Q.Realize the half adder using only NAND gates and write the Boolean expression at the output of each gate.

A half adder can be built entirely from NAND gates — four NANDs form the sum (X-OR) and a fifth NAND inverts to give the carry. …

[!TLDR]

Five NAND gates give a half adder: four form the sum $A\oplus B$ and the fifth inverts $\overline{AB}$ to the carry $AB$.

Realization: The sum of a half adder is $A\oplus B$, which needs four NAND gates, and the carry is $AB$, obtained by inverting the output of the first NAND.

• Gate G1 (inputs $A, B$): $G1 = \overline{AB}$.
• Gate G2 (inputs $A, G1$): $G2 = \overline{A\cdot G1} = \overline{A\,\overline{AB}} = \overline{A}+B$.
• Gate G3 (inputs $B, G1$): $G3 = \overline{B\cdot G1} = \overline{B\,\overline{AB}} = A+\overline{B}$.
• Gate G4 (inputs $G2, G3$): $Sum = \overline{G2\cdot G3} = A\overline{B}+\overline{A}B = A\oplus B$. …