ECE465: Lecture #12

Digital Circuit Testing

Shantanu Dutt

UIC

Acknowledgement: Initial slides prepared by Huan Ren from Prof. Dutt’s Lecture Notes (significant modifications/additions made subsequently by Prof. Dutt).

Testing of Combinational Circuit• Complex submicron chips / multi-chip PCBs possibility of errors

in the manufacturing process physical faults in the digital circuits• Need to test at manufacturer before shipping out• Need to test periodically at user end as faults can also develop due

to various circuit stresses (high temp, moisture, impact, etc.)• Testing Scenarios

– Fault-detecting test set (FDTS): inputs (test vectors) detecting presence/absence of faults (under certain assumptions, e.g., single fault).

– Fault-locating test set (FLTS): inputs locating the fault(s).– CUT: Circuit under test .– Signature: Compaction of all O/Ps

• Fault Models– Types: permanent, intermittent, transient responses– Many possible physical faults: E.g., broken or shorted wires, defective

transistors, noise, improper power voltage

Wire break (s-a-0) Wire short to GND (s-a-0) Defective transistor (always on)

stuck-at-on

Vdd

Fault Models (Contd.)• A good fault model that can account for the logical behaviors of most faults is the “stuck at” fault model: s-a-0, s-a-1 for each

interconnect/wire in the circuit

• The s-a-0/1 fault model accounts for most physical fault types but not all. E.g., it does not account for ‘bridging’ faults) since we cannot say if either wire is s-a-0 or s-a-1—this situation will change dynamically based on the values (1 or 0) being driven on the wires and which driver is stronger

Bridging fault (2 signalwires shorted)

Wire break (s-a-0) Wire short to GND (s-a-0) Defective transistor (always on)

s-a-on

s-a-1

Vdd

Function Notations in Fault Circuits

f(x3)=x1x2+x3

For f(xn)=f(x1, x2,…,xn), let p be a wire in the circuit and d in {0,1}. Then fp/d(x1,…,xn) is the function for the faulty circuit with wire p s-a-d.

E.g., f3/0( )=x1x2, and f2/1( )=x1+x31

x1

x2

x3

f3/0

2

3

45 s-a-1

1x1

x2

x3

2

3

45 f

1x1

x2

x3

2

3

45 f2/1s-a-0

x3 x3

…

Fault Testing

• Circuits with r wires have 2r different single faults and 3r-1 possible single/multiple faults.

• Exhaustive testing is impractical for large circuits since it take O(2n) time for n-inputs.

CUTTest vector

…

O/Ps

Expected responses

Compare

Tests Response

x1 x2 x3 Ckt o/p

Exp.

resp

0 0 0 0 0

0 0 1 0 1

0 1 0 0 0

0 1 1 0 1

1 0 0 0 0

1 0 1 0 1

1 1 0 1 1

1 1 1 1 1• Need to determine a minimum set of inputs that can test for (detect/catch) any single fault. Single fault detection is a good goal as:

• Probability of multiple faults is much lower than single faults• Mutiple faults take much more time to detect

Test Generation • Function: f(x1,…,xn) (≡f(Xn)) represents the fault-free

circuit.• A test Ti for fault p/d is an input vector Xn

j to the circuit for which:

fp/d(Xnj) = f(Xn

j)

where j is the decimal value of the n bit binary input.• E.g., X3

1 =001 is a test for 3/0 for the earlier example: as f(X3

1)=1 and f3/0(x31)=0

• An alternative way of looking at this provides the basis for the EX-OR method of test generation:– The above condition implies Xn

j is a test for p/d iff: f(Xn

j) + fp/d(Xnj)=1

– The minterms (MTs) of Fp/d= f + fp/d are tests for p/d.

