Question 1

Memory taxonomy – persistence of stored data Why are ROM devices classified as nonvolatile memory in digital systems?

Accepted Answer

Introduction / Context: Computer memories are often grouped by volatility. Volatile memories require continuous power to retain stored information, while nonvolatile memories retain data without power. This classification is fundamental when selecting storage for firmware versus working RAM. Given Data / Assumptions: The device under discussion is ROM (Read-Only Memory) in its broad sense (mask ROM, EPROM, EEPROM, Flash used as ROM). We compare behavior with and without system power. Concept / Approach: Nonvolatile memory retains its state without external power. ROMs store firmware or constant data that must persist across power cycles. In contrast, volatile memory like DRAM and SRAM lose their contents when power is removed (DRAM also needs refresh even when powered). Step-by-Step Solution: Define nonvolatile: retains data without power.Confirm ROM behavior: contents remain intact across power cycles.Therefore the correct explanation is that ROM does not lose memory when power is removed. Verification / Alternative check: System boot sequences depend on code stored in ROM/Flash. If ROM were volatile, systems could not start without external reprogramming every power-up, which contradicts real-world behavior. Why Other Options Are Wrong: They lose memory when power is removed: This describes volatile RAM, not ROM. Blank options do not provide a technical statement and can be ignored. Common Pitfalls: Confusing “read-only” with “nonvolatile.” Read-only refers to typical use-mode; many ROM technologies can be rewritten under specific procedures, yet they are still nonvolatile. Final Answer: They do not lose memory when power is removed.

Question 2

Address space calculation – 2118 16K × 1 RAM How many binary address lines are required to uniquely select all memory locations in a 2118 RAM specified as 16K × 1 (i.e., 16,384 locations, 1 bit each)?

Accepted Answer

Introduction / Context: Sizing the address bus is a basic skill in memory interfacing. The number of address lines determines how many unique locations can be selected. Here, the device stores 1 bit at each of 16K locations. Given Data / Assumptions: Memory size: 16K × 1, which means 16,384 distinct addresses. One address selects one 1-bit cell. Each extra address line doubles the addressable locations. Concept / Approach: If a memory has N locations, the minimum number of address lines A must satisfy 2^A ≥ N. For exactly 16,384 locations, we solve 2^A = 16,384. Step-by-Step Solution: Compute the power: 2^14 = 16,384.Therefore A = 14 address lines are needed.No additional lines are required since data width is 1 bit; the data bus width is independent of address count. Verification / Alternative check: Recognize 1K = 1024 = 2^10, so 16K = 16 * 1024 = 2^4 * 2^10 = 2^14. Thus 14 lines are needed. Why Other Options Are Wrong: 8 or 10: 2^8 = 256 and 2^10 = 1024, far too small. 16: 2^16 = 65,536 addresses, more than required; 14 is sufficient and minimal. Common Pitfalls: Mistaking 16K as 16,000 rather than 16,384 (binary kilo). Confusing data width (×1) with address count, which are independent parameters. Final Answer: 14

Question 3

Digital fundamentals – terminology for 8 bits In standard digital electronics terminology, a group of eight binary bits is referred to as a:

Accepted Answer

Introduction / Context: Basic digital vocabulary is essential for interpreting datasheets and programming specifications. The term for eight bits is among the most frequently used in computing, microcontrollers, and communications. Given Data / Assumptions: We are dealing with binary groupings as used in common computer systems. The size in question is precisely eight bits. Concept / Approach: Historically, different systems used different word sizes, but the term for eight bits has long been standardized as “byte.” This is independent of the term “word,” which is architecture specific and may be 16, 32, 64 bits, etc. Step-by-Step Solution: Identify the required size: 8 bits.Match with the standard term: byte.Select “byte” from the options. Verification / Alternative check: Programming languages, file formats, and memory addressing commonly define one byte as eight bits; this is reflected in char size in many systems and in network protocol definitions. Why Other Options Are Wrong: Group: Vague, not a standard size. Word: Architecture dependent; not universally eight bits. Cell: Often used for a memory element, not a fixed eight-bit grouping. Common Pitfalls: Assuming “word” always equals two bytes; many systems use 32 or 64-bit words. Final Answer: byte.

