Unformatted text preview: 202 VLSI Test Principles and Architectures appearing on lines a , e 1 , e 2 , etc., would also be blocked due to implications of b = 1. The unobservable information could be propagated backwards until a fanout stem is reached at which the faults on the fanout stem may no longer be unobservable. The condition for checking if the fanout stem is unobservable is to see if the stem can reach any of the blocking conditions for each of the fanout branches. For instance, using the circuit illustrated in Figure 4.35, if a = 1 and c = 0, both fanout branchs b 1 and b 2 would be unobservable. However, because the fanout stem b cannot reach any of the conditions for blocking any of the branches ( i.e. , the blocking condition for b 1 is a = 1 and the blocking condition for b 2 is c = 0), stem b would still be unobservable. The complete set of faults that cannot be propagated due to b = 1 is: a/ a/ 1 e 1 / e 1 / 1 y/ y/ 1 e 2 / e 2 / 1 e/ e/ 1 c/ c/ 1 b 2 / b 2 / 1 Thus, S 1 is the union of the two sets computed above:...
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This note was uploaded on 05/16/2011 for the course ENGINEERIN mp108 taught by Professor Elbarki during the Spring '08 term at Alexandria University.
- Spring '08