Question 1

In a microprocessor-based memory system, address decoding for dynamic memory control can also be used for which important system-level function?

Accepted Answer

Introduction / Context: Address decoders translate processor address lines into device-select signals, enabling specific memory chips or I/O devices to respond to designated address ranges. Beyond raw chip enables, decoders play a central role in memory mapping, which defines how the processor’s address space is partitioned among RAM, ROM, and peripherals. Given Data / Assumptions: System includes a CPU, address bus, control lines, and memory devices. Dynamic memory requires refresh, but that is not the only use of decoding. We focus on high-level system function enabled by decoding logic. Concept / Approach: Memory mapping is the process of assigning blocks of addresses to different devices. The address decoder outputs chip-select signals based on address ranges, effectively implementing the memory map. While decoders can help generate control signals (e.g., read/write) or assist refresh logic, their primary system-level value is to define where in the address space each device resides. Step-by-Step Solution: Address lines enter the decoder (e.g., PLD, discrete gates, or integrated decoder).Decoder asserts one of many chip-select outputs corresponding to ranges.This selection maps specific regions to RAM, ROM, or memory-mapped I/O.Hence, decoders are fundamental to implementing memory mapping. Verification / Alternative check: Typical schematics show 74HC138/139 or CPLDs generating CS lines for multiple memory devices, clearly reflecting the address map in hardware equations or tables. Why Other Options Are Wrong: Controlling refresh circuits only: refresh is a DRAM-specific timing issue, not the principal system function.Read/write control only: R/W often comes from CPU control pins; decoders primarily map addresses.Chip selection and address location only: close, but memory mapping encompasses the broader, system-level allocation.Thermal throttling: unrelated to address decoding. Common Pitfalls: Assuming address decoding is only a low-level enable signal. In practice, it codifies the entire memory map and dictates system behavior and software visibility. Final Answer: Memory mapping (assigning address ranges to devices)

Question 2

In digital hardware design, a FIFO (First-In, First-Out) memory buffer is most commonly implemented using which arrangement of elements?

Accepted Answer

Introduction / Context: FIFOs are ubiquitous buffering structures used in digital systems to decouple producer and consumer rates. The earliest and simplest hardware FIFOs were built from shift registers, where data enters one end and emerges in the same order at the other end—hence First-In, First-Out behavior. Given Data / Assumptions: We focus on structural implementation, not protocol-level FIFOs. Small/medium FIFOs can be efficiently built from shift-register ICs or FPGA shift resources. Larger FIFOs often use dual-port RAM, but the fundamental principle remains order preservation. Concept / Approach: A shift-register FIFO uses clocked storage elements (flip-flops or LUT-based shift chains) that shift data forward with each clock. The oldest data moves toward the output and exits first, preserving order. Modern devices may implement FIFOs with dual-port RAM and read/write pointers, but the simplest canonical form is a shift-register arrangement. Step-by-Step Solution: Recognize FIFO's defining property: the first datum in must be first out.Map to hardware: a chain of registers shifting on each clock naturally preserves order.Conclude: the arrangement is shift registers in series.Note: alternative implementations use RAM plus pointers but are conceptually equivalent in function. Verification / Alternative check: Common parts (e.g., 74HC595/74HC164 based chains or FPGA SRL primitives) demonstrate shift-register FIFOs. Vendor IPs for FIFOs use RAM and pointers but documentation still describes FIFO order semantics. Why Other Options Are Wrong: Diodes/transistors alone: do not provide ordered, clocked storage by themselves.MOS CAM: supports associative lookups, not FIFO order.Ring oscillator chain: generates clocks, not ordered storage. Common Pitfalls: Confusing shift-register FIFOs with LIFOs (stacks) and overlooking latency growth with depth. For deeper buffers, RAM-based FIFOs are preferred for area efficiency. Final Answer: Shift registers arranged to pass data in order

Question 3

In random-access memory timing, the time from the start of a read cycle until valid data is available (ending at tACS or tAA) is called what?