Question 4

Memory construction – widening data width using multiple RAM ICs How many 1K × 4 RAM chips are required to build a 1K × 8 memory subsystem (same address depth, doubled data width)?

Accepted Answer

Introduction / Context: Designers often combine memory ICs to achieve a desired word width and depth. This question checks your ability to reason about arranging multiple narrow memories in parallel to widen the data bus while keeping the address space the same. Given Data / Assumptions: Available IC: 1K × 4 (1,024 addresses, 4 data bits each). Target memory: 1K × 8 (1,024 addresses, 8 data bits each). Address depth must remain 1K; only width doubles from 4 to 8 bits. Concept / Approach: To double data width, place identical RAMs in parallel on the same address lines. One chip provides the lower 4 bits (D3..D0), the other provides the upper 4 bits (D7..D4). The chip enable and write enable signals are shared so both devices respond to the same address and control. Step-by-Step Solution: Depth required: 1K → match with single chip depth 1K (address lines tied in common).Width required: 8 bits → combine two 4-bit devices in parallel.Total chips needed: 2. Verification / Alternative check: Block-diagramming shows two chips tied A0..A9 in common, with data busses concatenated to form D7..D0. This is a standard technique in memory expansion. Why Other Options Are Wrong: 4, 8, 16: These would oversupply width to 16, 32, or 64 bits respectively for a 1K depth, which is unnecessary. Common Pitfalls: Confusing widening (parallel) with deepening (cascading addresses), which would multiply the address space rather than the word width. Final Answer: 2

Question 5

Computer organization — word length vs. bytes A 64-bit word consists of how many bytes in mainstream digital systems (8 bits = 1 byte)?

Accepted Answer

Introduction / Context: Understanding how many bytes are contained in a given word size is foundational to computer architecture, memory addressing, and data alignment. A word is the natural unit of data used by a particular processor design. This question asks you to relate a 64-bit word length to its equivalent in bytes. Given Data / Assumptions: 1 byte = 8 bits (industry-standard definition). Word length under discussion = 64 bits. No exotic or legacy byte sizes are assumed; use modern conventions. Concept / Approach: To convert bits to bytes, use the direct proportionality: bytes = bits / 8. Word sizes such as 8, 16, 32, and 64 bits map to 1, 2, 4, and 8 bytes respectively. This mapping influences memory alignment, pointer sizes, and instruction encodings across contemporary 64-bit systems. Step-by-Step Solution: Identify the relationship: 1 byte = 8 bits.Compute bytes = 64 / 8.64 / 8 = 8.Therefore, a 64-bit word contains 8 bytes. Verification / Alternative check: Check against common architectures: x86-64, ARM64, RISC-V RV64 all define a 64-bit general-purpose register width, typically equated with an 8-byte word for loads/stores and ABI data types like 64-bit integers and pointers (platform dependent). Why Other Options Are Wrong: 4 bytes: That corresponds to 32 bits, not 64 bits. 10 bytes and 12 bytes: These imply nonstandard byte sizes or word definitions and do not match 64 / 8. Common Pitfalls: Confusing bits and bytes, especially when reading specifications that mix units. Assuming historical or nonstandard byte sizes; modern systems use 8-bit bytes. Final Answer: 8 bytes

Question 6

Nonvolatile memory technologies — mask ROM implementation Mask ROM (factory-programmed ROM) is typically implemented using which underlying technology?

