Question 1

Electromechanical relays terminology The abbreviation ‘‘NO’’ stands for ‘‘not operational’’. Is this statement correct?

Accepted Answer

Introduction / Context:Relay and switch terminology is foundational in electronics and control systems. Misinterpreting abbreviations like NO and NC can lead to wiring errors, unexpected behavior, and safety issues. Given Data / Assumptions: Statement under test: NO means ‘‘not operational’’. Context: electromechanical relays and standard panel switches. Industry-standard abbreviations are assumed. Concept / Approach:In relay/switch nomenclature, NO stands for ‘‘normally open’’ and NC stands for ‘‘normally closed’’. ‘‘Normally’’ refers to the device’s de-energized (rest) state. A normally open contact does not conduct in the rest state and closes (conducts) when actuated. Step-by-Step Solution:1) Define NO: normally open → contact open (infinite resistance) when coil is not energized or actuator not pressed.2) Define NC: normally closed → contact closed (low resistance) in the rest state.3) Compare with the statement ‘‘not operational’’: this is not a recognized meaning for NO.4) Conclude the statement is incorrect. Verification / Alternative check:Check a relay datasheet or a control wiring diagram legend; NO is universally documented as ‘‘normally open’’, never ‘‘not operational’’. Why Other Options Are Wrong:True / It depends / Not enough information / Only for solid-state relays: All incorrect because the abbreviation definition is standardized and does not vary by relay type. Common Pitfalls:Confusing ‘‘NO’’ with ‘‘OFF’’ or assuming ‘‘normally’’ means ‘‘most of the time in use’’. It specifically means the unpowered, rest condition as defined by the manufacturer. Final Answer:False

Question 2

Switching devices and ideal ON/OFF behavior Which device(s) exhibit nearly ideal low ON resistance and extremely high OFF resistance over typical operating ranges?

Accepted Answer

Introduction / Context:Different switching technologies display different ON and OFF resistance characteristics. Understanding which devices approximate the ideal open circuit and short circuit assists in selecting components for precision switching, isolation, and low-leakage applications. Given Data / Assumptions: Compare electromechanical relays, manual switches, and semiconductor switches. ‘‘Ideal’’ ON means very low contact resistance; ‘‘ideal’’ OFF means extremely high resistance with negligible leakage. Normal operating voltages and currents; no contact contamination or arcing damage assumed. Concept / Approach:Metal-to-metal mechanical contacts (relays, toggles, pushbuttons) can approximate a short when closed and an open when opened. Semiconductor devices, even when ‘‘off’’, exhibit leakage currents and finite OFF resistance, and when ‘‘on’’ they have a non-zero voltage drop (e.g., diode forward drop, MOSFET Rds(on)). Step-by-Step Solution:1) Evaluate relays: contacts provide milliohm-level ON resistance and megaohm to gigaohm OFF resistance → near-ideal behavior.2) Evaluate manual switches: similar mechanical contacts → near-ideal ON/OFF behavior.3) Evaluate semiconductors: off-state leakage (microamps to milliamps) and on-state voltage drops or Rds(on) → not ideal.4) Therefore, devices that best approximate ideal behavior are relays and manual switches. Verification / Alternative check:Datasheets show relay contact resistance typically < 100 mΩ and insulation resistance > 100 MΩ, while MOSFETs specify finite Rds(on) and leakage Idss or Igss, confirming the relative differences. Why Other Options Are Wrong:Electromechanical relays only / Manual switches only: Each alone ignores the other mechanical option that also behaves nearly ideally.Semiconductor devices: Significant on-state drop and off-state leakage prevent ‘‘ideal’’ ON/OFF.All of the above: Incorrect because semiconductors do not meet the near-ideal criterion. Common Pitfalls:Overlooking contact resistance rise due to wear/oxidation, or assuming MOSFETs are ideal due to low Rds(on); they still have leakage and body diodes that impact isolation. Final Answer:Electromechanical relays and manual switches

Question 3

TTL IC power pins Must TTL logic ICs be connected to VCC and GND even if all inputs and outputs are tied to valid signals?

