1
Combinational Logic Design
Design using SSI chips
i)
Arithmetic Circuits
ii)
Code Convertor
Combinational Logic Circuits


The term "combinational" comes to us from mathematics which
means an unordered set where nobody cares which order the
items came in
With combinational logic, the circuit always produces the same
output solely by a given set of inputs , regardless of the order
the inputs are changed


The inside of a combinational circuit is made of logic gates and
it does not store any information (i.e. memoryless)


If there are m outputs and n inputs then there are m Boolean
functions, one describing each outputs
There are circuits which depend on inputs and previous outputs.
These circuits are called sequential logic where we will discuss in
later chapter
Combinational logic circuits are important components of digital
systems
3
Combinational vs. Sequential
Logic
(a) Combinational
Output = F (In)
(b) Sequential
Output = F (In, Previous In)
4
Designing Combinational Circuits

In general we have to do following steps:
1. From the specifications of the circuit, determine
the required number of inputs and outputs and
assign a letter symbol to each
2. Derive the truth table that defines the relationship
between inputs and outputs
3. Obtain the simplified Boolean functions for each
output as a function of the input variables
4. Draw the logic diagram
5. Verify the correctness of the design
5
Arithmetic Circuits


Arithmetic circuits are the ones which
perform arithmetic calculations on binary
numbers like addition, subtraction,
multiplication, and division etc.
Arithmetic circuits can be built using logic
circuits and they are excellent examples of
combinational logic design

We will build arithmetic circuits by using the
design procedure as stated previously
6
Adder




Adders are the basic building blocks of all arithmetic
circuits
Adders add two binary numbers and give out sum
and carry as output
In modern computers adders reside in the arithmetic
and logic unit (ALU) where other operations are
performed
Basically we have two types of adders


Half Adder
Full Adder
7
Half Adder


Adding two single-bit binary values, X, Y produces a sum S bit and a
carry out C-out bit
This operation is called half addition and the circuit to realize it is called
a half adder
Half Adder Truth Table
Inputs
X
0
0
1
1
Y
0
1
0
1
Outputs
S
0
1
1
0
C-out
0
0
0
1
S(X,Y) = S (1,2)
S = XY + X Y
S = XY
C-out(X,Y) = S (3)
C-out = XY
X
Y
Sum S
C-out
8
Full Adder

Sum S
X
XY
Adding two single-bit binary
values, X, Y with a carry
input bit C-in produces a
sum bit S and a carry out Cout bit
00
C-in
0
01
2
0
1
1
1
11
10
6
1
3
7
4
1
5
1
C-in
Y
The S function is the three-bit XOR function (Odd Function):
S = X  Y  (C-in)
Carry C-out
X
XY
00
C-in
01
0
2
1
3
11
6
0
1
7
1
1
1
10
4
5
1
C-in
Y
C-out = XY + X(C-in) + Y(C-in)
9
Full Adder Circuit using XOR and
Basic Gates

Logical Implementation
10
Full Adder implemented by Two
Half Adders and an OR Gate

A ‘Full Adder’ can also be implemented using
two half adders and an ‘OR’ Gate as follows:
The sum S = X  Y  (C-in)
The carry out C _ out = XY + XC_ in + YC_ in
= XY + C _ in ( X + Y )
= XY + C _ in ( X + Y )( X + X )(Y + Y )
= XY + C _ in ( X + Y )( XY + X Y + X Y + X Y )
= XY + C _ in ( XY + X Y + X Y )
= XY + C _ in XY + C _ in ( X Y + X Y )
= XY (1 + C _ in ) + C _ in ( X Y + X Y )
= XY + C _ in ( X ⊕ Y )
11
Full Adder implemented by Two Half
Adders and an OR Gate (Cont.)

