Transmission Line Basics
Prof. Tzong-Lin Wu
NTUEE
1
Outlines
Transmission Lines in Planar structure.
Key Parameters for Transmission Lines.
Transmission Line Equations.
Analysis Approach for Z0 and Td
Intuitive concept to determine Z0 and Td
Loss of Transmission Lines
Example: Rambus and RIMM Module design
2
Transmission Lines in Planar structure
Homogeneous
Inhomogeneous

Coaxial Cable

Stripline
Microstrip line

Embedded Microstrip line
3
Key Parameters for Transmission Lines
1. Relation of V / I : Characteristic Impedance
2. Velocity of Signal: Effective dielectric constant
3. Attenuation:
Conductor loss
Dielectric loss
Z0
e
ac
ad
Lossless case
Z0
Vp
L
C
1
VpC
1
LC
c0
e
Td
C
1
Td
4
Transmission Line Equations
Quasi-TEM assumption
5
Transmission Line Equations
R0  resistance per unit length(Ohm / cm)
G0  conductance per unit length (mOhm / cm)
L0  inductance per unit length (H / cm)
C0  capacitance per unit length (F / cm)
KVL :
KCL :
dV
 ( R0  jwL0 ) I
dz
dI
  (G0  jwC0 )V
dz
Solve 2nd order D.E. for
V and I
6
Transmission Line Equations
Two wave components with amplitudes
V+ and V- traveling in the direction of +z and -z
V  V e  rz  V e  rz
1
I
(V e  rz  V e  rz )  I   I 
Z0
Where propagation constant and characteristic impedance are
r  ( R0  jwL0 )(G0  jwC0 )  a  j
V V
Z0 


I I
R0  jwL0
G0  jwC0
7
Transmission Line Equations
a and  can be expressed in terms of (R0 , L0 , G0 , C0 )
a 2   2  R0 G0   2 L0 C0
2a   ( R0 C0  G0 L0 )
The actual voltage and current on transmission line:
V ( z, t )  Re[(V e az e  jz  V e az e jz )e jwt ]
1
I ( z, t )  Re[ (V e az e  jz  V e az e jz )e jwt ]
Z0
8
Analysis approach for Z0 and Td (Wires in air)
C = ? (by Q=C V)
L = ? (by Ψ=L I)
9
Analysis approach for Z0 and Td (Wires in air):
Ampere’s Law for H field
H(r)=
I
d

I
2 r
c
2)  e   BT d s  
S
R2
r  R1
0 I
I R
dr  0 ln( 1 ) (in Wb)
2 r
2
R2
10
Analysis approach for Z0 and Td (Wires in air):
Ampere’s Law for H field
0 I R2
e 
ln( )
2
R1
L e / I
11
The per-unit-length Parameters (E):
Gauss’s Law
1) from gauss law

 D
 ET 
S
 ET d s  Qtotal
q  1m
 0  ds
S

q
2 0 r
2) V=  E T d   
r  R2
C

R1
q
2 0
ln
q
2 0 r
dr
R2
R1
12
The per-unit-length Parameters (E)
q
R2
V
ln( )
2 0
R1
C  Q /V
13
c. For example Determine the L.C.G.R of the two-wire line.
(note：homogeneous medium)
S
Inductance：
L=
e

e
I
I
rw2
rw1
I
where
0 I s-rw2 0 I s-rw1
e 
ln(
)+
ln(
)
2
rw1
2
rw2
0 I (s-rw2 )(s-rw1 )
=
ln(
)
2
rw1rw2
assume s
e
( 0 ,  0 )
rw1 , rw2
0
s2
 L=
ln(
)
2
rw1rw2
14
Capacitance：
1)
e
S
c   0 0
+
2 0
C 
s2
ln (
)
rw1rw2
+
rw1
-
+
+
+
q C/m
V
-
rw2
-
- -q C/m
s-rw2
s-rw1
q
2) V=
ln(
)+
ln(
)
2 0
rw1
2 0
rw2
q
(s-rw2 )(s-rw1 )
=
ln(
)
2 0
rw1rw2
q
s2

