Question 1

In transistor–transistor logic (TTL) families, an input that is left open or floating exhibits a characteristic default behavior due to internal biasing networks. When an open occurs on the input of a TTL device, how will the device interpret that i…

Accepted Answer

Introduction / Context: Understanding default input behavior is crucial when wiring TTL logic. Unlike CMOS, where floating inputs can drift unpredictably, classic TTL employs input structures that tend to bias an unwired input toward a logic HIGH. This knowledge helps diagnose faults and design with proper pull-ups or pull-downs to ensure deterministic operation. Given Data / Assumptions: Technology: standard TTL families (e.g., 74xx bipolar TTL). Scenario: one input is left open (floating). No external pull resistors are attached unless stated. Concept / Approach: TTL inputs are typically multi-emitter transistor bases with internal bias paths. An unconnected input sources a small bias current so that the input is treated as logic HIGH. Therefore, a floating TTL input is not equivalent to 0 V; instead, it behaves as a logical 1. This design quirk is why unused TTL inputs should not be left floating in practice; tie them explicitly HIGH or LOW using appropriate resistors to avoid noise sensitivity. Step-by-Step Solution: Identify family: TTL input stage has bias currents and internal structures favoring HIGH.Leave one input open: internal bias makes the gate sense a logic HIGH on that pin.Output response: the gate computes its function assuming that input is HIGH.Design implication: always terminate unused inputs; use pull-downs for a defined LOW if needed. Verification / Alternative check: Consult TTL datasheets: many specify that a floating input is HIGH-level; still, do not rely on it for robust designs due to susceptibility to noise and EMI. Why Other Options Are Wrong: “It goes LOW because there is no current”: Incorrect for TTL; the bias currents do not make an open act as 0.“Forces the output HIGH”: The output depends on the gate function, not merely on the presence of an open circuit.“Output remains correct anyway”: Floating inputs can cause erratic behavior; relying on other inputs is unsafe. Common Pitfalls: Applying CMOS intuition (where floating can be mid-level) to TTL devices.Leaving unused inputs floating instead of tying them to defined logic levels. Final Answer: It is interpreted as a HIGH input level

Question 2

Half-adder capability check in digital arithmetic: Which functional element is not provided by a standard half-adder circuit when adding two single-bit operands?

Accepted Answer

Introduction / Context: Adders are fundamental combinational blocks for arithmetic units in CPUs, DSPs, and ALUs. The two basic building blocks are the half-adder and the full-adder. Knowing the exact capabilities and limitations of each helps you choose the appropriate component when chaining bits into multi-bit adders or designing fast adder architectures. Given Data / Assumptions: A half-adder has two primary data inputs (A and B). Outputs are SUM and CARRY-OUT. No prior carry-in is assumed for a half-adder stage. Concept / Approach: A half-adder computes SUM = A XOR B and CARRY-OUT = A * B. It is “half” because it does not accept a carry-in from a less significant stage. In contrast, a full-adder has three inputs (A, B, and CARRY-IN) and two outputs (SUM and CARRY-OUT), enabling cascading across bit positions in multi-bit addition. Step-by-Step Solution: List half-adder I/O: inputs = A, B; outputs = SUM, COUT.Check for carry-in: not present in the half-adder definition.Identify the missing capability: lack of CARRY-IN support.Therefore, the half-adder does not have a carry-in input. Verification / Alternative check: Compare with full-adder equations: SUM = A XOR B XOR CIN; COUT = A*B + (A XOR B)*CIN. The extra input CIN is the distinguishing feature. Why Other Options Are Wrong: carry out: A half-adder does produce a carry-out (A*B).two inputs: It has exactly two inputs by definition.all of the above: Incorrect because only carry-in is missing. Common Pitfalls: Assuming a half-adder can be dropped into any bit position of a ripple adder; without CIN it cannot handle propagation from lower bits.Confusing half-adder (2 inputs) with full-adder (3 inputs). Final Answer: carry in

Question 3

In Karnaugh-map (K-map) simplification, certain input combinations are irrelevant to the final function (for example, invalid BCD codes in a BCD-to-decimal converter). What are these entries called, and how may they be treated during grouping to ach…

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Introduction / Context: Karnaugh maps (K-maps) are a visual method to minimize Boolean expressions by grouping adjacent 1s into powers-of-two rectangles. In many real systems, some input combinations never occur or have outputs that do not matter. Recognizing and using these conditions properly can lead to significantly simpler hardware implementations with fewer gates and less propagation delay. Given Data / Assumptions: Some input patterns are irrelevant (e.g., BCD codes 1010 to 1111 are invalid). These map cells can be assigned flexibly to help simplification. Goal: reduce logic by forming larger groups in the K-map. Concept / Approach: Such irrelevant combinations are called don’t-care conditions (often marked X or d). When minimizing, you are free to treat a don’t-care cell as either 1 or 0 in whichever way allows you to form larger groups (1, 2, 4, 8, … cells), thereby producing a simpler sum-of-products (SOP) or product-of-sums (POS) expression. After minimization, the implemented circuit need not produce a defined value for those unused combinations, which is acceptable by specification. Step-by-Step Solution: Identify invalid/irrelevant input codes for the target function.Mark their K-map cells as X (don’t-care).While grouping 1s (for SOP) or 0s (for POS), include X cells as 1 or 0 respectively to maximize group sizes.Derive the minimized expression consistent with these choices. Verification / Alternative check: Compare gate counts with and without using don’t-cares; the don’t-care–assisted solution should have fewer literals or gates. Why Other Options Are Wrong: “Spurious, treat as ANDs/ORs”: Not standard terminology nor a valid K-map method.“Duplicate, treat as 1s or 0s to verify”: They are not duplicates; verification is not the purpose of don’t-cares.“Spurious, treat as 1s or 0s to simplify”: Terminology is incorrect; the accepted term is don’t-care. Common Pitfalls: Forgetting to include don’t-care cells to enlarge groups.Accidentally forcing outputs on don’t-care combinations in later specifications. Final Answer: don’t care, treat as 1's or 0's to simplify

Question 4

Identify which Boolean expression is already in a minimal two-level form and therefore cannot be simplified further using basic Boolean algebra rules. Select the equation that cannot be further reduced in literal count or gate count (at the two-leve…

Accepted Answer

Introduction / Context: Boolean simplification aims to reduce literal count and gate depth while preserving functionality. Some expressions are already minimal in a two-level implementation (sum-of-products or product-of-sums). Recognizing them avoids unnecessary algebra and helps in mapping to standard cells efficiently (e.g., XOR/XNOR implementations). Given Data / Assumptions: We compare several common expressions for reducibility. “Further simplified” means fewer literals or fewer product/sum terms at the same two-level form. No exploitation of multi-level factorization beyond standard two-level simplification is intended. Concept / Approach: Some identities: A + AB = A (absorption). A + A'B = A + B (consensus/absorption). AB + AB' = A(B + B') = A (complementation). However, AB + A'B' represents equivalence (XNOR) between A and B, which in two-level SOP uses two distinct product terms and cannot be reduced to a single literal or a single product term without introducing an XOR/XNOR gate primitive or changing the level of logic. Step-by-Step Solution: Check option a: AB + AB' = A(B + B') = A → reducible.Check option b: A + AB = A → reducible by absorption.Check option d: A + A'B = (A + A')(A + B) = 1 * (A + B) = A + B → reducible.Option c: AB + A'B' corresponds to equivalence (A XNOR B) and is minimal as two product terms of two literals; not further reducible in two-level SOP. Verification / Alternative check: Build a truth table: AB + A'B' is 1 when A = B, validating XNOR behavior and showing no single-literal collapse is possible. Why Other Options Are Wrong: Each of a, b, and d simplifies to a single-literal or simpler expression using standard identities. Common Pitfalls: Attempting to reduce XNOR to fewer product terms without allowing XOR/XNOR primitives.Confusing AB + A'B' with AB' + A'B (which is XOR). Final Answer: AB + A'B'

Question 5

Cascading two 4-bit magnitude comparators (such as the 74x85 series) to compare two 8-bit numbers: How should the cascading inputs of the most significant (upper) 4-bit comparator be connected so the overall 8-bit comparison is correct?

