Address space calculation – 2118 16K × 1 RAM How many binary address lines are required to uniquely select all memory locations in a 2118 RAM specified as 16K × 1 (i.e., 16,384 locations, 1 bit each)?

Introduction / Context:Sizing the address bus is a basic skill in memory interfacing. The number of address lines determines how many unique locations can be selected. Here, the device stores 1 bit at each of 16K locations.

Given Data / Assumptions:

• Memory size: 16K × 1, which means 16,384 distinct addresses.
• One address selects one 1-bit cell.
• Each extra address line doubles the addressable locations.

Concept / Approach:If a memory has N locations, the minimum number of address lines A must satisfy 2^A ≥ N. For exactly 16,384 locations, we solve 2^A = 16,384.

Step-by-Step Solution:

Verification / Alternative check:Recognize 1K = 1024 = 2^10, so 16K = 16 * 1024 = 2^4 * 2^10 = 2^14. Thus 14 lines are needed.

Why Other Options Are Wrong:

• 8 or 10: 2^8 = 256 and 2^10 = 1024, far too small.
• 16: 2^16 = 65,536 addresses, more than required; 14 is sufficient and minimal.

Common Pitfalls:

• Mistaking 16K as 16,000 rather than 16,384 (binary kilo).
• Confusing data width (×1) with address count, which are independent parameters.

Final Answer:14