power divider, combiner and coupler by professor syed idris syed hassan sch of elect. & electron...
TRANSCRIPT
Power divider, combiner and coupler
By
Professor Syed Idris Syed Hassan
Sch of Elect. & Electron Eng
Engineering Campus USM
Nibong Tebal 14300
SPS Penang
Power divider and combiner/coupler
divider combinerP1
P2= nP1
P3=(1-n)P1
P1
P2
P3=P1+P2
Divide into 4 output
Basic
S-parameter for power divider/coupler
333231
232221
131211
SSS
SSS
SSS
SGenerally
For reciprocal and lossless network
jiforSSN
kkjki
0
1
*1
1
*
N
kki kiSS
1131211 SSS
1232221 SSS
1333231 SSS
0*2313
*2212
*2111 SSSSSS
0*3323
*3222
*3121 SSSSSS
0*3313
*3212
*3111 SSSSSS
Row 1x row 2
Row 2x row 3
Row 1x row 3
ContinueIf all ports are matched properly , then Sii= 0
0
0
0
2313
2312
1312
SS
SS
SS
S
For Reciprocalnetwork
For lossless network, must satisfy unitary condition
12
132
12 SS
1223
212 SS
1223
213 SS
012*23
SS
023*13 SS
013*12 SS
Two of (S12, S13, S23) must be zero but it is not consistent. If S12=S13= 0, then S23 should equal to 1 and the first equation will not equal to 1. This is invalid.
Another alternative for reciprocal network
332313
2312
1312
0
0
SSS
SS
SS
S
Only two ports are matched , then for reciprocal network
For lossless network, must satisfy unitary condition
12
132
12 SS
1223
212 SS
1233
223
213 SSS 013
*3312
*23 SSSS
023*13 SS
033*2313
*12 SSSS
The two equations show that |S13|=|S23|
thus S13=S23=0 and |S12|=|S33|=1
These have satisfied all
Reciprocal lossless network of two matched
S 21 =e j
S 12 =e j
S 33 =e j
1
3
2
j
j
j
e
e
e
S
00
00
00
For lossless network, must satisfy unitary condition
12
132
12 SS
12
232
21 SS
12
322
31 SS
032*31
SS
023*21 SS
013*12
SS
Nonreciprocal network (apply for circulator)
0
0
0
3231
2321
1312
SS
SS
SS
S
0312312 SSS
0133221 SSS
1133221 SSS
1312312 SSS
The above equations must satisfy the following either
or
Circulator (nonreciprocal network)
010
001
100
S
001
100
010
S
1
2
3
1
2
3
Four port network
44434241
34
24
14
333231
232221
131211
SSSS
S
S
S
SSS
SSS
SSS
SGenerally
For reciprocal and lossless network jiforSS
N
kkjki
0
1
*
11
*
N
kki kiSS
114131211 SSSS
124232221 SSSS
134333231 SSSS
0*2414
*2313
*2212
*2111 SSSSSSSS
0*4424
*4323
*4222
*4121 SSSSSSSS
0*3414
*3313
*3212
*3111 SSSSSSSS
R 1x R 2
R 2x R3
R1x R4
144434241 SSSS
0*4414
*4313
*4212
*4111 SSSSSSSS
0*3424
*3323
*3222
*3121 SSSSSSSS
0*4434
*4333
*4232
*4131 SSSSSSSS
R1x R3
R2x R4
R3x R4
Matched Four port network
0
0
0
0
342414
34
24
14
2313
2312
1312
SSS
S
S
S
SS
SS
SS
S
The unitarity condition become
1141312 SSS
1242312 SSS
1342313 SSS
0*2414
*2313 SSSS
0*3423
*1412 SSSS
0*3414
*2312 SSSS
1342414 SSS
0*3413
*2412 SSSS
0*3424
*1312 SSSS
0*2423
*1413 SSSS
Say all ports are matched and symmetrical network, then
*
**
@
@@
#
##
To check validityMultiply eq. * by S24
* and eq. ## by S13* , and substract to obtain
0214
213
*14
SSS
Multiply eq. # by S34 and eq. @@ by S13 , and substract to obtain
0234
21223
SSS
%
$
Both equations % and $ will be satisfy if S14 = S23 = 0 . This means that no coupling between port 1 and 4 , and between port 2 and 3 as happening in most directional couplers.