Accepted Answer

Introduction / Context: Multi-bit magnitude comparison often uses cascaded comparators. Standard 4-bit comparator ICs (e.g., 74x85) provide “cascade” inputs (AB) and corresponding outputs so multiple units can be chained to handle wider operands. Correct wiring ensures that the decision made by lower-order bits properly influences the overall result once the upper bits are evaluated. Given Data / Assumptions: Two 4-bit comparators are used to compare 8-bit numbers A7..A0 and B7..B0. The least significant comparator evaluates A3..A0 vs. B3..B0. The most significant comparator evaluates A7..A4 vs. B7..B4. Concept / Approach: The cascade mechanism propagates the relationship found in lower bits to the higher-stage decision. The lower (LSB) comparator produces the intermediate AB outputs. These must feed the corresponding cascade inputs of the upper (MSB) comparator so the final outputs reflect the entire 8-bit comparison, not just the upper nibble independently. Step-by-Step Solution: Evaluate low nibble (LSB comparator): obtain signals LT_L, EQ_L, GT_L.Feed these to MSB cascade inputs: LT_in = LT_L, EQ_in = EQ_L, GT_in = GT_L.The MSB comparator uses its own nibble and the cascade inputs to produce final LT, EQ, GT for the full 8-bit numbers.Confirm polarity and pin names per datasheet (some families use active-low signals). Verification / Alternative check: Test boundary cases: when upper nibbles are equal, the MSB comparator defers to the cascaded equality from the LSB stage; when upper nibbles differ, MSB outputs dominate regardless of LSB state. Why Other Options Are Wrong: Tying to LSB inputs or ground ignores the lower nibble result, breaking correctness.Forcing A=B HIGH and others LOW defeats dynamic comparison and yields wrong outputs if MSB nibbles differ. Common Pitfalls: Mixing up active-high/active-low cascade pins; always check the exact device logic levels.Failing to route EQ correctly, which is crucial when upper nibbles match. Final Answer: Connect them to the cascade outputs from the least significant 4-bit comparator

Question 6

AND–OR–INVERT (AOI) standard cells and networks are commonly used to implement minimized sum-of-products efficiently. AOI gates are primarily designed to simplify the implementation of which canonical logic form?

Accepted Answer

Introduction / Context: Many standard-cell libraries include complex gates like AOI (AND–OR–INVERT) and OAI (OR–AND–INVERT) because they reduce transistor count and delay compared to building the same logic strictly from basic gates. Recognizing which canonical form each complex gate targets helps map minimized expressions directly to efficient hardware. Given Data / Assumptions: AOI topology: multiple AND gates feed an OR gate, followed by an inversion. Target expressions: minimized two-level SOP forms (sum of products). Goal: identify which canonical form AOI matches most naturally. Concept / Approach: A two-level SOP is of the form F = Σ(products) = (A*B) + (C*D) + … . AOI realizes the complemented SOP: F' = [(A*B) + (C*D) + …]'. In CMOS, AOI gates can be highly efficient because the pull-down (or pull-up) networks implement the inverted sum directly with fewer transistors. By placing or removing bubbles (De Morgan transformations), an AOI gate effectively implements the intended SOP with minimal staging, often as a NAND–NAND equivalent. Step-by-Step Solution: Start from SOP: F = P1 + P2 + … where Pi are product terms.Implement OR of ANDs feeding an inverter: X = (P1 + P2 + …)'.Use bubble pushing/De Morgan to match signal polarities so that X' = F is obtained at the desired node.Thus AOI maps directly to SOP with high efficiency. Verification / Alternative check: Compare gate/transistor counts for a typical 4-term SOP: AOI often reduces both delay and area versus discrete AND + OR + INV stages. Why Other Options Are Wrong: POS: OAI structures are typically better aligned with POS (sum terms feeding an AND followed by inversion).De Morgan’s theorem: It is a transformation tool, not a canonical target form.NAND-only logic: While AOI can be equivalent to NAND–NAND, the design goal is efficient SOP implementation. Common Pitfalls: Confusing AOI with OAI; remember AOI corresponds to SOP, OAI to POS.Forgetting that the final inversion in AOI is part of the cell and can be absorbed depending on signal polarity needs. Final Answer: SOP logic (sum of products)

Question 7

In a product-of-sums (POS) realization, the logic network forms OR terms (sums) first and then combines those terms to drive the final output. Which logic gate generates the final output stage in a canonical POS circuit?

