Class06 Transmission Line Basics

Transmission Line Basics II - Class 6 Prerequisite Reading assignment: CH2

Acknowledgements: Intel Bus Boot Camp: Michael Leddige

2

Real Computer Issues Dev a

Clk

data Dev b Signal Measured here

Switch Threshold

An engineer tells you the measured clock is nonmonotonic and because of this the flip flop internally may double clock the data. The goal for this class is to by inspection determine the cause and suggest whether this is a problem or not. Transmission Lines Class 6

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Agenda

 The Transmission Line Concept  Transmission line equivalent circuits

and relevant equations  Reflection diagram & equation  Loading  Termination methods and comparison  Propagation delay  Simple return path ( circuit theory, network theory come later) Transmission Lines Class 6

Two Transmission Line Viewpoints

 Steady state ( most historical view) Frequency domain

 Transient Time domain Not circuit element Why?

 We mix metaphors all the time Why convenience and history

Transmission Lines Class 6

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5

Transmission Line Concept Power Frequency (f) is @ 60 Hz Wavelength (λ ) is 5× 10 m 6

( Over 3,100 Miles)

Consumer Home

Transmission Lines Class 6

Power Plant

PC Transmission Lines Signal Frequency (f) is approaching 10 GHz Wavelength (λ ) is 1.5 cm ( 0.6 inches)

6

Integrated Circuit Stripline

Microstrip

T

PCB substrate Cross section view taken here

Stripline W Cross Section of Above PCB Copper Trace

Via FR4 Dielectric

Micro Strip

Signal (microstrip)

T

Copper Plane

Ground/Power Signal (stripline) Signal (stripline) Ground/Power Signal (microstrip)

W Transmission Lines Class 6

7

Key point about transmission line operation Voltage and current on a transmission line is a function of both time and position.

V = f ( z, t ) I = f ( z, t )

I2

I1

V1

V2

dz The major deviation from circuit theory with transmission line, distributed networks is this positional dependence of voltage and current! Must think in terms of position and time to understand transmission line behavior This positional dependence is added when the assumption of the size of the circuit being small compared to the signaling wavelength

Transmission Lines Class 6

Examples of Transmission Line Structures- I  Cables and wires (a) (b) (c) (d)

Coax cable Wire over ground Tri-lead wire Twisted pair (two-wire line)

 Long distance interconnects + -

+

(a)

-

-

+ (c)

-

(b)

+ (d)

-

Transmission Lines Class 6

8

Segment 2: Transmission line equivalent circuits and relevant equations  Physics of transmission line structures  Basic transmission line equivalent circuit  ?Equations for transmission line propagation

Transmission Lines Class 6

9

10

E & H Fields – Microstrip Case How does the signal move from source to load? Signal path

Y

Z (into the page) X Electric field Magnetic field Remember fields are setup given an applied forcing function. (Source) Ground return path

The signal is really the wave propagating between the conductors Transmission Lines Class 6

Transmission Line “Definition”  General transmission line: a closed system in which

power is transmitted from a source to a destination

 Our class: only TEM mode transmission lines

A two conductor wire system with the wires in close proximity, providing relative impedance, velocity and closed current return path to the source. Characteristic impedance is the ratio of the voltage and current waves at any one position on the transmission V line

Z0 =

I

Propagation velocity is the speed with which signals are transmitted through the transmission line in its surrounding medium. c

v=

εr

Transmission Lines Class 6

11

Presence of Electric and Magnetic Fields I

+

+

+

+

I + ∆I

H

I

I + ∆I

E V I

-

-

-

-

V + ∆V I + ∆I

V I

H

V + ∆V I + ∆I

 Both Electric and Magnetic fields are present in the transmission lines

These fields are perpendicular to each other and to the direction of wave propagation for TEM mode waves, which is the simplest mode, and assumed for most simulators(except for microstrip lines which assume “quasi-TEM”, which is an approximated equivalent for transient response calculations).

 Electric field is established by a potential difference between two conductors.

Implies equivalent circuit model must contain capacitor.

