Question 1

Write-cycle control signals in RAM: A typical static or dynamic RAM writes (stores data internally) only when the chip-select (CS) is active and the write-enable (WE) signal is also active for a write, according to the device’s active level. Evaluat…

Accepted Answer

Introduction / Context: Random-access memory devices control read and write behavior via chip-select (CS), write-enable (WE), and output-enable (OE) signals. Correctly understanding their active levels prevents accidental overwrites and bus conflicts. Given Data / Assumptions: Typical RAMs use active-LOW control pins labeled /CS, /WE, /OE. “Active” means the asserted level (often LOW). Write occurs only when both the device is selected and WE is asserted for writing. Concept / Approach: For active-LOW controls: to write, /CS = 0 and /WE = 0 during the valid address/data window. If /WE is inactive (i.e., HIGH for active-LOW WE), the device does not perform a write; it typically either reads (if /OE is active) or remains deselected/high-impedance. Thus the statement claiming writing when WE is inactive is incorrect. Step-by-Step Solution: 1) Identify control polarity from the datasheet: /CS, /WE, /OE.2) Write truth: write = (/CS asserted) AND (/WE asserted) with proper timing.3) Read truth: read = (/CS asserted) AND (/OE asserted) AND (/WE deasserted).4) Conclude that an inactive WE prevents writing, invalidating the claim. Verification / Alternative check: Timing diagrams explicitly show write-cycle waveforms where /WE goes active for a defined pulse width while addresses and data meet setup/hold times. Why Other Options Are Wrong: “Correct” reverses the necessary WE polarity. References to ROM or OE do not change the fundamental requirement that WE be asserted for writes. Common Pitfalls: Mixing up active-high vs active-low naming; neglecting setup/hold windows; assuming CS alone controls writes. Final Answer: Incorrect

Question 2

Volatility of RAM: Assess the statement: "RAM is nonvolatile." Clarify what volatile and nonvolatile mean in the context of memory technologies.

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Introduction / Context: Memory technologies are often classified as volatile or nonvolatile. This distinction affects system design, data retention, and power-failure behavior. Random-access memory (RAM) commonly refers to DRAM and SRAM used for working memory during operation. Given Data / Assumptions: Volatile memory loses its stored information when power is removed. Nonvolatile memory retains data without power (e.g., flash, EEPROM). Standard DRAM and SRAM require power to preserve state; DRAM additionally requires refresh. Concept / Approach: By definition, RAM in typical systems (main memory DRAM, caches SRAM) is volatile. When power is lost, stored bits are not preserved. Special cases exist, such as battery-backed SRAM or NVRAM technologies, but these are explicitly designed for retention and are not what “RAM” ordinarily denotes. Step-by-Step Solution: 1) Define volatility: data retention depends on continuous power (and for DRAM, periodic refresh).2) Identify mainstream RAM types: DRAM (needs refresh) and SRAM (latch-based, no refresh, but still requires power).3) Nonvolatile types include flash and EEPROM; these are not standard “RAM” despite random access capability in a broad sense.4) Conclude the statement “RAM is nonvolatile” is false. Verification / Alternative check: System behavior on power loss (cold boot) demonstrates the volatility of RAM: OS and applications must reload from nonvolatile storage. Why Other Options Are Wrong: “Correct” contradicts the definition. Limiting volatility to only one RAM type or citing battery-backed SRAM changes the premise; the general statement remains false. Common Pitfalls: Equating “random access” with “RAM” in a strict technology sense; ignoring special retention techniques that are not the default. Final Answer: Incorrect

Question 3

DRAM addressing technique: Dynamic memories such as the classic 2118 16K × 1 RAM multiplex the address bus into row and column portions to reduce pin count. Evaluate the correctness of this statement.

