Question 1

Magnetic media specification One of the most important density specifications for magnetic media (such as disks or tapes) is the:

Accepted Answer

Introduction / Context: Magnetic storage performance and capacity are influenced by several parameters. Among density metrics, the number of tracks per inch (TPI) is a key measure of how tightly data is packed radially on media. Given Data / Assumptions: Focus is on a specification intrinsic to media density. Applies to magnetic disks and tape systems. Concept / Approach: TPI directly reflects radial recording density. Higher TPI means more tracks on the surface, increasing potential capacity. While other specs like data transfer rate and RPM affect performance, TPI is a core density metric. Step-by-Step Solution: List candidate parameters.Identify which measure represents areal density in one dimension: TPI.Select “tracks per inch” as the answer. Verification / Alternative check: Storage datasheets commonly list TPI (radial) and bits per inch (linear) to derive areal density. Why Other Options Are Wrong: Rotation speed and data transfer rate are performance metrics, not density measures.Polarity reversal rate and spindle torque are not standard capacity indicators. Common Pitfalls: Confusing throughput (MB/s) with capacity density; both matter but measure different aspects. Final Answer: tracks per inch

Question 2

Identifying UV-erasable EPROM packages How can ultraviolet (UV) erasable PROMs (EPROMs) be visually recognized on a board or in inventory?

Accepted Answer

Introduction / Context: EPROM devices can be erased by exposing the silicon die to ultraviolet light. This requires a special package that allows UV to reach the die surface. Given Data / Assumptions: EPROMs are typically in ceramic packages with a quartz or glass window. UV light passes through the window to erase stored charge on floating gates. Other programmable memories (OTP PROM) lack the window and cannot be UV-erased. Concept / Approach: The unmistakable identifier of a UV-erasable EPROM is the transparent window. Labels are often placed over the window to prevent accidental erasure. Step-by-Step Solution: Look for a ceramic package with a circular/rectangular window.Recognize that markings or dots are not universal indicators.Select the option describing the window. Verification / Alternative check: Manufacturer packaging guides and component photos show the characteristic quartz window on UV EPROMs. Why Other Options Are Wrong: Colored dots or part-number prefixes are not standardized indicators.“Not identifiable” is incorrect; the window is explicit.Colored plastic packages are not used for UV erasure. Common Pitfalls: Confusing UV-erasable EPROM with OTP (one-time programmable) EPROM; OTP has no window. Final Answer: There is a small window on the chip.

Question 3

DRAM refresh interval (legacy specification) What is the maximum time interval allowed before a dynamic RAM must be refreshed to avoid data loss (choose the typical legacy textbook value)?

Accepted Answer

Introduction / Context: Dynamic RAM (DRAM) stores data as charge on tiny capacitors that leak over time. A refresh cycle periodically restores charge to prevent data loss. Given Data / Assumptions: Question refers to a typical legacy value taught in basic electronics/logic courses. Older DRAM generations often cited a refresh window in the 2–8 ms range for the entire array. We are asked for the maximum allowable interval before refresh. Concept / Approach: If a range such as 2–8 ms is considered typical, the maximum of that range is 8 ms. Waiting longer than the specified maximum can allow charge to decay below reliable sensing thresholds. Step-by-Step Solution: Recall typical legacy range: a few milliseconds (often 2–8 ms).Select the highest limit among given options that matches this range.Choose 8 ms as the maximum allowable interval. Verification / Alternative check: Datasheets vary by generation and temperature. Later devices specify longer intervals (for example, 64 ms at nominal conditions), but the foundational teaching example commonly uses 8 ms as the maximum in the basic set. Why Other Options Are Wrong: 2 ms and 4 ms are within the typical range but are not the maximum.10 ms exceeds the often-cited maximum for early DRAMs and risks data loss.16 ms is not among the original choices for many basic examples and is too long for the referenced range. Common Pitfalls: Mixing modern DRAM refresh specs (e.g., 64 ms) with classic textbook values targeted by this question. Final Answer: 8 ms

Question 4

Memory technologies – identifying how dynamic RAM (DRAM) stores one bit In dynamic semiconductor memory (DRAM), a single data bit is stored as charge on which circuit element?

