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
data
Dev a Dev b

Signal
Measured
Clk here Switch
Threshold

An engineer tells you the measured clock is non-monotonic
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

3

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

4

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

5

Transmission Line Concept
Power Frequency (f) is @ 60 Hz Power
Wavelength (λ ) is 5× 10 m
6
Plant
( Over 3,100 Miles)

Consumer
Home

Transmission Lines
Class 6

PC Transmission Lines
6

Signal Frequency (f) is
approaching 10 GHz Integrated Circuit
Stripline
Wavelength (λ ) is 1.5 cm
( 0.6 inches) Microstrip T

PCB substrate
Cross section view taken here
Stripline
W
Via
Micro-
FR4 Dielectric
Copper Trace
Cross Section of Above PCB
Strip

Signal (microstrip)

Ground/Power
T Copper Plane Signal (stripline)
Signal (stripline)
Ground/Power
Signal (microstrip)
Transmission Lines
Class W6

7

Key point about transmission line operation
Voltage and current on a transmission line is
a function of both time and position.
I2

V = f ( z, t )
I1

I = f ( z, t )
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
Transmission Lines
small compared to the signaling wavelength
Class 6

8

Examples of Transmission Line
Structures- I
 Cables and wires
(a) Coax cable
(b) Wire over ground
(c) Tri-lead wire
(d) Twisted pair (two-wire line)
 Long distance interconnects
+

+ -
(a) (b)
-

- + - Transmission Lines
+ -
(c)
Class 6 (d)

9

Segment 2: Transmission line equivalent
circuits and relevant equations

 Physics of transmission line structures
 Physics of transmission line structures
 Basic transmission line equivalent circuit
 Basic transmission line equivalent circuit
 ?Equations for transmission line propagation
 ?Equations for transmission line propagation

Transmission Lines
Class 6

10

E & H Fields – Microstrip Case

How does the signal move
Signal path

from source to load? Y

Z (into the page)
X

Electric field

Remember fields are setup given field
Magnetic

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”
11

 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
line V
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

12

Presence of Electric and Magnetic Fields
H
I I + ∆I I I + ∆I
+ + + +

E
V V + ∆V V V + ∆V
H
I I + ∆I I 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
Transmission Lines
Implies equivalent circuit model must contain inductor.
Class 6

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 lC0

Transmission Lines
Class 6

14

Ideal T Line
 Ideal (lossless) Characteristics of
Transmission Line lL0
Ideal TL assumes:
Uniform line lC0
Perfect (lossless) conductor (R0→ 0)
Perfect (lossless) dielectric (G0→ 0)
We only consider T0, Z0 , C0, and L0.
 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
15

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

16

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

Other “Rules of Thumb”
17

 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

When does a T-line become a T-Line?
18

 Whether it is a
bump or a
mountain depends
on the ratio of its
When do we need to size (tline) to the
use transmission line size of the vehicle
analysis techniques vs. (signal
lumped circuit wavelength)
analysis?
 Similarly, whether 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
Wavelength/edge rate Tline the rise/fall edge of
the signal
Transmission Lines
Class 6

Equations & Formulas

How to model & explain
transmission line behavior

20

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

21

Ideal Transmission Line Parameters
 Knowing any two out of Z0,
Td, C0, and L0, the other two
L0
can be calculated. Z0 = ; T d = L0 C0 ;
 C0 and L0 are reciprocal C0
functions of the line cross- T0
sectional dimensions and C0 = ; L0 = Z 0 T 0 ;
Z0
are related by constant me.
1
 ε is electric permittivity v0 = ; C0 L0 = µε;
ε 0= 8.85 X 10-12 F/m (free space) µε
ε ri s relative dielectric constant
µ = µr µ0 ; ε = εr ε0 .
 µ is magnetic permeability
µ 0= 4p X 10-7 H/m (free space)
µ r is relative permeability