[f(x3)=x1x2+x

3

, f3/0(x3)=x1x2]

EX-OR Method• MTs of Fp/d=f + fp/d are tests for p/d. Can be determined algebraically • E.g., In the previous example

F1/0=(x1x2+x3) + x3=x1x2x’3. Hence, test=110 (only MT for F1/0 ) for 1/0 F3/0=(x1x2+x3) + x1x2 = x’1x3+x’2x3. Hence, test=001, 011,101 (MTs of F3/0)• These tests (MTs of Fp/d) can also be determined in a tabular manner as

shown below, but this is more cumbersome

Input Vectors

Functions realizedby Faulty Circuits

The XOR’s (1 in a col. indicates i/p vector is a test for the corresp. fault)

x1 x2 x3 f f1/0 f2/1 f3/0 f + f1/0

f + f2/1

f + f3/0

0 0 0 0 0 0 0 0 0 0

0 0 1 1 1 1 0 0 0 1

0 1 0 0 0 0 0 0 0 0

0 1 1 1 1 1 0 0 0 1

1 0 0 0 0 1 0 0 1 0

1 0 1 1 1 1 0 0 0 1

1 1 0 1 0 1 1 1 0 0

1 1 1 1 1 1 1 0 0 0

[f(x3)=x1x2+x3]

Fault Table• Displays a set of faults (generally single faults) and a set of test inputs that test

them (i.e., detect them)—can be drived, for ex., from the XOR table• A “1” at the intersection of i/p vector Xn

j and fault p/d Xnj is a test for p/d

• A “0” at the intersection of i/p vector Xnj and fault p/d Xn

j is not a test for p/d

Tests Faults

x1 x2 x3 1/0 1/1 2/0 2/1 3/0 3/1 4/0 4/1 5/0 5/1

0 0 0 0 0 0 0 0 1 0 1 0 1

0 0 1 0 0 0 0 1 0 0 0 1 0

0 1 0 0 1 0 0 0 1 0 1 0 1

0 1 1 0 0 0 0 1 0 0 0 1 0

1 0 0 0 0 0 1 0 1 0 1 0 1

1 0 1 0 0 0 0 1 0 0 0 1 0

1 1 0 1 0 1 0 0 0 1 0 1 0

1 1 1 0 0 0 0 0 0 0 0 1 0

Determining a Minimal Test Set• A test set for a fault set S is a set of i/p vectors (also called test vectors [TVs]) which contains at

least one test for each fault in S• From the fault table, we need to determine a minimal test set that covers all faults in the table (the

set of faults in the table is the set S)• Determining a minimal test set lower testing cost in terms of

– Test time (this allows, for ex, product to get to market faster after manufacturing testing)– Power consumption during testing

• Determining a minimal test set is a minimal covering problem, just like in the PIT part of QM where we need to cover all MTs w/ a minimal set of PIs (this is especially true in PLA design, where each PI has a cost of 1).

– Here, TVs PIs and faults MTs– So a QM-type method can be used for covering all faults using a minimal set of TVs– TV cost = 1, and since there is no multi-function type issue as in QM for logic min., costs are not mentioned

Tests Faults

x1 x2 x3 1/0 1/1 2/0 2/1 3/0 3/1 4/0 4/1 5/0 5/1

0 0 0 0 0 0 0 0 1 0 1 0 1

0 0 1 0 0 0 0 1 0 0 0 1 0

0 1 0 0 1 0 0 0 1 0 1 0 1

0 1 1 0 0 0 0 1 0 0 0 1 0

1 0 0 0 0 0 1 0 1 0 1 0 1

1 0 1 0 0 0 0 1 0 0 0 1 0

1 1 0 1 0 1 0 0 0 1 0 1 0

1 1 1 0 0 0 0 0 0 0 0 1 0

1 1 1 123 3 3 3

1*(ETV)

2*(ETV)

3*(ETV)

6*(p-ETV)

4

• After all the essential TVs (ETVs) and their faults are deleted on fault 3/0 remains• In the reduced fault table (RFT), all TVs covering 3/0 cover each other & all but one can be deleted (dels not shown here for simplicity)• We choose to not delete TV 011, & it is thus chosen to cover 3/0 • Final minimal test set = {010, 011, 100, 110}

4

5

Path Sensitizing method• XOR method can be very cumbersome for leven

medium-size circuits.• An alternative more time-efficient method: Path

Sensitizing method1. Select a path from fault-site p to ckt. o/p(s).