Accepted Answer

Introduction / Context: This item checks core memory-technology knowledge: how dynamic random-access memory (DRAM) represents a single binary bit. Understanding the storage element explains why DRAM needs periodic refresh and how it differs from static RAM (SRAM) and register flip-flops. Given Data / Assumptions: The device class is DRAM (dynamic RAM). We are considering the physical circuit element that retains a 0 or 1. Binary information is represented by the presence or absence of electric charge. Concept / Approach: In DRAM, each cell pairs a MOS access transistor with a tiny capacitor. A stored 1 corresponds to charge on the capacitor; a 0 corresponds to little or no charge. Because the charge leaks over time, DRAM must be refreshed periodically. This is fundamentally different from SRAM, which uses cross-coupled inverters (flip-flop like) to hold state as long as power is applied. Step-by-Step Solution: Identify the technology: DRAM uses charge storage, not latch feedback.Determine the element that stores charge: the capacitor is the storage device.Conclude that the correct answer is the capacitor. Verification / Alternative check: Any DRAM cell schematic shows a single transistor connected to a storage capacitor; refresh specifications in datasheets further confirm charge-based storage. Why Other Options Are Wrong: Diode: Not used as the storage element in DRAM cells. Resistor: Not the charge-holding component; resistors conduct but do not store charge like capacitors. Flip-flop: Flip-flops are used in SRAM/registers; DRAM does not use a bistable latch per bit. Common Pitfalls: Confusing DRAM with SRAM and assuming a latch stores the bit. Overlooking the need for refresh, which is a direct consequence of capacitor leakage. Final Answer: capacitor

Question 5

Digital delay elements – definition of a multitap delay line What best describes a multitap digital delay line implemented with basic logic and passive components?

Accepted Answer

Introduction / Context: Multitap delay lines are used to create multiple well-defined timing references in digital systems (edge shaping, pulse stretching, time-stamping). Knowing their structure helps when analyzing propagation delay and jitter. Given Data / Assumptions: The delay line is “digital” and “multitap.” We consider simple gate-level realizations using passives for timing. Taps provide intermediate delayed versions of the same signal. Concept / Approach: A common discrete implementation chains inverters, each separated by an RC network to create incremental delays. Each node (tap) outputs a progressively delayed replica. Inverters are preferred because they provide restoration, squaring, and predictable propagation; RC networks introduce the controlled time constant. RL versions are uncommon in basic CMOS/TTL design, and using NANDs is unnecessary unless adding gating logic. Step-by-Step Solution: Identify the simplest delay element: inverter + RC forms a gate-level delay.Replicate stages and expose intermediate nodes as “taps.”Therefore, the best description is inverters with RC between each. Verification / Alternative check: Timing chains in many app notes show cascaded inverters with RC shaping or controlled delay cells; ECL/CMOS ICs also implement internal variants of this concept. Why Other Options Are Wrong: Inverters with RL: RL is not typical for logic-level delay; inductors are bulky and less predictable on PCBs. NAND + RC / NAND + RL: A NAND is not required for pure delay; it adds logic functionality rather than simple timing. Common Pitfalls: Thinking any multi-gate chain is a “multitap” even without RC timing control. Assuming inductive elements are common in digital delay blocks; they are not. Final Answer: a series of inverter gates with RC circuits between each one

Question 6

Memory width expansion – composing an 8 k × 8 from 8 k × 1 devices How many separate 8 k × 1 RAM chips are required to build a memory with word capacity 8 k and word length 8 bits (i.e., 8 k × 8)?

