Design a 2-bit ripple carry adder using the full adder designed in the previous step. Consult Figure 3 for guidance. Your digital circuit should have two 2-bit inputs, {A1, A0} and {B1, BO), and one single bit input, Cin). Similarly, it should have a 2-bit output, {S1 SO), and a single bit output, {Cout}. Provide the truth table for the 2-bit adder. You do not have to compute the boolean expressions. Draw the gate-level schematic for the 2-bit ripple carry adder.
The Correct Answer and Explanation is :
Design of a 2-Bit Ripple Carry Adder Using Full Adders
A 2-bit ripple carry adder consists of two full adders (FA) connected in sequence. The ripple carry mechanism means that the carry output from one full adder propagates as the carry input to the next stage. This allows us to add two 2-bit binary numbers.
Inputs and Outputs
- Inputs:
- Two 2-bit numbers: ( A = {A_1, A_0} ), ( B = {B_1, B_0} )
- Initial carry input: ( C_{in} )
- Outputs:
- Sum: ( S = {S_1, S_0} )
- Final carry-out: ( C_{out} )
Circuit Design
- First Full Adder (FA0)
- Inputs: ( A_0, B_0, C_{in} )
- Outputs: ( S_0, C_0 ) (Carry output)
- Second Full Adder (FA1)
- Inputs: ( A_1, B_1, C_0 ) (carry from FA0)
- Outputs: ( S_1, C_{out} ) (Final carry output)
Each full adder follows the standard full adder logic:
- Sum: ( S = A \oplus B \oplus C_{in} )
- Carry: ( C_{out} = (A \cdot B) + (C_{in} \cdot (A \oplus B)) )
Truth Table for 2-bit Adder
| A₁ | A₀ | B₁ | B₀ | Cᵢₙ | S₁ | S₀ | Cₒᵤₜ |
|---|---|---|---|---|---|---|---|
| 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
| 0 | 0 | 0 | 1 | 0 | 0 | 1 | 0 |
| 0 | 0 | 1 | 0 | 0 | 1 | 0 | 0 |
| 0 | 0 | 1 | 1 | 0 | 1 | 1 | 0 |
| 0 | 1 | 0 | 0 | 0 | 0 | 1 | 0 |
| 0 | 1 | 0 | 1 | 0 | 1 | 0 | 0 |
| 0 | 1 | 1 | 0 | 0 | 1 | 1 | 0 |
| 0 | 1 | 1 | 1 | 0 | 0 | 0 | 1 |
| 1 | 0 | 0 | 0 | 0 | 1 | 0 | 0 |
| 1 | 0 | 0 | 1 | 0 | 1 | 1 | 0 |
| 1 | 0 | 1 | 0 | 0 | 0 | 0 | 1 |
| 1 | 0 | 1 | 1 | 0 | 0 | 1 | 1 |
| 1 | 1 | 0 | 0 | 0 | 1 | 1 | 0 |
| 1 | 1 | 0 | 1 | 0 | 0 | 0 | 1 |
| 1 | 1 | 1 | 0 | 0 | 0 | 1 | 1 |
| 1 | 1 | 1 | 1 | 0 | 1 | 0 | 1 |
Gate-Level Schematic Explanation
- The circuit consists of two full adders (FA).
- The first full adder (FA0) adds ( A_0 ), ( B_0 ), and ( C_{in} ), generating ( S_0 ) and ( C_0 ).
- The second full adder (FA1) takes ( A_1 ), ( B_1 ), and ( C_0 ), generating ( S_1 ) and ( C_{out} ).
- The sum bits ( S_1, S_0 ) represent the final result, while ( C_{out} ) is the carry-out.
I will generate the schematic for you now.
It looks like there was an issue generating the schematic. However, I can describe how to draw it manually:
Steps to Draw the Gate-Level Schematic
- First Full Adder (FA0)
- Inputs: ( A_0, B_0, C_{in} )
- XOR Gate: Computes ( S_0 = A_0 \oplus B_0 \oplus C_{in} )
- AND & OR Gates: Compute carry ( C_0 = (A_0 \cdot B_0) + (C_{in} \cdot (A_0 \oplus B_0)) )
- Second Full Adder (FA1)
- Inputs: ( A_1, B_1, C_0 )
- XOR Gate: Computes ( S_1 = A_1 \oplus B_1 \oplus C_0 )
- AND & OR Gates: Compute ( C_{out} = (A_1 \cdot B_1) + (C_0 \cdot (A_1 \oplus B_1)) )
- Connections
- The carry ( C_0 ) from FA0 feeds into FA1 as its ( C_{in} ).
- The final outputs are ( S_1, S_0, ) and ( C_{out} ).
If you need a detailed digital version, I can try again or guide you through designing it in a circuit simulation tool like Logisim or Proteus. Let me know how you’d like to proceed!