Termination techniques for high-speed digital circuitsntuemc.tw/upload/file/20120419205522ce0f9.pdf · · 2012-04-19The γcan be solved with two possible solutions ... • Propagation

1

Crosstalk

Tzong-Lin Wu, Ph.D.

Department of Electrical Engineering

National Taiwan University

2

Topics:

Introduction

Transmission lines equations for coupled lines

symmetrical lines and symmetrically driven

homogeneous medium

symmetrical lines in homogeneous medium

symmetrical lines and asymmetrical driven

Qualitative description of crosstalk

Terminations

Measurement of crosstalk parameters

How crosstalk noise can be reduced

3

Introduction

Three characteristics: 1. Polarity of noise at near end

is opposite to that at the far end. 2. Noise duration at near end is

larger than that at the far end. 3. Noise magnitude at near end

is smaller than that at the far end.

4

Introduction : Physical Geometry of Coupled Lines

Two kind of coupling: 1. Inductive coupling 2. Capacitive coupling

5

Introduction: cross-talk mechanism of the inductive coupling

Qualitative understanding of the coupling mechanism

• How about the cross-talk mechanism of the capacitive coupling ??

6

Lm Cm Lm Lm Lm Cm Cm

A B

C D

Driving signal

Forward

coupling

Reverse

coupling

Tp

Tr

2Tp

A

B

D

C

Tp

Tr

2Tp

A

B

D

C Rise and fall times

same as input

Derivative of input

signal (negative)

Total coupled areas

are the same

Inductive Capacitive

7

Coupled lines equations

V z V z z L zt

I z L zt

I z

I z I z z CtV z z C

tV z z V z z

m

m

1 1 1 1 2

1 1 1 1 1 2

( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) [ ( ) ( )]'

8


dV z

dzL

tI z

dI z

dzC

tV z

( )( )

( )( )

Ll l

l l

L L

L L

Cc c

c c

C C C

C C C

C C

C C

m

m

m m

m m

m

m

LNM

OQP LNM

OQP

LNM

OQP

LNM

OQP

LNM

OQP

11 12

21 22

1

2

11 21

12 22

1

2

1

2

'

'

• Matrix forms

where

L 1, 2 : self-inductance per unit length of line 1 and 2 when isolated

Lm : mutual inductance between line 1 and 2

C’ 1,2 : capacitance (to GND) of line 1 and 2 when isolated **

Cm : mutual capacitance between line 1 and 2

9


11 12

21 22

Capacitance matrix=C C

C C

11 1 12gC C C

11 12

21 22

inductance matrix=L L

L L

11 12 1

21 22

1

Capacitance matrix=

N

N NN

C C C

C C

C C

11 12 1

21 22

1

Inductance matrix=

N

N NN

L L L

L L

L L

For multi-conductor coupled lines

10


1 10

2 20

1 10

2 20

( )

( )

( )

( )

j t z

j t z

j t z

j t z

V z V e

V z V e

I z I e

I z I e

0 0

0 0

( ) ( )

( ) ( )

V z j LI z

I z j CV z

2

02( ) 0A V

1 1 2 1

1 2 2 2

[ ]m m m m

m m m m

L C L C L C L CA LC

L C L C L C L C

2

2det( ) 0A

The γcan be solved with two possible solutions

for nontrivial solutions

11

Symmetrical lines and symmetrical driven

( )( )L L C Cm m1 1

Conditions:

• L1 = L2 (balanced lines)

C1 = C2

• The balanced lines are driven by

common mode or differential mode only

Characteristics:

• Propagation delay per-unit length:

• Characteristic impedance

1

0 1

1

10 1

1

1

1

1

1

m me

m c

m mo

m c

L L kZ Z

C C k

L L kZ Z

C C k

where

magnetic coupling coefficient

capacitive coupling coefficient

kL

L

kC

C

mm

cm

1

1

:

:

12

Symmetrical lines and symmetrical driven

Z e0 ,

Z o0 ,

13

Symmetrical lines and symmetrical driven (Microstrip example)