Accepted Answer

Introduction / Context: Mask ROM is a read-only memory whose contents are defined during semiconductor fabrication via photolithographic masks. Recognizing its base technology helps differentiate it from programmable alternatives like PROM, EPROM, and Flash. Given Data / Assumptions: Modern integrated memories are fabricated on silicon using MOS (metal–oxide–semiconductor) processes. Historical, board-level diode matrices existed but are not the mainstream IC implementation. Mask ROM stores bits by the presence/absence of connections or transistors laid out at manufacture time. Concept / Approach: The term “mask ROM” derives from the photomask patterning used to create fixed data. In current practice, these arrays are realized as MOS transistor structures or hardwired connections formed within a MOS process node. Diode matrices are educational or very old implementations and not reflective of standard IC mask ROM today. Step-by-Step Solution: Identify mainstream IC process: CMOS/MOS.Recognize mask-defined connectivity in MOS arrays.Conclude that mask ROM is implemented using MOS technology. Verification / Alternative check: Fabrication flow and foundry process design kits (PDKs) for ROM macros target CMOS/MOS rules. Vendor datasheets for embedded ROM in microcontrollers also list CMOS processes and geometries (e.g., 180 nm, 90 nm, etc.). Why Other Options Are Wrong: Diode or resistor–diode technology: Not typical for monolithic, high-density modern ROM ICs. DROM technology: Not a standard term for fabrication technology. Common Pitfalls: Confusing legacy diode matrices with integrated mask ROM implementations. Assuming “ROM” implies a specific device package or programming method instead of a fabrication-defined array. Final Answer: MOS technology

Question 7

Memory volatility — selecting the volatile technology Which of the following memory types is volatile (loses data when power is removed)?

Accepted Answer

Introduction / Context: Computer systems use both volatile and nonvolatile memories. Volatile memories require continuous power to retain data and are typically used for working storage. Identifying which category a memory belongs to is crucial for system design and data integrity planning. Given Data / Assumptions: Volatile: RAM (SRAM, DRAM) loses data when power is removed. Nonvolatile: ROM, PROM/EPROM/EEPROM/Flash retain data without power. “EROM” is not a standard mainstream term; assume it is not the intended volatile choice. Concept / Approach: RAM stands for Random Access Memory and is used for temporary, high-speed storage during program execution. Dynamic RAM needs refresh even while powered, and all RAM loses content on power loss, making it volatile by definition. Step-by-Step Solution: List categories: RAM (volatile) vs. ROM/Flash (nonvolatile).Determine which loses data without power: RAM.Therefore, select RAM as the volatile memory. Verification / Alternative check: System behavior during reboot confirms this: contents of main memory vanish, while firmware stored in ROM/Flash persists to boot the machine again. Why Other Options Are Wrong: ROM and Flash: Nonvolatile; retain contents across power cycles. EROM: Not a standard volatile memory classification; even if interpreted as erasable ROM, it would still be nonvolatile when unpowered. Common Pitfalls: Assuming “random access” implies nonvolatility; it describes access method, not retention. Confusing EEPROM/Flash erasability with volatility; they keep data without power. Final Answer: RAM

Question 8

DRAM controllers — essential roles In a typical dynamic RAM (DRAM) subsystem, which two core functions are handled by the DRAM controller?

Accepted Answer

Introduction / Context: Dynamic RAM relies on a controller to manage row/column addressing and periodic refresh. Without these, the capacitive cells would quickly lose charge and the memory would be unusable. Knowing the controller’s duties clarifies why DRAM differs from SRAM in interface complexity. Given Data / Assumptions: DRAM uses multiplexed address buses: row address (RAS) and column address (CAS). Stored charge leaks with time; all rows must be refreshed periodically. “Data selection” is typically handled by bus interfaces or MUX/DEMUX logic outside the DRAM core protocol. Concept / Approach: The controller breaks a full address into row and column halves, asserting RAS and CAS signals to access the correct cell. It also issues refresh commands (auto-refresh or self-refresh modes) to maintain data integrity over time (e.g., full-array refresh within about 64 ms under normal temperatures). Step-by-Step Solution: Recognize DRAM addressing requires multiplexing: split address across time.Acknowledge leakage → periodic refresh needed.Therefore, the controller’s key roles are address multiplexing and refresh. Verification / Alternative check: DRAM datasheets and memory controller IP documentation specify RAS/CAS timing, tREFI scheduling, and refresh cycles, confirming these responsibilities. Why Other Options Are Wrong: Address multiplexing and data selection: Omits refresh, which is essential. Data selection and refresh: Ignores multiplexed addressing central to DRAM. Data selection and CPU accessing: Vague; not the distinctive DRAM controller duties. Common Pitfalls: Confusing DRAM refresh with cache coherence or DMA operations; they are unrelated. Assuming DRAM behaves like SRAM; DRAM requires controller-managed timing and refresh. Final Answer: Address multiplexing and the refresh operation

