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Chapter4: Combinational Logic
Lecture5- Design Decimal Adder and Binary Multiplier Circuits
Engr. Arshad Nazir, Asst Prof
Dept of Electrical Engineering
SEECS 1
Fall 2022
Objectives
• Design Decimal Adder and Binary Multiplier Circuits
2
Fall 2022
Decimal Adder
• Computers or calculators that perform arithmetic operations directly
in the decimal number system represent decimal number in binary
coded form (BCD).
• An adder for such a computer must employ arithmetic circuits that
accept binary coded decimal numbers and present results in the same
code
• To add two BCD digits, we require
9 inputs: eight inputs for two BCDs and one carry-in. Since four
bits are required to code each decimal digit
5 outputs: four outputs for one BCD and one carry-out
3
Fall 2022
BCD Adder
• To design a BCD adder we require a truth table with 2^9 entries (since 9
inputs). This may become too difficult to work with so we device some
easy alternative to design a BCD adder. This is shown in table 4-5
• Each digit doesn’t exceed 9 and so the output sum can’t exceed, 9+9+1 =
19 (two BCD digits and input carry)
• Suppose we apply two BCD digits to a 4-bit binary adder
• The adder will form the sum in binary and produces a result that ranges
from 0 through 19
• These binary numbers are listed in table 4-5 and are labeled by symbols
K, Z8, Z4, Z2 and Z1. (K is the output carry). But the output sum must be
represented in BCD (not binary) and should appear in the form listed in
the columns C, S8, S4, S2 and S1 under BCD sum
• The problem is to find a rule by which binary sum is converted to
corresponding BCD sum
4
Fall 2022
5
Fall 2022
Correction for BCD Adder
• From the table it is apparent that when binary sum is equal to or less
than 1001 (decimal 9), the corresponding BCD number is identical and
therefore no correction is needed.
• Modifications are needed if the sum is greater than 1001 as we get
non-valid BCD representation. The addition of binary 6 (0110) to
binary sum converts it to the correct BCD representation and also
produces an output carry as required (refer to section 1-7)
• The logic circuit that detects the necessary correction can be derived
from the table entries. Correction (adding decimal 6 or binary 0110) is
needed when the binary sum has an output
K = 1
Z8Z4 = 1
Z8Z2 = 1
6
Fall 2022
Map Simplification of the Function C
1 1 1 1
1 1
1 1 1 1
d d d d
d d d d
d d d d
Z8Z4
00 01 11 10
00
01
11
10
k=0
Z2Z1
Z8Z4
00 01 11 10
00
01
11
10
k=1
C(K,Z8,Z4,Z2,Z1)= K+ Z8Z4 +Z8Z2
Z2Z1
7
Fall 2022
8
Fall 2022
Correction for BCD Adder
• This condition for correction and an output carry can be expressed as
 C = K + Z8Z4 + Z8Z2
• In figure 4-14, the output carry generated from the bottom adder can
be ignored since this is already available at the output carry terminal
• A decimal parallel adder that adds n decimal digits (in BCD form)
needs n BCD adder stages with output carry from one stage
connected to the input carry of next higher-order stage
9
Fall 2022
BCD Adder
Numbers that need
correction (add 6) are:
K Z8 Z4 Z2 Z1
0 1 0 1 0 (10)
0 1 0 1 1 (11)
0 1 1 0 0 (12)
0 1 1 0 1 (13)
0 1 1 1 0 (14)
0 1 1 1 1 (15)
1 0 0 0 0 (16)
1 0 0 0 1 (17)
1 0 0 1 0 (18)
1 0 0 1 1 (19)
C = K + Z8Z4 +Z8Z2 10
Fall 2022
Binary Multiplier
• Multiplication of the binary number is performed in the same way as
in decimal numbers. The multiplicand is multiplied by each bit of the
multiplier starting from the least significant bit
• Each such such multiplication forms a partial product. Successive
partial products are shifted one position to the left. The final product
is obtained from the sum of the partial products
• Consider the binary multiplication of 2-bit numbers where B1 and B2
are multiplicand bits and A1 and A0 are multiplier bits. C3 C2 C1 C0 are
the product bits
11
Fall 2022
Binary Multiplier
• The first partial product is formed by multiplying A0 by B1 B0. The
multiplication of any two bits produces a 1 if both bits are 1,
otherwise it produces a 0. This is identical to AND operation.
