si bootcamp : crosstalkweb.cecs.pdx.edu/~muilenta/psev/lecture_10.pdf · intel confidential 2 si...

Intel Confidential 1

SI Bootcamp: Crosstalk00030687 Rev. 01 (03/16/2007)

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Si Si BootcampBootcamp: Crosstalk : Crosstalk

00030687 00030687

Rev #: Rev #: 0202

Date: 03/16/2007Date: 03/16/2007

Intel Confidential 2SI Bootcamp: Crosstalk00030687 Rev. 01 (03/16/2007)

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Chapter 3Crosstalk

Author: Stephen Hall, Garrett Hall, Maynard Falconer


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Signal Integrity BootcampModules

1. Introduction – Bus design Overview

2. Transmission Line Fundamentals

3. Crosstalk

4. Non-ideal Interconnect

5. Modeling Buffers

6. Differential Signaling

7. Timing Fundamentals


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Signal Integrity Boot CampModules (Page 2)

8. Equalization Overview

9. NIRPS/SSO/Pwr Delivery

10. Connectors/Packages/Vias

11. Frequency Domain Analysis

12. High Speed Measurements

13. Peak Distortion Analysis

14. Empirical Modeling - Bus

15. Jitter Analysis


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What is Crosstalk?

Crosstalk Induced Noise

Effect of crosstalk on transmission line parameters

Modal analysis

Overview


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What is crosstalk

Crosstalk is coupling of energy from one signal to another

Crosstalk occurs from the interaction of electric and magnetic fields between transmission lines

Crosstalk affects the electrical parameters of a line

– Impedance, Z

– Propagation delay, TD

– Equivalent Inductance and Capacitance

Crosstalk affects timing, noise, ringback, overshoot, and even EMI

As physical dimensions shrink, components are forced closer to each other, which tends to increase crosstalk


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Crosstalk Induced Noise

Key Topics:

Mutual Inductance and capacitance

Coupled noise

Circuit Model

Transmission line matrices


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The mechanisms of crosstalk:

– Mutual capacitance (electric field)

– Mutual inductance (magnetic field)

Mutual Inductance and CapacitanceMutual Inductance and Capacitance

Mutual Capacitance, Cm Mutual Inductance, Lm

Cm

Lm

Zo

Zo

far

Zs

Zo

near

Zs

Zo

near

Zo

Zo

far


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The mechanism that governs this transfer of energy are the following familiar equations

Mutual Inductance and Capacitance Mutual Inductance and Capacitance ““Mechanism of couplingMechanism of coupling””

dt

dILV mLm =

dt

dVCI mCm =

The mutual inductance will induce current on the victim line opposite of the driving current

The mutual capacitance will pass current through the mutual capacitance that flows in both directions on the victim line


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The near and far end victim line currents sum to produce the near and the far end crosstalk noise

Crosstalk Induced NoiseCrosstalk Induced Noise““Coupled CurrentsCoupled Currents””

ICmLm ILm

LmCmfarLmCmnear IIIIII −=+=

Zs

Zo

Near end

Zo

Zo

Far end

Zs

Zo

Near end

Zo

Zo

Far end

Driver


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Near end crosstalk is always positive for a rising edge

– Currents from Lm and Cm always add and flow into the node

For PCB’s, the far end crosstalk is “usually” negative for a rising edge

– more on this later

Crosstalk Induced NoiseCrosstalk Induced Noise““Voltage Profile of Coupled NoiseVoltage Profile of Coupled Noise””

Driven Line

Un-driven Line“victim”

Driver

Near End

Far End

Zs

Zo

Zo

Zo


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Graphical ExplanationGraphical Explanation

TD

2TD

~Tr

~Tr

far end crosstalk

Near end crosstalk

Zo

V

Time = 2TD

Zo

V

Time = 0

Zo

Near end crosstalk pulse at T=0 (Inear)

Far end crosstalk pulse at T=0 (Ifar)

Zo

Zo

V

Time= 1/2 TD

Zo

V

Time= TD

ZoFar end of current terminated at T=TD

Near end current terminated at T=2TD


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Approximate Crosstalk Approximate Crosstalk EquationsEquations

