VLSI Testing Power Aware Serial Scan Virendra Singh Associate - - PowerPoint PPT Presentation
VLSI Testing Power Aware Serial Scan Virendra Singh Associate Professor C omputer A rchitecture and D ependable S ystems L ab Department of Electrical Engineering Indian Institute of Technology Bombay http://www.ee.iitb.ac.in/~viren/ E-mail:
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Associate Professor Computer Architecture and Dependable Systems Lab Department of Electrical Engineering Indian Institute of Technology Bombay
http://www.ee.iitb.ac.in/~viren/ E-mail: viren@ee.iitb.ac.in
Lecture 25 (21 March 2013)
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To Bus Leading to Primary O/P From Combinational Logic CLK Row Col To combinational Logic
Row Decoder Column Decoder
lines
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N
lines
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2
{log }
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Address N
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Mudlapur, VDAT’05
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CTR CTR CTR
Test Control
Single Ended Read Circuits and MISR
CTR CTR CLK CLK CLK CLK CLK CLK CLK AOI D Q BIT Test Enable CLK READ
Row Decoder Col Decoder Address lines
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N lines
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Architecture Functional Mode Test Application & Simultaneous Read Reading Response
1 1 1 RA CA 1 1 3 Feb 28, 2012
Gandhi, ETS’10
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To Bus Leading to Primary O/P To combinational Logic From Combinational Logic CLK Row Col CLK D D Q Q CTR CTR
TFF Based Scan Cell MTFF Based Scan Cell
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Circuit #FF #Gates GOV_SS GOVTFF_ RAS GOVMTFF _RAS S5378 179 2779 17.47% 25.3% 20.3% S9234 228 5597 12.04% 17.44% 13.99% S13207 669 7951 21.36% 30.75% 24.6% S15850 597 9772 16.7% 24.15% 19.36% S35932 1728 16065 25.36% 36.4% 29.1% S38417 1636 22179 19.3% 27.8% 22.2% S38584 1452 19253 19.7% 28.3% 22.6%
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Abhishek, ISCAS’10
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– Test power – Test application time – Test data volume
– ATPG for low test power consumption Test power Test length – Reducing scan clock frequency Test power Test application time – Scan-chain re-ordering (with additional logic insertion) Test power/time Design time – Test Compression Test time/data size Has limited capability for Compacted test
– Random access scan instead of Serial-scan – Hardware overhead? Silicon cost << Testing cost
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sff1 sff2 sff3 sff4 sff5 Combinational Block
PI PO Shift Cycle Capture Cycle 8
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sff1 sff2 sff3 sff4 sff5 Combinational Block
PI PO 1 1 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 1 0 0 1 0 1 1 0 0 1 Initial State Shift Cycle 0 1 1 0 0 Capture Cycle Responses 9
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Shift Cycle Capture Cycle Shift & Launch Why ?
➔Peak Power -- Causes IR-Drop, Cross talk. ➔Average Power -- Causes excess heat dissipation.
When ? 10
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How ? Test Pattern Pi Response Ri 1 X 1 0 0 1 1 0 1 0 1 0 1 0 1 0 1 1 1 0 1 1 0 1 P1 P2 R1 R2 P2 1 0 1 0 1 0 1 0 1 0 1 1 R1
➔Reduce Intra Pattern Transition ➔Reduce Inter Pattern Transition
1 X 1 0 0 1 P1 X 1 11
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Problem Statement : Determination of the minimum achievable peak power during test for a given test set and find out a test vector sequence which can support the minimum achievable peak power.
