Enter two valid BCD numbers. Show the result in seven segment display and LED How to...

Enter two valid BCD numbers. Show the result in seven segment display and LED

How to do this using the components dip switch, Two BCD adders 74ls83, And gates, OR gates, 74 ls47 decoder, 7 segment display and LED

Solutions

Expert Solution

Step1:- Two binary adders(74ls83),AND gates and OR Gates are used to do the BCD addition

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 S3S2S1S0 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=S3S2+S3S1
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 B3=B1+0 as they are connected to ground permanently. This makes B3B2B1B0 = 0110 if Y' = 1.
The sum outputs of adder-1 are applied to A3A2A1A0 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.

Step2:-Resultant BCD is given to BCD to 7-Segment Decoder (74ls47) and as for the second digit is concerned it will be either one or zero,So a LED is used to represent '1' or '0'

Manojponduru answered 1 year ago

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