Animated Solution for Physics - Semiconductors: Which one of the following will be the output of the given circuit ?

When faced with a combination of logic gates, the best approach is to break the circuit down into smaller, manageable pieces. Let's trace the journey of our inputs, $A$ and $B$, as they travel through this network.

First, we observe the structure. The circuit consists of three distinct logic gates. The top gate has a D-shape with a small inversion bubble at its output, identifying it as a NAND gate. The bottom gate features a curved, shield-like shape, which is the standard symbol for an OR gate. Finally, the outputs of these two gates feed into a third gate—a standard D-shape without a bubble, making it an AND gate.

Let's determine the intermediate expressions generated by the first layer of gates. Both inputs $A$ and $B$ are routed directly into the top NAND gate. The NAND operation is essentially an AND operation followed by a NOT operation. Therefore, the output of this top gate is:

Simultaneously, the same inputs $A$ and $B$ are routed into the bottom OR gate. The OR gate performs logical addition, yielding the output:

Now, these two intermediate signals, $Y_1$ and $Y_2$, act as the inputs for the final AND gate. The AND gate performs logical multiplication on its inputs. By substituting our intermediate expressions, we can write the master Boolean equation for the entire circuit:

To simplify this expression, we need to break that stubborn bar over the $A\cdot B$ term. This is where De Morgan's Theorem comes to the rescue. De Morgan's law states that the complement of a product is the sum of the complements: $\overline{A\cdot B}=\overline{A}+\overline{B}$. Applying this to our equation, we get:

Now, we expand the brackets using standard algebraic distribution (FOIL method):

In Boolean algebra, a fundamental rule is that a variable ANDed with its own complement always results in zero ($X\cdot \overline{X}=0$). This is because a variable and its inverse can never be true simultaneously. Applying this rule, the terms $\overline{A}\cdot A$ and $\overline{B}\cdot B$ vanish:

We are left with a beautifully simplified expression:

If you are familiar with standard logic gates, you will immediately recognize this as the exact Boolean expression for an Exclusive-OR (XOR) gate, denoted mathematically as $A\oplus B$.

An XOR gate outputs a `1` (true) only when its inputs are different. This clever arrangement of a NAND, an OR, and an AND gate is a classic method for synthesizing the XOR function using more fundamental building blocks!