Question 9

Memory timing — typical tOE ranges by technology Considering common memory technologies, which set lists typical output-enable time (tOE) ranges?

Accepted Answer

Introduction / Context: The parameter tOE (output enable access time) indicates how quickly a memory's outputs respond after the output-enable control is asserted. Different process technologies exhibit different speed ranges. Recognizing realistic ranges helps you sanity-check datasheets and timing budgets. Given Data / Assumptions: Bipolar memories historically offered very fast switching (small tOE values). NMOS devices typically have moderate tOE values, slower than bipolar. CMOS devices vary, often between bipolar and NMOS depending on generation and design. Concept / Approach: Each option presents a plausible range for a given technology. Since all three sets are within realistic historical/typical ranges, the correct aggregate choice is that all are acceptable examples. Exact numbers vary by device family and year, but the orders of magnitude are consistent. Step-by-Step Solution: Evaluate bipolar: 10–20 ns → consistent with fast bipolar memories.Evaluate NMOS: 25–100 ns → consistent with classic NMOS speeds.Evaluate CMOS: 12–50 ns → plausible for many CMOS SRAM/ROM families.Therefore, accept all three → choose “All of the above”. Verification / Alternative check: Historical datasheets (e.g., fast bipolar SRAMs, NMOS EPROMs, CMOS SRAMs/ROMs) show tOE values in these neighborhoods, though exact numbers depend on process nodes and design targets. Why Other Options Are Wrong: Choosing only one technology ignores that each range is independently reasonable. Common Pitfalls: Confusing tOE with tACC (overall access time) or tRC (cycle time). They are related but distinct parameters. Assuming modern deep-submicron CMOS must always be faster; packaging and I/O design also influence tOE. Final Answer: All of the above

Question 10

Optical storage — how CD-R recording works Is the following statement about CD-R media correct? “A CD-R is created by heating special dye chemicals so they reflect less light than unburned areas, creating an effect equivalent to pits on a pressed CD…

Accepted Answer

Introduction / Context: CD-R (recordable compact disc) technology differs from factory-pressed CDs. Pressed discs have physical pits and lands molded into the substrate. CD-Rs simulate this difference in reflectivity using a photosensitive dye layer altered by a recording laser. This question verifies understanding of that mechanism. Given Data / Assumptions: Pressed CDs: physical pits (recessed) and lands (flat) change reflected light to encode data. CD-Rs: organic dye layer darkens where heated by the write laser. Optical pickup reads differences in reflected intensity rather than literal holes. Concept / Approach: When the CD-R's dye is locally heated, its optical properties change (absorption/reflectance). The altered spots reflect less light, mimicking the lower-reflectivity “pit” regions on a pressed disc. The drive's photodiode sees this contrast and recovers the data stream using run-length limited coding and clock recovery. Step-by-Step Solution: A recording laser targets dye areas along the spiral track.Heat causes a chemical/structural change, decreasing reflectivity.During playback, the reader detects reflectivity changes as pit/land transitions.Thus, burned dye regions emulate pits, confirming the statement. Verification / Alternative check: CD-R discs often have visible color tints (cyanine, phthalocyanine, azo dyes) that darken with recording. Technical references describe reflectivity targets for Red Book compatibility. Why Other Options Are Wrong: False: Would contradict the well-documented dye-based reflectivity mechanism of CD-R media. Common Pitfalls: Assuming CD-Rs physically form pits; they alter dye properties to emulate pits optically. Confusing CD-R with CD-RW, which uses a phase-change alloy rather than an organic dye. Final Answer: True

Question 11

Defining RAM — key property of access time Which statement best describes random-access memory (RAM) in terms of access time across locations?

