1. Glitches & Hazards

2. Glitches / Hazards Gates have an inherent delay
We have been ignoring this  but delays do exist
A glitch is an unwanted pulse at the output of a combinational circuit
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

3. 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

4. Static & Dynamic Hazards
Static hazards:
Dynamic hazards: 1 1 1 1 1 1

5. 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
F = AB + A'C + BC
No longer minimum form  but no hazard now
B C A 00 01 11 10 1

6. Hazard Removal: Example
F(A,B,C,D) = m(6,7,8,9,12,13,14,15)
The red prime implicant removes the static-1 hazard
C D A B 00 01 11 10 1

7. Dynamic Hazards
Dynamic hazards do not occur in two-level combinational circuits  only in multi-level circuits
One gate delay x 1 x , x , x b 2 3 4 x 1 a a f x 2 c d b x 3 x c 4 d f

8. 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 x 1 a f x 2 c x d 3 x 4

9. 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

10. Quine-McClusky Method

11. 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

12. 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
0000 4
0100 8
1000 10
1010 12
1100 11
1011 13
1101 15
1111

13. Forming 1-cubes
Compare each 0-cube in one group with each 0-cube in an adjacent group
Use combining property to form 1-cubes
1 cubes
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
0 cubes
0000 X 4
0100 8
1000 10
1010 12
1100 11
1011 13
1101 15
1111

14. Forming 2-cubes
Compare each 1-cube in one group with each 1-cube in an adjacent group
Use combining property to form 2-cubes
1 cubes
0,4
0x00 X
0,8
x000
8,10
10x0
4,12
x100
8,12
1x00
10,11
101x
12,13
110x
11,15
1x11
13,15
11x1
2 cubes
0,8,4,12
xx00
mark each 1-cube that contributes to a 2-cube

15. The Prime Implicants All “unchecked” cubes are prime implicants
They did not get combined into a larger cube
0 cubes
0000 X 4
0100 8
1000 10
1010 12
1100 11
1011 13
1101 15
1111
1 cubes
0,4
0x00 X
0,8
x000
8,10
10x0
4,12
x100
8,12
1x00
10,11
101x
12,13
110x
11,15
1x11
13,15
11x1
2 cubes
0,8,4,12
xx00
All prime implicants are:
P = { 10x0, 101x, 110x, x11, 11x1, xx00 }

16. Prime Implicant Cover Table
For each prime implicant, mark all minterms covered by that implicant
prime
minterms
implicants 4 8 10 11 12 13 15 p1
10x0 X p2
101x p3
110x p4
1x11 p5
11x1 p6
xx00

17. 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
minterms
implicants 4 8 10 11 12 13 15 p1
10x0 X p2
101x p3
110x p4
1x11 p5
11x1 p6
xx00

18. 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 13 15 p1
10x0 X p2
101x p3
110x p4
1x11 p5
11x1
p5 row dominates p3

19. 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
We could also have considered p2 and p5 as essential at this point
prime
minterms
implicants 10 11 13 15 p2
101x X p4
1x11 p5
11x1
column 15 dominates column 13

20. Final Cover The final cover is C = { p6, p2, p5 }
f = xx x + 11x1 = C'D' + AB'C + ABD
prime
minterms
implicants 10 13 p2
101x X p4
1x11 p5
11x1

21. Another Example f(A,B,C,D) = m(0,2,4,5,6,7,8,10,11,12,15) 0 cubes
0000 2
0010 4
0100 8
1000 5
0101 6
0110 10
1010 12
1100 7
0111 11
1011 15
1111

22. Cubes All prime implicants are:
0,2
00x0 X
0,4
0x00
0,8
x000
2,6
0x10
2,10
x010
4,5
010x
4,6
01x0
4,12
x100
8,10
10x0
8,12
1x00
5,7
01x1
6,7
011x
10,11
101x
7,15
x111
11,15
1x11
0 cubes
0000 X 2
0010 4
0100 8
1000 5
0101 6
0110 10
1010 12
1100 7
0111 11
1011 15
1111
2 cubes
0,2,4,6
0xx0
0,2,8,10
x0x0
0,4,8,12
xx00
4,5,6,7
01xx
All prime implicants are:
P = { 101x, x111, 1x11, xx0, x0x0, xx00, 01xx }