LCXTD =

+=

C

C

L

LVA MMinput

4

−−=

C

C

L

L

T

LCXVB MM

r

input

2

Terminated Victim TD

2TD

Tr ~Tr Tr

AB

Driven Line


Zs

Zo

Zo

Zo

Near end

Far end

X=line lengthX=line length


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Approximate Crosstalk Approximate Crosstalk EquationsEquations

+=

C

C

L

LVA MMinput

4

CB2

1=

−−=

C

C

L

L

T

LCXVC MM

r

input

Far end open victim

TD

2TD

Tr ~Tr ~Tr

AB

C

Driven Line


Zs

Zo

Zo


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Exercise

What will the Far End crosstalk look like?

–sketch

TD

2TD

Tr Tr Tr

A

3TD


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Answer

1. Examine the terminated case

2. Since the near end is open, it will be instantly reflected and doubled

3. The reflected near end crosstalk will propagate down the victim line and superimpose with the far end crosstalk

4. The far end crosstalk pulse will appear, and then the reflected near end pulse will instantly be superimposed upon it

+=

C

C

L

LVA MMinput

term4

+=

C

C

L

LVA MMinput

open2

TD ~Tr

B

Aopen

3TD


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AnswerAnswer

Driven Line


Driver

Zs

Zo

Zo

Near End

Far End

Near End Open Victim

TD

2TD

Tr Tr Tr

A

B

C

3TD

+=

C

C

L

LVA MMinput

2

−−=

C

C

L

L

T

LCXVB MM

r

input

2

+=

C

C

L

LVC MMinput

4

The Crosstalk noise characteristics are dependent on the termination of the victim line


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Creating a Crosstalk ModelCreating a Crosstalk Model““Equivalent CircuitEquivalent Circuit””

The circuit must be distributed into N segments as shown in the transmission line chapter

K1

L11(1)

L22(1)

C1G(1)

C12(1)K1

L11(2)

L22(2)

C1G(2)

C12(2)

C2G(2)C2G(1)

K1

L11(n)

L22(n)

C1G(n)

C12(n)

C2G(n)

C1G C2G

C12

2211

12

LL

LK =

Line 1

Line 2

Line 1 Line 2


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The transmission line Matrices are used to represent the electrical characteristics

The Inductance matrix is shown, where:

– Lnn = the self inductance of line N per unit length (all other lines open)

– Lmn = the mutual inductance between line M and N

Creating a Crosstalk ModelCreating a Crosstalk Model““Transmission Line MatricesTransmission Line Matrices””

Inductance Matrix =

NNN

N

LL

LL

LLL

1

2221

11211 ...


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The Capacitance matrix is shown, where:

– Cnn = the self capacitance of line N per unit length where:

– Cng = The capacitance between line N and ground

– Cmn = Mutual capacitance between lines M and N

Creating a Crosstalk ModelCreating a Crosstalk Model““Transmission Line MatricesTransmission Line Matrices””

Capacitance Matrix =

−

−

−−

NNN

N

CC

CC

CCC

1

2221

11211 ...

∑+= mutualsNGNN CCC

12111 CCC G +=

For example, for the 2 line circuit shown earlier:

(Cm’s are usually reported negative. Don’t get confused, ignore the sign)


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Example Calculate near and far end crosstalk-induced noise magnitudes and sketch the

waveforms of circuit shown below:

Vsource=2V, Trise = 100ps.Length of line is 2 inches. Assume all terminations are 70 Ohms. Assume the following capacitance and inductance matrix per unit inch:

1. Calculate the characteristic impedance and the delay.2. Is this a “matched” system?3. Calculate the peak crosstalk.

=

nHnH

nHnHL

869.9103.2

103.2869.9

−

−=

pFpF

pFpFC

051.2239.0

239.0051.2

vR1 R2


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Example (cont.)