14

Symmetrical lines and symmetrical driven (another definition for even

and odd mode)

Note:

Coupling Factor = 20log(Z12)

Z12 = (Ze – Zo)/(Ze + Zo)

15

Physical concept for the even-mode and odd mode impedance

Odd mode I1 = -I2 and V1 = -V2

odd 1 112g m mC C C C C 11 11 12odd mL L L L L

odd 11 12odd

odd 11 12

L L LZ

C C C

odd odd odd 11 12 11 12TD L C L L C C

16

Physical concept for the even-mode and odd mode impedance

Even mode I1 = I2 and V1 = V2

even 11 mL L L even 0 11 mC C C C

even 11 12even

even 11 12

=L L L

ZC C C

odd even even 11 12 11 12TD L C L L C C

17

Impedance trends for the even- and odd-modes

What impedance behaviors do you see ?

18

Impedance trends for the even- and odd-modes

1. High-impedance traces exhibit significantly more impedance variation than do the lower-impedance trace because the reference planes are much farther in relation to the signal spacing.

2. At smaller spacing, the single-line impedance is lower than the target. This is because the adjacent traces increase the self-capacitance of the trace and effectively lower its impedance even when they are not active.

3. Higher coupling the traces behave, the larger variation even- and odd- impedances will have.

19

Mutual coupling trends for the microstrip and stripline

Mutual parasitic fall off exponentially with trace-trace spacing

20

Example by simulation:

Please compare the coupling characteristics for the three cases: • Ze • Zo

21

22

Effect of switching patterns on transmission line

Crosstalk induced flight time and signal integrity variations

Switching pattern

What phenomena do you see ??

23

As the switching patterns are in even-mode:

Over-driven behavior is seen (Overshooting).

Longer propagation delay is seen.

As the switching patters are in odd-mode:

Under-driven behavior is seen (Undershooting).

Shorter propagation delay is seen.

Why ??


24


• Even-mode pattern impedance changed from Z0 to Ze > Z0 . overdriven when Ze > Zs. Slower propagation velocity.

• Odd-mode pattern impedance changed from Z0 to Zodd < Z0 . under driven when Zodd < Zs. Faster propagation velocity.

25


Simulating traces in a multiconductor system using a single-line equivalent model (SLEM)

Target trace is line 2. Please find the equivalent impedance and time delay

22 12 232,eff

22 12 23

L L LZ

C C C

2,eff 22 12 23 22 12 23TD L L L C C C

??

26


22 12 232,eff

22 12 23

L L LZ

C C C

2,eff 22 12 23 22 12 23TD L L L C C C

27

Example by TDR measurement:

28

Example by TDR measurement: question one ??

29

Example by TDR measurement: question one??

Differential impedance (D.I.) is 150Ω without GND plane And about 100Ωwith GND plane. Why ??

30

First half: two traces refer each other like a twisted line, L↑ and C↓

Second half: two traces refer the GND plane, mutual coupling is quite small (km ≒ kc ≒ 0), so D.I.=2*Zodd ≒ 2Z0=100

31

Example by TDR measurement: question two??

What’s the variation of the differential impedance

For the differential pair crossing the slot ?

32

Example by TDR measurement: question two??

The gap behaves like a high impedance discontinuity which Will cause reflections.

33

Homogeneous Medium

Conditions:

• Real (not Quasi-) TEM mode propagating

on the transmission lines

Characteristics:

• Propagation delay per-unit length:

2

0

1 2

1 2

1

where (Coupling coefficient)m c

mm

mc

k

k k k

Lk

L L

Ck

C C

: For mutual inductance coupling

: For mutual capacitance coupling

34

Symmetrical Lines on Homogeneous Medium

Conditions:

• L1 = L2 (balanced lines)

C1 = C2

• The balanced lines are driven by

common mode or differential mode only

•

Characteristics:

• Characteristic impedance

Z Z Z

ZL

C

e o0 0 1

2

11

1

where

35

Example: homogeneous coupled line

Please compare the coupling characteristics for the three cases: 1. K 2. Ze 3. Zo

36

37

A Quiz:

Please compare their Ze, Zo, Ve, Vo, and C.

