Q: TTL output stage design — purpose of the totem-pole Why does the TTL logic family use a totem-pole output structure on many gates?

Introduction / Context: Transistor–transistor logic (TTL) commonly employs a totem-pole output. Recognizing its purpose is key to understanding fan-out, rise/fall times, and why wired-OR is not permitted on standard TTL outputs. Given Data / Assumptions: Standard TTL gate with a push-pull (totem-pole) pair at the output. Comparison to open-collector style outputs. Focus on drive capability and switching speed. Concept / Approach: A totem-pole stage stacks two transistors so one sources current (active pull-up) while the other sinks current (active pull-down). Only one conducts at a time, yielding low output impedance in both logic states. This reduces rise and fall times markedly compared with passive pull-ups or a single active device. Step-by-Step Solution: When output = HIGH → upper transistor sources current → fast rising edges.When output = LOW → lower transistor sinks current → fast falling edges.Net effect: strong drive in both directions, improving speed and noise immunity. Verification / Alternative check: Contrast with open-collector outputs that require an external resistor for pull-up; these are slower due to RC charging and allow wired-OR. Totem-pole forbids tying outputs together because two active drivers would contend (short conflict). Why Other Options Are Wrong: Active pull-up only / active pull-down only: incomplete; totem-pole provides both. Enables wired-OR: incorrect; totem-pole prevents wired-OR due to contention. Eliminates output resistors: not the core reason; some series limiting may still exist. Common Pitfalls: Assuming all TTL is totem-pole; open-collector variants exist for wired-OR and level interfacing. Forgetting that bus sharing requires tri-state or open-collector, not totem-pole. Final Answer: It provides active pull-up and active pull-down.

Q: Reading digital waveforms — identifying the correct diagram Which graphical tool is used to show the time relationship (relative timing, phase, and duration) among two or more digital waveforms?

Introduction / Context: Debugging digital systems requires clear visualization of signals over time. Designers rely on specific diagrams to capture edges, widths, setup/hold, and sequencing among multiple lines (clocks, enables, data). Given Data / Assumptions: Multiple digital signals must be compared. We are interested in timing, not only static logic states. Waveforms include transitions, pulse widths, and relative alignment. Concept / Approach: A timing diagram plots voltage versus time for several signals on a shared horizontal time axis. It clarifies when one signal changes relative to another and is indispensable for verifying setup/hold times, propagation delays, and bus protocols. Step-by-Step Solution: Identify the need: show transitions and relative positions over time.Tool that does this: the timing diagram.Other listed items either depict static relationships or analog behaviors (e.g., load line). Verification / Alternative check: Oscilloscope captures and logic analyzer displays can be exported or redrawn as timing diagrams. Datasheets specify timing parameters directly on such diagrams, confirming their central role in digital design. Why Other Options Are Wrong: Reference diagram: generic term, not specific to time relationships. Voltage curve: typically a single analog signal vs. time; not multi-signal timing. Load line: DC analysis tool for transistor biasing, not time-domain logic. State transition table: enumerates logic states, not continuous timing. Common Pitfalls: Confusing block/state diagrams with timing diagrams; the latter are explicitly time-based. Ignoring scale: timing diagrams require accurate time axes to verify constraints. Final Answer: timing diagram

Q: CMOS vs. TTL — key advantage in typical digital designs What is the principal advantage of using CMOS logic families compared with TTL logic families under similar operating conditions?

Introduction / Context: Choosing a logic family affects speed, power, noise margins, and system cost. Complementary MOS (CMOS) and TTL have different transistor technologies and therefore different strengths. This item targets the most widely cited advantage of CMOS. Given Data / Assumptions: Standard low-power operation at moderate frequencies. Supply voltage appropriate to the family (e.g., 3.3 V or 5 V). Static versus dynamic power characteristics are considered. Concept / Approach: CMOS draws negligible static power when gates are not switching (idealized case: only leakage). Power consumption is primarily dynamic: P ≈ C_load * V^2 * f. TTL, by contrast, has higher static currents due to bipolar junction operation, leading to greater standby and overall power usage at comparable clock rates. Step-by-Step Solution: Assess steady-state: CMOS ≈ near-zero static power; TTL consumes steady current.Assess switching: CMOS power scales with V^2 and frequency; still typically lower than TTL for many applications.Conclusion: lower power is the key advantage. Verification / Alternative check: Datasheets for modern HC/HCT/AC CMOS show microampere quiescent currents, versus milliampere-level ICC for classic TTL at 5 V, confirming substantial power savings in CMOS designs. Why Other Options Are Wrong: It switches faster: not universally true; modern CMOS can be fast, but speed depends on subfamily. It is larger: CMOS ICs are not inherently larger physically. Cost: depends on family and vendor; not the universal advantage. No decoupling capacitors: all digital ICs need decoupling for transient currents. Common Pitfalls: Assuming CMOS is always faster than TTL; check timing specs for the specific subfamily. Ignoring input protection—CMOS is ESD-sensitive and requires proper handling. Final Answer: It uses less power.