Ω=== 4.69051.2

869.9

11

11

pF

nH

C

LZO

pspFnHinCLTD 284051.2869.921111 =⋅⋅=⋅=

=

nHnH

nHnHL

869.9103.2

103.2869.9

−

−=

pFpF

pFpFC

051.2239.0

239.0051.2

Therefore, the system is “matched”

Impedance:

Delay:


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VpF

pF

nH

nHV

C

C

L

LVV

input

near 082.0051.2

239.0

869.9

103.2

4

1

4 11

12

11

12 =

+=

+=

VpF

pF

nH

nH

ps

pFnHinchV

C

C

L

L

T

LCXVV

rise

input

far 137.0051.2

239.0

869.9

103.2

100*2

051.2*869.9*2*1

2

)(

11

12

11

12 −=

−=

−=

Near end crosstalk voltage amplitude

Far end crosstalk voltage amplitude:

100ps/div

20

0m

V/d

iv

Example (cont.)

Driving stepDriving step

Near endNear end

Far endFar end


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Effect of Crosstalk on Transmission line Parameters

Key Topics:

Odd and Even Mode Characteristics

Microstrip vs. Stripline

Modal Termination Techniques

Effect Switching Patterns

Single Line Equivalent Model (SLEM)

Modal Analysis


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Electromagnetic Fields between two driven coupled lines will interact with each other

These interactions will affect the impedance and delay of the transmission line

A 2-conductor system will have 2 propagation modes

– Even Mode (Both lines driven in phase)

– Odd Mode (Lines driven 180o out of phase)

The interaction of the fields will cause the system electrical characteristics to be directly dependent on patterns

Odd and Even Transmission ModesOdd and Even Transmission Modes

Even Mode

Odd Mode


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Potential difference between conductors increases of the effective Capacitance by the mutual capacitance (Cm)

Odd Mode TransmissionOdd Mode Transmission

Magnetic Field: Odd mode

Electric Field: Odd mode

+1 -1

+1 -1

Because currents are flowing in opposite directions, the total inductance is reduced by the mutual inductance (Lm)

V

-I

Lm

dt

dILmL

dt

IdLm

dt

dILV

)(

)(

−=

−+=

I

C2g

C1g Cm


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Odd Mode TransmissionOdd Mode Transmission““Derivation of Odd Mode InductanceDerivation of Odd Mode Inductance””

121111 LLLLL modd −=−=

dt

dIL

dt

dILV

dt

dIL

dt

dILV

mO

mO

122

211

+=

+=

2211LL

Lk m=

L11

L22

I2

I1

+ V2 -

+ V1 -

Since the signals for odd-mode switching are always opposite, I1 = -I2 and

V1 = -V2, so that:

dt

dILL

dt

IdL

dt

dILV

dt

dILL

dt

IdL

dt

dILV

mOmO

mOmO

2222

1111

)()(

)()(

−=−

+=

−=−

+=

Thus, since LO = L11 = L22,

Meaning that the equivalent inductance seen in an odd-mode environment

is reduced by the mutual inductance.

OLLL == 2211


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Odd Mode TransmissionOdd Mode Transmission““Derivation of Odd Mode CapacitanceDerivation of Odd Mode Capacitance””

mmgodd CCCCC +=+= 111 2

C2g

C1g Cm

V1

V2

C1g = C2g = CO = C11 – C12

So, dt

dVC

dt

dVCC

dt

VVdC

dt

dVCI

dt

dVC

dt

dVCC

dt

VVdC

dt

dVCI

mmOmO

mmOmO

121222

212111

)()(

)()(

−+=−

+=

−+=−

+=

And again, I1 = -I2 and V1 = -V2, so that:

dt

dVCC

dt

VVdC

dt

dVCI

dt

dVCC

dt

VVdC

dt

dVCI

mOmO

mgmO

22222

11

1111

)2())((

)2())((

+=−−

+=

+=−−

+=

Thus,

Meaning that the equivalent capacitance for odd mode switching increases.


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Odd Mode TransmissionOdd Mode Transmission““Odd Mode Transmission CharacteristicsOdd Mode Transmission Characteristics””

Impedance:

Thus the impedance for odd mode behavior is:

)2:(

1211

1211

oddaldifferenti

odd

oddodd

ZZNote

CC

LL

C

LZ

=

+

−==

and the propagation delay for odd mode behavior is:

))(( 12111211 CCLLCLTD oddoddodd +−==

Propagation Delay:


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Since the conductors are always at a equal potential, the capacitance is reduced by the mutual capacitance (Cm)

Even Mode TransmissionEven Mode Transmission

Because currents are flowing in the same direction, the inductance is increased by the mutual inductance (Lm)