23. Cover Table C = { p6, p7 } since p6 and p7 are essential prime
minterms
implicants 2 4 5 6 7 8 10 11 12 15 p1
101x X p2
x111 p3
1x11 p4
0xx0 p5
x0x0 p6
xx00 p7
01xx

24. Row Dominance Row p5 dominates row p4 Row p3 dominates row p2 prime
minterms
implicants 2 10 11 15 p1
101x X p2
x111 p3
1x11 p4
0xx0 p5
x0x0

25. Column Domination Column 10 dominates column 2
Final min cover is C = { p6, p7, p3, p5 }
f = xx xx + 1x11 + x0x0
= C'D' + A'B + ACD + B'D'
prime
minterms
implicants 2 10 11 15 p1
101x X p3
1x11 p5
x0x0

26. 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
0000 1
0001 2
0010 8
1000 3
0011 10
1010 7
0111 11
1011 14
1110 15
1111

27. Cubes All prime implicants are: P = { 00xx, x0x0, x01x, xx11, 1x1x }
0,1
000x X
0,2
00x0
0,8
x000
1,3
00x1
2,3
001x
2,10
x010
8,10
10x0
3,7
0x11
3,11
x011
10,11
101x
10,14
1x10
7,15
x111
11,15
1x11
14,15
111x
0 cubes
0000 X 1
0001 2
0010 8
1000 3
0011 10
1010 7
0111 11
1011 14
1110 15
1111
2 cubes
0,1,2,3
00xx
0,2,8,10
x0x0
2,3,10,11
x01x
3,7,11,15
xx11
10,11,14,15
1x1x
All prime implicants are:
P = { 00xx, x0x0, x01x, xx11, 1x1x }

28. Cover Table Include only required minterms in the initial cover table
There are no essential prime implicants
prime
minterms
implicants 3 10 15 p1
00xx X p2
x0x0 p3
x01x p4
xx11 p5
1x1x

29. Branching There are no dominant rows and no dominant columns
Require a branching strategy
Backtracking type algorithm
Select C = { p1 }
prime
minterms
implicants 3 10 15 p1
00xx X p2
x0x0 p3
x01x p4
xx11 p5
1x1x

30. 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 = = 9
prime
minterms
implicants 10 15 p2
x0x0 X p3
x01x p4
xx11 p5
1x1x

31. Backtrack Now backtrack and eliminate p1 to find covers prime minterms
implicants 3 10 15 p1
00xx X p2
x0x0 p3
x01x p4
xx11 p5
1x1x
prime
minterms
implicants 3 10 15 p2
x0x0 X p3
x01x p4
xx11 p5
1x1x
Column 10 dominates column 0

32. Min Cover That Excludes p1
Row p4 dominates rows p3 and p5
Min cover C = { p2, p4 }
f = x0x0 + xx11 = B'D' + CD
Cost = = 9
Same cost as other solution!
prime
minterms
implicants 3 15 p2
x0x0 X p3
x01x p4
xx11 p5
1x1x

33. Factoring and Decomposition

34. 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
Requires two 6 input ANDs, one 2 input OR
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 one 4 input AND, two 3 input ANDs, one 2 input OR
Requires 3 level logic  multi-level synthesis

35. Multi-level Synthesis
If fan-in is limited to at most 4, then the original two level circuit could be done as ….
The factored circuit could be done as …
cost = = 21
cascading ANDs
cost = = 16

36. 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

37. Example Functional Decomposition
SOP form: f = A'BC + AB'C + ABD + A'B'D
cost = = 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'

38. Implemented Decomposition
cost = inverters + 3 fan-in = 24

39. Another Example 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'
Y Z W X 00 01 11 10 1
Y Z W X 00 01 11 10 1
V = 0
V = 1

40. Implementation cost = 11 + 19 = 30
minimal SOP cost = 55 (includes inverters)