EPROM address decoding: A 16 KB EPROM memory space is to be built using 4K × 8 devices. Does this system require a 1-to-8 (3-to-8) address decoder for chip select?

Introduction / Context:Designing an EPROM subsystem requires determining how many physical chips are needed and how to decode their chip-select lines. The number of devices dictates the width of the decoder used to select one device at a time within the mapped address range.

Given Data / Assumptions:

• Total memory required: 16 KB (16K bytes).
• Device capacity: 4K × 8 per EPROM.
• Each device provides 4 KB, data width 8 bits.
• Goal: find the required decoder size for chip selection.

Concept / Approach:The number of devices needed is Total / Per-device = 16 KB / 4 KB = 4 devices. Selecting among 4 devices requires a 2-to-4 decoder (often called 1-of-4). A 1-of-8 (3-to-8) decoder is only needed if 8 devices must be selected, which would be the case with 2K × 8 devices for 16 KB.

Step-by-Step Solution:

Verification / Alternative check:Address bit budgeting: 4K needs 12 address lines per device; the next 2 address lines (A12–A13) decode which of the 4 chips is active. No third select bit is required.

Why Other Options Are Wrong:

• Correct: Overstates decoder width; would leave 4 outputs unused.
• Other options reference unrelated conditions (bank switching, different device size).

Common Pitfalls:Mixing up “KB” with “K locations”; ignoring that ×8 devices already match the data width, so only address-space selection is needed.

Final Answer:Incorrect