Decoder sizing How many 3-line-to-8-line decoders are required to realize a 1-of-32 decoder (select exactly one output out of 32 lines), assuming any necessary small predecode or gating can be handled outside these decoder ICs?

Introduction / Context:Large decoders are commonly constructed from smaller standard decoder ICs. Understanding how to hierarchically combine decoders helps with resource estimation and practical PCB design. A 1-of-32 decoder asserts exactly one of 32 outputs for a 5-bit input address.

Given Data / Assumptions:

• Target: 1-of-32 decode (5 input address bits select one of 32 outputs).
• Available building blocks: 3-to-8 decoders (8 outputs each, enable inputs available).
• Additional small glue logic (e.g., a simple 2-to-4 predecode) can be implemented outside the count if needed.

Concept / Approach:Partition the 5-bit address into a 2-bit high field and a 3-bit low field. Use a 2-to-4 predecoder (implemented with gates or other logic not counted here) to generate four enables. Feed each enable to one 3-to-8 decoder that handles the lower 3 address bits. Only the enabled decoder will assert one of its eight outputs, yielding a total of 4 * 8 = 32 distinct lines.

Step-by-Step Solution:

Verification / Alternative check:Count outputs: 4 decoders * 8 outputs/decoder = 32 outputs, matching specification. Timing-wise, this yields a two-level decode (predecode + 3-to-8) typical in memory address decoding.

Why Other Options Are Wrong:

• 1 or 2: Insufficient outputs; 2 decoders provide at most 16 lines.
• 8: Would over-provision to 64 outputs when combined similarly.

Common Pitfalls:Assuming an extra 3-to-8 is required to realize the 2-to-4 predecode. While some designs may count that as a fifth decoder, the question explicitly allows small predecode outside the count, leaving four 3-to-8 decoders sufficient.

Final Answer:4