Animated Solution for Physics - Semiconductors: The logic circuit shown above is equivalent to

Imagine you are looking at a complex puzzle made of logic gates. Our goal is to simplify this puzzle into a single, elegant piece. Let's break down the given circuit step by step.

We start with two input signals, $A$ and $B$. The very first thing that happens to these signals is that they pass through their own individual NOT gates. A NOT gate is the simplest logic gate—it just flips the signal. If it's a 1, it becomes a 0, and if it's a 0, it becomes a 1. Mathematically, we represent this by putting a bar over the variable. So, the output of the first NOT gate is $\bar{A}$, and the output of the second NOT gate is $\bar{B}$.

Now, these two inverted signals, $\bar{A}$ and $\bar{B}$, don't just stop there. They are fed together into a NOR gate.

What exactly is a NOR gate? You can think of it as an OR gate that has a NOT gate attached to its output. It first adds the inputs together (the OR part) and then inverts the entire result (the NOT part).

So, let's write down the master equation for our final output $C$. First, we OR the two inputs: $\bar{A}+\bar{B}$. Then, we apply a giant bar over the whole thing to represent the final NOT operation. This gives us:

This equation looks a bit intimidating with all those bars. But don't worry, we have a powerful mathematical wand called De Morgan's Theorem.

De Morgan's Theorem gives us a simple rule to break down long bars: "Break the line, change the sign." If you have a long bar over an OR operation ($+$), you can break the bar into two separate bars, but you must change the OR ($+$) into an AND ($\cdot$).

Let's apply this magic to our equation. We break the giant top bar over the plus sign. The plus sign magically transforms into a dot, and we are left with:

We are almost there! Notice that we now have double bars over both $A$ and $B$. In Boolean algebra, a double negation is like doing a 360-degree turn—you end up exactly where you started. Flipping a switch twice leaves it in its original state. Therefore, $\overline{\overline{A}}$ simplifies to just $A$, and $\overline{\overline{B}}$ simplifies to just $B$.

Substituting this back into our equation, we get our final, beautifully simple result:

And what does this expression represent? It is the fundamental Boolean expression for an AND gate! This means that despite all the NOT gates and the NOR gate, the entire circuit behaves exactly like a single, simple AND gate.

If you ever find yourself stuck and unable to recall De Morgan's Theorem during an exam, remember that you can always build a Truth Table. By plugging in all four possible combinations of 0s and 1s for $A$ and $B$, you can trace the logic through the circuit and arrive at the exact same conclusion. It's a fantastic, foolproof backup strategy!