Microwave Circuit Design
By
Professor Syed Idris Syed Hassan
Sch of Elect. & Electron Eng
Engineering Campus USM
Nibong Tebal 14300
SPS Penang
Introduction
• 10 weeks lecture + 4 weeks ADS simulation
• Assessments :8 tests + 2 ADS assignments
+ 1 final examination
• Class : 9.00- 10.30 lecture
10.30-11.00 rest (tea break)
11.00-12.30 lecture
12.30- 1.00 test
Dates
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06/04/02 Morning
20/04/02 Morning
27/04/02 Morning
04/05/02 Morning
11/05/02 Morning
18/05/02 Morning
25/05/02 Morning
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08/06/02 Morning
15/06/02 Morning
22/06/02 Morning
29/06/02 morning
06/07/02 Morning
20/07/02 Morning
27/07/02 Morning
Syllabus
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Transmission lines
Network parameters
Matching techniques
Power dividers and combiners
Diode circuits
Microwave amplifiers
Oscillators
Filters design
Applications
Miscellaneous
References
• David M Pozar ,Microwave Engineering- 2nd Ed.,
John Wiley , 1998
• E.H.Fooks & R.A.Zakarevicius, Microwave
Engineering using microstrip circuits, Prentice
Hall,1989.
• G. D. Vendelin, A.M.Pavio &U.L.Rohde, Microwave
circuit design-using linear and Nonlinear
Techniques, John Wiley, 1990.
• W.H.Hayward, Introduction to Radio Frequency
Design, Prentice Hall, 1982.
Transmission Line
z
Equivalent Circuit
R
L
R
L
C
G
Lossy line
R  j L
Zo 
G  j C

Lossless line
L
Zo 
C
   LC
R  j L G  j C 
Analysis
I  z
R  j L
I ( z, t )
G  j C
V ( z, t )
dI

z   RI  z  L  z
dt

dV
dI

 z  RI  z  L  z
dz
dt

dV
dI
 RI  L
dz
dt
V  z
dV
dz
Kirchoff current law
From Kirchoff Voltage Law
dV

V  V 

dz

dI
dz
dI
dV


I   I   z   GV  z  C
 z
dz
dt



(a)
dI
dV
 z  GV  z  C
 z
dz
dt

dI
dV
 GV  C
dz
dt
(b)
Analysis
Let’s V=Voejt , I = Ioejt
then
dV
 j V
dt
dI
 j I
dt
Therefore
a
dV

 R  j L I
dz
b

dI
 G  j C V
dz
Differentiate with respect to z
d 2V
dI
 2  R  j L 
d 2I
dV
 2  G  j C 
dz
dz
d 2V
 R  j L G  j C V
2
dz
d 2V
2


V
dz 2
d 2I
 R  j L G  j C I
2
dz
dz
dz
d 2I
2


I
2
dz
Analysis
The solution of V and I can be written in the form of
z
z
and
Ae 
V  Ae z  Be 
 R  j L 
Zo  

 G  j C 
    j 
I
where
z
 Be 
Zo
c
d
R  j LG  j C 
Let say at z=0 , V=VL , I=IL and Z=ZL
VL  A  B
and
VL
 ZL
IL
e
A B
IL 
Zo
Therefore
f
Analysis
Solve simultaneous equations ( e ) and (f )
V  I L Zo
B L
2
V  I L Zo
A L
2
Inserting in equations ( c) and (d) we have
 e  z  e  z 
 e  z  e  z 
V ( z )  VL 
  I L Zo 

2
2




 e  z  e  z  VL
I ( z)  I L 

2

 Z o
 e  z  e  z 


2


Analysis
 e z  e  z 
 and
But cosh(  z )  
2


 e z  e  z 
sinh(  z )  

2


Then, we have
V ( z )  VL cosh( z )  I L Z o sinh(  z )
and
*
VL
**
I ( z )  I L cosh(  z ) 
sinh(  z )
Zo




V ( z )  VL cosh(  z )  I L Z o sinh(  z ) 
Z ( z) 

I ( z )  I cosh(  z )  VL sinh(  z ) 
 L

Z
o


Analysis
or
 Z L cosh(  z )  Z o sinh(  z ) 

Z ( z )  Z o 
 Z o cosh(  z )  Z L sinh(  z ) 
Or further reduce
 Z L  Z o tanh(  z ) 

Z ( z )  Z o 
 Z o  Z L tanh(  z ) 
For lossless transmission line , = j since 0
cosh( j z )  cos(  z )
sinh( j z )   j sin(  z )
 Z L  jZo tan(  z ) 

Z ( z )  Z o 
 Z o  jZ L tan(  z ) 
Analysis
Standing Wave Ratio (SWR)
node
antinode
Reflection coefficient

Be  z
Aez
Ae-z
Bez
Voltage and current in term of reflection coefficient
VL  Ae z  Be z  Ae z 1   
Ae z  Be  z Ae z
1   
IL 

