EE3A1 Computer Hardware and Digital Design Lecture 13 Detecting faults in Digital Systems

Introduction  From concept to saleable product: 1. Get the specification right 2. Translate to a manufacturable design 3. Check design fulfils the specification 4. Manufacture the design 5. Check that manufactured units are fault-free  For complex systems, step 5 can be very difficult, unless u The designer builds special features into the design that make the circuit easy to test u Design for testability

Basic testing procedure  Initialise the circuit into a known state  Apply the test inputs  Compare outputs with expectation

Example  Simple example circuit ABCE

 One of the units has a fault  Node D has been short-circuited to a voltage supply rail  This node’s value is stuck at logic 0 Example 0 1 ABCE

 We cannot directly probe the node (it is inside the chip)  How do we detect this?  One of the units has a fault  Node D has been short-circuited to a voltage supply rail  This node’s value is stuck at logic 0 Example 0 1 ABCEFaulty E

 Apply all possible combinations of A, B, C  These inputs are called test vectors  Compare output of unit-under-test with expectation  If we find a difference, then the unit is faulty  This circuit requires 8 test vectors  (Truth table has 2 3 =8 rows) Exhaustive Testing 0 1 ABCEFaulty E

Exhaustive test of combinational logic  Generalisation:  n-input combinational system requires 2 n test vectors  Example: system with 32 inputs. u Needs 2 32 = 4  10 9 test inputs. u Using a standard 100 MHz tester would take almost a minute for just one unit u May not be acceptable if we have a high throughput manufacturing line n

Exhaustive test of sequential logic  If system contains state memory, things get much worse.  The sequence in which we apply the inputs becomes significant.  System with n inputs and m state flip flops  2 n distinct inputs  2 m distinct sequences that we can apply these inputs in  Needs 2 n+m test vectors for exhaustive test. n

Exhaustive test of sequential logic  Example: system with 32 inputs, containing 32 state bits  Need = 2 64 = 2x10 19 test vectors  100 MHz tester requires 6,000 years for one unit

Non-Exhaustive Testing 0 1 ABCEFaulty E  We cannot afford to apply every possible test input  Instead we must select a few  Suppose we can use only two test inputs  Will we detect the fault?

Non-Exhaustive Testing 0 1 ABCEFaulty E  We cannot afford to apply every possible test input  Instead we must select a few  Suppose we can use only two test inputs.  Will we detect the fault? This is OKOutput of faulty unit is not identical to output of good unit: Fault is detected

Non-Exhaustive Testing 0 1 ABCEFaulty E  We cannot afford to apply every possible test input  Instead we must select a few  Suppose we can use only two test inputs.  Will we detect the fault?  OK if we choose a set that includes ABC=111  Otherwise fault is not detected This is no goodOutput of faulty unit is identical to output of good unit: Fault is missed

Non-Exhaustive Test  If we can’t use all possible test vectors, we may miss faults  The percentage of faults that is detected by a set of test vectors is the fault coverage  If we can’t use all possible test vectors, fault coverage may be less than 100%.  Some choices of test vectors will have high fault coverage  Other choices will have low fault coverage  How do we choose test vectors that maximise fault coverage? (Ideally to 100%)

Fault modelling  Need some idea of the failure modes of our system (i.e. what a fault looks like)  Then we can do some detective work, and check whether any of these failure modes is present  The fault with our system was a stuck-at-one ( s-a-0 ) fault.  Very common.  Node is shorted to high voltage rail, is also common.  stuck-at-one ( s-a-1 ) fault.  Simple test procedures assume only s-a-0 and s-a-1 faults can occur, and check each node for these faults

Path sensitisation methods  For each node in the circuit (A,B,C,D,E) u For each possible fault (s-a-0, s-a-1) u Find a set of inputs that will u test that node for the fault u steer the result of the test to an output

Path sensitisation methods  For each node in the circuit (A,B,C,D,E) u For each possible fault (s-a-0, s-a-1) u Find a set of inputs that will u test that node for the fault u steer the result of the test to an output Is D s-a-0? 1 1 Get result of test to output Set inputs to try to drive this node to 1 1 if good 0 if faulty