Q: Converting period to frequency for a digital clock If a clock waveform has a period of 1.25 µs, what is the corresponding frequency in hertz?

Introduction / Context: Clock specification conversions between period (time per cycle) and frequency (cycles per second) are routine in digital design, timing analysis, and interface selection. Given Data / Assumptions: Clock period T = 1.25 microseconds. Frequency f = 1 / T. Unit conversions: 1 µs = 10^-6 s. Concept / Approach: The inverse relationship f = 1/T directly gives frequency once the period is expressed in seconds. Carefully convert microseconds to seconds to avoid thousand- or million-fold mistakes. Step-by-Step Solution: T = 1.25 µs = 1.25 × 10^-6 sf = 1 / T = 1 / (1.25 × 10^-6) HzCompute: 1 / 1.25 = 0.8; and 1 / 10^-6 = 10^6 → f = 0.8 × 10^6 HzTherefore f = 0.8 MHz. Verification / Alternative check: Check scale: a period near 1 µs corresponds to near 1 MHz. At 1.25 µs, the frequency should be slightly less than 1 MHz, consistent with 0.8 MHz. Why Other Options Are Wrong: 8 kHz / 0.8 kHz: off by factors of 10^2–10^3 due to incorrect microsecond conversion. 8 MHz: corresponds to a 0.125 µs period, not 1.25 µs. 12.5 MHz: corresponds to 0.08 µs, far smaller period. Common Pitfalls: Mixing micro (10^-6) with milli (10^-3). Inverting incorrectly: using f = T instead of f = 1/T. Final Answer: 0.8 MHz

Q: Serial vs. parallel — key wiring advantage What is the primary advantage of serial data transmission compared with parallel transmission for interconnections between digital systems?

Introduction / Context: As data rates increased and links grew longer, serial interconnects (e.g., USB, PCIe, SATA, Ethernet) displaced wide parallel cables in many applications. This question emphasizes the wiring economy of serial links. Given Data / Assumptions: Comparison at system or inter-board level, not inside a short PCB bus. Modern differential serial signaling (twisted pair) is assumed. Clock recovery or embedded clock is available for serial. Concept / Approach: Serial links reduce conductor count drastically because data travels one bit at a time over a differential pair (or even a single wire in some protocols). With proper encoding and equalization, serial channels can run at very high bit rates over relatively long distances with fewer wires and smaller connectors. Step-by-Step Solution: Parallel needs N conductors for N-bit width plus grounds and control lines.Serial needs just one pair (or a few) regardless of data width; the width is temporal, not spatial.Conclusion: “Only one pair of wires is required” captures the key interconnect advantage. Verification / Alternative check: Compare a classic 25-pin parallel cable to a two-pair Ethernet cable; or PCI (parallel) to PCIe (serial lanes). The reduction in pins and improved signal integrity are compelling reasons for serial adoption. Why Other Options Are Wrong: It is slower: modern serial links are extremely fast; “slower” is not inherent. More people use it: not a technical reason. It is faster: sometimes true vs. a given parallel link, but the defining advantage is wiring simplicity. No clock reference: serial links use embedded clocks or explicit references. Common Pitfalls: Assuming serial is always faster; bandwidth depends on protocol and encoding. Ignoring latency; some serial protocols add link-layer overhead. Final Answer: Only one pair of wires is required.

Q: Parallel transfer timing — 8-bit word at 2.25 MHz The hexadecimal number 4B (8-bit word) is sent in parallel . If the system clock is 2.25 MHz, how long does one parallel word transfer take?

