74LS138 enable logic — The decoder’s outputs become active (LOW) when which enable-input combination is present?

Question

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

Accepted Answer

Introduction / Context: Before the 74LS138 decodes A/B/C into one of eight outputs, its enable pins must admit operation. Misunderstanding enable polarity leads to design errors where no output ever asserts. This item consolidates the correct enabling combination for active-LOW outputs to respond to A/B/C.Given Data / Assumptions: One active-HIGH enable (G1) and two active-LOW enables (G2A, G2B). Outputs are active-LOW; inactive state is HIGH. “Outputs become active” means the selected Y line can go LOW according to A/B/C. Concept / Approach: The chip is enabled only when its enable logic is satisfied: G1 must be asserted HIGH, and both G2 inputs must be asserted LOW. Any deviation disables the device, forcing all outputs HIGH regardless of address inputs.Step-by-Step Solution: Set G1 = 1 (assert active-HIGH enable).Set G2A = 0 and G2B = 0 (assert both active-LOW enables).Apply A/B/C; the selected Y_n line will be driven LOW. Verification / Alternative check: Datasheet truth tables show decoding active only for G1=1 and G2A=G2B=0; otherwise all outputs remain HIGH. Why Other Options Are Wrong: G1 = LOW, G2A = HIGH, G2B = HIGH: All disables; no outputs active.G1 = HIGH, G2A = HIGH, G2B = HIGH: G2s not asserted; device disabled.G1 = LOW, G2A = LOW, G2B = LOW: G1 not asserted; device disabled. Common Pitfalls: Treating G2 pins as active-HIGH; they are active-LOW.Omitting one of the enables, leading to “stuck” outputs. Final Answer: G1 = HIGH, G2A = LOW, G2B = LOW

74LS138 enable logic — The decoder’s outputs become active (LOW) when which enable-input combination is present?

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