Accepted Answer

Introduction / Context:Digital integrated circuits require proper power and ground connections to bias internal transistors and reference logic levels. Forgetting these connections is a common assembly error that leads to nonfunctional or damaged devices. Given Data / Assumptions: Family: TTL (Transistor-Transistor Logic), typically powered at 5 V (VCC) with a ground reference. All inputs/outputs may appear ‘‘valid’’ when viewed schematically, but internal circuitry still needs bias power. Normal operating environment and standard TTL thresholds. Concept / Approach:TTL internal multi-emitter input transistors, bias networks, and output totem-pole stages require VCC and GND to establish the logic-level thresholds and drive capability. Without power pins, inputs cannot be interpreted, output stages cannot source/sink current, and parasitic paths may latch up or float. Step-by-Step Solution:1) Identify required pins: VCC (often pin 14 for small DIP TTL) and GND (often pin 7).2) Recognize function: Power pins provide biasing and reference levels for all internal gates.3) Conclude: Even with externally driven inputs and tied outputs, the IC will not operate without proper power and ground.4) Therefore, the statement that TTL chips must be powered regardless of I/O wiring is true. Verification / Alternative check:Consult any TTL datasheet pinout and absolute maximum ratings; operation is specified only with VCC applied and decoupled to GND. Why Other Options Are Wrong:False / Only for CMOS / Only if outputs unused / Only for high-speed: All ignore the fundamental need for internal transistor biasing and reference rails. Common Pitfalls:Neglecting decoupling capacitors on VCC-GND, which can cause noise-induced malfunctions; always place a small ceramic capacitor close to each TTL IC. Final Answer:True

Question 4

Plotting voltage versus time on axes In a V–t graph, which quantity is placed on which axis?

Accepted Answer

Introduction / Context:Oscilloscope displays, timing diagrams, and lab plots commonly present how a signal changes over time. Understanding axis conventions ensures correct interpretation of waveforms and measurements. Given Data / Assumptions: We are plotting voltage as a function of time (V vs. t). Standard Cartesian plot conventions apply. No special logarithmic or parametric plotting is implied. Concept / Approach:By convention, the independent variable is plotted on the horizontal axis (x-axis) and the dependent variable on the vertical axis (y-axis). When analyzing waveforms, time is the independent variable and the voltage value depends on time, so time goes on x and voltage goes on y. Step-by-Step Solution:1) Identify independent variable: time (t) → x-axis.2) Identify dependent variable: voltage (V) → y-axis.3) State the mapping: voltage on y, time on x.4) This matches oscilloscope practice: time base horizontal, vertical deflection proportional to voltage. Verification / Alternative check:Review any oscilloscope front panel: the horizontal control is the time/div setting, and the vertical control is volts/div, reflecting the same axis convention. Why Other Options Are Wrong:Time on y and voltage on x: Reverses the standard relationship.Either way; no convention: Incorrect—there is a widely accepted convention for functional plots and oscilloscopes.Power on y: Not relevant for a V–t graph. Common Pitfalls:Confusing specialized plots (e.g., X-Y mode on scopes for Lissajous figures) with standard time-domain displays; here the question explicitly states voltage versus time. Final Answer:voltage on the y-axis; time on the x-axis

Question 5

In practical electronics, a diode is often placed directly in parallel (reverse-biased with respect to the supply) across a relay coil. What is the primary purpose of this diode in the relay driver circuit?

Accepted Answer

Introduction / Context:Relays are inductive loads. When the driving switch or transistor turns off, the relay coil's magnetic field collapses and the inductor attempts to keep current flowing, producing a large voltage spike (inductive kick). A diode placed across the coil is commonly called a flyback (freewheel, snubber) diode and is essential for protecting the driving device and reducing contact arcing. Given Data / Assumptions: Inductive element: relay coil modeled as an inductor. Driver opens current path abruptly. Diode is wired in reverse polarity across the coil under normal operation (non-conducting when powered). Goal: identify the diode’s main protective function. Concept / Approach:Inductor voltage follows v = L * di/dt. When current is interrupted, di/dt is large and positive in magnitude, so the inductor generates a high voltage of opposite polarity to sustain current. The flyback diode provides a low-impedance loop for the decaying coil current, clamping the voltage near the diode drop and safely dissipating energy in the coil resistance and diode. Step-by-Step Solution:At turn-off, the coil current cannot instantly go to zero.Without a diode, the coil drives the switch node to a high voltage, causing arcing across mechanical contacts or overstress of a transistor.With the diode, current freewheels through the diode and coil, limiting voltage to approximately the supply plus one diode drop.Lower voltage stress means reduced arcing at the opening contacts and protection of semiconductor drivers. Verification / Alternative check:Scope measurements show that with no diode, switch node voltages can exceed hundreds of volts for a 12 V coil. With the diode installed, the node clamps near 12–13 V and the current decays exponentially with time constant L/R. Why Other Options Are Wrong:Regulate input voltage: Regulation is not the diode’s function.Protect the coil from arcing: The coil does not arc; the opening contacts or driver do.Ensure most current passes through the coil: That is determined by circuit design, not the diode.Increase pull-in force: The diode only conducts on turn-off, not during energizing. Common Pitfalls:Placing the diode in the wrong polarity (then it shorts the supply), omitting it entirely, or assuming a Zener is required in low-speed relay applications (a simple diode is usually enough). Final Answer:to reduce arcing across the relay contacts

