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74LS138 enables — The octal decoder’s outputs are enabled (active LOW) only for which specific enable-input combination?

Difficulty: Easy

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
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