Question 1

Address multiplexing (DRAM context): Evaluate the statement: “Address multiplexing is used to reduce the number of address lines.”

Accepted Answer

Introduction / Context: As memory densities grew, packaging a separate pin for every address bit became impractical. DRAM solved this by multiplexing row and column addresses over the same physical pins, dramatically cutting package pin counts and cost. This question checks whether the learner recognizes the purpose of address multiplexing. Given Data / Assumptions: We consider classic asynchronous DRAMs using RAS and CAS strobes. Row address and column address share the A0..An pins at different times. Goal: minimize package pins while still addressing large arrays. Concept / Approach: With N multiplexed pins, you can send R row bits and C column bits using the same pins: total addressability is 2^(R+C) while only max(R,C) pins are needed, not R+C pins. RAS latches the row; CAS latches the column. This halves the address pin count for square arrays and reduces it substantially for rectangular organizations, enabling high-density memories in economical packages. Step-by-Step Solution: Present row address on A0..An; assert RAS to latch it.Present column address on the same pins; assert CAS to latch it.Access proceeds using the combined row/column to select one cell or a page (for page-mode). Verification / Alternative check: Compare pin counts: a 4M x 1 DRAM might use 11 address pins multiplexed (row 11 + column 11 → 22 effective bits) rather than 22 separate lines, confirming the reduction. Why Other Options Are Wrong: Incorrect/only SRAM/parity/ROMs only: DRAMs are the canonical example; SRAMs typically use non-multiplexed addresses; parity is unrelated. Common Pitfalls: Assuming multiplexing is a timing optimization only; its primary benefit is pin reduction, with timing trade-offs handled by RAS/CAS cycles. Final Answer: Correct

Question 2

ROM usage characteristic (non-volatile storage): Evaluate: “Read-only memories (ROMs) are used to store data that generally cannot be easily changed.”

Accepted Answer

Introduction / Context: ROM technologies (mask ROM, PROM/OTP, EPROM, EEPROM, Flash) exist along a spectrum of mutability. What unifies them is non-volatility and relative resistance to casual change during normal operation. This item checks conceptual understanding of why ROMs are chosen for fixed code/data storage. Given Data / Assumptions: “ROM” is used broadly to encompass non-volatile read-mostly memories. “Easily changed” refers to in-circuit modification during normal system use. We acknowledge erasable families exist but with controlled procedures. Concept / Approach: ROMs are intended for content that should persist across power cycles and not be altered accidentally: bootloaders, lookup tables, microcode, calibration constants. Although EPROMs and Flash can be reprogrammed, doing so requires special voltages, sequences, or modes, not casual writes like RAM. Thus the statement—phrased as “generally cannot be easily changed”—is accurate. Step-by-Step Solution: Identify non-volatile need (survive power-down).Choose ROM/Flash family appropriate to update frequency.Recognize that runtime software cannot freely overwrite content without explicit programming steps. Verification / Alternative check: Compare with RAM: any write strobe updates RAM immediately, but ROM devices ignore normal write strobes or lack them; reprogramming paths are restricted. Why Other Options Are Wrong: Incorrect/only OTP/voltage-dependent/mask-only: Overly narrow. Even reprogrammable ROMs are not “easily changed” during normal operation. Common Pitfalls: Equating “reprogrammable” with “easy at runtime.” In embedded design, update workflows are deliberately guarded to prevent corruption. Final Answer: Correct

Question 3

Memory operation terminology: Evaluate the statement: “A write operation may also be referred to as a ‘fetch’ operation.”

Accepted Answer

Introduction / Context: Precision in memory-system vocabulary matters when reading datasheets, writing device drivers, or debugging bus cycles. “Fetch” has a specific meaning in computer architecture, and conflating it with “write” leads to misunderstandings about control signals and performance events. Given Data / Assumptions: We refer to standard CPU and memory terminology. Bus signals differentiate read vs write (e.g., WE vs OE, or read/write lines). Instruction fetch is a canonical read transaction. Concept / Approach: A write operation stores data into memory. A fetch operation is a read of an instruction (and, by extension, can mean any read of data/instruction when context is broad). On many CPUs, “instruction fetch” refers to the pipeline stage that reads opcodes from memory into the instruction cache or decode logic. Therefore, calling a write a “fetch” is incorrect; the terms denote different bus directions and control strobes. Step-by-Step Solution: Define write: transfer from CPU/peripheral to memory.Define fetch: memory to CPU read, typically for instructions.Conclude: write ≠ fetch. Verification / Alternative check: Examine timing diagrams: WE asserted for write cycles, OE asserted for read cycles; instruction fetches deassert WE. Performance counters also distinguish loads/stores and instruction fetches. Why Other Options Are Wrong: Correct/architecture/cache/DMA caveats: None change the fundamental directionality difference. Common Pitfalls: Using “fetch” generically for any memory access; in technical contexts, keep the distinction sharp. Final Answer: Incorrect

Question 4

ROM persistence claim: “ROMs are used to store data on a permanent basis.” Evaluate this statement with respect to non-volatility and practical updateability.