Directional coupler
00
0
00
00
0
3424
34
24
13
12
1312
SS
S
S
S
S
SS
S
If all ports matched , symmetry and S14=S23=0 to be satisfied
The equations reduce to 6 equations
11312 SS
12412 SS
13413 SS
13424 SS
0*3413
*2412 SSSS
0*3424
*1312 SSSS
2413 SS By comparing these equations yield
*
*
**
**
By comparing equations * and ** yield 3412 SS
Continue
00
0
00
00
0
j
j
j
j
S
Simplified by choosing S12= S34= ; S13=e j and S24= e j
Where + = + 2n
00
0
00
00
0
S
1. Symmetry Coupler = = /2
2. Antisymmetry Coupler =0 , =
2 cases
Both satisfy 2 +2 =1
Physical interpretation
|S13 | 2 = coupling factor = 2
|S12 | 2 = power deliver to port 2= 2 =1- 2
Characterization of coupler
Directivity= D= 10 log
dBP
P log203
1 Coupling= C= 10 log
dBSP
P
144
3 log20
Isolation = I= 10 log dBSP
P14
4
1 log20
I = D + C dB
1
4 3
2Input Through
CoupledIsolated
For ideal case |S14|=0
Practical couplerHybrid 3 dB couplers
Magic -T and Rat-race couplers
= = /2
010
1
0
00
001
10
2
1
j
j
j
j
S
0110
1
1
0
001
001
110
2
1S
=0 , =
= = 1 / 2
= = 1 / 2
T-junction power divider
E-plane TH-plane T
Microstrip T
T-model
jB
Z 1
Z 2
V o
Y in
21
11
ZZjBYin
21
11
ZZYin
Lossy line
Lossless line
If Zo = 50,then for equally divided power, Z1 = Z2=100
Example• If source impedance equal to 50 ohm and the
power to be divided into 2:1 ratio. Determine Z1 and Z2
ino PZ
VP
3
1
2
1
1
2
1
ino PZ
VP
3
2
2
1
2
2
2 752
32
oZZ
15031 oZZ
o
oin Z
VP
2
2
1 50// 21 ZZZo
Resistive divider
V 2
V 3
V 1
Z o
Z o
P 1
P 2
P 3
Z o V
oo ZZ
Z 3
Zo/3Zo/3
Zo/3
ooo
in ZZZ
Z 3
2
3
VVZZ
ZV
oo
o
3
2
3/23/
3/21
VVVZZ
ZVV
oo
o
2
1
4
3
3/32
oin Z
VP
21
2
1
in
oP
Z
VPP
4
12/1
2
12
132
Wilkinson Power Divider
50
50
50
100 70.7
70.7
/4
Z o
/2 Z o
/2 Z o
2Z o
Z o
Z o
/4
2
2Te Z
Zin
oT ZZ 2
For even mode
Therefore
For Zin =Zo=50
7.70502TZ
And shunt resistor R =2 Zo = 100
Analysis (even and odd mode)
2
2
1
1
Port 1
Port 2
Port 3
V g2
V g3
Z
Z
4
+V 2
+V 3
r/2
r/2
4
For even mode Vg2 = Vg3 and for odd mode Vg2 = -Vg3. Since the circuit is symmetrical , we can treat separately two bisection circuit for even and odd modes as shown in the next slide. By superposition of these two modes , we can find S -parameter of the circuit. The excitation is effectively Vg2=4V and Vg3= 0V.
For simplicity all values are normalized to line characteristic impedance , I.e Zo = 50
Even modeVg2=Vg3= 2V
Looking at port 2 Zin
e= Z2/2Therefore for matching
2Z
then V2e= V since Zin
e=1 (the circuit acting like voltage divider)
2
1
Port 1
Port 2
2V
Z
4
+V 2e
r/2+V 1e
O.CO.C outinZZZ 2
Note:
2ZIf
To determine V2e , using transmission line equation V(x) = V+ (e-jx + e+jx) , thus
VjVVV e )1()4(2
1
1)1()0(1
jVjVVV e
Reflection at port 1, refer to is
22
22
2Z
Then 21 jVV e
Odd modeVg2= - Vg3= 2V
2
1
Port 1
Port 2
2V
Z
4
+V 2o
r/2+V 1o
At port 2, V1o =0 (short) ,
/4 transformer will be looking as open circuit , thus Zin
o = r/2 . We choose r =2 for matching. Hence V2
