ECE260B – CSE241A Winter 2005 Testing

ECE260B – CSE241AWinter 2005Testing Website: http://vlsicad.ucsd.edu/courses/ece260b-w05

Outline • Defects and Faults • ATPG for Combinational Circuits • ATPG for Sequential Circuits

Fault models

Fault Models • Most Popular - “Stuck - at” model Covers many other occurring faults, such as opens and shorts. • a, g : x1 sa1 • b : x1 sa0 or • x2 sa0

Fault Models a • To detect an and-bridging • Detect a s-a-0 and b = 0 • Detect b s-a-0 and a = 0 • To detect a transition fault • Pattern 1: c = 1 • Pattern 2: detect c s-a-1 and b c good 1 bad

Problem with stuck-at model: CMOS open fault Sequential effect Needs two vectors to ensure detection! Other options: use stuck-open or stuck-short models This requires fault-simulation and analysis at the switch or transistor level - Very expensive!

Problem with stuck-at model: CMOS short fault Causes short circuit between Vdd and GND for A=C=0, B=1 Possible approach: Supply Current Measurement (IDDQ) but: not applicable for gigascale integration

Exhaustive Algorithm • For n-input circuit, generate all 2n input patterns • Infeasible, unless circuit is partitioned into cones of logic, with < 15 inputs • Perform exhaustive ATPG for each cone • Misses faults that require specific activation patterns for multiple cones to be tested

Random Pattern Generation • Flow chart for method • Use to get tests for 60-80% of faults, then switch to D-algorithm or other ATPG for rest

Fault simulation

Boolean Difference Symbolic Method (Sellers et al.) g = G (X1, X2, …, Xn) for the fault site fj = Fj (g, X1, X2, …, Xn) 1 jm Xi = 0 or 1 for 1 in

Boolean Difference (Sellers, Hsiao, Bearnson) • Shannon’s Expansion Theorem: F (X1, X2, …, Xn) = X2F (X1, 1, …, Xn) + X2F (X1, 0, …, Xn) • Boolean Difference (partial derivative): Fj g • Fault Detection Requirements (for g s-a-0): G (X1, X2, …, Xn) = 1 Fj g = Fj (1, X1, X2, …, Xn) Fj (0, X1, …, Xn) = Fj (1, X1, X2, …, Xn) Fj (0, X1, …, Xn) = 1

Basic Terms, Path Sensitization • Controllability: the ease of controlling the state of a node in the circuit • Observability: the ease of observing the state of a node in the circuit Goals: Determine input pattern that makes a fault controllable (triggers the fault, and makes its impact visible at the output nodes) sa0 1 Fault enabling 1 1 1 1 1 0 Fault propagation 0 Techniques Used: D-algorithm, Podem

5-Value Logic • 0 – binary 0 in both good and fault circuit • 1- binary 1 in both good and fault circuit • X – don’t care • D – binary 1 in good circuit, 0 in bad circuit • D – binary 0 in good circuit, 1 in bad circuit

Primitive D-Cube of Failure • Models circuit faults: • Stuck-at-0 • Stuck-at-1 • Bridging fault (short circuit) • Arbitrary change in logic function • AND Output sa0: “1 1 D” • AND Output sa1: “0 X D ” “X 0 D ” • Wire sa0: “D” • Propagation D-cube – models conditions under which fault effect propagates through gate

Forward Implication • Results in logic gate inputs that are significantly labeled so that output is uniquely determined • AND gate forward implication table:

Backward Implication • Unique determination of all gate inputs when the gate output and some of the inputs are given

Path Sensitization Method Circuit Example • Fault Sensitization • Fault Propagation • Line Justification

Path Sensitization Method Circuit Example • Try path f – h – k – L blocked at j, since there is no way to justify the 1 on i 1 D D D D 1 D 0 1 1

Path Sensitization Method Circuit Example • Try simultaneous paths f – h – k – L and g – i – j – k – L blocked at k because D-frontier (chain of D or D) disappears 1 D D 1 1 D D D 1

Path Sensitization Method Circuit Example • Final try: pathg – i – j – k – L – test found! 0 0 D D 1 D D D 1 1