V

I

Lmdt

dILmL

dt

IdLm

dt

dILV

)(

)(

+=

+=

I

Electric Field:

Even mode

Magnetic Field:

Even mode

+1 +1

+1 +1

C2g

C1g Cm


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Even Mode TransmissionEven Mode TransmissionDerivation of even Mode Effective InductanceDerivation of even Mode Effective Inductance

121111 LLLLL meven +=+=

2211LL

Lk m=

L11

L22

I2

I1

+ V2 -

+ V1 -

Again, consider the circuit:

Since the signals for even-mode switching are always equal and in the same

direction so that I1 = I2 and V1 = V2, so that:

dt

dIL

dt

dILV

dt

dIL

dt

dILV

mO

mO

122

211

+=

+=

dt

dILL

dt

IdL

dt

dILV

dt

dILL

dt

IdL

dt

dILV

mOmO

mOmO

2222

1111

)()(

)()(

+=+=

+=+=

Thus,

Meaning that the equivalent inductance of even mode behavior increases

by the mutual inductance.

OLLL == 2211


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Even Mode TransmissionEven Mode TransmissionDerivation of even Mode Effective CapacitanceDerivation of even Mode Effective Capacitance

meven CCCC −== 110

Mutual Capacitance:Again, consider the circuit:

C2g

C1g Cm

V1

V2

dt

dVC

dt

VVdC

dt

dVCI

dt

dVC

dt

VVdC

dt

dVCI

OmO

OmO

22222

11111

)(

)(

=−

+=

=−

+=

Thus,

Meaning that the equivalent capacitance during even mode behavior

decreases.


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Even Mode TransmissionEven Mode Transmission““Even Mode Transmission CharacteristicsEven Mode Transmission Characteristics””

Impedance:

Thus the impedance for even mode behavior is:

1211

1211

CC

LL

C

LZ

even

eveneven

−

+==

and the propagation delay for even mode behavior is:

))(( 12111211 CCLLCLTD eveneveneven −+==

Propagation Delay:


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Odd and Even Mode Comparison Odd and Even Mode Comparison for Coupled for Coupled MicrostripsMicrostrips

Input waveformsEven mode (as seen on line 1)

Odd mode (Line 1)

v2

v1

Probe point

Delay difference due to modal velocity differences

Impedance difference

V1

V2

Line 1

Line2

Time

Volt

s


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Question:

When would odd and even mode delays differ?

))(( 12111211 CCLLCLTD eveneveneven −+==

))(( 12111211 CCLLCLTD oddoddodd +−==

+1 +1

Electric Field: Odd mode

+1 -1

Electric Field: even mode

Answer: For odd mode, more of the electric fields are fringing though the air, reducing the εr and increasing the propagation velocity


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Microstrip vs. Stripline CrosstalkMicrostrip vs. Stripline CrosstalkCrosstalk Induced Velocity ChangesCrosstalk Induced Velocity Changes

Odd and Even mode electric fields in a microstrip will have different percentages of the total E field fringing through the air which will change the effective Er

+1 +1 +1 -1

The equivalent dielectric constant for a coupled microstrip system will change with the driving pattern

– Induces both impedance an velocity variations

Er=4.2

Er=1.0

Er=4.2

Er=1.0

equivr

cvelocity

_ε=

equivr

oC

LZ

_

1

ε∝=

Higher % of E fields in the air Lower Er_effLower % of E fields in the air Higher Er_eff


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Microstrip vs. Stripline CrosstalkMicrostrip vs. Stripline CrosstalkCrosstalk Induced Velocity ChangesCrosstalk Induced Velocity Changes

Subsequently, if the transmission line is implemented in a homogeneous dielectric, the velocity must stay constant between even and odd mode patterns

If the dielectric is homogeneous (i.e., buried microstrip or stripline) , the effective dielectric constant will not change because the electric fields will never fringe through air

+1 +1 +1 -1

Er=4.2Er=4.2

Stripline E field patterns


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Microstrip vs. Stripline CrosstalkMicrostrip vs. Stripline CrosstalkCrosstalk Induced NoiseCrosstalk Induced Noise

The constant velocity in a homogeneous media (such as a stripline) forces far end crosstalk noise to be zero

