Animated Solution for Physics - Semiconductors: The output of an OR gate is connected to both the inputs of a NAND gate. The combination will serve as a

From OR to NOR

Logic gates are the fundamental building blocks of digital electronics. By combining simple gates, we can create entirely new logical operations. In this problem, we are tasked with finding the equivalent logic gate when an OR gate is cascaded into a specially configured NAND gate. Let's break down the circuit step-by-step.

The OR Gate

We begin with a standard OR gate. Let's assume its two inputs are $A$ and $B$. The primary function of an OR gate is to perform logical addition.

If we denote the intermediate output of this OR gate as $y^{\prime }$, we can write its Boolean expression as:

This means the output $y^{\prime }$ will be high (1) if either $A$ or $B$, or both, are high.

The Universal NAND Gate

The problem states that this intermediate output $y^{\prime }$ is fed into both inputs of a NAND gate. This is a classic trick in digital logic design. A NAND gate is a "universal gate," meaning it can be configured to mimic any other basic gate.

When you tie both inputs of a NAND gate together and feed them the same signal (in our case, $y^{\prime }$), the NAND gate evaluates $\overline{y^{\prime }\cdot y^{\prime }}$. According to the idempotent law of Boolean algebra, $y^{\prime }\cdot y^{\prime }=y^{\prime }$. Therefore, the expression simplifies to:

By tying the inputs together, we have effectively transformed the NAND gate into a simple NOT gate (an inverter).

Combining the Logic

Now that we know the second stage simply inverts the signal from the first stage, we can substitute our initial expression for $y^{\prime }$ into the final output equation.

This final Boolean expression, $\overline{A+B}$, is the exact mathematical definition of a NOR gate.

Conclusion: By taking the logical addition of $A$ and $B$ (OR) and subsequently inverting the result (NOT), the entire cascaded circuit functions perfectly as a single NOR gate.