class20 crosstalkii calculations

7/31/2019 Class20 CrosstalkII Calculations

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Crosstalk

Calculation and SLEM


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Crosstalk Calculation

Topics

Crosstalk and Impedance

Superposition

Examples

SLEM


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Cross Talk and Impedance

Impedance is an electromagnetic parameterand is therefore effected by theelectromagnetic environment as shown inthe preceding slides.

In the this second half, we will focus on

looking at cross talk as a function ofimpedance and some of the benefits ofviewing cross talk from this perspective.


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Using Modal Impedances forCalculating Cross Talk

Any state can be described as asuperposition of the system modes.

Points to Remember:Each mode has an impedance and velocityassociated with it.

In homogeneous medium, all the modalvelocities will be equal.


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Super Positioning of Modes

Positive

Going

Voltage

Negative

Going

Voltage

Odd Mode Switching

Positive

Going

Voltage

Positive

Going

Voltage

Even Mode Switching

Even States

Single Bit States

Rising Edge

Falling Edge

0 No Change(Line stays high or low,

no transition occurs)

Odd States

,0 , 0

,Dont Care State 0 0

Digital States that can occur

in a 2 conductor systemTotal of 9 states

= Single

bit stateV

Time

V

Time

1.0

Line 1 Line 2

Even

Mode

Odd

Mode

0.5V

Time

0.5V

Time

0.5V

Time

-0.5

V

Time

For a two line case, there are two modes

+


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Two Coupled Line Example (Cont..)

First one needs the [L] and [C] matrices and then I need the modal

impedances and velocities.The following [L] and [C] matrices were created in HSPICE.

Lo = 3.02222e-007

3.34847e-008 3.02222e-007

Co = 1.67493e-010

-1.85657e-011 1.67493e-010

Zodd 38.0 [Ohms]

Vodd 1.41E+08 [m/s]

Zeven 47.5 [Ohms]

Veven 1.41E+08 [m/s]

H=4.5 mils

t=1.5 mils

W=7mils

Er=4.5

S=10mils

Sanity Check:

The odd and even

velocities are the same

30[Ohms] 50[inches]


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Now I deconvolve the the input voltage into the even

and odd modes:

= Single bit

stateV

Time

V

Time

1.0

Line A Line B

Even

Mode

Odd Mode

0.5V

Time

0.5V

Time

0.5V

Time

-0.5

V

Time

Line A Line B

This allows one tosolve four easy

problems and

simply add the

solutions together!

Case i Case ii

Case iii Case iv


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Zodd 38.0 [Ohms]

Vodd 1.41E+08 [m/s]

Zeven 47.5 [Ohms]Veven 1.41E+08 [m/s]

30[Ohms] 50[inches]Case i and Case iiare really the same:

A 0.5[V] step into a

Zeven=47.5[ ] line:Line A Line B

0.5V

Time

0.5V

Time

Case i Case iiTd=len*Veven=8.98[ns]Vinit=0.5[V]*Zeven/(Zeven+30[Ohms])

Vinit=.306[V]

Vrcvr=2*Vinit=.612[V]

0.000[V]

Driver (even)

0.0[ns] 9.0[ns]

0.306[V]

0.612[V]

0.000[V]

Receiver (even)

0.0[ns] 9.0[ns]

0.306[V]

0.612[V]


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Zodd 38.0 [Ohms]

Vodd 1.41E+08 [m/s]


30[Ohms] 50[inches]Case iii is -0.5[V]step into a

Zodd=38[ ] line:

Line A

Td=len*Vodd=8.98[ns]Vinit=-0.5[V]*Zodd/(Zodd+30[Ohms])

Vinit=-.279[V]

Vrcvr=2*Vinit=-.558[V]

Driver (odd)

0.000[V]9.0[ns]

0.279[V]

0.558[V]

-.558[V]

-.279[V]

Receiver (odd)

0.000[V]9.0[ns]

0.279[V]

0.558[V]

-.558[V]

-.279[V]

-0.5

V

Time

Case iii


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Zodd 38.0 [Ohms]

Vodd 1.41E+08 [m/s]


30[Ohms] 50[inches]Case iv is 0.5[V]step into a

Zodd=38[ ] line:

Td=len*Vodd=8.98[ns]Vinit=0.5[V]*Zodd/(Zodd+30[Ohms])

Vinit=.279[V]

Vrcvr=2*Vinit=.558[V]

0.000[V]

Driver (odd)

0.0[ns] 9.0[ns]

0.279[V]

0.558[V]