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Jaynarayan Tudu [ETS 2009]
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1 1
Next bit to shifted in Test Pattern Pi Response Ri 1 0 1 0 0, 1 0 1 0 1 1 0 1 0 0, 1 0 1 1 0 P1 P2 R1 R2 Computed Transition P2 1 1 1 1 1 0 1 0 3 0 1 1 0 1 4 1 0 1 1 0 4 0 1 0 1 1 4 1 0 1 0 1 4 Peak Power = Max V (total transition per clock) = 4
clock
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Graph theoretic problem formulation
P1 P2
Wij = Edge weight = Peak Power = MaxV (total transition per clock)
clock
W12 = 4
V1 V2
Direction of edge is mapped from the order of applying the test vectors Nodes are mapped from corresponding test vector
Vi
A Complete directed weighted graph can be formed by applying test vectors in all possible way 14
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Formulation of Problem1 (Minimal Test time) Given a complete weighted directed graph Dc, find out a Hamiltonian path which has minimum Path- weight which will be the minimum possible peak power Pth1 under minimal test time Problem Statement: Path-weight is defined as the maximum of weight of each edge in the path in a digraph Definition: Example:
V4 V2 V3 V5
3 8 10 10
path-weight = 10 15
V6
7 V1
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Observation:
V0 V1
V2 V3 V4
7 12 7 8
Can we reduce peak power below Pth1 ?
V5 V6
9 10 6 5
Pth1 = 12
V0 V1 V2 V3 V4
7 12 7 8
V5 V6
9 10 6 5
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Observation: There can be a walk which visits every node at least
Definition : walk-weight is defined as the maximum edge-weight in a walk
V0 V1 V2 V3 V4
7 12 7 8
walk-weight = 10 Can we reduce peak power below Pth1 ?
V5 V6
9 10 6 5
Conclusion:
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Formulation of Problem2 (Reapplication of vectors) Given an weighted directed graph D, find out a walk which visits each and every nodes at least once and having minimum walk-weight which corresponds to the peak power Pth2 Problem Statement: Definition: A walk in D is a non-empty alternating sequence of vertices and edges in D 18
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How to reduce further below the Pth2 ? Observation:
V0 V1 V2 V3 V4
7 12 7 8
V5 V6
9 10 6 5
Pth2 = 10
V0 V1 V2 V3 V4
7 7 8
V5 V6
9 6 5
Maximum edge-weight = 9
➔ Further reduction will cause a graph to be disconnected
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How to reduce further below the Pth2 ? Observation:
V0 V1 V2 V3 V4
7 7 8
V5 V6
9 6 5
Maximum edge-weight = 9
V0 V1 V2 V3 V4
7 8
V5 V6
9 6 5
Finding out disjoint paths Path 1 Path 2 20
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Formulation of Problem3 (Insertion of new vectors) Problem Statement: Find out a path cover for a given directed graph D and join those disjoint paths of path cover in such a way that a complete walk containing all the vertices can be formed which have minimum possible walk-weight less than Pth2. Definition: A path cover of a directed graph D is a set of disjoint paths in D which together contain all the vertices of D 21
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Algorithm1 Objective:
To find out a hamiltonian path which has minimum Path- weight which will be the minimum possible peak power Pth1 under minimal test time for a given complete weighted directed graph Dc,
Working principle: Algorithm1 first transform the weighted complete graph into unweighted graph and then it find out the hamiltonian path on unweighted graph which assure the minimum peak power Pth1. 22
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Example: P1 P2 P3 P4 1 1 1 1 1 1 1 0 1 0 1 0 1 1 1 1 1 1 1 1 1 1 1 1 0 1 0 1 0 1 1 1 1 1 1 1 1 1 1 1 1 0 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 R1 R2 R3 R4 1 1 1 0 1 0 1 0 1 1 1 1 0 0 0 1 0 1 0 1 0 1 1 1 1 1 1 1 1 1 1 0 1 0 1 1 1 1 1 1 1 1 0 0 1 1 1 0 1 1 1 1 1 1 1 1 0 1 1 0
Di V1 V2 V3 V4
7
14 12 7 12 10 8 8 6 7 7 6 5 3
11 8 23
Do 6 10 10 7 5
