Lecture 12 Advanced Combinational ATPG Algorithms

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Lecture 12 Advanced Combinational ATPG Algorithms FAN – Multiple Backtrace (1983) TOPS – Dominators (1987) SOCRATES – Learning (1988) Legal Assignments (1990) EST – Search space learning (1991) BDD Test generation (1991) Implication Graphs and Transitive Closure (1988 - 97) Recursive Learning (1995) Test Generation Systems Test Compaction Summary Copyright 2001, Agrawal & Bushnell VLSI Test: Lecture 12

FAN -- Fujiwara and Shimono(1983) FANout – oreineted test generaton New concepts: Immediate assignment of uniquely-determined signals Unique sensitization Stop Back trace at head lines Multiple Back trace -Rather than stopping at PIs, back tracing in FAN may stop at internal lines. (head lines) - Rather than trying to satisfy one objective, FAN uses a multiple-back trace procedure that attempts to simultaneously satisfy a set of objectives Copyright 2001, Agrawal & Bushnell VLSI Test: Lecture 12

PODEM Fails to Determine Unique Signals Behavior of PODEM Backtracing operation fails to set all 3 inputs of gate L to 1 Causes unnecessary search Copyright 2001, Agrawal & Bushnell VLSI Test: Lecture 12

FAN -- Early Determination of Unique Signals Determine all unique signals implied by current decisions immediately Avoids unnecessary search Copyright 2001, Agrawal & Bushnell VLSI Test: Lecture 12

PODEM Makes Unwise Signal Assignments Blocks fault propagation due to assignment J = 0 Copyright 2001, Agrawal & Bushnell VLSI Test: Lecture 12

Unique Sensitization of FAN with No Search Path over which fault is uniquely sensitized FAN immediately sets necessary signals to propagate fault Copyright 2001, Agrawal & Bushnell VLSI Test: Lecture 12

Headlines HL FL BL Headlines H and J separate circuit into 3 parts, for which test generation can be done independently Copyright 2001, Agrawal & Bushnell VLSI Test: Lecture 12

Bound Line: is a line, that is reachable from at least one stem. Free Line : is a line that is not bound. Head Line : is a free line that directly feeds a bound line , this line can be justified to logic 0 or 1 from the headline back to the circuit PIs. Stem A, B, C, E, F,G are free lines Fig.6.32 Copyright 2001, Agrawal & Bushnell VLSI Test: Lecture 12

Contrasting Decision Trees FAN decision tree PODEM decision tree Copyright 2001, Agrawal & Bushnell VLSI Test: Lecture 12

Multiple Backtrace PODEM – Depth-first search 6 times FAN – breadth-first passes – 1 time PODEM – depth-first passes – 6 times Copyright 2001, Agrawal & Bushnell VLSI Test: Lecture 12 PODEM – Depth-first search 6 times

AND Gate Vote Propagation [5, 3] [0, 3] [5, 3] [0, 3] [0, 3] AND Gate X Easiest-to-control Input – # 0’s = OUTPUT # 0’s # 1’s = OUTPUT # 1’s All other inputs -- # 0’s = 0 Copyright 2001, Agrawal & Bushnell VLSI Test: Lecture 12

Multiple Backtrace Fanout Stem Voting [5, 1] P is a stem not reachable From the fault Site ; If n0(p)>0 , and n1(p)>0 There is a conflict On P FAN sets P to 0 if n0(p)>n1(p) and otherwise sets P=1 FAN immediately does a forward implication from P , and find a signal conflict much sooner than a single backtrace procedure. [1, 1] [3, 2] [18, 6] [4, 1] [5, 1] Fanout Stem -- # 0’ s = S Branch # 0’s, # 1’s = S Branch # 1’s Copyright 2001, Agrawal & Bushnell VLSI Test: Lecture 12

Multiple Backtrace Algorithm repeat remove entry (s, vs) from current_objectives; If (s is head_objective) add (s, vs) to head_objectives; else if (s not fanout stem and not PI) vote on gate s inputs; if (gate s input I is fanout branch) vote on stem driving I; add stem driving I to stem_objectives; else add I to current_objectives; Copyright 2001, Agrawal & Bushnell VLSI Test: Lecture 12

Rest of Multiple Backtrace if (stem_objectives not empty) (k, n0 (k), n1 (k)) = highest level stem from stem_objectives; if (n0 (k) > n1 (k)) vk = 0; else vk = 1; if ((n0 (k) != 0) && (n1 (k) != 0) && (k not in fault cone)) return (k, vk); add (k, vk) to current_objectives; return (multiple_backtrace (current_objectives)); remove one objective (k, vk) from head_objectives; Copyright 2001, Agrawal & Bushnell VLSI Test: Lecture 12