Therefore S = X ⊕Y ⊕ C_ in and C_ out = XY + C_ in ( X ⊕Y )
HA1
HA2
X
Y
S
C-in
C-in
S2
C-out
S
HA2

X
S1
HA1
Y
C1
C2
C-out
Block diagram representation of
a full adder using two half
adders
12
Adding Two Binary Numbers





We want to add two 4-bit
binary numbers
We use a series of half and
full adders
Since there is no carry into
the first stage, we can use a
half adder there
For each other stage we use
a full adder and just carry in
from the previous stage
The carry out at the end
indicates that the sum is too
large to be represented by 4
bits
13
Subtraction



If we want to do subtraction,
the circuit is very similar
Looking at the truth table for
A-B
The only difference between a
half adder and a half subtractor
is the borrow



In order to perform this
subtraction a digit has to be
borrowed from the next highest
column of the subtraction
When it borrows, it gets a 2 and
the operation then becomes 2 - 1
=1
The Borrow output is set to
indicate that borrow operation is
required
14
Half Subtractor


Subtracting a single-bit binary value Y from anther X (i.e. X - Y)
produces a difference bit D and a borrow out bit B-out
This operation is called half subtraction and the circuit to realize
it is called a half subtractor
D(X,Y) = S (1,2)
= X'Y + XY'
Thus, D = X  Y
B-out(X,Y) = S (1)
Thus, B-out = X'Y
X
Y
Difference
D
B-out
Symbol
15
Half Subtractor circuit


Again, we shall represent the half subtractor using a
box
Since the two inputs of a subtractor are different, we
indicate the subtrahend with a ‘-’
16
Full Subtrator


Subtracting two single-bit binary
values, Y, B-in from a single-bit
value X produces a difference
bit D and a borrow out B-out bit
This is called full subtraction
D = X  Y  (B-in)
Note: D= X – Y – B-in
D(X,Y, B-in) = S (1,2,4,7)
B-out(x, y, B-in) = S (1,2,3,7)
B-out = X'Y + X'(B-in) + Y(B-in)
= X'Y + B-in(X  Y)'
(In the same way as we did with the full adder)
17
Full Subtractor Circuit
Implementation
X-OR gate
implementation
of half subtactor
Full subtractor using two half
subtractors and an OR gate
HS1
X
Y
B_in
Difference (D)
X
Y
HS2
D
X
Y
B_out
B_in
X
B_in
B_out
Text
Y
B_in
18
Parallel Adder IC

As shown in the pin layout diagram, the IC 7483







Accepts two 4-digit numbers A4A3A2A1, B4B3B2B1 and a carry-in Ci as inputs
Produces a 4-digit sum output S4S3S2S1 and a carry-out Co
If the SUM of the two inputs plus the carry-in is between 0 and 15, the
SUM appears in the S outputs and the carry-out is '0‘
If the SUM is between 16 and 31, carry-out C0 becomes ‘1‘ and the S
outputs are 16 less than the SUM
Example (not using 2’s complement)
A4A3A2A1 = 0111, B4B3B2B1 = 1010, Ci = 1
So, result should be 10010 (18)
As expected, the ICs produces
A4 A3
Co = 1 and SSSS
Co
A2 A1
B4 B3 B2 B1
74283
Ci
S4 S3 S2 S1
19
Cascading the 74283 ICs




A 74283 IC only accept 4-bits Boolean variables at its
input
So, how to do addition which involve 5-bits or more
Find another suitable IC! Or, by cascading 74283 ICs
To add two 8 bit numbers X8X7X6X5X4X3X2X1 and
Y8Y7Y6Y5Y4Y3Y2Y1 , connect two 74283 as below
X4X3 X2X1
X8X7X6X5
Y8Y 7Y6Y5
A4 A3 A2 A1
B4 B3 B2 B1
74283
Co
S4
S3
S2
Ci
S1
2nd stage
produce MSB
A4 A3 A2 A1
Y4Y3 Y2Y1
B4
B3
74283
Co
S4
S3
S2
B2
B1
Ci
S1
1st stage
produce LSB
20
n-bit Subtractors