Therefore partial product can be implemented with an AND gates as
shown in figure 4-15
• The second partial product is formed by multiplying A1 by B1 B0 and
shifted one position to the left
• The two partial products are added with two half adder (HA) circuits.
If there are more bits in the partial products then we use full adders
to produce the sum of the partial products
• The least significant bit of the partial product doesn’t have to go
through an adder since it is formed by the output of the first AND gate
12
Fall 2022
13
Fall 2022
K-bit by J-bit Binary Multiplier
• A combinational circuit for binary multiplier with more bits can be
constructed in a similar way. A bit of the multiplier is ANDed with
each bit of the multiplicand in as many levels as there are bits in the
multiplier
• The binary output in each level of AND gate is added with the partial
product of the previous level to form a new partial product. The last
level produces the product
• For J multiplier bits and K multiplicand bits we need (J x K) AND
gates and (J – 1) K bit adders to produce a product of (J + K) bits
14
Fall 2022
4-bit by 3-bit binary multiplier
• Consider a multiplier circuit that multiplies a binary number of four
bits by a number of three bits.
• Let the multiplicand is represented by B3 B2 B1 B0 and the
multiplier by A2 A1 A0
• Since K = 4 and J = 3, we need 12 AND gates and two 4 – bit adders
to produce a product of seven bits
B3 B2 B1 B0 Multiplicand
X A2 A1 A0 Multiplier
0 A0B3 A0B2 A0B1 A0B0 Partial Product 0
A1B3 A1B2 A1B1 A1B0 Partial Product 1
Cout S3 S2 S1 S0
A2B3 A2B2 A2B1 A2B0 Partial Product 2
C6 C5 C4 C3 C2 C1 C0 Final Product
15
Fall 2022
16
Fall 2022
The End
17
Fall 2022

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Lecture5 Chapter4- Design Decimal Adder and Binary Multiplier Circuits.pdf

  • 1. Chapter4: Combinational Logic Lecture5- Design Decimal Adder and Binary Multiplier Circuits Engr. Arshad Nazir, Asst Prof Dept of Electrical Engineering SEECS 1 Fall 2022
  • 2. Objectives • Design Decimal Adder and Binary Multiplier Circuits 2 Fall 2022
  • 3. Decimal Adder • Computers or calculators that perform arithmetic operations directly in the decimal number system represent decimal number in binary coded form (BCD). • An adder for such a computer must employ arithmetic circuits that accept binary coded decimal numbers and present results in the same code • To add two BCD digits, we require 9 inputs: eight inputs for two BCDs and one carry-in. Since four bits are required to code each decimal digit 5 outputs: four outputs for one BCD and one carry-out 3 Fall 2022
  • 4. BCD Adder • To design a BCD adder we require a truth table with 2^9 entries (since 9 inputs). This may become too difficult to work with so we device some easy alternative to design a BCD adder. This is shown in table 4-5 • Each digit doesn’t exceed 9 and so the output sum can’t exceed, 9+9+1 = 19 (two BCD digits and input carry) • Suppose we apply two BCD digits to a 4-bit binary adder • The adder will form the sum in binary and produces a result that ranges from 0 through 19 • These binary numbers are listed in table 4-5 and are labeled by symbols K, Z8, Z4, Z2 and Z1. (K is the output carry). But the output sum must be represented in BCD (not binary) and should appear in the form listed in the columns C, S8, S4, S2 and S1 under BCD sum • The problem is to find a rule by which binary sum is converted to corresponding BCD sum 4 Fall 2022