11

12

11

12

1112121112111112

1211121112111211 ))(())((

C

C

L

L

CLCLCLCL

CCLLCCLL

TDTD evenodd

=

+−=+−

−+=+−

=

02

)_(11

12

11

12 =

−−=

C

C

L

L

T

LCXVstriplinefarCrosstalk

r

input

Since far end crosstalk takes the following form:

Far end crosstalk is zero for a homogeneous εεεεr


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Termination TechniquesTermination TechniquesPi and T networksPi and T networks

Single resistor terminations do not work for coupled lines

3 resistor networks can be designed to terminate both odd and even modes

T Termination

-1

R1

R2

R3

+1Odd Mode

Equivalent-1

R1

R2

Virtual Ground

in center

+1Even Mode

Equivalent+1

R1

R2

2R3

2R3

oddZRR == 21

( )oddeven ZZR −=2

13

+1


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Termination TechniquesTermination TechniquesPi and T networksPi and T networks

The alternative is a PI termination

PI Termination

+1Odd Mode

Equivalent

R1

R2

R3

-1

½ R3

½ R3

+1Even Mode

Equivalent +1

R1

R2

evenZRR == 21

oddeven

oddeven

ZZ

ZZR

−= 23

R1

R2


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Single Line Equivalent Model

The effects of crosstalk can be estimated by incorporating the effects of crosstalk into a single line.

This allows quick estimates of the effects of crosstalk early without running fully coupled simulations

SLEM quickly estimates crosstalk effects on– Impedance variation

– TD variation

– Ringback/Overshoot

This allows a quick initial estimate line spacing tradeoffs


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Single Line Equivalent Model (SLEM) Using modal impedances, a single transmission line’s impedance can approximate:

–Even, Odd, or other state coupling

In this case, the crosstalk will introduce ~+/- 35% variation in the impedance at 10 mils for 2 line system

Impedance Variation for a Three Conductor Stripline

(Width=5[mils])

0

20

40

60

80

100

120

5 10 15 20Edge to Edge Spacing [mils]

Imp

ed

an

ce[

Oh

ms]

Z single bit states Z odd statesZ even states

30[Ohms] Zo=90[ΩΩΩΩ]

30[Ohms] Zo=40[ΩΩΩΩ]

Equiv to

Even State

Coupling

Equiv to

Odd State

Coupling

Zeven, TDeven

Zodd, TDodd


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Single Line Equivalent Model (SLEM)

How to create a SLEM model?

– One way is to use the formulas for even and odd modes

SLEM is only an approximation for more then 2 lines

– Works well for three lines, but exact for two lines

These models could then be used to estimate the impedance and velocity variations due to crosstalk. ie. In odd mode, even mode, at various spacing.

Vinit=Vin(Zstate/(Rin+Zstate))

1222

1222

1222

1222

2

2

2

2

CC

LLZodd

CC

LLZeven

⋅+

⋅−=

⋅−

⋅+=


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Single Line Equivalent Model (SLEM) Limitations of SLEM

– SLEM assumes the transmission line is in a particular state (odd or even) for it’s entire segment length

– This means that the edges are in perfect phase

– It also means one can not simulate random bit patterns properly with SLEM (e.g. Odd -> Single Bit -> Even state)

The edges maybe in

phase here, but not here

Three coupled lines, two with serpentining

V2

Time

V1

Time

V3

Time

1

2

3

1

2

3

1 2 3

Odd EvenSingle

Bit


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Example: Trading Off Tolerance vs. Spacing

Assume you perform simulations with no coupling and you find a solution space with an impedance range between ~35[Ω] to ~100[Ω]

– What are some possible design solutions?

