© H. Heck 2008Section 2.41 Module 2:Transmission Lines Topic 4: Parasitic Discontinuities OGI EE564...

© H. Heck 2008 Section 2.4 1

Module 2: Transmission LinesTopic 4: Parasitic Discontinuities

OGI EE564

Howard Heck


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Where Are We?

1. Introduction

2. Transmission Line Basics1. Transmission Line Theory

2. Basic I/O Circuits

3. Reflections

4. Parasitic Discontinuities

5. Modeling, Simulation, & Spice

6. Measurement: Basic Equipment

7. Measurement: Time Domain Reflectometry

3. Analysis Tools

4. Metrics & Methodology

5. Advanced Transmission Lines

6. Multi-Gb/s Signaling

7. Special Topics


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Contents

Capacitive Load: Qualitative Look Capacitive Load: Step Response Analytical Solution Discrete Capacitive and Inductances: Delays and

Reflections Distributed Loads: Loaded Velocity and Impedance Summary References Appendix: Reflections Due to Capacitive Load Appendix: Inductive Load Step Response


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Capacitive Loading: Qualitative Look

What does the signal at z=l, t=dl+ look like? The driver circuit sends a voltage step at t=0. The step travels along the line (as a wave) to z=l. The capacitor at z=l sees a voltage step input.

How does the capacitor respond to this input? When the incoming wave arrives, the capacitor initially acts like

a short. At steady state, the capacitor is an open circuit. The transmission line looks resistive to the capacitor. So, perhaps we should expect to get an RC circuit-like

(exponential) response.

Z0, d

VS(t) CL

RS

z = 0

z = l

VS(t) = VS·u(t)0

VS

t = 0


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Quantitative Analysis

Start at z=0, t=0+:

Z0, d

VS(t) CL

RS

z = 0

z = l

VS(t) = VS·u(t)0

VS

t = 0

V

Z

Z RV

SS0 0 0

0

, [2.5.1]

IV

Z Z RV

SS0 0

0 0 1

0 0

,,

[2.5.2]

The wave propagates to z=l, t=d:

ltlVVRZ

ZltlVltlVltlV dS

Sddd

,,,,

0

0�� [2.5.3]

000

,1,

1,,,

Z

ltlVV

RZltlIV

RZltlIltlIltlI d

SS

dSS

ddd

��

[2.5.4]


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Quantitative Analysis #2

ltlVV

RZ

Z

dt

dC

Z

ltlVV

RZltlI dS

SL

dS

Sd ,

,1,

0

0

00

��

[2.5.5]

S

LSL

dd VCRZCZ

ltlV

dt

ltlVd

00

1,, ��

[2.5.6]

We have a linear ordinary differential equation. Find the form of the solution by solving the homogeneous portion:

dV l t

dt

V l t

Z CL

� �, ,

0

0Homogenous equation: [2.5.7]

�V l t Ae

tZ CL,

0Solution form: [2.5.8]

000

,1,

1,,,

Z

ltlVV

RZltlIV

RZltlIltlIltlI d

SS

dSS

ddd

��

[2.5.4]

For a capacitor, . Apply it to [2.5.4]:I C dVdt

Note the “Z0C” time constant.


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S

LSL

dd VCRZCZ

ltlV

dt

ltlVd

00

1,, ��

[2.5.6]

General solution to O.D.E.:

General solution:

Boundary condition: . Apply it to [2.5.6]:0tdt

dV

�V l

Z C Z R CV

L S LS

,

0 0

1[2.5.9]

SS

VRZ

ZlV

0

0,�

[2.5.10]

�V l t Ae

Z

Z RV

tZ C

SS

L,

0 0

0[2.5.11]


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Substitute [2.5.13] into general solution [2.5.11]:

Initial condition: V(z=l,t=dl)=0.

