S–R NAND latch behavior If both inputs of an S–R NAND latch are driven LOW simultaneously, what happens at the outputs?

Difficulty: Easy

Correct Answer: The output would become unpredictable (forbidden state).

Explanation:


Introduction / Context:
NAND-based S–R latches interpret LOW as the active assertion of Set or Reset. Applying LOW to both inputs simultaneously produces a forbidden condition. Understanding this case prevents design errors and metastability in larger systems where latches are sub-elements.


Given Data / Assumptions:

  • NAND S–R latch (active-LOW S and R).
  • Both inputs forced LOW at the same time.
  • Standard cross-coupled configuration.


Concept / Approach:

For a NAND gate, any LOW input forces a HIGH output. With both inputs LOW, each NAND output goes HIGH, momentarily giving Q = 1 and Q̄ = 1, which violates complementarity. When inputs return HIGH, slight asymmetries decide the final latched state, making it unpredictable or indeterminate.


Step-by-Step Explanation:

Apply S = 0 and R = 0 → both NAND outputs forced HIGH.This creates Q = 1 and Q̄ = 1 (not complementary → invalid).Release inputs back HIGH → due to unequal delays, either side may win → unpredictable final state.


Verification / Alternative check:

Timing diagrams show both outputs driven HIGH during the forbidden input, then a potentially random resolution after inputs are released, consistent with the latch truth table warning.


Why Other Options Are Wrong:

  • Toggle (b) implies deterministic alternation, which does not occur.
  • Reset (c) or no change (d) depend on valid inputs, not on the forbidden case.
  • Complementary outputs (e) contradict the actual simultaneous HIGH outputs during the violation.


Common Pitfalls:

  • Confusing NAND and NOR versions; the forbidden combination differs.
  • Assuming hardware will always settle to a preferred state; it may vary each time.


Final Answer:

The output would become unpredictable (forbidden state).

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