Accepted Answer

Introduction / Context: Designers often describe ROM contents as “permanent.” While that word can be nuanced—since some ROM families are erasable—the intent is to contrast ROM with volatile RAM. This question tests whether you understand ROM's role as non-volatile storage for fixed code and tables that must survive power cycles. Given Data / Assumptions: “Permanent” is interpreted as “retained without power and not changeable during normal operation.” Mask ROM is physically fixed at fabrication; EPROM/EEPROM/Flash require special procedures to alter. We are not considering extreme failure modes (e.g., over-erase, data retention limits after decades). Concept / Approach: ROM devices are non-volatile: their contents persist across power-downs. In most systems, ROM code (firmware) is not intended to change frequently; updating, if supported, involves controlled programming steps. Thus, describing ROM usage as “permanent” is accurate in the practical engineering sense, even though technology variants allow reprogramming. Step-by-Step Solution: Identify storage need that must survive reset and power loss.Select ROM/Flash to host this content.Note that updates require specialized programming interfaces, not routine writes. Verification / Alternative check: Boot processes rely on immutable code in ROM/Flash to initialize DRAM, peripherals, and loaders. Consumer devices ship firmware in non-volatile memories specifically to ensure persistence. Why Other Options Are Wrong: Incorrect/mask-only/battery-only/temp-limited: Overly restrictive; non-volatility is the key trait across ROM families. Common Pitfalls: Interpreting “permanent” as “physically impossible to change.” Engineers should read it as “not normally changed and retained without power.” Final Answer: Correct

Question 5

Firmware in ROM (application and terminology): Evaluate the statement: “The most widespread use of ROM is storing data and program code that must persist across power-down in microprocessor-based systems; such code is called firmware because it is f…

Accepted Answer

Introduction / Context: From embedded controllers to PCs, systems require trusted code at power-on to initialize hardware, configure clocks, and load operating software. This immutable (or tightly controlled) code is typically stored in ROM or Flash and is referred to as firmware. The item checks whether learners link ROM's non-volatility to the practical need for persistent program storage. Given Data / Assumptions: We consider microcontroller and microprocessor systems broadly (IoT, appliances, automotive, PCs). “ROM” here includes reprogrammable non-volatile memories (Flash/EEPROM) used for code storage. Power cycling should not erase critical code. Concept / Approach: Firmware is software stored in non-volatile memory on or near the hardware it controls. It must be available immediately at reset to bootstrap the system. ROM is ideal for this role because it retains contents without power. Even when firmware resides in Flash (reprogrammable), it still functions as ROM at runtime and remains present after power-down. The term “firmware” emphasizes its close coupling with hardware and relative infrequency of updates. Step-by-Step Solution: At reset, CPU fetches from a fixed address mapped to ROM/Flash.Boot code initializes memory controllers, clocks, and peripherals.Optional loader then copies or maps further code into RAM for execution. Verification / Alternative check: Examine any microcontroller datasheet: it boots from internal Flash/ROM. PC BIOS/UEFI resides in SPI Flash, persisting across power cycles and updated only via explicit flashing procedures. Why Other Options Are Wrong: Incorrect/PC-only/mask-only/obsolete: All conflict with widespread embedded practice. Common Pitfalls: Assuming “firmware” means “never updatable.” Many devices allow controlled updates but still rely on non-volatile storage. Final Answer: Correct

Question 6

Specific DRAM pinout claim (TMS44100 4M × 1): Evaluate: “The Texas Instruments TMS44100 4M × 1 DRAM does not provide a dedicated Chip Select (CS) input.”

