Glitches & Hazards

Glitches / Hazards

Gates have an inherent delay


A glitch is an unwanted pulse at the output of a
combinational circuit



We have been ignoring this  but delays do exist
Glitches can result from gate delays
Glitches depend on the input patterns  glitches may not
occur if the input pattern that would cause the glitch never
occurs
A circuit with the potential for a glitch has a hazard

A circuit with a hazard may or may not glitch

Glitch Example

F = AB + A'C: assume all gate delays = 10ns

An input change from ABC = 111 to 011 results in a
glitch  output changes from 1  0  1 again

A static hazard: result should have stayed statically at 1

Static & Dynamic Hazards

Static hazards:
1
0
1 1

0 0
Dynamic hazards:
1
0
1 0
0 1

Removing Static Hazards

Glitch occurs in moving from one implicant to
another within a cover



When two adjacent 1's are not covered by a single implicant
Solution: add an extra implicant to provide that coverage
BC
A
00
01
11
10
0
0
1
1
0
1
0
0
1
1
F = AB + A'C + BC

No longer minimum form  but no hazard now

Hazard Removal: Example


F(A,B,C,D) = m(6,7,8,9,12,13,14,15)
CD
AB
00
01
11
10
00
0
0
0
0
01
0
0
1
1
11
1
1
1
1
10
1
1
0
0
The red prime implicant removes the static-1 hazard

Dynamic Hazards

Dynamic hazards do not occur in two-level
combinational circuits  only in multi-level circuits
One gate delay
x1
x2,x3,x4
b
x1
f
a
x2
c
x3
x4
d
a
b
c
d
f

Causes of Dynamic Hazards

For an output to change 0101 (i.e. 3 times) in
response to a single input change, there must be at
least 3 paths of different length in the circuit



2 gate delays: x1  b  f
3 gate delays: x1  a  b  f
4 gate delays: x1  a  c  d  f
b
x1
f
a
x2
c
x3
x4
d

Removing Dynamic Hazards

Removing dynamic hazards is very difficult

Even detecting dynamic hazards is very difficult

Stick to two level designs to ensure combinational
circuits do not have dynamic hazards

Sequential circuits do not have this problem
Quine-McClusky Method

Quine-McClusky Method

An algorithm that CAD tools can and do use

Uses tables to:




Compute all prime implicants
Identify essential prime implicants
Select the minimum number of prime implicants for a cover
Is based on the combining property again

x y + x y' = x (y + y') = x

Grouping the Minterms




f(A,B,C,D) = m(0,4,8,10,11,12,13,15)
Arrange all minterms according to the number of 1’s
in the binary representation
Sort and group them
These are called 0-cubes

Like vertices of a 4D cube
0
0000
4
0100
8
1000
10
1010
12
1100
11
1011
13
1101
15
1111

Forming 1-cubes

Compare each 0-cube in one group with each 0cube in an adjacent group

Use combining property to form 1-cubes
1 cubes
0 cubes
0
0000
X
4
0100
X
8
1000
X
10
1010
X
12
1100
X
11
1011
X
13
1101
X
15
1111
X
0,4
0x00
0,8
x000
8,10
10x0
4,12
x100
8,12
1x00
10,11
101x
12,13
110x
11,15
1x11
13,15
11x1

Forming 2-cubes

Compare each 1-cube in one group with each 1cube in an adjacent group

Use combining property to form 2-cubes
1 cubes
2 cubes
0,4
0x00
X
0,8
x000
X
8,10
10x0
4,12
x100
X
8,12
1x00
X
10,11
101x
12,13
110x
11,15
1x11
13,15
11x1
0,8,4,12
xx00
mark each 1-cube that
contributes to a 2-cube

The Prime Implicants

All “unchecked” cubes are prime implicants

They did not get combined into a larger cube
0 cubes
1 cubes
2 cubes
0
0000
X
0,4
0x00
X
4
0100
X
0,8
x000
X
8
1000
X
8,10
10x0
10
1010
X
4,12
x100
X
12
1100
X
8,12
1x00
X
11
1011
X
10,11
101x
13
1101
X
12,13
110x
15
1111
X
11,15
1x11
13,15
11x1
0,8,4,12
xx00
All prime implicants are:
P = { 10x0, 101x, 110x,
1x11, 11x1, xx00 }