Accepted Answer

Introduction / Context: Memory data sheets define several timing parameters to ensure reliable reads and writes. For reads, designers care about how long after a read begins until the data outputs are valid. This is encapsulated in the access time, often specified as tAA (address access time) and tACS (chip select access time). Given Data / Assumptions: We are examining read timing only. tAA and tACS are common parameters for asynchronous SRAM/ROM and historic DRAM specs. Read cycle start is defined by address/chip select assertion, depending on device. Concept / Approach: Access time is the interval from the point a read is initiated to when the memory device outputs valid data within specified voltage thresholds. Whether measured from address valid (tAA) or from chip select asserted (tACS), the endpoint is the arrival of valid, stable data. Step-by-Step Solution: Identify the read initiation event (address stable and CS asserted).Trace the device timing until data lines settle into valid logic levels.That interval is reported as tAA or tACS; both are forms of access time.Therefore, the time described is access time. Verification / Alternative check: Compare with typical SRAM datasheets: timing diagrams label tAA/tACS as access-time metrics culminating in data valid windows. Why Other Options Are Wrong: Data hold time: describes how long data must remain valid after a control edge, not time-to-valid.Read cycle time: full duration from one read to the next; larger than access time.Write enable time: pertains to writes, not reads.Setup time: requirement before a clock/control event, not data-out delay. Common Pitfalls: Confusing the start reference for access time and mixing up hold/setup with access. Always check the diagram legends in the datasheet. Final Answer: Access time

Question 4

Which statement best describes static memory (SRAM) devices compared with dynamic memory (DRAM)?

Accepted Answer

Introduction / Context: Static RAM (SRAM) and Dynamic RAM (DRAM) are the two fundamental types of semiconductor main memory. Understanding their retention behavior is critical for system design, performance, and power considerations. Given Data / Assumptions: We are comparing SRAM vs. DRAM characteristics. Power is assumed to remain applied to the device. No magnetic or optical technologies are in scope. Concept / Approach: SRAM stores bits in bistable flip-flop circuits; as long as power is applied and noise margins are maintained, the data remains valid without refresh. DRAM stores bits as charge on capacitors that leak and must be periodically refreshed even when powered. Thus, the defining characteristic of SRAM is “no refresh required with power applied.” Step-by-Step Solution: Identify storage mechanism: SRAM uses cross-coupled inverters (flip-flops), DRAM uses capacitors.Analyze retention: SRAM holds state indefinitely with power; DRAM leaks and needs refresh.Conclude the correct description: data retained as long as power is applied; no refresh for SRAM. Verification / Alternative check: Device datasheets and textbooks list “no refresh” as a key SRAM advantage, offset by larger cell size and lower density compared to DRAM. Why Other Options Are Wrong: Magnetic/optical statements: not semiconductor main memory devices.Constant refresh with power applied: describes DRAM, not SRAM. Common Pitfalls: Assuming “static” means non-volatile without power; SRAM still loses data when power is removed. “Static” only means no refresh needed while powered. Final Answer: Semiconductor devices in which stored data is retained as long as power is applied (no refresh)

Question 5

On a RAM device, to which control pin does the external address decoder connect to indicate that a particular memory chip is selected for access?

Accepted Answer

Introduction / Context: In systems with multiple memory chips, an address decoder asserts a unique select signal so that only one chip responds to a given address range. Understanding which pin receives this selection is fundamental to building reliable memory interfaces. Given Data / Assumptions: Standard asynchronous SRAM/DRAM-like pin naming is assumed. Decoder outputs one of many mutually exclusive select lines. The selected chip must drive or accept data; the others must remain inactive. Concept / Approach: The chip enable (CE) or chip select (CS) pin tells a specific memory chip that it is being addressed. OE controls the direction/enable of the data outputs (for reads), while WE controls writing. Address inputs carry address bits to all chips in parallel, but only the chip with CE/CS asserted is allowed to respond. Step-by-Step Solution: Decoder examines high-order address lines.It asserts one CE/CS line corresponding to the target address block.The selected chip is enabled for read/write per OE/WE status; others ignore the cycle.Therefore, the decoder connects to CE/CS. Verification / Alternative check: Typical memory timing diagrams show CE/CS gating the device’s internal access, with OE and WE further qualifying read versus write behavior. Why Other Options Are Wrong: Address input: all chips receive addresses in parallel; not the selection pin.OE and WE: control read/write direction and timing but do not select which chip is active.Data input: carries data, not selection. Common Pitfalls: Leaving CE permanently enabled on multiple chips leads to bus contention; always decode CE/CS uniquely for each device region. Final Answer: Chip enable (CE) / Chip select (CS)

