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EKT 441
MICROWAVE COMMUNICATION
CHAPTER 2:
PLANAR TRANSMISSION LINES
1
Transmission Lines

A device used to transfer energy from one point to another point
“efficiently”

Efficiently = minimum loss, reflection and close to a perfect
match as possible (VSWR = 1:1)

Important to be efficient at RF and microwave frequencies, since
freq used are higher than DC and low freq applications

as freq gets higher, any energy loss is in transmission lines are
more difficult and costly to be retrieved
2
Some Types of Transmission Lines

Some types of transmission lines are listed as follow:

Coaxial transmission line

transmission line which a conductor completely surrounds the
other

Both shares the same axis, separated by a continuous solid
dielectric or dielectric spacers

Flexible – able to be bent without breaking
3
Some Types of Transmission Lines

Waveguides

Hollow-pipe structure, in which two distinct conductor are not present

Open space of the waveguide is where electromagnetic energy finds the
path of least resistance to propagate
Do not need any dielectric medium as it uses air as medium of energy
transfer
Our focus in this
Planar transmission lines
chapter
 Planar – looks like a coaxial cable that have been run over and flattened



Usually made up of a layer of dielectric, and one or several ground
(metallic planes)

Four types of planar lines discussed in this chapter; (1) Stripline, (2)
microstrip (3) dielectric waveguide (4) Slotline
4
But Why Planar?


Waveguides

High power handling capability

low loss

bulky

expensive
Coaxial lines

high bandwidth,

convenient for test applications

difficult to fabricate complex microwave components in
the medium
5
But Why Planar?

Planar Transmission Lines

Compact

Low cost

Capability for integration with active devices such as
diodes, transistors, etc
6
STRIPLINE
Figure 3.1: Stripline transmission line (a) Geometry (b) Electric and
magnetic field lines.
7
STRIPLINE

Also known as “sandwich line” – evolved from “flattened”
coaxial transmission line

The geometry of a stripline is shown in Figure 3.1.

Consist of a; (1) top ground plane, (2) bottom ground plane and
(3) a center conductor

W is the width of thin conducting strip (centered between two
wide conducting ground planes).

b is the distance of ground planes separation.

The region between the ground planes is filled with a dielectric.

Practically, the centered conductor is constructed of thickness b/2.
8
STRIPLINE
Figure 3.2: Photograph of a stripline circuit assembly.
9
STRIPLINE

The phase velocity is given by:
vp  1
c
0 0 r
r
[3.1]
Thus, the propagation constant of the stripline is:


vp
   0 0 r
  r k0
[3.2]
10
STRIPLINE
From equation [3.1], c = 3 x 108 m/sec is the speed of light in
free-space.
 The characteristic impedance of a transmission line is given by:

L
LC
1
Z0 


C
C
v pC
[3.3]
L and C are the inductance and capacitance per unit length of the
line. There is a solution as explained in [M. Pozar’ book].
The resulting formula for the characteristic impedance is:
30
b
Z0 
 r We  0.441b
[3.4]
11
STRIPLINE
Where We is the effective width of the center conductor given by:
0
We W 


2
b
b 0.35  W b 
W
for
 0.35
b
W
for
 0.35
b
[3.5]
These formulas assume a zero strip thickness, and are quoted as
being accurate to about 1 % of the exact results.
It is seen from equation [3.4] and [3.5] that the characteristic
impedance decreases as the strip width W increase.
12
STRIPLINE

When designing stripline circuits, one usually needs to find the
strip width, W.
 By given characteristic impedance (and height b and
permittivity εr), the value of W can be find by the inverse of the
formulas in equation [3.4] and [3.5].
 The useful formulas is:
x
for  r Z 0  120
W 

b 0.85  0.6  x for  r Z 0  120
Where,
30
x
 0.441
 r Z0
[3.6]
[3.7]
13
STRIPLINE

The attenuation due to dielectric loss is:
k tan 
d 
Np / m
2
[3.8]
The attenuation due to the conductor loss:
 2.7 10 3 Rs r Z 0 A

