How many 128 × 8 bit RAMs are required to design 32 K × 32 bit RAM?

Concept

A larger memory is built by expanding width and depth independently. Width expansion places chips in parallel, while depth expansion uses additional groups to increase the number of addressable words.

The required chip count equals (target depth ÷ chip depth) × (target word width ÷ chip width). Here, K denotes 1024.

Application

1. Convert the target depth: 32K = 32 × 1024 = 32,768 addressable words.

2. Find the depth-expansion factor: 32,768 ÷ 128 = 256 groups.

3. Find the width-expansion factor: 32 ÷ 8 = 4 chips per group.

4. Multiply the independent expansion factors: 256 × 4 = 1024 chips.

Cross-check

Each chip stores 128 × 8 = 1,024 bits. Therefore, 1024 chips store 1,048,576 bits, equal to (32 × 1024) × 32 = 1,048,576 bits required by the target RAM.

Result

The 32K × 32-bit RAM requires 1024 chips.