2007 IEEE International Test Conference 2007
DOI: 10.1109/test.2007.4437578
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Diagnose compound scan chain and system logic defects

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Cited by 17 publications
(19 citation statements)
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“…After the flush test, a SA-fault can be narrowed down to two symmetric nodes (one in the top part and the other in the bottom part) or one node (in the middle part); a missing-edge fault can be narrowed down to some symmetric parts (for example part (2,4,5,8) or (3,6,7,9) in Fig.1); a missing-clock fault can be narrowed down to one candidate, but we cannot distinguish it from the SA-fault in the same position. Therefore, we first give the scheme to distinguish two SA-faults in the symmetric nodes and the missing-clock fault in the top part.…”
Section: A Non-timing Faultsmentioning
confidence: 99%
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“…After the flush test, a SA-fault can be narrowed down to two symmetric nodes (one in the top part and the other in the bottom part) or one node (in the middle part); a missing-edge fault can be narrowed down to some symmetric parts (for example part (2,4,5,8) or (3,6,7,9) in Fig.1); a missing-clock fault can be narrowed down to one candidate, but we cannot distinguish it from the SA-fault in the same position. Therefore, we first give the scheme to distinguish two SA-faults in the symmetric nodes and the missing-clock fault in the top part.…”
Section: A Non-timing Faultsmentioning
confidence: 99%
“…1 as an example, if nodes (2,4,5,8) or (3,6,7,9) are inaccessible, only the first 6 bits during the unloading process for a pattern are captured response bits, while the other four bits are error bits. Based on the values of the 6 observed response bits, we can determine which part (nodes (2,4,5,8) or (3,6,7,9)) is inaccessible. Therefore, for a DTS with inaccessible candidate parts, P1 and P2, we first calculate the expected response of P1 and P2 for each pattern.…”
Section: A Non-timing Faultsmentioning
confidence: 99%
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