Impedance Variation for a Three Conductor Stripline

(Width=5[mils])

0

20

40

60

80

100

120

5 10 15 20Edge to Edge Spacing [mils]

Imp

ed

an

ce[

Oh

ms]

Z single bit states Z odd statesZ even states

Answer:Answer:

Option1: Option1:

10 mil spacing with +/10 mil spacing with +/--5% 5%

manufacturing tolerance gives manufacturing tolerance gives

40*0.95<Z<95*1.0540*0.95<Z<95*1.05

= 38<Z<99.8= 38<Z<99.8

However, 5% tolerance is usually too However, 5% tolerance is usually too

expensive for high volume designsexpensive for high volume designs

Option 2: Option 2:

15 mil spacing with +/15 mil spacing with +/--15% tolerance15% tolerance

50*.85<Z< 81*1.15 = 42.5<Z<93.150*.85<Z< 81*1.15 = 42.5<Z<93.1


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Modal analysisModal analysis breaks the signals down into orthogonal combinations of components– Any signal can be described as a sum of orthogonal modes

– For instance, the line voltage is a sum of modal voltages

– The L or C is a combination of modal L’s or C’s

Points to remember– Each mode will have its own impedance and velocity

– Remember odd and even modes for n=2?

– All signals are a sum of modes

Why is modal analysis useful in bus design?– Bounds worst case impedance and delay variations due to switching patterns

– Similar to SLEM model, but exact calculation


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Super-positioning of Modes For a 2 line case there are 2 modes (odd and even)

Single bit states are a superposition of ½ odd and ½even mode


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A different way to look at far end crosstalk

In a 2 signal conductor microstrip system, the odd and even modes will have different velocities

Far end crosstalk results from the difference between odd and even mode delays

Line 1 Line 2

Even

Odd

Total

½

½

-½

-½

-½

1

0

TDeven-TDodd

Far end crosstalk Pulse

A

B

Line 1

Line 2“victim”

Driver

Near End

Far End

Zs

Zo

Zs

Zo

Zo

Zo

Zo

Zo


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Questions

Why is the Far End crosstalk usually negative?

Does Far End crosstalk pulse width change with line length?

Does far end crosstalk magnitude change with line length?

–zero rise time

–nonzero rise time

When will far end crosstalk be largest in magnitude?


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Modal analysis

Odd and Even modes correspond to specific switching patterns for n=2 signal case

For n>2 it gets more complicated

–Modes do not correspond to digital states someplace in between

Even Mode

Odd Mode

State = 1,1

State = -1,1

Combination of 3 modes State = 1,1,1


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N Conductor Considerations

Even States

Single Bit States

Rising Edge

Odd

Falling Edge

0 No Change

(Line stays high or low,

no transition occurs)

2 Bit Even States

2 Bit Odd States

Odd States

,

, , , 0

0

0 0 0

0 0 0 0 , , ,

0 , 0 0

,

… , ,

…

…

Odd and Even nomenclature is only correct for a 2

signal system

The number of modes in a multi-conductor system are equal to n, where n is the number of signals

3 modes ,many states. Don’t confuse modes with states

Switching states for 3 signal conductors


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Modal decomposition for N conductors

The circuit derived transmission line equations for a loss free transmission line system are shown, where z=line length

[ ] [ ] [ ]

[ ] [ ] [ ]),(),(

),(),(

tzVt

CtzIz

tzIt

LtzVz

∂

∂−=

∂

∂

∂

∂−=

∂

∂ If a sinusoidal signal is assumed, then

ωω

jz

e tj

=∂

∂

Subsequently:[ ] [ ] [ ]

[ ] [ ] [ ])()(

)()(

zVCjzIz

zILjzVz

⋅−=∂

∂

⋅−=∂

∂

ω

ω


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Modal decomposition (cont.)

We want an equation in voltage and current only

– Take the second order derivative with respect to z of the original equations and substitute

[ ] [ ] [ ] [ ] [ ] [ ]

[ ] [ ] [ ] [ ] [ ] [ ]),(),(),(

),(),(),(

2

2

2

2

tzIt

LCjtzVz

CjtzIz

tzVt

CLjtzIz

LjtzVz

∂

∂⋅−=

∂

∂−=

∂

∂

∂

∂⋅−=

∂

∂−=

∂

∂

ωω

ωωVoltage equation:

Current Equation:

If a transformation matrix [T] can be found so that LC and CL can be diaganolized, the modal equations can be uncoupled.

