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

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:

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:

Common Pitfalls:Mixing up “k” (depth) scaling with width scaling, or forgetting that width expansion multiplies device count per bank.

Final Answer:16