 Magnetic field induced by current flowing on the line Implies equivalent circuit model must contain inductor. Transmission Lines Class 6

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13

T-Line Equivalent Circuit  General Characteristics of Transmission Line

Propagation delay per unit length (T0) { time/distance} [ps/in] Or Velocity (v0) {distance/ time} [in/ps] Characteristic Impedance (Z0) Per-unit-length Capacitance (C0) [pf/in] Per-unit-length Inductance (L0) [nf/in] Per-unit-length (Series) Resistance (R0) [Ω/in] Per-unit-length (Parallel) Conductance (G0) [S/in]

lR0

lL0

lG0

Transmission Lines Class 6

lC0

14

Ideal T Line  Ideal (lossless) Characteristics of Transmission Line

Ideal TL assumes: Uniform line Perfect (lossless) conductor (R0→ 0) Perfect (lossless) dielectric (G0→ 0) We only consider T0, Z0 , C0, and L0.

lL0 lC0

 A transmission line can be represented by a cascaded network (subsections) of these equivalent models.

The smaller the subsection the more accurate the model

The delay for each subsection should be no larger than 1/10th the signal rise time. Transmission Lines Class 6

Signal Frequency and Edge Rate vs. Lumped or Tline Models In theory, all circuits that deliver transient power from one point to another are transmission lines, but if the signal frequency(s) is low compared to the size of the circuit (small), a reasonable approximation can be used to simplify the circuit for calculation of the circuit transient (time vs. voltage or time vs. current) response.

Transmission Lines Class 6

15

T Line Rules of Thumb So, what are the rules of thumb to use?

May treat as lumped Capacitance Use this 10:1 ratio for accurate modeling of transmission lines

Td < .1 Tx May treat as RC on-chip, and treat as LC for PC board interconnect

Td < .4 Tx Transmission Lines Class 6

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Other “Rules of Thumb”  Frequency knee (Fknee) = 0.35/Tr (so if Tr is   

1nS, Fknee is 350MHz) This is the frequency at which most energy is below Tr is the 10-90% edge rate of the signal Assignment: At what frequency can your thumb be used to determine which elements are lumped? Assume 150 ps/in

Transmission Lines Class 6

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When does a T-line become a T-Line?

18

 Whether it is a

bump or a mountain depends on the ratio of its size (tline) to the size of the vehicle (signal wavelength)

When do we need to use transmission line analysis techniques vs. lumped circuit analysis?

 Similarly, whether

Wavelength/edge rate

Tline

Transmission Lines Class 6

or not a line is to be considered as a transmission line depends on the ratio of length of the line (delay) to the wavelength of the applied frequency or the rise/fall edge of the signal

Equations & Formulas How to model & explain transmission line behavior

Relevant Transmission Line Equations Propagation equation

γ = ( R + jωL)(G + jωC ) = α + jβ α is the attenuation (loss) factor β is the phase (velocity) factor Characteristic Impedance equation

( R + j ωL ) Z0 = (G + jωC ) In class problem: Derive the high frequency, lossless approximation for Z0 Transmission Lines Class 6

20

Ideal Transmission Line Parameters  Knowing any two out of Z0, 



Td, C0, and L0, the other two can be calculated. C0 and L0 are reciprocal functions of the line crosssectional dimensions and are related by constant me. ε is electric permittivity ε 0= 8.85 X 10-12 F/m (free space) ε ri s relative dielectric constant

 µ is magnetic permeability

Z0 =

L0 ; C0

T0 ; Z0 1 = v0 ; µε

C0 =

µ = µr µ0 ;

T d = L0 C0 ; L0 = Z 0 T 0 ; C0 L0 = µε;

ε = εr ε0 .

