Arithmetic circuits
Binary addition Binary Subtraction Unsigned binary numbers Sign-magnitude numbers 2’S Complement
representation 2’S Complement arithmetic Arithmetic building blocks
Powers of 2Powers of 2
20 21
22
23
24
25
26
27 28 29 210 211 212 213 214 215
216
Decimal Equivalent1 248
163264
128 256512
1,024 2,048 4,096 8,192 16,38432,76865,536
Abbreviation
1K 2K 4K8K
16K32K64K
Decimal-Binary EquivalencesDecimal
1 37153163
127 255 511
1,0232,047 4,095 8,191 16,38332,76765,535
Binary1
11 111
11111 1111
11 1111 111 1111
1111 1111 1 1111 1111
11 1111 1111 111 1111 1111
1111 1111 1111 1 1111 1111 1111
11 1111 1111 1111
111 1111 1111 1111
1111 1111 1111 1111
Hexadecimal
137F
1F3F7FFF
1FF3FF7FFFFF
1FFF3FFF7FFFFFFF
Binary addition 0 + 0 = 0 0 + 1 = 1 1 + 0 = 1 1 + 1 = 10 = 0 + carry of 1 into next
position 1 + 1 + 1 = 11 = 1 + carry of 1 into next
positionA B SUM CO
0 0 0 0
0 1 1 0
1 0 1 0
1 1 0 1
HALF ADDER
A
B
SUM
CO
Carry-Out =
SUM =
(AB)
(AB) + (AB)
Binary addition
Carry-Out =
SUM =
1-bit 8 Strings Full Adder with Carry-In and Carry-Out
CI A B SUM CO
0 0 0 0 0
0 0 1 1 0
0 1 0 1 0
0 1 1 0 1
1 0 0 1 0
1 0 1 0 1
1 1 0 0 1
1 1 1 1 1
FULL ADDER
A
BSUM
COCI
(A B)CI + (A B)CI + +
(AB)CI + (A+B)CI
1-bit 8 Strings Full Adder with Carry-In and Carry-Out
SUM =
FULL ADDER
A
BSUM
COCI
(A B)CI + (A B)CI + +
Carry-Out = (AB)CI + (A+B)CI
Binary Subtraction0 - 0 = 01 - 0 = 11 - 1 = 00 - 1 = 1 ต้องยมืจากหลักท่ีสงูกวา่มา
1
A B SUB BO
0 0 0 0
0 1 1 1
1 0 1 0
1 1 0 0
HALF Subtracto
r
A
B
SUB
BO
Borrow-Out =
SUB =
Binary Subtraction
Borrow-Out =
SUB =
1-bit 8 Strings Full Subtractor with Borrow-In and Borrow -Out
BI A B SUB BO
0 0 0 0 0
0 0 1 1 1
0 1 0 1 0
0 1 1 0 0
1 0 0 1 1
1 0 1 0 1
1 1 0 0 0
1 1 1 1 1
FULL Subtracto
r
A
BSUB
BOBI
REPRESENTING REPRESENTING UNUNSIGNED NUMBERSSIGNED NUMBERS((Absolute valueAbsolute value))
0 0 0 0 0 0 0 0A7 A6 A5 A4 A3 A2 A1 A0
=00H
1 1 1 1 1 1 1 1B7 B6 B5 B4 B3 B2 B1 B0
=FFH
REPRESENTING SIGNED NUMBERSREPRESENTING SIGNED NUMBERSin in sign-magnitudesign-magnitude form. form.
0 0 1 1 0 1 0 0A7 A6 A5 A4 A3 A2 A1 A0
=+5210
SIGN BITMagnitude = 5210
1 0 1 1 0 1 0 0B7 B6 B5 B4 B3 B2 B1 B0
=-5210
SIGN BITMagnitude =
5210
REPRESENTING SIGNED NUMBERSREPRESENTING SIGNED NUMBERSin the in the 22’’ S-complement S-complement system. system.