Multiple trace generating Conflicting value on a stem Example 6.11 for following circuit `execute multiple trace algo. Starting with (I,1) and (J,0) as current objectives. 1 (I, 1) (J, 0 ) Copyright 2001, Agrawal & Bushnell VLSI Test: Lecture 12

1 Step 1 1 Current Objectives processed entry stem objectives Head objectives (I,1), (J,0) (I,1) 1 Step 2 Current Objectives processed entry stem objectives Head objectives (G,0), (J,0) (J,0) Copyright 2001, Agrawal & Bushnell VLSI Test: Lecture 12

1 Step 3 1 Current Objectives processed entry stem objectives Head objectives (G,0), (H,1) (G,0) 1 1 1 Step 4 1 Current Objectives processed entry stem objectives Head objectives (H,1), (A1,1) , (E1,1) (H,1) Copyright 2001, Agrawal & Bushnell VLSI Test: Lecture 12

1 1 1 Step 5 1 1 1 Current Objectives processed entry stem objectives Head objectives (A1,1), (E1,1), (E2,1),(C,1) (A1,1) A 1 1 1 1 Step 6 1 1 1 Current Objectives processed entry stem objectives Head objectives (E1,1), (E2,1) , (C,1) (E1,1) A,E Copyright 2001, Agrawal & Bushnell VLSI Test: Lecture 12

1 1 1 Step 7 1 1 1 Current Objectives processed entry stem objectives Head objectives (E2,1),(C,1) (E2,1) A,E 1 1 1 Step 8 1 1 1 Current Objectives processed entry stem objectives Head objectives (C,1) (C,1) A,E C Copyright 2001, Agrawal & Bushnell VLSI Test: Lecture 12

1 1 1 Step 9 1 1 1 Current Objectives processed entry stem objectives Head objectives Ø A,E C 1 1 1 Stem objectives 1 Step 10 1 1 1 Current Objectives processed entry stem objectives Head objectives (E,1) (E,1) A C Copyright 2001, Agrawal & Bushnell VLSI Test: Lecture 12

1 1 1 Step 11 1 1 1 1 Current Objectives processed entry stem objectives Head objectives (A2,0) (A2,0) A C 1 1 1 1 1 Step 12 1 (A, v) V ε [0,1] 1 1 Current Objectives processed entry stem objectives Head objectives Ø A C Copyright 2001, Agrawal & Bushnell VLSI Test: Lecture 12

TOPS – Dominators Kirkland and Mercer (1987) Tops found even more assignments than FAN Using Dominators A Dominator is a circuit signal through witch the Fault effect has to pass in order to be detected at A particular PO An absolute dominator is a dominator through the Fault effect has to pass to be detected at any PO Copyright 2001, Agrawal & Bushnell VLSI Test: Lecture 12

l and n are absolute dominator of A k and n are absolute dominator of B g,k and n are absolute dominator of C k and n are absolute dominator of D m and n are absolute dominator of E Relative – dominates only paths to a given PO If dominator of fault becomes 0 or 1, backtrack

SOCRATES Learning (1988) Static and dynamic learning: a = 1 f = 1 means that we learn f = 0 a = 0 by applying the Boolean contrapositive theorem Set each signal first to 0, and then to 1 Discover implications Learning criterion: remember f = vf only if: f = vf requires all inputs of f to be non-controlling A forward implication contributed to f = vf Copyright 2001, Agrawal & Bushnell VLSI Test: Lecture 12

Improved Unique Sensitization Procedure When a is only D-frontier signal, find dominators of a and set their inputs unreachable from a to 1 Find dominators of single D-frontier signal a and make common input signals non-controlling(b=1) Copyright 2001, Agrawal & Bushnell VLSI Test: Lecture 12

Constructive Dilemma [(a = 0) (i = 0)] [(a = 1) (i = 0)] (i = 0) If both assignments 0 and 1 to a make i = 0, then i = 0 is implied independently of a Copyright 2001, Agrawal & Bushnell VLSI Test: Lecture 12

Modus Tollens and Dynamic Dominators (f = 1) [(a = 0) (f = 0)] (a = 1) Dynamic dominators: Compute dominators and dynamically learned implications after each decision step Too computationally expensive Copyright 2001, Agrawal & Bushnell VLSI Test: Lecture 12