µ 0= 4p X 10-7 H/m (free space) µ r is relative permeability

Don’t forget these relationships and what they mean! Transmission Lines Class 6

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Parallel Plate Approximation  Assumptions

TC

TEM conditions

ε

Uniform dielectric (ε ) between conductors TC<< TD; WC>> TD

 T-line characteristics are function of:

Material electric and magnetic properties Dielectric Thickness (TD) Width of conductor (WC)

 Trade-off

22

TD ; C0 , L0 , Z0  WC ; C0 , L0 , Z0 

TD

WC

ε * PlateArea Base C= d equation C0

WC  F  ε⋅ ⋅  TD  m 

WC  pF  8.85 ⋅ε r ⋅ ⋅  TD  m 

L0

TD  F  µ⋅ ⋅  WC  m 

T D  µH  0.4 ⋅π ⋅µ r ⋅ ⋅  WC  m 

Z0

377 ⋅

TD WC

⋅

µr εr

⋅Ω

To a first order, t-line capacitance and inductance can be approximated using the parallel plate approximation. Transmission Lines Class 6

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Improved Microstrip Formula  Parallel Plate Assumptions + 

WC

Large ground plane with zero thickness To accurately predict microstrip impedance, you must calculate the effective dielectric constant.

Z0 ≈

F=

ε

 5.98TD  ln  εr + 1.41  0.8WC + TC  87

εr + 1 εe = + 2

εr − 1 12TD 2 1+ WC

for

for

WC TD WC TD

TD

From Hall, Hall & McCall: Valid when: 0.1 < WC/TD < 2.0 and 1 <

+ F − 0.217( εr − 1)

2 WC   0.02(εr −1)1 −  T D   0

TC

r

< 15

TC WCTD

<1

>1

Transmission Lines Class 6

You can’t beat a field solver

24

Improved Stripline Formulas  Same assumptions as used for microstrip apply here

TD1

WC

ε

TC TD2

From Hall, Hall & McCall: Symmetric (balanced) Stripline Case TD1 = TD2

  4(TD1 + TD1)  Z 0 sym ≈ ln  0 . 67 π ( 0 . 8 W C + T C ) εr   60

Valid when WC/(TD1+TD2) < 0.35 and TC/(TD1+TD2) < 0.25 Offset (unbalanced) Stripline Case TD1 > TD2

You can’t beat a field solver

Z 0 sym(2 A, WC , TC , εr ) ⋅ Z 0 sym(2 B, WC , TC , εr ) Z 0offset ≈ 2 Z 0 sym(2 A,WC , TC , εr ) + Z 0 sym(2 B,WC , TC , εr ) Transmission Lines Class 6

Refection coefficient  Signal on a transmission line can be analyzed by 

keeping track of and adding reflections and transmissions from the “bumps” (discontinuities) Refection coefficient

Amount of signal reflected from the “bump” Frequency domain ρ=sign(S11)*|S11| If at load or source the reflection may be called gamma (ΓL or Γs) Time domain ρ is only defined a location The “bump”

Time domain analysis is causal. Frequency domain is for all time. We use similar terms – be careful

 Reflection diagrams – more later Transmission Lines Class 6

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Reflection and Transmission Incident

ρ

1+ρ

Transmitted

Reflecte d

Reflection Coeficient Transmission Coeffiecent ρ

Zt − Z0

τ

(1 + ρ)

"" → ""

Zt + Z0 τ

Transmission Lines Class 6

2⋅ Zt Zt + Z0

τ

1+

Zt − Z0 Zt + Z0

26

27

Special Cases to Remember A: Terminated in Zo Zs Vs

Zo

Zo

− ρ = Zo Zo = 0 Zo + Zo

B: Short Circuit Zs Vs

Zo

− ρ = 0 Zo = −1 0 + Zo

C: Open Circuit Zs Vs

Zo

Transmission Lines Class 6

ρ=

∞ − Zo =1 ∞ + Zo

Assignment – Building the SI Tool Box Compare the parallel plate approximation to the improved microstrip and stripline formulas for the following cases: Microstrip: WC = 6 mils, TD = 4 mils, TC = 1 mil, εr = 4 Symmetric Stripline: WC = 6 mils, TD1 = TD2 = 4 mils, TC = 1 mil, εr = 4 Write Math Cad Program to calculate Z0, Td, L & C for each case. What factors cause the errors with the parallel plate approximation? Transmission Lines Class 6

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Transmission line equivalent circuits and relevant equations  Basic pulse launching onto transmission lines  Calculation of near and far end waveforms for classic load conditions

Transmission Lines Class 6

29

Review: Voltage Divider Circuit  Consider the

simple circuit that contains source voltage VS, source resistance RS, and resistive load RL.

