Transmission Line Basics II - Class 6 Prerequisite Reading assignment: CH2 Acknowledgements: Intel Bus Boot Camp: Michael Leddige
Nov 15, 2014
Transmission Line Basics II - Class 6
Prerequisite Reading assignment: CH2
Acknowledgements: Intel Bus Boot Camp: Michael Leddige
2
Transmission Lines Class 6
Real Computer Issues
Dev a Dev b
Clk Switch Threshold
Signal Measured here
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.
data
3
Transmission Lines Class 6
Agenda
The Transmission Line ConceptTransmission line equivalent circuits
and relevant equationsReflection diagram & equationLoadingTermination methods and comparisonPropagation delaySimple return path ( circuit theory,
network theory come later)
4
Transmission Lines Class 6
Two Transmission Line Viewpoints
Steady state ( most historical view)Frequency domain
TransientTime domainNot circuit element Why?
We mix metaphors all the timeWhy convenience and history
5
Transmission Lines Class 6
Transmission Line Concept
PowerPlant
ConsumerHome
Power Frequency (f) is @ 60 Hz
Wavelength () is 5 106 m
( Over 3,100 Miles)
6
Transmission Lines Class 6
PC Transmission Lines
Integrated Circuit
Microstrip
Stripline
Via
Cross section view taken herePCB substrate
T
W
Cross Section of Above PCB
T
Signal (microstrip)
Ground/PowerSignal (stripline)Signal (stripline) Ground/Power
Signal (microstrip)
Copper Trace
Copper Plane
FR4 Dielectric
W
Signal Frequency (f) is approaching 10 GHz
Wavelength () is 1.5 cm ( 0.6 inches)
Micro-Strip
Stripline
7
Transmission Lines Class 6
Key point about transmission line operation
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 behaviorThis positional dependence is added when the assumption of the size of the circuit being small compared to the signaling wavelength
tzfI
tzfV
,
,
V1 V2
dz
I2I1
Voltage and current on a transmission line is a function of both time and position.
8
Transmission Lines Class 6
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)
(c) (d)
+
-
+
+ +-
- -
-
9
Transmission Lines Class 6
Segment 2: Transmission line equivalent circuits and relevant equations
Physics of transmission line structures
Basic transmission line equivalent circuit
?Equations for transmission line propagation
Physics of transmission line structures
Basic transmission line equivalent circuit
?Equations for transmission line propagation
10
Transmission Lines Class 6
Remember fields are setup given an applied forcing function.
(Source)
How does the signal move from source to load?
E & H Fields – Microstrip Case
The signal is really the wave propagating between the conductors
Electric field
Magnetic field
Ground return path
X
Y
Z (into the page)
Signal path
Electric field
Magnetic field
Ground return path
X
Y
Z (into the page)
Signal path
11
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 linesA 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
Propagation velocity is the speed with which signals are transmitted through the transmission line in its surrounding medium.
I
VZ 0
r
cv
12
Transmission Lines Class 6
Presence of Electric and Magnetic Fields
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 lineImplies equivalent circuit model must contain inductor.
V
I
I
E
+
-
+
-
+
-
+
-
V + DV
I + DI
I + DI
V
IH
IH
V + DV
I + DI
I + DI
13
Transmission Lines Class 6
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) [/in]
Per-unit-length (Parallel) Conductance (G0) [S/in]
T-Line Equivalent Circuit
lL0lR0
lC0lG0
14
Transmission Lines Class 6
Ideal T Line Ideal (lossless) Characteristics of
Transmission LineIdeal TL assumes:
Uniform line
Perfect (lossless) conductor (R00)
Perfect (lossless) dielectric (G00)
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.
lL0
lC0
15
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.
16
Transmission Lines Class 6
T Line Rules of Thumb
Td < .1 Tx
Td < .4 Tx
May treat as lumped Capacitance Use this 10:1 ratio for accurate modeling of transmission lines
May treat as RC on-chip, and treat as LC for PC board interconnect
So, what are the rules of thumb to use?
17
Transmission Lines Class 6
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
18
Transmission Lines Class 6
When does a T-line become a T-Line?
