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Draw a neat circuit of BCD adder using IC 7483 and explain.

Mumbai University > Electronics and Telecommunication Engineering > Sem 3 > Digital Electronics

Marks: 10M

Year: May 2016

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1. A BCD adder adds two BCD digits and produces a BCD digit. BCD number cannot be greater than 9.

2. The two given BCD numbers are to be added using the rules of binary addition.

3. If sum is less than or equal to 9 and $carry = 0$ then correction is necessary. The sum is correct and in the true BCD form.

4. But if sum is invalid BCD or $carry = 1$, then the result is wrong and needs correction.

5. The wrong result can be corrected by adding six (0110) to it.

6. The 4 bit binary adder IC 7483 can be used to perform addition of BCD numbers.

7. In this, if the four-bit sum output is not a valid digit, or if a carry $C_3$ is generated then decimal 6 (0110 binary) is to be added to the sum to get the correct result.

8. Fig1 shows a 1-digit BCD adders can be cascaded to add numbers several digits long by connecting the carry-out of a stage to the carry-in of the next stage.

9. The output of combinational circuit should be 1 if the sum produced by adder 1 is greater than 9 i.e. 1001. The truth table is as follows

• The output of the combinational circuit should be 1 if Cout of adder-1 is high. Therefore Y is ORed with Cout of adder 1 as shown in fig1.

• The output of combinational circuit is connected to 〖 B〗_1 B_2inputs of adder-2 and 〖 B〗_(3 )= 〖 B〗_1=0 as they are connected to ground permanently.This make 〖 B〗_(3 ) 〖〖 B〗_2 B〗_1 〖 B〗_(0 ) = 0 1 1 0 if Y’ = 1.

• The sum outputs of adder-1 are applied to A_(3 ) A_2 A_1 A_0 of adder-2. The output of combinational circuit is to be used as final carry and the carry output of adder-2 is to be ignored

Operation: Case1: Sum ≤ 9 and carry = 0

• The output of combinational circuit Y’ = 0. Hence 〖 B〗_(3 ) 〖〖 B〗_2 B〗_1 〖 B〗_(0 ) = 0 0 0 0 for adder-2.

Case2: Sum >9 and carry = 0

• If S_3 S_2 S_1 S_0 of adder -1 is greater than 9, then output Y’ of combinational circuits becomes 1.

• Therefore 〖 B〗_(3 ) 〖〖 B〗_2 B〗_1 〖 B〗_(0 ) = 0 1 1 0 (of adder-2).

• Hence six (0 1 1 0) will be added to the sum output of adder-1.

• We get the corrected BCD result at the output of adder-2.