 Hello everyone, I am Prashant S. Malge, Assistant Professor, Department of Electronics Engineering Walshan Institute of Technology, Swalapur. Today we will discuss the path sensitizing method for generating the test set for testing a circuit, learning outcomes. At the end of the session, student will be able to explain the path sensitizing technique. Also he will be able to derive the test set using path sensitizing method, contains the path sensitizing and then detection of specific fault. First we will discuss the path sensitizing method. In the previous lecture, we have seen how to derive a test set considering the individual faults on each and every wire in a circuit. But the test set by considering these individual faults on all individual wires in a circuit is not attractive from the practical point of view because there are too many wires and too many faults. So when the circuit becomes larger, the method is not practical. So the better alternative is to deal with several wires that form a path. So for this, we will activate a path so that the changes in the signal that propagates along a path have a direct impact on the output signal. Now consider this particular circuit. We have four inputs W1, W2, W3 and W4. Consider a path from W1 that is through A, B, C and F. Now to activate this particular path, we have to have input W2 equal to 1 so that the output of the AND gate only depends on the input on W1. Similarly for this NOR gate, this W3 has to be 0 so that the output depends on only this particular path. And for this AND gate, W4 must be 1 so that output depends on only this path. So therefore, to sensitize the path W1, B, C, F, we need to apply W2, 1, W3, 0 and W4, 1. So in general, to sensitize the path through an input of an AND or AND gate, all other inputs of that gate must be set to 1 whereas to sensitize the path through an input of OR or NOR gate all the inputs must be set to 0. Now consider this particular circuit. Think on what input is to be applied on W2, W3 and W4 to sensitize the path W1, C, F. Pause the video for a minute and write down your answer. You might have written the answer and see consider a path W1, C, F. So here we must, your output must depend only on this particular path. So the other two inputs of this OR gate must be 0 that is W4 must be 0 and D must be 0. And now to activate this particular path, your W2 must be set to 1 so that the output of this gate depends on W1. And when WD is 1, B turns out to be 0 and as B turns out to be 0 irrespective of the value of W3, your value of D is 0. So therefore, to activate a path W1, C, F, we can have W2 equal to 1, W3 may be 0 or 1 and W4 must be 0. So the input to activate this path is W2, W3, W4 must be 1x0 where x is a don't care. So therefore, now by applying a test when the input W1 is 0, when W1 is 0, your output must be 0. But if the circuit is faulty then the output may be 1 and this output 1 is due to fault on W1 if it is stuck to 1 or C stuck as 1 or F stuck at 1. So this particular test 0, 1 and 0 will determine the faults W1 stuck at 1, C stuck at 1 and F stuck at 1. On same lines when the input W1 is 1, the output is 1, but if we get output 0, that may be due to W1 stuck at 0, C stuck at 0 and F stuck at 0. On same lines now we have to consider the next path that is W2 CF. Once again for W2 CF we must have D and W4 must be 0 and W1 must be 1. So D will be 0 when W3 is 0, W4 must also be 0, W1 1. So 1 0 0 that is W1, W3, W4, 1 0 0 will activate this part and therefore, this 1 0 0 0 and 1 1 0 0 will determine all the faults on this particular part W2 CF. Same lines for W2 BDF, we need this C must be now 0 and W4 must be 0, for C0 we can make W1 0. So W1 0 makes C0, now for this W3 must be 1. So to activate this particular path we have W1 W3 W4 as 0 1 0 and the test 0 0 1 0 and 0 1 1 0 will determine all the faults on W2 BDF. Same way for next path that is W3 DF. So we must have C and W4 must be 0. So for and to activate this path B must be 1. So if we make W2 0, B will become 1, making W2 0, C will become 0. So therefore, your W1 may be 0 or 1 and therefore, to activate a path W4 DF we may have W1 either 0 or 1 that is do not care W2 0 and W4 0. So therefore, W1 W2 and W4 must be do not care 0 0 and therefore, the two inputs test that is do not care 0 0 0 and do not care 0 1 0 will determine all the faults on the wires W3 DF. On similar way for the last path that is W4 F, C and D must be 0. So making 0 W2 0 your C will become 0 therefore, W2 may be anything 0 or 1 and W3 making 0 your D will become 0. So therefore, to sensitize this particular path we may have 0 do not care 0 and the two inputs do not 0 do not care 0 0 and 0 do not care 0 1 will determine the faults on W4 and F. So here we have seen that in this particular circuit there are 5 paths and these 10 test set vectors are sufficient to determine the faults on all the wires. But that too it is not necessary to have all these tests. So if you look at the test 1 0 0 0 and do not care 0 0. So this do not care 3 0s will be covered in 1000. Same way 1 1 0 0 and 1 1 do not care 0. So this 1 1 do not care 0 will be considered by 1 1 0 0. Now 0 0 1 0 and 0 do not care 0 1 0. So 0 0 1 0 will consider 0 do not care 0 0. So on same lines 0 1 do not care 0 and 0 1 1 0. So 0 1 1 0 will consider 0 1 do not care 0 and therefore, the test 0 1 1 0 1 1 0 1000 0 1 1 0 0 1 0 0 do not care 0 0 and 0 do not care 0 1. These 6 tests are sufficient for testing this particular circuit. So out of these 10 test vectors these 6 tests are sufficient. So most of the cases the number of paths in a Gaiwan circuit is likely to be much smaller than the number of individual wires. And therefore, this method that is the method of determining the test set by path sensitizing is better compared to the method in which we determine the faults on each and everywhere. So therefore, this method is more attractive than the previous method. Thank you.