# Copyright 2001, Agrawal & BushnellVLSI Test: Lecture 13/12alt1 Lecture 13 Sequential Circuit ATPG Time-Frame Expansion (Lecture 12alt in the Alternative.

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Copyright 2001, Agrawal & BushnellVLSI Test: Lecture 13/12alt1 Lecture 13 Sequential Circuit ATPG Time-Frame Expansion (Lecture 12alt in the Alternative Sequence) n Problem of sequential circuit ATPG n Time-frame expansion n Nine-valued logic n ATPG implementation and drivability n Complexity of ATPG n Cycle-free and cyclic circuits n Asynchronous circuits n Summary and Exercise

Copyright 2001, Agrawal & BushnellVLSI Test: Lecture 13/12alt2 Sequential Circuits n A sequential circuit has memory in addition to combinational logic. n Test for a fault in a sequential circuit is a sequence of vectors, which n Initializes the circuit to a known state n Activates the fault, and n Propagates the fault effect to a primary output n Methods of sequential circuit ATPG n Time-frame expansion methods n Simulation-based methods

Copyright 2001, Agrawal & BushnellVLSI Test: Lecture 13/12alt3 Example: A Serial Adder FF AnAn BnBn CnCn C n+1 SnSn s-a-0 1 1 1 1 1 X X X D D Combinational logic

Copyright 2001, Agrawal & BushnellVLSI Test: Lecture 13/12alt4 Time-Frame Expansion AnAn BnBn FF CnCn C n+1 1 X X SnSn s-a-0 1 1 1 1 D D Combinational logic S n-1 s-a-0 1 1 1 1 X D D Combinational logic C n-1 1 1 D D X A n-1 B n-1 Time-frame -1 Time-frame 0

Copyright 2001, Agrawal & BushnellVLSI Test: Lecture 13/12alt5 Concept of Time-Frames n If the test sequence for a single stuck-at fault contains n vectors, n Replicate combinational logic block n times n Place fault in each block n Generate a test for the multiple stuck-at fault using combinational ATPG with 9-valued logic Comb. block Fault Time- frame 0 Time- frame Time- Frame - n+1 Unknown or given Init. state Vector 0Vector – 1 Vector – n +1 PO 0 PO – 1 PO – n +1 State variables Next state

Copyright 2001, Agrawal & BushnellVLSI Test: Lecture 13/12alt6 Example for Logic Systems FF2 FF1 A B s-a-1

Copyright 2001, Agrawal & BushnellVLSI Test: Lecture 13/12alt7 Five-Valued Logic (Roth) 0,1, D, D, X A B X X X 0 s-a-1 D A B X X X 0 D FF1 FF2 D D Time-frame -1 Time-frame 0

Copyright 2001, Agrawal & BushnellVLSI Test: Lecture 13/12alt8 Nine-Valued Logic (Muth) 0,1, 1/0, 0/1, 1/X, 0/X, X/0, X/1, X A B X X X 0 s-a-1 0/1 A B 0/X 0/1 X s-a-1 X/1 FF1 FF2 0/1 X/1 Time-frame -1 Time-frame 0

Copyright 2001, Agrawal & BushnellVLSI Test: Lecture 13/12alt9 Implementation of ATPG n Select a PO for fault detection based on drivability analysis. n Place a logic value, 1/0 or 0/1, depending on fault type and number of inversions. n Justify the output value from PIs, considering all necessary paths and adding backward time-frames. n If justification is impossible, then use drivability to select another PO and repeat justification. n If the procedure fails for all reachable POs, then the fault is untestable. n If 1/0 or 0/1 cannot be justified at any PO, but 1/X or 0/X can be justified, the the fault is potentially detectable.

Copyright 2001, Agrawal & BushnellVLSI Test: Lecture 13/12alt10 Drivability Example d(0/1) = 4 d(1/0) = (CC0, CC1) = (6, 4) s-a-1 (4, 4) (10, 15) (11, 16) (10, 16) (22, 17) (17, 11) (5, 9) d(0/1) = 9 d(1/0) = d(0/1) = 109 d(1/0) = d(0/1) = 120 d(1/0) = 27 d(0/1) = d(1/0) = 32 (6, 10) 8 8 8 8 FF d(0/1) = d(1/0) = 20 8 CC0 and CC1 are SCOAP combinational controllabilities d(0/1) and d(1/0) of a line are effort measures for driving a specific fault effect to that line

Copyright 2001, Agrawal & BushnellVLSI Test: Lecture 13/12alt11 Complexity of ATPG  Synchronous circuit -- All flip-flops controlled by clocks; PI and PO synchronized with clock:  Cycle-free circuit – No feedback among flip-flops: Test generation for a fault needs no more than dseq + 1 time- frames, where dseq is the sequential depth.  Cyclic circuit – Contains feedback among flip-flops: May need 9 Nff time-frames, where Nff is the number of flip- flops.  Asynchronous circuit – Higher complexity! Time- Frame 0 Time- Frame max-1 Time- Frame max-2 Time- Frame -2 Time- Frame S0S1 S2 S3 Smax max = Number of distinct vectors with 9-valued elements = 9 Nff

Copyright 2001, Agrawal & BushnellVLSI Test: Lecture 13/12alt12 Cycle-Free Circuits n Characterized by absence of cycles among flip- flops and a sequential depth, dseq. n dseq is the maximum number of flip-flops on any path between PI and PO. n Both good and faulty circuits are initializable. n Test sequence length for a fault is bounded by dseq + 1.