EST – Dynamic Programming (Giraldi & Bushnell) Equivalent STate hashing E-frontier – partial circuit functional decomposition Equivalent to a node in a BDD Cut-set between circuit part with known labels and part with X signal labels EST learns E-frontiers during ATPG and stores them in a hash table Dynamic programming – when new decomposition generated from implications of a variable assignment, looks it up in the hash table Avoids repeating a search already conducted Terminates search when decomposition matches: Earlier one that lead to a test (retrieves stored test) Earlier one that lead to a backtrack Accelerated SOCRATES nearly 5.6 times Copyright 2001, Agrawal & Bushnell VLSI Test: Lecture 12

Fault B sa1 X Copyright 2001, Agrawal & Bushnell VLSI Test: Lecture 12

Fault h sa1 X Copyright 2001, Agrawal & Bushnell VLSI Test: Lecture 12

Implication Graph ATPG Chakradhar et al. (1990) Model logic behavior using implication graphs Nodes for each literal and its complement Arc from literal a to literal b means that if a = 1 then b must also be 1 Extended to find implications by using a graph transitive closure algorithm – finds paths of edges Made much better decisions than earlier ATPG search algorithms Uses a topological graph sort to determine order of setting circuit variables during ATPG Copyright 2001, Agrawal & Bushnell VLSI Test: Lecture 12

Example and Implication Graph Copyright 2001, Agrawal & Bushnell VLSI Test: Lecture 12

Transitive Closure D E Given a digraph G, the transitive closure of G is the digraph G* such that G* has the same vertices as G if G has a directed path from u to v (u  v), G* has a directed edge from u to v The transitive closure provides reachability information about a digraph B G C A D E B C A G* Directed Graphs

Graph Transitive Closure When d set to 0, add edge from d to d, which means that if d is 1, there is conflict Can deduce that (a = a , F=F , b=b) a=b=f=1 When d set to 1, add edge from d to d Copyright 2001, Agrawal & Bushnell VLSI Test: Lecture 12

Consequence of F = 1 Boolean false function F (inputs d and e) has deF For F = 1, add edge F F so deF reduces to d e To cause de = 0 we add edges: e d and d e Now, we find a path in the graph b b So b cannot be 0, or there is a conflict Therefore, b = 1 is a consequence of F = 1 Copyright 2001, Agrawal & Bushnell VLSI Test: Lecture 12

f=1 A=b =c=1 Copyright 2001, Agrawal & Bushnell VLSI Test: Lecture 12

Related Contributions Larrabee – NEMESIS -- Test generation using satisfiability and implication graphs Chakradhar, Bushnell, and Agrawal – NNATPG – ATPG using neural networks & implication graphs Chakradhar, Agrawal, and Rothweiler – TRAN --Transitive Closure test generation algorithm Cooper and Bushnell – Switch-level ATPG Agrawal, Bushnell, and Lin – Redundancy identification using transitive closure Stephan et al. – TEGUS – satisfiability ATPG Henftling et al. and Tafertshofer et al. – ANDing node in implication graphs for efficient solution Copyright 2001, Agrawal & Bushnell VLSI Test: Lecture 12

Recursive Learning Kunz and Pradhan (1992) Applied SOCRATES type learning recursively Maximum recursion depth rmax determines what is learned about circuit Time complexity exponential in rmax Memory grows linearly with rmax Copyright 2001, Agrawal & Bushnell VLSI Test: Lecture 12

Recursive_Learning Algorithm for each unjustified line for each input: justification assign controlling value; make implications and set up new list of unjustified lines; if (consistent) Recursive_Learning (); if (for all signals f with same value V for all consistent justifications) learn f = V; make implications for all learned values; if (all justifications inconsistent) learn current value assignments as consistent; Copyright 2001, Agrawal & Bushnell VLSI Test: Lecture 12

Recursive Learning i1 = 0 and j = 1 unjustifiable – enter learning a1 h h1 a2 b2 e2 f2 c2 g2 i2 j = 1 d2 h2 k Copyright 2001, Agrawal & Bushnell VLSI Test: Lecture 12

Justify i1 = 0 Choose first of 2 possible assignments g1 = 0 a1 a b b1 d d1 h h1 a2 b2 e2 f2 c2 g2 i2 j = 1 d2 h2 k Copyright 2001, Agrawal & Bushnell VLSI Test: Lecture 12

Implies e1 = 0 and f1 = 0 Given that g1 = 0 a1 a e1 = 0 b b1 c1 c j = 1 d2 h2 k Copyright 2001, Agrawal & Bushnell VLSI Test: Lecture 12

