BCD-to-decimal decoder with active-LOW outputs — identify the asserted output. A BCD-to-decimal decoder has active-HIGH inputs and active-LOW outputs. Which decimal output line goes LOW when the inputs are 1001 (i.e., 9)?
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A0
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B3
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C9
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DNone. All outputs are HIGH.
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E8
Answer
Correct Answer: 9
Explanation
Introduction / Context:BCD-to-decimal decoders (e.g., 7442/7443 types) take a 4-bit BCD input and assert exactly one of ten outputs corresponding to digits 0–9. In many parts, the outputs are active-LOW, meaning the selected line is driven LOW while all others remain HIGH.
Given Data / Assumptions:
- Inputs are BCD with active-HIGH logic.
- Outputs are active-LOW (asserted LOW, deasserted HIGH).
- Input pattern 1001 corresponds to decimal 9.
Concept / Approach:With active-LOW outputs, the selected decimal line is the one matching the BCD input value and will be 0 (LOW); all other decimal output lines stay at 1 (HIGH). Thus, for 1001, the 9 output goes LOW.
Step-by-Step Solution:
Interpret input: 1001₂ (BCD) ⇒ digit 9.Active-LOW outputs: selected line is LOW; others HIGH.Therefore, output '9' is LOW, all others HIGH.Verification / Alternative check:Consult any active-LOW decoder truth table: for input 9, Y9 = 0 and Y0–Y8,Y10… = 1. Simulations or bench tests with LEDs to ground (since outputs sink current when LOW) also demonstrate this behavior clearly.
Why Other Options Are Wrong:
- 0, 3, 8: incorrect digit selections for BCD input 9.
- None. All outputs HIGH: would correspond to no valid input or an enable not asserted, not the stated valid BCD input.
Common Pitfalls:
- Forgetting the output polarity and expecting a logic HIGH on the selected line.
- Confusing binary 1001 (9) with an invalid BCD digit; it is valid.
Final Answer:9