Introduction / Context: Parallel buses transfer an entire word per clock. Understanding the relationship between clock frequency and transfer time is essential for bus bandwidth calculations and timing budgeting. Given Data / Assumptions: Data width is 8 bits (hex 4B), but width does not affect time per transfer in parallel—one word per clock. Clock frequency f_clk = 2.25 MHz. Assume one clock per transfer (no wait states). Concept / Approach: In parallel transfer, the time to send one word typically equals one clock period. Period T = 1 / f_clk. Convert units carefully from megahertz to seconds per cycle. Step-by-Step Solution: f_clk = 2.25 MHz = 2.25 × 10^6 HzT = 1 / f_clk = 1 / (2.25 × 10^6) sNumeric: 1 / 2.25 ≈ 0.444... microsecondsTherefore, T ≈ 0.444 µs = 444 ns per word transfer. Verification / Alternative check: Quick scale check: A 1 MHz clock has T = 1 µs. At 2.25 MHz, period should be ~0.44 µs, matching 444 ns. Why Other Options Are Wrong: 444 µs / 3.55 ms: off by factors of 10^3–10^6 due to unit mistakes. 3.55 µs: corresponds to ~281 kHz, not 2.25 MHz. 2.25 ns: that would be a 444 MHz clock, not 2.25 MHz. Common Pitfalls: Confusing serial timing (bits per second) with parallel word timing. Including protocol overhead; the problem states a direct per-clock transfer. Final Answer: 444 ns

Q: For a P–N diode to conduct in the forward direction, which polarity must be applied to the anode and to the cathode?

Introduction / Context: Understanding diode biasing is essential for rectifiers, clippers, clampers, and switching circuits. Forward biasing a diode allows current to flow easily from anode to cathode, while reverse biasing blocks current (except for a small leakage current). Given Data / Assumptions: Ideal P–N junction diode model is assumed. Forward bias condition lowers the junction barrier. Real silicon diodes exhibit approximately 0.7 V drop when forward-conducting at moderate current. Concept / Approach: A diode conducts when its P-side (anode) is at a higher electric potential than its N-side (cathode). Equivalently, the cathode must be more negative than the anode so that carriers overcome the junction barrier and the device conducts. Step-by-Step Solution: 1) Identify terminals: anode (P-type), cathode (N-type).2) Forward bias rule: anode at higher potential than cathode.3) Translate to polarities: positive on anode, negative on cathode.4) Conclude that the correct statement is “negative on the cathode and positive on the anode.” Verification / Alternative check: In rectifier circuits, the diode conducts when the anode side of the AC cycle goes positive with respect to the cathode. This practical observation matches the forward-bias rule. Why Other Options Are Wrong: Negative on the anode and positive on the cathode: this is reverse bias.Positive on both terminals: does not guarantee forward bias; polarity difference matters.Negative on both terminals: again, no forward conduction is guaranteed.High-frequency alternation only: frequency does not change the basic biasing requirement. Common Pitfalls: Confusing the arrow on diode symbols (used for transistors) or thinking the stripe mark (cathode) should be tied to positive for conduction. It is the potential difference that matters: anode must be more positive than cathode. Final Answer: Negative on the cathode and positive on the anode (forward bias)

Question 1

Transistor switches vs. electromechanical relays — why are transistors preferred in most applications? Select the best overall reason set for using transistor switches instead of relays in electronic control.

Accepted Answer

Introduction / Context: Switching elements are everywhere in electronics. While relays still have niche uses for isolation and high-voltage/AC loads, semiconductor switches (BJTs, MOSFETs) dominate low-voltage digital and power electronics. This question asks for the comprehensive benefits that make transistors the default choice. Given Data / Assumptions: Application context: low-voltage DC or logic-level switching. Comparison dimensions: power, speed, size/noise. Concept / Approach: Relays are mechanical; they require coil current, have millisecond switching times, audible clicks, contact bounce, wear, and larger size. Transistors are solid-state; they consume little control power, switch in nanoseconds to microseconds, are silent, compact, and highly reliable with no moving parts. Step-by-Step Solution: Evaluate power: transistor gate/base drive is minimal compared to coil power → less power.Evaluate speed: solid-state devices switch orders of magnitude faster than relay contacts.Evaluate acoustics/size: transistor switches are silent and tiny compared to relays.Therefore the most complete choice is “all of the above.” Verification / Alternative check: Datasheets: a small MOSFET gate charge vs. a relay coil steady current confirms lower drive power; turn-on/off delay figures show massive speed advantage for semiconductors. Why Other Options Are Wrong: Single-benefit options omit other key advantages. Relays still excel for galvanic isolation or when switching AC mains; however, the question asks in most applications, where transistors win across multiple metrics. Common Pitfalls: Ignoring contact bounce and lifetime limitations of relays. Overlooking that MOSFETs need appropriate gate drive and protection (e.g., flyback diodes for inductive loads). Final Answer: all of the above

Question 2

Why is a NOT-gate circuit called an “inverter”? In digital electronics, explain the reason this basic gate is termed an inverter.

