Question 1

DRAM pin-count reduction technique: To reduce the number of pins on high-capacity DRAM chips, what address technique is used so a single set of pins can carry both row and column addresses at different times?

Accepted Answer

Introduction / Context: Dynamic RAM devices require many address lines, but package pin counts must be constrained for cost and size. The industry solution is to time-multiplex the address bus, sending row and column addresses sequentially over the same pins. Given Data / Assumptions: High-capacity DRAM with large address space. Desire to limit package pin count. Standard DRAM control pins: RAS (row address strobe) and CAS (column address strobe). Concept / Approach: Address multiplexing uses the same physical address pins twice per access: first the controller presents the row address and asserts RAS; then it presents the column address and asserts CAS. Internally, DRAM latches each portion separately, allowing many address bits with fewer external pins. Step-by-Step Solution: Place row address on the address pins and assert RAS.Place column address on the same pins and assert CAS.DRAM uses both to select the target cell(s).Result: fewer external pins without sacrificing address space. Verification / Alternative check: Pinouts and timing diagrams for common DRAMs show A0..An used with RAS/CAS to latch row and column portions of the address. Why Other Options Are Wrong: conversion/programming/firmware: These terms do not describe hardware pin reuse for addressing. Common Pitfalls: Confusing multiplexing with bank interleaving or burst modes; overlooking the role of RAS/CAS strobes in timing. Final Answer: multiplexing

Question 2

DRAM storage element: Dynamic RAMs store information primarily by using which physical mechanism at the cell level?

Accepted Answer

Introduction / Context: DRAM technology is based on storing electrical charge that leaks over time, which is why refresh is required. Identifying the core storage element clarifies differences from SRAM and nonvolatile memories. Given Data / Assumptions: We focus on the fundamental DRAM cell. Compare to alternatives like flip-flops (SRAM) and magnetic storage (disks, MRAM). Consider volatility and refresh behavior. Concept / Approach: A DRAM bit cell consists of one access transistor and one capacitor. A logic 1 is represented by stored charge; a logic 0 by the absence of charge (or vice versa depending on convention). Because charge leaks, periodic refresh restores the correct level. None of the other mechanisms listed match DRAM’s underlying physics. Step-by-Step Solution: Access transistor connects the capacitor to bit-line during reads/writes.Sense amplifiers detect small charge differences and restore levels.Refresh cycles periodically rewrite stored charge across all rows.Conclusion: the storage element is the capacitor. Verification / Alternative check: Cell schematics and timing diagrams in memory textbooks show the 1T1C DRAM cell and explain the need for refresh. Why Other Options Are Wrong: Magnetism: Magnetic media or MRAM, not DRAM. Flip-flops/latches: Characterize SRAM or register logic, not DRAM. Common Pitfalls: Assuming DRAM retains data indefinitely; forgetting that reads are destructive and require restore operations. Final Answer: capacitors

Question 3

Legacy DRAM pinout fact check: For the TMS44100 DRAM device, how many address inputs (shared via row/column multiplexing) are provided on the package?

Accepted Answer

Introduction / Context: Historical DRAM parts commonly multiplexed row and column addresses to limit package pins. Recognizing typical address-pin counts helps in decoding maps and designing compatible memory controllers for legacy systems or educational labs. Given Data / Assumptions: Device: TMS44100 (legacy DRAM). Standard DRAM practice: present row and column addresses over the same address pins. Question asks for the number of external address inputs (pins), not total logical address bits stored internally. Concept / Approach: Many classic high-density DRAMs expose 10 external address pins that are time-multiplexed by RAS and CAS strobes (10 for row and the same 10 for column internally). This supports large address spaces without requiring 20 physical address pins. Thus, the package typically lists A0–A9 for addresses, for a total of 10 external address inputs. Step-by-Step Solution: Identify external address pins in the package pinout (A0..A9).Understand multiplexing: same pins carry row then column addresses.Relate to controller: two timing strobes select row and column.Conclude there are 10 address inputs exposed. Verification / Alternative check: Typical datasheets for 1 Mbit-era DRAMs show 10 address inputs, consistent with multiplexing and common package pin budgets of that generation. Why Other Options Are Wrong: 11/12/13: Do not match the standard pinout practice for this device class; more pins would counter the purpose of address multiplexing. Common Pitfalls: Confusing internal logical address width with the count of external address pins; overlooking multiplexing’s impact on pin budgeting. Final Answer: 10