  • 6. Correction for BCD Adder • From the table it is apparent that when binary sum is equal to or less than 1001 (decimal 9), the corresponding BCD number is identical and therefore no correction is needed. • Modifications are needed if the sum is greater than 1001 as we get non-valid BCD representation. The addition of binary 6 (0110) to binary sum converts it to the correct BCD representation and also produces an output carry as required (refer to section 1-7) • The logic circuit that detects the necessary correction can be derived from the table entries. Correction (adding decimal 6 or binary 0110) is needed when the binary sum has an output K = 1 Z8Z4 = 1 Z8Z2 = 1 6 Fall 2022
  • 7. Map Simplification of the Function C 1 1 1 1 1 1 1 1 1 1 d d d d d d d d d d d d Z8Z4 00 01 11 10 00 01 11 10 k=0 Z2Z1 Z8Z4 00 01 11 10 00 01 11 10 k=1 C(K,Z8,Z4,Z2,Z1)= K+ Z8Z4 +Z8Z2 Z2Z1 7 Fall 2022
  • 9. Correction for BCD Adder • This condition for correction and an output carry can be expressed as  C = K + Z8Z4 + Z8Z2 • In figure 4-14, the output carry generated from the bottom adder can be ignored since this is already available at the output carry terminal • A decimal parallel adder that adds n decimal digits (in BCD form) needs n BCD adder stages with output carry from one stage connected to the input carry of next higher-order stage 9 Fall 2022
  • 10. BCD Adder Numbers that need correction (add 6) are: K Z8 Z4 Z2 Z1 0 1 0 1 0 (10) 0 1 0 1 1 (11) 0 1 1 0 0 (12) 0 1 1 0 1 (13) 0 1 1 1 0 (14) 0 1 1 1 1 (15) 1 0 0 0 0 (16) 1 0 0 0 1 (17) 1 0 0 1 0 (18) 1 0 0 1 1 (19) C = K + Z8Z4 +Z8Z2 10 Fall 2022
  • 11. Binary Multiplier • Multiplication of the binary number is performed in the same way as in decimal numbers. The multiplicand is multiplied by each bit of the multiplier starting from the least significant bit • Each such such multiplication forms a partial product. Successive partial products are shifted one position to the left. The final product is obtained from the sum of the partial products • Consider the binary multiplication of 2-bit numbers where B1 and B2 are multiplicand bits and A1 and A0 are multiplier bits. C3 C2 C1 C0 are the product bits 11 Fall 2022
  • 12. Binary Multiplier • The first partial product is formed by multiplying A0 by B1 B0. The multiplication of any two bits produces a 1 if both bits are 1, otherwise it produces a 0. This is identical to AND operation. Therefore partial product can be implemented with an AND gates as shown in figure 4-15 • The second partial product is formed by multiplying A1 by B1 B0 and shifted one position to the left • The two partial products are added with two half adder (HA) circuits. If there are more bits in the partial products then we use full adders to produce the sum of the partial products • The least significant bit of the partial product doesn’t have to go through an adder since it is formed by the output of the first AND gate 12 Fall 2022
  • 14. K-bit by J-bit Binary Multiplier • A combinational circuit for binary multiplier with more bits can be constructed in a similar way. A bit of the multiplier is ANDed with each bit of the multiplicand in as many levels as there are bits in the multiplier • The binary output in each level of AND gate is added with the partial product of the previous level to form a new partial product. The last level produces the product • For J multiplier bits and K multiplicand bits we need (J x K) AND gates and (J – 1) K bit adders to produce a product of (J + K) bits 14 Fall 2022
  • 15. 4-bit by 3-bit binary multiplier • Consider a multiplier circuit that multiplies a binary number of four bits by a number of three bits. • Let the multiplicand is represented by B3 B2 B1 B0 and the multiplier by A2 A1 A0 • Since K = 4 and J = 3, we need 12 AND gates and two 4 – bit adders to produce a product of seven bits B3 B2 B1 B0 Multiplicand X A2 A1 A0 Multiplier 0 A0B3 A0B2 A0B1 A0B0 Partial Product 0 A1B3 A1B2 A1B1 A1B0 Partial Product 1 Cout S3 S2 S1 S0 A2B3 A2B2 A2B1 A2B0 Partial Product 2 C6 C5 C4 C3 C2 C1 C0 Final Product 15 Fall 2022