Question

6. (5pts) Using four 1-bit full adders only to design a four-bit combinational Excess-3 to BCD converter. Show the block diagram and label all inputs and outputs.

Answer 1

We will get Excess - 3 code by adding decimal 3 [or BCD0011) to Bco code So we will reverse the procedure to get BCD code out of excess 3 code input det excess_3_input be ist Binary and Binary Number Number CIN V2 A₂ A2 Al Ao B3 B2B, Bo 74L5 83 4-BIT Full Adder CINK COUT S3 S2 Si So Summed output COUT To get BCD code Sz S, S, So from the Excess 3 code then we can subtract 3 from it. We will carry out the subtraction using 2's complement method.

Az A,A,A, -3) 10 = Az A2A, A0 + 18 complement 310 of El A 3 12 A, Ao +(1100) + (!), 1 So in adder we will assign В, В, В, В. (1100)2 and Cin =12 and output S3 S₂ S, So will give BCD code. The other way to do is using 4 bit Adder / subtractor circuit B3 B2 BI Bo LADO | SUBTRACT IC 7486 A3 Az AI AO cout IC 7483 cin S3 S2 si So

Here XOR gates are used as controlled inverter and will be used for is complement of B . So Az A, A, AO -(3), AgA, A, Ao ti's complement of 310 +1 Sa By B, B, Bo = (3710 (0011), and Cin =1

Please comment below for any doubts. I am happy to help you. Have a good day. ?