198_pdfsam_VLSI TEST PRINCIPLES &amp; ARCHITECTURES

# 198_pdfsam_VLSI TEST PRINCIPLES &amp; ARCHITECTURES - n...

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Test Generation 167 generation, the target fault must be excited. In this example, the logic value on primary input y must be logic 1 to excite the fault y/ 0. Putting these two conditions together, the following equation is obtained: y · f±y = 1 ² f±y = 0 ² = 1 ³ (4.1) Note that f±y = 1 ² f±y = 0 ² indicates the exclusive-or operation on the two functions f±y = 1 ² and f±y = 0 ² ; it evaluates to logic 1 if and only if the two functions evaluate to opposing values. In terms of ATPG, this is synonymous to propagating the fault effect at node y to the primary output f . Therefore, any input vector on primary inputs x , y , and z that can satisfy Equation (4.1) is a valid test vector for fault y/ 0: y · f±y = 1 ² f±y = 0 ² = y · ±x = y · ±x z + xz² = xy z + xyz In this running example, the two vectors xyz = ´ 110 µ 011 are candidate test vectors for fault y/ 0. Formally, f±y = 1 ² f±y = 0 ² is called the Boolean difference of f with respect to y and is often written as: df dy = f±y = 1 ² f±y = 0 ²³ In general, if f is a function of x 1 µx 2 µ···µx n , then: df dx i = f±x 1 µx 2 µ···µx i µ···µx
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Unformatted text preview: n ² ⊕ f±x 1 µx 2 µ· · · µ x i µ· · · µx n ² In terms of test generation, for any target fault on some fault ¸/v , the set of all vectors that can propagate the fault-effect to the primary output f is then those vectors that can satisfy: df d¸ = 1 (Note that this is independent of the polarity of the fault, whether it is stuck-at-0 or stuck-at-1.) Next, the constraint that the fault must be excited, ¸ set to value v , must be added. Subsequently, the set of test vectors that can detect the fault becomes all those input values that can satisfy the following equation: ±¸ = v² · df d¸ = 1 (4.2) Consider the same circuit shown in Figure 4.5 again. Suppose the target fault is w/ 0. The same analysis can be performed for this new fault. The set of test vectors that can detect w/ 0 is simply: w · df dw = 1 ⇒ w · ±f±w = 1 ² ⊕ f±w = ²² = 1...
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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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