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Let following be given complete digraph Step1: Sort the edges in increasing order. Sorted order of edge weight will be: 3,5,6,7,8,10,11,12,14 24
Di V1 V2 V3 V4
7
14 12 7 12 10 8 8 6 7 7 6 5 3
11 8
Do 6 10 10 7 5
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Removing the edges whose weight is greater than next-weight to transform the given complete weighted graph into unweighted graph. Step2: next-weight=8 is chosen using binary search from sorted edge array 3,5,6,7,8,10,11,12,14 25
Di V 1 V 2 V 3 V 4
7
14 12 7 12 10 8 8 6 7 7 6 5 3
11 8
D
10 10 7 5 Di V 1 V 2 V 3 V 4
7
7 8 8 6 7 7 6 5 3
8
D
7 5
=>
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Attempt to find out Hamiltonian path. If path exist then store the path and next-weight value in temp_path and temp_weight and modify the next-weight from left half of array otherwise from right half of array. Go back to Step2. Step3: next-weight=11 => 3,5,6,7,8,10,11,12,14 No Hamiltonian path exist 26
Di V 1 V 2 V 3 V 4
7
7 8 8 6 7 7 6 5 3
8
D
7 5 Di V 1 V 2 V 3 V 4 D
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Removing the edges whose weight is greater than next-weight to transform the given complete weighted graph into unweighted graph. Step2: next-weight=11 => 3,5,6,7,8,10,11,12,14 Sorted edge array: 27
Di V 1 V 2 V 3 V 4
7
14 12 7 12 10 8 8 6 7 7 6 5 3
11 8
D
10 10 7 5 Di V 1 V 2 V 3 V 4
7
7 10 8 8 6 7 7 6 5 3
11 8
D
10 10 7 5
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Attempt to find out Hamiltonian path. If path exist then store the path and next-weight value in temp_path and temp_weight and modify the next-weight from left half of array otherwise from right half of array. Go back to Step2. Step3: next-weight=10 => 3,5,6,7,8,10,11,12,14 Hamiltonian path exist 28
Di V 1 V 2 V 3 V 4
7
7 10 8 8 6 7 7 6 5 3
11 8
D
10 10 7 5 Di V 1 V 2 V 3 V 4 D
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Removing the edges whose weight is greater than next-weight to transform the given complete weighted graph into unweighted graph. Step2: next-weight=10 => 3,5,6,7,8,10,11,12,14 Sorted edge array: 29
Di V 1 V 2 V 3 V 4
7
14 12 7 12 10 8 8 6 7 7 6 5 3
11 8
D
10 10 7 5 Di V 1 V 2 V 3 V 4
7
7 10 8 8 6 7 7 6 5 3
8
D
10 10 7 5
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Attempt to find out Hamiltonian path. If a path exists then store the path and next-weight value in temp_path and temp_weight and modify the next-weight from left half of array otherwise from right half of array. Go back to Step2. Step3: => 3,5,6,7,8,10,11,12,14 Hamiltonian path found and Binary search process exhausted, hence algorithm will stop here. 30
Di V 1 V 2 V 3 V 4
7
7 10 8 8 6 7 7 6 5 3
8
D
10 10 7 5 Di V 1 V 2 V 3 V 4 D
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Hamiltonian path found is:
Pth1 = 10
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Di V 1 V 2 V 3 V 4 7 8
8
D
5
And corresponding peak power is:
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Algorithm2 Objective: Working principle: Algorithm2 first transform the weighted complete graph into unweighted graph and then it find out the walk on unweighted graph which assure the minimum peak power Pth2. Find out a walk which visits each and every nodes at least once and having minimum walk-weight which corresponds to the peak power Pth2, for a given weighted directed graph D, 32
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Let following be given complete digraph: Step1: Given a sorted edge weights : 3,5,6,7,8,10,11,12,14. 33 Let Pthtemp = Pth1= 10
Di V 1 V 2 V 3 V 4
7
14 12 7 12 10 8 8 6 7 7 6 5 3
11 8
D
10 10 7 5
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Remove the edges whose weight is greater than or equal to
Pthtemp to transform the given complete weighted graph into
unweighted graph Du. Step2:
Note: Rest of the work onwards will be done in unweighted graph only
=> Pthtemp=10 34
Di V 1 V 2 V 3 V 4
7
14 12 7 12 10 8 8 6 7 7 6 5 3
11 8
D
10 10 7 5 Di V 1 V 2 V 3 V 4
7