(a) (b)

38

Symmetrical Lines and asymmetrically driven

R R Z

RZ Z

Z Z

Z R R

e

e o

e o

o

1 2 0

30 0

0 0

0 1 3

2

1

2

(terminate even mode)

terminate odd mode)(

( ( / / ))

P-Terminations:

V V V

V V V

05

05

1 2

1 2

. ( )

. ( )

Asymmetrical driven

39

Symmetrical Lines and asymmetrically driven

R R Z

R Z Z

R Z

o

e o

e

1 2 0

3 0 0

3 0

1

2

2

(terminate odd mode)

(terminate even mode)

( R 2

( )

)

T - terminations

40

Imperfect Termination for Differential Transmission Line

Bridge Termination

AC Termination

Single ended termination

1. What are their values of the resistance? 2. What are their advantages and disadvantages ?

41


Bridge Termination

AC Termination

Single ended termination

2Zo

Zo

Zo Zo

Zo

42


Bridge: 1. Odd mode is completely absorbed 2. Even mode is completely reflected 3. Only one resistance is used. Single-ended 1. Odd mode is completely absorbed 2. Even mode has the reflection coefficient T= (Zo–Ze)/(Zo+Ze) 3. The extra damping of the even mode costs an additional resistance. AC • Odd mode is completely absorbed • Even mode is completely reflected at low frequency, but has the reflection T = (Zo–Ze)/(Zo+Ze) at high frequency.

3. Compared to bridge type, even mode is attenuated without increasing static power dissipation. 4. Needing three components.

43

Example: Termination for Differential Transmission Line

44


A simple bridge termination

45


Differential signal is still good

Common signal on each line is oscillating

A simple bridge termination, but Unbalance driving

46


Common-mode is damped

A Pi-termination for unbalance driving

47


Common-mode resonance Differential pair excited by common-mode signal

48


Single-ended termination for Unbalanced driving

Better than bridge termination

49


AC T-termination for Unbalanced driving

50


Bridge termination at the far end and Series termination at the near end

1. Power saving 2. Idea of LVDS

51


Bridge termination at the far end and Series termination at the near end For LVDS case

52

Example: Effects of Termination Networks on Signal-Induced EMI from

the Shields of Fibre Channel Cables Operating in the Gb/s Regime

2000 IEEE EMC Symposium

53



Measurement of S11 for both differential mode and common mode

Termination effect is reduced for high frequency range

54



differential

Why the termination effect is degraded at high frequency range ?

EMI

55

Quantitative description of crosstalk

Capacitive coupling Inductive Coupling

V x t K xd

dtV t xf f in( , ) ( )

V x t K V t x V t x Tb b in in d( , ) [ ( ) ( )] 2

KL

ZC Z

KL

ZC Z

fm

m

bm

m

1

2

1

4

1

1

1

1

( )

( )

(ns / cm) : Forward coupling coefficient

(dimensionless) : Backward coupling coefficient

If the coupled lines are symmetrical

(see appendix 7.1)

56

Quantitative description of crosstalk

In the loosely coupled system (km, kc << 1)

ZL

CL C1

1

1

1 1 and is accurate enough

KL

Z

C Zk k

KL

ZC Z k k

fm m

m c

bm

m m c

2 2

1

4

1

4

1

1

1

1

( ) ( )

( ) ( )

1. It is clear that the backward coupling can be reduced by decreasing the mutual coupling (Lm and Cm) or by increasing the coupling to ground(L1, L2, C1, and C2).

