Transcript Transmission lines

Transmission Line Basics II - Class 6
Prerequisite Reading assignment: CH2
Acknowledgements: Intel Bus Boot Camp:
Michael Leddige
Real Computer Issues
Dev a
Clk
2
data
Dev b
Signal
Measured
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
Agenda
3
 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
4
5
Transmission Line Concept
Power Frequency (f) is @ 60 Hz
Wavelength (l) is 5
106 m
( Over 3,100 Miles)
Consumer
Home
Transmission Lines Class 6
Power
Plant
PC Transmission Lines
Signal Frequency (f) is
approaching 10 GHz
Wavelength (l) is 1.5 cm
( 0.6 inches)
Microstrip
6
Integrated Circuit
Stripline
T
PCB substrate
Cross section view taken here
Stripline
W
Cross Section of Above PCB
Copper Trace
Via
FR4 Dielectric
MicroStrip
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)
-
+
(d)
(b)
-
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
r
Transmission Lines Class 6
11
Presence of Electric and Magnetic Fields
I
+
+
+ +
E
V
I
-
-
-
-
I + DI
V + DV
I + DI
H
I
V
I
H
I + DI
V + DV
I + DI
 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) [W/in]
Per-unit-length (Parallel) Conductance (G0) [S/in]
lR0
lL0
lG0
Transmission Lines Class 6
lC0
Ideal T Line
14
 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
16
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
17
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
Transmission Lines Class 6
Tline
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
  (R  jL)(G  jC)    j
 is the attenuation (loss) factor
 is the phase (velocity) factor
Characteristic Impedance equation
( R  jL)
Z0 
(G  jC )
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.
  is electric permittivity
0= 8.85 X 10-12 F/m (free space)
ri s relative dielectric constant
 m is magnetic permeability
Z0 
L0
;
C0
T0
C0  ;
Z0
1
v0 
;
m
m  mr m0 ;
T d  L0 C0 ;
L0  Z 0 T 0 ;
C0 L0  m;
  r 0 .
m0= 4p X 10-7 H/m (free space)
mr is relative permeability
Don’t forget these relationships and what they mean!
Transmission Lines Class 6
21
Parallel Plate Approximation
 Assumptions
TC
TEM conditions

Uniform dielectric ( )
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
TD ; C0 , L0 , Z0 
WC ; C0 , L0 , Z0 
22
TD
WC
 * PlateArea Base
C
d
C0
L0
Z0
WC F
 

 
TD  m 
TD
F
m
 
WC  m 
377 
TD
WC

mr
r
equation
WC p F
 
8.85  r 
 
TD  m 
T D  mH 
0.4  m r 


WC  m 
W
To a first order, t-line capacitance and inductance can
be approximated using the parallel plate approximation.
Transmission Lines Class 6
23
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.
TC

From Hall, Hall & McCall:
TD
 5.98TD 
Valid when:
ln

0.1 < WC/TD < 2.0 and 1 < er < 15
r  1.41  0.8WC  TC 
r  1
r  1
TC
e 

 F  0.217r  1
2
12TD
WCTD
2 1
WC
You can’t beat
2
W
WC 

for
1
0.02r  11 

a field solver
T
TD
87
Z0 
C
F
0


D
for
WC
TD
1
Transmission Lines Class 6
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Improved Stripline Formulas
 Same assumptions as
used for microstrip
apply here
WC

TD1
TC
TD2
From Hall, Hall & McCall:
Symmetric (balanced) Stripline Case TD1 = TD2


4(TD1  TD1)

Z 0 sym 
ln

0
.
67

(
0
.
8
W
C

T
C
)
r 

60
Valid when WC/(TD1+TD2) < 0.35 and TC/(TD1+TD2) < 0.25
Offset (unbalanced) Stripline Case TD1 > TD2
Z 0offset
You can’t beat a
field solver
Z 0 sym(2 A,WC , TC , r )  Z 0 sym(2 B,WC , TC , r )
2
Z 0 sym(2 A,WC , TC , r )  Z 0 sym(2 B,WC , TC , r )
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 r=sign(S11)*|S11|
If at load or source the reflection may be called gamma (GL
or Gs)
Time domain r 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
r
1r
Transmitted
Reflected
Reflection Coeficient Transmission Coeffiecent
r
Zt  Z0

1  r 
""  ""
Zt  Z0

Transmission Lines Class 6
2 Zt
Zt  Z0

1
Zt  Z0
Zt  Z0
26
Special Cases to Remember
27
A: Terminated in Zo
Zs
Zo
Vs
Zo

r  Zo Zo  0
Zo  Zo
B: Short Circuit
Zs
Zo
Vs

r  0 Zo  1
0  Zo
C: Open Circuit
Zs
Vs
Zo
Transmission Lines Class 6
r
  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, 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?
Transmission Lines Class 6
28
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 + RS
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:
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.
Transmission Lines Class 6
31
Determining Launch Voltage
32
TD
Vs
0
Rs A
B
Zo
Vs
Rt
(initial voltage)
t=0, V=Vi
Vi = VS
Z0
Z0 + 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
Voltage Delivered to the Load
TD
Vs
Rs A
Zo
Vs
0
B
Rt
(initial voltage)
t=0, V=Vi
rB 
t=2TD,
rArB)(Vi )
V=Vi
 +Zo
Rt+ rB(Vi)
Rt  Zo
(signal is reflected)
t=TD, V=Vi +rB(Vi )
Vreflected = rB (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
33
Voltage Reflected Back to the Source
Vs
0
Rs A
Vs
B
Zo
rA
rB
Rt
TD
(initial voltage)
t=0, V=Vi
(signal is reflected)
t=2TD,
V=Vi + rB (Vi) + rAr B )(Vi )
Transmission Lines Class 6
t=TD, V=Vi + rB (Vi )
34
Voltage Reflected Back to the Source
rA
 Zo
Rs

