Question:easy

Total number of NAND gates required to implement half adder are

Show Hint

Memorize the "Universal Gate Count" for basic circuits: - Half Adder: 5 NAND or 5 NOR. - Full Adder: 9 NAND or 9 NOR. - Half Subtractor: 5 NAND or 5 NOR.
Updated On: Jul 1, 2026
  • 5
  • 4
  • 3
  • 2
Show Solution

The Correct Option is A

Solution and Explanation

1. Logic Expressions for Half Adder: For two inputs $A$ and $B$:

Sum ($S$) = $A \oplus B = A\bar{B} + \bar{A}B$

Carry ($C$) = $A \cdot B$

2. Implementing with NAND Gates: Since NAND is a universal gate, any logic function can be built using it. To implement a half adder efficiently:

• The first NAND gate takes $A$ and $B$ to produce $\overline{AB}$.

• This output $\overline{AB}$ is fed into two separate NAND gates along with the original $A$ and $B$ respectively.

• The outputs of these two gates are fed into a fourth NAND gate to produce the Sum ($A \oplus B$).

• The output of the first NAND gate ($\overline{AB}$) is fed into a fifth NAND gate (configured as an inverter) to produce the Carry ($AB$).

3. Summary: It takes exactly

5 NAND gates to produce both the Sum and Carry outputs of a half adder. If only the Sum (XOR) was required, it would take 4 NAND gates.
Was this answer helpful?
0