by professor syed idris syed hassan sch of elect. & electron eng engineering campus usm nibong...
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By
Professor Syed Idris Syed Hassan
Sch of Elect. & Electron Eng
Engineering Campus USM
Nibong Tebal 14300
SPS Penang
Microwave Circuit Design
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
• 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
• 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 • 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
Equivalent CircuitR RL L
C
G
z
C
LZo
CjG
LjRZo
Lossy line
Lossless line
CjGLjR
LC
AnalysisLjR ),( tzI
dz
dIzI
),( tzV CjG dz
dVzV
zdt
dILzRIz
dz
dV
dt
dILRI
dz
dV
zdt
dVCzGVz
dz
dI
dt
dVCGV
dz
dI
From Kirchoff Voltage Law Kirchoff current law
zdt
dVCzGVz
dz
dIII
z
dt
dILzRIz
dz
dVVV
(a) (b)
Analysis
Let’s V=Voejt , I = Ioejt
Therefore
Vjdt
dV Ijdt
dI then
ILjRdz
dV VCjGdz
dI a b
Differentiate with respect to z
dz
dILjR
dz
Vd 2
2
VCjGLjRdz
Vd 2
2
Vdz
Vd 22
2
dz
dVCjG
dz
Id 2
2
ICjGLjRdz
Id 2
2
Idz
Id 22
2
AnalysisThe solution of V and I can be written in the form of
z zBe Ae V o
z z
Z
Be AeI
where
CjG
LjRZo
Let say at z=0 , V=VL , I=IL and Z=ZL
Therefore
B A VL o
LZ
B AI
and LL
L ZI
V
e f
c d
CjGLjRj and
AnalysisSolve simultaneous equations ( e ) and (f )
2o L LZ I V
A
2
o L LZ I VB
Inserting in equations ( c) and (d) we have
22)(
zz
oL
zz
Lee
ZIee
VzV
22)(
zz
o
Lzz
Lee
Z
VeeIzI
Analysis
2)cosh(
zz eez
But and
2)sinh(
zz eez
Then, we have
)sinh()cosh()( zZIzVzV oLL
)sinh()cosh()( zZ
VzIzI
o
LL
)sinh()cosh(
)sinh()cosh(
)(
)()(
zZV
zI
zZIzV
zI
zVzZ
o
LL
oLL
and
*
**
Analysis
)sinh()cosh(
)sinh()cosh()(
zZzZ
zZzZZzZ
Lo
oLo
)tanh(
)tanh()(
zZZ
zZZZzZ
Lo
oLo
Or further reduce
or
For lossless transmission line , = jsince
)tan(
)tan()(
zjZZ
zjZZZzZ
Lo
oLo
)cos()cosh( zzj
)sin()sinh( zjzj
AnalysisStanding Wave Ratio (SWR)
node
antinode
Ae-zBez
z
z
Ae
Be
1zzzL AeBeAeV
1o
z
o
zz
L Z
Ae
Z
BeAeI
Reflection coefficient
Voltage and current in term of reflection coefficient
oL
LL Z
I
VZ
1
1
1
1
o
L
Z
Zor
AnalysisFor loss-less transmission line = jBy substituting in * and ** ,voltage and current amplitude are
2/12)2cos(21)( zAzV
2/12)2cos(21)( z
Z
AzI
o
Voltage at maximum and minimum points are
)1(max AV )1(min AVand
1
1sVSWRTherefore
For purely resistive load o
L
Z
Zs
g
h
Analysis
oL
oL
ZZ
ZZ
oo sZZI
VZ
1
1
min
maxmax s
ZZ
I
VZ o
o
1
1
max
minmin
Other related equations
From equations (g) and (h), we can find the max and min points
,4,2,02 z
,3,2 z
Maximum
Minimum
Important Transmission line equations
tanh
tanh
Lo
oLoin jZZ
jZZZZ
oL
oL
ZZ
ZZ
1
1SWR
ZoZL
Zin
Various forms of Transmission Lines
Two wirecable Coaxial
cable
Microstripeline
Rectangularwaveguide
Circularwaveguide
Stripline
Parallel wire cable
daforadoradL /ln2/cosh 1
daforad
orad
C /ln2/cosh 1
adZo 2/cosh1 1
Where a = radius of conductor d = separation between conductors
Coaxial cable
abC
/ln
2
abL /ln2
abZo /ln2
1
Where a = radius of inner conductor b = radius of outer conductor c = 3 x 108 m/s
r
cc
ckf
2
bakc
2
ab
Micro strip
whe r
t
t=thickness of conductor
Substrate
Conducted strip
Ground
Characteristic impedance of Microstrip line
hwwhZhwFor eeeff
o /25.0/8ln60
1/
444.1/ln667.0393.1/
3771/
hwhwZhwFor
eeeffo
25.0 /104.0/1212
1
2
1hwwh ee
rreff
5.0/1212
1
2
1
err
eff wh
1
2ln
t
htww e
e thhe 2Where
w=width of striph=height and t=thickness
Microstrip width
rr
rroZA
11.023.0
1
1
2
1
60
2/1
roZB
260
2)2exp(
)exp(8/
A
AhW
rr
r BlbBBhW
61.039.01
2
112ln1
2/
For A>1.52
For A<1.52
Simple Calculation
2
377
hw
Z
r
o
2377
/ orZ
hw
Approximation only
Microstrip components
• Capacitance
• Inductance
• Short/Open stub
• Open stub
• Transformer
• Resonator
Capacitance
Zo ZoZoc
1cZC
oc
12sin
2
cZC
oc
smc
r/
103 8
1
For 8
For8
Inductance
Zo ZoZoL
1c
ZL oL
1sin
c
ZL oL
smc
r/
103 8
1
For 8
For8
Short Stub
Zo
Z
Zo
Zo ZL
oL ZX /tan 1eff
o
360
tanoLsc jZXZ
Open stub
Zo
Z
Zo
Zo ZL
oc ZX /cot 1eff
o
360
cotococ jZXZ
Quarter-wave transformer
Zo ZoZT
Zmx/min
ZL
x
oL
oL
ZZ
ZZ
2
x in radian sZxZ omax)(
1
1s
sZsZZZZZ ooomxinT .
