Calculation and SLEM Crosstalk

Crosstalk

Calculation and SLEM

Topics

 Crosstalk and Impedance

 Superposition

 Examples

 SLEM

Cross Talk and Impedance

Impedance is an electromagnetic parameter and is therefore effected by the

electromagnetic environment as shown in the preceding slides.

In the this second half, we will focus on

looking at cross talk as a function of

impedance and some of the benefits of

viewing cross talk from this perspective.

Using Modal Impedance’s for Calculating Cross Talk

 Any state can be described as a

superposition of the system modes.

 Points to Remember:

Each mode has an impedance and velocity associated with it.

In homogeneous medium, all the modal

velocities will be equal.

Super Positioning of Modes

Odd Mode Switching Even Mode Switching

Even States Single Bit States

Rising Edge Odd Falling Edge 0 No Change

(Line stays high or low, no transition occurs)

Odd States

,

0

,

,

Don’t Care State 0 0

Digital States that can occur in a 2 conductor system

= Single

bit state ^V

Time V

Time 1.0

Line 1 Line 2

½ Even Mode

½ Odd Mode

V 0.5

Time V 0.5

Time

V 0.5

Time

-0.5 V

Time

For a two line case, there are two modes

+

Two Coupled Line Example

Calculate the waveforms for two coupled lines when one is driven from the low state to the high and the other is held low.

H=4.5 mils

t=1.5 mils W=7mils

Er=4.5

S=10mils ^30[Ohms]

50[inches]

Input

V

Time

V

Time 1.0

Line A

Line B

Output?

V

Time

?

V

Time

?

Line A

Line B

At Driver At Receiver

V

Time

?

V

Time

?

Two Coupled Line Example

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]

Two Coupled Line Example

Now I deconvolve the the input voltage into the even and odd modes:

= Single bit

state ^V

Time

V

Time 1.0

Line A Line B

½ Even Mode

½ Odd Mode

V 0.5

Time

V 0.5

Time V 0.5

Time

-0.5 V

Time

Line A Line B

This allows one to solve four easy problems and simply add the solutions together!

Case i Case ii

Case iii Case iv

Two Coupled Line Example

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 ii are really the same:

A 0.5[V] step into a

Zeven=47.5[] line:

V 0.5

Time

V 0.5

Time

Case i Case ii

Td=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]

Two Coupled Line Example

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 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

Two Coupled Line Example

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 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]

V 0.5

Time

Line B

Case iv

Two Coupled Line Example

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]

Two Coupled Line Example

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] 0.5

Vinit(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:

Assignment1

 Use PSPICE and perform previous

simulations

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

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

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

Crosstalk Calculation

HSPICE Results of Microstrip

Vodd 176724383

Veven 163801995.6

length[in] 50

length[m] 1.27

delay 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

Calculation

Assignment 2 and 3

 Use PSPICE and perform previous

simulations

Modal Impedance’s for more than 2 lines

 So far we have looked at the two line crosstalk

case, however, most practical busses use more than two lines.

 Points to Remember:

For ‘N’ signal conductors, there are ‘N’ modes.

There are 3^N digital states for N signal conductors

Each mode has an impedance and velocity associated with it.

In homogeneous medium, all the modal velocities will be equal.

Any state can be described as a superposition of the modes

Three 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}

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

, , …

,

…

…

There are 3

digital states for N signal conductors

There are 3

digital states for N signal conductors

Three Coupled Microstrip Example

H=4.5 mils

t=1.5 mils W=7mils Er=4.5

S=10mils S=10mils

From HSPICE:

Lo = 3.02174e-007

3.32768e-008 3.01224e-007

9.01613e-009 3.32768e-008 3.02174e-007 Co = 1.15088e-010

-4.03272e-012 1.15326e-010

-5.20092e-013 -4.03272e-012 1.15088e-010

Three Coupled Microstrip Example

Zmode

56.887 50.355 46.324



v

1.60910 ⁸ 1.71810 ⁸ 1.78910 ⁸



Tv

0.53 0.663

0.53

0.707 1.52410 ¹⁵

0.707

0.467 0.751

0.467



Using the approximations gives: Actual modal info:

Zeven L_{2 2} 2 L _{1 2}

C_{2 2} 2 C _{1 2} Ut Zeven 58.692

Zodd L_{2 2} 2 L _{1 2} C_{2 2} 2 C _{1 2} Ut

Zodd 43.738

Veven 1.0

L_{2 2} 2 L _{1 2}  C_{2 2} 2 C _{1 2}

Vodd 1.0

L_{2 2} 2 L _{1 2}  C_{2 2} 2 C _{1 2}

Veven 1.59210  ⁸

Vodd 1.85610  ⁸

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.

