Address space implied by an 8-bit address bus With an 8-bit address bus, what is the accessible address range expressed in hexadecimal?

Introduction / Context:
An address bus of width N bits can uniquely select 2^N addresses. Embedded developers must translate bus widths into concrete ranges to reason about memory mapping and peripheral decoding. This question checks the ability to map an 8-bit bus to the correct hexadecimal range.

Given Data / Assumptions:

• Address bus width is 8 bits.
• Addresses are zero based and inclusive at both ends when stated as ranges.
• Hexadecimal notation is used.

Concept / Approach:

Two raised to the power of eight equals 256. The lowest address is 00H and the highest is FFH, giving 256 total unique locations. Wider buses such as 16-bit would yield 0000H to FFFFH.

Step-by-Step Solution:

Verification / Alternative check:

Compare with a 16-bit case: 2^16 = 65536 addresses from 0000H to FFFFH, confirming the pattern.

Why Other Options Are Wrong:

0000–FFFFH: corresponds to a 16-bit bus.

000–FFFH: corresponds to a 12-bit bus.

0–FH: corresponds to a 4-bit bus.

Common Pitfalls:

Confusing data bus width with address bus width, or assuming decimal endpoints rather than hexadecimal.

Final Answer:

00 to FFH