Prime Implicant Cover Table

For each prime implicant, mark all minterms covered
by that implicant
prime
implicants
p1
10x0
p2
101x
p3
110x
p4
1x11
p5
11x1
p6
xx00
minterms
0
4
8
10
X
X
X
11
12 13 15
X
X
X
X
X
X
X
X
X
X
X

Find Essential Prime Implicants

p6 is an essential prime implicant  minterms 0 and
4 must be covered by p6

Minimum cover C = { p6 } at the moment
prime
implicants
p1
10x0
p2
101x
p3
110x
p4
1x11
p5
11x1
p6
xx00
minterms
0
4
8
10
X
X
X
11
12 13 15
X
X
X
X
X
X
X
X
X
X
X

Row Dominance

p2 row dominates p1 since p2 covers everything that
p1 covers

p1 can be eliminated

Note: we remove the dominated row
prime
minterms
implicants
10 11
p1
10x0
X
p2
101x
X
p3
110x
p4
1x11
p5
11x1
13 15
X
p5 row dominates p3
X
X
X
X
X

Column Dominance

Column 11 column dominates column 10 since
everything that covers column 10 also covers
column 11

Column 11 can be eliminated

Note: we remove the dominating column
prime

minterms
implicants
10 11
p2
101x
X
p4
1x11
p5
11x1
13 15
X
X
X
X
column 15 dominates
column 13
X
We could also have considered p2 and p5 as
essential at this point

Final Cover
prime
minterms
implicants
10
p2
101x
X
p4
1x11
p5
11x1
13
X

The final cover is C = { p6, p2, p5 }

f = xx00 + 101x + 11x1 = C'D' + AB'C + ABD

Another Example

f(A,B,C,D) = m(0,2,4,5,6,7,8,10,11,12,15)
0 cubes
0
0000
2
0010
4
0100
8
1000
5
0101
6
0110
10
1010
12
1100
7
0111
11
1011
15
1111
Cubes

1 cubes
0 cubes
0,2
00x0
X
0,4
0x00
X
0,8
x000
X
2,6
0x10
X
0
0000
X
2,10
x010
X
2
0010
X
4,5
010x
X
4
0100
X
4,6
01x0
X
8
1000
X
4,12
x100
5
0101
X
8,10
6
0110
X
10
1010
12
2 cubes
0,2,4,6
0xx0
0,2,8,10
x0x0
X
0,4,8,12
xx00
10x0
X
4,5,6,7
01xx
8,12
1x00
X
X
5,7
01x1
1100
X
6,7
011x
X All prime implicants are:
X
7
0111
X
10,11
101x
11
1011
X
7,15
x111
15
1111
X
11,15
1x11
P = { 101x, x111, 1x11,
0xx0, x0x0, xx00, 01xx }

Cover Table
prime
implicants

minterms
0
2
p1
101x
p2
x111
p3
1x11
p4
0xx0
X
X
p5
x0x0
X
X
p6
xx00
X
p7
01xx
4
5
6
7
8
10
11
X
X
12 15
X
X
X
X
X
X
X
X
X
X
X
X
X
X
C = { p6, p7 } since p6 and p7 are essential
X

Row Dominance
prime
implicants


minterms
2
p1
101x
p2
x111
p3
1x11
p4
0xx0
X
p5
x0x0
X
10
11
X
X
15
X
X
X
Row p5 dominates row p4
Row p3 dominates row p2
X

Column Domination
prime
implicants




p1
101x
p3
1x11
p5
x0x0
minterms
2
10
11
X
X
15
X
X
X
X
Column 10 dominates column 2
Column 11 dominates column 15
Final min cover is C = { p6, p7, p3, p5 }
f = xx00 + 01xx + 1x11 + x0x0
= C'D' + A'B + ACD + B'D'

Handling Don't Cares



f(A,B,C,D) = m(0,3,10,15) + d(1,2,7,8,11,14)
Don't care conditions are
used for creating the prime
implicants
Don't care conditions are
not used for computing the
minimum cover
0 cubes
0
0000
1
0001
2
0010
8
1000
3
0011
10
1010
7
0111
11
1011
14
1110
15
1111

