transgalactic

i need to implement a circuit that inputs two 4bit numbers (y3 y2 y1 y0 x3 x2 x1 x0)
and it has an output of (c3 c2 c1 c0 s3 s2 s1 s0) where "c" is the carry
and "s" is the sum

i was given an example in which X=0101 (5) Y=1000 (8) and it outputs
S=1010 (10) and C=0011 (3)
5+8=13

i dont know what is the role of C and S
how do they split the numbers???

i was asked to implement it using two 4 bit adders

here is the answer:
http://s290.photobucket.com/albums/ll279/t...nt=IMG_8831.jpg

i cant understand this circuit
????????

Pommie

Consider each bit individually. For each bit there are 3 inputs (X,Y,Ci) and 2 outputs (S,Co). The carry gets carried to the next significant bit, Carry Out (Co) becomes Carry In (Ci) for the next digit.

You example is wrong. X=0101(5) + Y=1000(8) = S=1101(13) and C=0000

A better example is X=0111(7) + Y=0110(6) = S=1101(13) and C=0110

To get the next digit of the answer you add the X and Y bits and the carry bit from the previous bit addition. Adding 2 bits can result in an answer from 00 to 11(3) - the left bit is the carry the right is the output.

Mike.

transgalactic

how can you explain that
S binary column is bigger by 6 then DEC number
and when the DEC number goes bigger then 10
then it has "1" and a normal number
the dec number is supposed to equal to the binay
noy to be smaller by 6
???

and in the EX3 column we each number is bigger by 3 then DEC number
and when the DEC number is smaller then 10
we dont have another 4 bits
?????????

Pommie

I don't understand what you are saying, 7+6=13 or 0111+0110=1101.

The carry is just a record of where a carry took place. Think of, 123+389 = 512 and the carries are 011. I.E 3+9 = 12 = 2 + carry, 8+2+C = 11 = 1 + carry, 1+3+C = 5.

Mike.

transgalactic

i understood you
but its about the sum of two EX3 number
and a output of two numbers one represents the number of "tens"
and the other represents the "ones"

there is a table in the link i was given

http://s290.photobucket.com/albums/l...t=IMG_8831.jpg

S binary column is bigger by 6 then DEC number
and when the DEC number goes bigger then 10
then it has "1" and a normal number
the dec number is supposed to equal to the binay
not to be smaller by 6
???

and in the EX3 column we each number is bigger by 3 then DEC number
and when the DEC number is smaller then 10
we dont have another 4 bits
?????????