Accepted Answer

Introduction / Context: Canonical logic forms—sum-of-products (SOP) and product-of-sums (POS)—guide how we implement Boolean functions using two-level logic. Understanding which gate finishes the computation in each form is essential for reading schematics, performing K-map minimization, and mapping functions to AOI/OAI cells in VLSI flows. Given Data / Assumptions: POS means an AND of sum terms. Each sum term is an OR of literals (variables or complements). We consider the straightforward, non-inverted canonical form. Concept / Approach: By definition, a POS expression has the structure F = (S1) * (S2) * … where each Si = (l1 + l2 + …). The network first creates each OR term (sum) and then combines all sum terms using an AND gate. In practical CMOS libraries, an OAI (OR–AND–INVERT) gate may implement the complement efficiently, but the canonical POS form’s final combining operation is still the AND of those sums. Step-by-Step Solution: Write function in POS: F = (A + B)(C + D’)(E + F + G’).Implement each parenthesized sum with OR gates.Combine all sums with an AND gate to produce F.If an inverted output is desired, append an inverter or use an OAI variant accordingly. Verification / Alternative check: Cross-check with SOP: SOP ends with an OR gate; POS ends with an AND gate—these are duals. Why Other Options Are Wrong: OR: That is the final stage for SOP, not POS.NOR / NAND: These are inverted versions; though useful in CMOS, the canonical POS definition’s last combining operator is AND. Common Pitfalls: Mixing up the end-stage gate between SOP and POS.Assuming the presence of inversion in the canonical definition; inversion pertains to specific implementations like OAI. Final Answer: an AND

Question 8

Karnaugh map minimization rule: When simplifying a Boolean function with a Karnaugh map, you must use the ________ number of loops (groups), while making each loop as large as possible.

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Introduction / Context: Karnaugh maps (K-maps) are a visual method for minimizing Boolean expressions. A key rule is to cover all 1-cells (or 0-cells for POS) using the fewest number of groups, and to make each group as large as allowed by adjacency and the power-of-two constraint. This question checks your understanding of the “fewest groups, largest size” principle. Given Data / Assumptions: K-maps arrange truth table outputs in Gray-code order to expose adjacencies. Groups (loops) must contain 1, 2, 4, 8, … cells (powers of two). Groups may wrap around edges and may overlap when it reduces literal count. Concept / Approach: Minimization is achieved by maximizing group size (reduces literals per term) and minimizing group count (reduces number of terms). Thus, the correct strategy is to use the minimum number of groups, each being as large as possible within constraints. Step-by-Step Solution: List all output cells that are 1 (for SOP).Form the largest possible power-of-two groups that cover these 1s.Allow wrapping and overlapping to enlarge groups and share coverage.Ensure the total number of groups is minimized while still covering all 1s. Verification / Alternative check: Compare the resulting expression's literal count with one formed from smaller or more numerous groups; the minimal grouping yields fewer literals and fewer terms, verifying optimality for two-level logic. Why Other Options Are Wrong: maximum: Maximizing the number of loops increases terms and is non-minimal. median: There is no “median” strategy in K-maps. Karnaugh: Not a valid quantitative choice. power-of-two only (count unspecified): True about size, but the rule asked for the count; it must be minimized. Common Pitfalls: Forgetting wrap-around adjacencies; avoiding overlap even when it reduces terms; making many small groups instead of fewer large ones. Final Answer: minimum

Question 9

2's-complement subtraction refresher: To subtract a signed subtrahend from a signed minuend using the 2's-complement method, the minuend is ________.

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Introduction / Context: The 2's-complement method converts subtraction into addition, which hardware adders perform efficiently. Knowing which operand to complement is crucial to avoid sign errors and overflow confusion. Given Data / Assumptions: We are subtracting: result = minuend − subtrahend. Operands are in 2's-complement form with equal bit width. Standard adder with carry-in as needed. Concept / Approach: In 2's-complement arithmetic, subtraction A − B is implemented as A + (2's complement of B). That is, complement and add one to the subtrahend, not the minuend. The adder adds the minuend unchanged to the complemented subtrahend. Step-by-Step Solution: Compute B' = 2's-complement(B) = (~B) + 1.Add: A + B' using the n-bit adder.Ignore the final carry-out for signed arithmetic; check overflow via sign bits.The minuend A remains uncomplemented throughout. Verification / Alternative check: Try A = 5, B = 3 (8-bit). 3 → 00000011; B' = 11111101 + 1 = 11111110? (careful) Correctly: ~00000011 = 11111100; +1 → 11111101 (−3). Then 00000101 + 11111101 = 00000010 with carry out discarded → 2, as expected. Why Other Options Are Wrong: Complementing the minuend (always or conditionally) is incorrect; you would compute the wrong value. Overflow anticipation does not dictate which operand to complement; the rule is fixed: complement the subtrahend. Common Pitfalls: Complementing the wrong operand; mixing up 1's vs 2's complement; forgetting to ensure equal bit widths; misinterpreting the carry-out as overflow for signed results. Final Answer: never complemented

Question 10

Odd parity generation: An 8-bit binary word is applied to an odd-parity generator. The parity bit will be 1 only if ________.

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Introduction / Context: Parity bits provide a simple error-detection mechanism. With odd parity, the total count of 1s across data bits plus the parity bit must be odd. Understanding when the parity bit becomes 1 is essential for designing and checking parity logic. Given Data / Assumptions: 8 data bits (any pattern of 1s and 0s). Odd-parity convention: total ones (data + parity) is odd. Single parity bit added to the word. Concept / Approach: Let k be the number of 1s in the 8-bit word. Odd parity sets parity_bit so that k + parity_bit is odd. Therefore parity_bit = 1 when k is even, and parity_bit = 0 when k is odd. The numeric value of the word is irrelevant; only the count of 1s matters. Step-by-Step Solution: Let k = count_ones(data).If k is even → choose parity_bit = 1 to make k + 1 odd.If k is odd → choose parity_bit = 0; odd + 0 remains odd.Thus the parity bit equals 1 only when the number of 1s is even. Verification / Alternative check: Example: data = 10110000 has k = 3 (odd) → parity_bit = 0; total ones = 3 (odd). Another: data = 11000000 has k = 2 (even) → parity_bit = 1; total ones = 3 (odd). Why Other Options Are Wrong: “number is odd/even”: The arithmetic value of the byte does not determine parity; the Hamming weight does. “number of 1s is odd”: For odd k, parity bit is 0, not 1. “MSB is 1”: Parity depends on all bits, not only the MSB. Common Pitfalls: Confusing odd vs even parity definitions; focusing on numeric value rather than bit count; omitting the parity bit in the final odd-count requirement. Final Answer: the number of 1s in the word is even

Question 11

Logic family prefixes: The ________ prefix on integrated circuits denotes a broader (often military-grade) operating temperature range and is commonly used for defense/aerospace applications.

Accepted Answer

Introduction / Context: Classic 54/74-series numbering distinguishes environmental qualifications. Selecting the correct series ensures reliable operation under the expected temperature extremes and conditions. Given Data / Assumptions: 54-series traditionally indicates extended temperature range (e.g., around −55 °C to +125 °C). 74-series is typically commercial/industrial, depending on suffix. 2N is a discrete transistor prefix, not a logic family series. Concept / Approach: Recognize that “54” corresponds to military/extended temperature logic, whereas “74” is commercial/industrial logic families (TTL/CMOS variants). “TTL” is a technology class, not a prefix level; HCT denotes CMOS with TTL-compatible thresholds but uses 74HCTxx part numbers. Step-by-Step Solution: Identify the environmental requirement (mil/extended).Choose the 54xx family that meets the specified range.Verify in datasheet the exact min/max temperatures and screening.Confirm pin/function equivalence to 74xx counterparts as needed. Verification / Alternative check: Compare datasheets of 54HCxx vs 74HCxx: operating ranges and screening differ, but logical function/pinouts are typically aligned. Why Other Options Are Wrong: 2N: Prefix for discrete transistors. 74: Commercial/industrial range, not the broader military range. TTL: Technology class, not a series prefix. HCT: A subfamily (74HCTxx), not the temperature-grade prefix in question. Common Pitfalls: Assuming 74 always means 0–70 °C; industrial variants exist, but “54” is the canonical military/extended indicator. Final Answer: 54

Question 12

XNOR truth table completion: For a two-input XNOR gate with inputs (A,B) ordered as 00, 01, 10, 11, the corresponding outputs w, x, y, z are ____, ____, ____, and ____, respectively.

Accepted Answer

Introduction / Context: The exclusive-NOR (XNOR) function outputs HIGH when its inputs are equal and LOW when they differ. Completing a truth table tests conceptual understanding of equality detection in digital logic. Given Data / Assumptions: Two-input XNOR gate. Row order: (A,B) = 00, 01, 10, 11. Positive logic convention. Concept / Approach: XNOR is the complement of XOR. XOR is 1 when inputs differ; therefore XNOR is 1 when inputs are the same: 00 or 11. It is 0 for 01 and 10. This maps directly to the requested sequence w, x, y, z. Step-by-Step Solution: For 00: inputs equal → XNOR = 1 → w = 1.For 01: inputs differ → XNOR = 0 → x = 0.For 10: inputs differ → XNOR = 0 → y = 0.For 11: inputs equal → XNOR = 1 → z = 1. Verification / Alternative check: Boolean identity: XNOR = A′B′ + AB. Evaluate each row to confirm 1,0,0,1 in the specified order. Why Other Options Are Wrong: 0,1,0,1 or 1,1,1,0 etc.: These patterns do not match the equality-only HIGH behavior; they either assert on differing inputs or on too many cases. Common Pitfalls: Swapping XOR and XNOR; mixing row order; assuming a different input ordering, which would permute the output sequence. Final Answer: 1, 0, 0, 1

Question 13

Equality detection: Which gate provides a HIGH output exactly when its two inputs have equal logic levels (both 0 or both 1)?