Copyright 2001, Agrawal & BushnellVLSI Test: Lecture 13/12alt13 Cycle-Free Example F1 F2 F3 Level = 1 2 F1 F2 F3 Level = 1 2 3 3 dseq = 3 s - graph Circuit All faults are testable in this circuit.

Copyright 2001, Agrawal & BushnellVLSI Test: Lecture 13/12alt14 Cyclic Circuit Example F1 F2 CNT Z Modulo-3 counter s - graph F1 F2

Copyright 2001, Agrawal & BushnellVLSI Test: Lecture 13/12alt15 Modulo-3 Counter n Cyclic structure – Sequential depth is undefined. n Circuit is not initializable. No tests can be generated for any stuck-at fault. n After expanding the circuit to 9 Nff = 81, or fewer, time-frames ATPG program calls any given target fault untestable. n Circuit can only be functionally tested by multiple observations. n Functional tests, when simulated, give no fault coverage.

Copyright 2001, Agrawal & BushnellVLSI Test: Lecture 13/12alt16 Adding Initializing Hardware F1 F2 CNT Z Initializable modulo-3 counter s - graph F1 F2 CLR s-a-0 s-a-1 Untestable fault Potentially detectable faults

Copyright 2001, Agrawal & BushnellVLSI Test: Lecture 13/12alt17 Benchmark Circuits Circuit PI PO FF Gates Structure Seq. depth Total faults Detected faults Potentially detected faults Untestable faults Abandoned faults Fault coverage (%) Fault efficiency (%) Max. sequence length Total test vectors Gentest CPU s (Sparc 2) s1196 14 18 529 Cycle-free 4 1242 1239 0 3 0 99.8 100.0 3 313 10 s1238 14 18 508 Cycle-free 4 1355 1283 0 72 0 94.7 100.0 3 308 15 s1488 8 19 6 653 Cyclic -- 1486 1384 2 26 76 93.1 94.8 24 525 19941 s1494 8 19 6 647 Cyclic -- 1506 1379 2 30 97 91.6 93.4 28 559 19183

Copyright 2001, Agrawal & BushnellVLSI Test: Lecture 13/12alt18 Asynchronous Circuit n An asynchronous circuit contains unclocked memory often realized by combinational feedback. n Almost impossible to build, let alone test, a large asynchronous circuit. n Clock generators, signal synchronizers, flip-flops are typical asynchronous circuits. n Many large synchronous systems contain small portions of localized asynchronous circuitry. n Sequential circuit ATPG should be able to generate tests for circuits with limited asynchronous parts, even if it does not detect faults in those parts.

Copyright 2001, Agrawal & BushnellVLSI Test: Lecture 13/12alt19 Asynchronous Model Clocked Flip-flops Feedback delays Synchronous PIs Synchronous POs System Clock, CK Fast model Clock, FMCK CK Feedback-free Combinational Logic C Combinational Feedback Paths: Feedback set Modeling circuit is Shown in orange. PPOPPI

Copyright 2001, Agrawal & BushnellVLSI Test: Lecture 13/12alt20 Time-Frame Expansion Time-frame k Time-frame -k+1 Time-frame -k-1 C FMCK C FMCK C FMCK C CK Asynchronous feedback stabilization PI PO Feedback set PPI PPO Feedback set Vector k

Copyright 2001, Agrawal & BushnellVLSI Test: Lecture 13/12alt21 Asynchronous Example s-a-0 s-a-1 1010 1111 0000 0101 Vectors 1 2 3 4 101101 XX0XX0 101101 101101 Outputs 1 2 3 4 Gentest results: Faults: total 23, detected 15, untestable 8 (shown in red), potentially detectable none Vectors: 4 Sparc 2 CPU time: test generation 33ms, fault simulation 16ms

Copyright 2001, Agrawal & BushnellVLSI Test: Lecture 13/12alt22 Summary n Combinational ATPG algorithms are extended:  Time-frame expansion unrolls time as combinational array  Nine-valued logic system  Justification via backward time n Cycle-free circuits:  Require at most dseq + 1 time-frames  Always initializable n Cyclic circuits:  May need 9 Nff time-frames  Circuit must be initializable  Partial scan can make circuit cycle-free (Chapter 14) n Asynchronous circuits:  High complexity  Low coverage and unreliable tests  Simulation-based methods are more useful (Section 8.3)

Copyright 2001, Agrawal & BushnellVLSI Test: Lecture 13/12alt23 Exercise n Which type of circuit is easier to test? Circle one in each: n Combinational or sequential n Cyclic or cycle-free n Synchronous or asynchronous n What is the maximum number of input vectors that may be needed to initialize a cycle-free circuit with k flip-flops?

Copyright 2001, Agrawal & BushnellVLSI Test: Lecture 13/12alt24 Answers to Exercise n Which type of circuit is easier to test? Circle one in each: n Combinational or sequential n Cyclic or cycle-free n Synchronous or asynchronous n What is the maximum number of input vectors that may be needed to initialize a cycle-free circuit with k flip-flops? k vectors. Because that is the maximum sequential depth possible. An example is a k bit shift register.

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