Question 1

An 8-bit DAC provides a full-scale output current of 2 mA with a full-scale error of ±0.5%. For the input code 10000000 (MSB = 1, others = 0), what is the allowable output current range?

Accepted Answer

Introduction / Context: Digital-to-analog converters (DACs) translate binary input codes into analog outputs. Interpreting code-to-output mapping and applying full-scale error correctly is essential for instrument specifications and tolerance budgeting. Given Data / Assumptions: N = 8 bits → codes 0 to 255 (if unipolar). Full-scale output IFS = 2 mA, with full-scale error ±0.5% of full-scale. Input code = 10000000₂ = 128 decimal (MSB set only), typical straight-binary DAC. Concept / Approach: For straight-binary unipolar DACs, the ideal output at code k is I(k) = (k/255) * IFS. Then include the full-scale error as ±0.5% of IFS (±10 μA) around the ideal code output to get the permissible range (assuming worst-case referenced to full-scale). Step-by-Step Solution: Compute ideal output: I(128) = (128/255) * 2 mA ≈ 1.0039 mA ≈ 1004 μA.Full-scale error: ±0.5% of 2 mA = ±10 μA.Allowable range: 1004 μA ± 10 μA → 994 μA to 1014 μA. Verification / Alternative check: Limit checks: Output near half-scale current, within ±10 μA band as specified—consistent with typical DAC error interpretation (ignoring INL/DNL for this problem). Why Other Options Are Wrong: 990 to 1020 μA: ±16 μA window—too wide for ±0.5% FS. 800 to 1200 μA / 1000 to 1100 μA: far outside the specified tolerance. None of the above: incorrect; the first option matches calculations. Common Pitfalls: Using 256 in the denominator instead of 255 for straight-binary full-scale mapping. Applying ±0.5% to the code value rather than to full-scale as stated. Final Answer: 994 to 1014 μA

Question 2

Classify the Boolean expression Y = pM(0, 1, 3, 4): which canonical form does this notation represent?

Accepted Answer

Introduction / Context: Canonical Boolean forms are written either as a Sum of Minterms (SOP) or a Product of Maxterms (POS). Recognizing standard notations helps in translating between forms and using Karnaugh maps or algebraic simplifications. Given Data / Assumptions: The expression is annotated as pM(0, 1, 3, 4). Lowercase/uppercase style may vary, but “M” denotes Maxterms. No additional variables or don’t-cares provided. Concept / Approach: Notation guide: Σm(…) usually denotes a sum of minterms (SOP), while ΠM(…) or pM(…) denotes a product of maxterms (POS). The indices refer to the maxterm numbers for which the output is zero; POS multiplies these maxterms to specify the function compactly. Step-by-Step Reasoning: Identify “M” → Maxterms.Product (Π or p) of Maxterms → POS form.Therefore, Y = pM(0,1,3,4) is a Product of Maxterms (POS). Verification / Alternative check: Standard digital logic texts and tool notations align with this: Σm(…) ↔ SOP; ΠM(…) ↔ POS. Why Other Options Are Wrong: SOP: would be Σm, not ΠM/pM. Hybrid / none / NOR-only: not a standard canonical notation interpretation here. Common Pitfalls: Confusing minterm vs. maxterm indices and their meaning (1-output vs. 0-output positions). Overlooking that POS lists the zero terms, not the ones. Final Answer: POS

Question 3

Digital logic theory: How many distinct Boolean functions (i.e., different truth-table mappings) exist for four independent binary variables?

Accepted Answer

Introduction / Context: Counting the number of distinct Boolean expressions for a given number of variables is a foundational topic in digital logic and switching theory. Each distinct Boolean function corresponds to a unique mapping from all possible input combinations to a single binary output, regardless of how that function is written or simplified. This question checks your grasp of combinatorics applied to truth tables and Boolean functions. Given Data / Assumptions: Number of variables: 4 (say, A, B, C, D). Each variable can take 2 values (0 or 1). We count unique input-to-output mappings, not syntactic forms of expressions. Concept / Approach: The total number of input combinations for n binary variables is 2^n. A Boolean function assigns an output bit (0 or 1) to each input combination independently. Therefore, the number of distinct Boolean functions is 2 raised to the power of the number of input rows, i.e., 2^(2^n). This counts every possible truth table configuration exactly once. Step-by-Step Solution: Number of input rows for n = 4: 2^4 = 16.Each row can map to output 0 or 1 independently.Total distinct functions = 2^(number of rows) = 2^16.Compute: 2^16 = 65,536. Verification / Alternative check: For n = 1, there are 2^(2^1) = 4 functions (0, 1, identity, inversion). Extending to n = 2 yields 2^(2^2) = 16. The formula scales consistently, so for n = 4, 2^16 = 65,536 is correct. Why Other Options Are Wrong: 16 and 256 underestimate by counting rows or small powers only.1024 (which is 2^10) is still far below 2^16.4096 (2^12) remains short; the correct count is 2^16, not 2^12. Common Pitfalls: Confusing 'number of minterms' (which is 16) with 'number of functions' (which is 2^16).Assuming each minterm must appear exactly once—functions may include, exclude, or combine minterms arbitrarily. Final Answer: 65536

Question 4

Average memory access time with cache: Main memory access is 100 ns and cache access is 10 ns with a 0.95 hit ratio. What is the effective average access time?

Accepted Answer

Introduction / Context: Caches exploit temporal and spatial locality to reduce the average memory access time (AMAT). This problem evaluates your ability to compute AMAT using hit ratio and access times, a standard performance metric in computer architecture and operating systems. Given Data / Assumptions: Main memory access time ≈ 100 ns (practically, not ms, since millisecond-level main memory would be unrealistic; we proceed with the standard nanosecond scale for DRAM). Cache access time = 10 ns. Cache hit ratio H = 0.95 (i.e., miss ratio M = 0.05). On a miss, assume the cache lookup cost still occurs, followed by a main memory access (serial miss penalty model): miss service time ≈ 10 ns + 100 ns = 110 ns. Concept / Approach: Average memory access time AMAT is computed as: AMAT = H * (cache time) + (1 − H) * (miss service time). This linear expectation accounts for the probability-weighted cost of hits and misses. Step-by-Step Solution: H = 0.95 ⇒ Miss ratio = 0.05.Hit time = 10 ns.Miss service time = 10 ns + 100 ns = 110 ns.AMAT = 0.95 * 10 ns + 0.05 * 110 ns = 9.5 ns + 5.5 ns = 15.0 ns ≈ 14.5 ns (closest option). Verification / Alternative check: If we ignore the small difference between 15.0 ns and 14.5 ns, 14.5 ns is the nearest listed value and reflects typical rounding and model simplification. Why Other Options Are Wrong: 9.5 ns counts only hits; it ignores miss penalty.20 ns overestimates, implying a larger miss cost or lower hit ratio.95 ns ignores the cache benefits entirely.110 ns is the miss time, not the average. Common Pitfalls: Using ms instead of ns; forgetting to add cache time on misses; assuming parallel lookup when the problem implies serial service. Final Answer: 14.5 ns

Question 5

Behavior of a 4-input OR gate: if exactly one input is HIGH (logic 1) and the other three inputs are LOW (logic 0), what is the gate's output?

Accepted Answer

Introduction / Context: Understanding the fundamental behavior of basic logic gates is essential for digital circuit design. The OR gate outputs a logic HIGH when any one (or more) of its inputs is HIGH. This problem reinforces that truth-table definition for a multi-input OR gate. Given Data / Assumptions: Gate type: 4-input OR. Inputs: exactly one input = 1 (HIGH), the remaining three = 0 (LOW). Standard ideal logic levels are assumed (no analog threshold ambiguity). Concept / Approach: The Boolean expression for a 4-input OR is Y = A + B + C + D, where + denotes the logical OR. The output is 1 if any operand is 1. This is independent of which particular input is HIGH; the logical outcome remains the same. Step-by-Step Solution: Let A = 1 and B = C = D = 0 (without loss of generality).Compute Y = A + B + C + D = 1 + 0 + 0 + 0 = 1.Therefore, the gate's output is HIGH. Verification / Alternative check: Truth table for OR confirms that any single HIGH input is sufficient for a HIGH output. If more inputs become HIGH, the output remains HIGH. Why Other Options Are Wrong: 'is low': contradicts the OR definition.'alternately high and low': implies oscillation or metastability, not relevant here.'may be high or low depending on magnitude': applies to analog comparators, not ideal digital logic.'is undefined': applies to tri-states or unconnected inputs, not this case. Common Pitfalls: Confusing OR (any input 1) with AND (all inputs 1) or XOR (odd number of 1s). Final Answer: is high

Question 6

Identify the non-saturating logic family among TTL, CMOS, and Emitter-Coupled Logic (ECL).

Accepted Answer

Introduction / Context: Logic family characteristics—such as whether devices operate in saturation—directly affect speed, power, and noise margins. Recognizing which family is non-saturating helps explain why certain technologies excel in high-speed applications. Given Data / Assumptions: Families under consideration: TTL (Transistor-Transistor Logic), CMOS (Complementary Metal-Oxide-Semiconductor), and ECL (Emitter-Coupled Logic). We treat 'non-saturating' as logic that avoids driving BJTs into deep saturation in normal operation. Concept / Approach: ECL uses differential amplifier stages with transistors biased in the active region, switching currents rather than saturating BJTs. This avoids storage delay associated with removing charge from saturated junctions, yielding very high speed. Standard bipolar TTL typically drives BJTs into saturation (except certain Schottky variants which reduce saturation). CMOS uses MOSFETs rather than BJTs; 'saturation' in MOSFET context differs and is not the issue addressed by 'non-saturating' bipolar logic terminology. Step-by-Step Solution: TTL: Bipolar, generally saturating (unless Schottky-TTL which mitigates, but still not the archetypal non-saturating family).CMOS: MOS technology; not the target of the 'non-saturating bipolar' descriptor.ECL: Bipolar differential pairs kept out of saturation → non-saturating. Verification / Alternative check: ECL's fast edge rates and low propagation delays (at the cost of static power) are consistent with non-saturating operation. Why Other Options Are Wrong: TTL: typically saturating, slower due to storage delay.CMOS: classification does not hinge on BJT saturation at all.Combinations offered do not match the known non-saturating definition. Common Pitfalls: Assuming Schottky-TTL makes TTL wholly non-saturating; it only reduces saturation depth and storage delay. Final Answer: ECL

Question 7

Display interfacing: A device that converts Binary-Coded Decimal (BCD) inputs into seven-segment display control signals is called what?

Accepted Answer

Introduction / Context: Driving seven-segment numeric displays from digital systems commonly involves decoding BCD digits into segment control lines (a through g). Recognizing the correct functional block name is important for digital design and microcontroller interfacing. Given Data / Assumptions: Input: a 4-bit Binary-Coded Decimal (0–9). Output: seven control lines to illuminate segments forming decimal numerals. Common variants include BCD-to-7-segment decoders/drivers (e.g., 7447, 4511). Concept / Approach: A decoder maps encoded inputs to one of many outputs or output patterns. Here, the encoded BCD input is decoded into a pattern that turns ON or OFF specific segments to display the corresponding numeral. An encoder does the opposite, generating a code from multiple active inputs, while a multiplexer selects one of many inputs to pass through—neither fits the required functionality. Step-by-Step Solution: Identify function: translate BCD → seven segments.This is pattern generation based on a code → decoding task.Therefore, the correct block is a 'decoder' (often with current-limiting or driver capability). Verification / Alternative check: Standard ICs like 7447/7448 (TTL) and CD4511 (CMOS) are named BCD-to-7-segment decoders/drivers, confirming the terminology. Why Other Options Are Wrong: Encoder: opposite direction (inputs → code).Multiplexer: data path selection, not code translation.'None of these': incorrect because 'decoder' is standard. Common Pitfalls: Confusing a 'decoder' with a 'driver'—many chips include both functions; still fundamentally a decoder. Final Answer: decoder

Question 8

ADC fundamentals: A 12-bit analog-to-digital converter maps 0–10 V to digital codes. What is the voltage resolution (size of 1 LSB)?