0 0 1 0 1 1 0 1A7 A6 A5 A4 A3 A2 A1 A0
=+4510
SIGN BITTrue binary
1 1 0 1 0 0 1 1B7 B6 B5 B4 B3 B2 B1 B0
=-4510
SIGN BIT2’s complement
Range of Sign-Magnitude NumbersRange of Sign-Magnitude Numbers
0 0 0 0 0 0 0 1A7 A6 A5 A4 A3 A2 A1 A0
=+110
SIGN BIT
0 1 1 1 1 1 1 1B7 B6 B5 B4 B3 B2 B1 B0
=+1271
0
1 0 0 0 0 0 0 1A7 A6 A5 A4 A3 A2 A1 A0
=-12710
1 1 1 1 1 1 1 1B7 B6 B5 B4 B3 B2 B1 B0
=-110
Range of Sign-Magnitude NumbersRange of Sign-Magnitude Numbers
0 0 0 0 0 0 0 1A7 A6 A5 A4 A3 A2 A1 A0
=+110
SIGN BIT
0 1 1 1 1 1 1 1B7 B6 B5 B4 B3 B2 B1 B0
=+1271
0
1 0 0 0 0 0 0 1A7 A6 A5 A4 A3 A2 A1 A0
=-12710
1 1 1 1 1 1 1 1B7 B6 B5 B4 B3 B2 B1 B0
=-110
การคอมพลีเมนต์เลขฐานการคอมพลีเมนต์เลขฐานสองสอง
แบง่ออกเป็น คอมพลีเมนต์ 1 (1’s complement) คอมพลีเมนต์ 2 (2’s complement) การคอมพลีเมนต์เลขฐานสองนี้นำาไปใชเ้ก่ียวกับ
การคำานวณทางไมโครคอมพวิเตอรม์าก เพราะวา่จะใชใ้นลักษณะการลบด้วยวธิกีารบวกด้วยคอมพลีเมนต์
สรุป การลบด้วยการบวกด้วยคอมพลีเมนต์นัน้จะทำานองเดียวกับการคอมพลีเมนต์เลขฐานสบิ
การคอมพลีเมนต์เลขฐานการคอมพลีเมนต์เลขฐานสองสอง
X3X2X1X0 = 1000
1’s complementX3X2X1X0 = 0111
2’s complement2’s complement = 1’s complement + 1
X3
X3
X2
X2
X1
X1
X0
X0
Positive and Negative NumbersPositive and Negative Numbers
-8 -7 -6 -5 -4 -3 -2 -1 0 +1 +2 +3 +4 +5 +6 +7
1000 1001 1010 1011 1100 1101 1110 1111 0000 0001 0010 0011 0100 0101 0110 0111
Magnitude Positive Negative12345678
0001001000110100010101100111
-
11111110110111001011101010011000
22’’ S-complement representation S-complement representation summarysummary
Positive numbers always have a sign bit of 0, and negative numbers always have a sign bit of 1.
Positive numbers are stored in sign-magnitude form.
Negative numbers are stored as 2’s complements. Taking the 2’s complement is equivalent to a sign
change.
Example :
Binary contents Hexadecimal contents
Decimal contents
0001 0100____ ________ ________ ____
1001 1110____ ________ ________ ____
___ ___ ___ ___
14HDDH___HBDH___H70H___H6EH
_____H
+20___+47_________
-125___
-19,750
1101 1101 -35 0010 1111 2F 1011 1101 -67 9E -98 0111 0000 +112 1000 0011 83 0110 1110 110 1011 0010 1101 1010 B2DA
CASE 4 Both negative. -43 -78
ADDITIONCASE 1 Both positive.
+83+16
2’s complement arithmetic2’s complement arithmetic
0101 00110001 0000
83 0101 0011+16 +0001 0000 99 0110 0011
CASE 2 Positive and smaller negative.
+125 -68
0111 11011011 1100
125 0111 1101+(-68) +1011 1100 57 1 0011 1001
CASE 3 Positive and larger negative.
+37 -115
37 0010 0101+(-115) +1000 1101 -78 1011 0010
1101 01011011 0010
-43 1101 0101 +(-78) +1011 0010 -121 1 1000 0111
0010 01011000 1101
SUBTRACTIONCASE 1 Both positive.
+83+16
2’s complement arithmetic2’s complement arithmetic
0101 00110001 0000
CASE 2 Positive and smaller negative.