An n-bit subtracor used to subtract an n-bit number
Y from another n-bit number X (i.e. X - Y) can be
built by using an n-bit adder and n inverters:
1. Find two’s complement (1's complement plus 1) of Y by:
Inverting all the bits of Y using the n inverters
2. Adding 1 by setting the carry in of the least significant
position to 1

The original subtraction (X - Y) now becomes an
addition of X to two’s complement of Y using the nbit adder
21
4-bit Subtractor using 4-bit Adder
22
Adding and Subtracting


Since the circuits are so
similar, it is reasonably easy
to combine them together
into a single circuit, with a
control line used to select
either addition or subtraction
The only difference between
the half adder and the half
subtractor is the NOT
A
A


The above table gives a
truth table of what we
require
What logic gate can fulfill
this truth table?
23
Adding a Control Line



If we define a 0 on our control signal to mean Add then the
input to the AND Gate should be the same as our A signal
When the control signal is 1, we want to do a subtraction, so we
require the input to the AND Gate to be the NOT A signal
We make our full adder/subtractor just as before, with all the
control inputs tied together
A
Sum /
Difference
Control
B
Carry /
Borrow
24
Adder and Subtractor Using
74283 IC

When the switch is closed, 74283 is an
adder




A '0' is applied to Ci and the EX-OR
gates
So, there is no Carry-in and
Addend pass through the EX-OR
unchanged (1  0 = 1, 0  0 = 0)
When the switch is opened, 74283
becomes a subtractor




A ‘1' is applied to Ci and the EX-OR
gates
So, there is a Carry-in (adding 1) and
Addend pass through the EX-OR
inverted (1  1 = 0, 0  1 = 1)
So, 74283 acts as an subtractor, by
adding (2’s complement of Y4Y3Y2Y1)
instead of (Y4Y3Y2Y1)
Y4Y3Y2Y1
X4X3X2X1
A4
A3
A2
V cc
A1
B4
B3
74283
Co
S4
S3
S2
B2
B1
Ci
S1
25
BCD Adder



When the sum of two digits is less than or
equal to 9 then the ordinary 4-bit adder can
be used
But if the sum of two digits is greater than 9
then a correction must be added “I.e adding
0110”
We need to design a circuit that is capable of
doing the correct addition
Examples :
1. CASE I : Sum <= 9 & carry = 0.

Add BCD digits 3 & 4
1.
0011
+0100
--------0111
Answer is valid BCD number = (7)BCD & so addition of 0110
is not added.
2. CASE II : Sum > 9 & carry = 0.

Add BCD digits 6 & 5
1.
0110
+0101
----------1011
Invalid BCD (since sum > 9) so 0110 is to be added
2.
1011
+0110
----------1 0001
(1

1)BCD
Valid BCD result = (11) BCD
3. CASE III : Sum < = 9 & carry = 1.
Add BCD digits 9 & 9


1001
+ 1001
----------1 0010
Invalid BCD ( since Carry = 1 ) so 0110 is to be added
1 0010
+ 0110
-----------1 1000
(1
8)BCD
Valid BCD result = (18) BCD
Design of BCD Adder



Design of BCD adder :
To execute first step i. e. binary addition of two
4 bit numbers we will use IC 7483 ( with Cin =
0 ), which is 4 bit binary adder.
We need to design a digital circuit which will
sense sum & carry of IC 7483 & if sum exceeds
9 or carry = 1, this digital circuit will produce
high output otherwise its output will be zero.
30
Design of BCD Adder

The cases where the sum of two 4-bit
numbers is greater than 9 are in the following
table:
S4
S3
S2
S1
S0
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
BCD Adder
Whenever S4=1 (sums greater than 15)
 Whenever S3=1 and either S2 or S1 or both
are 1 (sums 10 to 15)
 The previous table can be expressed as:
X = S4 + S3 ( S2 + S1)
So, whenever X = 1 we should add a correction
of 0110 to the sum.