Accepted Answer

Introduction / Context: Designers often widen memory data paths by placing multiple narrow RAM ICs in parallel. This question tests the principle of data width expansion without changing address depth. Given Data / Assumptions: Available IC: 8 k × 1 (8,192 addresses, 1 data bit each). Target memory: 8 k × 8 (8,192 addresses, 8 data bits each). All devices share the same address and control lines to maintain identical depth. Concept / Approach: To increase word length from 1 bit to 8 bits while keeping the same number of addresses, connect eight identical 8 k × 1 chips in parallel on A0..A12 and control pins. Each chip supplies one bit of the 8-bit data bus (D7..D0). Step-by-Step Solution: Depth required = 8 k; each chip provides 8 k → match.Width required = 8 bits; each chip contributes 1 bit.Number of chips = 8. Verification / Alternative check: A quick block diagram shows eight chips with common address and enable signals, outputs OR rather “wired” onto separate data lines forming an 8-bit bus. Why Other Options Are Wrong: Four / Two / One: Would result in 4, 2, or 1 data bits per word, not the required 8. Common Pitfalls: Confusing “depth increase” (cascading addresses) with “width increase” (parallel data lines). Final Answer: Eight

Question 7

Hard disk metrics – identifying non-parameters of performance Which of the following is not a recognized hard disk performance parameter?

Accepted Answer

Introduction / Context: Disk performance is commonly summarized by seek time, rotational latency, and overall access time, which together affect throughput and responsiveness. Recognizing the correct terminology avoids confusion when reading storage specifications. Given Data / Assumptions: We compare listed terms with standard HDD metrics. Traditional HDDs use moving heads and spinning platters. Concept / Approach: Seek time measures how long the head takes to move to the right track. Latency (rotational latency) is the waiting time for the sector to rotate under the head. Access time is the sum (seek + latency, sometimes plus controller overhead). “Break time” is not a recognized HDD performance metric. Step-by-Step Solution: List recognized parameters: seek, latency, access.Identify the imposter term: break time.Select “Break time” as not a performance parameter. Verification / Alternative check: Vendor datasheets and benchmarks always cite seek, rotational latency, and access time; “break time” does not appear as a formal metric. Why Other Options Are Wrong: Seek time: Standard metric. Latency period: Standard metric (rotational latency). Access time: Standard composite metric. Common Pitfalls: Confusing “throughput” (MB/s) with time-based metrics; both are important but distinct. Final Answer: Break time

Question 8

Storage media classification – identifying non-magnetic technologies Which of the following is not a magnetic storage device?

Accepted Answer

Introduction / Context: Digital storage technologies are often grouped by their recording physics: magnetic, optical, and solid-state. Knowing the differences helps when discussing durability, rewrite cycles, and performance. Given Data / Assumptions: We compare representative devices against the “magnetic” category. Optical media encode data via pits/lands and reflectivity changes. Concept / Approach: Magnetic disk (HDD) and magnetic tape store data by magnetizing domains. Magneto-optical (MO) uses a laser to heat spots while a magnetic field sets polarity—hybrid but fundamentally uses magnetization to write. A pure optical disk (CD/DVD) uses physical/reflective features without magnetic fields; thus it is not magnetic storage. Step-by-Step Solution: Identify which option lacks magnetic recording: optical disk.Confirm others rely on magnetization.Choose “Optical disk.” Verification / Alternative check: Technical descriptions of CDs/DVDs/Blu-ray emphasize pits, lands, and reflectivity changes, not magnetic domains. Why Other Options Are Wrong: Magnetic disk: Magnetic storage. Magnetic tape: Magnetic storage. Magneto-optical disk: Uses magnetic writing with optical assistance, therefore magnetic-based. Common Pitfalls: Confusing “optical” with “magneto-optical.” They are different categories. Final Answer: Optical disk

Question 9

Hard disk assembly – recognizing mechanical components Which of the following is not a component found inside a typical hard disk drive (HDD)?