[2.5.13] SS

dd VRZ

ZllVllV

0

0,, �

0,,, llVllVllV ddd �

[2.5.12]

[2.5.14] SS

CZl

SS

d VRZ

ZAeV

RZ

ZllV L

d

0

0

0

0 0,

�

L

dCZ

l

SS

eVRZ

ZA 0

0

02

[2.5.15]

The expression for becomes: dtlV ,�

SS

CZl

CZl

SS

d VRZ

ZeeV

RZ

ZtlV L

d

L

d

0

0

0

0 002

,

�

[2.5.16]

L

dCZ

lt

SS

d eVRZ

ZltlV 021,

0

0

�

[2.5.17]


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Equation [2.5.18] is valid for dl t <3dl.

If RS=Z0, [2.5.18] is valid for dl t.

Solution:

L

dCZ

lt

SS

SS

ddd eVRZ

ZV

RZ

ZllVllVltlV 021,,,

0

0

0

0

�

[2.5.17]

L

dCZ

lt

SS

d eVRZ

ZltlV 01

2,

0

0

[2.5.18]


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Quantitative Analysis Check

Let RS=Z0. The following boundary conditions must be

satisfied:

llVllVllV ddd ,,0,�

The wave components are:

SS

SS

d

VRZ

ZlV

VRZ

ZllV

0

0

0

0

,

,

SS

SS

d

VRZ

ZlV

VRZ

ZllV

0

0

0

0

,

,

�

SSS

d

VVRZ

ZlV

llV

0

02,

0,

,,, lVlVVlV S

�

This checks out.


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time [ns]0 1 2 3 4

0.00

0.25

0.50

0.75

1.00

1.25

1.50

1.75

2.00v

olt

ag

e [

V]

Waveforms

CalculatedCalculated

SimulatedSimulated

V(z=0)

V(z=l)

0 1 2 3 4time [ns]

0.00

0.25

0.50

0.75

1.00

1.25

1.50

1.75

2.00

vo

lta

ge

[V

]

V(z=0)

V(z=l)

50, 1.0 ns2.0V 5.0 pF

50

z = 0

z = l

VS(t) = VS·u(t)0VS

t = 0

tr = 1 ps


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Capacitive Reflections & DelayZ0, d

VS(t) CL

RS

z = 0

z = l

VS(t) = VS·u(t)0

VS

t = 0

Refer back to equation [2.5.18] :

L

dCZ

lt

SS

d eVRZ

ZltlV 01

2,

0

0

[2.5.18]

Assume that 50% of the full signal swing is required for switching:

Solve for t (assume dl = 0 for simplicity):

L

dCZ

lt

S

eV

V015.0

[2.5.19]

LLC CZCZtL 00 7.02ln [2.5.20]

Many sources give the following: LC CZ

L 0 [2.5.21]


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Capacitive Reflections & Delay #2

With the load at an intermediate point, the capacitor is charged to half as much voltage as for a capacitor at the end of the line.

20 L

C

CZL [2.5.22]

Z0, d

VS(t) CL

RS

z = 0

z = lZ0, d

Reflections due to the capacitor (see the appendix):Mid-line

End

Vri

r

t

CZ

V

V 2

0 [2.5.23]

Vri

r

t

CZ

V

V 0 [2.5.24]


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Inductive Loads

Z0, d

VS(t)

RS LL

Z0, d

Z0, d

VS(t)

RS LL

02Z

LL [2.5.25]

Vr

L

i

r

tZ

L

V

V 02 [2.5.26]

0Z

LL [2.5.27]

r

iLr tZ

VLV

0

[2.5.28]


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Distributed Capacitance

Z0, d Z0, d Z0, d Z0, d Z0, d

CL

RS

Z0, d

CL

Z0, d Z0, d Z0, d Z0, d

CL CL CLVS(t)

Bi-directional network:

Each receiver presents a capacitive load:

The loads are uniformly distributed, separated by length l. The effective capacitance of each transmission line segment is:

lCCC L

eff 0 [2.5.24]

C0 is the transmission line capacitance per unit length.