Question 4

Memory system integration: When multiple memory ICs are combined to provide the full addressable space, what technique identifies which specific IC is being accessed for a given address range?

Accepted Answer

Introduction / Context: Large memory maps often require several chips in parallel (for width) or in banks (for depth). The CPU or memory controller must assert a unique chip-select to the correct device for each address window. This selection mechanism is fundamental digital design knowledge. Given Data / Assumptions: Multiple memory ICs share data and address buses. Each device must respond only to its own address range. Objective: name the technique used for per-chip selection. Concept / Approach: Address decoding derives chip-select signals from upper address bits using logic gates or decoders (e.g., 2-to-4, 3-to-8). Decoding maps address subranges to one-hot chip-select lines, ensuring exactly one device is enabled at a time (except for deliberate mirroring or interleaving designs). Step-by-Step Solution: Partition the global address space into ranges.Design combinational logic that asserts CS_n for the chosen range.Verify non-overlap and full coverage of intended ranges.Integrate CS lines with OE/WE timing to complete access cycles. Verification / Alternative check: Truth tables for decoders (e.g., 74HC138) and PLD/FPGA implementations illustrate straightforward mapping from address bits to chip selects. Why Other Options Are Wrong: Memory refresh: Maintains DRAM data; not selection. Data encoding: Relates to data representation, not device selection. Memory paging: A software/OS technique; distinct from hardware address decoding. Common Pitfalls: Overlapping decode windows causing bus contention; forgetting to include write-protect or byte-enable logic where needed. Final Answer: address decoding

Question 5

FIFO behavior and use-cases: Which of the following best characterizes a first-in/first-out (FIFO) memory device used in digital systems?

Accepted Answer

Introduction / Context: FIFOs are queue-like memory structures frequently used at clock-domain boundaries, in serial interfaces, and for rate matching between producers and consumers of data. Understanding their defining properties helps when selecting buffering strategies for robust digital designs. Given Data / Assumptions: FIFO preserves ordering (first-in/first-out discipline). Many FIFOs support independent write and read clocks (asynchronous FIFOs). Buffering can smooth bursty arrivals into steadier outputs. Concept / Approach: By decoupling input and output timing, a FIFO allows systems running at different or variable rates to exchange data safely. Its queue property preserves temporal order, while its depth absorbs bursts, preventing overrun/underrun when average rates match but instantaneous rates differ. Step-by-Step Solution: Write side pushes data entries into the FIFO at its own clock rate.Read side pops entries at another clock rate (same or different).Internal pointers and status flags (full/empty/almost) manage flow control.The order of items is maintained from input to output. Verification / Alternative check: Application notes for asynchronous FIFOs detail Gray-coded pointers and synchronizers to ensure reliable operation across clock domains while preserving sequence order. Why Other Options Are Wrong: Each individual statement (A, B, C) is true; therefore “All of the above” best summarizes FIFO capabilities. Common Pitfalls: Ignoring metastability risks at clock-domain crossings; not provisioning enough depth for worst-case bursts; misusing synchronous FIFOs where clocks differ. Final Answer: All of the above

Question 6

Solid-state storage trend: Because flash memory modules are nonvolatile, fast to access relative to disks, low in power, and have no moving parts, they have become feasible alternatives to magnetic disk storage. Choose the best match.