2. The forward crosstalk is zero for two identical lines in homogeneous medium. (not easy in practical circuits)

57

Crosstalk for a ramp step driver

V t V f t f t Tin m r( ) [ ( ) ( )]

58

Crosstalk for a ramp step driver

Forward crosstalk at the far end:

V l t K ld

dtV t T

K lV

TT t T T

f f in d

fm

r

d d r

( , ) ( )

= ,

= 0 elsewhere

Backward crosstalk at the near end

• If Tr < 2Td

V t K V t V t Tb b in in d( , ) [ ( ) ( )]0 2

V Kb b,max

• If Tr > 2Td

Vl

TK V

K

Tb

r

b mb

r

,max ( ) 2

59

SPICE simulation of crosstalk

60

Measurement of coupling coefficient (Frequency domain)

: (example) RAMBUS RIMM Connector

•Test module design

61


: RAMBUS RIMM Connector

• Calibration (Open, short, load )

62



• Theory (mutual capacitance)

Low freq. assumption

The slope of S21(f) at low frequency is taking to compute the Cm

Open circuit !!

63



• Theory (mutual inductance)

Short circuit !!

The slope of S21(f) at low frequency is taking to compute the Lm

64



• Measurement setup

65

Differential Impedance Measurement by TDR/TDT

They have good EMS (Immunity) and EMI (Interference) Performance.

66


Equivalent circuits of the differential interconnects: (Type I)

67


Equivalent circuits of the differential interconnects: (Type II)

Advantage is it is a rigorous equivalent circuit. Disadvantage is only valid in three conductor interconnects.

Negative impedance (not practical)

68


Equivalent circuits of the differential interconnects: (Type III)

69


Measurement Principle

70


Comparison of the equivalent model by the SPICE

71

A Final Question:

Can you explain why

72

How to reduce crosstalk noise

• Space out signal routing

• Route-void channels adjacent to critical nets

• Increase coupling to ground

• Run orthogonal rather than parallel

• Provide the homogeneous medium

• Using the slowest risetime and falling time

73

§10.1 Three conductor lines and crosstalk

Time Domain Crosstalk VS Frequency-domain Crosstalk

a. Two conductors transmission lines has no crosstalk problem. In order to have crosstalk, we need to have three or more conductors.

b.

)(tVs

NERFER

SRLR

_NEV

_FEV

Generator

Receptor

),( tzIG

),( tzIR

z0z Lz

74

c. The generator circuit consists of the generator conductor and reference conductor and has current IG(z,t) along the conductors and voltage VG(z,t) between them. The IG and VG will generate EM fields and interact with the receptor circuit.

d. The objective in a crosstalk analysis is to determine the near-end voltage VNE(t) and far-end voltage VFE(t) given the cross-sectional dimension and the termination characteristics VS(t), RS, RL, RNE, RFE.

75

e. Two type of crosstalk analysis

(1) Time-domain crosstalk :

(2) Frequency-domain crosstalk :

f. (1) For simplicity, we will assume that the medium surrounding the

conductors for these configurations is homogeneous.

(2) Closed-form expressions for the per-unit-length parameters for the lines in inhomogeneous medium, such as microstrip lines, are difficult to determine.

).( )( )( tVgiventVandtVpredict SNEFE

).cos()( )(ˆ )(ˆ wtVtVgivenjwVandjwVpredict SSNEFE

76

Transmission Line Equations

),( tzIG

),( tzIR

),(),( tzItzI RG

zRG

zRR

zRO

zLG

zLRzGm

zGR zCR

zCm zGG zCG

)( zzIG

)( zzIR

_

)( zzVR

_

)( zzVG

z

a. Assume only TEM mode present on the materials, so line voltages

VG(z,t) and VR(z,t), as well as line currents IG(z,t) and IR(z,t) can be

uniquely defined.

b. Equivalent circuit of TEM-mode on 3-conductor material :

77

In matrix terms :

t

tzVCC

t

tzVCtzVGGtzVG

z

tzI

t

tzVC

t

tzVCCtzVGtzVGG

z

tzI

t

tzIL

t

tzILtzIRRtzIR

z

tzV

t

tzIL

t

tzILtzIRtzIRR

z

tzV

RmR

GmRmRGm

R

Rm

GmGRmGmG

G

RR

GmRORGO

R

Rm

GGROGOG

G

),(),(),(),(

),(

),(),(),(),(

),(

),(),(),(),(

),(

),(),(),(),(

),(

t

tztz

z

tz

t

tztz

z

tz

),(),(

),(

),(),(

),(

VCGV

I

ILRI

V

78

c. Solutions

(1)In time domain, it is a difficult problem.

