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1 MICROELETTRONICA Logical Effort and delay Lection 4

1 MICROELETTRONICA Logical Effort and delay Lection 4

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Page 1: 1 MICROELETTRONICA Logical Effort and delay Lection 4

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MICROELETTRONICA

Logical Effort and delay

Lection 4

Page 2: 1 MICROELETTRONICA Logical Effort and delay Lection 4

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Outline

• Introduction

• Delay in a Logic Gate

• Multistage Logic Networks

• Choosing the Best Number of Stages

• Example

• Summary

Page 3: 1 MICROELETTRONICA Logical Effort and delay Lection 4

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Introduction

• The designer faces many choices:

– What is the best topology for a function in order to have the least delay?

– How wide should the transistor be?

• The logical effort method helps to make these decisions– Uses a simple linear model for the delay– Allows simple calculations and comparison of

alternatives

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Delay in a Logic Gate

• Express delays in process-independent unit

absdd

τ= 3RC

12 ps in 180 nm process 40 ps in 0.6 μm process

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Delay in a Logic Gate

• Express delays in process-independent unit

• Delay has two components

• Effort delay f = gh (. stage effort)

• Parasitic delay p

absdd

d pf

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Delay in a Logic Gate

• Effort delay f = gh (stage effort)– Has two components

• g: logical effort– Measures relative ability of gate to deliver current– g 1 for inverter

• h: electrical effort = Cout / Cin

– Ratio of output to input capacitance– Sometimes called fanout

• Parasitic delay p– Represents delay of gate driving no load– Set by internal parasitic capacitance

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Delay Plots

d = f + p = gh + p

Electrical Effort:h = C

out / C

in

Nor

mal

ized

Del

ay: d

Inverter2-inputNAND

g = 1p = 1d = h + 1

g = 4/3p = 2d = (4/3)h + 2

Effort Delay: f

Parasitic Delay: p

0 1 2 3 4 5

0

1

2

3

4

5

6

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Computing Logical Effort

• DEF: Logical effort is the ratio of the input capacitance of a gate to the input capacitance of an inverter delivering the same output current.

• Measure from delay vs. fanout plots• Or estimate by counting transistor widths

A YA

B

YA

BY

1

2

1 1

2 2

2

2

4

4

Cin = 3g = 3/3

Cin = 4g = 4/3

Cin = 5g = 5/3

Page 9: 1 MICROELETTRONICA Logical Effort and delay Lection 4

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Catalog of Gates• Logical effort of common gates

Gate type Inputs

1 2 3 4 n

Inverter 1

NAND 4/3 5/3 6/3 (n+2)/3

NOR 5/3 7/3 9/3 (2n+1)/3

Tristate-mux 2 2 2 2 2

Page 10: 1 MICROELETTRONICA Logical Effort and delay Lection 4

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Catalog of Gates

Gate type Number of inputs

1 2 3 4 n

Inverter 1

NAND 2 3 4 n

NOR 2 3 4 n

Tristate / mux

2 4 6 8 2n

Parasitic delay of common gatesIn multiples of pinv (1)

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Example: Ring Oscillator

Estimate the frequency of an N-stage ring oscillator

Logical Effort: g = 1Electrical Effort: h = 1Parasitic Delay: p = 1Stage Delay: d = 2Frequency: fosc = 1/(2*N*d) =1/4N

31 stage ring oscillator in 0.6 mm process has frequency of ~ 200 MHz

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Example: FO4 Inverter

Estimate the delay of a fanout-of-4 (FO4) inverter

Logical Effort: g = Electrical Effort: h =Parasitic Delay: p =Stage Delay: d =

d

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Example: FO4 Inverter

dEstimate the delay of a fanout-of-4 (FO4) inverter

Logical Effort: g = 1Electrical Effort: h = 4parasitic Delay: p = 1Stage Delay: d = 5

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Multistage Logic Networks

10x y z

20g1 = 1h

1 = x/10

g2 = 5/3h

2 = y/x

g3 = 4/3h

3 = z/y

g4 = 1h

4 = 20/z

out-path

in-path

CH

C

iG g

i i iF f g h

Logical effort generalizes to multistage networks

Path Logical Effort

Path Electrical Effort

Path Effort

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Paths that Branch

Is it possible write F=GH?No! Consider paths that branch:

G = 1H = 90 / 5 = 18GH = 18h1 = (15 +15) / 5 = 6h2 = 90 / 15 = 6F = g1g2h1h2 = 36 = 2GH

5

15

1590

90

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Branching Effort

on path off path

on path

C Cb

C

iB b

Introduce branching effortAccounts for branching between stages in path

Now we compute the path effortF = GBH

ih BH

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Designing Fast Circuits

i FD d D P

1ˆ Ni if g h F

1ND NF P

•Delay is smallest when each stage bears same effort

•Thus minimum delay of N stage path is

•This is a key result of logical effort Find fastest possible delay Doesn’t require calculating gate sizes

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Gate Sizes

• How wide should the gates be for least delay?