Accepted Answer

Introduction / Context: Many classic DRAMs omit a dedicated Chip Select input. Instead, selection is achieved by gating the Row Address Strobe (RAS) and Column Address Strobe (CAS) signals. Recognizing such conventions prevents wiring errors and clarifies why memory controllers often generate qualified RAS/CAS rather than a separate CS line. Given Data / Assumptions: Device: TI TMS44100 series, organized as 4,194,304 × 1. We refer to the manufacturer’s pin nomenclature. Active control pins include RAS, CAS, and W (write enable), with address lines A0..A10, D (data-in), and Q (data-out). Concept / Approach: If the datasheet pin list omits CS, then chip enabling is handled implicitly: when RAS (and subsequently CAS) are asserted with valid addresses, the device becomes active. System designs often implement bank selection by decoding addresses to generate distinct RAS lines to each DRAM, effectively serving the role of CS. Step-by-Step Solution: Check pin nomenclature: A0–A10, RAS, CAS, W, D, Q, VCC, VSS, NC.Observe absence of any CS/CE pin in the pin list and top-view diagrams.Conclude that the device lacks a dedicated CS; selection is via RAS/CAS. Verification / Alternative check: The TMS44100 datasheet pin list shows exactly these pins and no CS input; controller logic qualifies RAS/CAS per chip to select devices. :contentReference[oaicite:0]{index=0} Why Other Options Are Wrong: Incorrect/package/grade/refresh-mode caveats: The absence of CS is a device-family characteristic, not dependent on package, grade, or refresh mode. Common Pitfalls: Assuming every memory has “CS” like SRAMs or ROMs; DRAM ecosystems commonly rely on RAS/CAS gating instead. Final Answer: Correct

Question 7

Storage devices – does a compact disc (CD) player provide random access capability (for example, jump directly to a track/time position without scanning all prior data)?

Accepted Answer

Introduction / Context: Random access means the device can seek directly to a desired location without stepping through all intervening data. This question checks whether an optical compact disc (CD) player fits that definition, contrasting it with strictly sequential media like magnetic tape. Given Data / Assumptions: The medium is a standard audio/data CD with a table of contents and physical addressing. The player includes a laser pickup on a sled with a tracking/servo system. We are evaluating access behavior, not audio quality or file-system specifics. Concept / Approach: A CD stores data along a spiral track, but the mechanism can reposition the optical pickup rapidly to a new radius and lock to the requested logical block address or track index. Because the controller can compute the target position and move there directly, the access mode is classified as random (seekable), not purely sequential. This stands in contrast to tape, which must wind past intervening data to reach a later point. Step-by-Step Solution: 1) Define random access: reach item N without visiting items 1..N−1.2) Examine the CD mechanism: sled servo moves the pickup to the radius for the target block.3) Controller uses block addresses or track/time indices to lock accurately.4) Conclusion: the device performs direct seeks and thus supports random access. Verification / Alternative check: Practical behavior: users select track 10 instantly from track 1; data CD drives “seek” to files by logical block address. Typical firmware briefly references the table of contents (TOC) to map positions but does not scan intervening audio frames to reach the destination. Why Other Options Are Wrong: “Incorrect” confuses spiral layout with access mode. “Only on rewritable CDs” is irrelevant; read-only discs are also seekable. “Only after full TOC scan each time” misstates: TOC read is quick metadata, not sequential traversal. “Only if disc is in RAM mode” mixes up packet writing with access capability. Common Pitfalls: Equating spiral physical layout with sequential access; ignoring that mechanical seeking enables direct positioning similar to hard drives. Final Answer: Correct

Question 8

Commercial terminology – is “sequential-access memory” the term often used to refer to read/write memory (i.e., RAM) in products and documentation?

Accepted Answer

Introduction / Context: Clear terminology prevents design mistakes. “Read/write memory” in computing commonly means RAM (random access memory), whereas “sequential access” describes media that must be traversed in order, like tape. This item tests whether you can separate these terms in a commercial/engineering context. Given Data / Assumptions: Read/write RAM allows both reading and writing at arbitrary addresses. Sequential access implies access must follow physical order. Marketing names sometimes blur categories, but engineering definitions remain consistent. Concept / Approach: The industry-standard term for read/write main memory is “RAM,” specifically DRAM or SRAM. “Sequential-access memory” is not used as a synonym for RAM. Instead, it characterizes devices requiring ordered traversal (for example, tape libraries). Therefore, the statement equating read/write memory with sequential access is incorrect. Step-by-Step Solution: 1) Define RAM: read/write at any address in roughly uniform time.2) Define sequential access: reach record N only after passing 1..N−1.3) Compare definitions: they are mutually exclusive access models.4) Conclude the commercial term for read/write memory is RAM, not sequential-access memory. Verification / Alternative check: Textbooks and datasheets label commodity memory as DRAM/SRAM (RAM). Tape and some streams are described as sequential-access devices. Product literature for PCs lists “RAM: 8 GB,” not “sequential-access memory.” Why Other Options Are Wrong: “Correct” contradicts standard usage. “Only in magnetic tape systems” confuses device class with naming. “Only for FIFO buffers” describes queue behavior, not general memory. Cache hierarchy does not alter access classification. Common Pitfalls: Assuming “sequential” refers to how software processes data; access mode is a hardware capability, not an application pattern. Final Answer: Incorrect

