1 Answer
To convert a **NAND gate** to an **OR gate**, we need to use **De Morgan's Theorems** and understand how basic logic gates relate to each other.
---
### Step-by-step explanation:
#### **1. Understand the Logic Functions**
* **NAND Gate**:
The output is the **negation** of the AND gate.
**Y = (A · B)'**
* **OR Gate**:
The output is high (1) if **any input is high**.
**Y = A + B**
---
### **Using De Morgan’s Theorem**
De Morgan’s Theorems:
1. **(A · B)' = A' + B'**
2. **(A + B)' = A' · B'**
We can reverse the first identity to help us convert NAND to OR:
> **A + B = (A' · B')'**
This means:
> **OR gate output = NAND of inverted inputs**
---
### Step-by-step conversion using NAND gates:
We want to build an OR gate using **only NAND gates**.
#### Required expression:
> **Y = A + B = (A' · B')'**
This is a NAND gate with inverted inputs.
---
### ️ Implement using NAND gates only
You’ll need **3 NAND gates**:
#### Step 1: Invert A using a NAND gate
Connect both inputs of a NAND gate to A:
**A' = (A · A)'**
#### Step 2: Invert B using a NAND gate
Connect both inputs of a NAND gate to B:
**B' = (B · B)'**
#### Step 3: Apply NAND to inverted A and B
Use the third NAND gate to compute:
**Y = (A' · B')' = A + B**
---
### Final Circuit
| NAND Gate # | Inputs | Output |
| ----------- | ------ | ------ |
| 1 | A, A | A' |
| 2 | B, B | B' |
| 3 | A', B' | A + B |
So, **the output of the third NAND gate is the same as the OR gate output**.
---
### ✅ Final Expression
**OR using NANDs**:
> **A + B = ((A · A)' · (B · B)')'**
---
### Summary
To convert a **NAND gate to an OR gate**, you:
* Invert each input using NAND gates (as NOT gates),
* Feed the inverted inputs into another NAND gate.
This works due to **De Morgan’s Law**.
Would you like a diagram showing this circuit with gates and connections?
---
### Step-by-step explanation:
#### **1. Understand the Logic Functions**
* **NAND Gate**:
The output is the **negation** of the AND gate.
**Y = (A · B)'**
* **OR Gate**:
The output is high (1) if **any input is high**.
**Y = A + B**
---
### **Using De Morgan’s Theorem**
De Morgan’s Theorems:
1. **(A · B)' = A' + B'**
2. **(A + B)' = A' · B'**
We can reverse the first identity to help us convert NAND to OR:
> **A + B = (A' · B')'**
This means:
> **OR gate output = NAND of inverted inputs**
---
### Step-by-step conversion using NAND gates:
We want to build an OR gate using **only NAND gates**.
#### Required expression:
> **Y = A + B = (A' · B')'**
This is a NAND gate with inverted inputs.
---
### ️ Implement using NAND gates only
You’ll need **3 NAND gates**:
#### Step 1: Invert A using a NAND gate
Connect both inputs of a NAND gate to A:
**A' = (A · A)'**
#### Step 2: Invert B using a NAND gate
Connect both inputs of a NAND gate to B:
**B' = (B · B)'**
#### Step 3: Apply NAND to inverted A and B
Use the third NAND gate to compute:
**Y = (A' · B')' = A + B**
---
### Final Circuit
| NAND Gate # | Inputs | Output |
| ----------- | ------ | ------ |
| 1 | A, A | A' |
| 2 | B, B | B' |
| 3 | A', B' | A + B |
So, **the output of the third NAND gate is the same as the OR gate output**.
---
### ✅ Final Expression
**OR using NANDs**:
> **A + B = ((A · A)' · (B · B)')'**
---
### Summary
To convert a **NAND gate to an OR gate**, you:
* Invert each input using NAND gates (as NOT gates),
* Feed the inverted inputs into another NAND gate.
This works due to **De Morgan’s Law**.
Would you like a diagram showing this circuit with gates and connections?