Question 6

Within a typical Flash memory architecture, which element initiates and controls internal program/erase/read operations by interpreting host requests?

Accepted Answer

Introduction / Context: Flash memories execute complex internal algorithms—such as program, erase, and verify—coordinated by an internal controller or command/state machine. The host (processor) communicates intent by issuing specific command codes via the memory’s interface. Given Data / Assumptions: We are considering NOR/NAND Flash command interfaces. The device exposes an address/data bus and control pins (CE, OE, WE). Programming and erasing require multi-step internal sequences. Concept / Approach: The host writes a command code (or sequence) to special addresses, which the internal command/state machine interprets to launch the appropriate operation. The floating-gate MOSFET is the storage element but does not “manage” operations; the I/O pins are a conduit only. Program-verify is a phase within programming, not the manager of all functions. Step-by-Step Solution: Host issues command code(s) over the bus.Internal command/state machine decodes the command and starts program/erase/read flow.Device performs verify loops, status reporting, and timing autonomously.Therefore, management originates from the interpreted command code inside the device. Verification / Alternative check: Manufacturer datasheets (e.g., for NOR Flash) document command tables and status registers that are parsed by an internal state machine governing all algorithms. Why Other Options Are Wrong: I/O pins: physical interface only; do not interpret or manage.Floating-gate MOSFET: storage device; not a controller.Program verify code: a sub-step outcome, not the manager.Row decoder: selects rows; does not orchestrate high-level operations. Common Pitfalls: Confusing storage physics with control logic. The command/state machine is the “brains,” while arrays, decoders, and sense amps perform lower-level functions. Final Answer: Command code (interpreted by the internal command/state machine)

Question 7

Which of the following is not considered a standard operational mode of Flash memory devices?

Accepted Answer

Introduction / Context: Flash memory supports a defined set of operations controlled by device commands. The canonical modes are read, program (write), and erase, often with status polling or ready/busy signaling. This question asks you to identify the option that is not a typical Flash operation mode. Given Data / Assumptions: We refer to generic NOR/NAND Flash behavior. Modes correspond to top-level operations listed in datasheets. Vendor-specific performance features are not the focus. Concept / Approach: Flash arrays require erase-before-program and support read access. “Burst” is a term commonly associated with burst read modes in DRAM/SRAM or certain NOR read acceleration, but it is not a fundamental operation like read/program/erase. Even when some devices offer “burst read,” it is a performance enhancement within the read category, not a separate core mode comparable to program or erase. Step-by-Step Solution: List core operations: Read, Program, Erase.Compare choices: only “Burst” is not a core Flash operation; it is either absent or a sub-feature of reading on certain devices.Therefore, the correct answer is “Burst.” Verification / Alternative check: Flash datasheets enumerate command sets with read/program/erase/status operations; “burst” is not universally defined as a primary mode across vendors. Why Other Options Are Wrong: Read: fundamental for retrieving data.Erase: necessary to clear blocks/sectors before programming.Programming: writes data into the erased cells.Read-while-write: a device capability on some architectures, still tied to core modes. Common Pitfalls: Confusing memory marketing terms with architectural modes. Always rely on the command set table for definitive operational modes. Final Answer: Burst

Question 8

Memory technologies – comparing EEPROM and EPROM Which of the following statements correctly highlights the key advantages of an EEPROM (Electrically Erasable Programmable Read-Only Memory) over an EPROM (Erasable PROM using ultraviolet light)?