o= 1V (looking as a voltage divider)
S-parameters
S11= 0 (matched Zin=1 at port 1)
S22 = S33 = 0 (matched at ports 2 and 3 both even and odd modes)
S12 = S21 = 2/
22
11 jVV
VV
oe
oe
S13 = S31 = 2/j
S23 = S32 = 0 ( short or open at bisection , I.e no coupling)
Example
Design an equal-split Wilkinson power divider for a 50 W system impedance at frequency fo
The quarterwave-transformer characteristic is
7.702 oZZ
1002 oZR
r
o
4The quarterwave-transformer length is
Wilkinson splitter/combiner application
/4
100
70.7
50
matchingnetworks
/4
100 50
70.7
70.7
70.7
Splittercombiner
Power Amplifier
Unequal power Wilkinson Divider
3
2
031
K
KZZ o
)1( 203
202 KKZZKZ o
KKZR o
1
R 2=Z o/K
R
R 3=Z o/K
Z 02
Z 03
Z o
2
3
2
32
portatPower
portatPower
P
PK
1
2
3
Parad and Moynihan power divider
4/1
2011
K
KZZ o
2
3
2
32
portatPower
portatPower
P
PK
KKZR o
1 4/124/302 1 KKZZ o
4/5
4/12
031
K
KZZ o
KZZ o04 K
ZZ o05
Z o
Z o
Z oZ 05
Z o4Z o2
Z o3
Z o1
R1
2
3
Cohn power divider
VSWR at port 1 = 1.106VSWR at port 2 and port 3 = 1.021Isolation between port 2 and 3 = 27.3 dBCenter frequency fo = (f1 + f2)/2Frequency range (f2/f1) = 2
1
2
3
Couplers
/4
/4
Y o Y o
Y oY o
Y se
Y sh Y sh
Branch line coupler 2sh
2se Y1Y
2se
2sh
sh
2
3
YY1
2Y
E
E
20
1
3 10E
E x
x dB coupling
23
22
21 EEE
2
1
3
2
1
2
E
E
E
E1
or
E1E2
E3
Couplers
input
isolate
Output3dB
Output3dB 90o out of phase
3 dB Branch line coupler
/4
/4
Z o
Z o
Zo
Z o2/Z o
2/Z o
Z o Z o
32 EE
1Ysh
2Y1Y 22se sh
1.414Yse
50oZ
50shZ
5.35seZ
Couplers9 dB Branch line coupler
355.010 209
1
3
E
E
22
1
2 355.01
E
E
935.0355.01 2
1
2
E
E
38.0935.0
355.0
2
3
E
E
8.0shYLet say we choose
38.08.01
8.02
1
22222
sesesh
sh
YYY
Y
962.136.038.0
6.1seY
500Z
5.628.0/50shZ
5.25962.1/50seZ
Note: Practically upto 9dB coupling
Couplers
/4
/4
/4
/4
Input
Output in-phase
Output in-phase
isolated
1
2
3
4
•Can be used as splitter , 1 as input and 2 and 3 as two output. Port is match with 50 ohm.•Can be used as combiner , 2 and 3 as input and 1 as output.Port 4 is matched with 50 ohm.
Hybrid-ring coupler
OC
1
21
2
OC
1/2
1/2
2
2
2
2
2
2
/8
/8
/4
/4
/8
/8
T e
T o
e
o
AnalysisThe amplitude of scattered wave
oeB 2
1
2
11
oe TTB2
1
2
14
oe TTB2
1
2
12
oeB 2
1
2
13
Couple lines analysis
Planar Stacked
Coupled microstrip
bw ws
ws
w ws
b
d
r
r
r
The coupled lines are usually assumed to operate in TEM mode. The electrical characteristics can be determined from effective capacitances between lines and velocity of propagation.
Equivalent circuits
+V +V
H-wall
+V -V
E-wall
C 11C 22C 11
C 22
2C 122C 12
Even mode Odd mode
C11 and C22 are the capacitances between conductors and the ground respectively. For symmetrical coupled line C11=C22 . C12 is the capacitance between two strip of conductors in the absence of ground. In even mode , there is no current flows between two strip conductors , thus C12 is effectively open-circuited.
