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Copyright 2001, Agrawal & BushnellVLSI Test: Lecture 261 Lecture 26 Logic BIST Architectures n Motivation n Built-in Logic Block Observer (BILBO) n Test.

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Presentation on theme: "Copyright 2001, Agrawal & BushnellVLSI Test: Lecture 261 Lecture 26 Logic BIST Architectures n Motivation n Built-in Logic Block Observer (BILBO) n Test."— Presentation transcript:

1 Copyright 2001, Agrawal & BushnellVLSI Test: Lecture 261 Lecture 26 Logic BIST Architectures n Motivation n Built-in Logic Block Observer (BILBO) n Test / clock systems n Test / scan systems n Circular self-test path (CSTP) BIST n Circuit initialization n Loop-back hardware n Test point insertion n Summary

2 Copyright 2001, Agrawal & BushnellVLSI Test: Lecture 262 Motivation n Complex systems with multiple chips demand elaborate logic BIST architectures  BILBO and test / clock system n Shorter test length, more BIST hardware  STUMPS & test / scan systems n Longer test length, less BIST hardware  Circular Self-Test Path n Lowest hardware, lower fault coverage n Benefits: cheaper system test, Cost: more hdwe. n Must modify fully synthesized circuit for BIST to boost fault coverage  Initialization, loop-back, test point hardware

3 Copyright 2001, Agrawal & BushnellVLSI Test: Lecture 263 Built-in Logic Block Observer (BILBO) n Combined functionality of D flip-flop, pattern generator, response compacter, & scan chain  Reset all FFs to 0 by scanning in zeros

4 Copyright 2001, Agrawal & BushnellVLSI Test: Lecture 264 Example BILBO Usage n SI – Scan In n SO – Scan Out n Characteristic polynomial: 1 + x + … + x n n CUTs A and C: BILBO1 is MISR, BILBO2 is LFSR n CUT B: BILBO1 is LFSR, BILBO2 is MISR

5 Copyright 2001, Agrawal & BushnellVLSI Test: Lecture 265 BILBO Serial Scan Mode n B1 B2 = “00” n Dark lines show enabled data paths

6 Copyright 2001, Agrawal & BushnellVLSI Test: Lecture 266 BILBO LFSR Pattern Generator Mode n B1 B2 = “01”

7 Copyright 2001, Agrawal & BushnellVLSI Test: Lecture 267 BILBO in D FF (Normal) Mode n B1 B2 = “10”

8 Copyright 2001, Agrawal & BushnellVLSI Test: Lecture 268 BILBO in MISR Mode n B1 B2 = “11”

9 Copyright 2001, Agrawal & BushnellVLSI Test: Lecture 269 Test / Clock System Example n New fault set tested every clock period n Shortest possible pattern length  10 million BIST vectors, 200 MHz test / clock  Test Time = 10,000,000 / 200 x 10 6 = 0.05 s  Shorter fault simulation time than test / scan

10 Copyright 2001, Agrawal & BushnellVLSI Test: Lecture 2610 Test / Scan System n New fault tested during 1 clock vector with a complete scan chain shift n Significantly more time required per test than test / clock  Advantage: Judicious combination of scan chains and MISR reduces MISR bit width  Disadvantage: Much longer test pattern set length, causes fault simulation problems n Input patterns – time shifted & repeated  Become correlated – reduces fault detection effectiveness  Use XOR network to phase shift & decorrelate

11 Copyright 2001, Agrawal & BushnellVLSI Test: Lecture 2611 STUMPS Example n SR1 … SRn – 25 full-scan chains, each 200 bits n 500 chip outputs, need 25 bit MISR (not 5000 bits)

12 Copyright 2001, Agrawal & BushnellVLSI Test: Lecture 2612 STUMPS n Test procedure: 1. Scan in patterns from LFSR into all scan chains (200 clocks) 2. Switch to normal functional mode and clock 1 x with system clock 3. Scan out chains into MISR (200 clocks) where test results are compacted n Overlap Steps 1 & 3 n Requirements:  Every system input is driven by a scan chain  Every system output is caught in a scan chain or drives another chip being sampled

13 Copyright 2001, Agrawal & BushnellVLSI Test: Lecture 2613 Alternative Test / Scan Systems

14 Copyright 2001, Agrawal & BushnellVLSI Test: Lecture 2614 BILBO vs. STUMPS vs. ATE n LSSD: Level-sensitive scan design n ATE rate: 325 MHz n P = # patterns n CP = clock period = 10 -9 s Self-test speed LSSD tester speed n Test times – BILBO: P x CP STUMPS: P x L x CP ATE: P x L x CP x k External test & ATE: 307 x longer than BILBO STUMPS: 100 x longer than BILBO n Due to extra scan chain shifting System clock rate: 1 GHz L = max. scan chain length  k = = 3.07692