Accepted Answer

Introduction / Context: Nonvolatile memory families such as EPROM and EEPROM are widely used for firmware storage. Understanding how they are erased and programmed is crucial for in-circuit updates, field servicing, and device lifecycle management. Given Data / Assumptions: EPROM requires ultraviolet (UV) light exposure through a quartz window to erase the entire chip. EEPROM supports electrical erase and program operations via standard supply pins. Both devices retain data without power (nonvolatile). Concept / Approach: EPROM erasure is bulk-only and off-board: you must physically remove the device and expose it to UV for minutes. EEPROM allows electrical erasure in-system, typically byte-by-byte or word-by-word, enabling selective updates and field reprogramming without removing the IC. Step-by-Step Solution: Identify erase method: EPROM uses UV light; EEPROM uses electrical signals on pins.Determine granularity: EPROM erases the whole array; EEPROM commonly supports byte/word erase and program.Infer maintenance benefit: EEPROM enables in-circuit, selective updates, minimizing downtime. Verification / Alternative check: Datasheets state that EPROM needs UV erasure and EEPROM provides electrical erase/program functionality, often with byte-level granularity and write/erase endurance specifications. Why Other Options Are Wrong: EPROM UV erasure is not “much less time” than EEPROM; EEPROM writes/erases electrically and quickly in-system.Stating only one EEPROM advantage (in-circuit or per-word) is incomplete; both apply.Claiming EEPROM must be removed is the opposite of its key benefit. Common Pitfalls: Confusing EEPROM with flash; while both are electrically erasable, flash typically erases in sectors/blocks, whereas EEPROM commonly allows byte/word operations. Final Answer: the EEPROM can be erased and reprogrammed without removal from the circuit, and can erase and reprogram individual words

Question 9

Shared digital buses – naming the error condition The condition that occurs when two or more devices attempt to drive (write to) a common bus at the same time is called:

Accepted Answer

Introduction / Context: In digital systems, multiple ICs often share a common data bus. Proper arbitration and output-enable control are required to avoid electrical conflict. Recognizing the correct term for this fault is important in debugging. Given Data / Assumptions: At least two devices can drive the bus. They are enabled simultaneously without tri-state isolation. The bus lines are common conductors with finite impedance. Concept / Approach: When two outputs drive different logic levels onto the same line, a short-like condition forms between supply rails through the drivers. This is known as bus contention and can cause excess current, signal corruption, and potential device damage. Step-by-Step Solution: Identify simultaneous drive: two output drivers active on one line.Recognize conflicting levels: one attempts HIGH, other LOW.Name the fault: bus contention. Verification / Alternative check: Oscilloscope traces often show rounded or mid-level voltages and overheating ICs during contention. System design uses tri-state, open-drain with pull-ups, or bus arbiters to prevent it. Why Other Options Are Wrong: Address decoding / multiplexing are design techniques, not fault conditions.“Bus collisions” is not the standard term; industry uses “bus contention.” Common Pitfalls: Leaving multiple devices enabled; forgetting pull-ups with open-drain; mis-timed enables during handover. Final Answer: bus contention

Question 10

Dynamic RAM (DRAM) refresh timing – typical interval Typically, how often must a DRAM array be refreshed to preserve stored data (order-of-milliseconds range)?

