Physics - Semiconductors Question with Solution | TestHub

From the given logic circuit, we can identify the intermediate outputs:

The top gate is a NOR gate with inputs $A$ and $B$. Its output is $\overline{A+B}$.

The bottom gate is a NAND gate with inputs $A$ and $B$. Its output is $\overline{A\cdot B}$.

These two outputs are fed into an AND gate. Therefore, the final output $Y$ is:

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

Using De Morgan's laws, we can expand the terms:
$\overline{A+B}=\bar{A}\cdot \bar{B}$
$\overline{A\cdot B}=\bar{A}+\bar{B}$

Substituting these into the expression for $Y$:

$Y=(\bar{A}\cdot \bar{B})\cdot (\bar{A}+\bar{B})$

Expanding the expression:

$Y=(\bar{A}\cdot \bar{B}\cdot \bar{A})+(\bar{A}\cdot \bar{B}\cdot \bar{B})$

Since $\bar{A}\cdot \bar{A}=\bar{A}$ and $\bar{B}\cdot \bar{B}=\bar{B}$, we get:

$Y=(\bar{A}\cdot \bar{B})+(\bar{A}\cdot \bar{B})$

Since $X+X=X$ in Boolean algebra:

$Y=\bar{A}\cdot \bar{B}$

Applying De Morgan's law again:

$Y=\overline{A+B}$

This is the Boolean expression for a NOR gate.

Answer: NOR gate