Accepted Answer

Introduction / Context: Comparing two binary signals is a common task. The equality function outputs HIGH when inputs match. Recognizing which primitive gate directly implements equality simplifies design work for comparators and parity logic. Given Data / Assumptions: Two-input standard logic gates under positive logic. Goal is equality detection (A = B). Concept / Approach: By definition, XOR is HIGH when inputs differ; therefore its complement, XNOR, is HIGH when inputs are equal. The XNOR truth table places 1s at (0,0) and (1,1) and 0s at (0,1) and (1,0). Step-by-Step Solution: Recall XOR identity: XOR = A′B + AB′ (different).XNOR = (XOR)′ = A′B′ + AB (equal).Thus, for equality detection, choose XNOR.No additional logic is required for the two-input case. Verification / Alternative check: Build a small truth table or evaluate A′B′ + AB for all four input pairs; outputs are 1 only on equality rows, confirming XNOR. Why Other Options Are Wrong: XOR: Detects inequality. NAND/NOR/AND: Do not implement equality; their truth tables do not match equality-only HIGH behavior. Common Pitfalls: Confusing XOR with XNOR; attempting to build equality with AND/OR and inverters when a single XNOR suffices. Final Answer: XNOR gate

Question 14

Inequality detection: The ________ logic function produces a HIGH output whenever its two inputs are unequal (different).

Accepted Answer

Introduction / Context: Detecting whether two signals differ is fundamental in adders, parity generators, and error checking. The gate that signals “different” with a HIGH output is the XOR (exclusive-OR). Given Data / Assumptions: Two-input standard logic functions. Positive logic convention. Concept / Approach: XOR outputs 1 when inputs are different (01 or 10) and 0 when they are equal (00 or 11). Its complement, XNOR, signals equality. Terms like “exclusive-AND” or “inexclusive-OR” are not standard logic functions. Step-by-Step Solution: Recall XOR identity: F = A′B + AB′.Evaluate for all four input pairs; F = 1 only for 01 and 10.Therefore, XOR detects inequality directly.Choose “exclusive-OR.” Verification / Alternative check: Implement using half-adder sum output; the sum bit equals A XOR B, which is 1 when inputs differ, further confirming behavior. Why Other Options Are Wrong: exclusive-NOR: Equality detection (opposite of required). exclusive-AND / inexclusive-OR: Nonstandard or meaningless in common logic nomenclature. NOR: 1 only when both inputs 0; not an inequality detector. Common Pitfalls: Swapping XOR and XNOR meanings; assuming NOR/AND derivatives can serve as direct inequality detectors without additional logic. Final Answer: exclusive-OR

Question 15

Logic-family compatibility: The ________ series of CMOS ICs are pin, function, and TTL-level compatible with the 74-series logic devices.

Accepted Answer

Introduction / Context: When mixing logic families, input threshold compatibility and pin/function equivalence are critical. Designers often select parts from families that are explicitly TTL-compatible to drop into 74-series designs without level translators. Given Data / Assumptions: 74HCTxx parts are CMOS internally but accept TTL-level inputs. Pinouts and functions mirror many 74xx devices. ALS is a TTL subfamily; 2N is for discrete transistors; ECL is a different logic technology altogether. Concept / Approach: The “HCT” designation stands for High-speed CMOS, TTL-compatible. These devices are designed to be drop-in compatible (functionally and often pin-wise) with standard 74-series TTL logic while offering CMOS power advantages. Step-by-Step Solution: Identify the need: TTL-level inputs yet CMOS implementation.Select the HCT family which guarantees TTL input thresholds.Confirm pin/function equivalence in the specific datasheet.Deploy without additional level shifting in mixed-TTL systems. Verification / Alternative check: Compare 74HC vs 74HCT input threshold specs; only HCT specifies TTL-compatible VIH/VIL, confirming compatibility with legacy TTL outputs. Why Other Options Are Wrong: ALS: A low-power Schottky TTL subfamily, not CMOS with TTL thresholds. CMOS (generic): 74HC is CMOS but not necessarily TTL-level compatible. 2N: Transistor prefix; not a logic family. ECL: Different logic technology with distinct levels and pinouts. Common Pitfalls: Assuming all CMOS (HC) inputs accept TTL levels interchangeably; ignoring VIH/VIL specs; overlooking pin compatibility across specific part numbers. Final Answer: HCT

Question 16

K-map grouping sizes: When grouping adjacent 1-cells in a Karnaugh map (for SOP), valid group sizes must be ________.

