Physics - Semiconductors Question with Solution | TestHub

A two-input XOR gate can be realized using four two-input NAND gates. The first two NAND gates produce $(\text{A}\cdot \text{B}{)}^{\prime }$ and $(\text{A}\cdot \text{B}{)}^{\prime }$. The third NAND gate takes inputs A and $(\text{A}\cdot \text{B}{)}^{\prime }$, and the fourth takes B and $(\text{A}\cdot \text{B}{)}^{\prime }$.

Their outputs are then fed into a final NAND gate to produce the XOR output: $\text{A}\oplus \text{B}=(\text{A}\cdot (\text{A}\cdot \text{B}{)}^{\prime }{)}^{\prime }\cdot (\text{B}\cdot (\text{A}\cdot \text{B}{)}^{\prime }{)}^{\prime }{)}^{\prime }$. This simplifies to $\text{A}\overline{\text{B}}+\overline{\text{A}}\text{B}$.