Accepted Answer

Introduction / Context: DRAM cells store data as charge on tiny capacitors, which leak over time. Refresh cycles periodically read and rewrite each row to restore charge. Understanding the order of magnitude of refresh intervals is fundamental to memory controller design. Given Data / Assumptions: Conventional single-data-rate and early-generation DRAMs. Refresh specified over all rows within a few milliseconds. Exact timing varies by generation and temperature, but typical teaching values are a few milliseconds. Concept / Approach: Manufacturers specify a maximum interval within which all rows must be refreshed (for example, every 2–8 ms or 4–16 ms historically). The key concept is that refresh is millisecond-scale, not microseconds per full array (though row refreshes themselves occur in microseconds or less). Step-by-Step Solution: Identify the correct scale: millisecond range for the entire array, not microseconds.Select the common textbook interval: 2 to 8 ms fits typical legacy specifications.Therefore, choose 2 to 8 ms as a representative “typical” value. Verification / Alternative check: Older DRAM standards often specified full-array refresh on the order of a few milliseconds; later devices may use 64 ms at normal temperature, but many basic electronics texts use 2–8 ms as the canonical range for examples. Why Other Options Are Wrong: 4 to 16 ms is plausible historically, but the more commonly cited “typical” in many entry texts is 2–8 ms.Microsecond options (8–16 µs, 1–2 µs) are far too short for full-array intervals.Hundreds of milliseconds would allow excessive leakage and data loss. Common Pitfalls: Confusing per-row refresh timing with full-array interval; not accounting for temperature-dependent refresh needs. Final Answer: 2 to 8 ms

Question 11

Memory capacity calculation – 8K × 8 organization A semiconductor memory device is specified as 8K × 8. How many bytes of data can this chip store in total?

Accepted Answer

Introduction / Context: Interpreting memory organization (number of addresses × bits per address) is a routine skill for selecting memory parts and mapping address spaces in microcontroller designs. Given Data / Assumptions: Organization: 8K × 8. “K” in memory convention means 1024 (2^10). 8 bits = 1 byte. Concept / Approach: The first number (8K) is the number of addressable locations. The second number (8) is the number of data bits per location. Total bytes equals number of locations multiplied by bytes per location. Step-by-Step Solution: Compute locations: 8K = 8 * 1024 = 8192 addresses.Each address stores 8 bits = 1 byte.Total capacity in bytes = 8192 * 1 = 8192 bytes. Verification / Alternative check: As a cross-check, total bits = 8192 * 8 = 65,536 bits. Dividing by 8 bits/byte returns 8192 bytes. Why Other Options Are Wrong: 8,000 uses decimal “K” and is not the binary memory convention.64,000 and 65,536 are totals in bits (approximate and exact respectively), not in bytes.32,768 equates to 32K bytes, which is unrelated here. Common Pitfalls: Confusing bits and bytes, and mixing decimal vs. binary definitions of K. Final Answer: 8,192

Question 12

Why ROM outputs are tri-stated in bused systems Most ROM ICs provide tri-state data outputs primarily to:

Accepted Answer

Introduction / Context: When multiple memory devices share a common data bus, only the selected device may drive the lines. Tri-state outputs solve this by letting nonselected devices disconnect electrically (Hi-Z), preventing bus contention. Given Data / Assumptions: One shared data bus services multiple ROMs. Chip-select logic enables only one ROM at a time. Tri-state outputs provide HIGH, LOW, or Hi-Z states. Concept / Approach: Tri-state outputs allow safe bus sharing. With CS inactive, the ROM outputs are Hi-Z, so another device can drive the same bus lines without conflict. This is the architectural reason most ROMs have tri-state data pins. Step-by-Step Solution: CS asserted → outputs enabled → ROM drives bus.CS deasserted → outputs disabled → Hi-Z, no interference.Therefore, tri-state enables multiple ROMs on one bus. Verification / Alternative check: ROM datasheets show output enable and chip select timing, with Hi-Z leakage specifications when deselected. Why Other Options Are Wrong: Tri-state is unrelated to “three input lines.”Bidirectional data typically applies to RAM, not ROM; ROM is read-only in-system.“Isolate registers during read” is too narrow; the core reason is multi-device bus sharing.Speed is not inherently increased by tri-state drivers. Common Pitfalls: Confusing tri-state with open-drain; both allow sharing but behave differently and require different bus biasing. Final Answer: permit the connection of many ROM chips to a common data bus.