Inputs:[A]=0101, [B]= 0011, Co=0
0011
0101
0
0
1
0
0
0
0
0
1000
1
1 0 0 0
0000
Inputs:[A]=0111, [B]= 0110, Co=0
0110
0111
0
1
1
1
0 1
1
1
1101
1
0 0 1 1
0110
Cascading BCD Adders




The previous circuit is used for adding two
decimal digits only. That is, “ 7 + 6 = 13”.
For adding numbers with several digits, a
separate BCD adder for each digit position
must be used.
For example:
BCD Adder
BCD Adder
2
4
7
+ 5
3
8
-------------------?
BCD Adder
Cascading BCD Adders
Example


Determine the inputs and the outputs when the
above circuit is used to add 538 to 247. Assume
a CARRY IN = 0
Solution:

Represent the decimal numbers in BCD
247 = 0010 0100 0111
538 = 0101 0011 1000
Put these numbers in registers [A] and [B]
[A] = 0010 0100 0111
[B] = 0101 0011 1000
Example
0 0 1 0
0
0 1 0 0
0
0 1 1 1
1
0
0111
0 1 0 1
1000
0 0 1 1
0101
1 0 0 0
BCD Subtractor

Rules for BCD subtraction :
1. Find 9’s complement of the subtrahend.
a) To find 9’s complement first find 1’s complement of
subtrahend.
b) Add (10)10 i.e. (1010)2 to it.
2. Add 9’s complement of the subtrahend to the minuend using
BCD rules of addition.
3. After BCD addition if MSD is 0, result is negative expressed in 9’s
complement form to get it in natural form find 9’s complement
of the LSD of the result.
4. But after BCD addition if MSD is 1 it indicates that result is
positive expressed in natural form & add ( 1 )10 i. e. ( 0001 )2
to it.
39
CASE I : carry = 1, answer is positive
Subtract BCD digit 2 from 8.

1. 9’s complement of (2)10 i.e. (0010)2
a) 1’s complement of (2)10 1 1 0 1
b) add (1010)2
+ 1010
--------1 0111

2.
+
1000
0111
---------1111
(discard carry).
(8)10
9’s comp. of (2)10
Invalid BCD so (0110)2 is added
1111
+ 0110
---------1 0101

Since MSD after BCD addition is 1, answer is positive expressed in natural form & to get final
answer carry is added at the end
.
0101
+
1
----------0 1 1 0 i. e. (6)10
CASE II : carry = 0, answer is negative

Subtract BCD digit 9 from 6.
1. 9’s complement of (9)10 i.e. (1001)2
a) 1’s complement of (9)10
b) add (1010)2
0110
+ 0000
---------00110
valid BCD with carry = 0


2.
0110
+ 1010
--------1
0000
(discard carry).
(6)10
9’s comp. of (9)10
Since MSD after BCD addition is 0, answer is negative expressed in 9’s compliment form & to get
final answer, 9’s compliment of answer is taken
.
a) 1’s complement of answer
b) add (1010)2
1001
+ 1010
---------1 0011
So final answer is –(3)10 i.e.- (0011)
(discard carry).
Design of BCD Subtractor
1. To execute first step i. e. to find 9’s compliment of subtrahend
a) First we find 1s compliment of subtrahend.
b) Then add (10)10 i.e. (1010)2 to 1’s compliment of subtrahend.
We use inverter to get 1’s compliment of subtrahend since subtrahend is 4 bit
we need 4 inverters and to add this to (1010)2, 4 bit binary adder IC 7483 is
used.

2. Circuit for BCD addition of 9’s compliment of subtrahend and minuend we
will use already designed BCD adder

3. Circuit to get answer in correct form.
If MSD after BCD addition is 0, answer is negative expressed in 9’s
compliment form & to get answer in correct form we have to find 9’s
compliment of the result.
If MSD after BCD addition is 1, answer is positive expressed in natural form
& to get final answer carry is added to the earlier answer.

