Animated Solution for Physics - Semiconductors: The truth table for the following logic circuit is

The Path to XNOR

Logic circuits can sometimes look like a tangled web of wires and gates, but the secret to mastering them is to trace the signal path methodically. Let's break down this interesting combination of NOT, AND, and NOR gates to reveal its true identity.

Imagine you are a digital signal traveling through this circuit. We start with two primary inputs, $A$ and $B$. Notice how both inputs split. One path goes directly into an AND gate, while the other path takes a detour through a NOT gate.

For input $A$, the NOT gate flips it to $\bar{A}$. Similarly, input $B$ is inverted to $\bar{B}$.

Now, let's look at the two AND gates. The top AND gate receives $A$ directly and $\bar{B}$ from the NOT gate. Therefore, its output is simply $A\cdot \bar{B}$.

The bottom AND gate receives $\bar{A}$ from the NOT gate and $B$ directly. Its output becomes $\bar{A}\cdot B$.

These two intermediate signals, $A\cdot \bar{B}$ and $\bar{A}\cdot B$, now march into the final stage: a NOR gate. A NOR gate performs an OR operation followed by a NOT operation.

So, we first add them together: $A\bar{B}+\bar{A}B$. Then, we invert the entire result. This gives us our master equation for the final output $Y$:

This expression might look a bit intimidating, but this is exactly where De Morgan's laws come to the rescue. Remember the golden rule: "Break the line, change the sign."

First, we break the large overline covering the sum, changing the OR ($+$) to an AND ($\cdot$):

Next, we apply De Morgan's law again to the individual terms inside the brackets. The product becomes a sum:

Now, it's time for some basic algebraic expansion. Let's multiply these two brackets together:

Here is the beautiful part of Boolean algebra: a variable ANDed with its complement is always zero ($A\cdot \bar{A}=0$). Why? Because they can never both be true at the same time!

Applying this logic, the terms $\bar{A}A$ and $B\bar{B}$ simply vanish, leaving us with:

This final simplified expression, $AB+\bar{A}\bar{B}$, is the standard Boolean signature of an XNOR gate (Exclusive-NOR).

An XNOR gate acts as an equality detector. It outputs a $1$ only when both inputs are identical (either both $0$ or both $1$). If the inputs are different, it outputs a $0$.

Let's map this out in a truth table:

Comparing our derived truth table with the given options, we can confidently conclude that option (b) is the perfect match. By patiently tracing the logic and applying fundamental Boolean rules, we've successfully decoded the circuit!