Hexadecimal decoder input width: does a hexadecimal (1-of-16) decoder select one of sixteen outputs based on an 8-bit binary input?

Introduction / Context:
Hexadecimal decoders are common in address decoding and display drivers. A “hex” quantity spans sixteen distinct symbols (0–F), which directly maps to 4 bits of binary information. The question probes whether such a decoder requires 8-bit inputs.

Given Data / Assumptions:

• Hexadecimal implies 16 distinct codes.
• Binary representation uses 4 bits for 16 states.
• One-of-16 output selection is assumed.

Concept / Approach:
Because 2^4 = 16, only 4 input bits are needed to uniquely select one of 16 outputs. An 8-bit input would encode 256 states, vastly exceeding the 16 outputs of a basic hexadecimal decoder. Some devices may include enables or polarity options, but the main data input width for hex selection is 4 bits.

Step-by-Step Solution:

Verification / Alternative check:
Datasheets for 4-to-16 decoders (e.g., 74HC154) specify four address inputs A–D and sixteen outputs.

Why Other Options Are Wrong:
Applies: misstates the necessary input width.
BCD and extra enables do not change that 4 bits encode 16 states; enables gate functionality but are not primary address bits.

Common Pitfalls:
Confusing “hexadecimal display decoder” (which may include segment mapping) with the address-width requirement; conflating control lines with data width.

Final Answer:
Does not apply