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Unformatted text preview: ECE 553: Testing and Testable Design of Digital Systems Department of Electrical and Computer Engineering University of WisconsinMadison ECE 553 : Testing and Testable Design of Digital Systems Fall 2011 Assignment #2 Date Tuesday, September 27, 2011 Due date Thursday, October 6, 2011 1. (10 points) Compute the total number of stuckat (single and multiple) faults for a logic in figure 1. For the circuit of Figure 1, we have Number of fault sites = PIs + gates + fanout branches = 15 Therefore, Number of single faults = 15 2 = 30 Number of single and multiple faults = 3 number of fault sites 1 = 3 15 1 = 14348906 The circuit has 14348906 single and multiple stuckat faults. Total number of transistors = 6 2 + 4 3 + 2 = 26 2. (Bushnell and Agrawal) Problem 4.5 (a) A twoinput NAND gate is shown in the above figure. The following table gives tests for transistor stuckopen (sop) faults: Test No. Fault Test: Vector 1, Vector 2 1 P1 sop 11, 01 2 P2 sop 11, 10 3 N1 sop 01, 11 or 10, 11 or 00, 11 4 N2 sop 01, 11 or 10, 11 or 00, 11 1 Fall 2011 (Lec: Saluja, TA: Millican) ECE 553: Testing and Testable Design of Digital Systems Ground DD V N2 N1 P2 Circuit for Problem 4.5. CMOS NAND gate. A B C P1 Logic NAND gate. A B C Notice that the sop faults of N1 and N2 have exactly the same tests. These two faults are equivalent. Equivalence of transistor faults is discussed in the following paper: M.L Flottes, C. Landrault and S. Provossoudovitch, Fault Modeling and Fault Equiv alence in CMOS Technology, J. Electronic Testing: Theory and Applications , vol. 2, pp. 229241, August 1991. (b) The following sequence of four vectors contains one vector pair for each fault in the above table: 11 , 01 , 11 , 10 Notice that this sequence also detects all single stuckat faults in the logic model of the NAND gate. (c) A stuckat fault in a signal affects two transistors in the twoinput NAND gate. For example, the fault A sa1 will mean that N1 remains permanently shorted (N1ssh) and P1 remains permanently open (P1sop). The following table gives all equivalences: Stuckat fault Equivalent transistor faults A sa1 N1ssh and P1sop B sa1 N1ssh and P2sop C sa1 (P1ssh or P2ssh) and (N1sop or N2sop) A sa0 N1sop and P1ssh B sa0 N2sop and P2ssh C sa0 N1ssh, N2ssh, P1sop and P2sop Notice that the three equivalent faults, A sa0, B sa0 and C sa0, are actually caused by different faulty transistors. They are detected by the same test (11)....
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