ln(
)
2 0
rw1rw2
q
q
C= 
V
2 0
s2
ln (
)
rw1rw2
if s
rw1 , rw2
 the same with 1) approach
15
The per-unit-length Parameters
Homogeneous structure
TEM wave structure is like the DC (static) field structure
LG  
LC  
So, if you can derive how to get the L, G and C can be
obtained by the above two relations.
16
The per-unit-length Parameters (Above GND )
2C
Why?
L/2
17
d. How to determine L,C for microstrip-line.
0 ,  0
1 , 1
1) This is inhomogeneous medium.
2) Nunerical method should be used to solve
the C of this structure, such as Finite element,
Finite Difference...
3) But e can be obtained by
e
C0  0 0 
e

 0 0
C0
where C0 is the capacitance when 1 medium
is replaced by  0 medium.
18
Analysis approach for Z0 and Td (Strip line)
Approximate electrostatic solution
y

1.

b
x
-a/2
2.
a/2
The fields in TEM mode must satisfy Laplace equation
 t ( x , y )  0
where  is the electric potential
The boundary conditions are
( x , y )  0 at x   a / 2
( x , y )  0 at y  0, b
2
19
Analysis approach for Z0 and Td
3.
Since the center conductor will contain the surface charge, so
n x
n y

A
cos
sinh
 n
a
a
n 1
 odd
 ( x, y )   
 B cos n x sinh n (b  y )
n

a
a
n 1
 odd
for 0  y  b / 2
for b / 2  y  b
Why?
4.
The unknowns An and Bn can be solved by two known conditions:
R|The potential at y  b / 2 must continuous
S|The surface charge distribution for the strip:   RS1
T0
T
s
for x  W / 2
for x  W / 2
20
Analysis approach for Z0 and Td
5.
R|V   E ( x  0, y)dy   ( x, y) / y( x  0, y)dy
z
|S z
||Q  z  ( x)dx  W (C / m)
T
b/2
b/2
y
0
w/ 2
0
s
 w/ 2
6.
Q
C 
V
W

2a sin(nW / 2a ) sinh(nb / 2a )

(n ) 2  0  r cosh(nb / 2a )
n 1
odd
r
1
Z0 

v pC
cC
7.
Answers!!
Td   r / c
21
Analysis approach for Z0 and Td (Microstrip Line)
y
1.
PEC
PEC
d

W

x
-a/2
2.
a/2
The fields in Quasi - TEM mode must satisfy Laplace equation
 t ( x , y )  0
where  is the electric potential
The boundary conditions are
( x , y )  0 at x   a / 2
( x , y )  0 at y  0, 
2
22
Analysis approach for Z0 and Td (Microstrip Line)
3.
Since the center conductor will contain the surface charge, so
R| A cos nx sinh ny
a
a
|
( x , y )  S
|| B cos nax e
T

n
n 1
odd

 ny / a
n
for 0  y  d
for d  y  
n 1
odd
4.
The unknowns An and Bn can be solved by two known conditions and
the orthogonality of cos function :
R|The potential at y  d must continuous
S|The surface charge distribution for the strip:   RS1
T0
T
s
for x  W / 2
for x  W / 2
23
Analysis approach for Z0 and Td (Microstrip Line)
5.
R|V   E ( x  0, y)dy   ( x, y) / y( x  0, y)dy   A sinh nd
z
a
|S z
||Q  z  ( x)dx  W (C / m)
T
b/2
b/2

y
n
0
0
n 1
odd
w/2
s
 w/ 2
6.
C
Q

V
W

4a sin(nW / 2a ) sinh(nd / 2a )