30

RS RL

VS

 The output

voltage, VL is easily calculated from the source amplitude and the values of the two series resistors.

VL = VS

Why do we care for? Next page…. Transmission Lines Class 6

RL RL + R S

VL

Solving Transmission Line Problems The next slides will establish a procedure that will allow you to solve transmission line problems without the aid of a simulator. Here are the steps that will be presented:

 Determination of launch voltage & final “DC” or “t =0” voltage

 Calculation of load reflection coefficient and voltage delivered to the load

 Calculation of source reflection coefficient and resultant source voltage

These are the steps for solving all t-line problems. Transmission Lines Class 6

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Determining Launch Voltage

32

TD Vs 0

Rs A

B Zo

Vs

Rt

(initial voltage) t=0, V=Vi

Vi = VS

Z0 Z 0 + RS

Vf = VS

Rt Rt + RS

Step 1 in calculating transmission line waveforms is to determine the launch voltage in the circuit.

 The behavior of transmission lines makes it

easy to calculate the launch & final voltages – it is simply a voltage divider! Transmission Lines Class 6

33

Voltage Delivered to the Load TD Vs

Rs A Zo

Vs

0

B Rt

(initial voltage) t=0, V=Vi

ρΒ =

t=2TD, ρA(ρB)(Vi ) V=Vi − +Zo Rt+ ρB(Vi)

Rt + Zo

(signal is reflected) t=TD, V=Vi +ρB(Vi )

Vreflected = ρ Β (Vincident) VB = Vincident + Vreflected

Step 2: Determine VB in the circuit at time t = TD 

The transient behavior of transmission line delays the arrival of launched voltage until time t = TD. 



VB at time 0 < t < TD is at quiescent voltage (0 in this case)

Voltage wavefront will be reflected at the end of the t-line 

VB = Vincident + Vreflected at time t = TD

Transmission Lines Class 6

Voltage Reflected Back to the Source Vs 0

Rs A

Vs

ρA

B Zo

ρB

Rt

TD

(initial voltage) t=0, V=Vi

(signal is reflected)

t=2TD, V=Vi + ρB (Vi) + ρA(ρ B )(Vi )

t=TD, V=Vi + ρB (Vi )

Transmission Lines Class 6

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Voltage Reflected Back to the Source − Zo ρ Α = Rs Rs + Zo

Vreflected = ρ Α (Vincident) VA = Vlaunch + Vincident + Vreflected

Step 3: Determine VA in the circuit at time t = 2TD  The

transient behavior of transmission line delays the arrival of voltage reflected from the load until time t = 2TD.  VA

at time 0 < t < 2TD is at launch voltage

 VA

= Vlaunch + Vincident + Vreflected at time t = 2TD

 Voltage

wavefront will be reflected at the source

In the steady state, the solution converges to VB = VS[Rt / (Rt + Rs)] Transmission Lines Class 6

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Problems

36

Solved Homework

 Consider the circuit

shown to the right with a resistive load, assume propagation delay = T, RS= Z0 . Calculate and show the wave forms of V1(t),I1(t),V2(t), and I2(t) for (a) RL= ∞ and (b) RL= 3Z0

RS

VS

I1

V1

Transmission Lines Class 6

Z0 ,Τ 0 l

I2

V2

RL

Step-Function into T-Line: Relationships

 Source matched case: RS= Z0 V1(0) = 0.5VA, I1(0) = 0.5IA Γ S = 0, V(x,∞ ) = 0.5VA(1+ Γ L)

 Uncharged line V2(0) = 0, I2(0) = 0

 Open circuit means RL= ∞ Γ L = ∞ /∞ = 1 V1(∞) = V2(∞) = 0.5VA(1+1) = VA I1(∞) = I2 (∞) = 0.5IA(1-1) = 0 Transmission Lines Class 6