Question 13

Acronym expansion in sensors and imaging In electronics and imaging, the widely used acronym CCD stands for:

Accepted Answer

Introduction / Context: CCD technology underpins many image sensors and precision charge-transfer circuits. Knowing the correct expansion of the acronym helps place the device’s operating principle in context. Given Data / Assumptions: CCD is an established term in optics, imaging, and signal processing. The device moves charge packets through potential wells under clock control. Concept / Approach: “Charge-coupled device” describes a structure that stores and transfers discrete packets of electric charge across the substrate via capacitive coupling controlled by clocked electrodes, mapping charge to voltage at an output amplifier. Step-by-Step Solution: Identify domain: imaging/sensing.Recall physical mechanism: charge transfer between adjacent wells.Select correct phrase: charge-coupled device. Verification / Alternative check: Introductory texts and sensor datasheets consistently expand CCD as “charge-coupled device.” Why Other Options Are Wrong: Other expansions are not standard terms and do not describe the coupled-charge transfer mechanism. Common Pitfalls: Confusing CCD with CMOS active pixel sensors; both capture images but use different architectures. Final Answer: charge-coupled device

Question 14

Mask ROM fundamentals – programmability Which statement best describes a mask ROM (Read-Only Memory) device?

Accepted Answer

Introduction / Context: ROM variants differ by how and when their contents are defined. Mask ROM is one of the earliest and most cost-effective approaches for high-volume, unchanging firmware. Given Data / Assumptions: Mask ROM contents are fixed by a photomask during wafer fabrication. No field-programmability is provided. Nonvolatile retention is expected. Concept / Approach: Because data is encoded at the silicon level, mask ROM offers low unit cost at volume but zero flexibility after production. It is ideal for stable products but unsuitable for frequent updates. Step-by-Step Solution: Establish programmability: fixed at manufacture.Determine volatility: nonvolatile by definition.Conclude: “permanently programmed during manufacturing” is correct. Verification / Alternative check: Standard memory taxonomy lists mask ROM as factory-programmed, contrasted with EPROM/EEPROM/flash which are field-programmable. Why Other Options Are Wrong: Volatile is false; ROM retains data without power.Mask ROM is not reprogrammable; UV erasure applies to EPROM, not mask ROM.“Extremely expensive” is context dependent; at high volume it is cheaper per unit than field-programmable options. Common Pitfalls: Assuming all ROMs are field-programmable; confusing mask ROM with PROM/EPROM. Final Answer: permanently programmed during the manufacturing process

Question 15

Why dynamic RAMs use a multiplexed address bus Most DRAMs present row and column addresses on shared pins (multiplexed). The primary reason for this design choice is:

Accepted Answer

Introduction / Context: DRAM density scales with the number of memory cells, but package pin count adds cost and size. Multiplexing addresses (using RAS/CAS strobes) reduces the number of external pins while still selecting a unique cell. Given Data / Assumptions: DRAM cells are organized in a 2D array with row and column selection. External pins are limited; sharing pins for row and column addresses halves the address pins needed. RAS (row address strobe) and CAS (column address strobe) latch the multiplexed halves. Concept / Approach: Provide the row address on the address pins, latch with RAS; then provide the column address on the same pins, latch with CAS. This minimizes package pins and cost, enabling higher densities per package. Step-by-Step Solution: Split N-bit address into row and column halves.Present row → assert RAS to latch.Present column → assert CAS to latch.Access cell at [row, column] while using fewer pins. Verification / Alternative check: DRAM datasheets and timing diagrams show multiplexed address cycles with RAS/CAS, a standard since early DRAM generations. Why Other Options Are Wrong: It is not the only possible way; it is a cost/pin optimization.Multiplexing does not inherently increase raw speed; it adds timing overhead.It does not remove the need for refresh; DRAM still requires periodic refresh. Common Pitfalls: Assuming multiplexing is a speed feature; it is primarily about pin economy and density. Final Answer: to keep the number of pins on the chip to a minimum

Question 16

SRAM vs. DRAM – key operational difference What is the major difference between Static RAM (SRAM) and Dynamic RAM (DRAM) in normal operation?