0.000[V]

Receiver (odd)

0.0[ns] 9.0[ns]

0.279[V]

0.558[V]

0.5V

Time

Line B

Case iv


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Line A (Receiver)

0.0[V]

0.5[V]

1.0[V]

-1.0[V]

-0.5[V]9.0[ns]

6.12-.558=

.0539[V]

Line B (Driver)

0.0[V]

0.5[V]

1.0[V]

-1.0[V]

-0.5[V]9.0[ns]

.306+.279=.585[V]

Line B (Receiver)

0.0[V]

0.5[V]

1.0[V]

-1.0[V]

-0.5[V]9.0[ns]

.612+.558=1.17[V]

0.0[V]

0.5[V]

1.0[V]

-1.0[V]

-0.5[V]9.0[ns]

Line A (Driver)

.306-.279=.027[V]

0.000[V]

Driver (even)

0.0[ns] 9.0[ns]

0.306[V]

0.612[V]

Driver (odd)

0.000[V]9.0[ns]

0.279[V]

0.558[V]

-.558[V]

-.279[V]

0.0[V]

0.5[V]

1.0[V]

-1.0[V]

-0.5[V]9.0[ns]

Line A (Driver)

.306-.279=.027[V]

0.000[V]

Driver (odd)

0.0[ns] 9.0[ns]

0.279[V]

0.558[V]

0.000[V]

Driver (even)

0.0[ns] 9.0[ns]

0.306[V]

0.612[V]

Line B (Driver)

0.0[V]

0.5[V]

1.0[V]

-1.0[V]

-0.5[V]9.0[ns]

.306+.279=.585[V]


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embebed

ustrip

L 3.02E-07

Lm 3.35E-08

C 1.67E-10

Cm 1.86E-11

Zodd 38.004847

Vodd 1.41E+08

Zeven 47.478047

Veven 1.41E+08

Tdelay 8.98E-09

Rin 30

Odd [V] 0.5

Even [V] 0.5Vinit(odd) 0.2794275

Vinit(even) 0.3063968

sum 0.5858243

diff 0.0269693

2xodd 0.558855

2x(odd+even) 1.1716485

2x(even-odd) 0.0539386

Simulating in HSPICE results are identical to

the hand calculation:


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Assignment1

Use PSPICE and perform previoussimulations


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Super Positioning of Modes

Continuing with the 2 line case, the following [L] and [C]matrices were created in HSPICE for a pair of microstrips:

Lo = 3.02222e-007

3.34847e-008 3.02222e-007

Co = 1.15083e-010

-4.0629e-012 1.15083e-010

Zodd=47.49243354 [Ohms]Vodd=1.77E+08[m/s]

Zeven=54.98942739 [Ohms]

Veven=1.64E+08 [m/s]

H=4.5 mils

t=1.5 mils

W=7mils

Er=4.5

S=10mils

Note:

The odd and even velocities

are NOT the same


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Microstrip Example

The solution to this problem follows the same

approach as the previous example with one

notable difference.

The modal velocities are different and result in

two different Tdelays:

Tdelay (odd)= 7.19[ns]

Tdelay (even)= 7.75[ns]

This means the odd mode voltages will arrive at

the end of the line 0.56[ns] before the even mode

voltages


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Microstrip Cont..

ustrip

L 3.02E-07

Lm 3.35E-08

C 1.15E-10

Cm 4.06E-12

Zodd 47.492434

Vodd 1.77E+08

Zeven 54.989427

Veven 1.64E+08

Td(odd) 7.19E-09

Td(even) 7.75E-09

Rin 30

Odd [V] 0.5

Even [V] 0.5

Vinit(odd) 0.3064327

Vinit(even) 0.3235075

sum 0.6299402

diff 0.0170747

2xodd 0.6128654

2x(odd+even) 1.2598803

2x(even-odd) 0.0341495

HSPICE Results:

Single Bit switching, two coupled microstrip example


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HSPICE Results of MicrostripVodd 176724383

Veven 163801995.6

length[in] 50

length[m] 1.27delay odd 7.18633E-09

delay even 7.75326E-09

delta[sec] 5.66932E-10

The width of the pulse is calculated from the mode

velocities. Note that the widths increases in 567[ps]increments with every transit

567[ps] 1134[ps] 1701[ps] 2268[ps]

Calculation


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Modal Impedances formore than 2 lines

So far we have looked at the two linecrosstalk case, however, most practicalbusses use more than two lines.