Rs  Zo
Vreflected = rA (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
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
35
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
Transmission Lines Class 6
I1
V1
Z0 ,T0
l
I2
V2
RL
Step-Function into T-Line: Relationships
 Source matched case: RS= Z0
V1(0) = 0.5VA, I1(0) = 0.5IA
GS = 0, V(x,) = 0.5VA(1+ GL)
 Uncharged line
V2(0) = 0, I2(0) = 0
 Open circuit means RL= 
GL =  / = 1
V1() = V2() = 0.5VA(1+1) = VA
I1() = I2 () = 0.5IA(1-1) = 0
Solution
Transmission Lines Class 6
37
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
38
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 ,T0
l
I2
V2
0.25IA
0
T
3T
4T Time (ns)
V1
V2
VA
Voltage (V)
2T
0.75VA
This is called
“reflected wave
switching”
0.5VA
0.25VA
0
T
3T
2T
4T Time (ns)
Transmission Lines Class 6
Solution
Open
40
Problem 1b: Relationships
 Source matched case: RS= Z0
V1(0) = 0.5VA, I1(0) = 0.5IA
GS = 0, V(x,) = 0.5VA(1+ GL)
 Uncharged line
V2(0) = 0, I2(0) = 0
 RL= 3Z0
GL = (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
Transmission Lines 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
Solution
Transmission Lines Class 6
Waveshapes for Problem 1b
I1
I2
Current (A)
IA
RS
0.75IA
0.5IA
VS
I1
V1
42
Z0 ,T0
l
I2
V2
0.25IA
0
T
2T
3T
I1
I2
VA
Voltage (V)
4T Time (ns)
Note that a
properly terminated
wave settle out at
0.5 V
0.75VA
0.5VA
0.25VA
Solution
0
T
2T
3T
4T Time (ns)
Transmission Lines Class 6
Solution
RL
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
43
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.
Vs
0
Vs
Zo
V(source)
Rs
TD = N ps
V(load)
Time V(source)
N ps
a
A’
A
b
B’
2N ps
3N ps
c
B
d
 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
Rt
rload
rsource
0
V(load)
4N ps
5N ps
Transmission Lines Class 6
C’
e
Lattice Diagram Analysis – Detail
r
r
source
load
V(load)
V(source)
0
45
Vlaunch
0
Time
Vlaunch
N ps
Vlaunch rload
Vlaunch(1+rload)
2N ps
Time
Vlaunch rloadrsource
Vlaunch(1+rload +rload rsource)
3N ps
Vlaunch r2loadrsource
Vlaunch(1+rload+r2loadrsource+ r2loadr2source)
4N ps
Vlaunch r2loadr2source
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
r source  0 . 2
0
Assume Zs=75 ohms
Zo=50ohms
Vs=0-2 volts
V(load)
r load  1
V(load)
Time V(source)
0.8v
Vinitial  Vs
r source 
0v
500 ps
rload 
0.8v
0.8v
1000 ps
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
0.16v
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
V(source)
2v
0
Zo
Zs
TD = 250 ps
Vs
rsource  0 . 3333
Time
Assume Zs=25 ohms
Zo =50ohms
Vs=0-2 volts
V(load)
V(load)
V(source)
0
rload  1
1.33v
0v
500 ps 1.33v
Vinitial Vs
 50 
Zo
 (2) 
 1.3333


Zs Zo
 25 50 


rsource  Zs Zo  25 50  0.33333
Zs  Zo

 50
rload  Zl Zo 
1
Zl  Zo
1.33v
2.66v
1000 ps
  50
Response from lattice diagram
-0.443v
3
1500 ps 2.22v
-0.443v
0.148v
Volts
2.5
1.77v
2000 ps
25  50
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
Zo1
TD
Vs
X
Zo2
TD
T3 T2
r 2 r3
r1
Rt
r4
Aa
B  acd
A'  b  e
B'  b  e  g  i
C  Ac  d  f  h
C'  b  e  g  i  k  l
a
TD A
2TD
3TD B
4TD
5TD C
vi  Vs
c
b
d
e
f
g
h
i
j
k
r1 
A’
r2 
B’
l
C’
Z o1
Rs  Z o1
Rs  Z o1
Rs  Z o1
Z o 2  Z o1
Z o 2  Z o1
Z Z
r 3  o1 o 2
Z o1  Z o 2
a  vi
b  aT2
c  ar 2
d  cr1
e  br 4
f  dr 2  eT3
g  er 3  dT2
h  fr1
Rt  Z o 2
r4 
Rt  Z o 2
i  gr 4
T2  1  r 2
j  hr 2  iT3
T3  1  r 3
k  ir 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
I
I
R I
Lattice diagram and
Z ,T
Z ,T
l
l
calculate reflection
V
V
V
V
coefficients at the
interfaces and show the
wave forms of V1(t), V2(t),
and V3(t).
S
1
02 02
01 01
S
 Check results with PSPICE
Transmission Lines Class 6
3
2
1
2
1
2
3
Short