The logic gate equivalent to the given logic circuit is
(a) NOR
(b) NAND
(c) OR
(d) AND

Q: Why do we use De Morgan's Law here?

De Morgan's Law is essential for breaking down complex complemented expressions, especially when dealing with NAND and NOR gates, making it easier to identify the equivalent basic gate.

Q: What if the final gate was a NOR gate instead of a NAND gate?

If it were a NOR gate, the expression would be $Y = \overline{\overline{A} + \overline{B}}$. Using De Morgan's Law, this would simplify to $Y = \overline{\overline{A}} \cdot \overline{\overline{B}} = A \cdot B$, which is an AND gate.

Q: Can we solve this using a truth table?

Absolutely! You can construct a truth table for all possible combinations of inputs A and B, find the intermediate outputs, and then the final output. Comparing the final output column with standard gate truth tables will give you the answer.

Q: What does a double bar over a variable mean?

A double bar represents the double complement or double negation of a variable. In Boolean algebra, negating a value twice brings it back to its original state, so $\overline{\overline{A}} = A$.

Q: How do I quickly identify a NAND gate from a diagram?

A NAND gate looks like an AND gate (straight back, rounded front) but has a small circle (bubble) at its output, which signifies the NOT operation applied to the AND output.

Question

The logic gate equivalent to the given logic circuit is

Options:

(A) NOR

(B) NAND

(C) OR

(D) AND

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are an electrical engineer tasked with simplifying a digital circuit. You are given a circuit with two inputs, $A$ and $B$.

Each of these inputs is first passed through its own NOT gate. The outputs from these NOT gates are then combined and fed into a single NAND gate to produce the final output, $Y$.

Our goal is to find a single, basic logic gate that performs the exact same operation as this entire combination.

### Tracing the Intermediate Signals

Let's break this down step by step. First, we need to determine the signals immediately after the NOT gates.

When input $A$ passes through the first NOT gate, it gets inverted. Let's call this intermediate output $Y_1$. Therefore, we can write:
$$Y_1 = \overline{A}$$

Similarly, when input $B$ passes through the second NOT gate, it also gets inverted. Let's call this second intermediate output $Y_2$. This gives us:
$$Y_2 = \overline{B}$$

### The Master Equation

Now, these two intermediate signals, $Y_1$ and $Y_2$, act as the inputs for the final NAND gate.

Remember the function of a NAND gate: it first performs an AND operation on its inputs, and then applies a NOT operation to the result.

So, the final output $Y$ can be expressed as the complement of the product of $Y_1$ and $Y_2$:
$$Y = \overline{Y_1 \cdot Y_2}$$

### Final Calculation and Simplification

Now, let's substitute the expressions we found for $Y_1$ and $Y_2$ back into our master equation.

Replacing $Y_1$ with $\overline{A}$ and $Y_2$ with $\overline{B}$, we get:
$$Y = \overline{\overline{A} \cdot \overline{B}}$$

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

De Morgan's Law states that the complement of a product is equal to the sum of the complements. Mathematically, $\overline{X \cdot Y} = \overline{X} + \overline{Y}$.

Applying this law to our expression, we break the long top bar and change the AND ($\cdot$) operation to an OR ($+$) operation:
$$Y = \overline{\overline{A}} + \overline{\overline{B}}$$

Finally, we apply the principle of double negation. The double complement of any variable is simply the variable itself ($\overline{\overline{A}} = A$).

Applying this, our expression simplifies beautifully to:
$$Y = A + B$$

This final equation is the standard Boolean expression for an **OR gate**. Therefore, the entire complex circuit is functionally equivalent to a simple OR gate!

Animated Solution for Physics - Semiconductors: The logic gate equivalent to the given logic circuit is

Select Answer:

Visualized Solution

Analyzing the Logic Circuit

Finding $Y_{1}$ and $Y_{2}$

Expression for Output $Y$

Substituting Intermediate Values

Applying De Morgan's Law

Final Equivalent Gate

The Sigma Insight: Logic Gates

Analyzing the Setup

Tracing the Intermediate Signals

The Master Equation

Final Calculation and Simplification

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