Accepted Answer

Introduction / Context: The NOT gate is one of the fundamental building blocks of digital logic. It is commonly called an inverter. Understanding why it carries this name reinforces the concept of logical complement and its role in creating more complex functions. Given Data / Assumptions: Binary logic levels: 0 (LOW) and 1 (HIGH). Single-input, single-output device. Concept / Approach: An inverter performs logical negation. It maps 0 to 1 and 1 to 0. In Boolean algebra terms, if the input is A, the output is X = A'. The action “inverts” or complements the input, hence the name inverter. This complement operation is essential in implementing De Morgan transformations, generating control signals, and constructing NAND/NOR-based universal logic. Step-by-Step Solution: Define behavior: if A = 0 then X = 1; if A = 1 then X = 0.Relate to Boolean notation: X = NOT(A).Interpret naming: output is the inverse of the input → inverter. Verification / Alternative check: Truth table check confirms the inversion for all input states. Timing diagrams show that whenever the input toggles, the output toggles to the opposite level after the device propagation delay. Why Other Options Are Wrong: “In phase with the input” and “the same as the input” describe buffers, not inverters. “There is no output” is incorrect; an inverter produces a well-defined output logic state. Common Pitfalls: Confusing an inverter with a Schmitt-trigger inverter; both invert, but Schmitt adds hysteresis for noisy inputs. Overlooking propagation delay; inversion is not instantaneous in real circuits. Final Answer: The output is the opposite (inverse) of the input.

Question 3

Digital communication basics — parallel vs. serial Why is parallel data transmission generally preferred over serial transmission in many short-distance digital system applications?

Accepted Answer

Introduction / Context: Parallel and serial are two fundamental methods of moving digital data between subsystems. This question checks conceptual understanding of why parallel transmission is often chosen inside a device or over very short distances, such as between a microprocessor and memory or peripheral buses on a PCB. Given Data / Assumptions: Comparison is for typical short-run interconnects on a board or backplane. Signal integrity and skew are manageable over short distances. Clocking or strobing is available to coordinate transfers. Concept / Approach: Parallel transmission uses multiple conductors to carry several bits simultaneously (often 8, 16, 32, or more). Serial sends bits sequentially over one pair. For a given edge rate and clock, parallel buses can deliver a higher aggregate throughput because multiple bits traverse per cycle. Step-by-Step Solution: Throughput_parallel = bus_width_bits * f_clockThroughput_serial = 1 * f_bit (bits per second over one line)At the same toggling speed, a 16-bit parallel bus moves 16 bits each tick, vastly exceeding one serial line at the same frequency.Within a chassis, added conductors and connectors are acceptable, so bandwidth per clock strongly favors parallel. Verification / Alternative check: Historical buses (e.g., classic memory buses, PCI) used wide parallel paths to achieve high bandwidth before differential high-speed serial links matured. Even today, internal memory interfaces (like DDR) remain parallel for extreme bandwidth per pin group. Why Other Options Are Wrong: It is much slower: incorrect; parallel is typically faster at short distances. It is cheaper: extra traces, pins, and connectors usually increase cost. More people use it: popularity does not determine technical preference. It avoids clocking altogether: parallel buses still require timing references or strobes. Common Pitfalls: Overgeneralizing: over long distances, crosstalk and skew make serial preferable. Ignoring signal integrity: as speeds rise, parallel timing skew can limit scalability. Final Answer: It is much faster.

Question 4

BJT operating regions — identifying cutoff in an NPN transistor Which condition best indicates that an NPN BJT is in cutoff (no conduction) between collector and emitter?

Accepted Answer

Introduction / Context: Understanding BJT operating regions—cutoff, active, and saturation—is essential for analog biasing and digital switching. This question asks you to identify the signature condition of cutoff in an NPN transistor. Given Data / Assumptions: NPN transistor in a typical biasing configuration. Cutoff means no base drive sufficient to forward-bias the base–emitter junction. Voltages are referenced in standard silicon BJT terms. Concept / Approach: In cutoff, the base–emitter junction is not forward biased; as a result, collector current (IC) is approximately zero and the device behaves like an open switch between collector and emitter. VCE typically rises toward the supply because there is no conduction path to drop voltage across the collector resistor. Step-by-Step Solution: Cutoff condition: VBE < ~0.6–0.7 V (for silicon) → IC ≈ 0.With IC ≈ 0, the collector node floats up (through its load) toward supply → device looks open C–E.Therefore, the best descriptive statement is that the collector–emitter path appears open. Verification / Alternative check: Compare to saturation: in saturation, VCE(sat) is low (typically 0.1–0.3 V) and IC is relatively large for the circuit’s load. In active region, VBE ≈ 0.7 V and IC = β * IB with VCE moderate. Neither matches cutoff behavior. Why Other Options Are Wrong: Collector current is maximum: describes saturation, not cutoff. VCE = 0 V: also indicates saturation (short-like behavior), not cutoff. VBE = 0.7 V: implies a forward-biased base–emitter junction (active or saturation), not cutoff. Base current very high: opposite of cutoff. Common Pitfalls: Confusing high VCE in cutoff with a fault; high VCE is normal when no current flows. Assuming VBE must be exactly 0.7 V; it varies with device and current. Final Answer: collector to emitter appears to be open

Question 5

TTL output stage design — purpose of the totem-pole Why does the TTL logic family use a totem-pole output structure on many gates?

Accepted Answer

Introduction / Context: Transistor–transistor logic (TTL) commonly employs a totem-pole output. Recognizing its purpose is key to understanding fan-out, rise/fall times, and why wired-OR is not permitted on standard TTL outputs. Given Data / Assumptions: Standard TTL gate with a push-pull (totem-pole) pair at the output. Comparison to open-collector style outputs. Focus on drive capability and switching speed. Concept / Approach: A totem-pole stage stacks two transistors so one sources current (active pull-up) while the other sinks current (active pull-down). Only one conducts at a time, yielding low output impedance in both logic states. This reduces rise and fall times markedly compared with passive pull-ups or a single active device. Step-by-Step Solution: When output = HIGH → upper transistor sources current → fast rising edges.When output = LOW → lower transistor sinks current → fast falling edges.Net effect: strong drive in both directions, improving speed and noise immunity. Verification / Alternative check: Contrast with open-collector outputs that require an external resistor for pull-up; these are slower due to RC charging and allow wired-OR. Totem-pole forbids tying outputs together because two active drivers would contend (short conflict). Why Other Options Are Wrong: Active pull-up only / active pull-down only: incomplete; totem-pole provides both. Enables wired-OR: incorrect; totem-pole prevents wired-OR due to contention. Eliminates output resistors: not the core reason; some series limiting may still exist. Common Pitfalls: Assuming all TTL is totem-pole; open-collector variants exist for wired-OR and level interfacing. Forgetting that bus sharing requires tri-state or open-collector, not totem-pole. Final Answer: It provides active pull-up and active pull-down.

Question 6

Reading digital waveforms — identifying the correct diagram Which graphical tool is used to show the time relationship (relative timing, phase, and duration) among two or more digital waveforms?

Accepted Answer

Introduction / Context: Debugging digital systems requires clear visualization of signals over time. Designers rely on specific diagrams to capture edges, widths, setup/hold, and sequencing among multiple lines (clocks, enables, data). Given Data / Assumptions: Multiple digital signals must be compared. We are interested in timing, not only static logic states. Waveforms include transitions, pulse widths, and relative alignment. Concept / Approach: A timing diagram plots voltage versus time for several signals on a shared horizontal time axis. It clarifies when one signal changes relative to another and is indispensable for verifying setup/hold times, propagation delays, and bus protocols. Step-by-Step Solution: Identify the need: show transitions and relative positions over time.Tool that does this: the timing diagram.Other listed items either depict static relationships or analog behaviors (e.g., load line). Verification / Alternative check: Oscilloscope captures and logic analyzer displays can be exported or redrawn as timing diagrams. Datasheets specify timing parameters directly on such diagrams, confirming their central role in digital design. Why Other Options Are Wrong: Reference diagram: generic term, not specific to time relationships. Voltage curve: typically a single analog signal vs. time; not multi-signal timing. Load line: DC analysis tool for transistor biasing, not time-domain logic. State transition table: enumerates logic states, not continuous timing. Common Pitfalls: Confusing block/state diagrams with timing diagrams; the latter are explicitly time-based. Ignoring scale: timing diagrams require accurate time axes to verify constraints. Final Answer: timing diagram

Question 7

CMOS vs. TTL — key advantage in typical digital designs What is the principal advantage of using CMOS logic families compared with TTL logic families under similar operating conditions?

Accepted Answer

Introduction / Context: Choosing a logic family affects speed, power, noise margins, and system cost. Complementary MOS (CMOS) and TTL have different transistor technologies and therefore different strengths. This item targets the most widely cited advantage of CMOS. Given Data / Assumptions: Standard low-power operation at moderate frequencies. Supply voltage appropriate to the family (e.g., 3.3 V or 5 V). Static versus dynamic power characteristics are considered. Concept / Approach: CMOS draws negligible static power when gates are not switching (idealized case: only leakage). Power consumption is primarily dynamic: P ≈ C_load * V^2 * f. TTL, by contrast, has higher static currents due to bipolar junction operation, leading to greater standby and overall power usage at comparable clock rates. Step-by-Step Solution: Assess steady-state: CMOS ≈ near-zero static power; TTL consumes steady current.Assess switching: CMOS power scales with V^2 and frequency; still typically lower than TTL for many applications.Conclusion: lower power is the key advantage. Verification / Alternative check: Datasheets for modern HC/HCT/AC CMOS show microampere quiescent currents, versus milliampere-level ICC for classic TTL at 5 V, confirming substantial power savings in CMOS designs. Why Other Options Are Wrong: It switches faster: not universally true; modern CMOS can be fast, but speed depends on subfamily. It is larger: CMOS ICs are not inherently larger physically. Cost: depends on family and vendor; not the universal advantage. No decoupling capacitors: all digital ICs need decoupling for transient currents. Common Pitfalls: Assuming CMOS is always faster than TTL; check timing specs for the specific subfamily. Ignoring input protection—CMOS is ESD-sensitive and requires proper handling. Final Answer: It uses less power.

Question 8

Converting period to frequency for a digital clock If a clock waveform has a period of 1.25 µs, what is the corresponding frequency in hertz?

Accepted Answer

Introduction / Context: Clock specification conversions between period (time per cycle) and frequency (cycles per second) are routine in digital design, timing analysis, and interface selection. Given Data / Assumptions: Clock period T = 1.25 microseconds. Frequency f = 1 / T. Unit conversions: 1 µs = 10^-6 s. Concept / Approach: The inverse relationship f = 1/T directly gives frequency once the period is expressed in seconds. Carefully convert microseconds to seconds to avoid thousand- or million-fold mistakes. Step-by-Step Solution: T = 1.25 µs = 1.25 × 10^-6 sf = 1 / T = 1 / (1.25 × 10^-6) HzCompute: 1 / 1.25 = 0.8; and 1 / 10^-6 = 10^6 → f = 0.8 × 10^6 HzTherefore f = 0.8 MHz. Verification / Alternative check: Check scale: a period near 1 µs corresponds to near 1 MHz. At 1.25 µs, the frequency should be slightly less than 1 MHz, consistent with 0.8 MHz. Why Other Options Are Wrong: 8 kHz / 0.8 kHz: off by factors of 10^2–10^3 due to incorrect microsecond conversion. 8 MHz: corresponds to a 0.125 µs period, not 1.25 µs. 12.5 MHz: corresponds to 0.08 µs, far smaller period. Common Pitfalls: Mixing micro (10^-6) with milli (10^-3). Inverting incorrectly: using f = T instead of f = 1/T. Final Answer: 0.8 MHz

Question 9

Relays vs. semiconductor switches — unique capability Compared with solid-state switching devices, what can an electromechanical relay provide between the control (coil) circuit and the switched load?

Accepted Answer

Introduction / Context: Choosing between a relay and a semiconductor switch involves tradeoffs in speed, isolation, lifetime, and current capability. This question focuses on capabilities relays uniquely provide or commonly exceed compared with typical semiconductor devices in basic applications. Given Data / Assumptions: Conventional electromagnetic relays with galvanic isolation between coil and contacts. Typical semiconductor switches (BJT, MOSFET, SSR) without transformers or opto-isolation added. General-purpose current and voltage levels. Concept / Approach: Relays offer galvanic isolation because the coil circuit is physically separate from the contact path. They also commonly switch higher voltages and currents (including AC) with robust contact ratings. Solid-state devices are faster and long-lived, but they do not inherently provide galvanic isolation unless paired with isolation components, and their current/voltage ratings depend on device selection. Step-by-Step Solution: Identify unique property: mechanical contacts separate from coil → total isolation.Typical advantage: many relays are rated for higher currents than small signal semiconductors.Thus the best choice is “total isolation and higher current rating.” Verification / Alternative check: Datasheets for power relays show dielectric strength between coil and contacts and contact ratings in amperes to tens of amperes. Standard logic-level MOSFETs can also handle high currents, but they do not provide galvanic isolation without an isolating driver stage. Why Other Options Are Wrong: Total isolation only: incomplete; higher current capability is also common. Faster: relays are slower than semiconductors. Higher current rating only: ignores the isolation advantage. Lower on-resistance than a MOSFET: false; modern MOSFETs have milliohm R_DS(on). Common Pitfalls: Ignoring AC switching; relays easily handle AC without polarity concerns. Overlooking contact wear; relays have finite mechanical life. Final Answer: total isolation and higher current rating

Question 10

Serial vs. parallel — key wiring advantage What is the primary advantage of serial data transmission compared with parallel transmission for interconnections between digital systems?

