Question: Design a 1 digit BCD adder using IC 7483 and explain the operation for
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## $(0111)_{BCD } + (1001)_{BCD}$.

Mumbai University > ELECTRO > Sem 3 > Digital Circuits and Designs

Marks: 10M

Year: May 2014

 modified 3.2 years ago  • written 3.2 years ago by Pooja Joshi • 740
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• A BCD adder adds two BCD digits and produces output as a BCD digit. A BCD or Binary Coded Decimal digit cannot be greater than 9.

• The two BCD digits are to be added using the rules of binary addition. If sum is less than or equal to 9 and carry is 0, then no correction is needed. The sum is correct and in true BCD form.

• But if sum is greater than 9 or carry =1, the result is wrong and correction must be done. The wrong result can be corrected adding six (0110) to it.

• For implementing a BCD adder using a binary adder circuit IC 7483, additional combinational circuit will be required, where the Sum output $S_3-S_0$ is checked for invalid values from 10 to 15. The truth table and K-map for the same is as shown:

• The Boolean expression is, $Y = S_3S_2 + S_3S_1$

• The BCD adder is shown below. 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.

• The output of combinational circuit is connected to B1B2 inputs of adder-2 and $B_3 = B_1 + 0$ as they are connected to ground permanently. This makes $B_3B_2B_1B_0$ = 0110 if Y' = 1.

• The sum outputs of adder-1 are applied to $A_3A_2A_1A_0$ of adder-2. The output of combinational circuit is to be used as final output carry and the carry output of adder-2 is to be ignored.

Operations of: $(011)_{BCD} + (1001)_{BCD}$

Thus,

Cout = 1

$S_3S_2S_1S_0 = 0000$

Hence, for adder, inputs will be $A_3A_2A_1A_0 = 0000$ and $B_3B_2B_1B_0 = 0110$

This will give final output as Cout $S_3S_2S_1S_0= 1 0110$.

Therefore, $(0111)_{BCD} + (1001)_{BCD}$ = $(0001 0110)_{BCD}$ .