To get an OR gate from a NAND gate we need

Options

A. Only 2 NAND gates.
B. Two NOT gates obtained from NAND gates and one NAND gate.
C. 4 NAND gates and 2 AND gates obtained from NAND gate.
D. 3 NAND gates and 3 NOT gates obtained from NAND gates.

Explanation

To derive an OR gate using NAND gates, we need to follow these steps:

### Step-by-Step Solution:

1. **Understanding the NAND Gate**: A NAND gate outputs a low signal (0) only when both of its inputs are high (1). The output is high (1) in all other cases. The truth table for a NAND gate is as follows:
- A = 0, B = 0 → Output = 1
- A = 0, B = 1 → Output = 1
- A = 1, B = 0 → Output = 1
- A = 1, B = 1 → Output = 0

2. **Using NAND Gates to Create NOT Gates**: We can use a NAND gate to create a NOT gate by connecting both inputs of the NAND gate to the same input. For example, if we connect A to both inputs of a NAND gate, the output will be NOT A (A').

3. **Creating the Required Configuration**:
- Connect input A to the first NAND gate (let's call it NAND1) with both inputs as A. The output will be A'.
- Connect input B to the second NAND gate (NAND2) with both inputs as B. The output will be B'.

4. **Combining the Outputs**: Now, we need to combine the outputs of NAND1 and NAND2 to get the OR function. To do this:
- Connect the outputs A' and B' to a third NAND gate (NAND3). The output of this gate will be (A' AND B')'.

5. **Applying De Morgan's Theorem**: According to De Morgan's theorem, the output of NAND3 can be simplified:
- (A' AND B')' = A + B
This means the output of NAND3 is equivalent to A OR B.

6. **Counting the NAND Gates**: We used:
- 2 NAND gates to create the NOT gates (NAND1 and NAND2).
- 1 NAND gate to combine the outputs (NAND3).
Therefore, we need a total of **3 NAND gates** to create an OR gate.

### Final Answer:
To get an OR gate from a NAND gate, we need **3 NAND gates**.

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