binary subtraction section 3-9 mano & kime. binary subtraction review from cse 171 two’s...

Binary Subtraction

Section 3-9 Mano & Kime

Binary Subtraction

• Review from CSE 171

• Two’s Complement Negative Numbers

• Binary Adder-Subtractors

• 4-bit Adder/Subtractor in VHDL

Half Subtractor

C A B D 0 0 0 1

0 0 0 00 1 1 11 0 1 01 1 0 0

A 0 B 0

D 0

C 1

Full Subtractor

0 0 0 0 00 0 1 1 10 1 0 1 00 1 1 0 01 0 0 1 11 0 1 0 11 1 0 0 01 1 1 1 1

Ci Ai Bi Di Ci+1

1 1

1 1

Ci

AiBi00 01 11 10

0

1

Di

Di = Ci $ (Ai $ Bi)

Same as Si in full adder

Full Subtractor

0 0 0 0 00 0 1 1 10 1 0 1 00 1 1 0 01 0 0 1 11 0 1 0 11 1 0 0 01 1 1 1 1

Ci Ai Bi Di Ci+1 Ci

AiBi00 01 11 10

0

1

1

1 11

Ci+1

Ci+1 = !Ai & Bi

# Ci !Ai & !Bi

# Ci & Ai & Bi

Full SubtractorCi+1 = !Ai & Bi

# Ci !Ai & !Bi

# Ci & Ai & Bi

Ci+1 = !Ai & Bi

# Ci & (!Ai & !Bi # Ai & Bi)

Ci+1 = !Ai & Bi # Ci & !(Ai $ Bi)

Recall:Si = Ci $ (Ai $ Bi)

Ci+1 = !Ai & Bi # Ci & !(Ai $ Bi)

Full Subtractor

A

B

D

C

C i+1

i

i

i

i

Si = Ci $ (Ai $ Bi)

Ci+1 = !Ai & Bi # Ci & !(Ai $ Bi)

half subtractorhalf subtractor

Binary Subtraction





Negative Numbers

Subtract by adding

73-35 38

10’s complement 73+65138

Ignore carry

Negative Numbers

10’s complement: Subtract from 100

100-35 65

Take 9’s complement and add 1

99-35 64 +1 65

Negative Numbers

2’s complement:

Subtract from

100000000 01001101 10110011

Take 1’s complement and add 1

11111111-01001101 10110010 +1 10110011

Finding 2’s Complement

0 1 0 1 1 0 0 0

Copy all bitsto first 1

2’s complement

0001

Complementremaining bits

0101

Negative NumberTake 2’s Complement

7510 = 4B16 = 01001011-7510 = B516 = 10110101

FF-4B B4 +1 B5

Negative NumberTake 2’s Complement

110 = 0116 = 00000001-110 = FF16 = 11111111

12810 = 8016 = 10000000-12810 = 8016 = 10000000

Table 2.2Positive and Negative Binary Numbers

Signed decimal Hex Binary Unsigned decimal-128 80 10000000 128-127 81 10000001 129-126 82 10000010 130

… … … …… … … …… … … …-3 FD 11111101 253-2 FE 11111110 254-1 FF 11111111 2550 00 00000000 01 01 00000001 12 02 00000010 23 03 00000011 3

… … … …… … …

… … …125 7D 01111101 125126 7E 01111110 126127 7F 01111111 127

Signed Numbers

4-bit: 8H = -8 to 7H = +7 1000 to 0111

8-bit: 80H = -128 to 7F = +127

16-bit: 8000H = -32,768 to 7FFFH = +32,767

32-bit: 80000000H = -2,147,483,648 to 7FFFFFFFH = +2,147,483,647

Binary Subtraction





Block Diagram of Binary Adder-Subtractor

Adder/SubtractorA 0 B 0

D 0

C 1

A 0 B 0

S 0

C 1

A 0

B 0 0

CB1 E

SD

Half adder Half subtractor

E = 0: Half adder

E = 1: Half subtractor

Adder/Subtractor-1

i+1

A B

D

C

C

i i

i

i

E

E = 0: Full adderE = 1: Full subtractor

0 0 0 0 00 0 1 1 00 1 0 1 00 1 1 0 11 0 0 1 01 0 1 0 11 1 0 0 11 1 1 1 1

Ci Ai Bi Si Ci+1

1 0 1 0 11 0 0 1 01 1 1 1 11 1 0 0 10 0 1 1 00 0 0 0 00 1 1 0 10 1 0 1 0

Ci Ai Bi Si Ci+1

Full AdderReorderedFull Adder

0 0 0 0 00 0 1 1 10 1 0 1 00 1 1 0 01 0 0 1 11 0 1 0 11 1 0 0 01 1 1 1 1

Ci Ai Bi Di Ci+1

FullSubtractor

NOT

Making a full adderfrom a full subtractor

Full Adder

A B

C C

D

i i

i+1 i

i

Adder/Subtractor-2

Full Adder

A B

C

0 0

1

0

Full Adder

A B

C

1 1

2

1

Full Adder

A B

C

2 2

3

2

Full Adder

A B

C SD

3 3

4 3 SD SD SD

E

E = 0: 4-bit adderE = 1: 4-bit subtractor

4-bit Subtractor: E = 1

Full Adder

A B

C

0 0

1

0

Full Adder

A B

C

1 1

2

1

Full Adder

A B

C

2 2

3

2

Full Adder

A B

C SD

3 3

4 3 SD SD SD

E

+1

Add A to !B (one’s complement) plus 1That is, add A to two’s complement of BD = A - B

Adder- Subtractor Circuit

Binary Subtraction





2's Complementer

2-to-1 MUX

4-bit Adder

E

A(3:0) B(3:0)

B1(3:0)

B2(3:0)

SD(3:0)

0 1

CB

A 4-Bit Adder-Subtractor

2's Complementer

2-to-1 MUX

4-bit Adder

E

A(3:0) B(3:0)

B1(3:0)

B2(3:0)

SD(3:0)

0 1

CB

addsub2

a(3:0)

b(3:0)

E

CB

SD(3:0)

addsub2.vhd

binary subtraction section 3-9 mano & kime. binary subtraction review from cse 171 two’s...

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