class20 crosstalk ii_calculations
TRANSCRIPT
Crosstalk
Calculation and SLEM
Crosstalk Calculation
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Topics
Crosstalk and ImpedanceSuperpositionExamplesSLEM
Crosstalk Calculation
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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.
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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.
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Super Positioning of Modes
Odd Mode SwitchingEven Mode Switching
Even States
Single Bit StatesRising Edge
OddFalling Edge
0 No Change(Line stays high or low,no transition occurs)
Odd States
,0 , 0
,Don’t Care State 0 0
Digital States that can occur in a 2 conductor system
Total of 9 states
= Single bit state
V
Time
V
Time
1.0
Line 1 Line 2
½ Even Mode
½ 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 ExampleCalculate 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 milsW=7mils
Er=4.5
S=10mils30[Ohms] 50[inches]
InputV
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
?
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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-007Co = 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 milsW=7mils
Er=4.5
S=10mils
Sanity Check:The odd and even velocities are the same
30[Ohms] 50[inches]
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Two Coupled Line Example (Cont..)
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
0.5V
Time
0.5V
Time
0.5V
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
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Two Coupled Line Example (Cont..)
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:
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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Two Coupled Line Example (Cont..)
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
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Two Coupled Line Example (Cont..)
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]
0.5V
Time
Line B
Case iv
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Two Coupled Line Example (Cont..)
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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Two Coupled Line Example (Cont..)
embebed ustrip
L 3.02E-07Lm 3.35E-08C 1.67E-10Cm 1.86E-11
Zodd 38.004847Vodd 1.41E+08Zeven 47.478047Veven 1.41E+08Tdelay 8.98E-09
Rin 30Odd [V] 0.5Even [V] 0.5Vinit(odd) 0.2794275Vinit(even) 0.3063968sum 0.5858243diff 0.02696932xodd 0.5588552x(odd+even) 1.17164852x(even-odd) 0.0539386
Simulating in HSPICE results are identical to the hand calculation:
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Assignment1
Use PSPICE and perform previous simulations
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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-007Co = 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 milsW=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-07Lm 3.35E-08C 1.15E-10Cm 4.06E-12
Zodd 47.492434Vodd 1.77E+08Zeven 54.989427Veven 1.64E+08Td(odd) 7.19E-09Td(even) 7.75E-09Rin 30Odd [V] 0.5Even [V] 0.5Vinit(odd) 0.3064327Vinit(even) 0.3235075sum 0.6299402diff 0.01707472xodd 0.61286542x(odd+even) 1.25988032x(even-odd) 0.0341495
HSPICE Results:Single Bit switching, two coupled microstrip example
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HSPICE Results of MicrostripVodd 176724383Veven 163801995.6length[in] 50length[m] 1.27delay odd 7.18633E-09delay even 7.75326E-09delta[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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Assignment 2 and 3
Use PSPICE and perform previous simulations
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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 3N digital states for N signal conductorsEach 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
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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 3N digital states for N signal conductorsThere are 3N digital states for N signal conductors
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Three Coupled Microstrip Example
H=4.5 milst=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-007Co = 1.15088e-010 -4.03272e-012 1.15326e-010 -5.20092e-013 -4.03272e-012 1.15088e-010
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Three Coupled Microstrip Example
Zmode
56.887
50.355
46.324
v
1.609108
1.718108
1.789108
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:
ZevenL
2 2 2 L1 2
C2 2 2 C
1 2
Ut Zeven 58.692
ZoddL
2 2 2 L1 2
C2 2 2 C
1 2
Ut Zodd 43.738
Veven1.0
L2 2 2 L
1 2 C
2 2 2 C1 2
Vodd1.0
L2 2 2 L
1 2 C
2 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 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 3N 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.
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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]
Imp
edan
ce[O
hm
s]
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 line2) The further the lines are spaced apart the the less the impedance changes
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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
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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 20Edge to Edge Spacing [mils]
Imp
edan
ce[O
hm
s]
Z single bit states Z odd statesZ even states
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 SLEMSLEM 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 phaseIt 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
123
123
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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 the approximationsUse 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)
ZevenL
2 2 2 L1 2
C2 2 2 C
1 2
Ut
ZoddL
2 2 2 L1 2
C2 2 2 C
1 2
UtVinit=Vin(Zstate/(Rin+Zstate))
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32Trading 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
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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 tolerance10[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 20Edge to Edge Spacing [mils]
Impe
danc
e[O
hm
s]
Z single bit states Z odd statesZ even states
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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 may help or hurt (recall reverse cross talk!)
Add Guard tracesOne needs to be careful to ground the guard traces sufficiently, otherwise you could actually increase the cross talkAt GHz frequency this becomes very difficult and should be 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 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