[ ] [ ] [ ][ ] [ ] [ ]MI

MV

ITzI

VTzV

⋅=

⋅=

)(

)(


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TrainingTraining

Modal decomposition (cont.) The matrices that accomplish the desired transformation are the Eigenvectors

– The eigenvectors allow you to separate the system into orthogonal components, or to diagonalize the matrix

[ ][ ]=

=

I

V

T

T

Writing the transmission line equations in matrix form

[ ][ ]

⋅

−

−=

^

^

^

^

0

0

I

V

Cj

Lj

I

V

dz

d

ω

ω

Using the modal transformation matrices, [Tv] and [TI], the voltage equation can be re-written in terms of the modal voltage

[ ] [ ] [ ] ]].[.[1

MIvm ITLTjVdz

d −−= ω

[ ] [ ]( )[ ] [ ]( )LC

CL

⋅

⋅EigenvectorEigenvector

EigenvectorEigenvector


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Modal decomposition (cont.)

The current can also be written in terms of the modal voltages

[ ] [ ]

[ ] [ ]

==

==

−

−

eN

e

e

IVe

eN

e

e

VIe

L

L

L

TLTL

C

C

C

TCTC

mod

2mod

1mod

1

mod

mod

2mod

1mod

1

mod

00

00

00

00

00

00

Subsequently, the [L] and [C] matrices can be converted to modal values

The modal impedance and delays are then calculated

[ ] [ ] [ ][ ][ ]mVIm VTCTjIdz

d..

1−−= ω

ieieie

ie

ie

ie CLTDC

LZ _mod_mod_mod

_mod

_mod

_mod ⋅==


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Effective Impedance In systems for more than 2 lines, the modal conditions will not necessarily correspond to a specific binary state

We need to calculate the equivalent impedance for a binary state, which is some combination of modes

[ ][ ]

[ ] [ ][ ] [ ]

[ ] [ ] [ ] 1

mod][−

⋅⋅=⋅

⋅== Iev

mI

mvo TZT

IT

VT

I

VZ

[ ]

−

−

=

1

1

1

lineV

[ ] [ ] [ ]lineoline VZI ⋅=−1

iline

iline

ieqivI

VZ

_

_

_ =

1 denotes a positive transition

-1 denotes a negative transition

The equivalent impedances are simply calculated from the line voltage and current for a given state


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TrainingTraining

Revisit of the single bit switching example

Tv matrix for a two line system (eigenvector of LC):

−

=

2

1

2

12

1

2

1

vT

[ ]

vvvVmTvVmTvVline

vvvVmTvVmTvVline

VmVm

VmVm

VmVmv

Vm

VmVmTv

v

vVlines

02

1

2

1

12

1

2

1

2

1

2

1

2

1

2

1

2

1

2

10

2

1

2

11

.

2

1

2

12

1

2

1

]].[[0

1

2221212

2121111

21

21

21

2

1

=−=+=

=+=⋅+⋅=+=

=

⋅+⋅−=

⋅+⋅=

−

==

=


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TrainingTraining

Zmode

56.887

50.355

46.324

=

v

1.609 108.

1.718 108.

1.789 108.

=

Tv

0.53

0.663

0.53

0.707

1.5241015.

0.707

0.467

0.751

0.467

=

Using the

approximations gives:Actual modal info:

ZevenL

2 2,2 L

1 2,.

C2 2,

2 C1 2,

.Ut. Zeven 58.692=

ZoddL

2 2,2 L

1 2,.

C2 2,

2 C1 2,

.Ut.

Zodd 43.738=

Veven1.0

L2 2,

2 L1 2,

. C2 2,

2 C1 2,

..

Vodd1.0

L2 2,

2 L1 2,

. C2 2,

2 C1 2,

..

Veven 1.592 108.=

Vodd 1.856 108.=

Modal velocities

The three mode vectors

Z[1,-1,1]=44.25[Ohms]

Z[1,1,1]=59.0[Ohms]

The Approx. impedances and velocities are pretty

close to the actual, but much simpler to calculate.

SLEM vs. Modal Analysis for 3 lines


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TrainingTraining

Why does SLEM work? To answer, let’s look at the 2 & 3 conductor transformation matrices

−−

−

=

467.0707.53.0

751.00663.0

467.0707.053.0

vT

−=

707.0707.0

707.0707.0vT

2 LinesOdd mode

(1,-1)

Even mode

(1,1)

3 LinesModes are NOT equivalent

to specific binary states, but

the extremes are “close” for

a 3-line system

Even state

similar to 1,1,1

(0.8, 1, 0.8)

Odd state

similar to -1,1,-1

(-0.62, 1, -0.62)

Modes are equivalent to

specific binary states (odd

and even)


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Question: What are the ways to reduce crosstalk?