When R=G=0 (lossless lines), there are exact solutions in SPICE.

(2)In frequency domain, the exact solutions for above material equations are possible, we will show the exact solution for lossless case.

ORO

OOG

R

G

R

G

RRR

RRR

tzI

tzItz

tzV

tzVtz

where

,

),(

),(),( ,

),(

),(),(

RIV

, ,

G m G m m G m m

m R m R m m R m

L L G G G C C C

L L G G G C C C

L G C

79

Note: In frequency domain, the equations =>

jwt

jwt

eztz

eztz

where

zdz

zd

zdz

zd

)(ˆRe),(

)(ˆRe),(

jwˆ

jwˆ

)(ˆˆ)(ˆ

)(ˆˆ)(ˆ

II

VV

CGY

LRZ

VYI

IZV

80

§ 10.3.1 The per-unit-length parameters

Internal parameters : RG, RR, RO and internal inductance are not dependent on the line configurations.

a. For material in homogeneous medium (μ,ε)

(1)

2

1 1

2

2

2 2

2 2

2 2

1

Example : Three-conductor line

1

( )

( )

( )

( )

G m m R m

m R m m GG R m

mm

G R m

RG m

G R m

GR m

G R m

v

C C C L L

C C C L Lv L L L

LC

v L L L

LC C

v L L L

LC C

v L L L

LC CL 1 C L L

1 G LG GL 1 L(2)

81

§ Wide-Separation Approximation for Wires

a. (1) Magnetic flux of a current-carrying wire through a surface

(2) voltage between two points for a charge-carrying wire

1

20 ln2 R

RIμ

1R

2Ra

b

1R

2Rq

baV

1

2

0

ln2 R

RqVba

82

b. Parameters of three-conductor transmission lines

(1) Inductance :

00

0

,

RG

GR

IG

R

IR

Gm

IR

RR

oIG

GG

R

G

Rm

mG

R

G

IIL

IL

IL

I

I

LL

LL

LI

83

dR

dGR receptorgenerator

GND WOr

WRrWGr

1.

2. Self inductance

WOWR

RR

WOWG

G

WO

G

WG

GG

rr

dL

rr

d

r

d

r

dL

20

2000

ln2

ln2

ln2

ln2

GI

GI

WGr

WOr

G

84

3. Mutual inductance

(2) Capacitance

dR

dGR

WOr

WRrWGr

GI

GI

R

WOGR

RG

WO

R

GR

Gm

rd

dd

r

d

d

dL

ln2

ln2

ln2

0

00

R

G

mRm

mG

V

V

CCC

CC

CVq

R

G

Rm

mG

q

q

pp

pP

pqqCV

1

85

0000

,

GRRR qR

G

qG

Rm

qR

RR

qG

GG

q

V

q

VP

q

VP

q

VP ,

1. Self capacitance

WOWG

G

WO

G

WG

GG

rr

d

r

d

r

dP

2

0

00

ln2

1

ln2

1ln

2

1

Gq

Gq

WGr

WOr

_V

86

2. Mutual capacitance

WOGR

RG

WO

R

GR

Gm

rd

dd

r

d

d

dP

ln2

1

ln2

1ln

2

1

0

00

dR

dGR

WOr

WRrWGr

Gq

Gqreference

_RV

87

2

020

31

3

2010

0

0

00

0

41ln4

ln2

)(;ln2

ln2

2ln

2

2ln

2ln

2

S

hh

S

S

SSS

S

S

S

IL

r

h

h

h

r

h

IL

RG

IG

Rm

WG

G

G

G

WG

G

IG

GG

R

R

c. Example

GI

GI

WGr

WRr

RR

hG

hG

hR

hR

S

S1

S2

S3

88

§10.1.4 Frequency-Domain (Steady State) Crosstalk

§ General Solution :

a.

b.

