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Unformatted text preview: 168 VLSI Test Principles and Architectures w 1 xy = 1 w xy = 1 w x+y = 1 wx + wy = 1 Now, w can be expanded from the circuit shown in the figure to be w = yz. Plugging this into the equation above gives us: wx+wy = 1 yzx+yzy = 1 xyz+yz = 1 yz = 1 Therefore, the set of vectors that can detect w/0 is 001 101 . 4.3.1 Untestable Faults If the target fault is untestable, it would be impossible to satisfy Equation 4.2. Consider the circuit shown in Figure 4.6. Suppose the target fault is z/0. Then the set of vectors that can detect z/0 are those that can satisfy: z df =1 dz z f z = 1 f z = 0 z xy xy = 1 z0 = 1 UNSATISFIABLE df = 1, indicating dz =1 In other words, there exists no input vectors that can satisfy z that the fault z/0 is untestable. x y f z FIGURE 4.6 Example circuit for an untestable fault. ...
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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.

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