Logic
Diagram
CIRCUIT DELAYS (1/5)

Given a logic gate with delay t. If inputs are stable at
times t1, t2, …, tn, then the earliest time in which the
output will be stable is:
max( t1, t2, …, tn ) + t
t1
t2
:
tn

:
Logic
Gate
max (t1, t2, ..., tn ) + t
To calculate the delays of all outputs of a combinational
circuit, repeat above rule for all gates.
CIRCUIT DELAYS (2/5)

As a simple example, consider the full adder circuit
where all inputs are available at time 0. Assume each
gate has delay t.
X
Y
Z

0
0
S
0
C
CIRCUIT DELAYS (3/5)

More complex example: 4-bit parallel adder.
Y4
X4
00
C
FA
Y3
C 0X03
Y2
C 0X02
Y1
C 0X01
4
3
2
FA
FA
FA
0 C
1
5
S4
S3
S2
S1
CIRCUIT DELAYS (4/5)


Analyse the delay for the repeated block.
Xi
where Xi, Yi are
0
Si
stable at 0t, while
Full
Yi 0
Ci is assumed to be
Adder
C
i+1
stable at mt.
Ci mt
Performing the delay calculation:
X
0
max(0,0)+t =
t
0
max(t,mt)+t
i
Y
t
i
Ci
mt
max(t,mt)+t
Si
max(t,mt)+2t
Ci+1
CIRCUIT DELAYS (5/5)

Calculating:
When
When
When
When

i=1,
i=2,
i=3,
i=4,
m=0;
m=3;
m=5;
m=7;
S1
S2
S3
S4
=
=
=
=
2t
4t
6t
8t
and
and
and
and
C2
C3
C4
C5
=
=
=
=
3t
5t
7t
9t
In general, an n-bit ripple-carry parallel adder will
experience the following delay times:
Sn = ?
Cn+1 = ?

Propagation delay of ripple-carry parallel adders is
proportional to the number of bits it handles.
FASTER CIRCUITS

Three ways of improving the speed of circuits:

Use better technology (eg. ECL faster than TTL gates)
BUT



Use gate-level designs to two-level circuits! (use sumof-products/product-of-sums) BUT



Faster technology is more expensive, needs more power,
lower-level of integrations
Physical limits (eg. speed of light, size of atom)
Complicated designs for large circuits
Product/sum terms need MANY inputs!
Use clever look-ahead techniques BUT

There are additional costs (hopefully reasonable).
LOOK-AHEAD CARRY ADDER (1/6)

Consider the FA, where
intermediate signals are
labelled as Pi and Gi:
Pi = Xi  Yi
Gi = Xi ∙Yi

Pi
Si
Gi
Ci+1
Ci
The outputs Ci+1, Si, in terms of Pi, Gi, Ci are:
Si = Pi  Ci
Ci+1 = Gi + Pi∙Ci

Xi
Yi
… (1)
… (2)
Looking at equation (2):
Gi = Xi ∙Yi is a carry generate signal, and
Pi = Xi  Yi is a carry propagate signal.
LOOK-AHEAD CARRY ADDER (2/6)

For 4-bit ripple-carry adder, the equations for the four
carry signals are:
Ci+1
Ci+2
Ci+3
Ci+4

=
=
=
=
Gi + Pi∙Ci
Gi+1 + Pi+1∙Ci+1
Gi+2 + Pi+2∙Ci+2
Gi+3 + Pi+3∙Ci+3
These formulae are
deeply nested, as
shown here for Ci+2:
Ci
Pi
Ci+1
Gi
Pi+1
Gi+1
Ci+2
4-level circuit for Ci+2 = Gi+1 +Pi+1.Ci+1
LOOK-AHEAD CARRY ADDER (3/6)


Nested formulae/gates cause more propagation delay.
Reduce delay by expanding and flattening the
formulae for carries. Example, for Ci+2:
Ci+2 = Gi+1 + Pi+1∙Ci+1
= Gi+1 + Pi+1∙(Gi + Pi∙Ci)
= Gi+1 + Pi+1∙Gi + Pi+1∙Pi∙Ci

