NAND behavior – is the output LOW only when all inputs are simultaneously HIGH?

Introduction / Context:
NAND gates are ubiquitous universal primitives. Knowing the exact condition for a LOW output is crucial when designing fault indicators, active-LOW control signals, and safety interlocks. This question verifies the canonical NAND condition for zero output.

Given Data / Assumptions:

• N-input NAND with ideal Boolean behavior.
• Active-HIGH convention for logic 1.
• No analog issues (timing or fan-in limits) change the logic definition.

Concept / Approach:
NAND is the negation of AND: Y = NOT(A * B * ... * N). The AND product equals 1 only when every input is 1. Therefore, Y becomes 0 only in that all-HIGH case. For any input being 0, the product is 0 and the NAND output is 1.

Step-by-Step Solution:

Verification / Alternative check:
Truth tables for 2-input, 3-input, and 4-input NANDs all share this property. Timing diagrams show brief hazards during transitions in real hardware, but the steady-state condition remains exact.

Why Other Options Are Wrong:
The gate count (two vs many inputs), technology family (CMOS vs TTL), and pull-ups do not alter Boolean function; they affect electrical characteristics only.

Common Pitfalls:
Confusing NAND with NOR (which is the complement of OR). Also, overlooking that any single LOW forces NAND HIGH, which designers exploit to build robust inhibit signals.

Final Answer:
Correct