CGY LRZ

VYI

I V IZV

jwjwwhere

zdz

zd

zI

zIz

zV

zVzz

dz

zd

R

G

R

G

ˆ,ˆ

)(ˆˆ)(ˆ

)(ˆ

)(ˆ)(ˆ,

)(ˆ

)(ˆ)(ˆ)(ˆˆ

)(ˆ

YZZY

YZII

ZYVV

,

)()(ˆ

)()(ˆ

2

2

2

2

generalin:Note

zdz

zd

zdz

zd

89

(1)

(2) ZY r

ZY T

rTZYT

T

ˆˆˆ

ˆˆˆ

ˆ0

0ˆˆˆˆˆˆ

:ˆ

2

2

21

ofseigenvaluetheis

ofrseigenvectotheiswhere

eddiagonalizr

r

thatsuchmatrixtiontransformaafindcanweif

R

G

)(ˆ)(ˆ)(ˆ

)(ˆ)(ˆ)(ˆ

)(ˆˆ)(ˆ

)(ˆˆ)(ˆ

)(ˆ

)(ˆ

ˆ0

0ˆ

)(ˆˆˆˆˆ)(ˆ

)(ˆ

)(ˆ)(ˆ)(ˆˆ

)(ˆˆˆˆ)(ˆˆ

ˆˆ

ˆˆ

2

2

2

2

1

2

2

1

11

2

2

zIezIezI

zIezIezI

zIrzI

zIrzI

zI

zI

r

r

zzdz

d

zI

zIentmodal currzzlet

zzdz

d

mRzr

mRzr

mR

mGzr

mGzr

mG

mRRmR

mGGmG

mR

mG

R

G

mm

mR

mG

m

RR

GG

ITZYTI

IIT

IZYTIT

90

(3)

c.

d.

mR

mG

mzr

zrzr

mzr

mzr

m

I

I

e

ewhere

formmatrixinz

G

G

ˆ

ˆˆ,

0

0

.ˆˆ)(ˆ

ˆ

ˆˆ

ˆˆ

I

e

IeIeI

)IeI(eTTrTZV

rTYrTZ rTZYT

)IeI(erTYIYV

IeIeTITI

mzr

mzr

mzr

mzr

mrz

mrz

m

z

by

zdz

dz

zz

ˆˆˆˆˆˆˆ)(ˆ

ˆˆˆˆˆˆˆˆˆˆˆ

ˆˆˆˆˆ)(ˆˆ)(ˆ

)ˆˆ(ˆ)(ˆˆ)(ˆ

ˆˆ11

1121

ˆˆ11

1 1 1

1

ˆ ˆ ˆ ˆ ˆ ˆ ˆˆ )

ˆ ˆ ˆ

Z Z Tr T Z( YZ

=Y YZ

c

see below for detail

91

e. Solving the four unknowns Im+ and Im

- by the

termination condition

(1)

11 1

1 2 1 1 1

1 1 1 1

1 1

11 1

ˆ ˆ ˆ ˆˆ

ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆˆ ˆ ˆ

ˆ ˆ ˆ ˆ ˆ ˆˆ ˆ

ˆ ˆ ˆ ˆˆ

ˆ ˆ ˆ ˆˆ

-2

Tr T YZ

T YZT r r T YZTr =1

Tr T YZ Tr T =1

YZ Tr T

Tr T = YZ

上式的

1ˆˆ ˆ ˆ ˆ( )ˆ

similar to the scalar results

zZ Z YZ

y

(0) (0)

( ) ( )

V V Z I

V Z I

S S

LL L

0 0

0 00

V , Z , Z

S LSS S L

NE FE

R RVwhere

R R

92

(2)

-m m

-m m

- -m m m m

ˆ ˆ -m m

ˆ ˆ -m m

ˆ ˆ ˆ-m m

0

ˆ ˆ ˆ ˆ ˆ(0) ( )

ˆ ˆ ˆ ˆ(0) ( )

ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ( ) ( )

ˆ ˆ ˆ ˆ ˆ( ) ( )