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TrainingTraining

Answers

1. Route traces as far apart as possible

2. Route traces orthogonal to each other on different layers when no ground/power plane separates the layers

3. Use differential lines (more on this in the differential chapter)

4. Place grounded shield lines between conductors (be careful here, in some cases, this can INCREASE crosstalk)

5. Route with striplines to eliminate far end crosstalk

6. Minimize parallelism between traces when possible


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In Summary: Crosstalk is unwanted signals due to coupling

Mutual capacitance and inductance between lines creates forward and backwards traveling waves on neighboring lines

Crosstalk can also be analyzed as a change in the transmission line’s impedance and delay

A SLEM approach can be used to quickly budget impedance tolerance and trace spacing

N orthogonal modes will exist for an N line system


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TrainingTraining

Multiple Choice1. Crosstalk is a function of the

A) Electric fields between conductors

B) Magnetic fields between conductors

C) Capacitance between conductors

D) Mutual inductance

E) All of the above

2. The near end crosstalk is due toa) Current coupled through the mutual inductance

b) Voltage coupled through the mutual capacitance

c) The sum of the current coupled through both the mutual inductance and capacitance

d) The difference between current coupled through mutual inductance and capacitance

3. The current induced by the mutual inductance on the victim line a) Is in the same direction as the driving current

b) Is in the opposite direction as the driving current

c) Travels in both directions

d) Is dependent on the electric field

Topic Quiz


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TrainingTraining

Multiple Choice (example)4. For the cross-section shown, and a rising edge on the

driving linea) The far end crosstalk pulse is positive

b) The far end crosstalk pulse is negative

c) The near end crosstalk pulse is negative

d) The far end crosstalk pulse zero

5. The crosstalk pulse shown on the graph isa) The near end crosstalk

b) The far end crosstalk

c) A reflection

d) Due to a modal velocity change

Topic Quiz

εεrr = 1 (air)= 1 (air)

εεrr = 4 (FR4)= 4 (FR4)

driverdriver victimvictim

timetime

volts

volts

DrivingDriving

stepstep

crosstalkcrosstalk


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TrainingTraining

Multiple Choice (example)6. For the given L and C matrices and a 4 inch line, the

quantity A on the graph is

a) 1276.8ps

b) 159.6ps

c) 319.2ps

d) 638.5ps

7. The odd mode impedance of the above example is

a) 52.16 ohms

b) 104.32 ohms

c) 54.8 ohms

d) 61.9 ohms

Topic Quiz

timetime

volts

volts

DrivingDriving

stepstep

crosstalkcrosstalk

AA

=

nHnH

nHnHL

1.9103.1

103.11.9

−

−=

pFpF

pFpFC

8.2139.0

139.08.2


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TrainingTraining

Multiple Choice (example)8. The crosstalk pulse on the following line is

a) The near end

b) The far end

c) A reflection

d) Not physically possible

9. For a 4 inch line and the following matrices, the quantity A is

a) 1276.8ps

b) 159.6ps

c) 319.2ps

d) 638.5ps

Topic Quiz

timetime

volts

volts

DrivingDriving

stepstep

crosstalkcrosstalk

AA

=

nHnH

nHnHL

1.9103.1

103.11.9

−

−=

pFpF

pFpFC

8.2139.0

139.08.2


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TrainingTraining

Multiple Choice (example)10. For a 3 inch line and the following matrices, the

quantity B is approximately

a) 133.38ps

b) 142.27ps

c) 41.4ps

d) 82.8ps

11. For a 3 inch line and the matrices above, the even mode impedance is

a) 69.4 ohms

b) 81.32 ohms

c) 58.24 ohms

d) 72.29 ohms

Topic Quiz

timetime

volts

volts

DrivingDriving

stepstep

crosstalkcrosstalk

BB

=

nHnH

nHnHL

869.9103.2

103.2869.9

−

−=

pFpF

pFpFC

051.2239.0

239.0051.2


DEG DEG

TrainingTraining

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