Accepted Answer

Introduction / Context: HDDs are precision electromechanical devices. Knowing their internal parts helps interpret specifications and failure modes. Given Data / Assumptions: A conventional HDD contains rotating media and a head positioning system. We identify which item does not belong to an HDD bill of materials. Concept / Approach: The spindle rotates the platters. Read/write heads fly over the surface to access data. A valve is not a standard HDD component; while HDDs include a breather/filter, there is no mechanical valve in the data path or motion system. Step-by-Step Solution: List typical HDD parts: platters, spindle motor, actuator/heads, PCB.Spot the odd term: valve.Select “Valve” as the non-component. Verification / Alternative check: Teardowns and manufacturer diagrams show spindles, platters, arms/heads, voice-coil actuators—no valves. Why Other Options Are Wrong: Spindle: Core rotating element. Platter: Magnetic medium. Read/write head: Magnetic transducer. Common Pitfalls: Confusing the breather hole/filter with a valve; they are passive vents with filters. Final Answer: Valve

Question 10

Address bus sizing – 512 × 8 organization from 4096 total bits How many address lines are required for a 4096-bit memory organized as 512 words × 8 bits per word?

Accepted Answer

Introduction / Context: Address line count is a fundamental design decision when interfacing memories. It depends on the number of addressable words (depth), not the number of bits per word (width). Given Data / Assumptions: Total memory: 4096 bits. Organization: 512 × 8 → depth = 512 words, width = 8 bits. Each unique address selects one of the 512 words. Concept / Approach: The required number of address lines A satisfies 2^A = number of addressable words. Here, words = 512. Since 512 = 2^9, we need 9 binary address lines. Width (8 bits) determines data bus size, not address count. Step-by-Step Solution: Compute words: 4096 bits / 8 bits per word = 512 words.Find A: 2^A = 512 → A = 9.Therefore, 9 address lines are required. Verification / Alternative check: Recall powers of two: 2^8 = 256, 2^9 = 512. Hence A = 9 is minimal and sufficient. Why Other Options Are Wrong: 2, 4, 8: 2^2 = 4, 2^4 = 16, 2^8 = 256, all too small for 512 addresses. Common Pitfalls: Multiplying by width when sizing address lines; address count depends only on depth. Final Answer: 9

Question 11

DRAM servicing policy: During DRAM maintenance, can a burst refresh operation and normal memory accesses be interspersed (i.e., mixed) with each other?

Accepted Answer

Introduction / Context: Dynamic RAM (DRAM) periodically requires refresh cycles to retain data in its capacitive cells. Controllers can schedule refresh using different policies, including burst refresh (many rows back-to-back), distributed/hidden refresh, or self-refresh. Whether normal reads/writes can occur during a burst refresh affects latency and system design. Given Data / Assumptions: Focus on the burst refresh policy, where multiple refresh cycles are issued consecutively. Normal CPU/DMAC memory operations contend for the same DRAM interface and banks during refresh. We use conventional synchronous DRAM behavior at a high level. Concept / Approach: During a burst refresh window, the controller dedicates the bus to issuing refresh commands for successive rows. Normal read/write commands are deferred until the burst completes. Hence they are not interleaved within that burst window. In contrast, a distributed refresh issues refresh commands periodically between normal commands, but that is a different policy from burst refresh. Step-by-Step Solution: Define burst refresh: schedule N refresh cycles contiguously.Observe bus ownership: refresh cycles occupy the DRAM interface.Result: normal commands must wait; no interspersing within the burst.After the burst, normal operation resumes between subsequent refresh windows. Verification / Alternative check: Timing diagrams from DRAM controllers show refresh “blocks” with no concurrent read/write activity. Hidden/auto-refresh features similarly block user commands for the duration of each refresh transaction. Why Other Options Are Wrong: Correct: Would imply mixing operations inside the burst, which is not how burst refresh is defined. Ambiguous / Applies only to SRAM: SRAM does not need refresh; the statement concerns DRAM and is not ambiguous under standard definitions. Common Pitfalls: Confusing burst refresh with distributed refresh; assuming QoS schedulers allow overlap when the DRAM command bus is actually dedicated to refresh cycles. Final Answer: Incorrect

Question 12

EPROM address decoding: A 16 KB EPROM memory space is to be built using 4K × 8 devices. Does this system require a 1-to-8 (3-to-8) address decoder for chip select?

