chp2-planar transmission lineswithexamples
DESCRIPTION
ppt on planar transmission lines, lecture for students of engineeringTRANSCRIPT
1
EKT 441
MICROWAVE Communications
CHAPTER 2:
PLANAR TRANSMISSION LINES
2
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
3
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
4
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
Planar transmission lines Planar – looks like a 3D line 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
Our focus in this chapter
5
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
6
But Why Planar?
Planar Transmission Lines
Compact
Low cost
Capability for integration with active devices such as
diodes, transistors, etc
7
STRIPLINE
Figure 3.1: Stripline transmission line (a) Geometry (b) Electric and magnetic field lines.
8
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.
9
STRIPLINE
Figure 3.2: Photograph of a stripline circuit assembly.
10
STRIPLINE
The phase velocity is given by:
r
rp
c
v
001
0
00
k
v
r
rp
Thus, the propagation constant of the stripline is:
[3.1]
[3.2]
11
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:
CvC
LC
C
LZ
p
10 [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:
bW
bZ
er 441.0
300
[3.4]
12
STRIPLINE
35.0
35.0
35.0
02
b
Wfor
b
Wfor
bWb
W
b
We
Where We is the effective width of the center conductor given by:
[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.
13
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:
120
120
6.085.0 0
0
Zfor
Zfor
x
x
b
W
r
r
441.030
0
Z
xr
[3.6]
[3.7]Where,
14
STRIPLINE
The attenuation due to dielectric loss is:
mNpk
d /2
tan
mNpZfor
Zfor
BbZ
Rtb
AZR
r
r
s
rs
c /120
12016.0
30
107.2
0
0
0
03
The attenuation due to the conductor loss:
[3.8]
[3.9]
15
STRIPLINE
With:
t
tb
tb
tb
tb
WA
2ln
121
t
W
W
t
tW
bB
4ln
2
1441.05.0
7.05.01
[3.10]
[3.11]
Where t is the thickness of the strip
16
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 0.026 Ω
17
SOLUTION [EXAMPLE 2.1]
Since and
Eq [3.6] gives the width as W = bx = (0.32)(0.830) = 0.266 cm. At 10 GHz, the wave number is
1202.74)50(2.20 Zr
830.0441.030
0
Z
xr
16.3102 m
c
fk r
18
SOLUTION [EXAMPLE 2.1]
The dielectric attenuation is
Surface resistance of copper at 10 GHz is Rs = 0.026 Ω. Then from eq [3.9]
since A = 4.74
The total attenuation constant is
mNptb
AZR rsc /122.0
)(30
107.2 03
mNpk
d /155.02
)001.0)(6.310(
2
tan
mNpdc /277.0
19
SOLUTION [EXAMPLE 2.1]
In dB;
At 10 GHz, the wavelength on the stripline is;
So in terms of the wavelength the attenuation is
mdBemdB /41.2log20)/(
cmf
c
r
02.2
/049.0)0202.0)(41.2()/( dBdB
20
SOLUTION [EXAMPLE 2.1]
But why do we need to convert Np/m to dB/m using this way?
loss to the transmission line is reflected by the attenuation constant. The amplitude of the signal decays as e-α.
The natural units of the attenuation constant are Nepers/meter, but we often convert to dB/meter in microwave engineering. To get loss in dB/length, multiply Nepers/length by 8.686.
mdBemdB mNp /41.2log20)/( )/(
21
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 cm. The surface resistance is 0.03 Ω.
22
SOLUTION [EXAMPLE 2.2]
The wave number at f = 8 GHz
120603.865030 Zr
6474.0441.0503
30441.0
30
0
Z
xr
mbxWxb
W003237.06474.0105.0 2
18
9
246.290103
310822
mc
fk r
23
SOLUTION (cont) [EXAMPLE 2.2]
The dielectric attenuation is
Np/m
The conductor attenuation is
29.0
2
002.0246.290
2
tan
kd
t
tb
tb
tb
tb
WA
2ln
121
2
2
2
2
2 10003.0
10003.0005.02ln
10003.0005.0
10003.0005.01
10003.0005.0
003237.021
A
)173.4(88.4A
24
SOLUTION (cont) [EXAMPLE 2.2]
Total loss:
Np/m
In dB:
dB/m
tb
AZR rsc
30
107.2 03
1259.0
10003.0005.030
88.450303.0107.22
3
c
4159.01259.029.0 cd
6123.3log20log20/ 4159.0 eemdB
25
SOLUTION (cont) [EXAMPLE 2.2]
The guided wavelength:
The attenuation in dB/λ:
02165.01083
1039
8
f
c
r
g
0782.002165.06123.3/ dB
26
STRIPLINE DISCONTINUITY
Figure 3.3: geometry of enclosed stripline
27
STRIPLINE DISCONTINUITY
By derivation found in M.Pozar’s book (page 141), the surface charge density on the strip at y = b/2 is:
2,2,
2,2,
0 byxEbyxE
byxDbyxD
yyr
yys
odd
nnr a
bn
a
xn
a
nA
10 2
coshcos2 [3.12]
The charge density on the strip line by uniform distribution:
2
2
0
1
Wxfor
Wxforxs [3.13]
28
STRIPLINE DISCONTINUITY
The capacitance per unit length of the stripline is:
mFd
abnn
abnaWnaW
V
QC
odd
n r
/
2cosh
2sinh2sin2
1 02
The characteristic impedance is then found as:
cCCvC
LC
C
LZ r
p
10
[3.14]
[3.15]
29
MICROSTRIP
Figure 3.3: Microstrip transmission line. (a) geometry. (b) Electric and magnetic field lines.
