Questions and Answers

1. Using path sensitization and Routh’s algebra generate test pattern for the s-a-0 fault in the circuit given below:

Answer:
To sensitize the fault, 1 is to be applied at the output of the AND gate. Also, there is only one path for fault propagation-- fault location to E. Now, to justify we need to apply a 0 at the second input of the OR gate. This leads to a collision. For justification we need a 0 in the second input of the OR gate and to sensitize the fault B is to be 1, which makes the second input of the OR gate a 1. This is illustrated in the figure below in terms of Routh’s algebra.

Now, since there is no other alternative path for fault propagation, this fault is not testable. In fact the answer is true. This fault is not testable. This example actually illustrates another utility of ATPG algorithm—finding redundancy in circuits. If we observer carefuuly, function represented by this circuit is: E=A.B+B. This can be simplified as B(A+1), which is B. So these two gates are redundant. So, if ATPG by path sensitization discovers that a fault is not testable, then there is redundancy in the circuit.