New faster circuit for Ci+2:
Ci
Pi
Pi+
1
Gi
Pi+
1
Gi+1
Ci+2
LOOK-AHEAD CARRY ADDER (4/6)

Other carry signals can be similarly flattened:
Ci+3 = Gi+2 + Pi+2∙Ci+2
= Gi+2 + Pi+2∙(Gi+1 + Pi+1∙Gi + Pi+1∙Pi∙Ci)
= Gi+2 + Pi+2∙Gi+1 + Pi+2∙Pi+1∙Gi + Pi+2∙Pi+1∙Pi∙Ci
Ci+4= Gi+3 + Pi+3∙Ci+3
= Gi+3 + Pi+3∙(Gi+2 + Pi+2∙Gi+1 + Pi+2∙Pi+1∙Gi + Pi+2∙Pi+1∙Pi∙Ci)
= Gi+3 + Pi+3∙Gi+2 + Pi+3∙Pi+2∙Gi+1 + Pi+3∙Pi+2∙Pi+1∙Gi +
Pi+3∙Pi+2∙Pi+1∙Pi∙Ci


Note that formulae gets longer with higher carries.
Also, all carries are two-level sum-of-products
expressions, in terms of the generate signals Gs, the
propagate signals Ps, and the first carry-in Ci.
LOOK-AHEAD CARRY ADDER (5/6)

We employ lookahead formula in
this lookahead-carry
adder circuit:
LOOK-AHEAD CARRY ADDER (6/6)


The 74182 IC chip
allows faster
lookahead adder to
be built.
Assuming gate delay
is t, maximum
propagation delay
for circuit is hence
4t



t to get generate and
propagate signals
2t to get the carries
t for the sum signals
Code Converter




Binary to Gray Code
Gray to Binary Code
BCD to Excess – 3 Code
Excess-3 to BCD Code

Binary To Gray Conversion

Gray To Binary Conversion
Binary To Gray Code Conversion:
INPUT (BINARY CODE)
Truth Table:
B
B
B
B
Binary To Gray Code Conversion
OUTPUT (GRAY CODE)
G3
G2
G1
G0
0
0
0
0
0
0
1
0
0
0
1
0
1
0
0
0
1
1
0
0
1
1
0
0
1
0
0
1
0
0
0
1
1
0
0
1
0
1
0
1
1
1
0
1
1
0
0
1
0
1
0
1
1
1
0
1
0
0
1
0
0
0
1
1
0
0
1
0
0
1
1
1
0
1
1
0
1
0
1
1
1
1
1
0
1
1
1
1
1
0
1
1
0
0
1
0
1
0
1
1
0
1
1
0
1
1
1
1
1
0
1
0
0
1
1
1
1
1
1
0
0
0
3
2
1
0
0
0
0
0
0
0

K-Map For Reduced Boolean Expressions
Of Each Output:
3) Circuit Diagram for Binary to Gray Code
Conversion
Gray To Binary Code Conversion:
Truth Table:
Gray To Binary Code Conversion
INPUT (GRAY CODE)
OUTPUT (BINARY CODE)
G3
G2
G1
G0
B3
B2
B1
B0
0
0
0
0
0
0
0
0
0
0
0
1
0
0
0
1
0
0
1
1
0
0
1
0
0
0
1
0
0
0
1
1
0
1
1
0
0
1
0
0
0
1
1
1
0
1
0
1
0
1
0
1
0
1
1
0
0
1
0
0
0
1
1
1
1
1
0
0
1
0
0
0
1
1
0
1
1
0
0
1
1
1
1
1
1
0
1
0
1
1
1
0
1
0
1
1
1
0
1
0
1
1
0
0
1
0
1
1
1
1
0
1
1
0
0
1
1
1
1
0
1
0
0
0
1
1
1
1

K-Map For Reduced Boolean Expressions Of
Each Output:
Circuit Diagram: