ENE 428
Microwave Engineering
Lecture 6 Transmission lines
problems and microstrip lines
RS
1
Review
• Input impedance for finite length line
– Quarter wavelength line
– Half wavelength line
• Smith chart
– A graphical tool to solve transmission line problems
– Use for measuring reflection coefficient, VSWR, input
impedance, load impedance, the locations of Vmax and Vmin
2
Ex1 ZL = 25+j50 , given Z0 = 50  and the line length is
60 cm, the wavelength is 2 m, find Zin.
Ex2 A 0.334 long TL with Z0 = 50  is terminated in a load
ZL = 100-j100 . Use the Smith chart to find
a) L
b) VSWR
c) Zin
d) the distance from load to the first voltage minimum
3
Ex3 ZL = 80-j100  is located at z = 0 on a lossless 50  line,
given the signal wavelength = 2 m, find
a) If the line is 0.8 m in length, find Zin.
b) VSWR
c) What is the distance from load to the nearest voltage
maximum
d) what is the distance from the input to the nearest point at
which the remainder of the line could be replaced by a pure
resistance?
4
Ex4 A 0.269- long lossless line with Z0 = 50  is terminated
in a load ZL = 60+j40 . Use the Smith chart to find
a) L
b) VSWR
c) Zin
d) the distance from the load to the first voltage maximum
5
Impedance matching
• To minimize power reflection from load
• Zin = Z0
• Matching techniques
1. Quarter - wave transformers Z S  Z 0 RL for real load
2. single - stub tuners
3. lumped – element tuners
• The capability of tuning is desired by having variable reactive elements
or stub length.
6
Quarter-Wave Transformer
Z S2
Z in 
RL
Z S  Z0 RL
7
Simple matching by adding reactive elements (1)
EX5, a load 11+j25  is terminated in a 50  line. In order
for 100% of power to reach a load, ZLoad must match with
Z0, that means ZLoad = Z0 = 50 . This value corresponds to
center of the chart.
8
Simple matching by adding reactive elements (1)
9
Distance d WTG = (0.188-0.076)
= 0.112 
to point 1+ j2 Therefore cut TL and insert a reactive
element that has a normalized reactance of -j2.
The normalized input impedance becomes
1+ j2 - j2 = 1
which corresponds to the center or the Smith chart.
10
Simple matching by adding reactive
elements (1)
EX5, a load 10-j25  is terminated in a 50  line. In order
for 100% of power to reach a load, ZLoad must match with
Z0, that means ZLoad = Z0 = 50 .
 Distance d WTG = (0.5-0.424) +0.189 
= 0.265 
to point 1+ j2.3. Therefore cut TL and insert a
reactive element that has a normalized reactance of
-j2.3.
 The normalized input impedance becomes
1+ j2.3 - j2.3 = 1
which corresponds to the center or the Smith chart.
11
Simple matching by adding reactive
elements (2)
 The value of capacitance can be evaluated by known
frequency, for example, 1 GHz is given.
1
  j 2.3  50   j115 
jC
1
C 
 1.38 pF
 j115
XC 
12
Single stub tuners
 Working with admittance (Y) (Yo = 1/Zo) since it is more
convenient to add shunt elements than series elements
 Stub tuning is the method to add purely reactive elements
13
Ex6 let zL = 2 + j1, what is the admittance?
y L  1 / z L  1 /(2  j1)  0.40  j 0.20
Where is the location of y on Smith chart?
We can easily find the admittance on the Smith chart by
moving 180 from the location of z.
Smith Chart is also a chart of normalized admittance. The
normalized load admittance is
YL
1
yL 

YO z L
14
15
Stub tuners on Y-chart (Admittance
chart) (1)

There are two types of stub tuners
1. Shorted end, y =  (the rightmost of the Y chart)
2. opened end, y = 0 (the leftmost of the Y chart)
l
l
Short-circuited shunt stub
Open-circuited shunt stub
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Short-circuited Shunt Stub
Z in  jZ O tan( d )
zin  j tan( d )
Purely reactive tuning element follows
the constant || circle along the
periphery of Smith chart (z = 0  jx).
Proper selection of the T-line length d
allows us to choose any value of
reactance that we want, whether it’s
capacitive or inductive.
The admittance of the short-circuited
starts on the right side of the chart (
yshort = ∞ + j∞)
17
Stub tuners on Y-chart (Admittance
chart) (2)