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Distributed Capacitance #2 Since the distributed capacitance has increased, the

propagation velocity and impedance are changed:

We can also express the loaded impedance and velocity in terms of the characteristic impedance and non-loaded velocity:

ZL is the “loaded” impedance of the transmission line.

vL is the “loaded” propagation velocity of the transmission line.

[2.5.28]v

v

ClC

LL

0

01

Z LC

L

C Cl

Leff L

0

[2.5.25]

v LC L C ClL eff

L

0 [2.5.26]

ZZ

ClC

LL

0

01 [2.5.27]


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Summary

Capacitive loads slow the rise/fall times and add delay.

Capacitive loads generate reflected waves of opposite polarity to the incoming wave.

Inductive loads also affect the delay and generate reflections.

Distributed capacitive loads increase the propagation delay of the transmission line and reduce the effective impedance.


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References S. Hall, G. Hall, and J. McCall, High Speed Digital System

Design, John Wiley & Sons, Inc. (Wiley Interscience), 2000, 1st edition.

W. Dally and J. Poulton, Digital Systems Engineering, Cambridge University Press, 1998.

Ramo, Whinnery, and Van Duzer, Fields and Waves in Communication Electronics, 1985.

R. Poon, Computer Circuits Electrical Design, Prentice Hall, 1st edition, 1995.

Ramo, Whinnery, and Van Duzer, Fields and Waves in Communication Electronics, 1985.

R.E. Matick, Transmission Lines for Digital and Communication Networks, IEEE Press, 1995

H.B.Bakoglu, Circuits, Interconnections, and Packaging for VLSI, Addison Wesley, 1990.

“Transmission Line Effects in PCB Applications,” Motorola Application Note AN1051, 1990.

W.R. Blood, MECL System Design Handbook, Motorola, Inc., 1988.


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Appendix: Reflections from Capacitive Load

Z0, d

CL

Z0, d

Vi

Vr

Recall:0

0

ZZ

ZZ

eff

eff

Combine [2.5.a1] and [2.5.a2]:

Where:0

00 ZZ

ZZZZZ

C

CCeff

[2.5.A1]

LC Cj

Z1

[2.5.A2]

0

0

0

0

0

0

111

1

ZCj

Z

Cj

ZCjCj

Z

ZCj

ZCjZ

L

L

L

L

L

Leff

[2.5.A3]


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Appendix: Reflections from Capacitive Load #2 Substitute [2.5.a3] into the expression for :

We can rewrite [2.5.a5] in terms of Vr and Vi:

0

0

0

00

0

00

00

0

00

0

21

111

11

1

1

ZCj

ZCj

ZCj

ZCjZZCj

ZCjZ

ZCZj

Z

ZZCj

Z

L

L

L

L

L

L

L

[2.5.A4]0

0

2 ZCj

ZCj

L

L

Assume . Then [2.5.a4] simplifies to:20 ZCj L

[2.5.A5]i

rL

V

VZCj

20

[2.5.A6] iL

r VjZC

V 2

0


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Appendix: Reflections from Capacitive Load #3 We need to make use of the following time domain

relationship:

For , the time derivative is , which must be satisfied for all .

jwteVt 00 tj

dt

td0

0

For our case, 0(t) is Vi. Substitute into [2.5.a6] :

[2.5.A7]dt

dVZCV iL

r 20

Letting , we get [2.5.a8]: r

ii

t

V

dt

dV

Limitation: We must satisfy Z0CL << tr. Why?