• Working backward, apply capacitance transformation to find input capacitance of each gate given load it drives.

• Check work by verifying input cap spec is met.

ˆ

ˆ

out

in

i

i

CC

i outin

f gh g

g CC

f

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Example: 3-stage path

• Select gate sizes x and y for least delay from A to B

8 x

x

x

y

y

45

45

A

B

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Example: 3-stage path

8 x

x

x

y

y

45

45

A

B

Logical Effort G = Electrical Effort H =Branching Effort B =Path Effort F =Best Stage EffortParasitic Delay P =Delay D =

Page 21: 1 MICROELETTRONICA Logical Effort and delay Lection 4

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Example: 3-stage path

8 x

x

x

y

y

45

45

A

B

Logical effort G = (4/3)*(5/3)*(5/3) = 100/27Electrical Effort H = 45/8Branching Effort B = 3 * 2 = 6Path Effort F = GBH = 125Best Stage EffortParasitic Delay P = 2 + 3 + 2 = 7Delay D = 3*5 + 7 = 22 = 4.4 FO4

3ˆ 5f F

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Example: 3-stage path

• Work backward for sizes

y =

x =

8 x

x

x

y

y

45

45

A

B

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Example: 3-stage path

• Work backward for sizes

y = 45 * (5/3) / 5 = 15

x = (15*2) * (5/3) / 5 = 10

8 x

x

x

y

y

45

45

A

B

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Best Number of Stages

• How many stages should a path use?– Minimizing number of stages is not always fastest

• Example: drive 64-bit datapath with unit inverter

D =

1 1 1 1

64 64 64 64

Initial Driver

Datapath Load

N:f:D:

1 2 3 4

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Best Number of Stages• How many stages should a path use?

– Minimizing number of stages is not always fastest• Example: drive 64-bit datapath with unit inverter

D = NF1/N + P= N(64)1/N + N

1 1 1 1

8 4

16 8

2.8

23

64 64 64 64

Initial Driver

Datapath Load

N:f:D:

16465

2818

3415

42.815.3

Fastest

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Derivation

• Consider adding inverters to end of path

– How many give least delay?

• Define best stage effort

11

11

N

n

i invi

D NF p N n p

N - n

1 Extra Inverters

Logic Block:n

1 Stages

Path Effort F

1 1 1

ln 0N N Ninv

DF F F p

N

1NF

1 ln 0invp

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Best Stage Effort

1 ln 0invp

has no closed-form solution

Neglecting parasitics (pinv = 0), we find ρ= 2.718 (e)

For pinv = 1, solve numerically for ρ = 3.59

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Sensitivity Analysis

• How sensitive is delay to using exactly the best number of stages?

• 2.4 < < 6 gives delay within 15% of optimal– We can be sloppy!– I like = 4

1.0

1.2

1.4

1.6

1.0 2.00.5 1.40.7

N / N

1.151.26

1.51

( =2.4)(=6)

D(N

) /D

(N)

0.0

Page 29: 1 MICROELETTRONICA Logical Effort and delay Lection 4

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Review of Definitions

Term Stage Pathnumber of stages l Nlogical effort gelectrical effort

branching effort

effort

effort delay fparasitic delay p

delay d=f+P

iG gout

in

CCh out-path

in-path

C

CH

on-path off-path

on-path

C C

Cb iB b

f gh F GBH

F iD fiP p

i FD d D P

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Method of Logical Effort

1) Compute path effort

2) Estimate best number of stages

3) Sketch path with N stages

4) Estimate least delay

5) Determine best stage effort

6) Find gate sizes

F GBH

4logN F

1ND NF P

1ˆ Nf F

ˆi

i

i outin

g CC

f

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Limits of Logical Effort

• Chicken and egg problem– Need path to compute G– But don’t know number of stages without G

• Simplistic delay model– Neglects input rise time effects

• Interconnect– Iteration required in designs with wire

• Maximum speed only– Not minimum area/power for constrained delay

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Summary

• Logical effort is useful for thinking of delay in circuits– Numeric logical effort characterizes gates– NANDs are faster than NORs in CMOS– Paths are fastest when effort delays are ~4– Path delay is weakly sensitive to stages, sizes– But using fewer stages doesn’t mean faster paths– Delay of path is about log4F FO4 inverter delays– Inverters and NAND2 best for driving large caps

• Provides language for discussing fast circuits– But requires practice to master