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Introduction / Context: Dynamic RAM achieves high density by arranging cells in arrays addressed by row and column. To limit package pins, DRAM devices multiplex the address bus: the same pins carry row addresses during one phase and column addresses during another, strobed by RAS and CAS signals. Given Data / Assumptions: A legacy example is the 2118 16K × 1 device. Address multiplexing reduces pin count compared to providing separate row and column pins simultaneously. Timing uses RAS (row address strobe) and CAS (column address strobe). Concept / Approach: On a memory access, the controller first places the row address on the address pins and asserts RAS to latch it. Next, it places the column address on the same pins and asserts CAS to latch the column. Internally, sense amps select the correct cell(s). This technique is universal to mainstream DRAM generations (FPM, EDO, SDR, DDR families) though details vary. Step-by-Step Solution: 1) Present row address → assert RAS to capture it.2) Present column address → assert CAS to capture it.3) DRAM core accesses cell(s) at [row, column].4) Data is read or written according to the cycle type. Verification / Alternative check: Timing diagrams in DRAM datasheets explicitly show row/column multiplexing; SDR/DDR data sheets maintain the RAS/CAS naming convention conceptually (even as signaling evolves). Why Other Options Are Wrong: “Incorrect” disagrees with standard DRAM operation. SRAM does not use row/column multiplexing; its arrays are accessed without RAS/CAS strobes. “CAS precedes RAS” refers to a specific refresh mode, not normal addressing. Common Pitfalls: Confusing DRAM’s multiplexed external addressing with internal bank/row/column abstractions in modern DDR; the multiplexing principle remains. Final Answer: Correct

Question 4

Memory timing terminology: The time from the beginning of a read cycle until the data output becomes valid at the device pins is commonly called the access time. Evaluate the statement: "This interval is called propagation delay."

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Introduction / Context: Precise timing terms are crucial when reading memory datasheets and designing bus interfaces. Two commonly encountered terms are access time and propagation delay. Although related, they describe different measurements and contexts. Given Data / Assumptions: We consider asynchronous memories for clarity; synchronous memories add clocked timing parameters. Read cycle begins with address (and CS/OE) assertion. The device outputs valid data after some finite time. Concept / Approach: Access time (tACC conceptually) is the interval from a defined starting condition—often address valid with chip selected—to the time when the data at the output pins is guaranteed valid. Propagation delay is a more generic term for the internal delay from an input transition to a corresponding output transition for a specific signal path. Datasheets use “access time” as the headline specification for memory speed in read operations. Step-by-Step Solution: 1) Establish start of read cycle: address asserted, CS/OE active.2) Measure until data lines achieve and hold valid logic levels.3) This interval is the access time parameter.4) Therefore, calling it “propagation delay” is imprecise and, in datasheet nomenclature, incorrect. Verification / Alternative check: Open any asynchronous SRAM/ROM datasheet: the key timing table lists tAA (address to data valid) or tCE/tOE paths under “Access Time.” Why Other Options Are Wrong: “Correct” conflicts with standard naming. The qualifiers about sync SRAM or OE polarity do not change the definition of the parameter describing read latency. Common Pitfalls: Using “prop delay” loosely for any delay; ignoring that multiple paths (address, CS, OE) each have their own access-related timing. Final Answer: Incorrect

Question 5

Flash memory erase granularity: Assess the statement: "Most flash chips use a bulk (chip) erase operation in which all cells on the chip are erased simultaneously." Consider how flash is typically erased in practice.

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Introduction / Context: Nonvolatile flash memory supports erase-before-write semantics. Understanding its erase granularity is important for firmware updates, wear leveling, and file-system design. The term “bulk erase” (chip erase) exists, but typical workflows rely on smaller erase units. Given Data / Assumptions: Modern flash families: NOR and NAND. Erase granularity options: sector/block erase is common; some devices also allow chip erase. Real systems rarely need to erase an entire chip at once. Concept / Approach: Most flash chips are designed to erase data in sectors/blocks (for NOR, often called sectors; for NAND, blocks comprising many pages). A “chip erase” command may be provided, but practical use emphasizes sector/block erase to limit data loss and wear. Therefore, saying “most flash chips use bulk erase” as the normal or defining operation mischaracterizes typical usage and capability. Step-by-Step Solution: 1) NOR flash: supports sector erase (e.g., 4 KB, 64 KB) and often chip erase; code storage commonly erases sectors only.2) NAND flash: supports block erase (much larger than a page); bulk chip erase is uncommon in normal operation.3) System software performs targeted erases to update regions while preserving others.4) Hence, the statement that “most chips use bulk erase” is inaccurate. Verification / Alternative check: Bootloaders, flash file systems, and datasheets emphasize sector/block erase cycles, endurance per block, and bad-block management rather than routine chip-erase procedures. Why Other Options Are Wrong: “Correct” exaggerates chip-erase usage. Tying the claim to EEPROM or a single flash type ignores that both NOR and NAND depend primarily on sector/block erases in practice. Common Pitfalls: Assuming “erase” means “erase everything”; overlooking endurance implications and the need for granular updates. Final Answer: Incorrect

