Memory organization and chip count calculation: A 2147 static RAM device is organized as 4K × 1, meaning 4096 storage locations with 1 bit per location. How many identical 2147 RAM chips are required to construct an 8K × 8 memory module (8,192 addresses, 8-bit word width)?

Introduction / Context:
Designers often need to build a larger memory from smaller integrated circuits. Understanding how to expand both word width (data bits per address) and depth (number of addresses) is fundamental to digital design and memory mapping. This problem asks how many 4K × 1 static RAM chips are needed to realize an 8K × 8 memory, a classic exercise that combines bit-width paralleling with depth expansion via banking or decoding.

Given Data / Assumptions:

• Available chip: 2147 static RAM organized as 4K × 1 (4096 × 1 bit).
• Target memory: 8K × 8 (8192 addresses, each 8 bits wide).
• All chips are identical; ideal decoding and control signals are available.
• Ignore timing skew and fan-out limits; focus on count and organization.

Concept / Approach:
To build a wider word (more bits per address), place 1-bit devices in parallel so their data outputs form the multi-bit word. To build deeper memory (more addresses), place complete words in banks selected by an address decoder or bank-select logic. Total chip count = chips for width * banks for depth. Use powers-of-two relationships to determine address expansion cleanly.

Step-by-Step Solution:
1) Width expansion: each 2147 provides 1 data bit; to get 8 bits per address, use 8 chips in parallel. These share all address lines; each contributes one bit (D0…D7).2) Depth expansion: each 2147 has 4K addresses; target is 8K, which is 2 × 4K. Therefore, create 2 banks of the 8-parallel set and select the active bank with a decoder.3) Chip count: width 8 chips per bank × depth 2 banks = 16 chips total.4) Control wiring: the decoder output drives the Chip Enable (or CS) of each bank; WE and OE fan out to all chips with proper buffering if needed.

Verification / Alternative check:
Compute in bits: target capacity = 8K × 8 = 8192 × 8 = 65,536 bits. One 4K × 1 chip stores 4096 bits. Number of chips needed = 65,536 / 4096 = 16, matching the bank-and-width reasoning.

Why Other Options Are Wrong:
“One” grossly underestimates capacity. “Four” equals 16,384 bits only. “Eight” yields 32,768 bits and provides either 8-bit width at only 4K depth or 8K depth at only 4-bit width, not the full 8K × 8.

Common Pitfalls:
Confusing width and depth expansion; forgetting that a 1-bit-wide chip requires 8 devices for an 8-bit word; trying to expand depth by wiring addresses incorrectly rather than using bank enables.

Final Answer:
Sixteen