Memory addressing calculation How many address lines are required to uniquely address all locations in a 2K × 4 memory chip?

Introduction / Context:
Determining the number of address lines needed for a given memory organization is a frequent task in digital design and microcontroller interfacing.

Given Data / Assumptions:

• Memory size: 2K × 4 (2K addresses, 4 data bits per address).
• K uses the binary convention: 1K = 1024.
• Each unique address requires a unique binary code.

Concept / Approach:
If there are N addresses, the number of unique address codes required is N. The number of address lines A must satisfy 2^A ≥ N. For N = 2048, we solve 2^A = 2048.

Step-by-Step Solution:
Compute addresses: 2K = 2 × 1024 = 2048.Find A such that 2^A = 2048.Since 2^11 = 2048, A = 11 address lines.

Verification / Alternative check:
Confirm powers of two: 2^10 = 1024, 2^11 = 2048. Therefore, 11 is correct.

Why Other Options Are Wrong:
8, 9, 10: 2^8, 2^9, 2^10 are 256, 512, 1024 addresses—insufficient.12: 2^12 = 4096 addresses—more than needed.

Common Pitfalls:
Confusing the data width (×4) with address lines; only the number of locations determines address line count.

Final Answer:
11