Accepted Answer

Introduction / Context: The defining characteristic of Random Access Memory (RAM) is that any location can be accessed in essentially constant time relative to any other location, subject to timing constraints. This feature distinguishes RAM from sequential-access devices like tape drives. Given Data / Assumptions: RAM: SRAM or DRAM accessed by address decoding and internal arrays. Random access means location N and location M have similar access latency. Sequential media require traversal through intermediate positions. Concept / Approach: “Random access” refers to uniform-time addressing, not to volatility or write endurance. In RAM, decoding logic selects rows/columns quickly, enabling direct reads/writes at arbitrary addresses. Consequently, CPUs and DMA engines rely on RAM for rapid, nonsequential instruction and data access. Step-by-Step Solution: Define random access: time to access is independent of address sequence.Compare with sequential access: must pass through preceding data.Therefore, the best description is that access time is the same for each location. Verification / Alternative check: SRAM timing diagrams and DRAM controller schedules show consistent per-access timing envelopes regardless of address (ignoring caching/banking subtleties), validating the random-access property. Why Other Options Are Wrong: Depends on location: describes mechanical or sequential media, not RAM. Write once, read many: describes PROM/OTP, not RAM. Mass memory/sequential access: not the RAM model. Common Pitfalls: Confusing “random access” with “nonvolatile.” RAM is typically volatile. Assuming cache effects invalidate the definition; caches change effective latency, not the underlying property. Final Answer: Access time is the same for each memory location

Question 12

RAM meaning and role — selecting the correct definition What does RAM stand for, and what is its primary role in a computer system?

Accepted Answer

Introduction / Context: RAM is central to system performance because it holds the working set of instructions and data while programs run. Knowing what RAM stands for and its functional role clarifies the distinction between primary memory and long-term storage such as SSDs or HDDs. Given Data / Assumptions: RAM expands to Random Access Memory. RAM is volatile; contents are lost when power is removed. Primary usage: short-term storage for active processes, buffers, and caches. Concept / Approach: “Random Access” emphasizes uniform access time to any address. RAM temporarily holds code and data being actively used because it is much faster than secondary storage. Operating systems manage allocation, paging, and caching to optimize use of RAM for current workloads. Step-by-Step Solution: Identify the correct acronym expansion: Random Access Memory.Associate the role: temporary, short-term working store.Eliminate distractors that misuse the acronym or imply nonstandard behavior. Verification / Alternative check: System behavior during boot and shutdown shows RAM contents are transient, while applications load from persistent storage into RAM for execution, confirming the role. Why Other Options Are Wrong: Readily Available/Resettable Automatic Memory: Not correct expansions and not industry terms. “Memory that can be reached by any subsystem at any time”: Too vague and does not highlight the temporary working-store role. Common Pitfalls: Equating RAM with cache or virtual memory; they are related but distinct mechanisms. Confusing RAM with ROM/Flash, which are nonvolatile. Final Answer: Random Access Memory; short-term, temporary data storage used by the computer

Question 13

CD-ROM physical encoding — naming the recessed features On a pressed CD-ROM, the recessed areas that represent a logical 0 are called what?

Accepted Answer

Introduction / Context: Optical discs like CD-ROM store data as microscopic features molded into the polycarbonate substrate. Understanding the terminology for these features helps explain how drives read data using changes in reflectivity. Given Data / Assumptions: Two key surface states exist: pits (recessed) and lands (flat). The optical pickup detects reflectivity and phase changes between pits and lands. Logical encoding uses transitions; however, many explanations informally associate pits with zeros and lands with ones in didactic contexts. Concept / Approach: Pits are tiny depressions formed during pressing; lands are the untouched flat areas. The interference of reflected laser light differs between the two, producing the contrast needed for the photodetector to recover the bitstream after decoding and clock extraction. Step-by-Step Solution: Identify recessed features: pits.Identify nonrecessed/flat features: lands.Match the question's “recessed areas” to pits.Therefore, select “Pits”. Verification / Alternative check: CD standards (Red Book) and technical diagrams label the depressions as pits and the flat regions as lands; photography of disc surfaces under microscopes confirms the naming convention. Why Other Options Are Wrong: Mounds/Holes: Not standard terms for CD features. Lands: Refer to the flat, nonrecessed areas. Common Pitfalls: Over-interpreting logical assignments; the drive actually detects transitions rather than static pit/land levels alone. Confusing CD-R dye spots with pressed pits; the reading principle is still reflectivity contrast. Final Answer: Pits

