In a look-ahead carry generator, the carry generate function G_i and the carry propagate function P_i for inputs A_i and B_i are given by: P_i = A_i XOR B_i and G_i = A_i * B_i The expressions for the sum bit S_i and the carry bit C_{i+1} of the look-ahead carry adder are given by: S_i = P_i XOR C_i and C_{i+1} = G_i + P_i * C_i, where C_0 is the input carry. Consider a two-level logic implementation of the look-ahead carry generator. Assume that all P_i and G_i are available for computation. The number of AND gates and OR gates (each having any number of inputs) needed to implement a 4-bit look-ahead carry generator are respectively: A. 6, 3 B. 10, 4 C. 6, 4 D. 10, 5
GATE 2007 · Digital Logic · Carry Generator · medium
Answer: B. 10, 4 — 10 AND gates and 4 OR gates are needed.
- Expand all four carry expressions: C_1 = G_0 + P_0*C_0 C_2 = G_1 + P_1*G_0 + P_1*P_0*C_0 C_3 = G_2 + P_2*G_1 + P_2*P_1*G_0 + P_2*P_1*P_0*C_0 C_4 = G_3 + P_3*G_2 + P_3*P_2*G_1 + P_3*P_2*P_1*G_0 + P_3*P_2*P_1*P_0*C_0
- Count AND gates (multi-literal product terms only): C_1: terms are G_0 (1 literal, no AND needed) and P_0*C_0 (2 literals, 1 AND gate) => 1 AND gate C_2: G_1 (no AND), P_1*G_0 (1 AND), P_1*P_0*C_0 (1 AND) => 2 AND gates C_3: G_2 (no AND), P_2*G_1 (1 AND), P_2*P_1*G_0 (1 AND), P_2*P_1*P_0*C_0 (1 AND) => 3 AND gates C_4: G_3 (no AND), P_3*G_2 (1 AND), P_3*P_2*G_1 (1 AND), P_3*P_2*P_1*G_0 (1 AND), P_3*P_2*P_1*P_0*C_0 (1 AND) => 4 AND gates Total AND gates = 1 + 2 + 3 + 4 = 10
- Count OR gates (one per carry output): Each of C_1, C_2, C_3, C_4 requires one OR gate to combine its product terms. Total OR gates = 4
- Final count: AND gates = 10, OR gates = 4 => Option B: 10, 4