At maximum point
Quarter-wave transformer
oL
oL
ZZ
ZZ
2
x in radian
sZxZ o /)( min
1
1s
sZsZZZZZ oooinT //.min
at minimum point
Resonator
• Circular microstrip disk
• Circular ring
• Short-circuited/2 lossy line
• Open-circuited /2 lossy line
• Short-circuited /4 lossy line
Circular disk/ring
a
r
oa
2
841.1
feeding
a
4a
* These components usually use for resonators
Short-circuited/2 lossy line
n/2
Zin Zo
oZR
o
oZL
2
LC
o2
1
22
R
LQ o
2
where
= series RLC resonant cct
Open-circuited /2 lossy line
n/2
Zin Zo
22
RCQ o
= parallel RLC resonant cct
oZR
oo ZC
2
CL
o21
2where
Short-circuited /4 lossy line
/4
Zin Zo
oZR
= parallel RLC resonant cct
oo ZC
4
CL
o21
24
RCQ o
2
where
Rectangular waveguide
ab
22
2
1
b
n
a
mfcmn
Cut-off frequency of TE or TM mode
222111
cog
o
gTEZ
g
oTMZ
mNpabba
Ro
og
sc /2 232
3
Conductor attenuation for TE10
2o
sR
Example
Given that a= 2.286cm , b=1.016cm and xS/m. What are the mode and attenuation for 10GHz?
22
2
1
b
n
a
mfcmn
Using this equation to calculate cutoff frequency of each mode
Calculation
GHzfc9
28
10 10562.6002286.02
103
TE10
a=2.286mm, b=1.016mm, m=1 and n=0 ,thus we have
Similarly we can calculate for other modes
Example
Mode fcmn
TE10 6.562 GHz
TE20 13.123GHz
TE01 14.764GHz
TE11 16.156GHz
TE10TE20 TE01
TE11
6.562GHz 13.123GHz14.764GHz 16.156GHz
Frequency 10Ghz is propagating in TE10.mode since this frequency is below the 13.123GHz (TE20) and above 6.561GHz (TE10)
continue
026.02o
sR
12
05.158
mao
mNpabba
Ro
og
sc /0125.02 232
3
mdBedB cc /11.0log20)( or
Evanescent modeMode that propagates below cutoff frequency of a wave guide is called evanescent mode
22
112
oc
222ock Wave propagation constant is
Where kc is referred to cutoff frequency, is referred to propagation in waveguide and is in space
j =attenuation =phase constant
When f0< fc , 222ock
But
Since no propagation then
The wave guide become attenuator
Cylindrical waveguide
a
ap
f nmcnm 2
,
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
a
pk nmcnm
,
22cnmonm k
TE modeDominant mode is TE11
o
gTEZ
1'211
22
11p
ka
R oc
go
sc
continue
a
ap
f nmcnm 2
a
pk nmcnm
22cnmonm k
TM mode
g
oTMZ
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
TM01 is preferable for long haul transmission
ExampleFind the cutoff wavelength of the first four modes of a circular waveguide of radius 1cm
Refer to tables
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 modes TM modes
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
1st mode
2nd mode
3rd &4th modes3rd &4th modes
Calculation
nmcnm p
a 2
ap
f nmcnm 2
1st mode Pnm= 1.841, TE11mc 0341.0
841.1
01.0211
2nd mode Pnm= 2.405, TM01
1st mode Pnm= 3.832, TE01 and TM11
mc 0261.0405.2
01.0201
mcc 0164.0832.3
01.021101
Stripline
wb
bW
bZ
ero 441.0
30
35.0/35.0 2 b
WbW
b
W
b
We
35.0b
WWWe
Continue
120441.030
oror
ZforZb
W
120441.030
6.085.0
or
orZfor
Zb
W
On the other hand we can calculate the width of stripline for a given characteristic impedance
Continue
12016.0
12030
107.2 3
oro
s
orors
c
ZforBbZ
R
ZforAtb
ZR
t
tb
tb
tb
tb
WA
2ln
21
t
W
W
t
tW
bB
4ln
2
1414.05.0
7.05.01
Where
t =thickness of the strip
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