Justify a1 = 0, 1st Possibility Given that g1 = 0, one of two possibilities a1 = 0 a e1 = 0 b b1 c1 c g1 = 0 i1 = 0 d d1 f1 = 0 h h1 a2 b2 e2 f2 c2 g2 i2 j = 1 d2 h2 k Copyright 2001, Agrawal & Bushnell VLSI Test: Lecture 12

Implies a2 = 0 Given that g1 = 0 and a1 = 0 a1 = 0 a e1 = 0 b b1 c1 c f1 = 0 h h1 a2 = 0 b2 e2 f2 c2 g2 i2 j = 1 d2 h2 k Copyright 2001, Agrawal & Bushnell VLSI Test: Lecture 12

Implies e2 = 0 Given that g1 = 0 and a1 = 0 a1 = 0 a e1 = 0 b b1 c1 c f1 = 0 h h1 a2 = 0 e2 = 0 b2 f2 c2 g2 i2 j = 1 d2 h2 k Copyright 2001, Agrawal & Bushnell VLSI Test: Lecture 12

Now Try b1 = 0, 2nd Option Given that g1 = 0 a1 a e1 = 0 b b1 = 0 c1 c f1 = 0 h h1 a2 e2 b2 f2 c2 g2 i2 j = 1 d2 h2 k Copyright 2001, Agrawal & Bushnell VLSI Test: Lecture 12

Implies b2 = 0 and e2 = 0 Given that g1 = 0 and b1 = 0 a1 a e1 = 0 b c1 c g1 = 0 i1 = 0 d d1 f1 = 0 h h1 a2 e2 = 0 b2 = 0 f2 c2 g2 i2 j = 1 d2 h2 k Copyright 2001, Agrawal & Bushnell VLSI Test: Lecture 12

Both Cases Give e2 = 0, So Learn That d d1 f1 = 0 h h1 a2 e2 = 0 b2 f2 c2 g2 i2 j = 1 d2 h2 k Copyright 2001, Agrawal & Bushnell VLSI Test: Lecture 12

Justify f1 = 0 Try c1 = 0, one of two possible assignments a1 a e1 = 0 d d1 f1 = 0 h h1 a2 e2 = 0 b2 f2 c2 g2 i2 j = 1 d2 h2 k Copyright 2001, Agrawal & Bushnell VLSI Test: Lecture 12

Implies c2 = 0 Given that c1 = 0, one of two possibilities a1 a e1 = 0 d d1 f1 = 0 h h1 a2 e2 = 0 b2 c2 = 0 f2 g2 i2 j = 1 d2 h2 k Copyright 2001, Agrawal & Bushnell VLSI Test: Lecture 12

Implies f2 = 0 Given that c1 = 0 and g1 = 0 a1 a e1 = 0 b b1 c1 = 0 c j = 1 d2 f2 = 0 h2 k Copyright 2001, Agrawal & Bushnell VLSI Test: Lecture 12

Try d1 = 0 Try d1 = 0, second of two possibilities a1 a e1 = 0 b b1 c1 g1 = 0 i1 = 0 d d1 = 0 f1 = 0 h h1 a2 e2 = 0 b2 f2 c2 g2 i2 j = 1 d2 h2 k Copyright 2001, Agrawal & Bushnell VLSI Test: Lecture 12

Implies d2 = 0 Given that d1 = 0 and g1 = 0 a1 a e1 = 0 b b1 c1 c f1 = 0 h h1 a2 e2 = 0 b2 f2 c2 g2 i2 j = 1 d2 = 0 h2 k Copyright 2001, Agrawal & Bushnell VLSI Test: Lecture 12

Implies f2 = 0 Given that d1 = 0 and g1 = 0 a1 a e1 = 0 b b1 c1 c j = 1 d2 = 0 f2 = 0 h2 k Copyright 2001, Agrawal & Bushnell VLSI Test: Lecture 12

Since f2 = 0 In Either Case, Learn f2 = 0 b b1 c1 c g1 = 0 i1 = 0 d d1 f1 h h1 a2 e2 = 0 b2 c2 g2 i2 j = 1 d2 f2 = 0 h2 k Copyright 2001, Agrawal & Bushnell VLSI Test: Lecture 12

Implies g2 = 0 a1 a e1 b b1 c1 c g1 = 0 i1 = 0 d d1 f1 h h1 a2 e2 = 0 j = 1 d2 f2 = 0 h2 k Copyright 2001, Agrawal & Bushnell VLSI Test: Lecture 12

