Validity of S–R latch input combinations: A basic S–R latch has a forbidden or invalid input combination (depending on NOR or NAND implementation). Assess the claim: “The S–R flip-flop has no invalid or unused state.”

Difficulty: Easy

Correct Answer: Incorrect

Explanation:


Introduction / Context:
The S–R latch is the quintessential bistable made from cross-coupled NOR or NAND gates. While simple, it includes a prohibited input combination that leads to ambiguity or both outputs at the same level, violating complementarity.


Given Data / Assumptions:

  • NOR-based S–R (active-high): S=1 and R=1 is invalid.
  • NAND-based S̄–R̄ (active-low): S̄=0 and R̄=0 is invalid.
  • Outputs Q and Q̄ should be complements in valid operation.


Concept / Approach:
In the forbidden case, both cross-coupled gates are forced into the same driven state. When inputs are released, final state can be unpredictable and may depend on tiny device mismatches, breaking determinism. Therefore, stating that there is “no invalid or unused state” is factually wrong.


Step-by-Step Solution:

Identify technology: NOR or NAND implementation.List valid control combinations (set, reset, hold).Highlight the forbidden input: NOR (1,1) or NAND (0,0).Explain that this leads to non-deterministic release and broken complementarity.


Verification / Alternative check:
Truth/characteristic tables in textbooks mark the forbidden row explicitly (often with an X). Simulators also flag undefined behavior when both controlling inputs are asserted concurrently in these structures.


Why Other Options Are Wrong:

  • Correct: Contradicted by standard tables.
  • “Valid only for edge-triggered” or “if outputs are not complemented”: Edge triggering and labeling do not eliminate the forbidden combination.
  • Temperature: Environmental factors do not turn an invalid input into a valid one.


Common Pitfalls:
Driving both inputs active “just to be safe”; failing to add gating to prevent overlap; misunderstanding active-high vs active-low conventions and thereby asserting the wrong pair simultaneously.


Final Answer:
Incorrect

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