Question 14

Acronyms in memory devices — EEPROM expansion In embedded systems and digital electronics, EEPROM stands for what full form?

Accepted Answer

Introduction / Context: Accurately expanding acronyms is essential for technical communication. EEPROM is a common nonvolatile memory used for configuration, calibration, and small data logs in embedded products. Given Data / Assumptions: EEPROM is nonvolatile: retains data without power. It supports electrical erase and program operations at the byte or page level. It differs from EPROM (UV erasable) and Flash (block/page erase). Concept / Approach: The components of the acronym explicitly describe the device: “Electrically” (erase/program via electrical signals), “Erasable” (contents can be cleared/changed), “Programmable” (user-writable), “Read-Only Memory” (historical term indicating nonvolatile storage rather than literal read-only behavior). Step-by-Step Solution: Break down each letter: E-E-P-R-O-M.Confirm “Electrically erasable” distinguishes it from UV-erasable EPROM.Confirm “Programmable ROM” reflects nonvolatile storage that users can program.Therefore choose “Electrically erasable programmable read-only memory”. Verification / Alternative check: Datasheets and textbooks list EEPROM as “Electrically Erasable Programmable Read-Only Memory,” with typical I²C/SPI interfaces and endurance specifications (for example, 100k+ write/erase cycles). Why Other Options Are Wrong: Encapsulated/Elementary variants are not industry-standard expansions. They omit “erasable” or misstate the nature of programmability. Common Pitfalls: Assuming “read-only” means cannot be rewritten; in ROM terminology it denotes nonvolatile class rather than absolute immutability. Confusing EEPROM with Flash; Flash typically erases in larger blocks. Final Answer: Electrically erasable programmable read-only memory

Question 15

Address lines and addressable locations in digital memory If a memory device provides 12 address lines (A0 through A11), how many unique storage locations can be addressed in total?

Accepted Answer

Introduction / Context: In digital memory systems, the number of address lines determines how many unique locations can be accessed. Each distinct binary combination on the address bus selects exactly one storage location. Understanding this power-of-two relationship is foundational for memory sizing and address decoding. Given Data / Assumptions: Number of address lines n = 12 (A0–A11). Each unique address corresponds to one storage location (regardless of data width). Binary addressing yields combinations equal to powers of two. Concept / Approach: The total number of unique addresses provided by n binary lines is 2^n. This comes from combinatorics: each line can be 0 or 1, and the combinations multiply (2 * 2 * … * 2, n times). The data bus width (for example, 8 bits or 16 bits) affects bits per location, not the count of locations. Step-by-Step Solution: Total locations = 2^n.Substitute n = 12 → 2^12.Compute: 2^10 = 1024; 2^12 = 1024 * 4 = 4096 locations. Verification / Alternative check: Quick mental check using nearby powers: 2^8 = 256, 2^16 = 65,536. Twelve is midway between, giving 4096, which is consistent with standard memory sizing (4K locations). Why Other Options Are Wrong: 144: not a power of two; does not correspond to any integer number of address lines. 512: equals 2^9 (would require 9 address lines, not 12). 2048: equals 2^11 (would require 11 address lines). Common Pitfalls: Confusing number of locations with total bits (which also depends on data width per location). Assuming decimal growth rather than exponential power-of-two growth with additional lines. Final Answer: 4096

Question 16

Mask ROM limitations in system design What is a major disadvantage of mask ROM when system requirements or stored data must change after manufacture?