Question 6

Register stack behavior: In a register-based stack implementation (e.g., CPU stack or hardware register stack), data can be pushed (written) and popped (read) so that items effectively move in and out in a last-in, first-out order. Evaluate the stat…

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Introduction / Context: Stacks are abstract data structures supporting push and pop operations. Hardware and low-level software realize stacks with a stack pointer and contiguous storage (registers or memory). Understanding stack directionality is essential for subroutine calls, interrupts, and local variable storage. Given Data / Assumptions: A stack follows LIFO semantics. Push adds an item at the top; pop removes the most recently pushed item. The term “move up” vs “move down” is informal; actual direction depends on implementation. Concept / Approach: In many CPUs, the stack pointer either decrements on push and increments on pop (descending stack) or the opposite (ascending stack). In either design, data entries are added and removed symmetrically. Saying data “moves up but not down” incorrectly implies one-way movement. A correct description is that the stack grows and shrinks as items are pushed and popped; the stack pointer changes accordingly in both directions over time. Step-by-Step Solution: 1) Define push: store data at SP (or SP ± 1), then update SP by ±1 entry.2) Define pop: read data from SP (or SP ± 1), then update SP in the opposite sense.3) Observe bidirectionality: over program execution the pointer moves both ways as functions call and return.4) Conclude the statement is false for a stack abstraction. Verification / Alternative check: Assembly-language documentation (e.g., x86, ARM) specifies push/pop behavior and SP changes in both directions across operations; microarchitectures may implement stacks in registers or memory but the LIFO behavior remains. Why Other Options Are Wrong: “Correct” contradicts LIFO semantics. Qualifiers about increment-only or FIFO confuse stack behavior with implementation details or different data structures (FIFO queues move only one way). Common Pitfalls: Equating physical address direction with logical capability; assuming a stack can only grow; confusing stacks with circular buffers or FIFOs. Final Answer: Incorrect

Question 7

Memory fundamentals — In static RAM (SRAM), the term "static" does not mean non-volatile. Evaluate the statement: "Static memory will maintain storage even if power is removed."

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Introduction: This item probes a common misconception about static random-access memory (SRAM). Many learners assume the word "static" implies that data is retained without power, but in digital electronics, "static" and "dynamic" refer to how data is maintained while power is applied. We will clarify why the statement "Static memory will maintain storage even if power is removed" is incorrect and contrast SRAM with non-volatile memories. Given Data / Assumptions: Static memory here refers to SRAM used in caches and embedded systems. Non-volatile examples include ROM, EPROM, EEPROM, and flash. We consider normal operation without auxiliary batteries or supercapacitors. Concept / Approach: In SRAM, each bit is stored by a bistable element (commonly a 6-transistor cell) that actively holds a 0 or 1 as long as Vcc is present. "Static" means the cell does not require periodic refresh cycles like DRAM does; it does not mean the memory is persistent without power. When power is removed, the transistors no longer bias the bistable nodes, and the stored state is lost. Persistence without power is the defining property of non-volatile memory, not SRAM. Step-by-Step Solution: Identify memory type: SRAM stores bits in cross-coupled inverters requiring a supply to maintain state.Consider power removal: with Vcc = 0, the inverters cannot sustain a logic level, so data vanishes.Contrast with NVM: EPROM/EEPROM/flash use floating-gate charge to retain bits without Vcc. Verification / Alternative check: Look at device datasheets: data retention specifications for SRAM always assume a valid supply (sometimes low-retention Vcc). Without Vcc, retention time is 0. Why Other Options Are Wrong: Correct: Would incorrectly equate SRAM with non-volatile memory.True only with a backup battery: A battery-backed SRAM retains data, but that is an added system feature, not an intrinsic property.True only below 1 kHz clock: Clock frequency is irrelevant to non-volatility; SRAM does not need a clock to hold data while powered. Common Pitfalls: Confusing "no refresh needed" with "no power needed".Assuming all memories inside microcontrollers are non-volatile. Final Answer: Incorrect

Question 8

Technology comparison — Evaluate the statement: "The main advantage of bipolar (TTL) memories over MOS memories is speed."