Procedure
1. Locate zL and then yL. From yL, move clockwise to
1  jb circle, at which point the admittance yd = 1  jb.
On the WTG scale, this represents length d.
2. For a short-circuited shunt stub, locate the short end at
0.250 then move to 0 jb, the length of stub is then
l and then yl = jb.
3. For an open-circuit shunt stub, locate the open end at 0,
then move to 0 jb.
4. Total normalized admittance ytot = yd+yl = 1.
18
Ex: Construct a short-circuited stub matching network for a
50- line terminated in a load ZL = 20 – j55 
First, normalize load impedance, zL = ZL/ZO = 0.4 – j1.1
Next, locate yL
Move along the constant-|| circuit in WTG (clockwise)
direction ‘till it intersects with the 1  jb circle (in this case 1 +
j2.0). We travel from 0.112g to 0.187g on the WTG scale, so
our through-line length (d) is 0.075g
Next, we insert the short-ckted shunt stub. Admittance is on
the right side of the chart at WTG = 0.250g. We move
clockwise until we reach the point 0 – j2.0, located at WTG =
0.324g. This give us stub length (l) = 0.324g- 0.250g =
0.074g.
19
20
Ex: We want to construct an open-ckted shunt-stub matching network for a
50- line terminated in a load ZL =150 + j100 .
Solution:
First, normalize load impedance,
zL = ZL/ZO = 3.0 + j2.0
Next, locate yL
Move along the constant-|| circuit in WTG (clockwise) direction ‘till it intersects with
the 1  jb circle (in this case 1 + j1.6). We travel from 0.474g to 0.178g on the WTG
scale. We add 0.500g to the end point, so our through-line length (d) is (0.500g) –
0.474g + 0.178g = 0.204g
Next, we insert the open-ckted shunt stub. On the admittance chart, the location of the
open-ckted is on the left side of the chart at WTG = 0.000g. We move clockwise until
we reach the point 0 – j1.6, located at WTG = 0.339g. This give us stub length
(l) = 0.339g.
21
First, normalize load impedance, zL = ZL/ZO = 0.4 – j1.1
Next, locate yL
Move along the constant-|| circuit in WTG (clockwise) direction ‘till it intersects with
the 1  jb circle (in this case 1 + j2.0). We travel from 0.112g to 0.187g on the WTG
scale, so our through-line length (d) is 0.075g
Next, we insert the short-ckted shunt stub. Admittance is on the right side of the chart
at WTG = 0.250g. We move clockwise until we reach the point 0 – j2.0, located at
WTG = 0.324g. This give us stub length (l) = 0.324g- 0.250g = 0.074g.
22
23
Homework
Prob. 2.41: A load impedance ZL = 25 + j90  is to be matched to a 50- line using
an open-ended shunt-stub tuner. Find the solution that minimizes the length of the
shorted stub.
24
Microstrip (1)
• The most popular transmission line since it can be fabricated using
printed circuit techniques and it is convenient to connect lumped
elements and transistor devices.
• By definition, it is a transmission line that consists of a strip
conductor and a grounded plane separated by a dielectric medium
25
Microstrip (2)
• The EM field is not contained entirely in dielectric so it is not pure
TEM mode but a quasi-TEM mode that is valid at lower microwave
frequency.
• The effective relative dielectric constant of the microstrip is related to
the relative dielectric constant r of the dielectric and also takes into
account the effect of the external EM field.
Typical electric field lines
Field lines where the air and
dielectric have been replaced by
a medium of effective relative
26
permittivity, eff
Microstrip (2.1)
 Some typical dielectric substrates are RT/Duroid® (a trademark of Rogers
Corporation, Chandler, Arizona), which is available with several values of εr
(e.g. ε = 2.23εo, ε = 6εo, ε = 10.5εo, etc.); quartz (ε = 3.7εo); alumina (ε = 9εo)
and Epsilam-109® (ε = 10εo).

Various substrate materials are available for the construction of microstrip
lines, with practical values of εr ranging from 2 to 10. The substrate material
comes plated on both sides with copper, and an additional layer of gold
plating on top of the copper is usually added after the ckt pattern is etched in
order to prevent oxidation. Typical plating thickness of copper is from ½ mils
to 2 mils (1 inch = 1000 mils).
 The value of εr and the dielectric
thickness (h) determine the width of
the microstrip line for a given Zo.
These parameters also determine
the speed of propagation in the line,
and consequently its length. Typical
thickness are 25, 30, 40, 50 and
100 mils.
27
Microstrip (3)
Therefore in this case
up 
c
 eff
2 f

up
and
Z O  L C and u p  1
1
 ZO 
u pC
LC
m/s
rad / m
up
0
g  
f
 eff
m.
The evaluation of up, Zo and λ in microstrip line requires the evaluation of εeff and
C. There are different methods for determining εeff and C and, of course, closedform expressions are of great importance in microstrip-line design. The evaluation
of εeff and C based on a quasi-TEM mode is accurate for design purposes at
lower microwave freq. However, at higher microwave freq, the longitudinal
components of the EM fields are significant and the quasi-TEM assumption is no
longer valid.
28
Evaluation of the microstrip configuration (1)
• Consider t/h < 0.005 and assume no dependence of frequency, the
ratio of w/h and r are known, we can calculate Z0 as
29
Evaluation of the microstrip configuration (2)
• Assume t is negligible, if Z0 and r are known, the ratio w/h can be
calculated as
for w / h  2,
w
8e A
 2A
h e 2
for w / h  2,
 r 1 
w 2
0.61  
  B  1  ln(2 B  1) 
ln( B  1)  0.39 

h 
2 r 
 r  
where
and
Z0  r  1  r  1
0.11
A

(0.23 
)
60
2
r 1
r
B
377
2Z 0  r
The value of r and the dielectric thickness (h) determines the
width (w) of the microstrip for a given Z0.
30
Characteristic impedance of the
microstrip line versus w/h
31
Normalized wavelength of the
microstrip line versus w/h
32
Ex8 A microstrip material with r = 10 and h =
1.016 mm is used to build a TL. Determine the
width for the microstrip TL to have a Z0 = 50 .
Also determine the wavelength and the effective
relative dielectric constant of the microstrip line.
33
34
35
Wavelength in the microstrip line
Assume t/h  0.005,
1/ 2


 
r

for w / h  0.6,   0 
 r 1  0.6(  1)( w )0.0297 
r
h


1/ 2


 
r

for w / h  0.6,   0 
 r 1  0.63(  1)( w )0.1255 
r
h


36
37
Attenuation (losses in microstrip lines)
 conductor loss
 dielectric loss
 radiation loss
tot  c  d
where c = conductor attenuation (Np/m)
d = dielectric attenuation (Np/m
38
Conductor attenuation
Rskin
c 
Zo w
( Np / m)
Rskin
 c  8.686
Zo w
Rskin 
1

(dB / m)

If the conductor is thin, then the more accurate skin
resistance can be shown as
Rskin
1

.
t / 
 (1  e )
39
Dielectric attenuation
2 f  r ( eff  1)
d 
tan 
c 2  eff ( r  1)
Np / m
40