Describe the derivation procedure of look ahead carry adder by an example with the help of block diagram.

Deriving the carry equations

A full adder adds the bits $A_i$ and $B_i$ and the carry $C_i$. Its carry out is:

$$C_{i+1} = A_iB_i + (A_i \oplus B_i)C_i$$

We give the two parts names. The generate term $G_i = A_iB_i$ means that the stage creates a carry by itself. The propagate term $P_i = A_i \oplus B_i$ means that the stage passes the carry on. The equations become:

$$C_{i+1} = G_i + P_iC_i \qquad S_i = P_i \oplus C_i$$

Now we replace each carry by the equation of the stage before it, one at a time.

$$C_1 = G_0 + P_0C_0$$

$$C_2 = G_1 + P_1C_1 = G_1 + P_1G_0 + P_1P_0C_0$$

$$C_3 = G_2 + P_2C_2 = G_2 + P_2G_1 + P_2P_1G_0 + P_2P_1P_0C_0$$

$$C_4 = G_3 + P_3G_2 + P_3P_2G_1 + P_3P_2P_1G_0 + P_3P_2P_1P_0C_0$$

After the substitution, no carry depends on another carry. Each one depends only on $G$, $P$ and $C_0$, so all can be built as two-level AND-OR circuits and are ready at the same time.

Block diagram

   A3 B3   A2 B2   A1 B1   A0 B0
    |  |    |  |    |  |    |  |
  [ P/G ] [ P/G ] [ P/G ] [ P/G ]   P = A xor B, G = A.B
    |  |    |  |    |  |    |  |
    |  +----+--+----+--+----+--+--> look-ahead carry unit
    |       |       |       |        (C1, C2, C3, C4 from
    |       |       |       |         all P, G and C0)
  [XOR]   [XOR]   [XOR]   [XOR] <--- C3, C2, C1, C0
    |       |       |       |
    S3      S2      S1      S0          C4 = carry out

Example. Add $A = 0111$ and $B = 0011$ with $C_0 = 1$.

The look-ahead equations give:

$$C_1 = G_0 + P_0C_0 = 1 + 0 = 1$$

$$C_2 = G_1 + P_1G_0 + P_1P_0C_0 = 1 + 0 + 0 = 1$$

$$C_3 = G_2 + P_2G_1 + P_2P_1G_0 + P_2P_1P_0C_0 = 0 + 1 + 0 + 0 = 1$$

$$C_4 = G_3 + P_3G_2 + \dots = 0$$

The sums are $S_i = P_i \oplus C_i$, so $S_0 = 0 \oplus 1 = 1$, $S_1 = 0 \oplus 1 = 1$, $S_2 = 1 \oplus 1 = 0$ and $S_3 = 0 \oplus 1 = 1$. The result is $S_3S_2S_1S_0 = 1011$ with carry out $0$. Therefore $7 + 3 + 1 = 11 = 1011_2$.