Accepted Answer

Introduction / Context: Storage media selection balances performance, durability, power, and cost. Solid-state drives (SSDs) and embedded flash modules have transformed storage by replacing or augmenting magnetic disks in many applications, from laptops to data centers and embedded systems. Given Data / Assumptions: Nonvolatility: retains data without power. High-speed access compared to seek/rotational delays in disks. Low power and no mechanical parts increase reliability and shock tolerance. Concept / Approach: Flash memory modules (NAND-based SSDs, eMMC, UFS, NVMe devices) provide rapid random access and high bandwidth without moving parts. While raw latency is higher than DRAM, it is far lower than disk seek times. Thus, flash-based modules are practical disk alternatives in many systems. Step-by-Step Solution: Identify attributes that make an alternative feasible: nonvolatile + fast + low power + no moving parts.Match these attributes to candidate technologies.CD-ROMs are read-only and slower for random access; ROM/EPROM are not capacity- or interface-appropriate as disk replacements.Conclude flash memory modules fit all criteria. Verification / Alternative check: Market trends show widespread adoption of SSDs over HDDs in client devices; embedded modules replace disks in mobile and IoT platforms due to size, power, and robustness advantages. Why Other Options Are Wrong: CD-ROMs: Optical, slow random access, read-only. ROMs/EPROMs: Intended for firmware storage; limited rewrite cycles and interfaces. Common Pitfalls: Assuming flash endurance equals DRAM or ignoring wear-leveling; overlooking that flash still has write/erase constraints despite being a strong disk alternative. Final Answer: flash memory modules

Question 7

RAM vs. ROM fundamentals: The key differences between random access memory (RAM) and read-only memory (ROM) are that ________.

Accepted Answer

Introduction / Context: Understanding the practical differences between RAM and ROM is foundational in digital systems. Designers choose between these technologies based on mutability, volatility, and how the memory connects to buses and control signals. This question asks which statements correctly distinguish RAM from ROM in typical systems. Given Data / Assumptions: RAM refers to volatile semiconductor memory (e.g., SRAM, DRAM) that supports read and write operations. ROM refers broadly to non-volatile memory (mask ROM, PROM/OTP, EPROM, EEPROM, Flash) that is read-mostly during normal operation. Both RAM and ROM are randomly addressable; neither inherently requires sequential access like tape. Concept / Approach: During normal operation, RAM exposes a read/write control (often WE for write enable) that permits frequent updates. ROM devices, by contrast, either cannot be altered or require special programming procedures, so a normal read/write control for everyday writes is absent. RAM loses its contents at power-down (volatile), while ROM retains data (non-volatile). Addressing for both is random: any valid address can be accessed in roughly constant time. Step-by-Step Solution: Check mutability: RAM is read/write; ROM is read-mostly in-system.Check volatility: RAM contents vanish on power loss; ROM persists.Check addressing: both are random access; sequential media are different technologies. Verification / Alternative check: Datasheets for SRAM/DRAM show WE and volatile storage; ROM/Flash parts document read-mostly operation and data retention across power cycles. Why Other Options Are Wrong: Option A: True but incomplete; volatility is also a key difference.Option B: True but incomplete; read/write controllability also differs.Option C: Incorrect; ROM is also random access.Option E: Incorrect because C is false. Common Pitfalls: Assuming “random access” means “RAM only,” or thinking ROM must be read sequentially. Final Answer: RAM has a read/write signal and ROM doesn't; RAM will lose data when the power is removed and ROM won't.

Question 8

DRAM vs. SRAM advantage: The major advantage of dynamic RAM (DRAM) over static RAM (SRAM) is ________.