2
n 1 ( n ) W 0 [sinh( nd / a )   r cosh( nd / a )]
odd
24
Analysis approach for Z0 and Td (Microstrip Line)
7.
To find the effective dielectric constant  e , we consider two cases of capacitance
1. C = capacitance per unit length of the microstrip line with the dielectric substrate  r  1
2. C 0 = capacitance per unit length of the microstrip line with the dielectric substrate  r  1
 e 
8.
C
C0
e
1
Z0 

v pC
cC
Td   e / c
25
Tables for Z0 and Td (Microstrip Line)
Z 0 ( )
20
28
40
50
75
90
100
 eff
3.8
3.68
3.51
3.39
3.21
3.13
3.09
L0
( nH / mm)
0.119
0.183
0.246
0.320
0.468
0.538
0.591
C0
( pF / mm)
0.299
0.233
0.154
0.128
0.083
0.067
0.059
6.54
6.41
6.25
6.17
5.99
5.92
5.88
T0
( ps / mm)
Fr4 : dielectric constant = 4.5
Frequency: 1GHz
26
Tables for Z0 and Td (Strip Line)
Z 0 ( )
20
28
40
50
75
90
100
 eff
4.5
4.5
4.5
4.5
4.5
4.5
4.5
L0
( nH / mm)
0.141
0.198
0.282
0.353
0.53
0.636
0.707
C0
( pF / mm)
0.354
0.252
0.171
0.141
0.094
0.078
0.071
T0
( ps / mm)
7.09
7.09
7.09
7.09
7.09
7.09
7.09
Fr4 : dielectric constant = 4.5
Frequency: 1GHz
27
Analysis approach for Z0 and Td (EDA/Simulation Tool)
1. HP Touch Stone (HP ADS)
2. Microwave Office
3. Software shop on Web:
4. APPCAD
(http://softwareshop.edtn.com/netsim/si/termination/term_article.html)
(http://www.agilent.com/view/rf or http://www.hp.woodshot.com )
28
Concept Test for Planar Transmission Lines
• Please compare their Z0 and Vp
(a)



(b)



29
(b)
(a)



(c)





30
(b)
(a)
(c)








31
32
Loss of Transmission Lines
Typically, dielectric loss is quite small -> G0 = 0. Thus
Z0 
R0  jwL0

jwC0
L0
(1  jx ) 1/ 2
C0
r  ( R0  jwL0 )( jwC0 )  a  j
where
R
x 0
wL0
• Lossless case : x = 0
• Near Lossless: x << 1
• Highly Lossy: x >> 1
w
R0
L0
Highly Lossy
Near Lossless
w
 w
33
Loss of Transmission Lines
•For Lossless case: • For Near Lossless case:
a 0
a
   L0 C0
R0
2 L0 / C0
L xO
   L C M1  P
N 8Q
L F
R I
Z 
1

j

G
J
C H
2 wL K
2
Z0 
L0
C0
Time delay T0 
0
L0 C0
0
0
0
0
0
Time delay T0 
0
L0
1

where C  2T0 / R0
C0
jwC
L0 C0
34
Loss of Transmission Lines
• For highly loss case: (RC transmission line)
a
wR0 C0
1
[1  ]
2
2x

wR0 C0
1
[1  ]
2
2x
Z0 
R0
1
[1  ]
2 wC0
2x
Nonlinear phase relationship with f
introduces signal distortion
That’s why telephone company terminate
the lines with 600 ohm
Example of RC transmission line:
AWG 24 telephone line in home
F R  iwL IJ
Z ( w)  G
H jwC K
0
1/ 2
 648(1  j )
where
R  0.0042 / in
L  10nH / in
C  1 pF / in
w  10,000rad / s(1600 Hz ) : voice band
35
Loss of Transmission Lines ( Dielectric Loss)
The loss of dielectric loss is described by the loss tangent
G
FR4 PCB tan  D  0.035
tan  D 
wC
a D 
GZ 0
 ( wC tan  D Z 0 ) / 2  f tan  D LC
2
36
Loss of Transmission Lines (Skin Effect)
• Skin Effect
DC resistance
AC resistance
37
Loss of Transmission Lines (Skin Effect)
s 
1
a

