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EE4271 VLSI Design Interconnect Optimizations Buffer Insertion

EE4271 VLSI Design Interconnect Optimizations Buffer Insertion. Moore’s law. Twice the number of transistors, approximately every two years, so double clock frequency accordingly. Interconnects Dominate. 300. 250. Interconnect delay. 200. 150. Delay (psec). 100. Transistor/Gate delay.

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EE4271 VLSI Design Interconnect Optimizations Buffer Insertion

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  1. EE4271 VLSI DesignInterconnect Optimizations Buffer Insertion

  2. Moore’s law Twice the number of transistors, approximately every two years, so double clock frequency accordingly

  3. Interconnects Dominate 300 250 Interconnect delay 200 150 Delay (psec) 100 Transistor/Gate delay 50 0 0.25 0.8 0.5 0.35 0.25 0.18 0.15 Technology generation (m) Source: Gordon Moore, Chairman Emeritus, Intel Corp. This is why Moore’s law is not true anymore. 3

  4. Objectives • What have we learned? • Compute circuit delay on wires and gates • Gate delay optimization • What are we going to learn? • Interconnect delay optimization: buffer insertion • Why reducing delay • How to perform it • This is the most important optimization in circuit design

  5. Why is this trend? 300 250 Interconnect delay 200 150 Delay (psec) 100 Transistor/Gate delay 50 0 0.25 0.8 0.5 0.35 0.25 0.18 0.15 Technology generation (m) Source: Gordon Moore, Chairman Emeritus, Intel Corp. 5

  6. G S D w S h l hs ls Ss ws A scaling primer • Ideal process scaling: • Device geometries shrink by S (= 0.7x) • Device delay shrinks by s • Wire geometries shrink by s • Unit resistance R/m: r/(ws.hs) = r/s2 • Unit coupling capacitance Cc/m : (hs)/(Ss) • Resistance doubled, capacitance roughly unchanged for unit length • How about the change in wire length?

  7. Technology scaling • Global (long) interconnect lengths don’t shrink • Global interconnect link cells far apart • Local (short) interconnect lengths shrink by s • Local interconnects link cells nearby

  8. Interconnect delay scaling • Delay of a wire of length l : tint= (rl)(cl) = rcl2 (a quadratic function of length) • Local interconnects : tint : (r/s2)(c)(ls)2 = rcl2 • Local interconnect delay unchanged • Global interconnects : tint : (r/s2)(c)(l)2 = (rcl2)/s2 • Global interconnect delay doubled – unsustainable! • Interconnect delay increasingly more dominant

  9. Buffer Insertion For Delay Reduction

  10. Elmore Delay for Wire x unit wire capacitance c unit wire resistance r C

  11. Elmore Delay for Buffer v u C Driving resistance Input capacitance

  12. Elmore Delay for A Circuit • Delay = all Ri all Cj downstream from Ri Ri*Cj • Elmore delay to n1 R(B)*(C1+C2) • Elmore delay to n2 R(B)*(C1+C2)+R(w)*C2 n1 n2 R(B) B R(w) C1 C2

  13. Buffers Reduce Wire Delay x/2 x/2 R C rx/2 R rx/2 cx/4 cx/4 cx/4 cx/4 C ∆t t_unbuf = R( cx + C ) + rx( cx/2 + C ) t_buf = 2R( cx/2 + C) + rx( cx/4 + C) t_buf – t_unbuf = RC – rcx2/4 x

  14. l l1 l2 l3 ln Buffered global interconnects: Intuition Interconnect delay = r.c.l2/2 Interconnect delay =  r.c.li2 /2< r.c.l2/2(where l = S lj ) sinceS (lj 2) < (S lj )2 (Of course, we need to consider buffer delay as well)

  15. … … L Rd – On resistance of inverter Cg – Gate input capacitance r,c – unit resistance and capacitance l Optimal Buffer Insertion on A Wire • Delay before buffer insertion = rcL2/2 • Assume N identical buffers with equal inter-buffer length l(L = Nl) • For minimum delay,

  16. Optimal interconnect delay • Substituting lopt back into the interconnect delay expression: Delay grows linearly with L (instead of quadratically)

  17. 80 clk-buf 70 buf 60 tot-buf 50 40 % cells used to buffer nets 30 20 10 0 90nm 65nm 45nm 32nm Total buffer count • Ever-increasing fractions of total cell count will be buffers • 70% in 32nm • 25% is widely observed

  18. Feature size (nm) Relative 250 180 130 90 65 45 32 delay 100 Gate delay Local interconnect (M1,2) Global interconnect with repeaters Global interconnect without repeaters 10 1 Source: ITRS, 2003 0.1 ITRS projections

  19. Exercise 1 • Given a wire of length 10 with r=2, c=2, what is its delay? • Given a buffer with Rd =10, Cg=20, after optimally buffering the wire, what is the delay? • What if wire length is 100? • Any conclusion?

  20. Exercise 2 • Relationship with gate sizing • If we can size the buffer, what is the best buffer size? • Let R0 denote the unit size buffer driving resistance, and C0 denote the unit size buffer input capacitance. Thus, Rd=R0/h and Cg=C0h • What is best h leading to smallest delay?

  21. Analogy

  22. Analogy • Advancing technology = period of city expansion, more transistors = larger city • Interconnects = streets • Buffers = gas stations • Signal delay (timing) = time to cross the city • Buffer insertion = gas station construction

  23. Previous Result is Only Theoretical: Discrete Buffer Locations Candidate buffer locations

  24. RAT: Required Arrival Time RAT = 100 AT = 0 RAT = 100 AT = 0 Wire delay = 80 Wire delay = 80 AT = 80 RAT = 20

  25. Slack: RAT - AT RAT = 100 AT = 0 Wire delay = 80 AT = 80 RAT = 20 Slack = 20 Slack = 20 Minimizing circuit delay = maximizing RAT at driver = maximizing slack at driver

  26. Motivation for Problem Formulation RAT = 300 AT = 350 Slack = RAT-AT= -50 slack = -50 RAT = 700 AT = 600 Slack = 100 RAT = Required Arrival Time Slack = RAT - AT RAT = 300 AT = 250 Slack = 50 Decouple capacitive load from critical path slack= 50 RAT = 700 AT = 400 Slack = 300 We need to maximum slack or RAT at driver

  27. Timing Driven Buffering Problem Formulation • Given • A Steiner tree • RAT at each sink • A buffer type • RC parameters • Candidate buffer locations • Find buffer insertion solution such that the slack (or RAT) at the driver is maximized

  28. An Example for Buffer Insertion C Q • r = 1, c = 1 • Rb = 1, Cb = 1 • Rd = 1 2 2 (v1, 1, 20) Add wire (v2, 3, 16) (v2, 1, 13) v1 v1 Insert buffer Add wire Add wire (v3, 5, 8) (v3, 3, 9) v1 v1 slack = 3 slack = 6 Add driver Add driver

  29. Candidate Buffering Solution • Definition • Each candidate solution is associated with • vi: a node • ci: downstream capacitance • qi: RAT vi is a sink ciis sink capacitance vis an internal node

  30. Van Ginneken’s Algorithm Candidate solutions are propagated toward the source

  31. Solution Propagation: Add Wire • c2 = c1 + cx • q2 = q1 – rcx2/2 – rxc1 • r: wire resistance per unit length • c: wire capacitance per unit length x (v1, c1, q1) (v2, c2, q2)

  32. Solution Propagation: Insert Buffer (v1, c1, q1) (v1, c1b, q1b) • c1b = Cb • q1b = q1 – Rbc1 • Cb: buffer capacitance • Rb: buffer resistance

  33. Solution Propagation: Add Driver (v0, c0, q0) (v0, c0d, q0d) • q0d = q0 – Rdc0 • Rd: driver resistance • Pick solution with max slack

  34. Exercise (20,400) 2 2 2 Unit Wire Cap = 5 Unit Wire Res = 3 Buffer C=5, R=1 Perform buffer insertion to maximize the slack at driver

  35. Exponential Runtime 2 solutions 4 solutions 8 solutions 16 solutions n candidate buffer locations lead to 2n solutions

  36. Solution Pruning • Two candidate solutions • (v, c1, q1) • (v, c2, q2) • Solution 1 is inferior if • c1 > c2 : larger load • and q1 < q2 : tighter timing

  37. An Analogy - I Faster -> Smaller Delay -> Larger RAT (since RAT = RAToutput - Delay) Larger Load -> Larger Capacitance LOAD

  38. An Analogy - II Faster & smaller load (larger RAT, smaller capacitance): Good END LOAD Slower & larger load (smaller RAT, larger capacitance): Inferior LOAD

  39. An Analogy - III END Who will be the winner? Cannot tell at this moment, so keep both of them.

  40. An Analogy - IV END Who will be the winner? Cannot tell at this moment, so keep both of them.

  41. Pruning When Insert Buffer They have the same load cap Cb, only the one with max q is kept

  42. (1) (2) (3) Generating Candidates From Dr. Charles Alpert

  43. (3) (b) (a) Both (a) and (b) “look” the same to the source. Throw out the one with the worse slack (4) Pruning Candidates

  44. (4) (5) Candidate Example Continued

  45. (5) At driver, compute which candidate maximizes slack. Result is optimal. Candidate Example Continued After pruning

  46. Example 2 2 2 Unit Wire Cap = 5 Unit Wire Res = 3 Buffer C=5, R=1 (20,400) (30,250) (5, 220) (20,400) (30,250) (5, 220) (40, 40) (5, 0) (15,160) (5, 145) (20,400)

  47. Example Cont’d (30,250) (5, 220) (40, 40) (5, 0) (15,160) (5, 145) (20,400) (5,0) is inferior to (5,145). (45,40) is inferior to (15,160) (30,250) (5, 220) (15,160) (5, 145) (5,15) (5,70) (20,400) Pick solution with largest slack, follow arrows to get solution

  48. Exercise • Without pruning, there will be exponential number of candidate solutions (i.e., given n candidate buffer locations, there will be 2n solutions). With pruning, how many solutions will we have?

  49. Exercise • Continue the following buffer insertion process. Assume that all partial candidate buffering solutions are as shown. 2 2 (10,40) (8,50) (5,10) (15,40) (7,10) (9,30) (12,20) Unit Wire Cap = 1 Unit Wire Res = 1 Buffer C=1, R=1

  50. Summary • Interconnect delay increases with technology scaling • Linear interconnect delay with buffer insertion • Buffer insertion with candidate buffer locations • Pruning for accelerating buffer insertion technique

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