Three Coupled Microstrip Example

Single Bit Example: HSPICE Result

Points to Remember

 The modal impedances can be used to hand calculate crosstalk waveforms

 Any state can be described as a superposition of the modes

 For ‘N’ signal conductors, there are ‘N’ modes.

 There are 3

digital states for N signal conductors

 Each mode has an impedance and velocity associated with it.

 In homogeneous medium, all the modal velocities

will be equal.

Crosstalk Trends

Key Topics:

 Impedance vs. Spacing

SLEM

 Trading Off Tolerance vs. Spacing

Impedance vs Line Spacing

Impedance Variation for a Three Conductor Stripline (Width=5[mils])

0 20 40 60 80 100 120

5 10 15 20

Edge to Edge Spacing [mils]

Impedance[Ohms]

Z single bit states Z odd states Z 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

Single Line Equivalent Model ^(SLEM)

 SLEM is an approximation that allows some cross talk effects to be modeled without running fully coupled

simulations

 Why would we want to avoid fully coupled simulations?

Fully coupled simulations tend to be time consuming and dependent on many

assumptions

Single Line Equivalent Model ^(SLEM)

 Using the knowledge of the cross talk impedances, one can change a single

transmission line’s impedance to approximate:

Even, Odd, or other state coupling

Impedance Variation for a Three Conductor Stripline (Width=5[mils])

0 20 40 60 80 100 120

5 10 15 20

Edge to Edge Spacing [mils]

Impedance[Ohms]

Z single bit states Z odd states Z even states

30[Ohms] Zo=90[]

30[Ohms] Zo=40[]

Equiv to Even State Coupling

Equiv to Odd State Coupling

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

Single Line Equivalent Model ^(SLEM)

 How does one create a SLEM model?

There are a few ways

Use the [L] and [C] matrices along with the approximations

Use the [L] and [C] matrices along with Weimin’s MathCAD program

Excite the coupled simulation in the desired state and back calculate the equivalent impedance (essentially TDR the simulation

)

Zeven L_{2 2} 2 L _{1 2} C_{2 2} 2 C _{1 2} Ut

Zodd L_{2 2} 2 L _{1 2} C_{2 2} 2 C _{1 2} Ut

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

Trading Off Tolerance vs. Spacing

 Ultimately in a design you have to create guidelines specifying the trace spacing and specifying the tolerance of the motherboard

impedance

i.e. 10[mil] edge to edge spacing with 10% impedance variation

 Thinking about the spacing in

terms of impedance makes this

much simpler

Trading Off Tolerance vs. Spacing

 Assume you perform simulations with no

coupling and you find a solution space with an impedance range of

Between ~35[  ] to ~100[  ]

Two possible 65[  ] solutions are

15[mil] spacing with 15% impedance tolerance 10[mil] spacing with 5% impedance tolerance

Impedance Variation for a Three Conductor Stripline (Width=5[mils])

0 20 40 60 80 100 120

5 10 15 20

Edge to Edge Spacing [mils]

Impedance[Ohms]

Z single bit states Z odd states Z even states

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 may help or hurt (recall reverse cross talk!)

 Add Guard traces

One needs to be careful to ground the guard traces sufficiently, otherwise you could actually increase the cross talk

At GHz frequency this becomes very difficult and should be avoided

 Route on different layers and route orthogonally

In Summary:

 Cross talk is unwanted signals due to coupling or leakage

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

 Cross talk can also be analyzed as a change in the transmission line’s impedance

 Reverse cross talk is often the dominate cross talk in a design

(just because the forward cross talk is small or zero, does not mean you can ignore cross talk!)

 A SLEM approach can be used to budget impedance

tolerance and trace spacing