2
w
1  length 
w
R( w)  

  area 

NOTE: In the near lossless region ( R / wL  1),
the characteristic impedance Z 0 is not much
affected by the skin effect
 R(w)  w
 R( w) / wL  (1 / w )  1
38
Loss of Transmission Lines (Skin Effect)
100
200
400
800
1200
1600
2000
6.6um
4.7um
3.3um
2.4um
1.9um
1.7um
1.5um
R s ()
2.6m
ohm
3.7m
ohm
5.2m
ohm
7.4m
ohm
9.0m
ohm
10.8m
ohm
11.6m
ohm
Trace
resistance
1.56
ohm
2.22
ohm
3.12
ohm
4.44
ohm
5.4
ohm
6.48
ohm
7.0
ohm
f (MHz)
s 
1
f
Skin depth resistance R s =
6mil
Cu
f
()

 = 4  10 -7 H / m
17um
 (Cu) = 5.8  10 7 S / m
Length of trace = 20cm
39
Loss Example: Gigabit differential transmission lines
For comparison: (Set Conditions)
1.
2.
3.
4.
Differential impedance = 100
Trace width fixed to 8mil
Coupling coefficient = 5%
Metal : 1 oz Copper
Question:
1.
2.
Which one has larger loss by skin effect?
Which one has larger loss of dielectric?
40
Loss Example: Gigabit differential transmission lines
Skin effect loss
41
Loss Example: Gigabit differential transmission lines
Skin effect loss
Why?
42
Loss Example: Gigabit differential transmission lines
Look at the field distribution of the common-mode coupling
Coplanar structure has more surface for current flowing
43
Loss Example: Gigabit differential transmission lines
How about the dielectric loss ? Which one is larger?
44
Loss Example: Gigabit differential transmission lines
The answer is dual stripline has larger loss. Why ?
The field density in the dielectric between the trace and GND is
higher for dual stripline.
45
Loss Example: Gigabit differential transmission lines
Which one has higher ability of rejecting common-mode noise ?
46
Loss Example: Gigabit differential transmission lines
The answer is coplanar stripline. Why ?
47
48
49
50
51
Intuitive concept to determine Z0 and Td
•How physical dimensions affect impedance and delay
Sensitivity is defined as percent change in impedance
per percent change in line width, log-log plot shows sensitivity directly.
Z 0 is mostly influenced by w / h,
the sensitivity is about 100%.
It means 10% change in w / h will
cause 10% change of Z 0
The sensitivity of Z 0 to changes in  r
is about 40%
52
Intuitive concept to determine Z0 and Td
•Striplines
impedance
Delay
53
Ground Perforation:
BGA via and impedance
54
Ground Perforation:
Cross-talk (near end)
55
Ground Perforation :
Cross-talk (far end)
56
Example(II): Transmission line on non-ideal GND
Reasons for splits or slits on GND planes
57
Example(II): Transmission line on non-ideal GND
58
Example(II): Transmission line on non-ideal GND
59
Example(II): Transmission line on non-ideal GND
60
Example(II): Transmission line on non-ideal GND
61
Example(II): Transmission line on non-ideal GND
62
Example(II): Transmission line on non-ideal GND
63
Input side
64
Output side
65
66
67
68
Example: Rambus RDRAM and RIMM Design
RDRAM Signal Routing
69
Example: Rambus RDRAM and RIMM Design
•Power:
VDD = 2.5V, Vterm = 1.8V, Vref = 1.4V
•Signal:
0.8V Swing: Logic 0 -> 1.8V, Logic 1 -> 1.0V
2x400MHz CLK: 1.25ns timing window, 200ps rise/fall time
Timing Skew: only allow 150ps - 200ps
•Rambus channel architecture:
(30 controlled impedance and matched transmission lines)
Two 9-bit data buses (DQA and DQB)
A 3-bit ROW bus
A 5-bit COL bus
CTM and CFM differential clock buses
70
Example: Rambus RDRAM and RIMM Design
RDRAM Channel is designed for 28 +/- 10%
Impedance mismatch causes signal reflections
Reflections reduce voltage and timing margins
PCB process variation -> Z0 variation -> Channel error
71
Example: Rambus RDRAM and RIMM Design
• Intel suggested coplanar structure
Ground flood & Stitch
• Intel suggested strip structure
Ground flood & Stitch
72
Example: Rambus RDRAM and RIMM Design
PCB Parameter sensitivity:
• H tolerance is hardest to control
• W & T have less impact on Z0
73
Example: Rambus RDRAM and RIMM Design
•
•
How to design Rambus channel in RIMM Module
with uniform Z0 = 28 ohm ??
How to design Rambus channel in RIMM Module
with propagation delay variation in +/- 20ps ??
74
Example: Rambus RDRAM and RIMM Design
Impedance Control: (Why?)
Loaded trace
Unloaded trace
Connector
75
Example: Rambus RDRAM and RIMM Design
Multi-drop Buses
Unloaded Z0
Stub
Equivalent loaded ZL
ZL 
L0

