Let's break down how to **implement a 4-bit subtraction circuit using full adders**, step by step. --- ## **1. Understanding Binary Subtraction with Full Adders** A **full adder** performs the addition of three bits: \( A \), \( B \), and \( C_{in} \) (carry input). It outputs a **sum** and a **carry out**. But for **subtraction**, specifically \( A - B \), you can use the concept of **2's complement**: \[ A - B = A + (\text{2's complement of } B) \] The **2's complement of B** is obtained by: 1. Inverting all bits of B (**1's complement**). 2. Adding 1. --- ## **2. Step-by-Step Implementation** ### **Step 1: Inputs** - Two 4-bit numbers: \( A_3A_2A_1A_0 \) and \( B_3B_2B_1B_0 \) ### **Step 2: Invert B (Find 1's Complement)** - Each bit of B is inverted using a NOT gate: - \( B_0' = \overline{B_0} \) - \( B_1' = \overline{B_1} \) - \( B_2' = \overline{B_2} \) - \( B_3' = \overline{B_3} \) ### **Step 3: Add A, Inverted B, and 1 (2's Complement Addition)** - Set up **4 full adders** in series, just like a 4-bit ripple carry adder. - Connect: - **A inputs:** \( A_0, A_1, A_2, A_3 \) - **B inputs:** \( B_0', B_1', B_2', B_3' \) (the inverted B bits) - **Carry-in for the lowest bit (C\(_0\))**: Set to 1 (to complete 2's complement) - Carry-in for the next bits: Use the carry-out from the previous stage. ### **Step 4: Diagram** ``` A3 ----->| |<----- B3' | Full | A2 ----->| Adder |<----- B2' | (FA3) | A1 ----->| Chain |<----- B1' | (FA2-FA1) | A0 ----->| |<----- B0' |___________| |C0=1 (initial carry-in) ``` - Each full adder outputs a **sum bit** (\( S_i \)) and a **carry-out**. ### **Step 5: Output** - The sum outputs \( S_3S_2S_1S_0 \) give the result of \( A - B \). - The final carry-out can be interpreted as a **borrow** indicator (if needed). --- ## **3. Final Circuit Connections** | Full Adder | A input | B input (inverted) | Carry-in | Sum output | Carry-out | |------------|---------|--------------------|---------------|------------|-----------| | FA0 | A0 | NOT B0 | 1 (from user) | S0 | C1 | | FA1 | A1 | NOT B1 | C1 | S1 | C2 | | FA2 | A2 | NOT B2 | C2 | S2 | C3 | | FA3 | A3 | NOT B3 | C3 | S3 | Carry-out | --- ## **4. **Explanation** - **Inverting B** and adding 1 converts subtraction to addition. - Full adders sum A, inverted B, and the initial carry-in (1) for the 2's complement. - The 4-bit output is the difference \( A - B \). --- ## **5. Final Answer (Summary Table)** **To implement a 4-bit subtractor using full adders:** 1. **Invert** each bit of B (use NOT gates). 2. **Connect** each bit of A and the inverted B to the 4 full adders. 3. **Set** the initial carry-in (to FA0) to 1 (for two's complement). 4. **Cascade** the carry-outs to the next full adder. 5. **Sum outputs** give the 4-bit result \( A - B \). --- ### **Conclusion** **A 4-bit subtractor can be implemented by:** - Inverting all bits of B with NOT gates, - Feeding A and the inverted B to a 4-bit ripple-carry adder (made of 4 full adders), - Setting the initial carry-in to 1. **The output is \( S_3S_2S_1S_0 = A - B \) (in 2's complement form).**