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

Parallel Plate Approximation
22

 Assumptions TC

TEM conditions ε TD

Uniform dielectric (ε )
between conductors WC
TC<< TD; WC>> TD
ε * PlateArea Base
 T-line characteristics are C=
function of: d equation
Material electric and WC  F  WC  pF 
magnetic properties C0 ε⋅ ⋅  8.85 ⋅ε r ⋅ ⋅ 
TD  m  TD  m 
Dielectric Thickness (TD)
TD  F  T D  µH 
Width of conductor (WC)
L0 µ⋅ ⋅  0.4 ⋅π ⋅µ r ⋅ ⋅ 
 Trade-off WC  m  WC  m 
TD ; C0 , L0 , Z0  TD µr
Z0 377 ⋅ ⋅ ⋅Ω
WC ; C0 , L0 , Z0  WC εr

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

23

Improved Microstrip Formula
 Parallel Plate Assumptions +
WC
Large ground plane with
TC
zero thickness
ε TD
 To accurately predict
microstrip impedance, you
must calculate the effective
dielectric constant. From Hall, Hall & McCall:
87  5.98TD 
Z0 ≈ ln  Valid when:
εr + 1.41  0.8WC + TC  0.1 < WC/TD < 2.0 and 1 < r < 15

εr + 1 εr − 1 TC
εe = + + F − 0.217( εr − 1)
2 12TD WCTD
2 1+
WC
2 You can’t beat
 WC 
0.02(εr −1)1 − a field solver
WC
 for
TD
<1
F=  TD 
0 WC
Transmission Lines
for
T
>1
D

Class 6

24

Improved Stripline Formulas
 Same assumptions as WC TD1
used for microstrip ε
TC

apply here TD2

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

60 4(TD1 + TD1) 
Z 0 sym ≈ ln 
 0.67π (0.8WC + TC ) 
εr  
Valid when WC/(TD1+TD2) < 0.35 and TC/(TD1+TD2) < 0.25
You can’t beat a
Offset (unbalanced) Stripline Case TD1 > TD2 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

25

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

Reflection and Transmission
26

Incident 1+ρ Transmitted

ρ Reflected

Reflection Coeficient Transmission Coeffiecent
Zt − Z0
ρ
Zt − Z0 τ (1 + ρ) "" → "" τ 1+
Zt + Z0
Zt + Z0
2⋅ Zt
τ
Zt + Z0

Transmission Lines
Class 6

27

Special Cases to Remember
A: Terminated in Zo
Zs −
Zo Zo ρ = Zo Zo = 0
Vs Zo + Zo

B: Short Circuit
Zs −
Zo ρ = 0 Zo = −1
Vs 0 + Zo

C: Open Circuit
Zs ∞ − Zo
Zo ρ= =1
Vs ∞ + Zo

Transmission Lines
Class 6

28

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
Transmission Lines
plate approximation?
Class 6

29

Transmission line equivalent circuits and
relevant equations

 Basic pulse launching onto transmission lines
 Basic pulse launching onto transmission lines
 Calculation of near and far end waveforms for
classic load conditions
classic load conditions

Transmission Lines
Class 6

Review: Voltage Divider Circuit
30

 Consider the
simple circuit that RS
contains source
voltage VS, source RL
VS VL
resistance RS, and
resistive load RL.
 The output
voltage, VL is
RL
easily calculated VL = VS
from the source RL + R S
amplitude and the
values of the two
series resistors.
Why do we care for?
Why do we care for?
Next page….
Transmission Lines
Next page….
Class 6

31

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:
1. Determination of launch voltage &
final “DC” or “t =0” voltage
2. Calculation of load reflection coefficient and
voltage delivered to the load
3. Calculation of source reflection coefficient
and resultant source voltage
These are the steps for solving
These are the steps for solving
all t-line problems.
all t-line problems.
Transmission Lines
Class 6

Determining Launch Voltage
32

TD
Rs A B
Vs
Zo
0 Vs Rt

(initial voltage)
t=0, V=Vi

Z0 Rt
Vi = VS Vf = VS
Z 0 + RS 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 –
Transmission Lines
it is simply a voltage divider!
Class 6