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Introduction: Historically, semiconductor memories were fabricated in different device technologies, notably bipolar (TTL/ECL families) and metal-oxide-semiconductor (MOS). This question asks whether bipolar memories have speed as their main advantage over MOS. Understanding the speed-power-density trade-offs is central to computer architecture history and embedded design choices. Given Data / Assumptions: Bipolar technology includes TTL and especially ECL for very high speed. MOS technology underlies SRAM, DRAM, EPROM/EEPROM, and flash. Comparison concerns typical devices from the same era and process generation. Concept / Approach: Bipolar transistors switch faster than MOSFETs of comparable vintage due to stronger transconductance and lower intrinsic capacitances at the operating point, leading to shorter propagation delays. However, bipolar devices consume more static power and offer lower density than MOS. Therefore, the principal advantage of bipolar memories has historically been speed, while MOS wins in density, cost, and power efficiency. Step-by-Step Solution: Identify property: speed (access time / cycle time) is typically smaller for bipolar memories.Note counter-properties: power consumption is higher; bits per die are lower.Conclude that speed is indeed the primary advantage attributed to bipolar memory over MOS. Verification / Alternative check: Compare historical datasheets: ECL RAMs show sub-10 ns access when contemporary MOS parts are slower; at the cost of power and price. Why Other Options Are Wrong: Incorrect: Neglects well-documented performance leadership of bipolar devices of the time.True only at cryogenic temperatures: Bipolar speed advantage does not require cryogenic operation.Depends only on word width: Access speed is not determined solely by word width. Common Pitfalls: Assuming modern CMOS always wins in speed; at equal nodes, specialized bipolar/ECL historically led.Ignoring that speed came with significant power and cost penalties. Final Answer: Correct

Question 9

Computer architecture — Cache memory is intentionally used in high-speed systems to reduce average memory access time. Evaluate the statement.

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Introduction: Cache memory places a small, fast memory between the CPU and slower main memory to reduce average access time. This principle underlies nearly all modern high-speed processors and microcontrollers. The question asks whether caches are used in high-speed systems, which tests fundamental architectural awareness rather than implementation details. Given Data / Assumptions: Cache is faster but smaller than main memory. Programs exhibit locality of reference (temporal and spatial locality). We consider typical write policies and hierarchies (L1, L2, L3). Concept / Approach: By exploiting locality, caches store recently or nearby accessed blocks, so subsequent accesses hit the cache, avoiding long-latency main memory. Even embedded systems employ cache (or tightly coupled memory) to accelerate critical code paths. Thus, caches are indeed a hallmark of high-speed designs. Step-by-Step Solution: Identify latency gap: CPU cycle time is often far shorter than DRAM access time.Introduce cache: a fast SRAM-based level buffers hot data/instructions.Result: higher hit rate reduces average access time and increases throughput. Verification / Alternative check: Performance counters in modern CPUs show high cache hit rates strongly correlate with achieved IPC and low stall cycles. Why Other Options Are Wrong: Incorrect: Contradicted by universal practice across CPUs and many MCUs.Only true for write-through caches: Write-back caches are also ubiquitous and effective.True only for Harvard architecture: Von Neumann CPUs also rely on caches. Common Pitfalls: Confusing cache presence with guaranteed speedup in cache-miss-heavy workloads.Overlooking real-time determinism issues; some systems disable cache for critical I/O regions. Final Answer: Correct

Question 10

Non-volatile memory — Evaluate the statement: "Information stored in an EPROM can be erased by prolonged exposure to ultraviolet light."

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Introduction: Erasable Programmable Read-Only Memory (EPROM) is a classic non-volatile memory technology that can be erased using ultraviolet (UV) radiation through a quartz window in the package. This question validates awareness of EPROM erase mechanisms versus other non-volatile memories like EEPROM and flash. Given Data / Assumptions: Device type: EPROM with quartz window (e.g., 27xx families). Erase mechanism: UV light exposure of sufficient energy and duration. Programming: high-voltage charge injection to a floating gate. Concept / Approach: EPROM stores bits by trapping charge on a floating gate inside a MOS transistor. UV photons with adequate energy discharge the trapped electrons via photoelectric effects, returning threshold voltages to their erased state. The process requires specific wavelength (commonly near 253.7 nm from mercury lamps) and exposure time (minutes) per datasheet recommendations. Step-by-Step Solution: Confirm technology: EPROM uses floating-gate MOS with a quartz window.Apply UV light for the recommended duration and distance.Verify erasure by reading the device; all bits return to the erased state (usually logic 1). Verification / Alternative check: Manufacturers specify erase times and wavelengths; device programmers include UV erasers designed for EPROMs. Why Other Options Are Wrong: Incorrect: Ignores the well-known UV erase property of EPROMs.Only true for EEPROM: EEPROM is electrically erasable, not UV-erasable.Only true below 300 nm wavelength: While germicidal 254 nm is standard, the key is sufficient photon energy per datasheet; it is not a strict single cutoff statement. Common Pitfalls: Confusing EPROM with OTP (one-time programmable) versions that lack a quartz window and cannot be UV-erased after final packaging.Assuming flashlight UV is adequate; intensity and exposure time matter. Final Answer: Correct

Question 11

EEPROM capability — Evaluate the statement: "EEPROMs can be electrically erased and reused."