CT
Electric pitch L
L0
CL
L
 02
L Z0
Device input
Capacitance Cd
A Multidrop Bus
 28 (for Rambus design)
where C T is the per - unit - length equivalent capacitance at length L,
including the loading capacitance and the unloaded trace capacitance
C L is the loading capacitance including the device input capacitance C d ,
the stub trace capacitance, and the via effect.
76
Example: Rambus RDRAM and RIMM Design
In typical RIMM module design
stub
via
Device input capacitance
If C L  0.2 pF + 0.1pF + 2.2pF, and
If you design unloaded trace Z 0  56
the electric pitch L = 7.06mm to reach loaded Z L  28
 L0  Z 0  56  6.77 psec / mm = 379 pH / mm = 9.5 pH / mil
C
L
2.5 pF
379 pH / mm
CT  L  02 

 0.475 pF / mm
L Z0
7.06mm
56 2
 ZL 
L0
 28.3
CT
77
Example: Rambus RDRAM and RIMM Design
• Modulation trace
Device pitch = Device height +
Device space
Electrical pitch L is designed as
L

Z0
CL Z L
2
(Z0  Z L )
2
2
If device pitch > electric pitch, modulation
trace of 28ohm should be used.
Modulation trace length
= Device pitch – Electric pitch
78
Example: Rambus RDRAM and RIMM Design
• Effect of PCB parameter variations on three key module
electric characteristics
79
Example: Rambus RDRAM and RIMM Design
• Controlling propagation delay:
•Bend compensation
•Via Compensation
•Connector compensation
Bend Compensation
•
•
Rule of thumb: 0.3ps faster delay of every bend
Solving strategies:
1. Using same numbers of bends for those critical traces(difficult)
2. Compensate each bend by a 0.3ps delay line.
80
Example: Rambus RDRAM and RIMM Design
Via Compensation (delay)
For a 8 layers PCB, a via with 50mil length can be modeled as
(L, C) = (0.485nH, 0.385pF).
 Delay T0 
LC  13.7 psec
1
Impedance Z 0 =
 38
LC
Inductive
Rule of thumb: delay of a specific via depth can be calculated by scaling
the inductance value which is proportional to via length.
30mil
 10.6 p sec
50mil
This delay difference can be compensated by adding a 1.566mm to the
 30mil via has delay  13.7 
unloaded trace (56)
81
Example: Rambus RDRAM and RIMM Design
Via Compensation (impedance)
82
Example: Rambus RDRAM and RIMM Design
Connector Compensation
83
Example: EMI resulting from a trace near a PCB edge
Experiment setup and trace design
84
Example: EMI resulting from a trace near a PCB edge
Measurement Setup
85
Example: EMI resulting from a trace near a PCB edge
EMI caused by Common-mode current : magnetic coupling
Measured by current probe
86
Example: EMI resulting from a trace near a PCB edge
EMI measured by the monopole : E field
Low effect at high frequency
87
Example: EMI resulting from a trace near a PCB edge
Trace height effect on EMI
88
Reference
1.
2.
3.
4.
5.
Howard W. Johnson, “High-speed digital design”, Prentice-Hall, 1993
Ron K. Poon, “Computer Circuits Electrical Design”, Prentice-Hall, 1995
David M. Pozar, “Microwave Engineering”, John Wiley & Sons, 1998
William J. Dally, “Digital System Engineering”, Cambridge, 1998
Rambus, “Direct Rambus RIMM Module Design Guide, V. 0.9”, 1999
89