Accepted Answer

Introduction / Context: Mask ROM (Read-Only Memory) is programmed during semiconductor fabrication by customizing photolithographic masks. It is excellent for high-volume, fixed-content applications but poses challenges when data or firmware needs to be updated after production. Given Data / Assumptions: Mask ROM content is physically set by the manufacturer before packaging. After fabrication, the stored data cannot be changed in the field. Updates would require an entirely new mask set and re-fabrication run. Concept / Approach: Because the bit pattern is “hardwired,” mask ROM devices are non-reprogrammable. Any change implies creating new masks (with significant non-recurring engineering cost) and re-manufacturing, which is impractical for small changes or low volumes. Step-by-Step Solution: Identify the core limitation: post-fabrication immutability of content.Evaluate options: “cannot be reprogrammed” precisely captures this limitation.Therefore, option C is correct. Verification / Alternative check: Compare with EPROM/EEPROM/Flash that allow erasure and reprogramming. Mask ROM lacks this ability entirely. Why Other Options Are Wrong: Time consuming or very expensive may be true indirectly, but the absolute limitation is non-reprogrammability. Short life expectancy is incorrect; ROM content is stable and long-lived. Common Pitfalls: Confusing mask ROM with OTP EPROM (one-time programmable) which can be programmed once by the user. Final Answer: cannot be reprogrammed if stored data needs to be changed

Question 17

Static RAM (SRAM) cell structure In a static RAM device, what circuit element forms the fundamental storage cell for each bit?

Accepted Answer

Introduction / Context: SRAM and DRAM store bits differently. Understanding the SRAM storage element explains why SRAM is fast and does not require refresh, yet is less dense than DRAM. This is crucial for cache memory design and performance analysis. Given Data / Assumptions: SRAM holds data statically as long as power is applied. Typical SRAM cells use cross-coupled inverters forming a bistable latch (flip-flop behavior). No periodic refresh is required for data retention under power. Concept / Approach: An SRAM bit cell is essentially a tiny flip-flop (often implemented with six transistors in CMOS: two cross-coupled inverters and two access transistors). It stays in one of two stable states, representing 0 or 1, while enabled by word lines during read/write operations. Step-by-Step Solution: Identify storage mechanism: bistable latch.A latch/flip-flop provides static retention without capacitive refresh.Hence the correct choice is “flip-flop.” Verification / Alternative check: Contrast with DRAM: a capacitor plus a transistor stores charge temporarily and needs refresh. With SRAM, the latching action holds the state indefinitely (under power). Why Other Options Are Wrong: Diode/resistor alone cannot provide bistability for storage. Capacitor is DRAM’s storage element, not SRAM’s. Common Pitfalls: Assuming “flip-flop” must mean a full edge-triggered FF; in SRAM, the cell is a minimalist latch with similar bistable behavior. Final Answer: flip-flop

Question 18

Address multiplexing in DRAM — primary benefit What is the principal advantage of using multiplexed row/column addresses with DRAM memory devices?

Accepted Answer

Introduction / Context: Most DRAMs use multiplexed addressing: the row address is presented first (latched by RAS), followed by the column address (latched by CAS). This technique enables large arrays with fewer external pins, which impacts cost and board routing complexity. Given Data / Assumptions: Row/column multiplexing means the same physical pins carry row and column bits at different times. Pin count is a major constraint in IC packaging and PCB layout. Refresh behavior is independent of address multiplexing. Concept / Approach: Without multiplexing, a 16-bit address would require 16 dedicated pins. With multiplexing, only 8 address pins are needed (plus RAS/CAS control), cutting the pin count roughly in half for the same address width. This directly reduces package size and cost, and simplifies routing. Step-by-Step Solution: Identify the effect of multiplexing → share pins between row and column addresses.Shared pins → fewer total address pins → smaller package.Therefore, option C is correct. Verification / Alternative check: DRAM datasheets list address pins as A0–Ax with RAS/CAS timing to latch row then column; SRAMs typically use non-multiplexed addresses and therefore more pins for comparable depth. Why Other Options Are Wrong: Access time: mostly determined by internal cell/array timing, not pin multiplexing. Refresh requirement: unaffected by multiplexing; refresh is needed regardless. Chip-select elimination: DRAMs still use one or more chip-select/enable signals; multiplexing does not remove that need. Common Pitfalls: Assuming multiplexing speeds the memory; it is primarily a packaging/routing optimization. Final Answer: reduced pin count and decrease in package size