ˆ ˆ ˆ ˆ( ) ( )

ˆ ˆ ˆ ˆ ˆ ˆ ˆ( ) (

V Z T I I

I T I I

Z T I I V Z T I I

V Z T I I

I T I I

Z T I I Z T I

C

C S S

rL rLC

rL rL

rL rL rLC L

at z

at z L

L e e

L e e

e e e

ˆ -m m

m

-ˆ ˆm

- -m m m m

ˆ )

ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ

ˆˆ ˆ ˆ ˆ ˆ ˆ 0

ˆ ˆ ˆ ˆ, , .

I

Z Z T Z Z T I V

IZ Z T Z Z T

I I I I

rL

C S C SS

rL rLC L C L

e

e e

and can be solved

93

§ Exact Solution for Lossless Lines in Homogeneous Media

a. for lossless line :

b. It can be shown that :

2 2 2

ˆ ˆ0 ,

ˆ ˆ

hom .

ˆ ˆ,

R G Z L Y C

YZ LC 1 1

LC 1

T 1 r 1 1

and jw jw

w w diagonal matrix

where for ogeneous transmission lines

wj j

v

2

2

2ˆ ˆ ˆ(0)

1

21 ˆ

1

ˆ ˆ ˆ ˆ( )

NER NE m LG GDC

en NE FE

NE FEm GDC

NE FE LG

NE FEFER FE m GDC m GDC

en NE FE NE FE

j LRS

V V jwL L C S ID R R k

j LR R

jwC L C S VR R k

R RRSV L V jwL LI jwC LV

D R R R R

94

2 2 2 2

:

1 11

1 1

sin( )cos( ),

( ) , 1

( )

(

SG LR LG SR

en G R G R

SR LR SG LG

m m

G R G m R m

G S LG G m

S L S L

R

where

D C S w k jwCS

LC L S

L

L Ck coupling coefficient k

L L C C C C

L L R Rtime constant C C L

R R R R

time const

) NE FERR m

NE FE NE FE

R RL Lant C C L

R R R R

95

2

2

ˆˆ ˆ ˆ,

ˆ 1

ˆ 1

, , ,

1 .

1

SLGDC S GDC

S L S L

GCG G

G m

RCR R

R m

S NEL FESG LG SR LR

CG CG CR CR

VRV V I DC excitation at generator circuit

R R R R

LZ vL k

C C

LZ vL k

C C

R RR R

Z Z Z Z

if low impedance load

high

. impedance load

Characteristic impedances

of each circuit in the presence

of the other circuit.

96

§ Inductive and Capacitive coupling

a. Two reasonable assumptions

(1)

(2)

. . .

cos 1, 1

The line is electrically short i e L

C L

.

1

(1 )(1 )

, 1

en G R

en

The generator and receptor circuits are weakly coupled

k

D jw jw

if w is small then D

97

b.

(1)

(2)

ˆ ˆ ˆ

ˆ ˆ ˆ

NE NE FENE m GDC m GDC

NE FE NE FE

NE FEFEFE m GDC m GDC

NE FE NE FE

R R RV jwL LI jwC LV

R R R R

R RRV jwL LI jwC LV

R R R R

NERFER

_NEV

_FEV

ˆm GDCjwL LI

ˆm GDCjwC LV

ˆ ˆ( )

ˆ ˆ( ) ( )

.

m GDC m GDC

m GDC m GDC

djwL LI L L I emf

dt

d djwC LV C L V CV

dt dt

which is the independent current source due

to the voltage of generator

98

(3) Crosstalk is the superposition of two components; One is inductive coupling, the other one is capacitive coupling.

c. Intuitively :

(1) When low-impedance load (high current, low voltage), the

inductive coupling is dominant.

(2) When high-impedance load (low current, high voltage), the

capacitive coupling is dominant.

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Termination techniques for high-speed digital circuitsntuemc.tw/upload/file/20120419205522ce0f9.pdf · · 2012-04-19The γcan be solved with two possible solutions ... • Propagation