Question 9

Definition check – if a memory lets you access any location directly without stepping through others first, is it called Random Access Memory (RAM)?

Accepted Answer

Introduction / Context: Memory devices are categorized by how data is accessed. This question targets the fundamental definition: RAM provides essentially uniform-time access to any address, while sequential devices do not. Recognizing this distinction is critical when designing address decoders and timing interfaces. Given Data / Assumptions: We refer to access behavior, not volatility. RAM includes SRAM and DRAM technologies. Direct addressing means the controller selects a location by its address lines immediately. Concept / Approach: Random access memory allows direct read/write to any location through address decoding, independent of the position accessed previously. Both SRAM and DRAM satisfy this property. Therefore, if a memory design allows access without sequencing through other locations, it meets the definition of RAM. Step-by-Step Solution: 1) Define RAM: direct, addressable, near-constant access time.2) Evaluate the described device: access without sequencing → direct addressing.3) Match to category: this behavior is RAM.4) Result: the statement is correct. Verification / Alternative check: DRAM timing diagrams show row/column selects to reach any cell; SRAM uses wordline/bitline selection. Neither requires scanning earlier addresses, unlike tape or linked serial EEPROMs with pointer stepping. Why Other Options Are Wrong: Limiting to SRAM excludes DRAM unnecessarily. Battery backing concerns retention, not access mode. Word alignment does not change random access capability. Common Pitfalls: Confusing volatility (RAM loses data without power) with access classification; not all nonvolatile memories are sequential and not all RAM must be volatile conceptually. Final Answer: Correct

Question 10

Portable storage – are flash memories frequently used in place of floppy disks or small-capacity hard drives in portable computers and devices?

Accepted Answer

Introduction / Context: Flash memory revolutionized portable storage by providing nonvolatile, shock-resistant, low-power alternatives to legacy magnetic media. This item asks whether flash has commonly replaced floppies and small hard drives in portable contexts such as laptops, handhelds, and embedded systems. Given Data / Assumptions: Flash includes NAND/NOR technologies, SD cards, USB drives, and SSDs. Target comparison: floppy disks and low-capacity HDDs historically used in portables. Criteria: practicality, prevalence, and typical usage scenarios. Concept / Approach: Flash is nonvolatile and has no moving parts, offering superior durability and power efficiency for mobile use. As capacities increased and cost/GB fell, flash devices displaced floppies for file transfer and small HDDs for system storage (solid-state drives). Therefore, the statement reflects mainstream practice across decades of portable computing evolution. Step-by-Step Solution: 1) Identify needs of portable systems: low power, robustness, adequate capacity.2) Evaluate flash: meets all three, bootable when used as SSD or embedded eMMC.3) Compare with floppy/small HDD: lower durability and higher power, mechanical fragility.4) Conclude: flash frequently replaces those media in portable devices. Verification / Alternative check: Market trends: laptops commonly ship with SSDs; removable flash (SD/USB) supplanted floppies for transfer. Embedded devices store firmware on NOR/NAND flash rather than micro-HDDs. Why Other Options Are Wrong: Read-only limitation is false; flash is rewritable. “Only below 128 MB” is outdated. Magnetic shielding is irrelevant to solid-state flash. Common Pitfalls: Assuming flash is always slower than HDDs; modern SSDs outperform small HDDs on latency and throughput for typical workloads. Final Answer: Correct

Question 11

Nonvolatile memory behavior – are EEPROM and Flash memory both electrically erasable technologies?

