Memory system sizing How many 2K × 8 ROM chips are required to construct a 16K × 8 memory system (same data width, larger address space)?

Introduction / Context:
Designers often build larger memories by banking multiple identical ICs. The key is matching data width and scaling address space appropriately.

Given Data / Assumptions:

• Each ROM chip: 2K × 8 (2048 addresses, 8-bit data).
• Target system: 16K × 8 (16384 addresses, 8-bit data).
• Same data width means chips are added in the address dimension, not in parallel for width.

Concept / Approach:
To expand from 2K to 16K at the same width, you need 16K / 2K chips in the address “stack.” The data width already matches at 8 bits, so you do not need to gang chips in parallel for width.

Step-by-Step Solution:
Compute: 16K / 2K = 8.Arrange 8 ROMs with chip-select decoding to cover all address ranges.Keep data buses in common and select one chip at a time.

Verification / Alternative check:
Total capacity check: 8 × (2K × 8) = 16K × 8, confirming the requirement.

Why Other Options Are Wrong:
2, 4: provide only 4K × 8 or 8K × 8 respectively.16 or 32: would exceed the target size at constant width.

Common Pitfalls:
Combining chips to increase width unnecessarily; forgetting chip-select decoding.

Final Answer:
8