7 8 8 6 7 7 6 5 3
8
D
7 5
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Step3 : Find out the walk in unweighted graph Du: Transform the unweighted graph Du into directed acyclic graph(DAG) by forming each cycle present in Du in to corresponding supernode 35
D i V 1 V 2 V 3 V 4 D
V 1 V 2 V43 D
i V 1 V432 D
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Find out a hamiltonian path in the DAG. If path exist store it in HAMtemp and assign Pthtemp= NextSmallEdgeWeight and goto Step2 else goto next step NextSmallEdgeWeight=8 Go to Step2... Given sorted array of edge weight: 3,5,6,7,8,10,11,12,14. 36
D i V 1 V432 D
i V 1 V432 D
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Go to next step... 37 Step 2 is repeated for edge-weight = 8. For this running example path does not exist for edge-weight=8
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Go to next step... If HAMtemp is empty then Print “no path below Pth1” else goto next step HAMtemp in running example is : 38
D i V 1 V432 D
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Form a final walk by unrolling (expanding) each of the supernode
Di V2 V4 V3 V4 V1
7 8 6 7 8 Hence the path given by Algo2 is:
Pth2 = 8 < Pth1=10
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Di V 1 V432 D
V4 V1 V 2 V3 V4 Do
=>
D
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Find out a path cover for a given directed graph D and join those disconnected paths of path cover in such a way that a complete walk containing all the vertices can be formed which has minimum possible walk-weight correspond to Pth3 less than Pth2. Objective: Algorithm3 Working principle: Algorithm3 first transform the weighted complete graph into unweighted graph and then it find out the walk on unweighted graph which assure the minimum peak power Pth3. 40
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Remove the edges having weight ≥ Pthtemp to transform the given complete weighted graph into unweighted graph Du. Given a sorted edge-weight: 3,5,6,7,8,10,11,12,14. Let Pthtemp = Pth2= 8 41
Di V 1 V 2 V 3 V 4
7
14 12 7 12 10 8 8 6 7 7 6 5 3
11 8
D
10 10 7 5 Di V 1 V 2 V 3 V 4
7
7 6 7 6 5 3 D
7 5
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Find a path cover in the unweighted graph Du 42
Di V 1 V 2 V 3 V 4 D
V 1 V 2 V 3 V 4 D
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Pth3 =6 < Pth2 = 8 < Pth1=10
43 Connect each paths to form a walk using node Va0 and Va1
Di V1 V 2 V 3 V 4 D
1 Va 1
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Summary:
Pth3 =6 < Pth2 = 8 < Pth1=10
Algorithm1 without increasing test time Algorithm2 by revisiting of some nodes with increase in certain amount of test time Algorithm3 by introduction of two new nodes with increase in certain amount of test time
Can peak power be reduced further ?
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Di V 1 V 2 V 3 V 4
7
14 12 7 12 10 8 8 6 7 7 6 5 3
11 8
D
10 10 7 5 Di V 1 V 2 V 3 V 4 D
Va 1
Maximum allowed test time:
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Theorem 1: The minimum achievable peak power for a test set is , Max i [Max{Min(EWa0i , EWa1i), Min(EWia0, EWia1)}] Lemma 1 : Let there be a test set of n vectors, and scan chain length be l. The worst case test time to achieve the minimum bound is (n – 1)(l + 1)+2(n-1)*l + l
Di Vi Vj Vk D
Va
1
Va
1
Va Va Va
1
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Di V 1 V 2 V 3 V 4 D
Va 1 Di V 1 V 2 V 3 V 4 D
Va 1
Reduced test time:
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Corollary 1 : The minimum achievable peak power under maximum test time (2n – 1) (l+1) + (n-1) * l + l is Max i [Max(EWa0i , EWa1i , EWia0, EWia1)]
V0 Vi Vj Vk Vn Va Va
1
Va
1
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b14 b22 s1196 s838 b07 s38417 s13207 2.5 5 7.5 10 12.5 15 17.5 20 22.5 25 27.5 30 32.5 35
Comparision of Results
Algo 1 Algo 2 Algo 3 Lower bound
Benchmark Cicuits % of reduction over random order
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