BEng Electronic Engineering
DIGITAL ELECTRONICS
Lecture 7 – Static Hazards (Part 1)
AIM
To start investigating the problem of
hazard conditions in digital circuits
Objectives
• Understand what static hazard conditions are
• To revisit the rules of boolean algebra and
understand how these are changed by hazard
condition
• To investigate a method of determining Static ‘1’
hazards
Hazard Conditions
Consider the following simple circuit
A
Z =A.A
A
A
Ā
Z
0
1
1
1
0
1
A
A
Z
STATIC ‘1’
HAZARD
Hazard Conditions
Also consider the following simple circuit
A
Z =A.A
A
A
Ā
Z
0
1
0
1
0
0
A
A
Z
STATIC ‘0’
HAZARD
Implications
The existence of these static hazards has implications
for our normal use of Boolean algebra
• In analysing a digital circuit for hazard conditions we must
treat a variable and its complement as two separate
variables
• Any reduction formula that makes use of a variable and it
complement can no longer be used for the reduction of
Boolean expressions
Boolean Algebra
This implies there are a number of Boolean
reduction equations we can no longer use
Useable
Not Useable
x.x = x
x+x = x
x.x = 0
x+x = 1
x + xy = x
x + xy = x + y
(x + y)(x + z) = x + yz
Association, Distribution,
Commutation
DeMorgans Theorem
xy + zx + yz = xy + zx
Detecting Static Hazards
Consider the following logic circuit
A
B
C
Z= A.B.B.C
Detecting Static Hazards
A
B
C
B
AB
CB
Z
Creating SoP Expressions
To examine this logic we need to convert it into an SoP
expression that can be plotted on a Karnaugh map
Z = A.B +. B.C
Apply DeMorgans
Z = A.B + B.C
Remove double negations
Z = A.B + B.C
SOP Function
Plotting the Function
These two groups can be plotted on a 3 variable Karnaugh Map
Z = A.B + B.C
• The transition occurs where two
'1's are in adjacent cells on the
Karnaugh map
Z
C
0
AB
1
00
01
11
• Where the '1's are not part of
the same group
1
10
1
1
1
Example Problem
Lets try these principles to detect static ‘1’
hazards in the following digital circuit
A
A.B.C
B
C
(A.B.C)+(A+C).(A+D)
F
A
A+C
C
A
D
(A+C).(A+D)
A+D
SoP Production
First convert the logic to SoP form
F = ABC + (A + D)(Ā + C̄)
Distribution
F = ABC + AĀ + AC̄ + ĀD + C̄D
The AA group cannot be reduced
Plotting Groups
Plotting these groups on a Karnaugh Map
F = ABC + AĀ + AC̄ + ĀD + C̄D
Indicates possible
static ‘0’ hazard
input A changes and B=1, C=1, D=1
F
CD
00
AB
00
01
01 11
1
1
1
1
1
10
input C changes and A=1, B=1, D=1
input C changes and A=1, B=1, D=0
11
1
1
10
1
1
1