Accepted Answer

Introduction / Context: Erasability determines how firmware and data can be updated in the field. Electrically erasable devices allow in-system programming without special lamps or ovens. This question distinguishes electrically erasable families from older EPROM technology that required ultraviolet light for erasure. Given Data / Assumptions: EEPROM stands for Electrically Erasable Programmable Read-Only Memory. Flash is a high-density variant using block/page erase operations. We compare erase mechanisms, not endurance or speed. Concept / Approach: Both EEPROM and Flash store charge on floating gates and remove or add charge via Fowler-Nordheim tunneling under controlled electric fields. EEPROM typically allows byte-level writes/erases; Flash erases larger blocks but is still electrically erasable. Therefore, both qualify as electrically erasable technologies. Step-by-Step Solution: 1) Identify storage element: floating-gate MOSFET in both devices.2) Identify erase method: electric field-induced tunneling removes charge.3) Note granularity difference: EEPROM (fine), Flash (block/page).4) Conclude both are electrically erasable. Verification / Alternative check: Device datasheets specify in-circuit erase/program protocols (I2C/SPI EEPROMs; on-board Flash controllers) with no UV window or exposure procedures, unlike classic EPROMs. Why Other Options Are Wrong: Claims requiring UV or heat describe EPROM or one-time PROMs, not EEPROM/Flash. Saying neither is erasable contradicts their core function. Common Pitfalls: Confusing EEPROM with EPROM due to similar acronyms; overlooking erase granularity differences that affect firmware update strategies. Final Answer: Correct

Question 12

EEPROM storage physics – is the floating-gate MOSFET the bit-cell, with charge retained on the floating gate for 10+ years unless electrically removed?

Accepted Answer

Introduction / Context: Understanding how nonvolatile memories store information helps explain retention and endurance specs. EEPROMs and Flash use a floating-gate transistor to store charge, which alters the cell’s threshold voltage and thereby encodes a bit. The statement summarizes this behavior and typical retention times. Given Data / Assumptions: Bit cell is a floating-gate MOSFET. Charge storage modifies threshold to represent 0/1. Typical retention guarantees exceed 10 years under nominal conditions. Concept / Approach: The floating gate is insulated by oxide, so injected electrons remain trapped for long periods. Reading senses the threshold shift; erasing/programming adds or removes charge via tunneling. Unless an erase/program pulse or high-temperature stress accelerates leakage, charge remains for many years, aligning with datasheet retention specs. Step-by-Step Solution: 1) Identify the storage mechanism: charge on an isolated conductor (floating gate).2) Correlate charge with threshold voltage and logic state.3) Note retention: oxide isolation yields multi-year stability (often 10–20+ years).4) Conclude the description is accurate. Verification / Alternative check: Vendor specifications specify retention at temperature (for example, 10 years at 85 C or longer at room temperature). Field-programmable microcontrollers store firmware in Flash for device lifetimes without continuous power. Why Other Options Are Wrong: The mechanism is common to EEPROM and Flash; continuous power is not required for charge retention; claiming hours contradicts practice. Common Pitfalls: Confusing retention (data persistence) with endurance (cycle count). Retention is long even as endurance limits write/erase frequency. Final Answer: Correct

Question 13

Per-cell data density – can a memory cell store more than one bit (for example, multi-level cell Flash storing 2–3–4 bits per cell)?

Accepted Answer

Introduction / Context: While textbook diagrams often depict “one cell = one bit,” advanced memories can encode multiple bits in a single physical cell by distinguishing several levels of a stored quantity. This question checks awareness of multi-level storage techniques that increase density at the cost of complexity and sometimes reliability. Given Data / Assumptions: Example: Flash MLC/TLC/QLC storing 2/3/4 bits per cell via multiple threshold levels. Other memories (PCM, MRAM variants) can also support multi-level operation. Error-correcting codes mitigate tighter sensing margins. Concept / Approach: Multi-level cells partition an analog parameter (threshold voltage, resistance) into several ranges. Each range maps to a bit pattern: 2 levels → 1 bit (SLC), 4 levels → 2 bits (MLC), 8 levels → 3 bits (TLC), 16 levels → 4 bits (QLC). Controllers perform precise program/verify steps and use ECC to preserve data integrity. Step-by-Step Solution: 1) Identify the measurable cell parameter (for example, Vt for Flash).2) Define multiple non-overlapping windows across that parameter.3) Map windows to bit patterns (00, 01, 10, 11, ...).4) Conclude: per-cell capacity can exceed one bit when reliably distinguished levels exist. Verification / Alternative check: Commercial SSDs widely advertise MLC/TLC/QLC technologies. Characterization data shows narrower margins and higher ECC strength with increased bits per cell. Why Other Options Are Wrong: Saying only DRAM or only magnetic disks can do this is incorrect; DRAM generally uses 1 bit per capacitor, while Flash commonly uses multi-level states. Parity encoding does not create physical multi-level storage. Common Pitfalls: Assuming “cell” always equals one bit; forgetting the controller’s role in managing program/verify and ECC for multi-level reliability. Final Answer: Correct

Question 14

After programming – are PROMs functionally equivalent to mask ROMs (i.e., fixed read-only contents)?

