ECE 570 Session 12 – IC 752-E Computer Aided Engineering for Integrated Circuits

Signal Distribution - basics

Objective: Introduction to interconnect modeling

Outline: 1. Modeling with perfect materials 2. Imperfect dielectrics 3. Imperfect conductors

4. Model of a lossless line 5. Two special design cases 6. Model of multi-conductor lines.

Appendix : Details of Signal Propagation in Multiconductor Lines

1

1. Modeling of structures with perfect materials

Computation of model parameters (expressed per unit of length) assuming TEM approximation

Concepts:

resistors - power equivalency

capacitors - equivalency of charge (or stored potential energy)

inductors - -equivalency of stored magnetic energy. Computational problems:

electrostatic field equations

electromagnetic field equations (sinusoidal steady state)

2

Prototypical structure

y

x z

m 2 1

ε r2

ε r1

Maxwell relations:

q c V c V c V

i m

i i i im m= + + +

=

1 1 2 2

1 2

.....

, ,...,

3

The matrix of Maxwell capacitances

C ci j i j

m=

=, , 1 symmetric, positive definite, diagonally dominant.

The inductances (per unit of length): the concept:

ϕ i i i im mi i i

i m

= + + +

=

1 1 2 2

1 2

.....

, ,...,

The matrix of inductances

L i j i j

m=

=, , 1 symmetric, positive definite.

4

2. Imperfect dielectrics There are two types of lossy substrates to consider a) substrates with polarization losses: organic materials such as FR-4

≈ 100[k cm b) substrates with ohmic losses high resistivity ( ]Ω ) silicon. These two types of loss result in different material behavior depending on frequency of transmitted signals. Lossy dielectric is characterized as follows:

1;"' −=−= jj rrr εεε or else

tan"

'δ εε

= r

r .

Effect of losses on modeling

equivalent conductance: g Cr o( ) "ω ωε=

and equivalent capacitance: C CL r oω εb g = '

5

3. Imperfect conductors

Resistances are defined using equivalency of power dissipated in the model and in the structure

rJ dS

Iii i

iS

i

i=z

ρ

2

2

zr

J dS

Iij j

jS

i

j= ρ

2

2

6

Skin effect: resistance is a function of frequency r r r= +

r D

wtconst

rwt

constD

d c a c

= +

= +

− −

total resistance

resistance per unit of length

ρ ω

ρ ω1

w

D

t

Basics:

E E ex o

z

=

−

−δ

δ

δ ρπ µ

skin depth

E - M analysis yields

=2 f

ρ

x

z

Eo

Ex

7

Sheet resistance (surface resistance) ρR

R f

S

S

=

=

σ

π µρ2

δ

Examples:

Cu R f

Al R f

Ag R f

S

S

S

= ⋅

= ⋅

= ⋅

−

−

−

2 61 10

326 10

2 52 10

7

7

7

.

.

.

8

4. Model of a lossless line D x

0

∂∂

∂∂

∂∂

∂∂

ux

it

u u x t

ix

c ut

i i x t

= − =

= − =

, ,

, ,

b g

b g

Unit time delay: τ = c (TOF ) D D c= =τ

9

Wave equations: voltage xb F Iu x t u tv

u t xv

u x t u x t

forward backwardwave wave

, , ,g b g b g= −HG KJ + +FHGIKJ = ++ − + −

current i x ib g b=t x t i x t, , ,g b g−+ −

Characteristic impedance: i x tu x t

Zi x t

u x tZo o

++

−−= =,

,, ,

,b g b g b g b g Z co = /

10

Terminations VA

e

rL

rG

+ -

Source injection coefficient: V ZZ r

eAo

o G

=+

Reflection coefficients: r Z−ρ ρ ρGG o

G oL

L o

L or Zr Zr Z

=+

=−+

− ≤ ≤∗, ; 1 1

11

5. Two special design cases 5.1. Point-to-point transmission in CMOS technology

“the first reflection switching”

VB VA

Z ZD o

D

==ρ 0

Z ZR o

R

>>=ρ 1

12

Transients in transmission - the “first reflection” switching

VLS e

t

0

0.5VLS

0 2tDtD t

VA

VLS

2tDtD t

VB

13

The driver output current (the line current at point A)

VZLS

o2

0 2tDtD

id

t

Exercise: Determine the transitions and sketch the voltages, VA , VB , in the following mismatch cases:

a) ZD<Zo - assume ZD=0.5Zo

b) ZD>Zo - assume ZD=2Zo .

