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)?

Electronics Combinational Logic Circuits Difficulty: Easy
Choose an option
  • A
    0
  • B
    3
  • C
    9
  • D
    None. All outputs are HIGH.
  • E
    8

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

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