8085-style microprocessor basics: In the same 8-bit microprocessor system context, how many bits wide is the address bus typically used to select memory locations?

Introduction / Context:The address bus width sets the maximum number of uniquely addressable locations. Many classic 8-bit processors pair an 8-bit data bus with a wider address bus to access a larger memory space. This question targets the typical address width for such systems (e.g., Intel 8085).

Given Data / Assumptions:

• 8-bit microprocessor class (e.g., 8085).
• We refer to the external address bus width.
• Linear addressing without bank switching.

Concept / Approach:The Intel 8085 exposes a 16-bit address bus (A0–A15), allowing 2^16 = 65,536 distinct addresses. When each address maps to a byte, this equals 64 KB of addressable memory. Many contemporary 8-bit processors followed this convention for simplicity and sufficient space for code and data of the era.

Step-by-Step Solution:Identify typical address bus width: 16 bits for 8085-class CPUs.Compute capacity: 2^16 = 65,536 locations.Relate to memory size: 65,536 bytes = 64 KB.Conclude the correct answer is 16 bits.

Verification / Alternative check:Pinout references show A0–A15 lines on the 8085, confirming a 16-bit address path.

Why Other Options Are Wrong:7, 8, 9: far too small to provide the conventional 64 KB space of classic 8-bit MPUs.

Common Pitfalls:Assuming data bus width equals address bus width. They are independent: one describes data quantity per transfer, the other defines addressable range.

Final Answer:16