D-Algorithm – Top Level • Number all circuit lines in increasing level order from PIs to POs; • Select a primitive D-cube of the fault to be the test cube; • D-drive (); • Consistency (); • return ();

D-Algorithm – Propagation • D-frontier: all gates whose output is X, at least one input is D or D • J-frontier: all gates whose output is defined, but is not implied by the input values

Example 7.2: Fault A sa0 • Step 1 – D-Drive – Set A = 1 D 1 D

Step 2 -- Example 7.2 • Step 2 – D-Drive – Set f = 0 0 D D 1 D

Step 3 -- Example 7.2 • Step 3 – D-Drive – Set k = 1 1 D 0 D D 1 D

Step 4 -- Example 7.2 • Step 4 – Consistency – Set g = 1 1 1 D 0 D D 1 D

Step 5 -- Example 7.2 • Step 5 – Consistency – f = 0 Alreadyset 1 1 D 0 D D 1 D

Step 6 -- Example 7.2 • Step 6 – Consistency – Set c = 0, Set e = 0 1 1 0 D 0 0 D D 1 D

D-Chain Dies -- Example 7.2 • Step 7 – Consistency – Set B = 0 • D-Chaindies X 1 1 0 D 0 0 0 D D 1 D • Test cube: A, B, C, D, e, f, g, h, k, L

Example 7.3 – Fault s sa1 • Primitive D-cube of Failure 1 D sa1

Example 7.3 – Step 2 s sa1 • Propagation D-cube for v 1 D 1 sa1 D 0 D

Example 7.3 – Step 2 s sa1 • Forward & Backward Implications 0 1 1 1 D 1 1 sa1 D 0 D

Example 7.3 – Step 3 s sa1 • Propagation D-cube for Z – test found! 0 1 1 1 D 1 1 sa1 D 0 D D 1

PODEM High-Level Flow • Assign binary value to unassigned PI • Determine implications of all PIs • Test Generated? If so, done. • Test possible with more assigned PIs? If maybe, go to Step 1 • Is there untried combination of values on assigned PIs? If not, exit: untestable fault • Set untried combination of values on assigned PIs using objectives and backtrace. Then, go to Step 2

Example 7.3 -- Step 1 s sa1 • Select path s –Y for fault propagation sa1

Example 7.3 -- Step 2 s sa1 • Initial objective: Set r to 1 to sensitize fault 1 sa1

Example 7.3 -- Step 3 s sa1 • Backtrace from r 1 sa1

Example 7.3 -- Step 4 s sa1 • Set A = 0 in implication stack 1 0 sa1

Example 7.3 -- Step 5 s sa1 • Forward implications: d = 0, X = 1 1 1 0 0 sa1

Example 7.3 -- Step 6 s sa1 • Initial objective: set r to 1 1 1 0 0 sa1

Example 7.3 -- Step 7 s sa1 • Backtrace from r again 1 1 0 0 sa1

Example 7.3 -- Step 8 s sa1 • Set B to 1. Implications in stack: A = 0, B = 1 1 1 0 0 1 sa1

Example 7.3 -- Step 9 s sa1 • Forward implications: k = 1, m = 0, r = 1, q = 1, Y = 1, s = D, u = D, v = D, Z = 1 1 1 0 0 0 1 sa1 D 1 1 1 D D 1

Backtrack -- Step 10 s sa1 • X-PATH-CHECK shows paths s – Y and s – u – v – Z blocked (D-frontier disappeared) 1 1 0 0 sa1

Step 11 -- s sa1 • Set B = 0 (alternate assignment) 1 0 0 sa1

Backtrack -- s sa1 • Forward implications: d = 0, X = 1, m = 1, r = 0, s = 1, q = 0, Y = 1, v = 0, Z = 1. Fault not sensitized 1 0 0 0 1 1 0 sa1 1 0 1 0 1

Step 13 -- s sa1 • Set A = 1 (alternate assignment) 1 1 sa1

Step 14 -- s sa1 • Backtrace from r again 1 1 sa1

Step 15 -- s sa1 • Set B = 0. Implications in stack: A = 1, B = 0 1 1 0 sa1

ECE260B – CSE241A Winter 2005 Testing