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 When do we need to use to use
transmission line transmission line analysis analysis
techniques vs. techniques vs. lumped circuit lumped circuit
analysis? analysis?
TlineWavelength/edge rate
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 the rise/fall edge of the signal
Equations & Formulas
How to model & explain transmission line behavior
20
Transmission Lines Class 6
Relevant Transmission Line Equations
Propagation equation
jCjGLjR ))((
)(
)(0
CjG
LjRZ
Characteristic Impedance equation
In class problem: Derive the high frequency, lossless approximation for Z0
is the attenuation (loss) factor
is the phase (velocity) factor
21
Transmission Lines Class 6
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 cross-sectional dimensions and are related by constant me.
is electric permittivity 0= 8.85 X 10-12 F/m (free space)
ri s relative dielectric constant
is magnetic permeability 0= 4p X 10-7 H/m (free space)
r is relative permeability
.;
;;1
;;
;;
00
000
0000
00
00d0
00
rr
LCv
TZLZ
TC
CLTC
LZ
Don’t forget these relationships and what they mean!Don’t forget these relationships and what they mean!
22
Transmission Lines Class 6
Parallel Plate Approximation Assumptions
TEM conditions
Uniform dielectric ( ) between conductorsTC<< TD; WC>> TD
T-line characteristics are function of:
Material electric and magnetic propertiesDielectric Thickness (TD)
Width of conductor (WC) Trade-off
TD ; C0 , L0 , Z0 WC ; C0 , L0 , Z0
TD
TC
WC
To a first order, t-line capacitance and inductance can be approximated using the parallel plate approximation.
d
PlateAreaC
* Base
equation
C0 WC
TD
F
m
8.85 r
WC
TD
pF
m
L0 TD
WC
F
m
0.4 r
TD
WC
H
m
Z0 377TD
WC
r
r
23
Transmission Lines Class 6
Improved Microstrip Formula Parallel Plate Assumptions
+Large ground plane with zero thickness
To accurately predict microstrip impedance, you must calculate the effectiveeffective dielectric constant.
TD
TC
WC
From Hall, Hall & McCall:
CC
D
r TW
TZ
8.0
98.5ln
41.1
870
DC
Cr
C
D
rre
TW
TF
W
T1217.0
1212
1
2
1
2
1102.0
D
Cr
T
WF
1D
C
T
Wfor
0 1D
C
T
Wfor
Valid when: 0.1 < WC/TD < 2.0 and 1 < r < 15
You can’t beat a field solver
24
Transmission Lines Class 6
Improved Stripline Formulas Same assumptions as used
for microstrip apply here
TD2
TC
WCTD1
From Hall, Hall & McCall:
)8.0(67.0
)(4ln
60 110
CC
DD
r
symTW
TTZ
Symmetric (balanced) Stripline Case TD1 = TD2
),,,2(),,,2(
),,,2(),,,2(2
00
000
rCCsymrCCsym
rCCsymrCCsymoffset
TWBZTWAZ
TWBZTWAZZ
Offset (unbalanced) Stripline Case TD1 > TD2
Valid when WC/(TD1+TD2) < 0.35 and TC/(TD1+TD2) < 0.25
You can’t beat a field solver
25
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 coefficientAmount of signal reflected from the “bump”Frequency domain =sign(S11)*|S11|If at load or source the reflection may be called gamma (L or s)Time domain 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
26
Transmission Lines Class 6
Reflection and Transmission
Incident
Reflected
Transmitted
Reflection CoeficientTransmission Coeffiecent
1 "" "" 1Zt Z0Zt Z0
Zt Z0Zt Z0
2 Zt
Zt Z0
27
Transmission Lines Class 6
Special Cases to Remember
1ZoZo
0ZoZo
ZoZo
100
ZoZo
Vs
ZsZo Zo
A: Terminated in Zo
Vs
ZsZo
B: Short Circuit
Vs
ZsZo
C: Open Circuit
28
Transmission Lines Class 6
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, r = 4
Symmetric Stripline:
WC = 6 mils, TD1 = TD2 = 4 mils, TC = 1 mil, r = 4
Write Math Cad Program to calculate Z0, Td, L & C for each case.
