Download - 8.login desing serial adder
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Cedar logic Simulation
SERIAL ADDER 구현하기
논리 설계 2 분반
학 번 이 름
20800577 장 호 상
21000599 임 유 빈
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1. 회로 overview
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2. 회로도
1)전체회로도
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2) 입 출력 라인
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3)입력부
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4)출력부
5)덧셈기
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3. 시뮬레이션
Case1) Overflow 가 발생하는 경우
1100(12) + 0111(7) -> X110 + X011 (S = 1, C = 0) -> XX11 + XX01 (S = 1, C = 0) ->
XXX1 + XXX0 (S = 0, C = 1) -> S=0, C=1
CLOCK A3,B3 A2,B2 A1,B1 A0,B0 S C
1
2
3
4
X
X
X
X
1,0
X
X
X
1,1
1,0
X
X
0,1
1,1
1,0
X
1
1
0
0
0
0
1
1
1100 + 0111 = (1) 0011 (19) : 맨 위가 LSB 이기 때문에 아래서부터 위로 읽는다. (1)은 S 가
ON 되어있는 것을 통해 알 수 있음
CLOCK Q3 Q2 Q1 Q0
1
2
3
4
5
X
1
1
0
0
X
X
1
1
0
X
X
X
1
1
X
X
X
X
1
Moore 머신이기 때문에 출력이 한 clock 씩 후에 나온다
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CASE2) OVERFLOW 없을 때
0110(6) + 0101(5) -> X011 + X010 (S = 1, C = 0) -> XX01 + XX01 (S = 1, C = 0)
-> XXX1 + XXX0 (S = 0, C= 1) -> S = 1, C= 0
CLOCK A3,B3 A2,B2 A1,B1 A0,B0 S C
1
2
3
4
X
X
X
X
0,0
X
X
X
1,1
0,0
X
X
1,0
1,1
0,0
X
1
1
0
1
0
0
1
0
0110 + 0101 = (0) 1011 (11)
CLOCK Q3 Q2 Q1 Q0
1
2
3
4
5
X
1
1
0
1
X
X
1
1
0
X
X
X
1
1
X
X
X
X
1
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![Page 11: 8.login desing serial adder](https://reader030.vdocuments.mx/reader030/viewer/2022020711/54c455da4a7959e53c8b45ba/html5/thumbnails/11.jpg)
CASE3) carry 가 발생하지 않을 때
1010 (10) + 0101 (5) -> X101 + X010 (S = 1, C = 0) -> XX10 + XX01 (S = 1, C = 0)
-> XXX1 + XXX0 (S = 1, C = 0) -> S = 1, C = 0
CLOCK A3,B3 A2,B2 A1,B1 A0,B0 S C
1
2
3
4
X
X
X
X
1,0
X
X
X
0,1
1,0
X
X
1,0
0,1
1,0
X
1
1
1
1
0
0
0
0
1010 + 0101 = 1111(15)
CLOCK Q3 Q2 Q1 Q0
1
2
3
4
5
X
1
1
1
1
X
X
1
1
1
X
X
X
1
1
X
X
X
X
1
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CASE4) CARRY 가 생길 때
0011 (3) + 0111 (7) -> X001 + X011 (S = 0, C = 1) -> XX00 + XX01 (S = 1, C = 1)
-> XXX0 + XXX0 (S = 0, C = 1) -> (S = 1, C = 0)
CLOCK A3,B3 A2,B2 A1,B1 A0,B0 S C
1
2
3
4
X
X
X
X
0,0
X
X
X
0,1
0,0
X
X
1,1
0,1
0,0
X
0
1
0
1
1
1
1
0
0011 + 0111 = (0) 1010 (10)
CLOCK Q3 Q2 Q1 Q0
1
2
3
4
5
X
0
1
0
1
X
X
0
1
0
X
X
X
0
1
X
X
X
X
0
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