--- title: "2-bit fast adder and its advantage" subject: "Digital Logic" chapter: 5 key: "semester_3/digital_logic:fast-adder-2bit" source: https://www.bctengineer.com/q/semester_3/digital_logic/fast-adder-2bit site: BCT Engineering --- # 2-bit fast adder and its advantage Digital Logic · Chapter 5: Arithmetic Circuits - Asked in: 2076 Asoj - Video: [Carry Lookahead Adder (Part 1) | CLA Generator](https://www.youtube.com/watch?v=6Z1WikEWxH0) — Neso Academy - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/digital_logic/fast-adder-2bit - Whole chapter: https://www.bctengineer.com/md/semester_3/digital_logic/chapter-5 ## Question Explain 2-bit fast Adder with its logical diagram and write the advantage of fast Adder. ## Short answer A 2-bit fast adder uses carry look-ahead: generate Gi = AiBi, propagate Pi = Ai⊕Bi; C1 = G0+P0C0, C2 = G1+P1G0+P1P0C0; sums S0 = P0⊕C0, S1 = P1⊕C1. Advantage: constant delay (~3–4 gates) independent of word length, eliminating ripple-carry propagation delay. ## Answer A **fast adder** (carry look-ahead adder, CLA) is a parallel adder that produces the carries for all bit positions at the same time, directly from the inputs, instead of letting the carry ripple from stage to stage. #### Principle For bit $i$: - **Carry generate** $G_i = A_iB_i$: a carry is produced regardless of $C_i$. - **Carry propagate** $P_i = A_i \oplus B_i$: an incoming carry is passed on. $$ C_{i+1} = G_i + P_iC_i, \qquad S_i = P_i \oplus C_i $$ #### 2-bit fast adder $$ \begin{aligned} C_1 &= G_0 + P_0C_0 \\ C_2 &= G_1 + P_1G_0 + P_1P_0C_0 \\ S_0 &= P_0 \oplus C_0 \\ S_1 &= P_1 \oplus C_1 \end{aligned} $$ #### Logic diagram ```text A0 B0 --+--[XOR]--> P0 +--[AND]--> G0 A1 B1 --+--[XOR]--> P1 +--[AND]--> G1 P0,C0 ----[AND]--+ G0 --------------+--[OR]--> C1 P1,G0 ----[AND]--+ P1,P0,C0 -[AND]--+--[OR]--> C2 (carry out) G1 --------------+ P0,C0 --[XOR]--> S0 P1,C1 --[XOR]--> S1 ``` **Working:** the P and G signals appear after one gate delay; $C_1$ and $C_2$ appear after two more gate delays (AND then OR); the sums after one more XOR delay. $C_2$ does not wait for $C_1$. **Example:** $A = 11$, $B = 01$, $C_0 = 0$: $G_0 = 1, P_0 = 0, G_1 = 0, P_1 = 1$; $C_1 = 1$, $C_2 = 0 + 1\cdot1 + 0 = 1$; $S_0 = 0$, $S_1 = 1 \oplus 1 = 0$. Result $C_2S_1S_0 = 100 = 4$ (3 + 1). Correct. #### Advantages of a fast adder 1. Much higher speed: carry delay is constant (about 3–4 gate delays), not $n$ times one stage delay as in a ripple adder. 2. Delay does not grow with word length within a block, so it suits fast ALUs. 3. Blocks can be cascaded with a second level of look-ahead (IC 74182) for wide adders. 4. Output settles quickly, with fewer glitches. (Cost: more gates and gates with many inputs, so very wide single-level CLAs are impractical.) ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. In a carry look-ahead adder, what do the carry generate G_i and carry propagate P_i signals equal? A. G_i = A_i B_i, P_i = A_i XOR B_i B. G_i = A_i B_i, P_i = C_i C. G_i = A_i XOR B_i, P_i = A_i B_i D. G_i = A_i + B_i, P_i = A_i B_i Answer: A. G_i = A_iB_i produces a carry regardless of C_i, while P_i = A_i XOR B_i passes an incoming carry on. (Key point: Generate and propagate) 2. Which expression gives the carry out C2 of the 2-bit fast adder directly from the inputs? A. C2 = G1 + P1 C1 only, using C1 from stage 0 B. C2 = G1 + P1 G0 + P1 P0 C0 C. C2 = G1 P1 + G0 P0 C0 D. C2 = P1 + G0 + C0 Answer: B. Expanding C2 = G1 + P1C1 with C1 = G0 + P0C0 gives G1 + P1G0 + P1P0C0, which needs no wait for C1. (Key point: Carry C2 expression) 3. Why is the fast adder faster than a ripple-carry adder? A. It uses fewer gates B. Its gates are physically faster C. It computes all carries at the same time from the inputs instead of passing them stage to stage D. It ignores the carry-in C0 Answer: C. Carries are produced in parallel by two-level AND-OR logic, so delay does not grow with each stage. (Key point: Why it is faster) 4. For A = 11, B = 01, C0 = 0, what is the result C2 S1 S0 of the fast adder? A. 011 B. 101 C. 110 D. 100 Answer: D. G0=1, P0=0, so C1=1; P1=1 gives C2=1; S0=0 and S1=1 XOR 1=0, so the result is 100 = 4 = 3 + 1. 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