OR gate behavior – is the output HIGH only when all inputs are HIGH?

Difficulty: Easy

Correct Answer: Incorrect

Explanation:

Introduction / Context: The OR gate forms the basis of sum-of-products design. Understanding its activation condition is crucial for alarms, enables, and multiplexing logic where any triggering input should set the output.

Given Data / Assumptions:

  • Standard OR gate with two or more inputs.
  • Active-HIGH logic convention.
  • No inversion at inputs or output (i.e., not NOR).

Concept / Approach: OR outputs 1 if at least one input is 1. Requiring “all inputs HIGH” would turn OR into an AND function, which contradicts the definition. Therefore, the claim is false.

Step-by-Step Solution:

1) Define OR: Y = A + B + C + ...2) If any input equals 1, Y = 1.3) The statement demands all-HIGH for HIGH output, which is the AND condition.4) Conclusion: the statement is incorrect.

Verification / Alternative check: Truth table check: OR(1,0)=1; OR(0,1)=1; OR(1,1)=1. Only AND requires all 1s. NOR is the inverted OR, giving 0 when any input is 1 and 1 only when all are 0, but that is a different gate entirely.

Why Other Options Are Wrong: “Correct” contradicts the OR definition. “Only for wired-OR buses” conflates wiring technique with logic function. “Only with positive logic” is irrelevant; OR semantics remain with consistent polarity. “Only for two-input devices” is false; the rule scales to N inputs.

Common Pitfalls: Confusing OR with AND or NOR due to symbol similarities. Remember: “+” denotes OR in Boolean algebra; “*” denotes AND.

Final Answer: Incorrect

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