Question:
How many NAND gates are required to implement a full adder?
How many NAND gates are required to implement a full adder?
Show Hint
NAND and NOR are called Universal Gates because every digital circuit can be implemented using only NAND gates or only NOR gates.
Updated On: Jun 25, 2026
- Eight
- Ten
- Six
- Nine
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The Correct Option is D
Solution and Explanation
Concept:
A Full Adder is a combinational circuit that adds three binary inputs:
\[
A,\quad B,\quad C_{in}
\]
and produces:
\[
\text{Sum}
\]
and
\[
C_{out}.
\]
Since NAND gates are universal gates, a full adder can be implemented entirely using NAND gates.
Step 1: Recall the Full Adder equations.
The Sum output is \[ S=A\oplus B\oplus C_{in}. \] The Carry output is \[ C_{out}=AB+BC_{in}+AC_{in}. \]
Step 2: Implement using NAND gates only.
Using NAND-only realization, both XOR and carry circuits are constructed using NAND networks. The standard implementation requires \[ 9 \] NAND gates.
Step 3: Select the answer.
Hence the number of NAND gates required is \[ \boxed{9}. \] Therefore option (D) is correct.
Step 1: Recall the Full Adder equations.
The Sum output is \[ S=A\oplus B\oplus C_{in}. \] The Carry output is \[ C_{out}=AB+BC_{in}+AC_{in}. \]
Step 2: Implement using NAND gates only.
Using NAND-only realization, both XOR and carry circuits are constructed using NAND networks. The standard implementation requires \[ 9 \] NAND gates.
Step 3: Select the answer.
Hence the number of NAND gates required is \[ \boxed{9}. \] Therefore option (D) is correct.
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