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Introduction: Electrically Erasable Programmable Read-Only Memory (EEPROM) is a non-volatile memory that supports electrical erase and reprogramming, typically at the byte or page granularity. The statement evaluates whether electrical erasure and reuse are defining features of EEPROMs. Given Data / Assumptions: EEPROM uses floating-gate transistors similar to flash. Erase/rewrite cycles are finite but large (e.g., 10^4 to 10^6 cycles depending on device). No UV exposure is required for erasure. Concept / Approach: EEPROMs permit in-circuit electrical erase and programming by applying controlled voltages to tunnel charges on or off the floating gate. This enables configuration storage, calibration constants, and data logging that survives power cycles and can be updated as needed, subject to endurance limits. Step-by-Step Solution: Acknowledge technology: floating-gate MOS with electrical tunneling for erase/program.Define operation: device supports erase then program cycles on selected locations.Conclude: electrical erase and reuse are core EEPROM capabilities. Verification / Alternative check: Datasheets specify write/erase protocols, endurance ratings, and retention times (often 10 years or more). Why Other Options Are Wrong: Incorrect: Conflicts with the definition of EEPROM.True only once per device: Endurance is limited but far greater than one cycle.True only if UV light is applied: UV applies to EPROM, not EEPROM. Common Pitfalls: Equating EEPROM with flash; flash usually erases in blocks, while EEPROM often supports finer granularity.Ignoring endurance management (wear leveling) in frequent-write applications. Final Answer: Correct

Question 12

Storage media classification — Evaluate the statement: "An optical disk is an example of magnetic storage."

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Introduction: Storage media are broadly categorized by their physical recording mechanisms. Magnetic storage (hard disks, tapes) encodes information using magnetic domains, while optical storage (CD, DVD, Blu-ray) uses pits and lands read by lasers. The prompt asks whether an optical disk is an example of magnetic storage, which tests understanding of these mechanisms. Given Data / Assumptions: Optical disks: CD/DVD/Blu-ray use optical reflectivity changes. Magnetic media: HDDs, tapes use magnetization direction changes. Hybrid media like magneto-optical exist but are distinct devices. Concept / Approach: Optical disks store data as physical or phase-change features that alter reflectivity. A laser detects the differences to reconstruct bits. No magnetic field changes are used in standard optical formats. Therefore an optical disk is not a magnetic storage device. Magneto-optical disks are a separate technology combining magnetic fields with laser heating, but the general term "optical disk" refers to purely optical schemes in mainstream products. Step-by-Step Solution: Identify recording principle: optical reflectivity/phase change vs magnetic orientation.Map examples: CD-R/RW, DVD-R/RW, BD-RE are optical; HDD is magnetic.Conclude that an optical disk is not magnetic storage. Verification / Alternative check: Product documentation and standards describe laser wavelengths and reflectivity thresholds, not coercivity or magnetic remanence, for optical disks. Why Other Options Are Wrong: Correct: Would conflate optical with magnetic mechanisms.True only for magneto-optical disks: MO is a specialized category, not the generic optical disk referenced.Depends on laser wavelength: Wavelength affects density, not whether the storage is magnetic. Common Pitfalls: Assuming all removable disks are magnetic like floppy disks.Confusing optical pickup assemblies with magnetic read/write heads. Final Answer: Incorrect

Question 13

DRAM behavior — Evaluate the statement: "One advantage of DRAMs is that they store data without needing periodic refresh of the memory contents."