33

Voltage Delivered to the Load
TD
Vs Rs A B
Vs Zo Rt
0

(initial voltage)
t=0, V=Vi

(signal is reflected)
t=2TD,

ρΒ = Rt − Zo
V=Vi + ρB(Vi) +ρA(ρB)(Vi ) t=TD, V=Vi +ρB(Vi )
Vreflected = ρ Β (Vincident)
Rt + Zo 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 + VTransmission Lines
reflected at time t = TD
Class 6

34

Voltage Reflected Back to the Source

Rs A B
Vs
Zo
0 Vs ρA 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

Voltage Reflected Back to the Source 35

− Zo Vreflected = ρ Α (Vincident)
ρ Α = Rs
Rs + Zo 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
 Voltage wavefront will be reflected at the source
 VA = Vlaunch + Vincident + Vreflected at time t = 2TD

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

36

Problems Solved Homework

 Consider the circuit
shown to the right
with a resistive load,
assume propagation RS I1 I2
Z0 ,Τ 0

delay = T, RS= Z0 . VS V1
l
V2 RL

Calculate and show
the wave forms of
V1(t),I1(t),V2(t), and
I2(t) for (a) RL= ∞
and (b) RL= 3Z0
Transmission Lines
Class 6

37

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 Solution
Class 6

38

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

Class 6

39

Waveshape:
Step-Function into T-Line with Open Ckt
I1
IA
I2
RS I1 I2
Current (A)

0.75IA Z0 ,Τ 0

l
0.5I A VS V1 V2 Open

0.25IA

0 Τ 2Τ 3Τ 4Τ Time (ns)

VA
V1 This is called
V2 “reflected wave
Voltage (V)

0.75VA
switching”
0.5VA

0.25VA

Transmission Lines (ns) Solution
0 Τ 2Τ 3Τ 4Τ Time
Class 6

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
Class 6

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

Class 6

Waveshapes for Problem 1b
42

I1
IA
I2 I1 I2
RS
0.75IA Z0 ,Τ 0
Current (A)

l
0.5IA VS V1 V2 RL

0.25IA

0 Τ 2Τ 3Τ 4Τ Time (ns)

I1
VA
I2
Note that a properly
0.75VA terminated wave
Voltage (V)

0.5VA
settle out at 0.5 V
0.25VA
Solution
0 Τ Transmission Time (ns)
2Τ 3Τ 4Τ Lines
Solution
Class 6

Transmission line step response
43

 Introduction to lattice diagram analysis
 Introduction to lattice diagram analysis
classic load impedances
classic load impedances

 Solving multiple reflection problems
Solving multiple reflection problems

Complex signal reflections at different types of
Complex signal reflections at different types of
transmission line “discontinuities” will be analyzed
transmission line “discontinuities” will be analyzed
in this chapter. Lattice diagrams will be introduced
in this chapter. Lattice diagrams will be introduced
as a solution tool.
as a solution tool.
Transmission Lines
Class 6

44

Lattice Diagram Analysis – Key Concepts
V(source) Zo V(load)
Vs Rs
The lattice diagram is a 0 TD = N ps
Vs Rt
tool/technique to simplify
the accounting of ρsource ρload
reflections and waveforms V(load)
Time V(source)
 Diagram shows the boundaries 0 a
(x =0 and x=l) and the reflection A’
coefficients (GL and GL )
N ps A
 Time (in T) axis shown b
vertically
 Slope of the line should 2N ps
c
B’
indicate flight time of signal
Particularly important for multiple
reflection problems using both 3N ps B
microstrip and stripline mediums. d
 Calculate voltage amplitude for C’
each successive reflected wave 4N ps
e
 Total voltage at any point is the
sum of all the waves that have 5N ps
reached that point Transmission Lines
Class 6

Lattice Diagram Analysis – Detail 45

ρ ρ
source load

V(source) V(load)