Question 6

Digital waveforms and timing terminology: True or False — The period of a waveform is the percentage of time the signal is HIGH.

Accepted Answer

Introduction / Context:Digital timing terms are often confused. Two common quantities are period (T) and duty cycle (D). Understanding the difference is critical when specifying clocks, PWM signals, or measuring timing with an oscilloscope. Given Data / Assumptions: Statement under test: “The period is the percentage of time a signal is HIGH.” Standard timing definitions apply to repetitive waveforms. We assume a stable, repeating waveform. Concept / Approach:The period T is the time for one complete cycle to repeat, measured in seconds. Frequency f = 1 / T. Duty cycle D is the fraction or percentage of one period during which the signal is HIGH (or active). They are distinct concepts: T has units of time; D is dimensionless (often in %). Step-by-Step Solution:Define period: T = time between identical points on successive cycles (e.g., rising edge to rising edge).Define duty cycle: D = (t_high / T) * 100%.The statement claims period equals a percentage; that is the definition of duty cycle, not period.Therefore, the statement is false. Verification / Alternative check:For a 10 kHz clock, T = 0.0001 s = 100 µs regardless of pulse width. If the HIGH time is 25 µs, then D = 25 µs / 100 µs = 25%. This confirms period (100 µs) is not a percentage. Why Other Options Are Wrong:“Only true for 50% duty-cycle clocks” or “True only for symmetric square waves”: Still incorrect—period is a time value, independent of duty cycle.“Undefined for pulse trains”: Period is defined for any repetitive pulse train. Common Pitfalls:Using “period” and “duty cycle” interchangeably, forgetting that period and frequency are reciprocals, and misreading oscilloscope cursors (time/div vs. % measurements). Final Answer:False

Question 7

What is the output voltage from the circuit in the given figure?

Accepted Answer

0 V

Question 8

Computer organization terminology: Internally, a computer's binary data are almost always moved on parallel channels called the ________.

Accepted Answer

Introduction / Context:Digital systems use dedicated signal groups to transfer information. Within a CPU or microcontroller, multiple parallel conductors carry binary words between registers, memory interfaces, and peripherals. Knowing the names and roles of these buses aids in debugging and system design. Given Data / Assumptions: We are referring to parallel channels inside a computer. The question asks for the common term used for the parallel paths that carry binary data. Concept / Approach:A typical system bus consists of three conceptual buses: data bus (carries data values), address bus (selects locations), and control bus (timing and control signals). The set of parallel lines dedicated to moving the actual data values is called the data bus. Step-by-Step Solution:Identify the role: moving binary data words between functional blocks.Select the bus dedicated to that role: the data bus.Therefore, the correct term is “data bus.” Verification / Alternative check:Textbook block diagrams and microprocessor datasheets label D[n:0] as “data bus,” A[n:0] as “address bus,” and various lines (RD, WR, CS, CLK) as “control.” Why Other Options Are Wrong:Parallel bus: A descriptive phrase, but the standard term is “data bus.”Serial bus: Carries data serially (e.g., SPI, I²C, UART), not the parallel internal channels implied here.Memory bus: A broader interface (often includes address, data, control); the specific parallel data lines are the data bus.Control bus: Not used to carry the data values; it carries timing and control signals. Common Pitfalls:Confusing any multi-line connection with a “bus” regardless of function, or using “memory bus” when the question specifically asks about the payload lines for data. Final Answer:data bus

Question 9

Integrated-circuit packaging: An IC package that has two parallel rows of through-hole pins on opposite sides is commonly called a ________.