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Introduction: Dynamic random-access memory (DRAM) stores data as charge on tiny capacitors. Because these capacitors leak, DRAM requires periodic refresh to maintain data integrity. The statement claims the opposite, so we must evaluate and correct it, reinforcing key memory-technology differences between DRAM and SRAM. Given Data / Assumptions: DRAM cells use 1T1C (one transistor, one capacitor) structures. Leakage currents cause charge decay over milliseconds. Memory controllers perform refresh operations at required intervals (e.g., all rows within tREFI constraints). Concept / Approach: DRAM's density advantage comes at the cost of mandatory refresh. Each row must be periodically accessed to restore charge (read and rewrite). Therefore a claimed advantage of "no refresh needed" is incorrect. "No refresh" is an attribute of SRAM, which maintains state statically while power is applied. Step-by-Step Solution: Understand storage: DRAM bit = charge on a capacitor; charge leaks over time.Controller action: refresh cycles access each row to restore lost charge.Conclusion: refresh is required; lack of refresh is not a DRAM advantage. Verification / Alternative check: JEDEC standards specify refresh intervals (e.g., 64 ms typical at room temperature) and timing parameters like tRFC, tREFI, proving refresh is intrinsic. Why Other Options Are Wrong: Correct: Would invert the defining behavior of DRAM.True only below 1 Hz clock: Refresh is independent of CPU clock; it is a memory requirement.Depends solely on bus width: Bus width does not remove the need for refresh. Common Pitfalls: Confusing DRAM with SRAM.Assuming self-refresh during low power eliminates refresh entirely; it merely automates it. Final Answer: Incorrect

Question 14

Device family — Evaluate the statement: "An EPROM is strictly a MOS device."

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Introduction: EPROMs (Erasable Programmable ROMs) are implemented using floating-gate MOS transistors. This question checks foundational device-technology knowledge: whether EPROM belongs to the MOS family. Understanding this helps distinguish EPROM from bipolar PROMs and other memory technologies. Given Data / Assumptions: EPROM cells use MOSFET structures with floating gates. Programming uses hot-carrier injection or tunneling to place charge on the gate. Erasure (for windowed EPROMs) uses ultraviolet light to remove charge. Concept / Approach: Since the storage mechanism is charge on a floating gate within a MOSFET, EPROM is a MOS technology by definition. Bipolar PROMs store data via fusible links or anti-fuses and are distinct. Package style (with or without quartz window) does not change the underlying MOS device physics. Step-by-Step Solution: Identify cell structure: floating-gate MOS transistor.Relate operations: electrical programming and UV erasure alter charge on the floating gate.Conclude: EPROM is a MOS device family member. Verification / Alternative check: Review technology descriptions in memory textbooks: EPROM and EEPROM/flash are categorized under MOS non-volatile memories. Why Other Options Are Wrong: Incorrect: Conflicts with the floating-gate MOS implementation.True only for EEPROM: Both EPROM and EEPROM are MOS-based.Depends on package type: Ceramic vs plastic does not change transistor technology. Common Pitfalls: Confusing bipolar PROM fuse technologies with EPROM.Assuming "windowed" package implies a different device physics family. Final Answer: Correct

Question 15

Magnetic performance — Evaluate the statement: "The highest-speed magnetic storage is achieved by using a floppy disk."

Accepted Answer

Introduction: Among magnetic storage devices, speed is influenced by rotational velocity, areal density, head technology, and interface bandwidth. Floppy disks, an older removable medium, operate at low rotational speeds and have limited density, yielding modest throughput and long seek times compared to hard disk drives (HDDs) and modern magnetic tape systems. The statement claims floppies are the fastest magnetic storage; we will evaluate this claim. Given Data / Assumptions: Floppies use flexible magnetic media with simple read/write heads. HDDs use rigid platters, high RPM, precision heads, and caches. Speed refers to sustained transfer rate and access latency. Concept / Approach: Floppy disks typically offer data rates in the tens to hundreds of kilobytes per second and high access latency. HDDs deliver orders-of-magnitude higher throughput (tens to hundreds of megabytes per second) and lower average latency due to faster rotation and advanced controllers. Therefore, the claim is incorrect: floppies are among the slowest magnetic media historically used in personal computing. Step-by-Step Solution: Define metric: compare sustained transfer and access time across devices.Observe technology differences: RPM, track density, multi-head designs, caches.Conclude: HDDs far exceed floppy performance; floppies are not highest speed. Verification / Alternative check: Typical figures: 3.5-inch floppy ~500 kbit/s raw rate vs HDDs with 100+ MB/s modern sustained rates (technology dependent). Why Other Options Are Wrong: Correct: Would invert established performance comparisons.True only for 8-inch floppies: Larger floppies still trail HDD performance by wide margins.Depends on laser alignment: Laser is unrelated to magnetic disk reading. Common Pitfalls: Confusing simplicity and portability with speed.Comparing across vastly different eras without normalizing technology generations. Final Answer: Incorrect

Question 16

RAM usage — Evaluate the statement: "Because they are easy to erase and reuse, magnetic memory devices are widely used for RAM."