+68 -27 83 0101 0011
+(-16) +1111 0000 67 1 0100 0011
0100 01001110 0101
68 0100 0100+(+27) +0001 1011 95 0101 1111CASE 3 Positive and
larger negative. +14 -108
14 0000 1110+(+108) +0110 1100 122 0111 1010
1101 01011011 0010
CASE 4 Both negative. -43 -78
-43 1101 0101 +(+78) +0100 1110 35 1 0010 0011
0000 11101001 0100
S8 S7 S6 S5 S4 S3 S2 S1 S0
B7 B6 B5 B4 A7 A6 A5 A4 B3 B2 B1 B0 A3 A2 A1 A0
SUB
S4 S3 S2 S1
A1A2A3A4B1B2B3B4CIN
COUT
S4 S3 S2 S1
A1A2A3A4B1B2B3B4CIN
COUT
INVERT
A7 A6 A5 A4 A3 A2 A1 A0
S8 S7 S6 S5 S4 S3 S2 S1 S0
B7 B6 B5 B4 A7 A6 A5 A4 B3 B2 B1 B0 A3 A2 A1 A0
SUB
S4 S3 S2 S1
A1A2A3A4B1B2B3B4CIN
COUT
S4 S3 S2 S1
A1A2A3A4B1B2B3B4CIN
COUT
Y7 Y6 Y5 Y4 Y3 Y2 Y1 Y0
001
A7-A0 0110 1110Y7-Y0 0110 1110
Y7-Y0 1001 0001
INV LOGIC
Controlled inverterControlled inverter
S8 S7 S6 S5 S4 S3 S2 S1 S0
B7 B6 B5 B4 A7 A6 A5 A4 B3 B2 B1 B0 A3 A2 A1 A0
SUB
S4 S3 S2 S1
A1A2A3A4B1B2B3B4CIN
COUT
S4 S3 S2 S1
A1A2A3A4B1B2B3B4CIN
COUT
ADD/SUB
A7 A6 A5 A4 A3 A2 A1 A0
S7 S6 S5 S4 S3 S2 S1 S0
S8 S7 S6 S5 S4 S3 S2 S1 S0
B7 B6 B5 B4 A7 A6 A5 A4 B3 B2 B1 B0 A3 A2 A1 A0
SUB
S4 S3 S2 S1
A1A2A3A4B1B2B3B4CIN
COUT
S4 S3 S2 S1
A1A2A3A4B1B2B3B4CIN
COUT
S8 S7 S6 S5 S4 S3 S2 S1 S0
B7 B6 B5 B4 A7 A6 A5 A4 B3 B2 B1 B0 A3 A2 A1 A0
SUB
S4 S3 S2 S1
A1A2A3A4B1B2B3B4CIN
COUT
S4 S3 S2 S1
A1A2A3A4B1B2B3B4CIN
COUT
S8 S7 S6 S5 S4 S3 S2 S1 S0
B7 B6 B5 B4 A7 A6 A5 A4 B3 B2 B1 B0 A3 A2 A1 A0
SUB
S4 S3 S2 S1
A1A2A3A4B1B2B3B4CIN
COUT
S4 S3 S2 S1
A1A2A3A4B1B2B3B4CIN
COUT
S8 S7 S6 S5 S4 S3 S2 S1 S0
B7 B6 B5 B4 A7 A6 A5 A4 B3 B2 B1 B0 A3 A2 A1 A0
SUB
S4 S3 S2 S1
A1A2A3A4B1B2B3B4CIN
COUT
S4 S3 S2 S1
A1A2A3A4B1B2B3B4CIN
COUT
S8 S7 S6 S5 S4 S3 S2 S1 S0
B7 B6 B5 B4 A7 A6 A5 A4 B3 B2 B1 B0 A3 A2 A1 A0
SUB
S4 S3 S2 S1
A1A2A3A4B1B2B3B4CIN
COUT
S4 S3 S2 S1
A1A2A3A4B1B2B3B4CIN
COUT
S8 S7 S6 S5 S4 S3 S2 S1 S0
B7 B6 B5 B4 A7 A6 A5 A4 B3 B2 B1 B0 A3 A2 A1 A0
SUB
S4 S3 S2 S1
A1A2A3A4B1B2B3B4CIN
COUT
S4 S3 S2 S1
A1A2A3A4B1B2B3B4CIN
COUT
S8 S7 S6 S5 S4 S3 S2 S1 S0
B7 B6 B5 B4 A7 A6 A5 A4 B3 B2 B1 B0 A3 A2 A1 A0
SUB
S4 S3 S2 S1
A1A2A3A4B1B2B3B4CIN
COUT
S4 S3 S2 S1
A1A2A3A4B1B2B3B4CIN
COUT
B7 B6 B5 B4 B3 B2 B1 B0
ADDITION A7 A6 A5 A4 A3 A2 A1 A0
+B7 B6 B5 B4 B3 B2 B1 B0
S7 S6 S5 S4 S3 S2 S1 S0
SUBTRACTION A7 A6 A5 A4 A3 A2 A1 A0
+B7 B6 B5 B4 B3 B2 B1 B0 +1 S7 S6 S5 S4 S3 S2 S1 S0
- - - - - - - -
Binary adder-subtractor diagramBinary adder-subtractor diagram
S8