Circuit to add 1011 and 1100

Two 4-bit numbers are added with a 4-bit parallel (ripple-carry) adder: four full adders, one for each bit position, with the carry of each stage feeding the next. The first stage's carry-in is 0 (a half adder could be used there).

Let $A = A_3A_2A_1A_0 = 1011$ and $B = B_3B_2B_1B_0 = 1100$.

Block diagram

   A3 B3     A2 B2     A1 B1     A0 B0
   1  1      0  1      1  0      1  0
   |  |      |  |      |  |      |  |
  +-----+   +-----+   +-----+   +-----+
C4<| FA3 |<C3| FA2 |<C2| FA1 |<C1| FA0 |<-C0=0
  +-----+   +-----+   +-----+   +-----+
     |         |         |         |
     S3        S2        S1        S0

Each full adder computes

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

Working, stage by stage

1. FA0 adds the LSBs 1 + 0 with $C_0 = 0$: sum 1, carry 0.
2. FA1 adds 1 + 0 + 0: sum 1, carry 0.
3. FA2 adds 0 + 1 + 0: sum 1, carry 0.
4. FA3 adds 1 + 1 + 0: sum 0, carry 1, which appears as $C_4$ (the fifth bit of the result).

Answer: $1011 + 1100 = 10111_2$ ($C_4 = 1$, $S_3S_2S_1S_0 = 0111$). Check: $11 + 12 = 23 = 10111_2$.

All bits are applied at the same time (parallel), but the final result is valid only after the carry has rippled through all four stages. IC 7483/74283 is a ready-made 4-bit parallel adder.