ContinueEven mode
The resulting capacitance Ce = C11 = C22
ee
e
eoe CC
LC
C
LZ
1
Therefore, the line characteristic impedance
Odd mode
The resulting capacitance Co = C11 + 2 C12 = C22 + 2 C12
Therefore, the line characteristic impedanceo
oo CZ
1
Planar coupled stripline
Refer to Fig 7.29 in Pozar , Microwave Engineering
Stacked coupled stripline
mFsb
b
sbsbC oWroWroWr /
4
2/2/ 2211
w >> s and w >> b
mFs
C oWr /12
mFsb
bCC oWr
e /4
2211
mFssb
bwCCC oro /
1222
221211
oor 1
ro
eoe
bw
sbZ
CZ
4
1 22
ssbbwZ
CZ
ro
ooo
/1/22
1122
Coupled microstripline
Refer to Fig 7.30 in Pozar , Microwave Engineering
Design of Coupled line Couplers
inputoutput
Isolated(can be matched)
Coupling
w
w
s
2
3 4
1
w c
/4
3 4
1 2
Z o
Z o Z o
Z o
Z ooZ oe
2V
+V 3
+V 2
+V 4
+V 1
I1
I4I3
I2Schematic circuit
Layout
Even and odd modes analysis
3 4
1 2
Z o
Z o Z o
Z o
Z oo
V
+V 3o
+V 2o
+V 4o
+V 1o
I1o
I4oI3
o
I2o
V
_
+
+
_
3 4
1 2
Z o
Z o Z o
Z o
Z oe
V
+V 3e
+V 2e
+V 4e
+V 1e
I1e
I4eI3
e
I2e
V_
+
+
_
I1e = I3
e
I4e = I2
e
Same excitation voltage
V1e = V3
e
V4e = V2
e
Even
I1o = -I3
o
I4o =- I2
o
V1o = -V3
o
V4o = -V2
o
Odd
Reverse excitation voltage
(100)
(99)
Analysis
ooin
oino
ZZ
ZVV
1
tan
tan
ooe
oeooe
ein jZZ
jZZZZ
oe
oe
inII
VV
I
VZ
11
11
1
1
Zo = load for transmission line = electrical length of the lineZoe or Zoo = characteristic impedance of the line
tan
tan
ooo
ooooo
oin jZZ
jZZZZ
By voltage division
oein
eine
ZZ
ZVV
1
ooin
o
ZZ
VI
1
oein
e
ZZ
VI
1
From transmission line equation , we have
where
(101)
(102)
(103)
(104)
(105)
(106)
(107)
continueSubstituting eqs. (104) - (107) into eq. (101) yeilds
o
oin
ein
oein
oin
oo
oin
ein
ooin
eino
ein
oin
inZZZ
ZZZZ
ZZZ
ZZZZZZZ
2
2
2
2
For matching we may consider the second term of eq. (108) will be zero , I.e
02 oein
oin ZZZ or 2
ooeooein
oin ZZZZZ
(108)
Let oeooo ZZZ
Therefore eqs. (102) and (103) become
tan
tan
oooe
oeoooe
ein
ZjZ
ZjZZZ
tan
tan
oeoo
oooeoo
oin
ZjZ
ZjZZZ
and (108) reduces to Zin=Zo
(110)(109)
continueSince Zin = Zo , then by voltage division V1 = V. The voltage at port 3, by substitute (99), (100) , (104) and (105) is then
ooin
oin
oein
einoeoe
ZZ
Z
ZZ
ZVVVVVV 11333 (111)
Substitute (109) and (110) into (111)
tan2
tan
oooeo
ooo
ooin
oin
ZZjZ
jZZ
ZZ
Z
tan2
tan
oooeo
oeo
oein
ein
ZZjZ
jZZ
ZZ
Z
Then (111) reduces to
tan2
tan3
oooeo
oooeZZjZ
ZZjVV
(112)
continueWe define coupling as
oooe
oooeZZ
ZZC
Then V3 / V , from ( 112) will become
oooe
oZZ
ZC
21 2
tan1
tan
tan2
tan
23
jC
jCV
ZZ
ZZj
ZZ
ZZZ
ZZj
VV
oooe
oooe
oooe
o
oooe
oooe
and
sincos1
1
2
2
222jC
CVVVV oe
022444 oeoe VVVVVSimilarly
V1=V
Practical couple line couplerV3 is maximum when = /2 , 3/2, ...
Thus for quarterwave length coupler = /2 , the eqs V2 and V3 reduce to
V1=V
04 V
VC
jC
jCV
jC
jCV
jC
jCVV
2223
11
)(
2/tan1
2/tan
22
2
2
2 11
2/sin2/cos1
1CjV
j
CV
jC
CVV
C
CZZ ooe
1
1
C
CZZ ooo
1
1
ExampleDesign a 20 dB single-section coupled line coupler in stripline with a 0.158 cm ground plane spacing , dielectric constant of 2. 56, a characteristic impedance of 50 , and a center frequency of 3 GHz.
