**1 Answer**

written 8.4 years ago by |

IC 7483 is a 4 bit parallel adder which consists of four interconnected full adders along with the look ahead carry circuit. The pin diagram of IC 7483 is shown above. It is a 16pin IC. The inputs to the IC are A, B and $C_{in0}$ while outputs are S and $C_{out3}$.

$A_3 A_2 A_1 A_0$ is a 4 bit input word 'A' and $B_3 B_2 B_1 B_0$ is the second 4 bit input word 'B'. $C_{in0}$ is the input carry. The IC adds the two four bit words along with input carry to prooduce a 4 bit sum and a one bit carry-out. $C_{out3}$ represents the output carry. $S_3,S_2,S_1,S_0$ represents sum output with S3 as the MSB.

In odrder to design an 8 bit adder, we require two IC 7483s cascaded as shown in the figure above.

Adder-1 is the LSB adder and it adds the four LSB bits of the two 8-bit input words ie $A_3-A_0$ and $B_3-B_0$. The carry input of first adder is supposed to be 0. Hence the carry in pin of LSB IC is connected to the ground. So the first IC adds the LSB bits of A and B, and produces $S_3 – S_0$ that is, LSB of sum, along with a carry out $C_{out3}$.

This $C_{out3}$ of adder-1 is connected to $C_{in0}$ input of Adder-2. The second adder adds this carry and the four MSB bits of numbers $A_3-A_0$ and $B_3-B_0$ to produce MSB sum $S_7 – S_4$ along with final carry out bit $C_{out7}$.

Thus adder-1 and adder-2 when cascaded as shown in the figure can add two 8-bit words. $C_{out7}$ of adder-2 acts as the final output carryand the sum output is from $S_7$ though $S_0$.