GATE CS 2007 Q35 — Computer Arithmetic

In a 4-bit look-ahead carry adder, the carry generator unit is responsible for generating the carry signals $C_1, C_2, C_3,$ and $C_4$ in parallel. The sum bits $S_i$ are generated outside this unit using XOR gates.

Using the recursive relation $C_{i+1} = G_i + P_i C_i$, we can write the two-level Sum-of-Products (SOP) expressions for each carry output as follows:

• For $C_1$: $$C_1 = G_0 + P_0 C_0$$
  • Number of AND gates = $1$ (for $P_0 C_0$)
  • Number of OR gates = $1$ (to sum $2$ terms)
• For $C_2$: $$C_2 = G_1 + P_1 G_0 + P_1 P_0 C_0$$
  • Number of AND gates = $2$ (for $P_1 G_0$ and $P_1 P_0 C_0$)
  • Number of OR gates = $1$ (to sum $3$ terms)
• For $C_3$: $$C_3 = G_2 + P_2 G_1 + P_2 P_1 G_0 + P_2 P_1 P_0 C_0$$
  • Number of AND gates = $3$ (for $P_2 G_1$, $P_2 P_1 G_0$, and $P_2 P_1 P_0 C_0$)
  • Number of OR gates = $1$ (to sum $4$ terms)
• For $C_4$: $$C_4 = G_3 + P_3 G_2 + P_3 P_2 G_1 + P_3 P_2 P_1 G_0 + P_3 P_2 P_1 P_0 C_0$$
  • Number of AND gates = $4$ (for $P_3 G_2$, $P_3 P_2 G_1$, $P_3 P_2 P_1 G_0$, and $P_3 P_2 P_1 P_0 C_0$)
  • Number of OR gates = $1$ (to sum $5$ terms)

Summing these up for the entire carry generator:

• Total AND gates: $1 + 2 + 3 + 4 = 10$
• Total OR gates: $1 + 1 + 1 + 1 = 4$

Therefore, the number of AND and OR gates needed are 10 and 4 respectively.