Accepted Answer

Introduction / Context: Choosing between DRAM and SRAM affects cost, capacity, power, and performance. DRAM cells use a single transistor and a capacitor, while SRAM uses multi-transistor latches. This architectural difference drives the primary advantages and trade-offs. Given Data / Assumptions: DRAM requires periodic refresh; SRAM does not. SRAM is faster at random accesses and simpler to interface (no refresh). We compare mainstream parts at similar technology nodes. Concept / Approach: Because a DRAM bit cell is much smaller than an SRAM bit cell, far more DRAM bits fit on a die area, yielding higher storage density and lower cost per bit. However, DRAM access times are typically longer and require refresh logic, so raw speed is not the primary DRAM advantage; capacity and price are. Step-by-Step Solution: Identify primary metrics: cost/bit and bits/mm^2.Relate to cell topology: 1T1C DRAM vs. 6T (typical) SRAM.Conclude DRAM leads in cost and density; SRAM leads in speed and simplicity. Verification / Alternative check: Market data and datasheets consistently show DRAM providing gigabits per chip at low cost, while SRAM targets caches/buffers where speed matters more than capacity. Why Other Options Are Wrong: Cost: True but incomplete; density is also a key reason.Speed: Generally an SRAM advantage.Storage density: True but incomplete without cost.Lower refresh complexity: DRAM actually requires refresh; SRAM does not. Common Pitfalls: Equating “popular in PCs” with “faster”; DRAM dominates main memory due to capacity and cost, not because it is faster than SRAM. Final Answer: cost and storage density

Question 9

Name the operation: The memory operation that places data on the outputs after a new address is applied is called ________.

Accepted Answer

Introduction / Context: Memory transactions are named by their data direction and control strobes. When the system wants the contents of a location to appear on the data outputs, it performs a read. Correctly identifying read versus write is critical for designing timing and control logic. Given Data / Assumptions: An address is driven onto the address bus and stabilized. Control signals select the memory and indicate read (e.g., WE inactive, OE active, CS active). After access time, the memory drives outputs with the stored value. Concept / Approach: A read cycle retrieves data from a specified address. In SRAM, this typically requires CS active, WE inactive, and OE active. In DRAM, after RAS/CAS latch row and column with WE high, the selected cell's data appears at the outputs. By contrast, write cycles store data into the addressed location, and refresh cycles restore DRAM charge without presenting new external data for general use. Step-by-Step Solution: Apply address and assert selection signals (CS and OE).Ensure write is not requested (WE inactive).Wait for access time; memory drives outputs with stored data. Verification / Alternative check: Timing diagrams label tACC and tOE associated with read, whereas write timing references tDW/tWP for data setup and write pulse width. Why Other Options Are Wrong: Write cycle: Direction is to memory, not from memory.Refresh cycle: Internal DRAM operation; not a general data readout.Chip select: An enabling signal, not the entire operation.Precharge interval: DRAM internal preparation, not data presentation. Common Pitfalls: Assuming that asserting CS alone produces data; OE/WE states also matter. Final Answer: a read cycle

Question 10

Address width: What is the minimum number of address lines required to address a 64K memory?

Accepted Answer

Introduction / Context: Address lines determine how many distinct locations a memory can select. The count follows a power-of-two relationship: with N address lines, you can uniquely address 2^N locations. This question applies the formula to a common capacity figure. Given Data / Assumptions: Memory size is 64K locations (65,536 locations). Each unique address selects one location. We need the minimum whole number of lines to reach at least 65,536 addresses. Concept / Approach: Use the identity 2^N = number of unique addresses. Solve for N such that 2^N ≥ 65,536. Because 2^16 = 65,536 exactly, N = 16 address lines are required. Step-by-Step Solution: Recall powers of two: 2^10 = 1,024; 2^16 = 65,536.Match target: 64K = 65,536 = 2^16.Therefore N = 16 address lines. Verification / Alternative check: Compute log2(65,536) = 16 exactly. Any fewer lines would address fewer than 65,536 locations. Why Other Options Are Wrong: 10/12/14: 2^10, 2^12, 2^14 are far below 65,536.18: Would over-provision to 262,144 addresses. Common Pitfalls: Confusing 64K bytes with 64K words and mixing address lines with data bus width; the address lines depend on number of locations, not data word size. Final Answer: 16

Question 11

Memory composition problem: How many 16k × 4 memory ICs are required to build a 128k × 8 memory?