Accepted Answer

Introduction / Context:Understanding package styles helps with PCB selection, breadboarding, and prototyping. One of the most common legacy through-hole packages has two rows of pins suited to sockets and breadboards. Given Data / Assumptions: Package has two parallel rows of pins. Through-hole form factor is implied. We identify the industry-standard name. Concept / Approach:DIP stands for Dual In-line Package. It features two rows of pins spaced typically at 2.54 mm (0.1 in) pitch, suitable for through-hole mounting and sockets. Other options listed are different families or surface-mount styles. Step-by-Step Solution:Match description: “two rows of parallel connecting pins” → dual in-line.Choose the standardized term → DIP.Confirm: DIPs are widely used in classic TTL/CMOS ICs (e.g., 7400, 555 timer). Verification / Alternative check:Compare with surface-mount packages: TSSOP (thin shrink small-outline) and QFP (quad flat package) are SMT, not through-hole; CMOS is a technology, not a package; BGA uses solder balls on the underside. Why Other Options Are Wrong:QFP/TSSOP/BGA: Surface-mount packages; pin layouts do not match dual in-line through-hole.CMOS: Refers to transistor technology, not a package style. Common Pitfalls:Mixing technology (TTL/CMOS) with package types, or confusing SOIC/TSSOP (two rows, but SMT gull-wing leads) with true through-hole DIP. Final Answer:DIP

Question 10

BJT operating biases (NPN): For normal forward-active operation, the base must be ________ with respect to the emitter, and ________ with respect to the collector.

Accepted Answer

Introduction / Context:Bias conditions determine a BJT’s operating region. For an NPN transistor used as an amplifier in forward-active mode, the base-emitter junction is forward biased and the collector-base junction is reverse biased. Given Data / Assumptions: Device: NPN BJT. Desired region: forward-active (normal operation for amplification). Polarity conventions referenced to the base node. Concept / Approach:Forward bias of the base-emitter junction requires V_BE > 0 (typically about +0.6 to +0.7 V for silicon). Reverse bias of the collector-base junction requires the collector to be at a higher potential than the base in an NPN, making the base negative relative to the collector. Step-by-Step Solution:Forward-active requires: base more positive than emitter → base-emitter junction forward biased.Simultaneously, collector more positive than base → base is negative with respect to the collector (C > B).Therefore, the base is positive with respect to the emitter and negative with respect to the collector. Verification / Alternative check:Typical bias example: E = 0 V, B ≈ +0.7 V, C ≈ +5 V. Here B is +0.7 V above E (forward bias), and B is 4.3 V below C (reverse bias of C-B junction), confirming forward-active conditions. Why Other Options Are Wrong:Positive, positive: Makes C-B not reverse biased if C is not sufficiently above B.Negative, positive / negative, negative / zero, zero: Would not forward-bias B-E properly and would not reverse-bias C-B as required. Common Pitfalls:Mixing up PNP vs. NPN polarities, or assuming saturation (both junctions forward-biased) is “normal” operation for linear amplification. Final Answer:positive, negative

Question 11

Clock timing calculation: What is the period of a clock waveform with frequency f = 15.4 kHz? (Give the closest option.)

Accepted Answer

Introduction / Context:Converting between frequency and period is a bread-and-butter skill in digital and analog electronics. Timers, baud rates, PWM, and sampling all rely on the relationship T = 1 / f. Given Data / Assumptions: Frequency f = 15.4 kHz. We want the period T in microseconds. Use exact reciprocal without rounding errors that change the order of magnitude. Concept / Approach:The fundamental relationship is T = 1 / f. Convert units carefully: 1 kHz = 10^3 Hz; 1 µs = 10^-6 s. Step-by-Step Solution:Convert frequency to Hz: f = 15.4 kHz = 15,400 Hz.Compute period: T = 1 / 15,400 s ≈ 0.000064935 s.Convert seconds to microseconds: 0.000064935 s * 10^6 µs/s ≈ 64.935 µs.Closest option listed: 64.9 µs. Verification / Alternative check:Check with rough mental math: 1/10 kHz = 100 µs; 1/20 kHz = 50 µs; 15.4 kHz should be between them, near 65 µs. This aligns with 64.9 µs. Why Other Options Are Wrong:649 µs: Off by a factor of 10.6.49 µs and 0.649 µs: Off by factors of 10 and 100 respectively (too fast).6.49 ms: Off by 100× in the other direction (too slow). Common Pitfalls:Missing unit conversions (kHz vs. Hz, s vs. µs) and misplacing decimal points in reciprocals. Final Answer:64.9 µs

Question 12

Determine the voltage on the collector Vc of the given transistor circuit.

