by the address lines. As a decoder, it asserts exactly one output (active state) based solely on the address, independent of time-varying data. This question asks what operating condition transforms the demultiplexer behavior into pure decode behavior.Given Data / Assumptions:Each 74139 section has one active-low enable (/G), two address lines (A1, A0), and four active-low outputs (Y0–Y3).Active output means logic LOW, consistent with the device’s low-true outputs.We desire “decoder mode,” where exactly one output is LOW determined by A1 A0.Concept / Approach:In demultiplexer use, the “data” is the enable input level being routed to one of the outputs. To force decoder behavior, fix the data to its active level so that the selected output alone is driven active, while all others remain inactive. Because the 74139’s outputs are active LOW and its enable is active LOW, holding the enable LOW while addressing with A1 A0 yields one LOW output (selected) and three HIGH outputs, i.e., a standard 1-of-4 decoder truth table.Step-by-Step Solution:Identify polarities: /G active LOW; outputs Yx active LOW.Force data active: set /G = 0 so “data” is constantly asserted.Let A1 A0 choose which Yx goes LOW; others remain HIGH.Verification / Alternative check:Compare to a 74xx139 datasheet decoder truth table: with /G = 0, exactly one output is LOW based on A1 A0. With /G = 1, all outputs idle HIGH, confirming the decoder-like behavior when /G is held LOW.Why Other Options Are Wrong:Enable HIGH or tying address lines improperly breaks the 1-of-4 active-output guarantee.Driving the address with data confuses roles; the decoder must use A1 A0 solely for selection.Common Pitfalls:Forgetting the active-low conventions and assuming HIGH is the active level; mixing demultiplexing (time-varying data) with decoding (static active level).Final Answer:Hold the data (enable) input LOW (active) while A1 and A0 select the output

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