Accepted Answer

Introduction / Context: Designers often scale memory depth and width by combining smaller ICs. This exercise practices two key ideas: stacking for width (more bits per word) and banking for depth (more addresses). Given Data / Assumptions: Target memory: 128k × 8 (128k words, 8 bits per word). Available chip: 16k × 4 (16k words, 4 bits per word). All chips share the same address width for a given depth. Concept / Approach: To reach the target depth, bank chips so that addresses select among 128k/16k = 8 banks. To reach the target width, place chips in parallel so that their data outputs concatenate. Each bank requires 2 devices in parallel to go from 4 bits to 8 bits. The total number of devices equals banks * devices per bank = 8 * 2 = 16. Step-by-Step Solution: Compute depth multiplier: 128k / 16k = 8 banks.Compute width multiplier: 8 bits / 4 bits = 2 devices per bank.Total devices: 8 * 2 = 16. Verification / Alternative check: Check interface: All chips in a bank share addresses and chip enable; the paired devices in parallel provide D[7:4] and D[3:0]. Why Other Options Are Wrong: 4 or 8 or 12: Under-provision either depth or width.24: Over-provision beyond required capacity. Common Pitfalls: Mixing up “k” (depth) scaling with width scaling, or forgetting that width expansion multiplies device count per bank. Final Answer: 16

Question 12

DRAM control convention: In DRAM operations, it is assumed that the R/W (write enable) signal is in its ________ state during a ________ operation.

Accepted Answer

Introduction / Context: Control-signal polarity matters in DRAM timing. Many DRAM families use active-low write enables (sometimes labeled W or WE). Understanding which state corresponds to read or write avoids bus-contention bugs and corrupted data. Given Data / Assumptions: Typical DRAM has RAS, CAS, and W/WE control pins. Write is requested when W is asserted (often LOW), while read occurs when W is deasserted (often HIGH). OE may qualify data outputs; internal details vary by family. Concept / Approach: During a read, the device must not sample data-in; instead, it should drive data-out. Thus W remains in the deasserted (HIGH) state for read. During a write, W goes LOW to enable storing D at the addressed location while outputs are typically not driven. Step-by-Step Solution: Set up row/column via RAS/CAS.For read: keep W HIGH (inactive) and enable outputs as specified.For write: drive W LOW (active) with valid input data and timing. Verification / Alternative check: Consult any classic DRAM timing: WE=HIGH indicates read; WE=LOW indicates write pulse width and data setup. Why Other Options Are Wrong: Hi-Z states do not apply to W; Hi-Z relates to outputs, not the control input level.HIGH, write contradicts active-low write convention.LOW, refresh: Refresh is driven by RAS/CAS cycling, not by forcing W alone. Common Pitfalls: Confusing output tri-state with control input levels or assuming all signals use active-high logic. Final Answer: HIGH, read

Question 13

Why “flash” memory is called flash: Flash memories are so named because of their rapid ________ times.

Accepted Answer

Introduction / Context: Flash memory evolved from EEPROM technology with improved block erase mechanisms. The term “flash” reflects that large blocks can be erased quickly, almost “in a flash,” enabling practical reprogramming compared with byte-erasable EEPROMs. Given Data / Assumptions: Flash supports block (sector/page) erase operations. Write (program) times per page/word are designed for practical firmware updates. Random reads are also fast, but erase is the historically notable improvement. Concept / Approach: Flash excels at rapid block erasure followed by programming, making it useful for code storage that needs occasional in-field updates. While read latency is important, what differentiates flash from older EEPROM is the efficient erase/program cycle for larger regions rather than slow, byte-at-a-time erasure. Step-by-Step Solution: Recognize the historical motivation: faster erase than EEPROM.Acknowledge that programming (write) is paired with erase to update contents.Identify the phrase “erase and write” as capturing the rapid update cycle. Verification / Alternative check: Device datasheets list sector erase times and page program times as key specs, emphasizing update speed compared to legacy EEPROM. Why Other Options Are Wrong: Read and write: Reads are fast but not the naming origin.Format and erase: “Format” is a filesystem notion, not a device operation.Erase and read: Ignores programming speed.Address decode and sense: Internal operations, not the naming reason. Common Pitfalls: Assuming “flash” refers primarily to read performance; the key differentiator is erasability/programming of large blocks. Final Answer: erase and write