Cubes
1 cubes
0 cubes
0,1
000x
X
0,2
00x0
X
2 cubes
0
0000
X
0,8
x000
X
0,1,2,3
00xx
1
0001
X
1,3
00x1
X
0,2,8,10
x0x0
2
0010
X
2,3
001x
X
2,3,10,11
x01x
8
1000
X
2,10
x010
X
3,7,11,15
xx11
3
0011
X
8,10
10x0
X
10,11,14,15
1x1x
10
1010
X
3,7
0x11
X
7
0111
X
3,11
x011
X
11
1011
X
10,11
101x
X
14
1110
X
10,14
1x10
X
15
1111
X
7,15
x111
X
11,15
1x11
X
14,15
111x
X
All prime implicants are:
P = { 00xx, x0x0, x01x,
xx11, 1x1x }

Cover Table

Include only required minterms in the initial cover
table
prime

minterms
implicants
0
3
p1
00xx
X
X
p2
x0x0
X
p3
x01x
X
p4
xx11
X
p5
1x1x
10 15
X
X
X
X
X
There are no essential prime implicants

Branching


There are no dominant rows and no dominant
columns
Require a branching strategy


Backtracking type algorithm
Select C = { p1 }
prime
minterms
implicants
0
3
p1
00xx
X
X
p2
x0x0
X
p3
x01x
X
p4
xx11
X
p5
1x1x
10 15
X
X
X
X
X

Min Cover That Includes p1

Row p5 dominates rows p2, p3, and p4
Min cover C = { p1, p5 }

f = 00xx + 1x1x = A'B' + AC


Cost = 3 + 6 = 9
prime
minterms
implicants
10
p2
x0x0
X
p3
x01x
X
p4
xx11
p5
1x1x
15
X
X
X

Backtrack

Now backtrack and eliminate p1 to find covers
prime
minterms
implicants
0
3
p1
00xx
X
X
p2
x0x0
X
p3
x01x
X
p4
xx11
X
p5
1x1x
prime
10 15
minterms
implicants
0
p2
x0x0
X
X
p3
x01x
X
X
p4
xx11
X
p5
1x1x
X
X
3
10 15
X
X
X
X
X
X
Column 10 dominates column 0

Min Cover That Excludes p1



Row p4 dominates rows p3 and p5
Min cover C = { p2, p4 }
f = x0x0 + xx11 = B'D' + CD


Cost = 3 + 6 = 9
Same cost as other solution!
prime
minterms
implicants
0
3
p2
x0x0
X
p3
x01x
X
p4
xx11
X
p5
1x1x
15
X
X
Factoring and Decomposition

Factoring

Two level logic is good for problems with only a few
variables and/or limited fan-in to all gates


Factoring can be used to reduce fan-in, but results in > 2
level logic
Ex:

f(A,B,C,D,E,F) = AB'CD'EF + ABC'D'E'F


If only 4 input AND gates are available then we can factor f
into f(A,B,C,D,E,F) = AD'F(B'CE + BC'E')


Requires two 6 input ANDs, one 2 input OR
Requires one 4 input AND, two 3 input ANDs, one 2 input OR
Requires 3 level logic  multi-level synthesis

Multi-level Synthesis

If fan-in is limited to at most 4, then the original two
level circuit could be done as ….
cost = 5 + 16 = 21
cascading ANDs

The factored circuit could be done as …
cost = 4 + 12 = 16

Functional Decomposition

Multi-level circuits may sometimes be preferred over
two-level circuits due to reduced cost


Done at the expense of longer delays
Decompose circuits into subcircuits with shared
functionality

Shared subcircuits provide reduced total cost

Example Functional Decomposition

SOP form: f = A'BC + AB'C + ABD + A'B'D


cost = 5 + 16 = 21 + 2 inverters + 2 fan-in = 25
factoring …
f = (A'B + AB') C + (AB + A'B') D
= g C + g' D where g = A'B + AB'
since g' = (A'B + AB')' = (A'B)'(AB')' = (A+B')(A'+B)
= AA' + B'A' + AB + B'B = AB + A'B'

Implemented Decomposition

cost = 6 + 12 + 3 inverters + 3 fan-in = 24

Another Example
YZ
WX
00
01
11
10
YZ
WX
00
01
11
10
00
1
0
0
0
00
0
0
0
0
01
0
1
1
1
01
1
1
1
1
11
1
0
0
0
11
0
0
0
0
10
0
1
1
1
10
1
1
1
1
V=0

V=1
The shaded region is g(X,W) = XW' + X'W

f(V,W,X,Y,Z) = g (V+Y+Z) + g' (V+Y+Z)' = g h + g' h'

Implementation

cost = 11 + 19 = 30

minimal SOP cost = 55 (includes inverters)