Accepted Answer

Introduction / Context: As data rates increased and links grew longer, serial interconnects (e.g., USB, PCIe, SATA, Ethernet) displaced wide parallel cables in many applications. This question emphasizes the wiring economy of serial links. Given Data / Assumptions: Comparison at system or inter-board level, not inside a short PCB bus. Modern differential serial signaling (twisted pair) is assumed. Clock recovery or embedded clock is available for serial. Concept / Approach: Serial links reduce conductor count drastically because data travels one bit at a time over a differential pair (or even a single wire in some protocols). With proper encoding and equalization, serial channels can run at very high bit rates over relatively long distances with fewer wires and smaller connectors. Step-by-Step Solution: Parallel needs N conductors for N-bit width plus grounds and control lines.Serial needs just one pair (or a few) regardless of data width; the width is temporal, not spatial.Conclusion: “Only one pair of wires is required” captures the key interconnect advantage. Verification / Alternative check: Compare a classic 25-pin parallel cable to a two-pair Ethernet cable; or PCI (parallel) to PCIe (serial lanes). The reduction in pins and improved signal integrity are compelling reasons for serial adoption. Why Other Options Are Wrong: It is slower: modern serial links are extremely fast; “slower” is not inherent. More people use it: not a technical reason. It is faster: sometimes true vs. a given parallel link, but the defining advantage is wiring simplicity. No clock reference: serial links use embedded clocks or explicit references. Common Pitfalls: Assuming serial is always faster; bandwidth depends on protocol and encoding. Ignoring latency; some serial protocols add link-layer overhead. Final Answer: Only one pair of wires is required.

Question 11

Parallel transfer timing — 8-bit word at 2.25 MHz The hexadecimal number 4B (8-bit word) is sent in parallel. If the system clock is 2.25 MHz, how long does one parallel word transfer take?

Accepted Answer

Introduction / Context: Parallel buses transfer an entire word per clock. Understanding the relationship between clock frequency and transfer time is essential for bus bandwidth calculations and timing budgeting. Given Data / Assumptions: Data width is 8 bits (hex 4B), but width does not affect time per transfer in parallel—one word per clock. Clock frequency f_clk = 2.25 MHz. Assume one clock per transfer (no wait states). Concept / Approach: In parallel transfer, the time to send one word typically equals one clock period. Period T = 1 / f_clk. Convert units carefully from megahertz to seconds per cycle. Step-by-Step Solution: f_clk = 2.25 MHz = 2.25 × 10^6 HzT = 1 / f_clk = 1 / (2.25 × 10^6) sNumeric: 1 / 2.25 ≈ 0.444... microsecondsTherefore, T ≈ 0.444 µs = 444 ns per word transfer. Verification / Alternative check: Quick scale check: A 1 MHz clock has T = 1 µs. At 2.25 MHz, period should be ~0.44 µs, matching 444 ns. Why Other Options Are Wrong: 444 µs / 3.55 ms: off by factors of 10^3–10^6 due to unit mistakes. 3.55 µs: corresponds to ~281 kHz, not 2.25 MHz. 2.25 ns: that would be a 444 MHz clock, not 2.25 MHz. Common Pitfalls: Confusing serial timing (bits per second) with parallel word timing. Including protocol overhead; the problem states a direct per-clock transfer. Final Answer: 444 ns

Question 12

In digital communications, the serial format for transmitting binary information uses how many conductors in the link, and what is the key idea behind this method?

Accepted Answer

Introduction / Context: Serial transmission is a foundational concept in digital electronics and data communications. This question checks whether you understand that, in a serial link, bits are sent one after another over a single signal path, rather than simultaneously over many paths. Given Data / Assumptions: We are discussing binary information transfer. Standard serial links (UART, SPI single-wire data line, I2C data line SDA, USB differential pair counted as a single logical channel) serialize bits in time. Physical layer specifics (electrical, optical, RF) may vary, but the logical idea remains: one channel carries the timed bit stream. Concept / Approach: Serial transmission sends data bit-by-bit in sequence across one communication path. This is contrasted with parallel transmission, where multiple conductors carry multiple bits at the same instant. Serial is favored for longer distances and high speeds because it reduces skew and cabling complexity. Step-by-Step Solution: 1) Define serial: a single logical channel conveys a timed sequence of bits.2) Count conductors for the data path: one signal conductor (plus a reference/return path) is sufficient for data.3) Recognize that higher layers (protocol/encoding) do not change the single-channel nature of serial data.4) Conclude that the correct choice is the one identifying a single conductor for the signal path. Verification / Alternative check: Compare with parallel buses (older printer ports, address/data buses) which require many conductors. Serial standards (RS-232 TX line, 1-Wire, half-duplex UART) still send bits on one signal conductor, confirming the principle. Why Other Options Are Wrong: Multiple conductors: describes a parallel bus, not serial.Infrared only: serial is not limited to IR; it can be electrical, optical, or RF.Fiber-optic ribbon per bit: that is a parallel optical concept, not serial. Common Pitfalls: Confusing the presence of a ground/return conductor with “multiple conductors for data.” The question focuses on the data signal path count, not the necessary return/reference path. Final Answer: A single conductor

