OR gate behavior – is the output HIGH only when all inputs are HIGH?
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ACorrect
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BIncorrect
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COnly for wired-OR buses
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DOnly with positive logic
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EOnly for two-input devices
Answer
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