Question 19

ROM capacity calculation from organization For a ROM organized as 1024 × 8 (1K locations by 8 bits per location), what is the total bit storage capacity?

Accepted Answer

Introduction / Context: Memory capacity is often specified as “number of locations × bits per location.” Converting this to total bits helps with system sizing, data buffering, and bandwidth calculations. Given Data / Assumptions: Organization: 1024 locations. Data width: 8 bits per location. Total capacity = locations * bits per location. Concept / Approach: Multiply the number of addresses by the data width to obtain the total number of stored bits. For 1K × 8, the “1K” is 1024 in binary terms. Step-by-Step Solution: Total bits = 1024 * 8.Compute: 1024 * 8 = 8192 bits.Therefore, the device stores 8192 bits in total. Verification / Alternative check: Sanity check: 1 kilobyte (8-bit bytes) equals 1024 bytes = 8192 bits, consistent with the organization given. Why Other Options Are Wrong: 1024, 2048, 4096 are intermediate values and do not reflect 1024 * 8. Common Pitfalls: Confusing bytes and bits; always multiply by 8 to convert bytes to bits. Final Answer: 8192

Question 20

Troubleshooting a memory with inactive outputs Given: Q0–Q3 outputs are always LOW. Address lines A0–A7 mostly pulse, except A3 is constantly HIGH and A7 is constantly LOW. Two select lines exist: one shows pulse activity and the other is stuck HIGH…

Accepted Answer

Introduction / Context: When a memory’s data outputs remain at a constant logic level, the fault may be within the device or in the external selection/enable logic. Understanding active-LOW chip-select behavior and how it gates the internal read path is key to isolating the issue quickly. Given Data / Assumptions: Outputs Q0–Q3 are always LOW. Address lines A0–A7 show activity except A3 is stuck HIGH and A7 is stuck LOW. Two select lines exist; one shows pulses, the other is stuck HIGH. Selection is via an active-LOW AND gate (both selects must be LOW to enable the device). Concept / Approach: If any input to an active-LOW select AND gate is held HIGH, the device never asserts its internal output drivers, regardless of address activity. Therefore, a stuck-HIGH select input can fully explain “outputs always LOW.” A practical test is to force both select lines LOW and observe whether Q0–Q3 now show activity; if they do, the device is likely fine and the fault lies upstream in the select-driving logic or code. Step-by-Step Solution: Recognize selection requirement: both active-LOW selects must be 0 to enable output.Observation: one select stuck HIGH → device never selected.Diagnostic: manually pull both selects LOW and monitor Q0–Q3 for pulses.If outputs now respond → external logic or firmware error is implicated, not the memory IC itself. Verification / Alternative check: Compare by probing the enable pins relative to datasheet truth table; if enable never satisfies the active condition, static outputs are expected. Additional check: inspect OE/CE/WE polarity and timing. Why Other Options Are Wrong: Address lines stuck levels (B): a stuck address bit alone would still produce changing data on other addresses; it does not force all outputs permanently LOW if the device is enabled. Defective output buffers (C): less likely; all outputs LOW with never-enabled chip is simpler explanation. X decoder failure (D): also possible in rare cases, but the stuck-HIGH select provides a direct cause consistent with the symptom. Common Pitfalls: Misinterpreting active-LOW enables—remember that LOW means “asserted.” Overlooking simple selection faults before replacing ICs. Final Answer: One select input stuck HIGH preventing chip enable; force both selects LOW to test. If outputs then respond, the external select-driving circuitry is at fault.

Memory and Storage Questions