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Introduction: The question asks whether magnetic memory devices are widely used for RAM in modern systems due to ease of erasure and reuse. Historically, magnetic core memory provided random access and non-volatility, but it has been obsolete for decades. Contemporary RAM is semiconductor-based (DRAM and SRAM), not magnetic media. Given Data / Assumptions: Modern main memory uses DRAM; caches and many buffers use SRAM. Magnetic media today primarily serve as storage (HDDs, magnetic tape), not volatile RAM. "Widely used" refers to mainstream contemporary computing. Concept / Approach: While magnetic storage is reusable and erasable, it is not used as RAM because its access time and bandwidth are far slower and its interfaces are block-oriented rather than true random, low-latency access at the word/byte level. RAM demands nanosecond-scale access with bus-level integration, which semiconductor memories provide. Therefore, the statement is incorrect for modern systems. Step-by-Step Solution: Define RAM requirements: nanosecond access, random addressing, high bandwidth, direct bus attachment.Compare with magnetic devices: millisecond access for disks, streaming/block interfaces for tape.Conclude: magnetic devices are unsuitable for RAM in current architectures; semiconductor RAM dominates. Verification / Alternative check: Survey system block diagrams: CPUs connect to DRAM modules (e.g., DDR generations) and use SRAM caches; no magnetic RAM in mainstream usage. Why Other Options Are Wrong: Correct: Would imply widespread current use of magnetic RAM, which is false.True only for core memory in the 1960s: That was historical, not modern widespread practice.Depends on UV eraser availability: UV erasure is irrelevant to magnetic media and RAM. Common Pitfalls: Confusing secondary storage (HDDs) with main memory (RAM).Assuming "erasable" implies suitability for RAM without considering latency and interface. Final Answer: Incorrect

Question 17

Flash memory operations and timing: Evaluate the statement: “Erasing or programming a flash memory device is a one-step operation.” Consider that flash typically uses block (sector) erase followed by page/word program sequences under controller supe…

Accepted Answer

Introduction / Context: Flash memory devices (NAND and NOR) are nonvolatile and widely used in embedded systems, SSDs, and firmware storage. Understanding their program/erase flow is essential for designing reliable storage stacks and meeting endurance, latency, and data integrity requirements. Given Data / Assumptions: The claim states erase/program is a single-step action. Flash devices organize data into blocks/sectors (erase units) and pages/words (program units). Standard flash timing includes setup, command sequences, status polling, and verification. Concept / Approach: Flash cells must be erased (typically to all 1s) before programming 0s into selected locations. Erase and program are distinct operations with different command codes, latencies, and endurance effects. Controllers orchestrate multi-step sequences: issue command, address, data, execute, poll status, and verify. Hence, describing them as “one-step” is inaccurate. Step-by-Step Solution: Issue an erase command to the targeted block; wait until erase-complete.Program data page-by-page or word-by-word with program commands.Poll status/busy flags and verify programmed data.If verification fails, apply retry or mark block as bad (NAND contexts). Verification / Alternative check: Device datasheets show separate timing diagrams and specifications for tBERS (block erase time) and tPROG (program time), confirming multi-step flows rather than a single atomic step. Why Other Options Are Wrong: Correct: Misrepresents flash operation. Applies only to EEPROM: EEPROM supports byte-wise electrical erase, but still involves multi-step command sequences. Ambiguous: Even with controllers, command sequences remain distinct steps. Common Pitfalls: Confusing “issuing one high-level API call” with the device’s internal multi-step process; overlooking mandatory erase-before-program constraints. Final Answer: Incorrect

Question 18

SRAM storage mechanism: Assess the statement: “Static RAMs (SRAMs) use internal capacitors as the basic storage elements.” Clarify by contrasting SRAM cells with DRAM cells.