Zo
Zo
VL  1   
Z o
ZL 
 
IL  1  
or Z L   1   
Z o  1   
Analysis
For loss-less transmission line
 = j
By substituting in * and ** ,voltage and current amplitude are

A

I ( z) 
1 
Z
2

 2  cos(2 z   )
V ( z )  A 1    2  cos( 2  z   )
2
1/ 2
g
1/ 2
o
h
Voltage at maximum and minimum points are
Vmax  A(1   )
and
Therefore VSWR  s  1  
1 
Vmin  A(1   )
ZL
For purely resistive load s 
Zo
Analysis
Other related equations
Z max
1  
Vmax
  sZ o

 Z o 

I min
1




Z min
 1    Zo
Vmin


 Z o 

I max
1   s
 Z L  Zo 

  
 Z L  Zo 
From equations (g) and (h), we can find the max and min points
Maximum
2  z    0,  2 ,  4 ,
Minimum
2  z     ,  3 ,
Important Transmission line equations
Zin
Zo
ZL
Z L  jZ o tanh  
Zin  Z o
Z o  jZ L tanh  
Z L  Zo

Z L  Zo
1 
SWR 
1 
Various forms of Transmission
Lines
Two wire
cable
Rectangular
waveguide
Coaxial
cable
Circular
waveguide
Microstripe
line
Stripline
Parallel wire cable
C

cosh1d / 2a 

L

Zo 
1
cosh
1

d / 2a 
or


ln d / a 

or  ln d / a 

for
for

cosh1d / 2a 

Where a = radius of conductor
d = separation between conductors
a  d
a  d
Coaxial cable
b
a
2 
C
ln b / a 
1 
Zo 
ln b / a 
2 

L
ln b / a 
2
fc 
ckc
2  r
2
kc 
ab
Where a = radius of inner conductor
b = radius of outer conductor
c = 3 x 108 m/s
Micro strip
Conducted strip
he
w
t
r
Substrate
Ground
t=thickness of conductor
Characteristic impedance of
Microstrip line
For
w / h  1 Zo 
 eff 
For
60
 eff
r  1 r 1
2

w / h  1 Zo 
 eff 
2
ln 8h / we  0.25 we / h 
1  12h / w 
0.5
e

377
 eff we / h  1.393  0.667 ln we / h  1.444 
r  1 r 1
2
 0.041  we / h 2

2
t  2he 
w

w

 1
Where e
 ln

t

1  12h / w  
0.5
e
he  h  2 t
w=width of strip
h=height and
t=thickness
Microstrip width
1/ 2
Zo   r  1 
A


60  2 
r 1
0.11 
 0.23 


r  1
r 
60 2
B
Zo  r
8 exp( A)
W /h 
exp(2 A)  2
For A>1.52
For A<1.52
r 1 
0.61
W / h  B  1  ln 2 B  1 
lbB  1  0.39 


2 r 
 r 
2
Simple Calculation
377
Zo 
w

r   2
h

377
w/ h 
2
 r Zo

Approximation only
Microstrip components
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Capacitance
Inductance
Short/Open stub
Open stub
Transformer
Resonator
Capacitance

Zoc
Zo
For

For  

8

8
C
Zo

Z occ1
  

C
sin 
Z oc  2c1 
2
c1 
3  108
r
m/ s
Inductance

ZoL
Zo
For

For  

8

8
Zo
Z oL
L
c1
Z oL   
L
sin  

 c1 
c1 
3  108
r
m/ s
Short Stub

Zo
Zo
ZL

Zo
Z
Z sc  X L   jZ o tan 
  tan
1
 X L / Zo 
 o

360  eff
Open stub

Zo
Zo
ZL

Zo
Z
Z oc  X c   jZ o cot 
1
  cot
 X c / Zo 
 o

360  eff
Quarter-wave transformer
/4
Zo
ZT
At maximum point
Z L  Zo

  
Z L  Zo

x
2
x
 in radian
Zo
ZL
Zmx/min
1 
s
1 
Z ( x) max  Z o s
ZT  Zin Zmx  Zo .Zo s  Zo s
Quarter-wave transformer
at minimum point
Z L  Zo