Accepted Answer

Introduction / Context: Karnaugh map groupings represent implicants whose sizes determine how many variables drop out of the resulting product term. Only specific group sizes are allowed to preserve adjacency rules derived from Gray-code ordering. Given Data / Assumptions: We are forming groups of adjacent 1s for SOP minimization. Adjacency includes edge wrap-around. Groups must be rectangular and uniform in size. Concept / Approach: Valid K-map group sizes are powers of two: 1, 2, 4, 8, 16, … This ensures every cell in a group differs from its neighbor by only one variable per step, enabling variable elimination in the product term. Arbitrary sizes like 3 or 5 break this property and are invalid. Step-by-Step Solution: Identify 1-cells to be covered.Form groups whose area equals 2^n for some integer n ≥ 0.Use wrapping to enlarge groups where possible (e.g., 2x2, 1x4, 4x1).Prefer larger groups to eliminate more variables and reduce literals. Verification / Alternative check: Translate a 2x2 group (size 4) into its product term and verify that two variables drop out; for a size-8 group, three variables drop out, confirming the power-of-two rule. Why Other Options Are Wrong: 2's / 4's: These are valid sizes but incomplete; not the full allowed set. 3's: Not a power of two; invalid. Any size up to map width: Violates adjacency/literal-elimination properties. Common Pitfalls: Forgetting that singletons (size 1) are allowed; refusing to overlap groups when it yields larger power-of-two coverage; overlooking wrap-around adjacencies that enable size-4 or size-8 groups. Final Answer: 1, 2, 4, 8, etc.

Question 17

Sum-of-products (SOP) implementation strategy: Which universal gate family most directly realizes SOP expressions with minimal complication, enabling two-level NAND–NAND realizations and easy inversion where needed?

Accepted Answer

Introduction / Context: The sum-of-products (SOP) form is one of the most common ways to describe Boolean functions for hardware realization. Engineers frequently translate an SOP expression into a two-level gate network. Choosing a universal gate family simplifies the build, reduces part types, and eases optimization. Given Data / Assumptions: SOP means an OR (sum) of multiple AND (product) terms, possibly with complemented literals. We aim for a uniform implementation using a single gate type where possible. De Morgan’s laws permit inversion bubbles to be moved between levels to match available gates. Concept / Approach: NAND gates are universal and particularly convenient for SOP. A product term can be created with a NAND followed by inversion; the final sum operation (OR) can be formed by a NAND whose inputs are the NANDed product terms with appropriate bubble placement. This yields a standard two-level NAND–NAND structure equivalent to AND–OR, but built entirely from NANDs. Step-by-Step Solution: Start with F = P1 + P2 + ... + Pn, where each Pi is an AND of literals.Realize each Pi using a NAND. The output is the complement of Pi (call it Pi’).Feed all Pi’ signals into a final NAND. By De Morgan, NAND of complements = OR of originals.Thus F is implemented with only NAND gates (two-level NAND–NAND network). Verification / Alternative check: Compare to a NOR–NOR approach. NOR is ideal for POS (product-of-sums) implementations. For SOP, NAND–NAND aligns naturally without extra inversion overhead. Why Other Options Are Wrong: NOR: Best matches POS, not SOP, for two-level realizations.AND: Not universal alone; needs separate OR and NOT for full functionality.DOOR: Not a logic gate; a distractor. Common Pitfalls: Mistaking NOR as equally suitable for SOP; forgetting bubble pushing when converting AND–OR to NAND–NAND; assuming inverters are always needed rather than using intrinsic NAND inversions. Final Answer: NAND

Question 18

Fault diagnosis with a current tracer: If a gate's output has an internal short, what will the current tracer reveal about the faulty point on the board?