for  r Z 0  120



30

b

t
c  
Np / m
0.16 Rs
for  r Z 0  120

B

Z 0b

[3.9]
14
STRIPLINE

With:
2W
1 b  t  2b  t 
A  1

ln 

bt  bt  t 
b
0.441t 1
4W 

B  1

ln
 0.5 

0.5W  0.7t  
W
2
t 
[3.10]
[3.11]
Where t is the thickness of the strip
15
STRIPLINE [EXAMPLE 2.1]
Find the width for a 50 Ω copper stripline conductor,
with b = 0.32 cm and εr = 2.20. If the dielectric loss
tangent (tan δ) is 0.001 and the operating frequency is
10 GHz, calculate the attenuation in dB/λ. Assume
the conductor thickness of t = 0.01 mm and surface
resistance, Rs of 0.028 Ω
16
SOLUTION [EXAMPLE 2.1]
Since
 r Z 0  2.2 (50)  74.2  120 and
x
30
 r Z0
 0.441  0.830
Eq [3.6] gives the width as W = bx = (0.32)(0.830) = 0.266
cm. At 10 GHz, the wave number is
k
2f  r
c
 310.6m 1
17
SOLUTION [EXAMPLE 2.1]
The dielectric attenuation is
k tan  (310.6)(0.001)
d 

 0.155 Np / m
2
2
Surface resistance of copper at 10 GHz is Rs = 0.026
Ω. Then from eq [3.9]
2.7  10 3 Rs  r Z 0 A
c 
 0.122 Np / m
30 (b  t )
since A = 4.74
The total attenuation constant is
   c   d  0.277 Np / m
18
SOLUTION [EXAMPLE 2.1]
In dB;
 (dB)  20 log e  2.41dB / m
At 10 GHz, the wavelength on the stripline is;
c

 2.02cm
f r
So in terms of the wavelength the attenuation is
 (dB)  (2.41)(0.0202)  0.049dB / 
19
STRIPLINE [EXAMPLE 2.2]
Find the width for 50 Ω copper stripline conductor
with b = 0.5 cm and εr = 3.0. The loss tangent is 0.002
and the operating frequency is 8 GHz. Calculate the
attenuation in dB/λ. Assume a conductor thickness of t
= 0.003 mm. The surface resistance is 0.03 Ω.
20
SOLUTION [EXAMPLE 2.2]
 r Z0  3  50  86.603  120
x
30
 r Z0
 0.441 
30
3  50

 0.441  0.6474

W
 x  W  bx  0.5  10  2 0.6474  0.003237m
b
The wave number at f = 8 GHz
k
2f  r
c
2 8  10 

9
3  108
3
 290.246m 1
21
SOLUTION (cont) [EXAMPLE 2.2]
The dielectric attenuation is
k tan  290.2460.002
d 

 0.29
2
2
Np/m
The conductor attenuation is
A  1
2W
1 b  t  2b  t 

ln 

bt  bt  t 
20.003237
1 0.005  0.003  10 2  20.005  0.003  10 2 

A  1

ln 
2
2
2
 0.005  0.003  10
0.005  0.003  10
0.003  10


A  4.88
22
SOLUTION (cont) [EXAMPLE 2.2]
2.7  10 3 Rs  r Z 0 A
c 
30 b  t 
2.7  10 3 0.033504.88
c 
 0.1259
2
30 0.005  0.003  10 
Total loss:
   d   c  0.29  0.1259  0.4159 Np/m
In dB:
 dB   20 log e  20 log e 0.4159  3.6123 dB/m
23
SOLUTION (cont) [EXAMPLE 2.2]
The guided wavelength:
c
3  108
g 


0
.
02165
r f
3  8  10 9
The attenuation in dB/λ:
 dB  3.61230.02165  0.0782
24
STRIPLINE DISCONTINUITY
Figure 3.3: geometry of enclosed stripline
25
STRIPLINE DISCONTINUITY

By derivation found in M.Pozar’s book (page 141), the surface
charge density on the strip at y = b/2 is:



 s  Dy x, y  b 2  Dy x, y  b 2

   E x, y  b 2  E x, y  b 2 

0 r
y

y
 n
 2 0 r  An 
n 1
 a

nx
nb

cosh
 cos
a
2a

[3.12]
odd
The charge density on the strip line by uniform distribution:
1 for x  W 2
 s x   
0 for x  W 2
[3.13]
26
STRIPLINE DISCONTINUITY

The capacitance per unit length of the stripline is:
Q
C

V
W
Fd / m

2a sin nW 2a sinh nb 2a 

n 2  0 r cosh nb 2a 
n 1
[3.14]
odd
The characteristic impedance is then found as:
r
L
LC
1
Z0 



C
C
v p C cC
[3.15]
27
MICROSTRIP
Figure 3.3: Microstrip transmission line. (a) geometry. (b)
Electric and magnetic field lines.
28
MICROSTRIP