Question 14

ROM checksum testing: Assume a ROM under test is compared to a known-good ROM. If the checksums are different, the ROM under test is ________.

Accepted Answer

Introduction / Context: Checksums and hashes detect corruption in stored data and firmware. Comparing a device's checksum to a known-good reference is a fast, practical way to screen for errors during manufacturing or field diagnostics. Given Data / Assumptions: The checksum algorithm is deterministic and covers the entire ROM image. Measurement conditions are reliable (no read timing violations causing transient errors). We compare against a trusted, correct reference checksum. Concept / Approach: If two checksums differ, at least one bit in the covered data differs with overwhelming probability (especially for strong checksums/hashes). While an unstable read could in theory cause a mismatch, in normal conditions a mismatch strongly indicates incorrect contents, hence “very likely to be bad.” Absolute certainty (“definitely bad”) requires ruling out instrumentation errors, but the diagnostic implication is clear. Step-by-Step Solution: Compute device checksum from all addresses.Compare with known-good checksum.If different, flag the unit as likely faulty or containing wrong firmware. Verification / Alternative check: Re-read the ROM, possibly at slower timing; if the mismatch persists, probability of bad contents approaches certainty. Why Other Options Are Wrong: Very likely/definitely good: Contradict evidence of mismatch.Definitely bad: Too strong without retest; instrumentation might err.Power-cycle: Not a valid corrective action for genuine data mismatch. Common Pitfalls: Assuming a simple checksum equals a cryptographic hash; weaker checksums have collision risks, but a mismatch still signals trouble. Final Answer: very likely to be bad

Question 15

CD-ROM storage medium: A CD-ROM is a form of read-only memory in which data are stored as ________.

Accepted Answer

Introduction / Context: Different read-only media use different physical storage mechanisms. CD-ROMs are optical media that encode data as a pattern of microscopic features on a reflective disk, which are sensed by a laser during readout. Given Data / Assumptions: Medium: Compact Disc Read-Only Memory (CD-ROM). Read mechanism: laser optics and photodetectors. Goal: identify the physical representation of bits. Concept / Approach: CD-ROMs encode data as pits and lands along a spiral track. Changes between pit and land edges modulate the reflected light, which the player detects and decodes into a digital bitstream. This is distinct from magnetic storage (which uses magnetized domains) and semiconductor ROM (which uses programmed cells or masks). Step-by-Step Solution: Recognize CD as optical, not magnetic or semiconductor charge-based.Identify the pit/land pattern as the data carrier.Select the option describing “pits on an optical disk.” Verification / Alternative check: CD standards (e.g., Red Book/Yellow Book) describe Eight-to-Fourteen Modulation and pit geometry used to store and retrieve data. Why Other Options Are Wrong: Magnetic bubbles/spots: Describe magnetic memories, not optical.Pinholes: Not how CDs encode bits; the substrate is continuous.Floating gates: Describe Flash/EEPROM, not optical disks. Common Pitfalls: Thinking the laser “burns holes”; pressed CD-ROMs have molded pits, while recordable CDs use dye changes, still read optically. Final Answer: "pits" on an optical disk

Question 16

DRAM refresh requirement (2118 device): All rows in a 2118 dynamic RAM must be refreshed ________.