Recall = 2f and f 0.35/tr. Apply to our assumption that . 20 ZCj L

r

L

i

r

t

ZC

V

V

20

[2.5.A8]


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Appendix: Mathematical Expression for

Inductive Load

Start at Start at zz=0, =0, tt=0+:=0+:

V

Z

Z RV

SS0 0 0

0

, [2.5.B1]

IV

Z Z RV

SS0 0

0 0 1

0 0

,,

[2.5.B2]

The wave propagates to The wave propagates to zz==ll, , tt==ddll::

ltlVVRZ

ZltlVltlVltlV dS

Sddd

,,,,

0

0��

[2.5.B3]

000

,1,

1,,,

Z

ltlVV

RZltlIV

RZltlIltlIltlI d

SS

dSS

ddd

��

[2.5.B4]

Z0, d

VS(t)

L

RS

z = 0

z = l

VS(t) = VSu(t)0

VS

t=0


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Inductive Loading #2

Rearrange [2.5.b6]:

000

,1,

1,,,

Z

ltlVV

RZltlIV

RZltlIltlIltlI d

SS

dSS

ddd

��

[2.5.B4]

For an inductor, . Apply it to [2.5.b4]:dtdILV

[2.5.B5] ltlVltlV

dt

ltldIL dd

d

,,

, �

[2.5.B6] ltlVV

RZ

Z

Z

ltlV

RZ

V

dt

dL dS

S

d

S

S

,

,

0

0

00

��

[2.5.B7] 0

,,

0

0

0

SS

dd V

RZ

Z

dt

ltlVd

Z

LltlV

�

�

Linear O.D.E.


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Homogeneous solution:Homogeneous solution:

[2.5.B7] 0

,,

0

0

0

SS

dd V

RZ

Z

dt

ltlVd

Z

LltlV

�

�

LtZ

AetlV0

,

�

[2.5.B9]

0

,,

0

dt

ltlVd

Z

LltlV d

d

�

�[2.5.B8]

General solution:General solution:

Use the boundary condition: .Therefore,

00

,0,

Z

llV

RZ

VllI d

S

Sd

�

00

,

Z

llV

RZ

V d

S

S �

[2.5.B10]

SS

d VRZ

ZllV

0

0,�

[2.5.B11]


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Inductive Loading #4 General solution is the sum of the homogeneous

solution and special solution:

[2.5.B12] LCZt

SS

AeVRZ

ZtlV 0

0

0,

�

Initial condition:Initial condition:

Apply the initial condition at t=dl (the inductor acts

like an open circuit): llVllVllVllV dddd ,2,,,

� [2.5.B13]

SS

dd VRZ

ZllVllV

0

0,, �

[2.5.B14]

Substitute into general solution:

LlZ

SS

SS

d

d

AeVRZ

ZV

RZ

ZllV

0

0

0

0

0,

�

[2.5.B15]

LlZ

SS

d

eVRZ

ZA

0

0

02

[2.5.B16]


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Equation [2.5.18] is valid for dl t <3dl.

If RS=Z0, [2.5.18] is valid for dl t.

Substitute for A in [2.5.B16]:

LtZ

LlZ

SS

SS

d eeVRZ

ZV

RZ

ZltlV

d 00

0

0

0

0 2,

�

[2.5.B17]

12,

0

0

0 LZtl

SS

d

d

eVRZ

ZltlV

�

Solution:Solution:

12,,,0

0

0

0

0 LZtl

SS

SS

ddd

d

eVRZ

ZV

RZ

ZllVllVltlV

�

LZtl

SS

d

d

eVRZ

ZltlV

0

0

02,

[2.5.B18]


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Check

Let RS=Z0. The following boundary conditions must be

satisfied:

The wave components are:

This checks out.

llVllVVllV ddSd ,,,�

,,0, lVlVlV�

2

,

2,

0

0

0

0

SS

S

SS

Sd

VV

RZ

ZlV

VV

RZ

ZllV

2

,

2,

0

0

0

0

SS

S

SS

Sd

VV

RZ

ZlV

VV

RZ

ZllV

�

0,

,

lV

VllV Sd

© H. Heck 2008Section 2.41 Module 2:Transmission Lines Topic 4: Parasitic Discontinuities OGI EE564...

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Transcript of © H. Heck 2008Section 2.41 Module 2:Transmission Lines Topic 4: Parasitic Discontinuities OGI EE564...