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- 1. VLSI Testing - Boolean difference method Dr. D. V. Kamath Professor, Department of E&C Engg., Manipal Institute of Technology, Manipal 1
- 2. 2 Boolean difference method Boolean difference is a type of deterministic method for finding TV BD gives all possible test vectors
- 3. 3 Conceptual view of ATPG Generate an input test vector that can distinguish the defect free circuit from the hypothetically defective one
- 4. 4 Boolean difference method Theoretical basis β Boolean difference The output of the circuit is given as π = π₯. π¦ + π¦ . z Let target fault be y s-a-0 . Under this condition, the output of the faulty circuit is shown to be ππ¦ = f/ y = 0 Generate an input test vector such that ππ¦ β ππ¦ = 1
- 5. 5 Boolean difference method ο§ ππ¦ β ππ¦ = 1 if and only if ππ¦ and ππ¦ result in opposing logic values ο§ Any TV that can set ππ¦ XOR ππ¦ = 1 is able to produce opposing values at the outputs of the fault-free and faulty circuits respectively ο§ ππ ππ¦ = ππ¦ β ππ¦ ο§ Now to test the fault say y at s-a-0, we need to initialize the node y to 1 (i.e., y = 1) and ππ ππ¦ = 1 i.e., y . π π π π = 1 ο§ Similarly, to test the fault say y at s-a-1 i.e., π . π π π π = 1
- 6. 6 Boolean difference example Find TV to test fault s-a-0 at node y using Boolean difference method y . ππ ππ² = y. π π² β π π² = y. π± β π³ = π± . y . z + x. y. π³ y . ππ ππ² = 1 will give the required TV TV will be x y z = {011, 110} π = π₯. π¦ + π¦ . z ; ππ¦ = π₯ ; ππ¦ = π§
- 7. 7 Boolean difference example Find TV to test fault s-a-0 at node w using BD method w . ππ ππ° = w. π π° β π π° = π² . z π β π±. π² = π² . z π± + π² = π± π² π³ + π² z =π² z w . ππ ππ° = 1 will give the required TV TV will be x y z = {x01} π = π₯. π¦ + π€ ; π€ = π¦ . z ; ππ€= 1 ; ππ€ = π₯. π¦
- 8. 8 Boolean difference example Find TV to test fault s-a-0 at node Z using BD method ο§ z . ππ ππ§ = z. ππ§ β ππ§ = 1 will give the required TV ο§ But, ππ§βππ§ = 0 ο§ The condition for testability ( ππ§βππ§ = 1 ) is not satisfiable ο§ Hence, the fault is undetectable ο§ Redundancy in the circuit is the cause for undetectable faults π = π₯. π¦ + π₯. π¦. π§ ; ππ§= π₯. π¦ ; ππ§ = π₯. π¦
- 9. 9 Boolean difference method Summary ο§ Given a circuit with output f and fault Ξ± s-a- b ο§ The set of test vectors that can detect this fault includes all the vectors that satisfy (Ξ± = π ). ππ πΞ± = 1
- 10. 10 β’ reachdvkamath@yahoo.com β’ dv.kamath@manipal.edu Contact

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