Accepted Answer

Introduction / Context: Read-only memories come in several manufacturing/programming styles. Mask ROMs are factory-programmed during fabrication; PROMs are programmed by the user once after fabrication. This question asks whether a PROM, once programmed, behaves like a mask ROM from a system’s perspective. Given Data / Assumptions: PROM = one-time programmable by blowing fuses or setting antifuses. Mask ROM = patterns defined in silicon masks at the factory. Focus on behavior after programming, not on how the pattern was created. Concept / Approach: After a PROM is programmed, its bit pattern is fixed and read-only. The access signals, timings, and logic-level behavior mirror ROM semantics. Therefore, at runtime, a programmed PROM and a mask ROM both present immutable data to the bus and are functionally indistinguishable to the rest of the system. Step-by-Step Solution: 1) Program PROM once (fuse/antifuse change establishes permanent connections).2) Use the device as read-only memory in circuit.3) Observe identical functional role to mask ROM: fixed lookup of addresses → data.4) Conclude equivalence in functional behavior post-programming. Verification / Alternative check: Datasheets list PROM read timing matching ROM interfaces; system schematics swap PROM/mask ROM footprints as drop-in options in some designs. Why Other Options Are Wrong: UV erasure applies to EPROM, not PROM. Bipolar vs CMOS technology does not change post-program read-only behavior. Refresh cycles are DRAM concerns, not ROM/PROM. Common Pitfalls: Confusing PROM with EPROM/EEPROM; forgetting PROM is one-time programmable and not erasable. Final Answer: Correct

Question 15

One-time programmable behavior – can a programmed PROM be changed later by applying a small electrical charge?

Accepted Answer

Introduction / Context: Memory families differ in whether content can be altered after initial programming. PROMs (programmable read-only memories) are designed for one-time programming, unlike EPROM/EEPROM/Flash which support erasure and rewriting. This item checks whether PROM contents can be changed electrically after programming. Given Data / Assumptions: PROM programming creates permanent physical changes (fuse blown or antifuse formed). No erase mechanism is provided by design. We are not discussing EPROM or EEPROM. Concept / Approach: Because PROM cells are altered irreversibly during programming, there is no subsequent electrical operation that restores the original state. Therefore, the assertion that a small electrical charge can change PROM contents later is false. Any device that can be erased by electrical means is not a classic PROM; it is an EEPROM or Flash device. Step-by-Step Solution: 1) Identify PROM cell type: fuse/antifuse architecture.2) Programming permanently changes connectivity at the cell.3) No erase path exists to undo that physical change.4) Conclude that post-program modification by a small charge is impossible. Verification / Alternative check: Datasheets describe PROMs as OTP (one-time programmable). In contrast, EPROMs mention UV windows; EEPROMs/Flash list electrical erase commands. PROMs lack such sections entirely. Why Other Options Are Wrong: UV erasers affect EPROM, not PROM. Reversing charge is inapplicable to fuse links. Temperature does not provide an erasure mechanism. Common Pitfalls: Mixing up acronyms (PROM vs EPROM vs EEPROM) and assuming all “programmable ROMs” are rewritable. Final Answer: Incorrect

Question 16

System organization – can DRAM chips be combined to build larger capacity and wider word sizes using the same methods applied to other memory types?

