Physics - Semiconductors Question with Solution | TestHub

Let the output of the first NAND gate be $Y_1$.

$Y_1=\overline{A\cdot B}$

The output of the upper NAND gate is $Y_2$.

$Y_2=\overline{A\cdot Y_1}=\overline{A\cdot \overline{A\cdot B}}=\overline{A\cdot (\overline{A}+\overline{B})}=\overline{A\overline{A}+A\overline{B}}=\overline{A\overline{B}}$

The output of the lower NAND gate is $Y_3$.

$Y_3=\overline{B\cdot Y_1}=\overline{B\cdot \overline{A\cdot B}}=\overline{B\cdot (\overline{A}+\overline{B})}=\overline{\overline{A}B+B\overline{B}}=\overline{\overline{A}B}$

The final output $Y$ is from the last NAND gate with inputs $Y_2$ and $Y_3$.

$Y=\overline{Y_2\cdot Y_3}=\overline{Y_2}+\overline{Y_3}$

Substituting $Y_2$ and $Y_3$:

$Y=\overline{\overline{A\overline{B}}}+\overline{\overline{\overline{A}B}}=A\overline{B}+\overline{A}B$

This is the Boolean expression for an X-OR gate.

Answer: X-OR gate