14
S. Reda EN160 SP’08 esign and Implementation of VLSI System (EN1600) Lecture 16: Interconnects & Wire Engineering

Lecture16

Embed Size (px)

DESCRIPTION

for more tutorials visit www.technofizi.in

Citation preview

Page 1: Lecture16

S. Reda EN160 SP’08

Design and Implementation of VLSI Systems(EN1600)

Lecture 16: Interconnects & Wire Engineering

Page 2: Lecture16

S. Reda EN160 SP’08

Interconnects introduce cross talk

• A capacitor does not like to change its voltage instantaneously.

• A wire has high capacitance to its neighbor.– When the neighbor switches from 1→ 0 or 0 → 1, the wire

tends to switch too.– Called capacitive coupling or crosstalk.

• Crosstalk has two harmful effects:1. Increased delay on switching wires

2. Noise on nonswitching wires

Page 3: Lecture16

S. Reda EN160 SP’08

1. Crosstalk impacts delay

• Assume layers above and below on average are quiet– Second terminal of capacitor can be ignored– Model as Cgnd = Ctop + Cbot

• Effective Cadj depends on behavior of neighbors– Miller effect

A BCadjCgnd Cgnd

Page 4: Lecture16

S. Reda EN160 SP’08

2.Crosstalk also creates noise

• Crosstalk causes noise on nonswitching wires• If victim is floating:

– model as capacitive voltage divider

Cadj

Cgnd-v

Aggressor

Victim

Vaggressor

Vvictim

adjvictim aggressor

gnd v adj

CV V

C C

Page 5: Lecture16

S. Reda EN160 SP’08

Crosstalk noise effects

• Usually victim is driven by a gate that fights noise– Noise depends on relative resistances– Victim driver is in linear region, agg. in saturation– If sizes are same, aggressor = 2-4 x Rvictim

1

1adj

victim aggressorgnd v adj

CV V

C C k

aggressor gnd a adjaggressor

victim victim gnd v adj

R C Ck

R C C

Cadj

Cgnd-v

Aggressor

Victim

Vaggressor

Vvictim

Raggressor

Rvictim

Cgnd-a

(time constant depends on the ratio of time constant of aggressor and victim)

Page 6: Lecture16

S. Reda EN160 SP’08

Simulating noise induced by couplingAggressor

Victim (undriven): 50%

Victim (half size driver): 16%

Victim (equal size driver): 8%

Victim (double size driver): 4%

t (ps)

0 200 400 600 800 1000 1200 1400 1800 2000

0

0.3

0.6

0.9

1.2

1.5

1.8

• Noise is less than the noise margin → nothing happens• Static CMOS logic will eventually settle to correct output even if

disturbed by large noise spikes• But glitches cause extra delay and power

• Large driver oppose the coupling sooner → smaller noise• Memories are more sensitive

Page 7: Lecture16

S. Reda EN160 SP’08

Wire Engineering

• Goal: achieve delay, area, power goals with acceptable noise

• Possible solutions:1. Width, Spacing, Layer

2. Shielding

3. Repeater insertion

4. Wire staggering and differential signaling

vdd a0a1gnd a2vdd b0 a1 a2 b2vdd a0 a1 gnd a2 a3 vdd gnd a0 b1

Page 8: Lecture16

S. Reda EN160 SP’08

1/2. Width, Spacing, Layer, Shielding

vdd a0a1gnd a2vdd b0 a1 a2 b2vdd a0 a1 gnd a2 a3 vdd gnd a0 b1

• Widening a wire reduces resistance but increases capacitance (but less proportionally) → RC delay product improves

• Spacing reduces capacitance → improves RC delay• Layers

• Coupling can be avoided if adjacent lines do not switch → shield critical nets with power or ground wires on one or both sides to eliminate coupling

Page 9: Lecture16

S. Reda EN160 SP’08

3. Repeater insertion

• R and C are proportional to l• RC delay is proportional to l2

– Unacceptably great for long wires

• Break long wires into N shorter segments– Drive each one with a repeater or buffer

source sink

buffer/repeater

Two questions:

A. What is the position that minimizes the delay?B. How many repeaters to insert to minimize the delay?

Page 10: Lecture16

S. Reda EN160 SP’08

What is the delay from source to sink without any repeaters?

source sinkL

Page 11: Lecture16

S. Reda EN160 SP’08

A. If you have one repeater, where is the optimal position to insert it?

source sink

x

L

buffer

Minimum delay is attained when

Makes sense to add a buffer only if D0-D1 > 0

Page 12: Lecture16

S. Reda EN160 SP’08

A. If there are multiple buffers, where are the optimal locations to insert them?

source sink

L/2

L

If Rbuf = Rsnk and Csnk = Cbuf

then the optimal location for the buffer is at distance L/2

If there are N buffers then minimum delay will occur when they are equally spaced, i.e., separation distance is L/(N+1)

L/2

Page 13: Lecture16

S. Reda EN160 SP’08

B. What is the optimal number N of repeaters?

source buffer

Page 14: Lecture16

S. Reda EN160 SP’08

4. Staggering and differential signaling

staggering

differential signaling