Accepted Answer

Introduction / Context: Volatile memory technologies differ in the circuitry that stores each bit. Knowing whether storage relies on capacitors or cross-coupled transistors is fundamental for understanding refresh requirements, speed, and power characteristics. Given Data / Assumptions: Claim: SRAM uses capacitors as storage elements. SRAM cells are commonly 6-transistor (6T) cross-coupled inverters plus access transistors. DRAM cells are 1-transistor/1-capacitor (1T1C) and require refresh. Concept / Approach: SRAM stores data in a bistable latch built from cross-coupled inverters. As long as power is applied, the latch holds its state without refresh. DRAM stores charge on a capacitor that leaks over time, hence periodic refresh is mandatory. Therefore, attributing capacitors to SRAM is incorrect; capacitors characterize DRAM storage. Step-by-Step Solution: Identify SRAM cell: cross-coupled inverter pair (latch) + access transistors.Identify DRAM cell: one access transistor + storage capacitor.Compare behaviors: SRAM no refresh vs DRAM periodic refresh.Conclude the statement is incorrect for SRAM. Verification / Alternative check: Any basic microelectronics text or memory datasheet illustrates 6T SRAM and 1T1C DRAM schematics, confirming the fundamental difference. Why Other Options Are Wrong: Correct: Would misclassify SRAM technology. Applies only to DRAM: DRAM indeed uses capacitors; the claim targeted SRAM. Valid for pseudo-static RAM: PSRAM wraps DRAM in a controller to emulate SRAM timing, but the underlying storage is DRAM, not SRAM. Common Pitfalls: Assuming all volatile memories need refresh; conflating PSRAM behavior with true SRAM cell structure. Final Answer: Incorrect

Question 19

ROM performance metric: The parameter “access time” (t_ac) measures how fast a ROM returns valid data after an address becomes stable. Evaluate this statement about access time being a measure of operating speed.

Accepted Answer

Introduction / Context: Nonvolatile memories such as ROM, PROM, EPROM, and flash specify timing parameters that dictate how quickly data can be retrieved. The most cited figure is access time, often shown as tAA or tACC, which directly influences system performance. Given Data / Assumptions: We consider read-only access, not programming operations. Access time is defined from address valid to data valid. System clocking and bus interfaces rely on this parameter to meet timing closure. Concept / Approach: Access time quantifies the delay through internal decoders, word-line/bit-line selection, and sense amplifiers until output drivers present stable data. Thus, smaller t_ac implies faster read performance. It is a primary operating-speed measure for random read operations in ROM-like devices. Step-by-Step Solution: Apply an address; wait at least t_ac.Sample data when the datasheet guarantees validity.Design bus cycles so t_cycle ≥ t_ac + margins.Conclude: t_ac reflects the device’s read speed. Verification / Alternative check: Datasheets specify maximum t_ac for speed grades (e.g., 70 ns, 55 ns). Lower numbers correspond to faster parts and higher sustainable bus frequencies during asynchronous reads. Why Other Options Are Wrong: Incorrect: Contradicts standard timing definitions. Ambiguous: Setup/hold apply to synchronous buses; t_ac remains the key asynchronous speed figure. Valid only for SRAM: Access time is defined for ROMs as well. Common Pitfalls: Confusing program/erase timing with read access; ignoring output enable delays (tOE) that may further constrain sampling. Final Answer: Correct

Question 20

ROM concept check: Read-only memory (ROM) is a memory type in which data are stored permanently or semi-permanently and are not lost when power is removed. Assess this statement.

Accepted Answer

Introduction / Context: Digital systems use both volatile and nonvolatile memories. ROM-class devices hold code and configuration across power cycles. Understanding permanence clarifies design choices for boot firmware and configuration storage. Given Data / Assumptions: ROM includes mask ROM, PROM, EPROM, and flash when used read-only. “Semi-permanent” recognizes erasable types (EPROM/flash) that retain data without power but can be reprogrammed by special procedures. We discuss data retention, not write mechanisms. Concept / Approach: In ROM families, bits remain stored without power (nonvolatility). Mask ROM is factory-programmed and fixed. PROM is one-time programmable by blowing fuses. EPROM and flash are field-programmable and erasable (UV or electrical) but still nonvolatile between power cycles. Hence the statement is accurate. Step-by-Step Solution: Classify ROM types and their programmability.Note that all retain data without power.Recognize that some are erasable—thus “semi-permanent.”Conclude the characterization matches standard usage. Verification / Alternative check: System block diagrams show ROM/flash holding firmware that persists across reboots, unlike DRAM/SRAM which clear on power loss. Why Other Options Are Wrong: Incorrect: Would deny nonvolatility. Only true for RAM / Applies only to EEPROM: RAM is volatile; EEPROM is a subset of erasable nonvolatile memories, not the only ROM type. Common Pitfalls: Using “ROM” to mean only mask ROM; ignoring that flash used read-only still qualifies functionally as ROM in many contexts. Final Answer: Correct

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