Points to Remember:For N signal conductors, there are N modes.There are 3N digital states for N signalconductorsEach mode has an impedance and velocity

associated with it.In homogeneous medium, all the modal velocitieswill be equal.Any state can be described as a superposition of

the modes


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Three Conductor Considerations

Even States

Single Bit States

Rising Edge

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

000

0 000, , ,0,0 0

,The remaining states can be fit into the 1 and 2 bits cases for 27 total cases

,,

There are 3N digital states for N signal conductors


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Three Coupled Microstrip Example

Zmode

56.887

50.355

46.324

v

1.609108

1.718108

1.789108

Tv0.53

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 22 L

1 2

C2 2

2 C1 2

Ut Zeven 58.692

ZoddL

2 22 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 1.592108

Vodd 1.856108

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.


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Three Coupled Microstrip ExampleSingle Bit Example: HSPICE Result


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Points to Remember

The modal impedances can be used to handcalculate crosstalk waveforms

Any state can be described as asuperposition of the modes

For N signal conductors, there are Nmodes.

There are 3N digital states for N signalconductors

Each mode has an impedance and velocityassociated with it.

In homogeneous medium, all the modalvelocities will be equal.


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Crosstalk Trends

Key Topics:

Impedance vs. Spacing

SLEM

Trading Off Tolerance vs. Spacing


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Impedance vs Line Spacing

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]

Impedance[Ohms]

Z single bit states Z odd statesZ even states

As we have seen in the preceding sections,1) Cross talk changes the impedance of the line

2) The further the lines are spaced apart the the

less the impedance changes


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Single Line Equivalent Model(SLEM)

SLEM is an approximation that allowssome cross talk effects to bemodeled without running fully coupledsimulations

Why would we want to avoid fullycoupled simulations?

Fully coupled simulations tend to be time

consuming and dependent on manyassumptions

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Single Line Equivalent Model(SLEM)

Using the knowledge of the cross talkimpedances, one can change a singletransmission lines impedance to approximate:

Even, Odd, or other state coupling


(Width=5[mils])

0

20

40

60

80

100

120


Impedance[O

hms]


30[Ohms] Zo=90[ ]

30[Ohms] Zo=40[ ]

Equiv to

Even State

Coupling

Equiv to

Odd State

Coupling

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Single Line Equivalent Model(SLEM) Limitations of SLEM

SLEM assumes the transmission line is in aparticular state (odd or even) for its entiresegment length

This means that the edges are in perfect phase

It also means one can not simulate random bit patternsproperly with SLEM (e.g. Odd -> Single Bit -> Evenstate)

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

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Single Line Equivalent Model(SLEM) How does one create a SLEM model?

There are a few waysUse the [L] and [C] matrices along with theapproximations

Use the [L] and [C] matrices along with WeiminsMathCAD program

Excite the coupled simulation in the desired state andback calculate the equivalent impedance (essentially

TDR the simulation)

Zeven

L2 2

2 L1 2

C2 2

2 C1 2

Ut

ZoddL

2 22 L

1 2

C2 2

2 C1 2

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

32T di Off T l S i


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Ultimately in a design you have to

create guidelines specifying thetrace spacing and specifying thetolerance of the motherboard

impedancei.e. 10[mil] edge to edge spacing with10% impedance variation

Thinking about the spacing interms of impedance makes thismuch simpler

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Assume you perform simulations with no

coupling and you find a solution space with animpedance range ofBetween ~35[W] to ~100[W]

Two possible 65[W] solutions are

15[mil] spacing with 15% impedance tolerance10[mil] spacing with 5% impedance tolerance


(Width=5[mils])

0

20

40

60

80

100

120


Impedance[O

hms]


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Reducing Cross Talk Separate traces farther apart

Make the traces short compared to the rise time Make the signals out of phase

Mixing signals which propagate in opposite directions mayhelp or hurt (recall reverse cross talk!)

Add Guard tracesOne needs to be careful to ground the guard traces

sufficiently, otherwise you could actually increase thecross talkAt GHz frequency this becomes very difficult and shouldbe avoided

Route on different layers and route orthogonally

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In Summary:

Cross talk is unwanted signals due to

coupling or leakage Mutual capacitance and inductance between

lines creates forward and backwardstraveling waves on neighboring lines

Cross talk can also be analyzed as a changein the transmission lines impedance

Reverse cross talk is often the dominatecross talk in a design

(just because the forward cross talk is small or zero, does notmean you can ignore cross talk!)

A SLEM approach can be used to budgetimpedance tolerance and trace spacing

class20 crosstalkii calculations

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