What factors cause the errors with the parallel plate approximation?
29
Transmission Lines Class 6
Transmission line equivalent circuits and relevant equations
Basic pulse launching onto transmission lines
Calculation of near and far end waveforms for
classic load conditions
Basic pulse launching onto transmission lines
Calculation of near and far end waveforms for
classic load conditions
30
Transmission Lines Class 6
Review: Voltage Divider Circuit Consider the
simple circuit that contains source voltage VS, source resistance RS, and resistive load RL.
The output voltage, VL is easily calculated from the source amplitude and the values of the two series resistors.
RS
RLVS VL
RSRL
RLVSVL
+=
Why do we care for? Next page….
Why do we care for? Next page….
31
Transmission Lines Class 6
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 all t-line problems.
These are the steps for solving all t-line problems.
32
Transmission Lines Class 6
Determining Launch Voltage
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!
VsZo
RsVs
0
TD
Rt
A B
t=0, V=Vi
(initial voltage)
RSZ0
Z0VSVi
+=
RSRt
RtVSVf
+=
33
Transmission Lines Class 6
Voltage Delivered to the Load
Vs ZoRsVs
0
TD
Rt
A B
t=0, V=Vi
t=TD, V=Vi + B(Vi )t=2TD, V=Vi + B(Vi) + AB)(Vi )
(signal is reflected)
(initial voltage)
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
ZoRtZoRt
Vreflected = (Vincident)
VB = Vincident + Vreflected
34
Transmission Lines Class 6
Voltage Reflected Back to the Source
VsZo
RsVs
0
TD
Rt
A B
t=0, V=Vi
t=TD, V=Vi + B (Vi )t=2TD,
V=Vi + B
(Vi) + A
B)(Vi )
(signal is reflected)
(initial voltage)
A B
35
Transmission Lines Class 6
Voltage Reflected Back to the Source
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)]
ZoRsZoRs
Vreflected = (Vincident)
VA = Vlaunch + Vincident + Vreflected
36
Transmission Lines Class 6
Problems
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
Z0 ,T0
V1 V2
l
I2I1
VS RL
RS
Solved Homework
37
Transmission Lines Class 6
Step-Function into T-Line: Relationships
Source matched case: RS= Z0
V1(0) = 0.5VA, I1(0) = 0.5IA
S = 0, V(x,) = 0.5VA(1+ L)
Uncharged lineV2(0) = 0, I2(0) = 0
Open circuit means RL= L = / = 1
V1() = V2() = 0.5VA(1+1) = VA
I1() = I2 () = 0.5IA(1-1) = 0Solution
38
Transmission Lines Class 6
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
39
Transmission Lines Class 6
Waveshape:Step-Function into T-Line with Open Ckt
Z0 ,T0
V1 V2
l
I2I1
VSOpen
RS
Cur
rent
(A)
2T Time (ns)3TT 4T0
0.5IA
0.25IA
IA
0.75IA
I1
I2
Vol
tage
(V)
2T Time (ns)3TT 4T0
0.5VA
0.25VA
VA
0.75VA
V1
V2
This is called This is called “reflected wave “reflected wave switching”switching”
Solution
40
Transmission Lines Class 6
Problem 1b: Relationships
Source matched case: RS= Z0
V1(0) = 0.5VA, I1(0) = 0.5IA
S = 0, V(x,) = 0.5VA(1+ L)
Uncharged lineV2(0) = 0, I2(0) = 0
RL= 3Z0
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
Solution
41
Transmission Lines Class 6
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
42
Transmission Lines Class 6
Waveshapes for Problem 1b
Z0 ,T0
V1 V2
l
I2I1
VS RL
RS
Cur
rent
(A)
2T Time (ns)3TT 4T0
0.5IA
0.25IA
IA
0.75IA
I1
I2
Vol
tage
(V)
2T Time (ns)3TT 4T0
0.5VA
0.25VA
VA
0.75VA
I1
I2Note that a Note that a properly properly terminated wave terminated wave settle out at 0.5 settle out at 0.5 VVSolution
Solution
43
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
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.