Microstrip line is one of the most popular types of planar
transmission line.
Easy fabrication processes.
Easily integrated with other passive and active microwave
devices.
The geometry of a microstrip line is shown in Figure 3.3
W is the width of printed thin conductor.
d is the thickness of the substrate.
εr is the relative permittivity of the substrate.
29
MICROSTRIP





The microstrip structure does not have dielectric above the strip
(as in stripline).
So, microstrip has some (usually most) of its field lines in the
dielectric region, concentrated between the strip conductor and
the ground plane.
Some of the fraction in the air region above the substrate.
In most practical applications, the dielectric substrate is
electrically very thin (d << λ).
The fields are quasi-TEM (the fields are essentially same as those
of the static case.
30
MICROSTRIP

The phase velocity and the propagation constant:
vp 
c
e
  k0  e
[3.16]
[3.17]
Where Єe is the effective dielectric constant of the microstrip line
used to compensate difference between the top and bottom of the
circuit line
The effective dielectric constant satisfies the relation:
1  e  r
and is dependent on the substrate thickness, d and conductor
width, W
31
MICROSTRIP

The effective dielectric constant of a microstrip line is given by:
e 
 r  1  r 1
2

2
1
1  12d W
[3.18]
The effective dielectric constant can be interpreted as the dielectric
constant of a homogeneous medium that replaces the air and
dielectric regions of the microstrip, as shown in Figure 3.4.
Figure 3.4: equivalent geometry of quasi-TEM microstrip line.
32
MICROSTRIP

The characteristic impedance can be calculated as:

60  8d W 
ln  


for W  1
W
4
d
e 


d
Z0  
120
for W  1

d
  e W d  1.393  0.667 ln W d  1.444
[3.19]
For a given characteristic impedance Z0 and the dielectric constant
Єr, the W/d ratio can be found as:

8e A

for W  2
W 
e2 A  2
d
 
d  2 B  1  ln 2 B  1   r  1 ln B  1  0.39  0.61 for W  2
d




2



r 
r 
 
[3.20]
33
MICROSTRIP

Where:
Z0  r  1  r 1 
0.11 
 0.23 

A

60
2
 r 1 
r 
377
B
2Z 0  r
Considering microstrip as quasi-TEM line, the attenuation due to
dielectric loss can be determined as
k0 r  e  1 tan 
d 
Np / m
2  e  r  1
[3.21]
Where tan δ is the loss tangent of the dielectric.
34
MICROSTRIP
This result is derived from Equation [2.37] by multiplying by a
“filling factor”:
 r  e  1
 e  r  1
Which accounts for the fact that the fields around the microstrip line
are partly in air (lossless) and partly in the dielectric.
The attenuation due to conductor loss is given approximately by:
Rs
c 
Np / m
Z 0W
[3.22]
Where Rs = √(ωμ0/2σ) is the surface resistivity of the conductor.
35
MICROSTRIP [EXAMPLE 2.3]
Calculate the width and length of a microstrip line for
a 50 Ω characteristic impedance and a 90° phase shift
at 2.5GHz. The substrate thickness is d = 0.127 cm,
with εr = 2.20.
Solution:
M.Pozar’ book (page: 146).
36
SOLUTION [EXAMPLE 2.3]
First we need to find W/d for Z0=50 Ω, and
initially guess that W/d > 2. From eq [3.20];
B = 7.895 and W/d = 3.081
Otherwise, we would use the expression for W/d<2
Then W = (3.081.d) = 0.391 cm. From eq [3.18];
εe = 1.87
37
SOLUTION [EXAMPLE 2.3]
The line length, l, for a 90o phase shift is found as;
  90 o  l   e k 0 l 
2f
k0 
 52.35m 1
c
  
90 
o 
180 

l
 2.19cm
 e k0
o
38
MICROSTRIP [EXAMPLE 2.4]
Design a microstrip transmission line for 70 Ω
characteristic impedance. The substrate
thickness is 1.0 cm, with εr = 2.50. What is the
guide wavelength on this transmission line if
the frequency is 3.0 GHz?
39
SOLUTION [EXAMPLE 2.4]
Z0
A
60
r 1 r 1 
0.11 
 0.23 