Accepted Answer

+10 V

Question 13

Determine if diodes D1 and D2 in the given figure are forward or reverse biased.

Accepted Answer

D1 forward and D2 reverse

Question 14

Serial shift timing: At a 1.5 MHz serial clock, how long does it take to shift out the 8-bit hexadecimal value C3₁₆?

Accepted Answer

Introduction / Context:Serial interfaces (shift registers, SPI, simple synchronous links) clock out one bit per clock period. The total time to transmit a fixed-width value equals the number of bits multiplied by the clock period. Given Data / Assumptions: Serial clock frequency f_clk = 1.5 MHz. Data size: one byte (C3₁₆ → 8 bits). Assume one bit is transferred per clock (no start/stop/parity overhead). Concept / Approach:Time per bit T_bit = 1 / f_clk. Total time T_total = N_bits * T_bit. With 8 bits and 1.5×10^6 Hz clock, compute precisely and convert to microseconds. Step-by-Step Solution:Compute T_bit: 1 / 1.5×10^6 s ≈ 0.6667 µs per bit.Number of bits N_bits = 8 (for one byte).Total time T_total = 8 * 0.6667 µs ≈ 5.333 µs.Therefore, the closest option is 5.33 µs. Verification / Alternative check:Convert to seconds: 8 / (1.5×10^6) = 5.333… × 10^-6 s = 5.33 µs. Matches the chosen option. Why Other Options Are Wrong:4.69 µs: Corresponds to a faster clock (~1.705 MHz) or fewer bits.10.7 µs: Would imply 16 bits at 1.5 MHz or 8 bits at ~0.75 MHz.Millisecond options (4.69 ms, 5.33 ms): Off by a factor of 1000 (wrong unit). Common Pitfalls:Forgetting protocol overhead (start/stop) when present; mixing up kHz/MHz; misinterpreting hex digits as decimal count of bits instead of 4 bits per hex digit. Final Answer:5.33 µs

Question 15

Semiconductor diodes — minimum forward voltage for conduction In practical silicon rectifier or signal diodes, what is the approximate minimum forward voltage required before significant current will flow from anode to cathode?

Accepted Answer

Introduction / Context:Understanding the forward conduction threshold of a pn-junction diode is fundamental in electronics. Designers often approximate a silicon diode as a switch that begins to conduct significantly after a certain forward voltage is reached. This question probes your grasp of that practical threshold, sometimes called the knee voltage or cut-in voltage. Given Data / Assumptions: Device: ordinary silicon diode (not Schottky, not germanium). Forward conduction of interest: macroscopic/sizable current region used in typical circuits. Room-temperature operation assumed. Concept / Approach:For a pn-junction diode, the Shockley relation ID = IS * (e^(VD/(n*VT)) - 1) shows current rises exponentially with forward voltage VD. In practice, engineers use a rule-of-thumb threshold where conduction becomes significant. For silicon diodes, this forward drop settles around 0.6–0.7 V at modest currents. Germanium diodes are typically ~0.2–0.3 V, and Schottky diodes ~0.2–0.4 V. Step-by-Step Solution:Recognize device type: standard silicon pn-junction.Recall practical knee/turn-on region: about 0.6–0.7 V for silicon.Select the closest standard value used in circuit calculations: 0.7 V. Verification / Alternative check:Datasheets of common silicon diodes (e.g., 1N4148, 1N400x) show typical forward drops near 0.7 V at tens to hundreds of milliamps. SPICE diode models also default to ~0.7 V as a convenient design estimate. Why Other Options Are Wrong: 0.07 V: Far too low; not characteristic of pn silicon (closer to on-resistance behavior of MOSFETs at tiny currents, not a diode knee). 4.3 V: Much too high for a single diode; may reflect Zener breakdown values, not forward conduction. 0.2 V: More typical of germanium or Schottky diodes, not standard silicon. Common Pitfalls: Confusing diode types (Schottky vs silicon) when estimating forward drop. Ignoring current dependence: VF rises with current and temperature effects reduce VF. Final Answer:0.7 V

Question 16

Surface-Mount Technology (SMT) — how are devices installed on the PCB? With modern SMT assembly practices, how should the devices be attached relative to the printed circuit board?

