Octal-to-binary encoder output width: An octal-to-binary encoder accepts eight distinct inputs (one active at a time). How many binary output bits are required to encode the active input line unambiguously?

Introduction / Context:
Encoders compress one-of-M input selections into a binary word. The number of output bits required is fundamentally tied to the number of unique input lines. Understanding this relationship is essential for sizing address buses, keypad scanners, and interrupt encoders.

Given Data / Assumptions:

• M = 8 input lines (octal domain).
• Exactly one input is active at a time.
• Unambiguous encoding is required for all input positions.

Concept / Approach:
The number of binary bits N needed to represent M unique states follows 2^N ≥ M. For M = 8, we need the smallest N such that 2^N ≥ 8. Since 2^3 = 8, three bits suffice. Two bits yield only four states, which is insufficient, while four bits would be more than necessary.

Step-by-Step Solution:
Compute: 2^2 = 4 (insufficient).Compute: 2^3 = 8 (exact fit).Thus, N = 3 bits are required.Conclude: an octal-to-binary encoder outputs three bits.

Verification / Alternative check:
Examine the 74148 priority encoder data: it maps eight inputs to a 3-bit binary code (active-low conventions aside). This confirms the theoretical derivation.

Why Other Options Are Wrong:
Two cannot represent eight states. Four or five are overprovisioned for eight states and are not minimal.

Common Pitfalls:
Forgetting the “power of two” relation; ignoring that priority encoders still output only N bits even when multiple inputs are asserted (priority resolves conflicts).

Final Answer:
three