NAND gate output condition for LOW Under which input condition does a NAND gate produce a LOW output?

Introduction / Context:
NAND gates are universal components used widely because any logic function can be synthesized from NANDs. Knowing exactly when a NAND output falls LOW helps when designing active-low enables, resets, and interlock logic.

Given Data / Assumptions:

• Positive logic and ideal gates.
• Two or more inputs may be present; the rule is the same.
• No additional inversion bubbles on the symbol.

Concept / Approach:

The NAND function is the logical AND followed by inversion: Y = (A · B · ... )’. The product A · B · ... equals 1 only when every input is 1. Inverting that product gives 0 only in that single all-HIGH condition; in every other case (at least one input 0), the output remains 1 (HIGH).

Step-by-Step Solution:

Verification / Alternative check:

Construct a small truth table for two or three inputs. Check the single row with all 1s; there Y is 0. All other rows give Y = 1. This confirms the rule for NAND outputs.

Why Other Options Are Wrong:

• “All inputs LOW” makes an AND product 0, which inverts to 1 at NAND output.
• “Any input LOW” similarly keeps the NAND output HIGH.
• “Any input HIGH” does not guarantee all inputs HIGH, so NAND stays HIGH unless all are HIGH.
• Toggling or clocking is irrelevant to combinational NAND behavior.

Common Pitfalls:

• Mixing up NAND and NOR. NAND goes LOW only when all inputs are HIGH; NOR goes HIGH only when all inputs are LOW.

Final Answer:

all inputs are HIGH