  
Z L  Zo
 
x
2
 in radian
1 
s
1 
Z ( x) min  Z o / s
ZT  Zin Zmin  Zo .Zo / s  Zo / s
Resonator
•
•
•
•
•
Circular microstrip disk
Circular ring
Short-circuited /2 lossy line
Open-circuited /2 lossy line
Short-circuited /4 lossy line
Circular disk/ring
feeding
a
a
1.841o
2  r
a

a
4
* These components usually use for resonators
Short-circuited /2 lossy line
Zin
R  Z o 
Zo
L
2o
C
Zo


 n/2
o L


Q


R
2  2
1
o L
2
= series RLC resonant cct
where

2

Open-circuited /2 lossy line
Zin
Zo


= parallel RLC resonant cct
 n/2
R
Zo


C
2o Z o
L
1
o C
2


Q  o RC 

2  2
where

2

Short-circuited /4 lossy line
Zin
Zo

R
Zo


C
4o Z o
L
1
o C
2


= parallel RLC resonant cct
/4


Q  o RC 

4  2
where

2

Rectangular waveguide
b
a
Cut-off frequency of TE or TM mode
2
ZTE
g

o
o

g
2
ZTM
 m   n 
fcmn 

 

2    a   b 
Conductor attenuation for TE10
1
1
g 2

1
o 2

1
c 2
c 
Rs
a b g
3

2b 2  a3  o2 

o
Rs 
o 
2
Np / m
Example
Given that a= 2.286cm , b=1.016cm and 5.8 x 107S/m.
What are the mode and attenuation for 10GHz?
Using this equation to calculate cutoff frequency of each mode
2
 m   n 
fcmn 

 

2    a   b 
1
2
Calculation
TE10
a=2.286mm, b=1.016mm, m=1 and n=0 ,thus we have
2
f c10
3 108 


9


  6.562 10 GHz
2
 0.002286 
Similarly we can calculate for other modes
Example
Mode
fcmn
TE10
6.562 GHz
TE20
13.123GHz
TE01
14.764GHz
TE11
16.156GHz
TE10
6.562GHz
TE20 TE01
13.123GHz14.764GHz
TE11
16.156GHz
Frequency 10Ghz is propagating in
TE10.mode since this frequency is
below the 13.123GHz (TE20) and
above 6.561GHz (TE10)
continue
2
 
  o     158.05m1
a
Rs 
c 
or
o 
 0.026 
2
Rs
a b g
3

2b 2  a3  o2   0.0125

Np / m
o
c (dB)  20 log ec  0.11 dB / m
Evanescent mode
Mode that propagates below cutoff frequency of a wave guide is
called evanescent mode
Wave propagation constant is
kc 2   2   o2
Where kc is referred to cutoff frequency,  is referred to
propagation in waveguide and  is in space
When f0< fc ,
But
   j
 2  kc 2   o2
=attenuation
Since no propagation then
  2
The wave guide become attenuator
=phase constant
1
c
2

1
o 2
Cylindrical waveguide
a
n
p'n1
p'n2
p'n3
0
3.832
7.016
10.174
1
1.841
5.331
8.536
2
3.054
6.706
9.97
TE mode
fcnm 
,
pnm
Dominant mode is TE11
2 a 
2
 nm   o2  kcnm
,
 g
pnm
ZTE 
kcnm 
o
a
2 

Rs

o 
 kc112 
c 
2
a o g 
p'11
1 
continue
a
n
pn1
pn2
pn3
0
2.405
5.520
8.654
1
3.832
7.016
10.174
2
5.135
8.417
11.620
TM mode
pnm
fcnm 
2 a 
2
 nm   o2  kcnm
 o
g
TM01 is preferable for long haul
transmission
pnm
kcnm 
a
ZTM 
Example
Find the cutoff wavelength of the first four modes of a circular waveguide
of radius 1cm
2nd mode
Refer to tables
TM modes
TE modes
n
p'n1
p'n2
p'n3
n
pn1
pn2
pn3
0
3.832
7.016
10.174
0
2.405
5.520
8.654
1
1.841
5.331
8.536
1
3.832
7.016
10.174
2
3.054
6.706
9.97
2
5.135
8.417
11.620
1st mode
3rd &4th
modes
Calculation
pnm
fcnm 
2 a 
1st mode Pnm= 1.841, TE11
2nd
mode Pnm= 2.405, TM01

cnm 
2a
pnm
2  0.01
 c11
 0.0341m
1.841
2  0.01
 c 01
 0.0261m
2.405
1st mode Pnm= 3.832, TE01 and TM11

c 01 
c11
2  0.01
 0.0164m
3.832
Stripline
b
w
30 b
Zo 
 r We  0.441b 
We  W
W
 0.35
b
We W
  0.35  W / b 2
b
b
W
 0.35
b
Continue
On the other hand we can calculate the width of
stripline for a given characteristic impedance
W
30

 0.441
b
 r Zo
 30

W
 0.85  0.6  
 0.441 
b
  r Zo

 r Z o  120
for
for
 r Z o  120
Continue
 2.7  10 3 Rs r Z o
A

30 b  t 

c  
 0.16 Rs B
 Z ob
Where
for  r Z o  120
2W
bt
 2b  t 
A 1

ln 

b  t  b  t   t 
for  r Z o  120
b
0.414t 1 4 W 

B 1

ln
 0.5 

0.5W  0.7t  
W
2
t 
t =thickness of the strip