Physics - Semiconductors Question with Solution | TestHub

From the given logic circuit diagram:

The output of gate $G_1$ (NOT gate) is $\overline{A}$.
The output of gate $G_2$ (NOT gate) is $\overline{B}$.

These are fed as inputs to gate $G_3$, which is a NOR gate. The output of $G_3$ is:
$Y^{\prime }=\overline{\overline{A}+\overline{B}}$

This output is then fed to gate $G_4$, which is another NOT gate. The final output $Y$ is:
$Y=\overline{Y^{\prime }}=\overline{\overline{\overline{A}+\overline{B}}}$
$Y=\overline{A}+\overline{B}$

Using De Morgan's theorem, we can write:
$\overline{A}+\overline{B}=\overline{A\cdot B}$

The boolean expression $Y=\overline{A\cdot B}$ represents a NAND gate.

Answer: NAND gate