Solution

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Step-Function into T-Line with Open Ckt

 At t = T, the voltage wave reaches load end

and doubled wave travels back to source end V1(T) = 0.5VA, I1(T) = 0.5VA/Z0

V2(T) = VA, I2 (T) = 0

 At t = 2T, the doubled wave reaches the source end and is not reflected V1(2T) = VA, I1(2T) = 0

V2(2T) = VA, I2(2T) = 0 Solution Transmission Lines Class 6

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39

Waveshape:

Step-Function into T-Line with Open Ckt I1 I2

Current (A)

IA

RS

0.75IA 0.5I A

VS

I1

V1

Z0 ,Τ 0 l

I2

V2

Open

0.25IA

0

Τ

2Τ

3Τ

V1 V2

VA Voltage (V)

4Τ Time (ns)

0.75VA 0.5VA

This is called “reflected wave switching”

0.25VA

0

Τ

2Τ

3Τ

4Τ Time (ns)

Transmission Lines Class 6

Solution

40

Problem 1b: Relationships

 Source matched case: RS= Z0 V1(0) = 0.5VA, I1(0) = 0.5IA Γ S = 0, V(x,∞ ) = 0.5VA(1+ Γ L)

 Uncharged line V2(0) = 0, I2(0) = 0

 RL= 3Z0 Γ L = (3Z0 -Z0) / (3Z0 +Z0) = 0.5 V1(∞) = V2(∞) = 0.5VA(1+0.5) = 0.75VA I1(∞) = I2(∞) = 0.5IA(1-0.5) = 0.25IA Transmission Lines Class 6

Solution

41

Problem 1b: Solution

 At t = T, the voltage wave reaches load end

and positive wave travels back to the source V1(T) = 0.5VA, I1(T) = 0.5IA

V2(T) = 0.75VA , I2(T) = 0.25IA

 At t = 2T, the reflected wave reaches the source end and absorbed V1(2T) = 0.75VA , I1(2T) = 0.25IA

V2(2T) = 0.75VA , I2(2T) = 0.25IA Solution Transmission Lines Class 6

Waveshapes for Problem 1b I1 I2

Current (A)

IA

RS

0.75IA 0.5IA

VS

I1

V1

42

Z0 ,Τ 0 l

I2

V2

RL

0.25IA

0

Τ

2Τ

3Τ

I1 I2

VA Voltage (V)

4Τ Time (ns)

0.75VA 0.5VA 0.25VA

0

Τ

2Τ

3Τ

Note that a properly terminated wave settle out at 0.5 SolutionV Solution 4Τ Time (ns)

Transmission Lines Class 6

Transmission line step response  Introduction to lattice diagram analysis  Calculation of near and far end waveforms for 

classic load impedances Solving multiple reflection problems

Complex signal reflections at different types of transmission line “discontinuities” will be analyzed in this chapter. Lattice diagrams will be introduced as a solution tool.

Transmission Lines Class 6

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44

Lattice Diagram Analysis – Key Concepts The lattice diagram is a tool/technique to simplify the accounting of reflections and waveforms

 Diagram shows the boundaries  

(x =0 and x=l) and the reflection coefficients (GL and GL ) Time (in T) axis shown vertically Slope of the line should indicate flight time of signal Particularly important for multiple reflection problems using both microstrip and stripline mediums.

 Calculate voltage amplitude for 

each successive reflected wave Total voltage at any point is the sum of all the waves that have reached that point

Vs 0

Vs

Zo V(source) Rs TD = N ps

V(load)

Time V(source)

N ps

a A

A’ b

2N ps 3N ps

Rt

ρload

ρsource 0

V(load)

c

B’

B d

4N ps 5N ps

Transmission Lines Class 6

C’ e

Lattice Diagram Analysis – Detail ρ

ρ

source

load

V(load)

V(source) 0

45

Vlaunch 0

Time

Vlaunch

N ps Vlaunch ρload Vlaunch(1+ρload)