Accepted Answer

Introduction / Context: DRAM cells store charge on tiny capacitors that leak over time. To preserve data, each row must be periodically refreshed by re-reading and re-writing it within a device-specified refresh interval. Classical 16K × 1 devices like the 2118 provide a representative requirement. Given Data / Assumptions: Device: 2118 DRAM (16K × 1 organization, classic generation). Refresh is row-based and must complete within a fixed window. We assume nominal operating conditions. Concept / Approach: Historical DRAMs often require that all rows be refreshed within approximately 2 ms (some families use 4 ms or similar, but the 2 ms figure is typical for this class). Controllers satisfy this by distributing refresh cycles across time, ensuring that each row is accessed before its charge decays below a safe threshold. Step-by-Step Solution: Determine required interval: complete a full set of refreshes every 2 ms.Schedule refreshes by cycling through row addresses.Verify that normal accesses do not starve refresh timing. Verification / Alternative check: Datasheets specify maximum refresh interval per row set; meeting or beating that interval guarantees retention across the array. Why Other Options Are Wrong: Every second/millisecond/50 ms/200 µs: Either far too long (data would be lost) or unnecessarily short relative to the stated device requirement. Common Pitfalls: Assuming refresh is optional; if missed, data corruption is inevitable in DRAM. Final Answer: every 2 milliseconds

Question 17

While a stored-program computer is executing a compiled program, the CPU will ________ specific memory locations as dictated by each instruction's address and operand fields.

Accepted Answer

Introduction / Context: A fundamental idea in computer architecture is the stored-program concept: instructions and data reside in memory, and the central processing unit (CPU) fetches, decodes, and executes those instructions. Each instruction specifies operands by referencing memory locations or registers. The essential CPU action with respect to memory is to address the correct location, not to arbitrarily format or scan memory. Given Data / Assumptions: Von Neumann or Harvard machine with a program counter and address bus. Instructions contain opcodes and operand specifiers (addresses/immediates). Standard fetch–decode–execute cycle with read/write transactions. Concept / Approach: Memory interaction begins by placing a destination or source address on the address bus. Control lines indicate read or write, and the data bus carries the transferred value. Without correctly addressing memory, neither instruction fetches nor data loads/stores can occur. Thus “addressing memory locations” is the precise, architecture-neutral verb for what the CPU does per instruction. Step-by-Step Solution: 1) Fetch: CPU addresses the next instruction using the program counter.2) Decode: determine required operands and effective addresses.3) Execute: initiate memory read/write by addressing the operand locations.4) Write-back: if needed, place result at the addressed destination. Verification / Alternative check: Any bus timing diagram shows an address phase followed by a data phase. Microarchitecture textbooks consistently describe addressing as the first step of memory access cycles. Why Other Options Are Wrong: “format” refers to initializing or low-level media prep; “scan” implies searching sequentially; “store data into” happens only for writes and still requires addressing; “verify” is a testing action, not routine execution. Common Pitfalls: Confusing “addressing” (pointing to a location) with the subsequent read/write data transfer; both occur, but addressing is the universal first requirement. Final Answer: address

Question 18

Read-only memory (ROM) – which statement best describes its user alterability and retention characteristics?

Accepted Answer

Introduction / Context: Read-only memory (ROM) refers to memory whose contents are fixed for normal operation. The data is established at manufacture (mask ROM) or by a one-time programming step (PROM) and is not changed by the system or user during routine use. Understanding ROM clarifies how firmware and lookup tables are delivered in embedded systems. Given Data / Assumptions: “User” means the system integrator or end user during normal use. ROM contrasts with RAM (read/write) and with EEPROM/Flash (rewritable nonvolatile). We discuss functional behavior, not interface timing details. Concept / Approach: The defining property of ROM is immutability from the user’s perspective. While various ROM technologies exist, they share that the stored pattern is fixed during normal device operation. ROM is also nonvolatile, retaining data without power. Therefore, the statement that a ROM’s data can never be altered by the user precisely captures the category. Step-by-Step Solution: 1) Identify the role: store fixed code/tables.2) Determine alterability: none during system use.3) Retention: contents persist without power (nonvolatile).4) Conclusion: the correct characterization is “cannot be altered by the user.” Verification / Alternative check: Datasheets for mask ROM/PROM specify read operations only and lack erase/write commands, confirming user immutability. Why Other Options Are Wrong: Unlimited capacity is untrue; refresh is a DRAM property; “maintains storage as long as power is applied” describes volatile RAM, not ROM; auto-erase on power-up is not a ROM behavior. Common Pitfalls: Confusing ROM with EPROM/EEPROM/Flash, which are erasable but not categorized as ROM in everyday usage when rewrite is possible. Final Answer: whose data can never be altered by the user