Accepted Answer

Introduction / Context:Surface-Mount Technology (SMT) revolutionized PCB assembly by eliminating the need for component leads to pass through holes. This question checks your understanding of how SMT parts are installed compared to older through-hole practices. Given Data / Assumptions: Context: standard SMT manufacturing with solder paste, reflow, and solder mask defined pads. Components: SMD packages such as 0603 resistors, SOIC, QFN, BGA. Concept / Approach:Through-hole components require drilled holes; SMT devices are placed directly onto copper pads on the PCB surface and soldered (often by reflow). No component lead insertion is needed. Mechanical retention and electrical connection come from the solder joints on the surface pads. Step-by-Step Solution:Identify SMT hallmark: components are placed on surface pads, not into plated-through holes.Consider manufacturing steps: apply solder paste to pads, place components, reflow to form joints.Therefore, SMT devices mount directly on the PCB surface. Verification / Alternative check:Inspect any SMT board: you will see flat solder joints and no leads protruding through the board. Packages like QFN/BGA can only be surface-mounted. Why Other Options Are Wrong: Utilize transistor outline connections: TO packages are typically through-hole or bolt-down, not SMT by default. Have parallel connecting pins: vague and not defining SMT assembly method. Require holes and pads: this describes through-hole technology, not SMT. Common Pitfalls: Assuming SMT always replaces through-hole; in mixed technology boards both may coexist. Confusing SMD package shapes with mounting method. Final Answer:mount directly

Question 17

TTL totem-pole outputs — intended transistor operation In a TTL totem-pole output stage, the upper and lower transistors are designed to operate such that they are:

Accepted Answer

Introduction / Context:Totem-pole outputs are a hallmark of TTL logic, providing active pull-up and active pull-down for fast edges and low output impedance. Safe, efficient operation requires careful timing to avoid destructive current spikes. This question focuses on that key design principle. Given Data / Assumptions: Device family: TTL with a push-pull (totem-pole) output structure. Two output transistors: an upper pull-up and a lower pull-down transistor. Concept / Approach:In a push-pull stage, if both transistors conduct simultaneously, current would shoot directly from VCC to ground (shoot-through), causing large instantaneous power dissipation and noise. TTL design includes cross-conduction prevention using internal biasing and transition shaping so that only one device conducts strongly at a time. Step-by-Step Solution:Consider logic HIGH: pull-up turns on, pull-down turns off.Consider logic LOW: pull-down turns on, pull-up turns off.During transitions, internal circuitry minimizes overlap conduction to prevent shoot-through. Verification / Alternative check:Examine TTL datasheets and timing diagrams showing output transistor states; they specify maximum transient currents and internal structures (e.g., emitter followers and phase-splitting stages) engineered to avoid simultaneous saturation of both devices. Why Other Options Are Wrong: Always on together: would create a near short between rails. Provide phase splitting: an internal earlier stage may provide this, but not the purpose of the output pair acting together. Provide voltage regulation: TTL outputs are logic drivers, not regulators. Common Pitfalls: Assuming brief crossover currents are the design goal; they are minimized, not desired. Confusing totem-pole with open-collector outputs which use passive pull-up. Final Answer:never on together

Question 18

Improving serial communication throughput — best practical method Which action most directly increases the bit rate of a synchronous or asynchronous serial link, assuming the medium is unchanged?