Accepted Answer

Introduction / Context: In digital troubleshooting, a current tracer is a handheld probe that senses the magnetic field around a conductor to show the direction and relative magnitude of current flow. When a logic output suffers an internal short within an IC, recognizing what the tracer can identify is crucial for efficient fault isolation. Given Data / Assumptions: One gate output is internally shorted inside its IC package. Board traces connect that output node to loads or other inputs. The circuit is powered so fault current flows along the net to the IC pin. Concept / Approach: A current tracer helps follow the path of abnormal current on a powered board. By tracing along the copper from the supply through series elements to the suspicious node, the technician can locate the component pin where current terminates into the fault. If the short is inside the gate, the current flow will “point” into that IC pin, uniquely identifying the defective gate rather than a downstream load node. Step-by-Step Solution: Power the board with appropriate current limiting.Use the tracer to follow elevated current from the affected net toward components.Observe current direction into a specific IC output pin.Conclude the internal short resides in that gate, identifying the bad IC/gate. Verification / Alternative check: Lift the suspect pin or remove the IC; if the abnormal current ceases, this confirms the gate as defective. Thermal imaging or freeze spray may corroborate (hot IC). Why Other Options Are Wrong: Show whether shorted to Vcc or ground: Polarity alone does not localize the failing gate.Probably cannot locate: Tracers are expressly used to localize current paths.Identify defective load node: With an internal IC short, the sink/source is the gate, not a passive load node. Common Pitfalls: Using the tracer on unpowered boards; misinterpreting bidirectional bus currents; ignoring current division across parallel paths; skipping current limiting during fault-finding. Final Answer: identify the defective gate

Question 19

8-to-1 multiplexer design capacity: What is the largest truth table size (number of rows) that can be implemented directly using a single 8-line-to-1-line MUX with 3 select inputs?

Accepted Answer

Introduction / Context: Multiplexers (MUXes) can implement combinational functions by steering one of many data inputs to the output based on select lines. The number of select lines determines how many input combinations are directly addressable, which in turn determines the size of the truth table that can be realized without extra logic. Given Data / Assumptions: An 8-to-1 MUX has 8 data inputs and 3 select lines. No additional external gating or input encoding is used (direct implementation). One output bit is produced. Concept / Approach: With 3 select lines, there are 2^3 = 8 possible select codes. Each select code chooses one of the 8 data inputs, which can be tied to constant 0/1 or other signals to implement the required function. Therefore, the directly realizable truth table has 8 distinct input cases along the select dimensions. Step-by-Step Solution: Identify number of select lines S = 3.Compute addressable combinations = 2^S = 2^3 = 8.Assign each MUX data input to the desired output value for its select code.Hence, maximum truth table rows (unique select combinations) = 8. Verification / Alternative check: General rule: an M-to-1 MUX where M = 2^k implements any 1-bit function of k variables directly. Here k = 3, so any function of three variables fits in 8 rows. Why Other Options Are Wrong: 3 rows / 4 rows: Understates the 2^3 capacity.16 rows: Would require 4 select lines or additional logic expansion (e.g., larger MUX or MUX tree). Common Pitfalls: Confusing data inputs with select inputs; assuming a MUX directly implements functions of more variables than it has select lines; forgetting that constants can be tied to unused data inputs. Final Answer: 8 rows

Question 20

CMOS floating input behavior: When an open (floating) condition occurs on the input of a CMOS gate, what best describes the resulting output state?

Accepted Answer

Introduction / Context: CMOS inputs have extremely high impedance. If an input pin is left unconnected (open), it can accumulate charge or pick up noise, leading to undefined logic levels. Understanding this behavior prevents intermittent faults and excessive power consumption. Given Data / Assumptions: Input of a CMOS gate is physically open (no defined drive). There is no internal pull-up or pull-down unless specified by the device. Environment introduces leakage, capacitive coupling, and noise. Concept / Approach: A floating CMOS input can drift to any voltage within the valid range due to leakage and coupling. The gate threshold region can be crossed unpredictably, causing the output to switch or even oscillate. This can also increase static current if input hovers in the transition region, stressing the device. Step-by-Step Solution: Recognize that an open input lacks a defined logic level.Due to high impedance, tiny leakage or noise can set the input voltage.The inverter pair inside CMOS responds non-deterministically to a drifting input.Therefore, the output is unpredictable and may be HIGH or LOW at different times. Verification / Alternative check: Datasheets recommend using defined pull-ups or pull-downs, or never leaving CMOS inputs floating. Simulation of an input with a large resistance to mid-supply shows unstable behavior and increased supply current. Why Other Options Are Wrong: Always LOW / always HIGH: Floating behavior is not deterministic.“As if HIGH”: Not consistently true; depends on leakage/coupling. Common Pitfalls: Leaving unused inputs floating; assuming TTL-like default highs; overlooking the power penalty of inputs stuck near threshold. Final Answer: be unpredictable; it may go HIGH or LOW

Combinational Logic Circuits Questions