Glitches & Hazards

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

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: Dynamic hazards: 1 ® 1 ® 1 1 1 ® ® 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 F = AB + A'C + BC No longer minimum form  but no hazard now B C A 00 01 11 10 1

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

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

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

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 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 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

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

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, 1x11, 11x1, xx00 }

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

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

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

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

Final Cover The final cover is C = { p6, p2, p5 } f = xx00 + 101x + 11x1 = C'D' + AB'C + ABD prime minterms implicants 10 13 p2 101x X p4 1x11 p5 11x1

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

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, 0xx0, x0x0, xx00, 01xx }

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

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

Column Domination Column 10 dominates column 2 Final min cover is C = { p6, p7, p3, p5 } f = xx00 + 01xx + 1x11 + x0x0 = C'D' + A'B + ACD + B'D' prime minterms implicants 2 10 11 15 p1 101x X p3 1x11 p5 x0x0

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

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 }

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

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

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 15 p2 x0x0 X p3 x01x p4 xx11 p5 1x1x

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

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 3 15 p2 x0x0 X p3 x01x p4 xx11 p5 1x1x

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 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

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 = 5 + 16 = 21 cascading ANDs 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 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

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