Question 13

For a P–N diode to conduct in the forward direction, which polarity must be applied to the anode and to the cathode?

Accepted Answer

Introduction / Context: Understanding diode biasing is essential for rectifiers, clippers, clampers, and switching circuits. Forward biasing a diode allows current to flow easily from anode to cathode, while reverse biasing blocks current (except for a small leakage current). Given Data / Assumptions: Ideal P–N junction diode model is assumed. Forward bias condition lowers the junction barrier. Real silicon diodes exhibit approximately 0.7 V drop when forward-conducting at moderate current. Concept / Approach: A diode conducts when its P-side (anode) is at a higher electric potential than its N-side (cathode). Equivalently, the cathode must be more negative than the anode so that carriers overcome the junction barrier and the device conducts. Step-by-Step Solution: 1) Identify terminals: anode (P-type), cathode (N-type).2) Forward bias rule: anode at higher potential than cathode.3) Translate to polarities: positive on anode, negative on cathode.4) Conclude that the correct statement is “negative on the cathode and positive on the anode.” Verification / Alternative check: In rectifier circuits, the diode conducts when the anode side of the AC cycle goes positive with respect to the cathode. This practical observation matches the forward-bias rule. Why Other Options Are Wrong: Negative on the anode and positive on the cathode: this is reverse bias.Positive on both terminals: does not guarantee forward bias; polarity difference matters.Negative on both terminals: again, no forward conduction is guaranteed.High-frequency alternation only: frequency does not change the basic biasing requirement. Common Pitfalls: Confusing the arrow on diode symbols (used for transistors) or thinking the stripe mark (cathode) should be tied to positive for conduction. It is the potential difference that matters: anode must be more positive than cathode. Final Answer: Negative on the cathode and positive on the anode (forward bias)

Question 14

For a periodic digital waveform, the HIGH time is measured in which units?

Accepted Answer

Introduction / Context: Timing metrics are fundamental in digital electronics. A waveform’s HIGH time describes how long the signal remains at the logic-HIGH level during each period. Selecting the correct units reinforces the distinction between time and frequency quantities. Given Data / Assumptions: A periodic digital signal with defined HIGH level and LOW level durations. Standard definitions: period T, frequency f, pulse width t_w (HIGH time for a positive pulse). Concept / Approach: The HIGH time is a duration. Durations are measured in time units such as seconds, milliseconds, or microseconds. Frequency (hertz) measures how often events repeat per second and is the reciprocal of period. Percent (%) indicates a ratio (duty cycle), not a time quantity by itself. Step-by-Step Solution: 1) Identify what “HIGH time” means: duration of logic-HIGH level within one cycle.2) Duration implies time units → seconds (or scaled units ms, µs, ns).3) Recognize that hertz is cycles per second, not a duration.4) Duty cycle is a percentage, not a time measure.5) Therefore, the correct unit is seconds. Verification / Alternative check: Formula check: duty cycle D = t_high / T. Here t_high must be in seconds to be compatible with period T in seconds. This confirms that HIGH time is a time quantity. Why Other Options Are Wrong: Hertz: a rate (1/s), not a time.Cycles: dimensionless count, not a duration.%: a ratio; you still need t_high in seconds to compute D.Volt-seconds: used in magnetics, not appropriate for HIGH time units. Common Pitfalls: Mixing up duty cycle (%) with pulse width. Also, confusing frequency with time due to their inverse relationship; they carry different units and meanings. Final Answer: Seconds (time units)

Question 15

The frequency of the clock waveform is defined as the reciprocal of the clock period.

Accepted Answer

True

Question 16

A plot of voltage versus time is called a timing diagram.

Accepted Answer

True

Question 17

Most modern digital systems are based on semiconductor technology.

Accepted Answer

True

Question 18

Precise timing is not important in digital signals.

Accepted Answer

False

Question 19

Serial communications cannot be sped up.

Accepted Answer

False

Question 20

Manual switches and relays have identical ON and OFF resistances.

Accepted Answer

False

Signals and Switches Questions