2
r 1 
r 
70 2.5  1 2.5  1 
0.11 
A

 0.23 

60
2
2.5  1 
2.5 
A  1.66
Initially, it is guessed that W/d < 2
A
1.66
W
8e
8e
 2A
 2(1.66)
 1.64
d e 2 e
2
40
SOLUTION cont [EXAMPLE 2.4]
Since the W/d < 2 assumption is valid;
W  1.64  1.0  10 2  0.0164m  1.64cm
We proceed to calculate εe
 r  1  r 1
1
e 

2
2
1  12d W
2.5  1 2.5  1
1
e 

 2.01
2
2
1  12(1.0 1.64)
41
SOLUTION cont [EXAMPLE 2.4]
Thus the guided wavelength is given by;
g 
c
e f

3.0  108
2.01  3  10
9
 7.05cm
42
MICROSTRIP [EXAMPLE 2.5]
Design a quarter wavelength microstrip
impedance transformer to match a patch
antenna of 80 Ω with a 50 Ω line. The system
is fabricated on a 1.6 mm substrate thickness
with εr = 2.3, that operates at 2 GHz.
43
MICROSTRIP [EXAMPLE 2.5]
From the quarter wave transformer equation;
Z 1  Z 1Z 0  80 x 50  63.246
44
SOLUTION [EXAMPLE 2.5]
Since W is not known, guess that W/d < 2
Z0
A
60
 r 1
2
 r 1 
0.11 
 0.23 


 r 1 
r 
63.246 2.3  1 2.3  1 
0.11 
A

 0.23 
  1.4635
60
2
2.3  1 
2.3 
377
377
B

 6.174
2Z 0  r 2(63.246) 2.3
45
SOLUTION cont [EXAMPLE 2.5]
A
1.4635
W
8e
8e
 2A
 2(1.4635)
 2.073
d e 2 e
2
From the calculation, the initial assumption of W/d <
2 is incorrect. The next formula (where W/d > 2) is
used
W 2
r 1 
0.61
  B  1  ln 2 B  1 
ln B  1  0.39 

d 
2 r 
r 
W
 2.0656
d
W  (2.0656)(1.6)  3.3mm
46
SOLUTION cont [EXAMPLE 2.5]
The next step is to find the effective dielectric
constant (εe)
e 
r  1 r  1
2

2
1
 1.8991
1  12d W
To determine the quarter wavelength of the line, the
guided wavelength λg need to be determined
c
3.0 108
g 

 10.88cm
9
e f
1.8991  2 10
47
SOLUTION cont [EXAMPLE 2.5]
Thus the quarter wave length of line is determined by
dividing the full wavelength by 4;
g
10.88

 2.72cm
4
4
48
MICROSTRIP
Figure 3.5: Geometry of a microstrip line with conducting sidewalls.
49
MICROSTRIP DISCONTINUITY

By derivation found in M.Pozar’s book (page 147), the surface
charge density on the strip at y = d is:
 s  Dy x, y  d    Dy x, y  d  



  0 E y x, y  d    0 r E y x, y  d 
 n
  0  An 
n 1
 a


nx 
nd
nd 

sinh
  r cosh
 cos

a 
a
a 

[3.23]
odd
The charge density on the microstrip line by uniform distribution:
s
1 for x  W 2
 
0 for x  W 2
[3.24]
50
MICROSTRIP DISCONTINUITY

The capacitance per unit length of the stripline is:
Q
C 
V
1
Fd / m

4a sin nW 2a sinh nd a 

2


n

W 0 sinh nd a    r cosh nd a 
n 1
odd
[3.25]
The characteristic impedance is then found as:
e
1
Z0 

v p C cC
[3.26]
51
DIELECTRIC WAVEGUIDE
Figure 3.6: Dielectric waveguide geometry.
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DIELECTRIC WAVEGUIDE
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The dielectric waveguide is shown in Figure 3.6.
εr2 is the dielectric constant of the ridge.
εr1 is the dielectric constant of the substrate.
Usually εr1 < εr2
The fields are thus mostly confined to the area around the
dielectric ridge.
Convenient for integration with active devices.
Very lossy at bends or junctions in the ridge line.
Many variations in basic geometry are possible.
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SLOTLINE
Figure 3.7: Geometry of a printed slotline.
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SLOTLINE
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The geometry of a slotline is shown in Figure 3.7.
 Consists of a thin slot in the ground plane on one side of a
dielectric substrate.
 The two conductors of slotline lead to quasi-TEM type of mode.
 The characteristic impedance of the line can be change by
changing the width of the slot.
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