Question 19

Magnetic media organization – floppy disks are arranged in concentric rings called ________ (as distinct from their wedge-shaped subdivisions).

Accepted Answer

Introduction / Context: Understanding the layout of legacy magnetic media helps when reading vintage documentation and file system structures. Floppy disks organize data radially and angularly: concentric circular paths and divisions along those paths. This question asks for the correct term for the circular paths. Given Data / Assumptions: Concentric geometry on rotating media. Common DOS/BIOS terminology still used in emulators and low-level tools. Terminology mirrors that of hard disks, though densities differ. Concept / Approach: Each concentric circular path on the disk surface is a track. Tracks are further divided into angular slices called sectors. Filesystems map logical blocks onto sector addresses (cylinder/head/sector or later LBA abstractions). The wording “concentric rings” cues “tracks” as the correct noun. Step-by-Step Solution: 1) Identify circular geometry → tracks.2) Note subdivisions along track angle → sectors.3) Confirm: “concentric rings” equals tracks, not sectors.4) Answer accordingly. Verification / Alternative check: Technical references consistently define tracks as concentric circles and sectors as arc segments along a track. Why Other Options Are Wrong: “arrays” and “cells” do not describe magnetic disk geometry; “sectors” are the wedge-shaped pieces of a track; “bands” is not standard floppy terminology. Common Pitfalls: Swapping “track” and “sector” because both describe disk structure; remember: track = ring, sector = slice. Final Answer: tracks

Question 20

EEPROM fundamentals – which statement accurately describes information stored in an EEPROM device used in digital systems?

Accepted Answer

Introduction / Context: Electrically Erasable Programmable Read-Only Memory (EEPROM) is common for configuration bytes, calibration constants, and firmware parameters. Distinguishing EEPROM from EPROM, Flash, and RAM avoids design errors in embedded systems and field updates. Given Data / Assumptions: Nonvolatile storage required (retains data without power). Random byte or small page access typical for EEPROM. Erase/program operations use internal charge pumps to apply elevated voltages safely. Concept / Approach: EEPROM stores charge on floating gates and supports electrical erase/program in-circuit. Unlike EPROM (UV-erasable) and PROM (one-time), EEPROM can be updated repeatedly at fine granularity. The erase/program sequence employs controlled high-voltage pulses (generated on-chip) to tunnel charge, but from a system viewpoint it is simply an electrical command over I2C/SPI/parallel interfaces. Step-by-Step Solution: 1) Identify nonvolatile and rewritable requirements → EEPROM fits.2) Confirm mechanism: electrical erase/program via charge tunneling.3) Note granularity: often byte-level writes with internal page buffers.4) Conclude the accurate statement is electrical erasability and reprogrammability. Verification / Alternative check: Vendor datasheets list write/erase command sets and endurance (for example, 100k+ cycles), with retention specs measured in years—no UV window is present. Why Other Options Are Wrong: Manufacturer-fixed refers to mask ROM; volatility describes RAM; UV erasure describes EPROM; forced sequential rewrite is not a universal EEPROM constraint. Common Pitfalls: Confusing EEPROM with Flash (block erase) or EPROM (UV erase). EEPROM’s electrical erase at small granularity is its hallmark. Final Answer: is electrically erasable and reprogrammable using controlled high-voltage pulses

Memory and Storage Questions