Accepted Answer

Introduction / Context: Designing an EPROM subsystem requires determining how many physical chips are needed and how to decode their chip-select lines. The number of devices dictates the width of the decoder used to select one device at a time within the mapped address range. Given Data / Assumptions: Total memory required: 16 KB (16K bytes). Device capacity: 4K × 8 per EPROM. Each device provides 4 KB, data width 8 bits. Goal: find the required decoder size for chip selection. Concept / Approach: The number of devices needed is Total / Per-device = 16 KB / 4 KB = 4 devices. Selecting among 4 devices requires a 2-to-4 decoder (often called 1-of-4). A 1-of-8 (3-to-8) decoder is only needed if 8 devices must be selected, which would be the case with 2K × 8 devices for 16 KB. Step-by-Step Solution: Compute devices: 16 KB / 4 KB = 4.Determine select lines: 4 devices ⇒ 2 select bits.Decoder choice: 2-to-4 (1-of-4) is sufficient.Therefore, a 1-of-8 decoder is unnecessary and incorrect for 4K × 8 parts. Verification / Alternative check: Address bit budgeting: 4K needs 12 address lines per device; the next 2 address lines (A12–A13) decode which of the 4 chips is active. No third select bit is required. Why Other Options Are Wrong: Correct: Overstates decoder width; would leave 4 outputs unused. Other options reference unrelated conditions (bank switching, different device size). Common Pitfalls: Mixing up “KB” with “K locations”; ignoring that ×8 devices already match the data width, so only address-space selection is needed. Final Answer: Incorrect

Question 13

Memory technology choices: Main computer memory is usually implemented with DRAM due to high density and low cost, while cache memory is typically SRAM due to higher speed. Assess this statement.

Accepted Answer

Introduction / Context: Computer memory hierarchies balance cost, capacity, latency, and bandwidth. Two dominant volatile memory technologies are DRAM and SRAM, each optimized for different points on this trade-off curve. Understanding which is used where is fundamental to systems design and performance analysis. Given Data / Assumptions: DRAM cells store charge on capacitors and require refresh. SRAM cells use cross-coupled transistors, no refresh required. Main memory refers to the large external memory pool; cache memory refers to smaller, fast buffers close to the CPU. Concept / Approach: DRAM achieves very high density and low cost per bit by using one-transistor/one-capacitor cells, trading off access latency and the overhead of refresh. SRAM, using more transistors per bit (commonly six), is faster with lower access latency, but at higher area and cost per bit, making it ideal for smaller, on-chip caches. Step-by-Step Solution: Identify performance needs: caches need lowest latency → SRAM.Identify capacity/cost needs: main memory needs large capacity at low cost → DRAM.Map technologies accordingly in standard computer architectures.Confirm that this mapping aligns with practical CPU and system designs. Verification / Alternative check: Block diagrams of modern CPUs show multiple levels of SRAM cache (L1/L2/L3) and external DRAM (DDR variants) as main memory. Embedded MCUs with integrated SRAM still commonly use external DRAM for larger memory footprints. Why Other Options Are Wrong: Incorrect: Contradicts a ubiquitous architectural choice. Ambiguous / MCU-only: The statement holds broadly across desktops, servers, and embedded systems where both memory types are present. Common Pitfalls: Confusing “faster” with “higher bandwidth”; caches improve effective bandwidth and latency via locality even though DRAM channels provide high raw bandwidth with higher latency. Final Answer: Correct

Question 14

Fault isolation scope: “Testing and troubleshooting only the address-decoding logic will not reveal problems with the memory chips themselves or their connections to the CPU buses.” Evaluate this statement.

