1RS
ENE 428Microwave
Engineering
Lecture 6 Transmission lines problems and microstrip lines
2
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
3
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
4
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 atwhich the remainder of the line could be replaced by a pureresistance?
5
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
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Impedance matching
• To minimize power reflection from load
• Zin = Z0
• Matching techniques1. Quarter - wave transformers for real
load2. single - stub tuners3. lumped – element tuners
• The capability of tuning is desired by having variable reactive elements or stub length.
0S LZ Z R
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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.
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Simple matching by adding reactive elements (2)
The value of capacitance can be evaluated by known frequency, for example, 1 GHz is given.
12.3 50 115CX j j
j C
11.38
115C pF
j
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Single stub tuners
Working with admittance (Y) since it is more convenient to add shunt elements than series elements Stub tuning is the method to add purely reactive elements 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.
Ex6 let z = 2+j2, what is the admittance?
10.25 0.25
2 2y j
j
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Stub tuners on Y-chart (Admittance chart) (1)
There are two types of stub tuners1. Shorted end, y = (the rightmost of the Y chart)2. opened end, y = 0 (the leftmost of the Y chart)
Short-circuited shunt stub Open-circuited shunt stub
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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 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 jb.
4. Total normalized admittance ytot = yd+yl = 1.
0
0
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Ex7 What about the open-circuited stub?
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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
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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 permittivity, eff
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Microstrip (3)
0
/
2/
.
p
eff
p
pg
eff
cu m s
frad m
u
um
f
Therefore in this case
and
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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
0
0
1 12
2 1 12
60 8/ 1, ln( )
4
1 120/ 1,
1.393 0.667 ln( 1.444)
r reff
eff
eff
hw
h wfor w h Z
w h
for w h Zw wh h
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Evaluation of the microstrip configuration (2)• Assume t is negligible, if Z0 and r are known, the ratio w/h
can be calculated as
2
0
8/ 2,
2
12 0.61/ 2, 1 ln(2 1) ln( 1) 0.39
2
1 1 0.11(0.23 )
60 2 1
A
A
r
r r
r r
r r
w efor w h
h e
wfor w h B B B
h
Zwhere A
0
377
2 r
and BZ
The value of r and the dielectric thickness (h) determines the width (w) of the microstrip for a given Z0.
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Characteristic impedance of the microstrip line versus w/h
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Normalized wavelength of the microstrip line versus w/h
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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.
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Wavelength in the microstrip line
1/ 2
0
0.0297
1/ 2
0
0.1255
/ 0.6,1 0.6( 1)( )
/ 0.6,1 0.63( 1)( )
r
rr
r
rr
for w hwh
for w hwh
Assume t/h 0.005,
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Attenuation
tot c d
conductor loss
dielectric loss
radiation loss
where c = conductor attenuation (Np/m) d = dielectric attenuation (Np/m
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Conductor attenuation
( / )
8.686 ( / )
1
skinc
o
skinc
o
skin
RNp m
Z w
RdB m
Z w
R
If the conductor is thin, then the more accurate skin resistance can be shown as
/
1.
(1 )skin tR
e
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Dielectric attenuation
( 1)2tan /
2 ( 1)r eff
deff r
fNp m
c