For fault p/d, z is the “fault value” (z = d if there is a fault p/d)

2. Forward Trace (FT):• Choose intermediate logic values at gate i/ps

along the path, so that logic values of lines/wires in the path are functions of z (i.e., z or z’).

• Also, choose the complement value d’ to the fault value to be “injected” at the fault line ( for no fault @ p, z = d’)

• The path is then said to be “sensitized” from the fault site to the o/p.

3. Backward Trace (BT):• Trace backwards and establish circuit inputs,

if possible, required for obtaining the logic values at different points in the sensitized paths determined during FT.

• If there is a “conflicting” values needed at any i/p or intermediate points/wires during BT thre is no test for p/d

• Don’t cares (X’s) at i/ps after BT w/ no conflict anywhere else multiple tests for p/d (<= 2k tests if k i/ps have X’s after BT)

4. This method may not yield all possible tests for a fault p/d (unlike the XOR method)

• This can compromise minimal test set determination

• If the reqd values shown above for wires p, p1, p2 and p3, can be “injected” into them via some TV at the ckt i/ps (such TV(s) are determined during BT) we have a test for p/d, since:

• If there is a fault p/d, z = d, and o/p = z’ = d’• If there is no fault @ p, z = d’ (d’ is injected at p) o/p = z’ = d• Thus o/p for p/d != o/p for no fault @ p

s-a-d

(p/d)

zd’

0

1

0

z

z

Wire p

p1

p2

p3

I/P cone of p3

Circ

uit

I/P

sBT

BT

BT

BT

z’s-a-d

(p/d)

zd’

0

1

0

z

z

Wire p

p1

p2

p3

FT

Path Sensitizing method (contd)• Examples of sensitized i/ps for different gates• Can also determine these for other gates/components (e.g., XOR, XNOR, AOI, MUXes)

x1=0x2=1

z1/1

x1=1x2=0

z1/0

x1=0x2=0

z1/1

x1=1x2=1

z1/0

Two input

OR gate w/ s-a-0 OR gate w/ s-a-1 AND gate w/ s-a-0 AND gate w/ s-a-1

x1=0x2=1x3=0x4=0

x1=0x2=0x3=0x4=0

2/0 2/1x1=0x2=1x3=0x4=0

2/0x1=0x2=0x3=0x4=0

2/1

x1=1x2=1x3=1x4=1

2/0

x1=1x2=0x3=1x4=1

2/1

x1=1x2=1x3=1x4=1

2/0x1=1x2=0x3=1x4=1

2/1

Four input

z z z z

zz

zz

zz z

z

zz z

zz zz z

Path Sensitizing method (contd)

1

2

3

45

Confirm w/ XOR method:F1/0=(x1x2+x3) + x3=x1x2x3

’

test for 1/0 = {110}

x1=1x2=1

x3=0

BTz

z

1

z

Forward trace

0

1 f

Forward trace

1

f23

4

5

3/1

Confirm: F3/1=(x1x2+x3) + 1 = (x1’+x2’)x3’ = x1’ x3’ + x2’x3’ tests for 3/1 = {000, 010, 100}

0

0z

z

Backward trace1/0

x3=0

x1=X/0

x2=0/Xz

Path Sensitizing method (contd)• Facts in fanout-free circuits (circuits in which each gate o/p feeds only 1 gate i/p

or is a ckt o/p) and circuits w/ fanout (at least 1 gate o/p feeds > 1 gate i/ps):– For a given sensitized path, each wire along the path for p/d is also tested for a s-a-0

or s-a-1 fault as determined as follows:• For each wire w on the sensitized path, if logic value on w is z, then w is tested for w/d,

otherwise (logic value on w is z’) it is tested for w/d’ • In Fig. 1 below, a test for 3/1 is also a test for 5/1 (the sensitized path is 3 5)• There is only 1 path from an i/p to an o/p each wire is on at least one (unique) path from

an i/p to an o/p• Thus fan-out free ckts can be tested for all possible single s-a-0 and s-a-1 faults by testing

each primary input for s-a-0 and s-a-1 faults

• For fanout ckts, if the faulty point has a fanout, then its test will test all sensitized paths for that test; but all fanout paths from a fault may not be sensitized; so a faulty point’s test is not necessarily a test for all fanout paths

– Thus should try to do multipath sensitization (either simultaneously or sequentially) if possible (it is not always!) in fanout ckts. For sensitization of simultaneous paths, when the paths are reconvergent, then only under special cases (given later) will tests for the primary fault be tests for wires on these paths

– If after tests for primary i/ps and their sensitized paths are determined for all possible faults, if some wires remain untested (because a path to them from primary i/ps cannot be sensitized), start the process from such wire(s) w/ the lowest level and find tests for them and all their sensitized paths

0

0

Forward trace

z

1

f23

45

3/1zx3=0

x1=X/0

x2=0/X

Backward trace

Fig. 2: Xnj is a test for p1/d, p2/d’, p3/d’, p4/d &

p5/d’ on path1 and p6/d, p7/d’ & p8/d on path 2 if sensitized non-simultaneouslyFig. 1: Test set for 3/1 & 5/1 = {000, 100, 010}

p1/d

z z’z’

z

z’

Tes

t X

nj

p2/d’p3/d’

p4/d

p5/d’

d’

z z’

zp6/d

p7/d’p8/d

2 sensitizedpaths for p1/d

Testing Circuits with Fanout• Path sensitization: a) Multiple or single paths can be sensitized.

• Single path sensitized by TV 110; also a test for 4/0, 6/0.

• Further, the path is sensitized also for fault 2/1, except for a 0 needed now to be injected on wire 2. Thus if wire 2 is a primary i/p and does not provide any of the “other” i/ps to any gate on the senstitized path, then just by changing the corresponding i/p from 1 to 0, we get a test vector 100 for 2/1 (and 4/1, 6/1). • Double path sensitized by TV 111

• However, this TV is not a test for 4/0, 5/0. E.g., consider fault 4/0. Since if 2 is non-faulty (nf)—we are testing for only single faults—wire 2 takes on value 1, which causes a 1 on wire 5, thus not sensitizing the path from wire 4 to the o/p 6.

• It is, however, a test for 6/0, since wire 6 occurs after the two paths reconverge (check out that 111 injects a 1 into 6)

• The corresp. double path sens. based test for 2/1 is 101 (as discussed above)

• TV 101 is also a test for 4/1, 5/1 and 6/1. E.g., consider 4/1. TV 101 injects a 0 into 4 (needed as the complement of the s-a-1 on 4), and on 5, needed for sensitizing the path from 4 to o/p 6

x1=0

x2=1

x3=1

2/0

0

zz

z

0

1

4

5

6x1=1

x2=1

x3=0

2/0

0z

zz

1

1

4

5

6

• Another single path sensitized by TV 011; also a test for 5/0 & 6/0.

• Further, as in the top path sensitization, the vector 001 is a TV for 2/1 (and 5/1, 6/1)

x1=1

x2=1

x3=1

2/0

zz

zz

1

1

1

4

5

6

Reconvergence point/gate of the 2 fan-out paths

Testing Circuits with Fanout---Multipath sensitization

• From the previous example, we can obtain (and prove) the following rules for determining when a test for an initial fault location p/d determined using multipath sensitization, is also a test for subsequent wires on these paths:

p1/d

z z’z’

z

z’

Tes

t X

nj

p2/d’p3/d’

p4/d

p5/d’

d’

z z’

zp6/dp7/d’

p8/d

2 sensitizedpaths for p1/d

o/p2

o/p1

• (a) If the multiple sensitized paths are not reconvergent, as shown above, then the wires pj on each path is also tested by the determined TV(s) for p1/d, based on whether the logic variable on wire pj is z—the logic variable on p1—(tested for pj/d) or z’ (tested for pj/d’)

Testing Circuits with Fanout---Multipath sensitization (contd)

