You have a 3-input NAND gate available but only need a 2-input NAND function. How should the unused input be handled to preserve correct two-input NAND behavior?

Introduction / Context:
Real designs frequently adapt available parts to required functions. Correctly fixing unused inputs on multi-input gates prevents unintended logic behavior and noise susceptibility.

Given Data / Assumptions:

• Device: 3-input NAND; required behavior: 2-input NAND.
• Inputs are standard digital levels with family-appropriate thresholds.
• No desire to alter the intended truth table for the two active inputs.

Concept / Approach:
A NAND outputs LOW only when all inputs are HIGH. To reduce a 3-input NAND to an effective 2-input NAND, force the third input to a benign value that preserves the 2-input truth table. Setting the unused input to logic 1 yields Y = NOT(A * B * 1) = NOT(A * B), which is exactly a 2-input NAND of A and B. Leaving inputs open is bad practice; tying to 0 changes behavior to Y = NOT(A * B * 0) = NOT(0) = 1 (trivial constant HIGH), which is incorrect.

Step-by-Step Solution:
1) Desired: Y = NOT(A * B).2) With a third input C fixed at 1: Y = NOT(A * B * 1) = NOT(A * B).3) Therefore tie the unused input to HIGH to preserve function.

Verification / Alternative check:
Check full truth table: with C = 1, the 3-input NAND reduces to the 2-input NAND truth table for A and B.

Why Other Options Are Wrong:
“connected to ground” (0) forces output always HIGH. “left open” risks undefined/floating behavior and noise. “any of the above” is incorrect because only tying HIGH is functionally correct and robust.

Common Pitfalls:
Leaving inputs floating; assuming tying LOW is harmless; ignoring family-specific recommendations about unused inputs.

Final Answer:
connected to a HIGH