Implies i2 = 0 and k = 1 a1 a e1 b b1 c1 c g1 = 0 i1 = 0 d d1 f1 h h1 j = 1 d2 f2 = 0 h2 k = 1 Copyright 2001, Agrawal & Bushnell VLSI Test: Lecture 12

Justify h1 = 0 Second of two possibilities to make i1 = 0 a1 a b b1 e1 g1 i1 = 0 d d1 h h1 = 0 a2 b2 e2 f2 c2 g2 i2 j = 1 d2 h2 k Copyright 2001, Agrawal & Bushnell VLSI Test: Lecture 12

Implies h2 = 0 Given that h1 = 0 a1 a b b1 e1 c1 f1 c g1 i1 = 0 d d1 h j = 1 d2 h2 = 0 k Copyright 2001, Agrawal & Bushnell VLSI Test: Lecture 12

Implies i2 = 0 and k = 1 Given 2nd of 2 possible assignments h1 = 0 a1 c1 f1 c g1 i1 = 0 d d1 h h1 = 0 a2 b2 e2 f2 c2 g2 i2 = 0 j = 1 d2 h2 = 0 k = 1 Copyright 2001, Agrawal & Bushnell VLSI Test: Lecture 12

Both Cases Cause k = 1 (Given j = 1), i2 = 0 Therefore, learn both independently a1 a b b1 e1 c1 f1 c g1 i1 = 0 d d1 h h1 a2 b2 e2 f2 c2 g2 i2 = 0 j = 1 d2 h2 k = 1 Copyright 2001, Agrawal & Bushnell VLSI Test: Lecture 12

Other ATPG Algorithms Legal assignment ATPG (Rajski and Cox) Maintains power-set of possible assignments on each node {0, 1, D, D, X} BDD-based algorithms Catapult (Gaede, Mercer, Butler, Ross) Tsunami (Stanion and Bhattacharya) – maintains BDD fragment along fault propagation path and incrementally extends it Unable to do highly reconverging circuits (parallel multipliers) because BDD essentially becomes infinite Copyright 2001, Agrawal & Bushnell VLSI Test: Lecture 12

Fault Coverage and Efficiency # of detected faults Total # faults Fault coverage = Fault efficiency # of detected faults Total # faults -- # undetectable faults = Copyright 2001, Agrawal & Bushnell VLSI Test: Lecture 12

Test Generation Systems Circuit Description Fault List Compacter SOCRATES With fault simulator Aborted Faults Test Patterns Backtrack Distribution Undetected Faults Redundant Faults Copyright 2001, Agrawal & Bushnell VLSI Test: Lecture 12

Test Compaction Fault simulate test patterns in reverse order of generation ATPG patterns go first Randomly-generated patterns go last (because they may have less coverage) When coverage reaches 100%, drop remaining patterns (which are the useless random ones) Significantly shortens test sequence – economic cost reduction Copyright 2001, Agrawal & Bushnell VLSI Test: Lecture 12

Static and Dynamic Compaction of Sequences Static compaction ATPG should leave unassigned inputs as X Two patterns compatible – if no conflicting values for any PI Combine two tests ta and tb into one test tab = ta tb using D-intersection Detects union of faults detected by ta & tb Dynamic compaction Process every partially-done ATPG vector immediately Assign 0 or 1 to PIs to test additional faults Copyright 2001, Agrawal & Bushnell VLSI Test: Lecture 12

Compaction Example t1 = 0 1 X t2 = 0 X 1 t3 = 0 X 0 t4 = X 0 1 Combine t1 and t3, then t2 and t4 Obtain: t13 = 0 1 0 t24 = 0 0 1 Test Length shortened from 4 to 2 Copyright 2001, Agrawal & Bushnell VLSI Test: Lecture 12

Summary Test Bridging, Stuck-at, Delay, & Transistor Faults Must handle non-Boolean tri-state devices, buses, & bidirectional devices (pass transistors) Hierarchical ATPG -- 9 Times speedup (Min) Handles adders, comparators, MUXes Compute propagation D-cubes Propagate and justify fault effects with these Use internal logic description for internal faults Results of 40 years research – mature – methods: Path sensitization Simulation-based Boolean satisfiability and neural networks Copyright 2001, Agrawal & Bushnell VLSI Test: Lecture 12

Problems 7.2 , 7.3, 7.5 , 7.7, 7.10,7.12, 7.21, 7,22 Copyright 2001, Agrawal & Bushnell VLSI Test: Lecture 12