Coupling factor is C = 10-20/20 = 0.1
Characteristic impedance of even and odd mode are
28.551.01
1.0150oeZ
23.451.01
1.0150
ooZ
4.88oer Z
4.72oor Z
From fig 7.29 , we have w/b=0.72 , s/b =0.34. These give us w=0.72b=0.114cms= 0.34b = 0.054cm
Then multiplied by r
Multisection Coupled line coupler (broadband)
V 1
V 3 V 4
V 2
input Through
IsolatedCoupled
C 1C N-2
C 3C 2 C NC N-1....
jejCj
jC
jC
jC
V
V
sintan1
tan
tan1
tan
21
3
jejC
C
V
V
sincos1
1
2
2
1
2
For single section , whence C<<1 , then
V4=0
and For =/ 2 then V3/V1= C and V2/V1 = -j
AnalysisResult for cascading the couplers to form a multi section coupler is
)1(2
1
212113
sin...
sinsin
Njj
N
jjj
eVejC
eVejCVejCV
)1(
)2(222
)1(2113
...
1sin
Nj
M
NjjNjj
eC
eeCeCejVV
M
jN
C
NCNCejV
2
1...
3cos1cossin2 211
Where M= (N+1)/2
For symmetry C1=CN , C2= CN-1 , etc
At center frequency 2/1
3
V
VCo
(200)
ExampleDesign a three-section 20 dB coupler with binomial response (maximally flat), a system impedance 50 , and a center frequency of 3 GHz .
Solution
For maximally flat response for three section (N=3) coupler, we require
2,10)(
2/
nforCd
dn
n
From eq (200) and M= (N+1)/2 =( 3+1)/2=2 , we have
21
1
32
12cossin2 CC
V
VC
sin)(3sinsinsin3sin 12121 CCCCC
(201)
(202)
ContinueApply (201)
0cos)(3cos32/
121
CCCd
dC
010sin)(3sin9 21
2/
1212
2
CCCCCd
Cd
Midband Co= 20 dB at =/2. Thus C= 10-20/20=0.1
From (202), we C= C2 - 2C1= 0.1 © ©
©
Solving © and © © gives us C1= C3 = 0.0125 (symmetry) and C2 = 0.125
continueUsing even and odd mode analysis, we have
63.500125.01
0125.0150
1
131 C
CZZZ ooeoe
38.490125.01
0125.0131 ooooo ZZZ
69.56125.01
125.0150
1
12 C
CZZ ooe
1.44125.01
125.012 ooo ZZ
continueLet say , r = 10 and d =0.7878mm
63.5031 oeoe ZZ 38.4931 oooo ZZ
69.562oeZ 1.442ooZ
Plot points on graph Fig. 7.30
We have , w/d = 1.0 and s/d = 2.5 , thus
w = d = 0.7878mm and s = 2.5d = 1.9695mm
Similarly we plot points
We have , w/d = 0.95 and s/d = 1.1 , thus
w = 0.95d = 0.748mm and s =1.1d = 0.8666mm
For section 1 and 3
For section 2
Couplers
Lange Coupler
Evolution of Lange coupler
1= input2=output3=coupling4=isolated
w
w
w
w
ws
s
ss
1
4 3
2
1
34
2
1
2
3
4
2
41
3
Analysis
1
4 3
2
1
34
2
C C
90 o
Z e4 Z o4
Z o4Z e4
1
4321
2C m
C exC ex
C
C ex C exC inC in
C mC mC m
Simplified circuit Equivalent circuit
mex
mexexin CC
CCCC
where
Continue/ 4 wire couplerEven mode
All Cm capacitance will be at same potential, thus the total capacitance is
inexe CCC 4
minexo CCCC 64
Odd mode
All Cm capacitance will be considered, thus the total capacitance is
Even and Odd mode characteristic impedance
44
1
ee C
Z
4
41
oo C
Z
lineontransmissiinvelocity
(300)
(301)
(302)
continueNow consider isolated pairs. It’s equivalent circuit is same as two wire line , thus it’s even and odd mode capacitance is
exe CC
mexo CCC 2
Substitute these into (300) and (301) , we have
oe
oeee CC
CCCC
3
4
mex
mexexin CC
CCCC
oe
eooo CC
CCCC
3
4
And in terms of impedance refer to (302)
oeoeoo
oeooe Z
ZZ
ZZZ
34
oooooe
oeooo Z
ZZ
ZZZ
34
continue
oooeoeoo
oeoooooeoeo ZZZZ
ZZZZZZZ
33
2
44
Characteristic impedance of the line is
oooeoooe
oooe
oe
oe
ZZZZ
ZZ
ZZ
ZZC
23
322
22
44
44
Coupling
The desired characteristic impedance in terms of coupling is
ooe ZCCC
CCZ
1/12
8934 2
ooo ZCCC
CCZ
1/12
8934 2
VHF/UHF Hybrid power splitter
50input
50output
50output
100 C
T1
T21
5
6
7
8
2
3
4
Guanella power divider (VHF/UHF)
R L
V 2
I2
I1
V 1
R g
V g I1
V 2
I2