2N ps

Time

Vlaunch ρloadρsource Vlaunch(1+ρload +ρload ρsource)

3N ps Vlaunch ρ2loadρsource Vlaunch(1+ρload+ρ2loadρsource+ ρ2loadρ2source)

4N ps Vlaunch ρ2loadρ2source 0

V(load) V(source) Zo Vs Rs TD = N ps Vs Rt

5N ps

Transmission Lines Class 6

Transient Analysis – Over Damped 2v 0

Vs

Zo V(source) Zs TD = 250 ps

ρ source = 0.2 0 500 ps

0.8v 0.8v

1000 ps

Assume Zs=75 ohms Zo=50ohms Vs=0-2 volts

V(load)

ρ load = 1 V(load)

Time V(source)

Vinitial = Vs

ρ source =

0v

ρ load = 0.8v 0.16v

46

Zo  50  = (2)  = 0.8 Zs + Zo  75 + 50 

Zs − Zo 75 − 50 = = 0.2 Zs + Zo 75 + 50

Zl − Zo ∞ − 50 = =1 Zl + Zo ∞ + 50

1.6v

Response fr om lattice diagram 2.5

1500 ps 1.76v

2

2000 ps 2500 ps

1.92v 0.032v

V olt s

0.16v

1.5 Sour ce

1

Load

0.5 0 0

2 50

500

750

Tim e , ps

Transmission Lines Class 6

1000

1250

Transient Analysis – Under Damped 2v 0

Zo

Zs TD = 250 ps

Vs

ρsource = −0 . 3333 Time

Assume Zs=25 ohms Zo =50ohms Vs=0-2 volts

V(load)

V(load)

V(source)

0

1.33v

500 ps 1.33v

1000 ps

ρload =

 50  Zo = (2)   =1.3333 + + Zs Zo  25 50 

Zs − Zo 25 −50 = = −0.33333 + + Zs Zo 25 50

Zl − Zo ∞ −50 = =1 + ∞ + Zl Zo 50

2.66v

Response from lattice diagram

3

-0.443v

2.5

1.77v

2000 ps

ρsource =

0v

1.33v

-0.443v

1500 ps 2.22v

ρload = 1

Vinitial =Vs

0.148v

Volts

V(source)

2 1.5

Source

1 0.5

2500 ps

1.92

Load

0

0.148v

0

250

500

750 1000 1250 1500 1750 2000 2250 Time, ps

2.07

Transmission Lines Class 6

47

Two Segment Transmission Line Structures X

Rs

X

Zo2 TD

Zo1 TD

Vs

T3 T2 ρ 2 ρ3

ρ1

Rt

ρ4

a TD A 2TD 3TD B 4TD 5TD C

A= a B = a+c+d C = A+ c+ d + f + h

c

b

Z o1 vi = Vs Rs + Z o1

d

e

ρ1 =

f

g

h

i

j

k

A’

ρ2 = B’

l C’

Rs − Z o1 Rs + Z o1 Z o 2 − Z o1 Z o 2 + Z o1

Z − Zo2 ρ 3 = o1 Z o1 + Z o 2

A' = b + e B' = b + e + g + i C' = b + e + g + i + k + l

a = vi b = aT2 c = aρ 2 d = cρ 1 e = bρ 4 f = dρ 2 + eT3 g = eρ 3 + dT2 h = fρ 1

Rt − Z o 2 ρ4 = Rt + Z o 2

i = gρ 4

T2 = 1 + ρ 2

j = hρ 2 + iT3

T3 = 1 + ρ 3

k = iρ 3 + hT2

Transmission Lines Class 6

48

49

Assignment 

Previous examples are the preparation Consider the two segment transmission line shown to the right. Assume RS= 3Z01 and Z02= 3Z01 . Use Lattice R I I I Z ,Τ Z ,Τ diagram and calculate l l V V V V reflection coefficients at the interfaces and show the wave forms of V1(t), S

1

01

S

3

2

1

V2(t), and V3(t).

 Check results with PSPICE Transmission Lines Class 6

02

01

02

2

1

2

3

Short