Accepted Answer

Introduction / Context:Serial communication speed is typically expressed as baud or bits per second. Improving throughput involves increasing symbol rate and/or bits per symbol, within the limits of channel bandwidth, noise, and protocol constraints. This question targets the most direct lever in basic digital links. Given Data / Assumptions: Physical medium unchanged (e.g., existing cable or PCB trace). Same modulation scheme and voltage levels. Focus on the immediate setting under digital designer control. Concept / Approach:For clocked serial links, the symbol rate is set by clock frequency. Raising the clock increases the number of symbols per second, thereby increasing throughput, provided the channel can support the higher bandwidth without excessive intersymbol interference or jitter. Step-by-Step Solution:Recognize throughput B ≈ symbols_per_second * bits_per_symbol.With the same coding, bits_per_symbol is constant.Increase symbols_per_second by raising the clock frequency → higher bit rate. Verification / Alternative check:Interfaces like SPI/UART clearly specify maximum clock/baud rates; increasing from 1 MHz to 5 MHz SPI, for example, yields a 5× raw throughput increase if timing margins are met. Why Other Options Are Wrong: Using silver/gold instead of copper: negligible speed gain; speed is limited by bandwidth, reflections, and signaling, not conductor material in typical digital runs. Adjusting duty cycle: in NRZ signaling, duty cycle follows data; changing it does not raise symbol rate. Combining precious metals with faster clocks: metals still do not materially change digital limit; the win comes from the clock increase itself. Common Pitfalls: Ignoring signal integrity; faster clocks require proper termination and layout. Confusing throughput with latency; protocol overhead also matters. Final Answer:using high-speed clock signals

Question 19

Logic families — which one uses bipolar junction transistors (BJTs) internally? Identify the logic family composed of bipolar junction transistors.

Accepted Answer

Introduction / Context:Digital ICs are built using different device technologies. Two major families historically are TTL and CMOS. Knowing which devices they are based on helps predict speed, power, and interface behavior. Given Data / Assumptions: TTL stands for Transistor-Transistor Logic. CMOS stands for Complementary Metal-Oxide-Semiconductor. DIP is a package type, not a logic family. Concept / Approach:TTL uses multiple-emitter BJTs and associated structures to implement logic gates. CMOS uses complementary MOSFET pairs (pMOS and nMOS). Therefore, if asked which logic family is composed of BJTs, the correct choice is TTL. Step-by-Step Solution:Match family names to device types: TTL → BJTs, CMOS → MOSFETs.Exclude non-families: DIP is a packaging form factor, not technology.Select TTL as the BJT-based family. Verification / Alternative check:Any TTL datasheet (74xx series) shows bipolar transistor schematics; classic 74LS, 74H, 74F families are all bipolar implementations. Why Other Options Are Wrong: CMOS: built with MOSFETs, not BJTs. DIP: Dual In-line Package is packaging only. BJT: a device type, not a standardized logic family. Common Pitfalls: Confusing the logic family name with the device type or package. Assuming all modern logic is CMOS; classic TTL remains widely referenced. Final Answer:TTL

Question 20

CMOS vs. TTL — identify a disadvantage of CMOS compared to TTL (classic families) In traditional comparisons at similar supply voltages, which is a disadvantage of CMOS relative to TTL?

Accepted Answer

Introduction / Context:Comparing classic CMOS and TTL logic families helps designers choose parts for speed, power, and integration. Historically, CMOS offered dramatic static power savings but at a cost in switching speed versus the fastest TTL families of the same era and voltage range. Given Data / Assumptions: Comparison refers to traditional 5 V logic families (e.g., 74HC/HCT vs 74F/74ALS). Modern advanced CMOS can be very fast, but the question concerns the classic generalization. Concept / Approach:CMOS uses complementary MOSFETs; static power is near zero except during transitions. TTL is bipolar and consumes more static power, but some TTL subfamilies (e.g., 74F) achieved very high speed. Thus, a standard textbook disadvantage of CMOS is lower switching speed compared to the fastest TTL, especially at older process nodes. Step-by-Step Solution:Identify traits: CMOS → low static power, high input impedance; TTL → faster in classic families.Eliminate benefits: “uses less power” and “smaller” are advantages of CMOS, not disadvantages.Choose “It switches slower.” as the disadvantage. Verification / Alternative check:Datasheet propagation delays: typical 74HC at 5 V often has tens of nanoseconds, while 74F TTL can be single-digit nanoseconds. This supports the general statement. Why Other Options Are Wrong: It uses less power: that is a CMOS advantage. It is smaller: CMOS integration scales well; also an advantage. cost: context-dependent, not an inherent technology disadvantage. Common Pitfalls: Projecting modern submicron CMOS speeds onto classic 5 V logic comparisons. Ignoring that HCT is pin-compatible with TTL levels but still CMOS internally. Final Answer:It switches slower.

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