• (c) If: (ii) the test is for p1/1, (ii) the multiple sensitized paths are reconvergent at an OR or NOR gate, and (iii) the i/ps to this gate are all z, as shown above, then the tests determined this way for p1/1 are also tests on subsequent wires pj on these paths for pj/d (d= 1, 0 for logic z, z’ resp. on pj) as shown above• (d) The above two situations are completely reversed when condition (iii) in (b)-(c) changes to: ( iii) the i/ps to the reconvergent gate are all z’ (the first 2 conditions in (b) remaining unchanged) as follows:Alternate of (b): All the p1/0 test(s) determined here are also test(s) for the intermediate wires s-a-0/1 (based on z/z’ var, resp. on them) on the sensitized paths. This is b/c for an intermediate wire p j, z’ on the other path(s) (the one(s) not containing p j) = 0 path through pj via the reconv. OR/NOR gate is sensitizedAlternate of (c): None of the the p1/1 test(s) determined here are tests on the intermediate wires on these paths before the reconvergent point (since z’ on the “other” path(s) = 1 thus not sensitizing the path through pj). They, are, however, tests for wires after the reconvergent gate.

• (b) If: (i) the test is for p1/0, (ii) the multiple sensitized paths are reconvergent at an OR or NOR gate, and (iii) the i/ps to this gate are all z, as shown above, then the tests determined this way for p1/0 are not tests for any pj/d (d= 0 or 1) for wires pj on each path from p1 to the reconverging gate. However, from the reconvergent point onwards, wires pk are also tested by the determined TV(s) for p1/0, based on whether the logic variable on wire pk is z—the logic variable on p1—(tested for pk/0) or z’ (tested for pk/1)

p1/0

z z’z’

z

z

Tes

t X

nj

p2

p3

p4

p5

1

z z’p6

p7

2 sensitizedpaths for p1/0

z’

p9/0 p10/1

No tests for these wires

z

Reconvergence point/gate

orzp8

p1/1

z z’z’

z

z

Tes

t X

nj

p5/1

0

z z’p6/1

2 sensitizedpaths for p1/1

z’

p9/1 p10/0

Reconvergence point/gate

orp7/0

p2/0p3/0

p4/1 zz

p8/1

Testing Circuits with Fanout---Multipath sensitization (contd)

• (f) If: (i) the test is for p1/0, (ii) the multiple sensitized paths are reconvergent at an AND or NAND gate, and (iii) the i/ps to this gate are all z, as shown above, then the tests determined this way for p1/0 are also tests on subsequent wires pj on these paths for pj/d (d= 0, 1 for logic value z, z’, resp., on p j) as shown above• (g) The above two situations are completely reversed when condition (iii) in (e)-(f) changes to: (iii) the i/ps to the reconvergent gate are all z’ (the first 2 conditions remaining unchanged) as follows:Alternate of (e): All of the p1/1 test(s) determined here are test(s) for the intermediate wires s-a-0/1 (based on z/z’ var, resp. on them) on the sensitized paths. This is b/c for an intermediate wire p j, z’ on the other path(s) (the one(s) not containing pj) = 1 path through pj via the recconv. AND/NAND gate is sensitized Alternate of (f): None of the the p1/0 test(s) determined here are tests on the intermediate wires on these paths before the reconvergent point (since z’ on the “other” path(s) = 0, thus not sensitzing the path through pj). They, are, however, tests for wires after the reconvergent gate.• (h) There cannot be simultaneously sensitized paths reconvergent on 2-i/p XOR/XNOR gates (the paths cannot be sensitized beyond such gates), & thus such gates are not considered here.