14

5.2. Point-to-point transmission in bipolar technology

“the first incident switching”

VB VA

Z ZD o

D

<<= −ρ 1

Z ZR o

R

==ρ 0

15

The excitation (the driver transition, the numbers are for typical ECL) The line voltage transitions (the “first incident” switching)

e

VLS

[V]

-1.7

-0.9

t tr

-1.7

VLS

VLS

tD

[V]

-1.7

-0.9

VB

t tr

[V]

-0.9

VA

t tr

tD+tr

16

The driver current (the numbers are for 50[Ω] line )

VZ

LS

o

[mA] id

6

22

t tr

Comments concerning resistive substrate losses: The resistive losses in the substrate decrease with frequency. The loss tangent is determined using the following formula

σtan( )δωε ε

=2 r o

where σ represents the conductivity of substrate layer, ω is an angular frequency, ε r is a magnitude of relative dielectric coefficient, and ε o is the permittivity of vacuum. This formula is derived for imperfect dielectrics with ohmic losses and is an approximation derived under the assumption that

σωε ε2

1r o

≤ , therefore it is not valid for very low frequencies, which for the materials involved would be under (20-40) MHz.

This formula, as stated above does not take into account the polarization effects and thus it is not valid for very high frequencies (THz range).

17

6. Model of multiconductor lines

Example of 2 conductor system

∂∂

∂∂x

uu t

ii

u L

1

2

11 12

21 22

1

2

LNMOQP = −LNM

OQPLNMOQP4

1 11 12 1

2 12 22 2

i C

i c c ui c cx t

∂ ∂∂ ∂ u

= −

In general or else in the form of the wave equation

∂∂

∂∂

∂∂

∂∂

ux

L it

ix

C ut

= −

= − ∂∂

∂∂

2

2

2

2u

xLC u

t=

18

6.1. Model properties Analogously to the single line case the solution may contain components traveling in the positive direction of x axis which are represented by the vector, u , and the components traveling in opposite direction designated by the vector, u , i.e.

u u u

+

−

= ++ − . Characteristic admittance matrix

The characteristic admittance matrix is defined as follows 1

Y L LCo =−1 2b g .

Matrices of reflection coefficients The reflection coefficient matrix at the receiver end

−b g b1ρ R o R oY Y Y Y= + − gR . The reflection coefficient matrix at the driver end as

−b g b1ρD o D oY Y Y Y= + − gD .

These relations can also be expressed using impedance matrix, . Z Yo o= −1

19

Terminating loads In the signal transmission we want to control reflections. In the case of single line we can control the reflection coefficient by load impedance. However, in the case of multi-conductor lines we do not have a complete control because in practical applications we can control m load values (m is the number of lines) only , but

there are m m( )+1

different reflection coefficients (the matrix 2

ρD is symmetric).

In the case of bipolar technology the lines are terminated at the receivers’ end, which yields first incident switching and the maximum operating speed. The drivers are designed to achieve minimum output impedance (impedance much smaller than the representative impedance of the lines) and thus the driver end is not properly terminated. Consequently the receivers’ input impedance must be carefully designed to minimize the reflections. In the case of CMOS technology the receivers’ input impedance is very high and thus the receivers’ end is practically open which results in perfect reflections. This phenomenon is incorporated into design of signal transmission, which aims at first reflection switching. In this technology the drivers’ impedance must be selected so that it is close to the representative impedance of lines in order to avoid multiple reflections and resulting problems such as overshoot or increased delay.

Diagonally matched loads The diagonal match consists of such a selection of terminating impedances that the main diagonal in the matrix of reflection coefficients contains elements of zero value

F I0

ρ =

H

GGGG K

JJJJ0

..

.

20

6.2. Cross-talk

em

rdm

+ -

e2

rd2

+ -

e1

rd1

+ -

m

2

1

.

.

.

rr1

rr2

rrm

m

2

1

.

.

.

21

Appendix Details of Signal Propagation in Multiconductor Lines A.1. Characteristic admittance matrix The characteristic admittance matrix is derived using the first of model equations rewritten in the form

∂∂

∂∂

it

L ux

= − −1

and the d’Alembert equation which is used to eliminate the derivative ∂∂

ux .

After differentiation, matrix manipulation, and integration we obtain the relation i L PTP u u− −1 1b g= −+ −

which defines the characteristic admittance matrix Y L= 1 1PTPo

− − .

It should be noted that the matrix T contains the square roots of eigenvalues of the matrix LC. The diagonalization of LC yields

P LC− P T=1 2

22

or else LC PT P= −2 1

which shows that

LC PTPb g12 1= −

. The last relation can be easily verified by a simple multiplication. Finally the characteristic admittance matrix is defined as follows

1

Y L LCo =−1 2b g .

This formula is quite practical because it shows that the admittance matrix can be computed using directly the matrix LC so that the complicated eigenanalysis can be avoided. The details of computation are given in the papers: 1. K. Reiss and O. A. Palusinski, " Procedure for Direct Calculation of Characteristic Admittance Matrix of Coupled Transmission Lines," IEEE Trans. on

Microwave Theory and Techniques, vol. 44, No. 1, January 1996, pp. 152-154.

