Animated Solution for Physics - Semiconductors: The combination of gates shown below yields

The world of digital electronics is built upon a foundation of simple logic gates, but the true magic happens when we start combining them. In this problem, we are presented with a fascinating circuit made entirely of NAND gates and asked to determine its overall behavior. This is a classic example of how universal gates can be used to construct any other logic function.

When we first look at the circuit diagram, we see three distinct gates. The first two gates are positioned in parallel, receiving the primary inputs $A$ and $B$. The outputs of these two gates then feed into a third, final gate, which produces the ultimate output $X$.

To solve this, we must break the circuit down into stages and analyze the Boolean logic at each point. This systematic approach ensures we don't get lost in the wiring!

The NOT Gate Disguise

Let's focus our attention on the first two gates. They are drawn as D-shaped symbols with a small bubble at the output, which is the standard representation of a NAND gate. However, there is a crucial detail: each of these gates only has a single input line connected to it.

In digital logic, when the two input terminals of a NAND gate are tied together and fed the same signal, the gate transforms. Let's think about the truth table for a NAND gate. If both inputs are $0$, the output is $1$. If both inputs are $1$, the output is $0$. This is exactly the behavior of a NOT gate (or inverter).

Therefore, the top gate takes the input $A$ and inverts it, producing an intermediate output of $\overline{A}$. Similarly, the bottom gate takes the input $B$ and inverts it, giving us $\overline{B}$.

The NAND Operation

Now we move to the second stage of our circuit. The intermediate signals, $\overline{A}$ and $\overline{B}$, travel along the wires and become the inputs for the third and final gate.

This third gate is a standard two-input NAND gate. The fundamental operation of a NAND gate is to take the logical AND (the product) of its inputs and then apply a logical NOT (an inversion) to the entire result.

Let's write this out mathematically. Our inputs are $\overline{A}$ and $\overline{B}$. First, we take their product: $\overline{A}\cdot \overline{B}$. Next, we invert the whole expression to get our final output $X$:

Simplifying the Expression

We now have a Boolean expression for our output, but it looks a bit complicated. To reveal the true nature of this circuit, we need to simplify it. This is where De Morgan's Laws come to our rescue.

De Morgan's Laws are powerful tools in Boolean algebra. The specific rule we need here states that the complement of a product is equal to the sum of the complements. A handy way to remember this is the phrase: "Break the line, change the sign."

Let's apply this to our expression:

When we "break the line" (the large overline covering the whole expression), we must "change the sign" (change the AND dot to an OR plus). The broken line then distributes to the individual terms:

Now, we apply the rule of double negation. Inverting a signal twice brings it back to its original state. Therefore, $\overline{\overline{A}}$ simplifies to just $A$, and $\overline{\overline{B}}$ simplifies to just $B$.

Substituting these back into our equation, we arrive at a beautifully simple final expression:

We have successfully distilled the complex-looking circuit down to a single, elegant Boolean equation: $X=A+B$.

In the language of digital logic, the plus sign ($+$) represents the logical OR operation. Therefore, this entire combination of three NAND gates behaves exactly like a single OR gate. If either input $A$ or input $B$ is high ($1$), the output $X$ will be high ($1$).

This problem perfectly illustrates the concept of universal gates. By cleverly arranging NAND gates, we were able to synthesize an OR gate, proving that you don't necessarily need a dedicated OR chip if you have enough NAND gates on hand!