Q.Draw the logic circuit for the expression $Y = AB + \overline{B}$ by using only the NAND gates.

The expression $Y = AB + \overline{B}$ can be realised with just two NAND gates, because the second product term $\overline{B}$ means its complement B is already available directly. …

[!TLDR]

Two NAND gates realise $Y = AB + \overline{B}$: NAND1(A, B) = $\overline{AB}$, then NAND2($\overline{AB}$, B) = $AB + \overline{B}$.

Since NAND is a universal gate, any SOP expression can be built as a two-level NAND-NAND network. For $Y = P_1 + P_2$ the standard realisation is $Y = \overline{\overline{P_1}\cdot\overline{P_2}}$, i.e. NAND the complemented product terms.

Here $P_1 = AB$ and $P_2 = \overline{B}$:

• $\overline{P_1} = \overline{AB}$ is produced by NAND gate 1 with inputs A and B.
• $\overline{P_2} = \overline{\,\overline{B}\,} = B$, which needs no gate — the uncomplemented input B is used directly.
• The output NAND gate 2 takes $\overline{AB}$ and B: $Y = \overline{\,\overline{AB}\cdot B\,}$. …