Accepted Answer

Introduction / Context: SRAM and DRAM are the two dominant volatile memory technologies. Designers must know how their internal storage mechanisms impact system timing and controller complexity. Given Data / Assumptions: SRAM uses cross-coupled latches (typically 6-transistor cells). DRAM uses capacitors to store charge in 1-transistor cells. Both lose data when power is removed (volatile). Concept / Approach: DRAM cell charge leaks, so each row must be refreshed (read and rewritten) within the specified interval. SRAM stores data as a stable latch state and does not require refresh while powered, simplifying timing at the cost of lower density per bit. Step-by-Step Solution: Identify storage mechanism: charge (DRAM) vs. latch (SRAM).Derive operational requirement: DRAM needs periodic refresh; SRAM does not.Conclude the major difference is the refresh requirement in DRAM. Verification / Alternative check: Memory controller specs include refresh cycles for DRAM. SRAM timing diagrams lack refresh signals and can support random access without periodic maintenance. Why Other Options Are Wrong: SRAM does not retain data without power; both are volatile.Drain/Source terminal choice is not the defining distinction.DRAMs are not “always active,” and SRAMs do not require resets between accesses. Common Pitfalls: Confusing nonvolatile memories (EEPROM/flash) with SRAM; believing DRAM reads are inherently destructive without rewrite (modern DRAM controllers automatically perform the necessary restore after reads). Final Answer: DRAMs must be periodically refreshed.

Question 17

Computer hardware – floppy disk storage fundamentals In a legacy floppy disk drive, data is written to and read from the rotating magnetic media by a magnetic ________ head mechanism. Choose the most appropriate term used in computer storage.

Accepted Answer

Introduction / Context: The question tests fundamental understanding of how floppy disk drives operate. Although floppies are now legacy devices, the terminology remains foundational in computer architecture and storage systems. Knowing the correct component name helps you distinguish between media geometry terms and the electromechanical head that performs the magnetic recording and playback. Given Data / Assumptions: The device is a floppy disk drive with rotating magnetic media. Data must be magnetically written and later read back. We are choosing the common term used for the head assembly that performs both actions. Concept / Approach: Floppy and hard disk drives use electromagnetic heads to change and sense magnetic domains on tracks. The same head assembly is typically used for both writing and reading, hence the industry term “read/write head.” Other terms in the options relate to storage geometry or file system concepts, not the head mechanism itself. Step-by-Step Solution: Identify the physical function needed: writing magnetic patterns and reading them later.Match that function to the correct industry term: read/write head.Eliminate distractors that refer to file system structures or geometric descriptors. Verification / Alternative check: In standard references for storage devices, the term “read/write head” is consistently used for both floppy and hard disk assemblies that magnetize and sense the medium. No alternate term from the choices correctly names this mechanism. Why Other Options Are Wrong: Cylinder: Refers to the set of tracks aligned across platters or surfaces in a disk system, not the head itself. Recordable: Describes a capability, not a component name. Cluster: A file system allocation unit; unrelated to the head mechanism. Common Pitfalls: Confusing terms from logical file systems (clusters) with physical hardware components. Assuming “cylinder” is a mechanical part rather than a geometric organization of tracks. Final Answer: read/write

Question 18

Optical storage basics – mapping bits on a CD-ROM surface On a CD-ROM, the raised, reflective areas that represent a logical 1 in the data encoding scheme are called ________.

