74LS138 decoder — Evaluate: “Output 6 is selected when A0 = 1, A1 = 1, A2 = 0 (with the device enabled).”

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Solve this multiple-choice question and choose the correct option.

Accepted Answer

Introduction / Context: The 74LS138 is a 3-to-8 decoder with active-LOW outputs Y0–Y7. It asserts exactly one output LOW based on the binary code on inputs A (LSB), B, and C (MSB) when properly enabled. The prompt tests correct output indexing for a specific input pattern.Given Data / Assumptions: Inputs: A0 ≡ A (LSB), A1 ≡ B, A2 ≡ C (MSB). Enable conditions satisfied: G1 = 1, G2A = 0, G2B = 0. Outputs are active LOW; “selected” means driven LOW. Concept / Approach: The output index equals the binary value ABC (with C as MSB, A as LSB). For A2 A1 A0 = 0 1 1, the binary value is 3, so Y3 goes LOW, not Y6. Therefore, saying “Output 6 is selected” for 0 1 1 is incorrect.Step-by-Step Solution: Write inputs in MSB→LSB order: C B A = 0 1 1.Compute index: 04 + 12 + 1*1 = 3.Map to output: Y3 is asserted LOW (active); Y6 remains HIGH (inactive). Verification / Alternative check: Truth tables in datasheets confirm A=1, B=1, C=0 → Y3 LOW. Why Other Options Are Wrong: Correct: Would wrongly claim Y6 is selected.True only if outputs are active HIGH: 74LS138 outputs are active LOW by design.Depends on enable levels only: Enables must be correct, but the output index still follows ABC coding. Common Pitfalls: Swapping bit significance (treating A as MSB).Forgetting that “selected” means LOW on an active-LOW decoder. Final Answer: Incorrect

74LS138 decoder — Evaluate: “Output 6 is selected when A0 = 1, A1 = 1, A2 = 0 (with the device enabled).”

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