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.
44
Transmission Lines Class 6
Lattice Diagram Analysis – Key Concepts
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
Rs
ZoV(source) V(load)
TD = N ps0
Vs
RtThe lattice diagram is a tool/technique to simplify the accounting of reflections and waveforms
Time V(source) V(load)
a
source load
bA
c
A’
B’
C’
dB
e
0
N ps
2N ps
3N ps
4N ps
5N ps
45
Transmission Lines Class 6
Lattice Diagram Analysis – Detail
V(source) V(load)
Vlaunch
source
load
Vlaunch load
Vlaunch
0
Vlaunch(1+load)
Vlaunch(1+load +load source)
Time
0
2N ps
4N ps
Vlaunch loadsource
Vlaunch loadsource
Vlaunch load
source
Vlaunch(1+load+loadsource+
loadsource)
Time
N ps
3N ps
5N psVs
RsZoV(source) V(load)
TD = N ps0Vs
Rt
46
Transmission Lines Class 6
Transient Analysis – Over DampedAssume Zs=75 ohms Zo=50ohmsVs=0-2 voltsVs
Zs
ZoV(source) V(load)
Time V(source) V(load)
150
50
2.05075
5075
8.05075
50)2(
ZoZl
ZoZl
ZoZs
ZoZs
ZoZs
ZoVsV
load
source
initial
0.8v
2.0source 1load
0.8v0.8v
0.16v
0v
1.6v
1.92v
0.16v1.76v
0.032v
TD = 250 ps
0
500 ps
1000 ps
1500 ps
2000 ps
2500 ps
0
2 v
Response fr om lattice diagram
0
0.5
1
1.5
2
2.5
0 2 50 500 750 1000 1250
Tim e , ps
Vo
lts
Source
Load
47
Transmission Lines Class 6
Transient Analysis – Under Damped
150
50
33333.05025
5025
3333.15025
50)2(
ZoZl
ZoZl
ZoZs
ZoZs
ZoZs
ZoVsV
load
source
initial
Assume Zs=25 ohms Zo =50ohmsVs=0-2 voltsVs
Zs
ZoV(source) V(load)
TD = 250 ps0
2 v
Time V(source) V(load)
1.33v
3333.0source
1load
1.33v1.33v
-0.443v
0v
2.66v
1.77v
-0.443v2.22v
0.148v
0
500 ps
1000 ps
1500 ps
2000 ps
2500 ps 1.920.148v
2.07
Response from lattice diagram
0
0.5
1
1.5
2
2.5
3
0 250 500 750 1000 1250 1500 1750 2000 2250
Time, ps
Vo
lts
Source
Load
48
Transmission Lines Class 6
Two Segment Transmission Line Structures
Vs
RsZo1 RtZo2
X X
a
bc
d e
f g
h i
j k
l
33
22
2
24
21
213
12
122
1
11
1
1
1
1
T
T
ZRt
ZRt
ZZ
ZZ
ZZ
ZZ
ZRs
ZRs
ZRs
ZVv
o
o
oo
oo
oo
oo
o
o
o
osi
23
32
4
1
23
32
4
1
2
2
hTik
iThj
gi
fh
dTeg
eTdf
be
cd
ac
aTb
va i
hfdcAC
dcaB
aA
lkigebC
igebB
ebA
'
'
'
A
B
C
A’
B’
C’
1 2 3 43T 2T
TD TD
TD
3TD
2TD
4TD
5TD
49
Transmission Lines Class 6
Assignment Consider the two segment
transmission line shown to the right. Assume RS= 3Z01
and Z02= 3Z01 . Use Lattice
diagram and calculate reflection coefficients at the interfaces and show the wave forms of V1(t), V2(t),
and V3(t).
Check results with PSPICE
Z01 ,T01
V1 V2
l1
I2I1
VS
RSZ02 ,T02
V3
l2
I3
Short
Previous examples are the preparation