Accepted Answer

Introduction / Context: Address-decoding logic ensures that a given address range selects the correct memory device. However, a functioning decoder does not guarantee that the memory ICs or their data/address/control bus connections are healthy. Effective troubleshooting must consider all stages in the memory access path. Given Data / Assumptions: We test only the decoder (e.g., chip-select generation). Memory devices and bus wiring are not tested by this decoder-only procedure. Goal: determine if decoder testing alone can expose all memory path faults. Concept / Approach: Decoder verification confirms that addresses map to the intended chip-select lines. But failures such as stuck data bits, broken address lines at the IC pins, timing violations, power/ground issues, or defective memory arrays will not be detected if only the decoder is exercised in isolation. Full memory testing requires actual reads/writes to the chips across the buses and validation of returned data patterns (e.g., walking ones/zeros, checkerboards, March tests). Step-by-Step Solution: Step 1: Verify decoder truth table → proves address-to-CS mapping.Step 2: Perform device-level tests (write/read) → proves IC and bus integrity.Step 3: Combine both to localize faults (decoder vs device/bus).Conclusion: Decoder-only tests cannot reveal chip or interconnect problems. Verification / Alternative check: Practical diagnostics differentiate “select” faults (wrong chip selected) from “data integrity” faults (wrong values read/written). The latter require memory test patterns, not just decoder toggling. Why Other Options Are Wrong: Incorrect: Implies decoder tests are sufficient, which is false. Ambiguous / ROM-only: The logic applies to RAM, ROM, and peripherals alike. Common Pitfalls: Assuming that seeing chip-select activity guarantees a healthy memory path; ignoring signal integrity, timing margins, or power anomalies that corrupt data independently of the decoder. Final Answer: Correct

Question 15

PROM programming method: Fusible-link PROMs are programmed by physically removing selected links with tools such as a microscope and tweezers. Is this characterization accurate?

Accepted Answer

Introduction / Context: Programmable read-only memories (PROMs) historically used fusible links to store bits. Understanding the correct programming mechanism is important for historical and educational contexts and prevents misconceptions about how one-time programmability is achieved in silicon. Given Data / Assumptions: Device type: fusible-link PROM (OTP PROM). Question claims physical removal of fuses with hand tools. We must decide whether this matches reality. Concept / Approach: Fusible-link PROMs are programmed electrically. A specialized programmer applies elevated current/voltage to selected links to blow (open) them internally. No manual mechanical operation takes place on the die. The statement describing a microscope and tweezers confuses semiconductor programming with physical rework, which is impractical and unnecessary for packaged ICs. Step-by-Step Solution: Place PROM in an electrical programmer.Programmer applies controlled pulses to targeted addresses.Pulses blow specific fuses inside the chip, permanently encoding data.Verify programmed bits by read-back at nominal voltage. Verification / Alternative check: Manufacturers specify programming algorithms and pulse parameters; no step mentions physical manipulation of links with tools. Photomicrographs show on-die links inaccessible post-packaging. Why Other Options Are Wrong: Correct: Would imply manual surgery on the die, which is not the practice. Ambiguous / Mask ROM: Mask ROM is defined at fabrication, not by fuses blown post-fab. Common Pitfalls: Confusing laser-trimmed resistors or post-fab fuse techniques in analog IC trim with user PROM programming; both are internal and controlled, not manual tweezing. Final Answer: Incorrect

Question 16

ROM taxonomy: A ROM that lets users program data by permanently opening fusible links is called what type of ROM?

Accepted Answer

Introduction / Context: Read-only memory technologies differ in how and when data is defined. The mechanism—fuse blowing, ultraviolet erasure, or electrical erase—determines the device class and how users interact with it. Accurately identifying these classes is essential in digital design and maintenance. Given Data / Assumptions: Statement labels a fusible-link user-programmable ROM as EPROM. We compare PROM, EPROM, and EEPROM mechanisms. We assume standard industry definitions. Concept / Approach: Fusible-link one-time programmable devices are called PROMs (OTP PROMs). EPROMs use floating-gate transistors that can be erased by ultraviolet light exposure through a quartz window and reprogrammed electrically. EEPROMs are electrically erasable and programmable in-circuit. Therefore, calling a fusible-link programmable ROM an “EPROM” is incorrect; it is a PROM. Step-by-Step Solution: Identify the mechanism: permanently opening fuses → PROM.Contrast with EPROM: UV-erasable floating-gate → reprogrammable, not one-time.Contrast with EEPROM: electrically erasable → reprogrammable in-system.Conclude the statement mislabels the technology. Verification / Alternative check: Vendor datasheets and textbooks clearly classify fusible-link devices as PROMs and describe EPROM erase procedures using UV lamps, not fuses. Why Other Options Are Wrong: Correct: Would endorse the wrong ROM type. Ambiguous / EEPROM-only: The statement is specific and unambiguous, and EEPROM uses a different mechanism. Common Pitfalls: Using “EPROM” generically for any user-programmable ROM; overlooking that EPROMs are erasable and not one-time like fusible PROMs. Final Answer: Incorrect