30
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.
31
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.
32
MICROSTRIP
The phase velocity and the propagation constant:
e
p
cv
ek 0
re 1
Where εe is the effective dielectric constant of the microstrip line used to compensate difference between the top and bottom of the circuit lineThe effective dielectric constant satisfies the relation:
[3.16]
[3.17]
and is dependent on the substrate thickness, d and conductor width, W
33
MICROSTRIP
The effective dielectric constant of a microstrip line is given by:
Wdrr
e121
1
2
1
2
1
[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.
34
MICROSTRIP
The characteristic impedance can be calculated as:
1
1
444.1ln667.0393.1
1204
8ln
60
0
dWfor
dWfor
dWdW
d
W
W
d
Z
e
e
[3.19]
For a given characteristic impedance Z0 and the dielectric constant Єr, the W/d ratio can be found as:
2
2
61.039.01ln
2
112ln1
22
82
dWfor
dWfor
BBB
e
e
d
W
rr
r
A
A
[3.20]
35
MICROSTRIP
Where:
r
rr
rr
ZB
ZA
0
0
2
377
11.023.0
1
1
2
1
60
mNpk
re
erd /
12
tan10
Considering microstrip as quasi-TEM line, the attenuation due to dielectric loss can be determined as
[3.21]
Where tan δ is the loss tangent of the dielectric.
36
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:
MICROSTRIP
1
1
re
er
This result is derived from Equation [2.37] by multiplying by a “filling factor”:
mNpWZ
Rsc /
0
Where Rs = √(ωμ0/2σ) is the surface resistivity of the conductor.
[3.22]
37
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 2.5 GHz. The substrate thickness is d = 0.127 cm, with εr = 2.20.
Solution: M.Pozar’ book (page: 146).
38
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<2Then W = (3.081.d) = 0.391 cm. From eq [3.18];
εe = 1.87
39
SOLUTION [EXAMPLE 2.3]
The line length, l, for a 90o phase shift is found as;
lkl eo
090
10 35.52
2 mc
fk
cmk
le
oo
19.2180
90
0
40
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?
41
Initially, it is guessed that W/d < 2
SOLUTION [EXAMPLE 2.4]
rr
rrZA
11.023.0
1
1
2
1
600
5.2
11.023.0
15.2
15.2
2
15.2
60
70A
66.1A
64.12
8
2
8)66.1(2
66.1
2
e
e
e
e
d
WA
A
42
SOLUTION cont [EXAMPLE 2.4]
Since the W/d < 2 assumption is valid;
We proceed to calculate εe
cmmW 64.10164.0100.164.1 2
Wdrr
e121
1
2
1
2
1
01.2)64.10.1(121
1
2
15.2
2
15.2
e
43
SOLUTION cont [EXAMPLE 2.4]
Thus the guided wavelength is given by;
cmf
c
e
g 05.710301.2
100.39
8
44
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.
45
MICROSTRIP [EXAMPLE 2.5]
246.6350x800 ZRZ LS
From the quarter wave transformer equation;
RLZ0ZS
Which is also Z0 of line
46
SOLUTION [EXAMPLE 2.5]
rr
rrZA
11.0
23.01
1
2
1
600
4635.13.2
11.023.0
13.2
13.2
2
13.2
60
246.63
A
174.63.2)246.63(2
377
2
377
0
rZB
Since W is not known, guess that W/d < 2
47
SOLUTION cont [EXAMPLE 2.5]
From the calculation, the initial assumption of W/d < 2 is incorrect. The next formula (where W/d > 2) is used
073.22
8
2
8)4635.1(2
4635.1
2
e
e
e
e
d
WA
A
rr
r BBBd
W 61.039.01ln
2
112ln1
2
mmWd
W
3.3)6.1)(0656.2(
0656.2
48
SOLUTION cont [EXAMPLE 2.5]
The next step is to find the effective dielectric constant (εe)
To determine the quarter wavelength of the line, the guided wavelength λg need to be determined
cmf
c
e
g 88.101028991.1
100.39
8
8991.1121
1
2
1
2
1
Wd
rr
e
49
SOLUTION cont [EXAMPLE 2.5]
Thus the quarter wave length of line is determined by dividing the full wavelength by 4;
cmg 72.24
88.10
4
50
MICROSTRIP
Figure 3.5: Geometry of a microstrip line with conducting sidewalls.
51
MICROSTRIP DISCONTINUITY
By derivation found in M.Pozar’s book (page 147), the surface charge density on the strip at y = d is:
dyxEdyxE
dyxDdyxD
yry
yys
,,
,,
00
odd
nrn a
dn
a
dn
a
xn
a
nA
10 coshsinhcos
[3.23]
The charge density on the microstrip line by uniform distribution:
2
2
0
1
Wxfor
Wxfors [3.24]
52
MICROSTRIP DISCONTINUITY
The capacitance per unit length of the stripline is:
mFd
adnadnWn
adnaWnaV
QC
odd
n r
/
coshsinh
sinh2sin41
1 02
The characteristic impedance is then found as:
cCCvZ e
p
10
[3.25]
[3.26]
53
DIELECTRIC WAVEGUIDE
Figure 3.6: Dielectric waveguide geometry.
54
DIELECTRIC WAVEGUIDE
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.
55
SLOTLINE
Figure 3.7: Geometry of a printed slotline.
56
SLOTLINE
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 characteristic impedance of the line can be change by
changing the width of the slot.