Sensitivity  We need to get the result of the test at D to the output E  If we choose C=0 u Then E=0 regardless of the value at D u E is insensitive to D  No good: can’t distinguish between faulty and good system Is D s-a-0? 1 1 Set inputs to try to drive this node to 1 1 if good 0 if faulty 0 0 Get result of test to output

Sensitivity  We need to get the result of the test at D to the output E  If we choose C=1 u Then E=1 if the circuit is good and 0 otherwise u E is sensitive to D  We can tell whether or not the circuit has this fault Is D s-a-0? 1 1 Set inputs to try to drive this node to 1 1 if good 0 if faulty Get result of test to output 1

Path sensitisation methods Is D s-a-0? if good 0 if faulty 1 FaultInputs: ABCFault free EFaulty E D s-a  For each node in the circuit (A,B,C,D,E) u For each possible fault (s-a-0, s-a-1)

Path sensitisation methods  For each node in the circuit (A,B,C,D,E) u For each possible fault (s-a-0, s-a-1) Is D s-a-1? Make E senstitive to D Set inputs to try to drive this node to 0 0 if good 1 if faulty or 1 FaultInputs: ABCFault free EFaulty E D s-a D s-a-1001 or 011 or 10101

Path sensitisation methods  For each node in the circuit (A,B,C,D,E) u For each possible fault (s-a-0, s-a-1) Is E s-a-0? Set inputs to try to drive this node to 1 1 if good 0 if faulty FaultInputs: ABCFault free EFaulty E D s-a E s-a D s-a-1001 or 011 or

Path sensitisation methods FaultInputs: ABCFault free EFaulty E A s-a B s-a C s-a D s-a-1001,101,01101 E s-a-1000,001,010,100,101,110,01101 A s-a B s-a C s-a D s-a E s-a  And so on…

Path sensitisation methods FaultInputs: ABCFault free EFaulty E A s-a B s-a C s-a D s-a-1001,101,01101 E s-a-1000,001,010,100,101,110,01101 A s-a B s-a C s-a D s-a E s-a  Find a set of test vectors that covers all possible faults  ABC = 111

Path sensitisation methods FaultInputs: ABCFault free EFaulty E A s-a B s-a C s-a D s-a-1001,101,01101 E s-a-1000,001,010,100,101,110,01101 A s-a B s-a C s-a D s-a E s-a  Find a set of test vectors that covers all possible faults  ABC = 111, 011

Path sensitisation methods FaultInputs: ABCFault free EFaulty E A s-a B s-a C s-a D s-a-1001,101,01101 E s-a-1000,001,010,100,101,110,01101 A s-a B s-a C s-a D s-a E s-a  Find a set of test vectors that covers all possible faults  ABC = 111, 011, 101

Path sensitisation methods FaultInputs: ABCFault free EFaulty E A s-a B s-a C s-a D s-a-1001,101,01101 E s-a-1000,001,010,100,101,110,01101 A s-a B s-a C s-a D s-a E s-a  Find a set of test vectors that covers all possible faults  ABC = 111, 011, 101, 110

Evaluation  Path sensitisation method u ABC = 111, 011, 101, 110 u 4 test vectors  Exhaustive test u ABC = 000, 001, 010, 011, 100, 101, 110, 111 u 8 test vectors

Generalisation  Exhaustive test u 2 number-of-inputs test vectors  Path sensitisation method u 2 x number-of-nodes

Generalisation  Example circuit u 32-bit combinational multiplier circuit u Number of inputs = 64 u Number of nodes = 7,200  Path-Sensitisation requires 2 x 7200 = 14,400 test vectors  On 100 MHz tester, takes 144  s u Batch of 10,000 chips tested in 14.4 seconds  Exhaustive test requires 2 64 = 1.8 x test vectors u On 100 MHz tester, takes 6,000 years u Batch of 10,000 chips tested in 60 million years

More about Sensitivity  Test for A s-a-0: u Drive A with 1 u Make E sensitive to A Is A s-a-0? 1 1 if good 0 if faulty

More about Sensitivity  Test for A s-a-0: u Drive A with 1 u Make E sensitive to A  What should B be? Is A s-a-0? if good 0 if faulty 0 Test result is lost: no good 0

More about Sensitivity  Test for A s-a-0: u Drive A with 1 u Make E sensitive to A  B must be 1  What about C? Is A s-a-0? if good 0 if faulty 0 0 Test result is lost: no good

More about Sensitivity  Test for A s-a-0: u Drive A with 1 u Make E sensitive to A  B must be 1  C must be 1 Is A s-a-0? if good 0 if faulty 1

More about Sensitivity  Test for A s-a-0: u Drive A with 1 u Make E sensitive to A  Intuitively BC=11  How do we handle more complicated cases, where answer is not obvious? Is A s-a-0? if good 0 if faulty 1

Derivative  Derivative is rate of change  Rate of change of distance x with respect to time t x t is bigis smallis zero  If derivative is zero, x is insensitive to t  If non-zero, then x is sensitive to t

 We want to know if E is sensitive to A  Boolean difference of E with respect to A is  Digital equivalent of derivative  If then E is sensitive to A  If then E is not sensitive to A Boolean Difference

Computation of Boolean Difference  Boolean relation between E and A is E=A.B.C  Work out value of this function with A set to 1  Work out value of this function with A set to 0  XOR the two together  This is the Boolean difference

 Boolean relation between E and A is E=A.B.C  Work out value of this function with A set to 1  Work out value of this function with A set to 0  XOR the two together  This is the Boolean difference  (1.B.C)  (0.B.C)  = B.C  0  = B.C.0 + B.C.0  = 0 + B.C.1  = B.C Computation of Boolean Difference Using identity X  Y=X.Y+X.Y

 Boolean relation between f and x i is  Work out value of this function with x i set to 1  Work out value of this function with x i set to 0  XOR the two together  This is the Boolean difference Computation of Boolean Difference x 1 x 2 x 3 x n z

Using Boolean Differences  Test for A s-a-0: u Drive A with 1 u Make E sensitive to A, i.e. make  But  So we need to make B=1 and C=1 Is A s-a-0? if good 0 if faulty

Enhancing Testability  Deeply embedded nodes are hard to test  Finding input values that will drive test onto NUT is hard u Too many logic gates between inputs and NUT  Sensitizing an output to the fault is also hard u Too many logic gates between NUT and outputs  We’d like to ensure that nodes do not have a large logic depth from the inputs or outputs InputsOutputs Node-Under-Test

System Partitioning  Most systems can be partitioned into sub-systems InputsOutputs Node-Under-Test

System Partitioning  Most systems can be partitioned into sub-systems Inputs Outputs S1 S3 S2 Node-Under-Test If we can gain direct access to these lines, then testing is much easier

Test Access Ports  Can we just copy the intermediate data to output pins? S1S2 Inputs Outputs S3  Adds many extra pins to chip  Makes chip very expensive Test Data Out 1 TDO2

Test Access Ports  Do parallel-to-serial conversion before copy-out S1S2 Inputs Outputs S3 Test Data Out 1 TDO2  Test Data Outputs are now only 1 bit wide  Doesn’t add much cost to chip  (But takes many clock cycles to shift data out)

Test Access Ports S1S2 Inputs Outputs S3 Test Data Out 1 TDO2  Add a way to get the test data in  Data is shifted in serially then converted to parallel  MUX controls whether S2 takes inputs from S1 or from TDI1 Test Data In 1TDI2

Test Access Ports S1S2 Inputs Outputs S3 Test Data Out 1 TDO2  Add a pin that controls whether the chip is in test mode or normal mode  In normal mode, the chip ignores the Test Access Ports  In test mode each stage receives input from TDI and sends output to TDO  Each node is now much easier to test Test Data In 1TDI2Normal/test

Summary  Efficient testing methods are based on u deciding what faults are possible and u testing circuit to see if any of the faults are present  Testability of design can be enhanced by incorporating special features at design stage

Questions  Find a test vector that will test for the fault D s-a-1.  Choose from ABC = (a) 000 (b) 001 (c) 010 (d) 011 (e) 100

Questions  What is the value of  (i.e. under what condition will a change in A cause a change in C) (a) a+b (b) 1 (c) a (d) b (e) b