Question 17

Bit and byte terminology (memory basics): Evaluate the statement: “A group of 6 bits is also known as 1 byte.”

Accepted Answer

Introduction / Context: A precise understanding of bit-group terminology is essential in digital electronics and computer architecture. The term “byte” is now universally used in hardware manuals, programming languages, and communication protocols, and it must be used unambiguously when sizing memories, defining data paths, or describing file formats. This question tests whether a learner can distinguish a generic grouping of bits from the standardized definition of a byte. Given Data / Assumptions: The claim states that 6 bits constitute 1 byte. We assume mainstream contemporary usage (computers, memories, buses, storage). We are not redefining “octet” or creating custom field sizes; we are checking the standard definition used in most systems. Concept / Approach: By modern convention, a byte equals 8 bits. Standards, operating systems, compilers, and memory chips encode and address data in 8-bit bytes (also called octets in networking standards). While historical systems sometimes used 6-bit or 7-bit character sets, these were not called “1 byte” in the modern sense. Therefore, equating 6 bits with a byte is incorrect in contemporary digital design practice. Step-by-Step Solution: Identify the definition in current standards: byte = 8 bits.Compare with the claim: “6 bits = 1 byte” → mismatch.Conclude the statement is incorrect for modern memory/processor conventions. Verification / Alternative check: Check typical RAM specifications (e.g., 1 MB = approximately 10^6 bytes, each 8 bits). Networking documentation uses “octet” explicitly to mean 8 bits. Instruction set manuals (x86, ARM, RISC-V) define byte-wide loads/stores as 8-bit transfers. All corroborate that a byte is 8 bits. Why Other Options Are Wrong: Correct: Conflicts with modern definition.Teleprinter/parity/legacy microcontrollers: Even where 6-bit or 7-bit codes existed, “byte” in current practice remains 8 bits.Depends on parity: Parity adds a bit to a byte in transmission but does not redefine byte size. Common Pitfalls: Confusing historical character width with the standardized byte, or assuming that transmission framing bits (start/stop/parity) change the definition of the stored byte. Final Answer: Incorrect

Question 18

RAM interface signals (read/write addressing): Complete the requirement: “Random-access memory (RAM) devices must have an address input, in addition to the data-in input, to select which location is written.” Evaluate the correctness of this require…

Accepted Answer

Introduction / Context: Writing to memory is fundamentally a two-part operation: you must indicate what data you are providing and where to store it. The “where” is specified through the address input(s). This question ensures that learners associate write operations with both data and addressing, rather than thinking of RAM as a simple latch without selection. Given Data / Assumptions: The device is a parallel RAM (static or dynamic) with random access. Control pins (e.g., Chip Select, Write Enable, Output Enable) exist in addition to address and data signals. We are not discussing shift-register memories or serial EEPROM streams where addressing may be embedded in a command sequence. Concept / Approach: In RAM, each storage cell is reached via an address. For SRAM, the address directly selects the word line; for DRAM, row/column multiplexing (A0..An via RAS/CAS) selects a single bit/word. During a write cycle, the memory captures the value on the data-in line(s) into the location specified by the active address while control signals (CS/OE/WE or RAS/CAS/W) define the operation. Thus, an address input is essential alongside data-in for any meaningful write. Step-by-Step Solution: Identify the operation: write requires a destination.Destination is specified by address lines (or multiplexed address phases in DRAM).Control pins qualify timing; data-in provides the value to store. Verification / Alternative check: Examine any SRAM/DRAM timing diagram: the address is valid (or latched) during the write window while WE is asserted; the data lines are sampled into the selected cell/word. Why Other Options Are Wrong: Incorrect / ROM only / tri-state exception / serial only: None removes the requirement that a destination be specified; tri-state affects bus sharing, not addressing. Serial memories still convey an address field before data. Common Pitfalls: Confusing “random access” with “no address needed,” or thinking Chip Select alone chooses a unique location. Final Answer: Correct