• (e) If: (i) the test is for p1/1, (ii) the multiple sensitized paths are reconvergent at an AND or NAND gate, and (iii) the i/ps to this gate are all z, as shown above, then the tests determined this way for p1/1 are not tests for any pj/d (d= 0 or 1) for wires pj on each path from p1 to the reconverging gate. However, from the reconvergent point onwards, wires pk are also tested by the determined TV(s) for p1/1, based on whether the logic variable on wire pk is z—the logic variable on p1—(tested for pk/1) or z’ (tested for pk/0)

p1/0

z z’z’

z

z

Tes

t X

nj

p5/0

1

z z’p6/0

2 sensitizedpaths for p1/0

z’

p9/0 p10/01

p7/1z

p8/0

p2/1p3/1

p4/0 z

Reconvergence point/gate

or

p1/1

z z’z’

z

z

Tes

t X

nj

p2

p3

p4

p5

0

z z’p6 p7

2 sensitizedpaths for p1/1

z’

p9/1 p10/0

No tests for these wires

z

Reconvergence point/gate

orzp8

Testing Circuits with Fanout (contd)• Path sensitization (contd)

– b) In some cases, only single path sensitization will work and multiple will not ( fault free O/P->O/P with fault at ckt O/P will be either 1->1 or 0->0, thus not detecting the fault.).

x1=1

x2=1

x3=0

1->0

Single path sensitized by TV 110

2/0 (1->0)

x1=0

x2=1

x3=1

0->1

z

Single path sensitized by TV 011

2/0

z

x1=1

x2=1

x3=1

1

Double path produced by 111 is not a test

2/0Not a function of z

z

z1

General case. 0

0

z

1->1z

z

z

z

z (1->1)

z

z

1

1

1

1

z z

p/0

1

Testing Circuits with Fanout• Path sensitization (contd)

– c) In other cases, single path sensitization will not work, and only multiple path will.

x1=1

p/1

z

1/z (either possible)

General case

zx2=0 &1

z

1

0->1

Forward trace OK, but backward trace fails.

x1=1

p/1

zx2=1

z

0->1 Double path produces a test (11)

Not possible to get a 0, but may be possible get a z at an intermediate gate on the path being sensitized in the FT

z

Conflict!

0

01

1

1

1

01

1

1/z

Need to get az instead of 1for sensitization

11

Summary of Path Sensitization• Overall Technique for Fanout-Free Circuits:

o Sensitize the unique path P(p,q) from a primary inp. p to an op. q, for p/d, and get corresponding tests for all wires along this path.

o Repeat process for p/d’ sensitization (sometimes very easy to derive from p/d sensitization as in prev. examples, but sometimes not, in which case have to do from scratch), and obtain the tests for the complement faults of wires on P(p,q)

• Overall Technique for Circuits with Fanout:o In turn, try single path sensitization from inp. p to op. q on each path from p q for p/d until

sensitization obtained. When this is attained, the tests for the successful path will be tests for corresponding faults on all wires on this path as explained earlier. (An option is not to stop at the 1st successful path, but to continue until all possible single paths from p q are sens. as this gives us more tests for p/d to choose from in the fault table and also provides tests for additional wires on these paths.)

o Repeat above process for p/d’o If single path sensitization is not possible for p/d or p/d’, then try all possible multiple paths

in turn from 2 paths onwards as explained earlier. If sensitization is obtained, then as explained earlier, the tests obtained for p/d or p/d’ (as the case may be), may or may not be a test for wires of these simultaneously sensitized paths

o If after tests obtained for all primary inputs via the above process, some internal wires still have no tests, then repeat the above steps for each such internal wire p, in order of increasing “levels” of such wires (i.e., starting from those at the smallest level, since tests for smaller level wires can yield test for higher level ones if on the same sensitized paths)

• Pros and Cons:o Adv over XOR method:

1. A test for p/d automatically becomes tests for various faults along all single sensitized paths from p (do not have to compute them separately), and in some cases on simultaneously sensitized paths

2. Generally, more time efficiento Disadv wrt XOR method:

1. Not all tests for a p/d may be found (if we stop before exploring all paths from p, in all combinations of simultaniety), thus the fault table is not exhaustive, and the least-cost test set may not be found

2. Worst-case more complex

Overview of Testing Phases

For each wire p and each d in {0,1}find tests for p/d by either:

(a) the XOR method or(b) the path-sensitizing

Construct the fault table from the above tests (Note: the path-sensitizingmethod may not find all tests, though it

will find at least 1 test for each p/d, if it exists)

Find a minimal test set from the faulttable using a min-cost covering

technique (e.g., similar to that used inthe PIT part of single-function QM)