Accepted Answer

Introduction / Context: This question evaluates your knowledge of optical storage terminology on compact discs (CD-ROM). Data is encoded as transitions between two surface features. Understanding which feature is raised and what it is called helps clarify how the optical pickup converts surface patterns into digital bits. Given Data / Assumptions: The medium is a CD-ROM with molded micro-features. The laser senses differences in reflectivity as the disc spins. We are asked to name the raised areas that correspond to a binary 1 in common descriptions. Concept / Approach: On a CD-ROM, “pits” are the recessed areas and “lands” are the raised, flat regions between pits. In simplified explanations, a binary 1 is associated with a transition between pit and land, but in many introductory treatments, the raised areas are referred to as lands and are commonly linked to logical 1 for identification, while pits are the depressions. Among the options, only “lands” correctly names the raised areas. Step-by-Step Solution: Identify which features are raised versus recessed.Recall: pits = recessed; lands = raised.Match raised feature to the correct term: lands. Verification / Alternative check: CD literature consistently uses “pit” for the depression and “land” for the flat raised area. Regardless of detailed channel coding, the physical term for raised areas is lands, which is the focus of this question. Why Other Options Are Wrong: Mounds: Not an industry term for CD features. Holes: Misleading; CDs do not have through-holes to represent bits. Pits: These are the recessed areas, not raised regions. Common Pitfalls: Confusing the logical representation rules with the physical naming of features. Assuming “pit” means any visible feature rather than specifically the depression. Final Answer: lands

Question 19

Memory organization – identifying the location of data In an addressable memory array, the specific location that identifies a unit of stored data is called its ________.

Accepted Answer

Introduction / Context: Memory arrays in digital systems are indexed so the processor can fetch or store data deterministically. The terminology for the locator of a memory cell is fundamental to computer architecture and microcontroller programming. Given Data / Assumptions: We are discussing a general memory array (ROM, RAM, cache, etc.). Each cell or word must be uniquely located for read/write operations. Concept / Approach: The identifier used to select one location among many is called an address. Address lines carry the binary value that selects the row/column/word line so that data can be read or written. Other terms in the options either describe memory types or the contents rather than the locator. Step-by-Step Solution: Recognize that processors place a value on address lines to select a location.The name of that location selector is the “address.”Therefore, the correct fill-in term is address. Verification / Alternative check: Any memory datasheet or CPU documentation will describe the address bus and the addressable range; these consistently use “address” for the location specifier. Why Other Options Are Wrong: Storage: A general concept, not the specific locator. RAM: A memory type (Random Access Memory), not the locator. Data: The content stored at the location, not its identifier. Common Pitfalls: Confusing the content (data) with the selector (address). Using memory type names where a structural term is required. Final Answer: address

Question 20

Troubleshooting SIMMs – capacity misreporting A 64-Mbyte SIMM is installed, but a memory test detects it as only 32-Mbyte. Which fault best explains this symptom in digital hardware troubleshooting?

Accepted Answer

Introduction / Context: When system firmware or diagnostic tools report half of an expected memory capacity, it often indicates an addressing issue rather than a complete module failure. Understanding how address lines map the accessible space helps quickly isolate the fault. Given Data / Assumptions: Expected capacity: 64 MB; observed: 32 MB. Module is otherwise detected and operates at least partially. Modern SIMMs map capacity by address line decoding across chip banks. Concept / Approach: If a most significant address (MSA) line is stuck at 0 or 1, the system can only access half the address space because it cannot toggle into the higher or lower half. Therefore the reported capacity becomes 1/2 of the true size. This is a classic symptom of a stuck MSB address line or a break in the address path. Step-by-Step Solution: Total accessible space is determined by the number of unique address combinations.A stuck MSB halves the number of combinations.Halving 64 MB yields 32 MB, matching the observation. Verification / Alternative check: Technicians often probe the address bus or perform boundary tests that flip the high address bit. If content mirrors every 32 MB block, it indicates the MSB cannot change state, confirming the diagnosis. Why Other Options Are Wrong: Not installed properly: Often results in no detection, intermittent errors, or parity/ECC faults, not a consistent 50% capacity. Incorrect voltage: Would likely cause instability or total failure rather than a clean 50% reduction. Faulty decoder on the SIMM: Possible but the hallmark symptom of exactly half capacity most directly points to a stuck MS address line in the system or module path. Common Pitfalls: Assuming power issues first; while important, a precise halving is a strong hint of addressing, not power. Overlooking simple continuity issues on connectors or traces affecting a single address pin. Final Answer: The most significant address line is stuck high or low.

Memory and Storage Questions