Difficulty: Easy
Correct Answer: Does not apply
Explanation:
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:
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:
1) Compute required width: m outputs need log2(m) input bits.2) For m = 16, log2(16) = 4, so four inputs are sufficient.3) Therefore, 8-bit input is not required for a 1-of-16 decoder.4) Conclude the statement is incorrect.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
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