Accepted Answer

Introduction / Context: Practical systems rarely use a single memory chip. Designers array multiple devices to meet total capacity and bus width targets. This question checks whether DRAM can be organized in banks and x-width groupings similarly to other memory families, acknowledging refresh considerations but focusing on organization principles. Given Data / Assumptions: Goal: increase total bytes and/or widen data bus (for example, from x8 devices to x64 bus). Standard memory controllers handle DRAM timing, refresh, and chip select signals. Comparable methods are used for SRAM/ROM arrays at the organization level. Concept / Approach: To widen the data path, place several x8 DRAMs in parallel (data lines aggregated) to create x16, x32, or x64 interfaces. To increase depth (capacity), use additional ranks/banks with separate chip selects or address multiplexing. These methods mirror how SRAM/ROM arrays are organized; the main difference is that DRAM requires periodic refresh and row/column multiplexed addressing, which the controller manages transparently. Step-by-Step Solution: 1) Choose device width (e.g., x8) and count needed to match bus (e.g., 8 chips → x64).2) Tie corresponding address/control lines; route each chip to its data byte lane.3) Add chip selects/ranks to scale capacity beyond a single set.4) Use a controller to schedule refresh and command sequences (ACT, RD/WR, PRE). Verification / Alternative check: DIMMs combine many DRAM chips into modules presenting standard widths (x64 with ECC x72). Reference designs for microcontrollers show similar organization with smaller DRAMs. Why Other Options Are Wrong: Claiming SRAM-only is false; DRAM modules exemplify the method. Refresh cannot be disabled; it is handled, not avoided. ECC is optional and does not determine combinability. Common Pitfalls: Forgetting byte-lane routing and DQ/DQS alignment; overlooking rank address mapping; assuming DRAM’s need for refresh prevents standard width/capacity scaling—it does not. Final Answer: Correct

Question 17

Memory terminology check in digital electronics: A "nibble" (sometimes spelled nybble) refers to a small fixed-width group of bits. Evaluate the statement: "A nibble is a group of eight bits." Provide the correct definition and reasoning.

Accepted Answer

Introduction / Context: Digital systems use consistent names for common bit widths so engineers and programmers can communicate precisely. The terms bit, nibble, byte, word, and double word appear in datasheets, assembly language manuals, and memory-mapped I/O documentation. Misdefining these units causes confusion when interpreting registers, encodings, and bus widths. Given Data / Assumptions: The statement claims that a nibble is a group of eight bits. Standard terminology is assumed: 1 byte = 8 bits in virtually all modern contexts. We contrast nibble with byte and related groupings. Concept / Approach: A nibble is defined as a group of 4 bits. It is exactly half of a byte. Because hexadecimal digits map neatly to 4-bit values (0000 to 1111 equals 0 to 15), the nibble is the natural unit for hex representation. For example, 0xAF is composed of two nibbles: A = 1010 and F = 1111. In contrast, eight bits form a byte (also called an octet), not a nibble. Step-by-Step Solution: 1) Recall standard sizes: bit (1), nibble (4), byte (8), word (implementation-dependent, often 16 or 32), double word (commonly 32 or 64).2) Map hexadecimal to binary: one hex digit ↔ one nibble (4 bits).3) Therefore, 8 bits is a byte, while 4 bits is a nibble.4) The claim “nibble = 8 bits” contradicts the accepted definition. Verification / Alternative check: Check any processor programming guide: status registers and immediate fields are often described in nibbles for hex readability. Tutorials consistently pair nibbles with hex digits and bytes with two hex digits. Why Other Options Are Wrong: “Correct” repeats the incorrect claim. “2 bits” and “16 bits” misstate the conventional size; neither aligns with standard usage. Common Pitfalls: Confusing historical variations of “byte” in very old systems with the modern, near-universal 8-bit byte; however, the nibble has consistently meant 4 bits in mainstream literature. Final Answer: Incorrect

Question 18

CPU–memory interaction during program execution: While a computer is running a program, the central processing unit (CPU) repeatedly fetches (1) instruction codes that specify the operations to perform and (2) the data operands on which to operate, …

Accepted Answer

Introduction / Context: Understanding how a CPU interacts with memory is foundational for computer architecture, embedded design, and performance tuning. Instruction fetch and data access are the two primary memory transactions that occur during normal execution of a program, regardless of instruction set specifics. Given Data / Assumptions: The CPU executes a sequence of instructions in memory. The CPU must fetch both instruction opcodes and associated operands. Memory hierarchy (caches, main memory) may be present but does not change the logical requirement to fetch code and data. Concept / Approach: Every instruction cycle includes an instruction fetch phase. Many instructions also require reading data operands (and possibly writing results) from/to memory or registers. Whether the architecture is von Neumann (unified instruction/data memory) or Harvard (separate instruction/data buses), the CPU conceptually “fetches” instructions and operands from designated memories. Step-by-Step Solution: 1) Instruction fetch: the program counter (PC) holds the address of the next instruction.2) Decode: the fetched opcode determines the operation and addressing mode.3) Operand access: the CPU reads data from registers or memory addresses computed by the addressing mode.4) Execute and write-back: results may be stored to registers or memory; PC updates to the next instruction (or branch target). Verification / Alternative check: Pipeline diagrams across architectures (e.g., fetch–decode–execute) explicitly show instruction fetch and data memory stages. Caches hide latency but do not eliminate the conceptual fetches. Why Other Options Are Wrong: “Incorrect” denies standard CPU behavior. Limiting correctness to interrupts or Harvard architectures misunderstands that ordinary instruction execution (not just interrupts) fetches code and data in all mainstream designs. Common Pitfalls: Confusing register-to-register instructions (which still fetch code) with memory operations; assuming caches remove the need to fetch—they only satisfy fetches faster. Final Answer: Correct

Question 19

Memory cell technology distinction: Evaluate the statement: "DRAM uses a cross-transistor configuration." Clarify the difference between dynamic RAM (DRAM) and static RAM (SRAM) cell structures.

