Physics - Semiconductors Question with Solution | TestHub

Let the intermediate outputs be $y_1$, $y_2$, and $y_3$.

The output of the first NAND gate is:
$y_1=\overline{A\cdot B}$

The outputs of the second and third NAND gates are:
$y_2=\overline{A\cdot y_1}=\overline{A\cdot \overline{AB}}=\bar{A}+AB=\bar{A}+B$
$y_3=\overline{B\cdot y_1}=\overline{B\cdot \overline{AB}}=\bar{B}+AB=A+\bar{B}$

The final output $Y$ from the fourth NAND gate is:
$Y=\overline{y_2\cdot y_3}=\overline{(\bar{A}+B)\cdot (A+\bar{B})}$

Expanding the expression inside the complement:
$y_2\cdot y_3=\bar{A}A+\bar{A}\bar{B}+BA+B\bar{B}=0+\bar{A}\bar{B}+AB+0=AB+\bar{A}\bar{B}$

Thus, the final output is:
$Y=\overline{AB+\bar{A}\bar{B}}=\overline{AB}\cdot \overline{\bar{A}\bar{B}}=(\bar{A}+\bar{B})\cdot (A+B)=\bar{A}A+\bar{A}B+\bar{B}A+\bar{B}B=\bar{A}B+A\bar{B}$

This is the Boolean expression for an XOR gate. The truth table for an XOR gate is:
A B Y
0 0 0
0 1 1
1 0 1
1 1 0

Answer: