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Lecture 30 Scale and Yield Mar. 24, 2003

Lecture 30 Scale and Yield Mar. 24, 2003. Scale. In library. Yield of fabrication process. 1. System Yield: = 0.95  2. Random Yield:  A = chip area D = defective density  3. Total Yield:. Example. Lectures 31 and 32 Routing and simulation Mar. 26,28, 2003. Topics.

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Lecture 30 Scale and Yield Mar. 24, 2003

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  1. Lecture 30 Scale and Yield Mar. 24, 2003

  2. Scale In library

  3. Yield of fabrication process 1. System Yield: = 0.95  2. Random Yield:  A = chip area D = defective density  3. Total Yield:

  4. Example

  5. Lectures 31 and 32 Routing and simulation Mar. 26,28, 2003

  6. Topics • Layouts for logic networks. • Channel routing. • Simulation.

  7. Standard cell layout • Layout made of small cells: gates, flip-flops, etc. • Cells are hand-designed. • Assembly of cells is automatic: • cells arranged in rows; • wires routed between (and through) cells.

  8. Standard cell structure pin VDD n tub pullups Feedthrough area Intra-cell wiring p tub pulldowns VSS pin

  9. Standard cell design • Pitch: height of cell. • All cells have same pitch, may have different widths. • VDD, VSS connections are designed to run through cells. • A feedthrough area may allow wires to be routed over the cell.

  10. height wire Horizontal track Vertical track Single-row layout design cell cell cell cell cell Routing channel cell cell cell cell cell

  11. Routing channels • Tracks form a grid for routing. • Spacing between tracks is center-to-center distance between wires. • Track spacing depends on wire layer used. • Different layers are (generally) used for horizontal and vertical wires. • Horizontal and vertical can be routed relatively independently.

  12. Routing channel design • Placement of cells determines placement of pins. • Pin placement determines difficulty of routing problem. • Density: lower bound on number of horizontal tracks needed to route the channel. • Maximum number of nets crossing from one end of channel to the other.

  13. Density = 3 Density = 2 Pin placement and routing a b c a b c b c a a c b before before

  14. Example: full adder layout • Two outputs: sum, carry. n1 x1 n4 n2 x2 sum n3 carry

  15. Layout methodology • Generate candidates, evaluate area and speed. • Can improve candidate without starting from scratch. • To generate a candidate: • place gates in a row; • draw wires between gates and primary inputs/outputs; • measure channel density.

  16. Density = 5 a x1 x2 n1 n2 n3 n4 b s cout c A candidate layout

  17. Improvement strategies • Swap pairs of gates. • Doesn’t help here. • Exchange larger groups of cells. • Swapping order of sum and carry groups doesn’t help either. • This seems to be the placement that gives the lowest channel density. • Cell sizes are fixed, so channel height determines area.

  18. Left-edge algorithm • Basic channel routing algorithm. • Assumes one horizontal segment per net. • Sweep pins from left to right: • assign horizontal segment to lowest available track.

  19. Example A B B C A B C

  20. ? aligned Limitations of left-edge algorithm • Some combinations of nets require more than one horizontal segment per net. A B B A

  21. Vertical constraints • Aligned pins form vertical constraints. • Wire to lower pin must be on lower track; wire to upper pin must be above lower pin’s wire. A B B A

  22. Dogleg wire • A dogleg wire has more than one horizontal segment. A B B A

  23. Rat’s nest plot • Can be used to judge placement before final routing.

  24. Simulation • Goals of simulation: • functional verification; • timing; • power consumption; • testability.

  25. Types of simulation • Circuit simulation: • analog voltages and currents. • Timing simulation: • simple analog models to provide timing but not detailed waveforms. • Switch simulation: • transistors as semi-ideal switches.

  26. Types of simulation, cont’d. • Gate simulation: • logic gates as primitive elements. • Models for gate simulation: • zero delay; • unit delay; • variable delay. • Fault simulation: • models fault propagation (more later).

  27. 1 0 1 Example: switch simulation + 0 c + d X X o b X a 1 0 c

  28. 0 1 0 Example, cont’d. + 0 c + d 1 0 o b 1 a 1 0 0 c

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