2. F. Szidarovszky and O. A. Palusinski, "Clarification of Decoupling Method for Multiconductor Transmission Lines," ," IEEE Trans. on Microwave Theory and Techniques, vol. 47, No. 5, May 1999.

23

A.2. Matrix of reflection coefficients Analogously to the case of single line the terminations may cause reflections. In practical cases each line is terminated by a load isolated from others. Assuming for simplicity resistive terminations it is possible to write the relation between the termination voltages and currents in the matrix form

I Y VL= . The transmission line equation at the termination end is

i Y u u= −b go + − . i I ,Equating the termination and the line quantities ( V u u= = −+ − )we obtain

b g bY u u Y u uo L+ − + −− = + g . This relation can be used to determine the matrices of reflection coefficients. For example at x=D we have ( u is a reflected voltage and− u is an incident voltage Y is the receiver loadR− + − −, )

u Y Y Y Y uo R o R−−

+= + −b g b g1

which defines the reflection coefficient matrix at the receiver end

24

ρ R o R oY Y Y Y= + −−b g b g1R .

Analogously we derive the reflection coefficient matrix at the driver end as −b g b g1ρD o D oY Y Y Y= + − D .

These relations can also be expressed in terms of impedance matrix, Zo, as follows −1Z Yo o= .

Diagonally matched loads The diagonal match consists of such a selection of terminating impedances that the main diagonal in the matrix of reflection coefficients contains elements of zero value

F I0

ρ =

H

GGGG K

JJJJ0

..

.

We have developed a robust, iterative algorithm for computation of impedances in the case of diagonal matching. The algorithm is based on a mathematical analysis which resulted in the convergence and solution existence conditions. We have demonstrated that in practical cases of interest the solution exists and the algorithm converges to the solution. The detail of the algorithm are described in F. Szidarovszky and O. A. Palusinski, "A Special Matrix Equation and Its Application in Microelectronics," Applied Mathematics and Computation, Elsevier Science Inc., Dec. 1994, pp.115-119.

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A.3. Cross-talk

The near-end cross-talk

The near-end cross-talk is discussed using the schematics shown below.

im

i2

i1

em

rdm

+ -

e2

rd2

+ -

e1

rd1

+ -

m

2

1

.

.

.

26

The circuit equations are u E Y iD= − −1

. The matrices and vectors in the circuit equation are defined as

u

uu

u

i

ii

i

E

ee

e

Y

r

r

rm m m

D

d

d

dm

=

L

N

MMMMMMM

O

Q

PPPPPPP

=

L

N

MMMMMMM

O

Q

PPPPPPP

=

L

N

MMMMMMM

O

Q

PPPPPPP

=

L

N

MMMMMMMM

O

Q

PPPPPPPP

1

2

1

2

1

21

2

1

1

1

.

.

.

,...

,...

,. .

The transmission line equations at the near end (x=0) are

u o t u u,b g = +

i o t Y u Y uo o,b g = −+ −

+ −

Considering that ,u o t u and i o t i,b g b g= = we obtain

b gu Y Y Y E Y Y Y Y uo D D o D o D+− −

−= + + + −b g b g1 1 .

27

If the drivers (ei I=1,2,…,m) are the only source of excitation of lines than the solution for 0 2≤ ≤t tD min is

b gu Y Y Y Eo D D+−= + 1

because during this time . u− ≡ 0This formula can be used to calculate an estimate of the cross-talk. The maximum cross-talk is expected to occur when the center line is quiescent (so called listening or victim line) and the surrounding lines are excited in an identical manner. The vector E has the zero entry corresponding to the position of quiescent line (assume

ith line) and the cross-talk is given by the ith component of the vector u . 2t

+

The computation of near-end cross-talk for time exceeding or computation of far-end cross-talk requires simulation based on transmission line equivalent circuit and is substantially more involved.

D min

The far-end cross-talk The far-end cross-talk can be computed assuming resistive terminations as shown below.

rr1

rr2

rrm

m

2

1

.

.

.

28

The result (quoted here without derivation) for limited time t t t tD r Dmax min+ < < 3 (multiple reflections are not included) is given by

b g bV Y Y Y Y Y Y EFE o R o o D D= + +− −2 1 1g where are the characteristic admittance matrix, the driver admittance matrix, and the vector of sources (driver voltages) respectively.

Y Y Eo D, , −

The matrix represents the receiver admittances and is defined as follows YR

L O

Y

r

r

r

R

r

r

rm

=

N

MMMMMMMM Q

PPPPPPPP

1

1

1

1

2. .

29