The following logic gate is equivalent to
(a) NOR Gate
(b) OR Gate
(c) AND Gate
(d) NAND Gate

Q: Why does a NAND gate with tied inputs act as a NOT gate?

If you look at the truth table for a NAND gate, when both inputs are 0, the output is 1. When both inputs are 1, the output is 0. By tying the inputs together, you force them to always be the same, effectively creating an inverter or NOT gate.

Q: What is De Morgan's Law and how do I remember it?

De Morgan's Law is a set of rules for simplifying Boolean expressions. A handy trick to remember it is 'break the line, change the sign'. If you break a bar over a dot (AND), it becomes a plus (OR), and vice versa.

Q: Why do we use multiple NAND gates instead of just buying a NOR gate?

In real-world digital design, NAND gates are considered 'Universal Gates'. It is often cheaper and more efficient to manufacture integrated circuits using thousands of identical NAND gates rather than mixing different types of gates.

Q: Could I have solved this problem using a truth table instead of Boolean algebra?

Absolutely! You can create a truth table with columns for A, B, the intermediate outputs, and the final output Y. Evaluating it for all four combinations (00, 01, 10, 11) will give you the exact truth table of a NOR gate.

Q: What would the output be if the final gate was removed?

If the final gate (which acts as a NOT gate) was removed, the output would just be the intermediate signal $Y_1 = A + B$. This means the first three gates together form an OR gate.

Question

The following logic gate is equivalent to

Options:

(A) NOR Gate

(B) OR Gate

(C) AND Gate

(D) NAND Gate

Sigma Percentile · Accepted Answer

### Decoding the Logic Circuit

Imagine you are an electronics engineer tasked with analyzing a mysterious black box of logic gates. You open it up and find a network of four identical gates. Your mission? To figure out what this entire combination does as a single unit.

This is a classic puzzle in digital electronics, and it relies on understanding how we can manipulate basic logic gates to perform complex operations. Let's break down the circuit step by step.

### The Input Stage: Tied NAND Gates

First, let's look at the two gates connected directly to our inputs, **A** and **B**. You'll notice they are NAND gates, but there's a twist—their two input terminals are tied together!

What happens when you feed the same signal into both inputs of a NAND gate? Let's think about the truth table. If the input is $0$, both terminals receive $0$. The AND operation gives $0$, and the NOT part of the NAND inverts it to $1$. If the input is $1$, both terminals receive $1$. The AND operation gives $1$, and the inversion makes it $0$.

This means a NAND gate with tied inputs behaves exactly like a **NOT gate**. Therefore, the signal emerging from the first gate is simply $\overline{A}$, and the signal from the second gate is $\overline{B}$.

### The Intermediate Stage: The Core NAND

Now, these inverted signals, $\overline{A}$ and $\overline{B}$, travel into the third gate. This is a standard NAND gate.

A NAND gate takes the dot product (AND) of its inputs and then applies a bar (NOT) over the entire result. So, our intermediate output, which we can call $Y_1$, is given by:

$$ Y_1 = \overline{\overline{A} \cdot \overline{B}} $$

This expression looks a bit intimidating, but we have a powerful mathematical tool to simplify it: **De Morgan's Law**.

### Applying De Morgan's Law

De Morgan's Law provides a simple rule for breaking down long inversion bars: *"Break the line, change the sign."*

When we break the long bar over our expression, the dot (AND) changes to a plus (OR). Let's apply this:

$$ Y_1 = \overline{\overline{A}} + \overline{\overline{B}} $$

We know that a double inversion cancels itself out (the NOT of a NOT is the original signal). Therefore, the expression simplifies beautifully to:

$$ Y_1 = A + B $$

This reveals a fascinating secret: the first three NAND gates working together are actually functioning as an **OR gate**!

### The Final Output

We are almost at the finish line. The signal $Y_1$ (which is $A + B$) now enters the fourth and final gate.

Just like the first two gates, this is a NAND gate with its inputs tied together. As we established earlier, this acts as a NOT gate. It will simply invert whatever signal comes into it.

Therefore, our final output $Y$ is the inversion of $Y_1$:

$$ Y = \overline{Y_1} $$
$$ Y = \overline{A + B} $$

### The Grand Conclusion

Take a look at that final expression: $Y = \overline{A + B}$. What does this represent?

It is the fundamental Boolean expression for a **NOR gate** (an OR gate followed by a NOT gate).

So, this entire intricate combination of four NAND gates is perfectly equivalent to a single NOR gate. This is a brilliant demonstration of why the NAND gate is called a **Universal Gate**—you can build any other logic function using only NAND gates!

Animated Solution for Physics - Semiconductors: The following logic gate is equivalent to

Select Answer:

Visualized Solution

$Input NOT Gates$

$Intermediate NAND Gate$

$De Morgan’s Law$

$Final Output Stage$

$Conclusion$

$The Way Forward$

The Sigma Insight: Logic Gates

Decoding the Logic Circuit

The Input Stage

The Intermediate Stage

Applying De Morgan's Law

The Final Output

The Grand Conclusion

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