Question 19

RAM acronym check (terminology): Evaluate the statement: “RAM stands for Readily Accessible Memory.”

Accepted Answer

Introduction / Context: Acronyms are foundational in digital electronics. Misstating an acronym often signals a gap in understanding that can snowball into design or documentation mistakes. Here, we evaluate the common misconception that RAM expands to “Readily Accessible Memory.” Given Data / Assumptions: The context is mainstream computer architecture and memory taxonomy. We treat RAM as a category that includes SRAM, DRAM, SDRAM, DDRx, etc. We compare the offered expansion with the standard one in textbooks and datasheets. Concept / Approach: RAM is universally defined as Random Access Memory. “Random access” means any addressable location can be accessed in approximately the same time, unlike sequential access media such as magnetic tape. While RAM tends to be faster (and thus “readily accessible” in a colloquial sense) compared with disk, “readily accessible” is not the correct or accepted expansion of the acronym. Step-by-Step Solution: Recall the canonical expansion: RAM → Random Access Memory.Compare with the claim: “Readily Accessible Memory” → nonstandard.Conclude that the given expansion is incorrect. Verification / Alternative check: Consult any memory datasheet or architecture text; the definition is consistent. For example, DRAM device overviews and JEDEC standards all expand RAM as Random Access Memory. Why Other Options Are Wrong: Correct/marketing/latency-dependent: None change the formal acronym. Static vs dynamic does not alter the meaning of the initialism. Common Pitfalls: Interchanging “random” with “arbitrary” or “fast.” Random access speaks to access pattern independence, not absolute speed. Final Answer: Incorrect

Question 20

Read cycle conditions for a typical RAM: A RAM places stored data on its outputs whenever Chip Select (CS) is active and Write Enable (WE) is inactive (read mode). Evaluate this statement.

Accepted Answer

Introduction / Context: Understanding RAM control pins is crucial for timing diagrams, bus arbitration, and glue logic. In most SRAMs and DRAMs (after row/column addressing for DRAM), a read is recognized when the chip is selected and write is not requested, causing data to appear on the output drivers subject to output-enable gating. Given Data / Assumptions: Conventional active-low control pins: CS (or CE), WE, and OE. During read: CS asserted, WE deasserted; OE typically asserted to enable the output buffers. Timing meets setup/hold and access-time constraints. Concept / Approach: For parallel SRAM: read cycle occurs when CS is active and WE is inactive; OE controls whether outputs drive the bus (tri-state when OE is inactive). For DRAM, after valid RAS/CAS sequences select the cell, read data is driven when write is not active. Hence, the simplified description is accurate for “placing data on outputs,” implicitly assuming OE is active as needed. Step-by-Step Solution: Assert chip select → device participates in the bus cycle.Keep WE inactive → operation is read, not write.Enable outputs via OE (often active-low) → data lines drive the stored value. Verification / Alternative check: Typical SRAM timing tables label tACC and tOE for access and output enable; both conditions are required for valid data appearance. DRAM read timing uses RAS/CAS with WE high (inactive) to produce data at Q/DQ pins. Why Other Options Are Wrong: Incorrect / DRAM-only / inverted logic: Misstate standard control-signal roles.“OE must also be active” is a nuance; many datasheets separate address access and output enable timing. The stem’s intent focuses on CS/WE; with OE considered, the statement remains operationally correct. Common Pitfalls: Forgetting that outputs may be tri-stated when OE is inactive, even though a read cycle is in progress. Final Answer: Correct

Memory and Storage Questions