74LS138 enables — The octal decoder’s outputs are enabled (active LOW) only for which specific enable-input combination?
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AG1 = 1, G2A = 0, G2B = 0
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BG1 = 0, G2A = 1, G2B = 1
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CG1 = 1, G2A = 1, G2B = 1
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DG1 = 0, G2A = 0, G2B = 0
Answer
Correct Answer: G1 = 1, G2A = 0, G2B = 0
Explanation
Introduction / Context: The 74LS138 has three enable inputs: one active-HIGH (G1) and two active-LOW (G2A, G2B). Only when the correct combination is applied are the eight active-LOW outputs allowed to reflect the A, B, C inputs. This question reinforces correct enable logic before decoding can occur.Given Data / Assumptions:
- Outputs are active LOW (selected output goes LOW).
- Enable logic: G1 must be HIGH; G2A and G2B must be LOW.
- If not properly enabled, all outputs remain HIGH (inactive).
Concept / Approach: Enabling conditions gate the decoder’s internal logic. The device truth table shows decoding is active only when G1=1 and both G2 inputs are 0. Any other combination disables the chip, forcing all outputs HIGH independent of A/B/C inputs.Step-by-Step Solution:
Check G1: must be 1 (active-HIGH enable).Check G2A and G2B: must both be 0 (active-LOW enables asserted).Confirm that other combinations disable the outputs.Verification / Alternative check:
Compare to datasheet truth tables listing enable conditions and output states.Why Other Options Are Wrong:
G1 = 0, G2A = 1, G2B = 1: All disables asserted; outputs inactive (HIGH).G1 = 1, G2A = 1, G2B = 1: G2s not asserted; chip disabled.G1 = 0, G2A = 0, G2B = 0: G1 not asserted; chip disabled.Common Pitfalls:
Forgetting that “active-LOW” means logic 0 is the asserted state.Assuming any single enable is sufficient; all three must be in the correct states.Final Answer:
G1 = 1, G2A = 0, G2B = 0