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S. Reda EN160 SP’08 esign and Implementation of VLSI System (EN1600) Lecture 15: Interconnects

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Page 1: Lecture15

S. Reda EN160 SP’08

Design and Implementation of VLSI Systems(EN1600)

Lecture 15: Interconnects

Page 2: Lecture15

S. Reda EN160 SP’08

Transistors + Wires = Circuits

• Wires (interconnects) are as important as transistors– Speed– Power– Noise

• Alternating

layers run orthogonally

Page 3: Lecture15

S. Reda EN160 SP’08

How interconnects contribute to delay and power?

• Interconnects have resistance, capacitance (and inductance)

• Interconnects increase circuit delay:– The wire capacitance adds loading to each gate– Long wires have significant resistance that further

contribute to the delay

• Interconnects increase dynamic power:– Because of the wire capacitance

Page 4: Lecture15

S. Reda EN160 SP’08

Wire geometry

• Pitch = w + s• Aspect ratio: AR = t/w

– Old processes had AR << 1– Modern processes have AR 2

• Pack in many skinny wires

Page 5: Lecture15

S. Reda EN160 SP’08

1. Wire Resistance

• ρ = resistivity (W*m)

• R = sheet resistance (Ω/) is a dimensionless unit(!)

Page 6: Lecture15

S. Reda EN160 SP’08

How does the kind of metal impact resistivity?

• Until 180 nm generation, most wires were aluminum• Modern processes often use copper

– Cu atoms diffuse into silicon and damage FETs– Must be surrounded by a diffusion barrier

Page 7: Lecture15

S. Reda EN160 SP’08

Contact and via resistance

• Contacts and vias also have 2-20 Ω• Use many contacts for lower R

– Many small contacts for current crowding around periphery

Page 8: Lecture15

S. Reda EN160 SP’08

2. Wire capacitance

• Wire has capacitance per unit length– To neighbors– To layers above and below

• Ctotal = Ctop + Cbot + 2Cadj

layer n+1

layer n

layer n-1

Cadj

Ctop

Cbot

ws

t

h1

h2

Page 9: Lecture15

S. Reda EN160 SP’08

Factors impacting the capacitance

• Parallel plate equation: C = A/d– Wires are not parallel plates, but obey trends– Increasing area (W, t) increases capacitance– Increasing distance (s, h) decreases capacitance

• Dielectric constant– = k0

• 0 = 8.85 x 10-14 F/cm• k = 3.9 for SiO2

• Processes are starting to use low-k dielectrics– k 3 (or less) as dielectrics use air pockets

Page 10: Lecture15

S. Reda EN160 SP’08

M2 capacitance data (180nm)

• Typical wires have ~ 0.2 fF/mm – Compare to 2 fF/mm for gate capacitance)

• Polysilicon has lower C but high R– Use sparingly for very short wires between gates

0

50

100

150

200

250

300

350

400

0 500 1000 1500 2000

Cto

tal (aF

/m

)

w (nm)

Isolated

M1, M3 planes

s = 320

s = 480

s = 640

s= 8

s = 320

s = 480

s = 640

s= 8

Page 11: Lecture15

S. Reda EN160 SP’08

Given R and C, how to calculate interconnect delay?

• Wires are a distributed system– Approximate with lumped element models

• 3-segment -model is accurate to 3% in simulation• L-model needs 100 segments for same accuracy!

C

R

C/N

R/N

C/N

R/N

C/N

R/N

C/N

R/N

R

C

L-model

R

C/2 C/2

R/2 R/2

C

N segments

-model T-model

Page 12: Lecture15

S. Reda EN160 SP’08

Interconnect delay: the lumped case

0V

Vm Vout

Page 13: Lecture15

S. Reda EN160 SP’08

Interconnect delay: ideal analysis

tpd~0.38RC

Ideally, modeling using diffusion equation;

Page 14: Lecture15

S. Reda EN160 SP’08

Interconnect delay: distributed Elmore delay

R1 R2 R3 RN

C1 C2 C3 CN

r = resistance per unit length

c = capacitance per unit length

Page 15: Lecture15

S. Reda EN160 SP’08

Delay calculations

Assuming ideal wires:

Realistic wire modeling:

Page 16: Lecture15

S. Reda EN160 SP’08

Layer stack

• AMI 0.6 m process has 3 metal layers• Modern processes use 6-10+ metal layers• Example:

Intel 180 nm process

• M1: thin, narrow (< 3)– High density cells

• M2-M4: thicker– For longer wires

• M5-M6: thickest– For VDD, GND, clk

Intel 180nm process

0

200

400

600

800

1000

1200

1400

1600

1800

2000

6 5 4 3 2 1

metal layer

t(nm)

w(nm)

Why do you think different metal layers have different widths/thickness?