2. 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. 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. 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. 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
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)
W
Transmission Lines Class 6
7. 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
small compared to the signaling wavelength
Transmission Lines Class 6
8. 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)
-
- + - + -
(c) (d)
Transmission Lines Class 6
9. 9
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
10. 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
11. 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. 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
Implies equivalent circuit model must contain inductor.
Transmission Lines Class 6
13. 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. 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
15. 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. 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
17. 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
18. 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
Wavelength/edge rate Tline the applied
frequency or the
rise/fall edge of the
Transmission Lines Class 6 signal
20. 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. 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!
Transmission Lines Class 6
22. 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
be approximated using the parallel plate approximation.
Transmission Lines Class 6
23. 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 for
WC
>1
TD
Transmission Lines Class 6
24. 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. 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
27. 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. 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
plate approximation?
Transmission Lines Class 6
29. 29
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
30. 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?
Next page….
Transmission Lines Class 6
31. 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:
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
32. 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 –
it is simply a voltage divider!
Transmission Lines Class 6
33. 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 + Vreflected at time t = TD
Transmission Lines Class 6
34. 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 B )(Vi )
(ρ t=TD, V=Vi + ρB (Vi )
Transmission Lines Class 6
35. 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. 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. 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
Solution
Transmission Lines Class 6
38. 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
Solution
Transmission Lines Class 6
39. 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
Solution
0 Τ 2Τ 3Τ 4Τ Time (ns)
Transmission Lines Class 6
41. 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
42. 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
0.75VA properly terminated
Voltage (V)
0.5VA
wave settle out at
0.5 V
0.25VA
Solution
0 Τ 2Τ 3Τ 4Τ Time (ns)
Solution
Transmission Lines Class 6
43. Transmission line step response
43
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
44. 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 C’
for each successive reflected 4N ps
e
wave
Total voltage at any point is the 5N ps
sum of all the waves that have
reached that point
Transmission Lines Class 6
45. 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
46. 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
Tim e , ps
Transmission Lines Class 6
47. 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
2.07
Transmission Lines Class 6
48. 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
T3 = 1 + ρ 3 k = iρ 3 + hT2
Transmission Lines Class 6
49. 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 I1 I2 I3
S
diagram and calculate Z01 ,Τ 01 Z02 ,Τ 02
l1 l2
reflection coefficients atV S
V1 V2 V3 Short
the interfaces and show
the wave forms of V1(t),
V2(t), and V3(t).
Check results with PSPICE
Transmission Lines Class 6