206_pdfsam_VLSI TEST PRINCIPLES & ARCHITECTURES

206_pdfsam_VLSI TEST PRINCIPLES & ARCHITECTURES - Test...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
Test Generation 175 a c b d ± FIGURE 4.11 Circuit with a constant circuit node. After these five calls to JustifyFanoutFree(), abc = 1 X 0 is an input vector that can justify g = 1. Note that in a fanout-free circuit, the JustifyFanoutFree() routine will always be able to set g to the desired value v and no conflict will ever be encountered. However, this is not always true for circuits with fanout structures. This is because in circuits with fanout branches, two or more signals that can be traced back to the same fanout stem are correlated , and setting arbitrary values on these correlated signals may not always be possible. For example, in the simple circuit shown in Figure 4.11, justifying d = 1 is impossible, as it requires both b = 1 and c = 1, thereby causing a conflict on a . Consider again the circuit shown in Figure 4.10. Suppose the objective is to set z = 0. Based on the JustifyFanoutFree() algorithm, it would first justify both g = 0 and h = 0. Now, for justifying g = 0, suppose it picks the signal
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

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.

Ask a homework question - tutors are online