Accepted Answer

Introduction / Context: Dynamic RAM (DRAM) and static RAM (SRAM) implement memory cells differently, trading density for speed and complexity. Knowing the cell structure helps explain refresh requirements, power behavior, and typical applications. Given Data / Assumptions: The claim says DRAM uses a “cross-transistor configuration.” We compare canonical DRAM and SRAM cell topologies. Terminology: cross-coupled (or cross-latched) generally refers to two inverters forming a bistable latch (SRAM). Concept / Approach: DRAM cells are typically 1T1C: one access transistor (the “T”) and one storage capacitor (the “C”). The capacitor holds charge to represent a bit and must be periodically refreshed because charge leaks over time. SRAM cells are commonly 6T in CMOS: two cross-coupled inverters (forming a stable latch) plus access transistors. Therefore, the phrase “cross-transistor configuration” fits SRAM, not DRAM. Step-by-Step Solution: 1) Identify DRAM cell: one transistor connects the storage capacitor to the bitline when the wordline is asserted.2) Recognize the need for refresh: stored charge decays → periodic read/restore.3) Contrast with SRAM: two cross-coupled inverters maintain state without refresh while powered.4) Conclude the claim about DRAM is incorrect; cross-coupled topology characterizes SRAM. Verification / Alternative check: Introductory VLSI texts depict DRAM 1T1C cells and SRAM 6T latch cells; timing diagrams show DRAM refresh cycles absent in SRAM. Why Other Options Are Wrong: “Correct” contradicts mainstream DRAM design. The 6T/4T options mis-attribute SRAM-like latch structures to DRAM and omit the capacitor central to DRAM. Common Pitfalls: Assuming all memories use latches; overlooking the capacitor in DRAM; conflating cell topology with sense-amplifier cross-coupling (a separate concept). Final Answer: Incorrect

Question 20

Shared bus behavior: When two or more devices attempt to drive their own digital levels onto a shared data bus at the same time (without proper arbitration or tri-state control), bus contention occurs. Assess the accuracy of this statement.

Accepted Answer

Introduction / Context: Digital systems often use a shared bus to connect multiple ICs. To prevent damage and logic errors, only one device should actively drive the bus at a time unless the bus uses a wired-OR/wired-AND topology with passive pull elements. The term “bus contention” describes the fault condition when multiple active drivers assert conflicting levels. Given Data / Assumptions: Multiple potential bus masters or slaves share lines. No tri-state, open-drain, or arbitration mechanism is engaged, or it is misused. Drivers can source and sink current strongly. Concept / Approach: On a push-pull (totem-pole) bus, two drivers fighting—one driving HIGH and another driving LOW—create excessive current, possible damage, and undefined logic levels. Proper design uses tri-state outputs with enable signals, open-drain/open-collector interfaces with pull-up resistors, bus transceivers, and arbitration protocols (e.g., grants, chip selects) to ensure only one active driver at a time. Step-by-Step Solution: 1) Identify the bus topology (push-pull vs open-drain).2) Ensure outputs are tri-stated when not selected (via OE or direction control).3) Use chip-select and arbitration so exactly one device drives at a time.4) If multiple devices drive simultaneously, contention results—confirming the statement. Verification / Alternative check: Bus standards (e.g., PCI, parallel memory buses, microcontroller GPIO buses) specify tri-state timing to avoid overlap; I2C avoids contention by open-drain wiring and level-dominance rules. Why Other Options Are Wrong: “Incorrect” ignores basic electrical behavior. Restricting to open-collector or both-HIGH cases misses that contention is specifically problematic when drivers oppose levels; open-drain buses are designed to avoid destructive contention. Common Pitfalls: Enabling two tri-state drivers at once; misunderstanding turnaround timing; forgetting bus-holder components can mask but not fix contention. Final Answer: Correct

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