15 Copyright 2001, Agrawal & BushnellVLSI Test: Lecture 2615 Circular Self-Test Path (CSTP) BIST n Combine pattern generator and response compacter into a single device n Use synthesized hardware flip-flops configured as a circular shift register  Non-linear mathematical BIST system  Superposition does not hold  Flip-flop self-test cell – XOR’s D with Q state from previous FF in CSTP chain n MISR characteristic polynomial: f (x) = x n + 1 n Hard to compute fault coverage

16 Copyright 2001, Agrawal & BushnellVLSI Test: Lecture 2616 CSTP System

17 Copyright 2001, Agrawal & BushnellVLSI Test: Lecture 2617 Examples of CSTP Systems n CSTP BIST for 4 ASICs at Lucent Technologies:  Tested everything on 3 of the 4, except for: n Input/Output buffers and Input MUX n BIST overheads: logic – 20 %, chip area – 13 % n Stuck-at fault coverage – 92 %

18 Copyright 2001, Agrawal & BushnellVLSI Test: Lecture 2618 Circuit Initialization n Full-scan BIST – shift in scan chain seed before starting BIST n Partial-scan BIST – critical to initialize all FFs before BIST starts  Otherwise we clock X’s into MISR and signature is not unique and not repeatable n Discover initialization problems by: 1. Modeling all BIST hardware 2. Setting all FFs to X’s 3. Running logic simulation of CUT with BIST hardware

19 Copyright 2001, Agrawal & BushnellVLSI Test: Lecture 2619 Circuit Initialization (continued) n If MISR finishes with BIST cycle with X’s in signature, Design-for-Testability initialization hardware must be added n Add MS (master set) or MR (master reset) lines on flip-flops and excite them before BIST starts n Otherwise: 1. Break all cycles of FF’s 2. Apply a partial BIST synchronizing sequence to initialize all FF’s 3. Turn on the MISR to compact the response

20 Copyright 2001, Agrawal & BushnellVLSI Test: Lecture 2620 Isolation from System Inputs n Must isolate BIST circuits and CUT from normal system inputs during test:  Input MUX  Blocking gates – n AND gate – apply 0 to 2 nd AND input, block normal system input n Note: Neither all of the Input MUX nor the blocking gate hardware can be tested by BIST  Must test externally or with Boundary Scan (covered later)

21 Copyright 2001, Agrawal & BushnellVLSI Test: Lecture 2621 Loop-Back Circuit n Loop back outputs into inputs:

22 Copyright 2001, Agrawal & BushnellVLSI Test: Lecture 2622 System Test with Loop- Back n Exercise entire system with loop-back circuit n Use Boundary Scan to test chip interconnects

23 Copyright 2001, Agrawal & BushnellVLSI Test: Lecture 2623 Test Point Insertion n BIST does not detect all faults:  Test patterns not rich enough to test all faults n Modify circuit after synthesis to improve signal controllability n Observability addition – Route internal signal to extra FF in MISR or XOR into existing FF in MISR

24 Copyright 2001, Agrawal & BushnellVLSI Test: Lecture 2624 0 and 1 Injection n Force b to 0 when TEST & S are 1 n Force b to 1 when TEST & S are 1

25 Copyright 2001, Agrawal & BushnellVLSI Test: Lecture 2625 Test Point Activation Four test epochs  0,  1,  2,  3 n Phase decoder: enables different parts at different phases n Apply specified test pattern count at each n Example:  g t = 0 in  1 &  2, so c 1 = 0  g t = 1 in  0 &  3, so c 1 = g  h t = 1 in  2 &  3, so c 2 = 1  h t = 0 in  0 &  1, so c 2 = h

26 Copyright 2001, Agrawal & BushnellVLSI Test: Lecture 2626 Test Point Activator

27 Copyright 2001, Agrawal & BushnellVLSI Test: Lecture 2627 Summary n Logic BIST system architecture --  Advantages: n Higher fault coverage n At-speed test n Less system test, field test & diagnosis cost  Disadvantage: Higher hardware cost n Architectures: BILBO, test / clock, test / scan n Needs DFT for initialization, loop-back, and test points

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