0 Vlaunch
0

Time Vlaunch N ps
Vlaunch ρload

Vlaunch(1+ρload) Time
2N ps
Vlaunch ρloadρsource

Vlaunch(1+ρload +ρload ρsource) 3N ps

Vlaunch ρ2loadρsource

Vlaunch(1+ρload+ρ2loadρsource+ ρ2loadρ2source)
4N ps
Vlaunch ρ2loadρ2source
V(source) Zo V(load)
Vs Rs 5N ps
0 Vs TD = N ps
Rt
Transmission Lines
Class 6

Transient Analysis – Over Damped 46

V(source) Zo V(load) Assume Zs=75 ohms
2v Zs Zo=50ohms
0 TD = 250 ps Vs=0-2 volts
Vs
Zo  50 
Vinitial = Vs = (2)  = 0.8
ρ source = 0.2 ρ load = 1 Zs + Zo  75 + 50 
Time V(source) V(load) Zs − Zo 75 − 50
0
ρ source = = = 0.2
0.8v Zs + Zo 75 + 50
0v
Zl − Zo ∞ − 50
500 ps 0.8v
ρ load = = =1
Zl + Zo ∞ + 50
0.8v

1000 ps 1.6v Response fr om lattice diagram
0.16v
2.5
1500 ps 1.76v 2
0.16v
1.5
V olt s
1.92v 1 Sour ce
2000 ps
0.032v 0.5
Load

0
2500 ps 0 2 50 500 750 1000 1250

Transmission Lines Tim e , ps

Class 6

Transient Analysis – Under Damped 47

V(source) Zo V(load) Assume Zs=25 ohms
2v Zs Zo =50ohms
0 Vs TD = 250 ps Vs=0-2 volts

Zo  50 
Vinitial =Vs = (2)   =1.3333
ρsource = −0 . 3333 ρload = 1 Zs + Zo  25 +50 
Time V(source) V(load)
Zs − Zo 25 −50
0 ρsource = = = −0.33333
1.33v Zs + Zo 25 +50
0v
Zl − Zo ∞ −50
500 ps 1.33v ρ = = =1
1.33v
load
Zl + Zo ∞ + 50

1000 ps 2.66v Response from lattice diagram
-0.443v

3
1500 ps 2.22v
-0.443v 2.5
2
Volts
1.77v 1.5
2000 ps
0.148v 1 Source
0.5 Load
2500 ps 1.92 0
0.148v 0 250 500 750 1000 1250 1500 1750 2000 2250

Time, ps

Transmission Lines
2.07

Class 6

Two Segment Transmission Line Structures
48

X X
Rs
Zo1 Zo2 Rt
Vs
TD TD A= a A' = b + e
T3 T2 B = a+c+d B' = b + e + g + i
ρ1 ρ 2 ρ3 ρ4 C = A+ c+ d + f + h C' = b + e + g + i + k + l
a a = vi
Z o1
vi = Vs
TD A c b Rs + Z o1 b = aT2

2TD Rs − Z o1 c = aρ 2
d e ρ1 =
Rs + Z o1
3TD B g
A’ d = cρ 1
f
Z o 2 − Z o1
4TD
ρ2 = e = bρ 4
h i Z o 2 + Z o1
B’ f = dρ 2 + eT3
5TD C j k Z − Zo2
ρ 3 = o1 g = eρ 3 + dT2
Z o1 + Z o 2
l
Rt − Z o 2 h = fρ 1
C’ ρ4 =
Rt + Z o 2 i = gρ 4

T2 = 1 + ρ 2 j = hρ 2 + iT3
Transmission Lines
T = 1+ ρ k = iρ 3 + hT2
Class 6 3 3

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 3
S 1 2

diagram and calculate Z ,Τ Z ,Τ
01 01 02 02

l l
reflection coefficients at
1 2
V V 1V V 3 Short
S 2

the interfaces and show
the wave forms of V1(t),
V2(t), and V3(t).
 Check results with PSPICE

Transmission Lines
Class 6

Class06 transmission line_basics

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