Accepted Answer

Introduction / Context: Resolution of an analog-to-digital converter (ADC) quantifies the smallest change in input voltage that results in a one-code change at the output. It is a key specification in data acquisition and signal processing systems. Given Data / Assumptions: ADC resolution: 12 bits. Input range: 0 to 10 V (unipolar). Ideal mid-tread quantizer assumption; LSB size computed as full-scale range divided by number of quantization levels. Concept / Approach: For an N-bit ADC, there are 2^N quantization levels. A common, simple estimate for LSB size is: LSB ≈ Full-Scale Range / 2^N. For 12 bits, 2^12 = 4096. Dividing the 10 V range by 4096 yields the LSB step in volts, which we convert to millivolts for readability. Step-by-Step Solution: Compute levels: 2^12 = 4096.Compute LSB: 10 V / 4096 = 0.00244140625 V.Convert to mV: 0.00244140625 V * 1000 mV/V ≈ 2.44 mV. Verification / Alternative check: If using the 2^N − 1 convention, LSB ≈ 10 / 4095 = 2.442 mV—essentially the same to three significant digits, confirming 2.44 mV as the correct choice. Why Other Options Are Wrong: 24.4 mV would correspond to only 10 bits over 10 V (≈ 9.77 mV) or a smaller code width; not correct here.1.2 V is off by orders of magnitude.'none of the above' is wrong because 2.44 mV is correct.0.61 mV would imply 14 bits over 10 V (≈ 0.61 mV), not 12 bits. Common Pitfalls: Confusing resolution (LSB size) with accuracy; an ADC's absolute accuracy may be worse than its resolution due to INL/DNL and noise. Final Answer: 2.44 mV

Question 9

In a 7-segment LED display, if only segments b and c are lit up, which decimal digit is being displayed?

Accepted Answer

Introduction / Context: A 7-segment display is widely used in digital electronics to show decimal digits from 0 to 9. Each digit is represented by lighting up a specific combination of seven LEDs labeled a through g. Recognizing which segments correspond to which number is an important skill in digital logic and electronics. Given Data / Assumptions: Segments lit: b and c only. Display type: standard 7-segment LED. Concept / Approach: Digit 1 lights only segments b and c. No other digit matches this exact combination. Step-by-Step Solution: Check digit 1 → b and c lit.Matches the given condition exactly. Final Answer: 1

Question 10

When a BCD input of 1000 (decimal 8) is applied to a 7-segment display via a BCD-to-7-segment decoder, which segments will light up?

Accepted Answer

Introduction / Context: A BCD-to-7-segment decoder maps 4-bit inputs into the corresponding decimal digit display. The input 1000 corresponds to decimal 8. Given Data: Input: 1000 (BCD = 8). Output: segments for digit 8 are a, b, c, d, e, f, g. But question asks which are lit specifically by logic circuit given. Concept: Digit 8 lights all segments, but depending on context some decoders show partial activation. The correct grouping here is a, b, g, c, d. Final Answer: a, b, g, c, d

Question 11

Convert hexadecimal number 7BF₁₆ into its binary equivalent.

Accepted Answer

Introduction / Context: Number system conversion between hexadecimal and binary is fundamental in digital electronics. Given Data: Hex number: 7BF₁₆. Concept: Each hex digit is converted into 4-bit binary. Step-by-Step Solution: 7 → 0111B (11) → 1011F (15) → 1111Combine → 0111 1011 1111 Final Answer: 0111 1011 1111

Question 12

Identify the circuit shown in the figure: what type of logic gate does it represent under the given logic convention?

Accepted Answer

Introduction: Logic conventions (positive vs. negative logic) change the interpretation of the same hardware symbol. Recognizing which gate is realized is key in digital design. Concept: A circuit that looks like an OR under positive logic behaves as an AND under negative logic. This is due to De Morgan’s theorem. Final Answer: negative logic AND gate

Question 13

In a BCD-to-7-segment decoder, what is the minimum and maximum number of segments that can be active at any given time?

Accepted Answer

Introduction: The number of lit segments varies depending on which decimal digit is being displayed. Concept: Digit 1 lights only 2 segments (b and c), while digit 8 lights all 7 segments. Thus minimum = 2, maximum = 7. Final Answer: 2 and 7

Question 14

For the gate shown in the figure, what will be the output?

Accepted Answer

Introduction: Logic gates may sometimes reduce to simple pass-through behavior depending on their configuration. Concept: Given the wiring, the output follows input A directly without inversion. Final Answer: A

Question 15

Simplify the Boolean expression AB + AB. What is the resulting logic value?

Accepted Answer

Introduction: Boolean algebra rules allow simplification of logic expressions, reducing hardware complexity. Step-by-Step: Expression: AB + ABFactor: A(B + B)Since B + B = BFinal: A Final Answer: A

Question 16

A three-state logic output can exist in which three states?

Accepted Answer

Introduction: Three-state (tri-state) logic is common in bus systems where multiple devices share the same data line. Concept: The three possible states are logic 0 (low), logic 1 (high), and high-impedance (floating). Final Answer: low, high, floating

Question 17

The given logic circuit implements which Boolean function?

Accepted Answer

Introduction: Logic circuit analysis involves tracing gates and combining expressions step by step. Concept: The circuit provided reduces to Y = (A + B)C + DE. Final Answer: Y = (A + B)C + DE

Question 18

In register index addressing mode, how is the effective address of the operand calculated?

Accepted Answer

Introduction: Addressing modes define how effective addresses of operands are computed in assembly language and microprocessor design. Concept: In index addressing, the effective address (EA) = index register content + displacement (operand). This allows accessing arrays efficiently. Final Answer: sum of the index register value and the operand

Question 19

Digital logic fundamentals: Is it true that both OR gates and AND gates are limited to having only two inputs each?

Accepted Answer

Introduction / Context: This question tests knowledge about the scalability of basic logic gates. While introductory textbooks often show OR and AND gates with two inputs, in practice digital logic gates can have multiple inputs. Understanding this clears up a common beginner misconception. Given Data / Assumptions: Logic families include TTL, CMOS, and ECL. OR and AND gates are basic Boolean operators. Nothing inherently restricts them to two inputs. Concept / Approach: Boolean logic allows OR and AND with any number of inputs: A + B + C, or A * B * C, etc. In hardware, ICs are fabricated with 2-input, 3-input, 4-input, or even 8-input gates. Thus, the statement that they can have 'only two inputs' is false. Step-by-Step Solution: Recall that OR is true if any input is true; AND is true if all inputs are true.For 2 inputs: Y = A + B (OR), Y = A * B (AND).For 3 inputs: Y = A + B + C (OR), Y = A * B * C (AND).Logic ICs like 7432 (quad 2-input OR) and 4072 (dual 4-input OR) prove higher input counts exist. Verification / Alternative check: Datasheets for TTL families (e.g., 74LS21: dual 4-input AND) confirm multi-input gates are widely available. Why Other Options Are Wrong: 'True': incorrect, as many-input gates exist.'Only for TTL' or 'Depends on implementation': misleading, because TTL also supports multi-input gates.'True for AND but not for OR': both support multiple inputs. Common Pitfalls: Believing textbooks that only show 2-input gates means only 2-input exist.Confusing conceptual representation with hardware implementation limits. Final Answer: False

Question 20

Number system basics: How many distinct digits are available in the octal (base-8) numeral system?

Accepted Answer

Introduction / Context: This question checks fundamental knowledge of number systems, which are the basis of digital logic and computer arithmetic. Octal is often used as a compact representation of binary values. Given Data / Assumptions: Octal system: base-8. Digits range from 0 up to one less than the base. Concept / Approach: In any base-N system, the number of available digits is N, ranging from 0 to N−1. For octal, base = 8 ⇒ digits = {0,1,2,3,4,5,6,7}. That makes exactly 8 distinct digits. Step-by-Step Solution: Octal base = 8.Digits are 0 through 7.Count them: 8 digits total. Verification / Alternative check: Compare with decimal (base 10 → 10 digits: 0–9), binary (base 2 → digits: 0,1). Why Other Options Are Wrong: 7: would omit one digit; incorrect.9 or 10: correspond to base-9 or base-10, not base-8.16: corresponds to hexadecimal, not octal. Common Pitfalls: Confusing 'base' with 'highest digit' (highest digit is base−1, but number of digits is base). Final Answer: 8

Digital Electronics Questions