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Power divider, combiner andcoupler
ByProfessor Syed Idris Syed HassanSch of Elect. & Electron EngEngineering Campus USM
Nibong Tebal 14300SPS Penang
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Power divider and combiner/coupler
divider combinerP1
P2= nP1
P3=(1-n)P 1
P1
P2
P3=P1+P2
Divide into 4 output
Basic
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S-parameter for power divider/coupler
333231
232221
131211
SSS
SSSSSS
SGenerally
For reciprocal and lossless network
ji for SS N
k kjki 0
1
*1
1
* N
k ki ki
SS
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
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ContinueIf all ports are matched properly , then S ii= 0
0
00
2313
2312
1312
SS
SSSS
S
For Reciprocalnetwork For lossless network, must satisfy unitary
condition
12
132
12 SS
1223
212 SS
1223
213 SS
012*23
SS
023*13SS
013
*
12SS
Two of (S 12, S13, S23) must be zero but it is not consistent. If S 12=S13= 0, thenS23 should equal to 1 and the first equation will not equal to 1. This is invalid.
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Another alternative for reciprocal network
332313
2312
13120
0
SSS
SS
SS
S
Only two ports are matched , then for reciprocal network
For lossless network, must satisfy unitarycondition
12
13
2
12SS
1223
212 SS
1233
223
213 SSS 013
*3312
*23 SSSS
023
*
13SS
033*2313
*12 SSSS
The two equations showthat |S 13|=|S23|
thus S 13=S23=0and |S 12|=|S33|=1
These have satisfied all
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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
ee
S00
0000
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For lossless network, must satisfy unitarycondition
1213
212 SS
12
23
2
21SS
1232
231 SS
032*31
SS
023
*
21SS
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
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Circulator (nonreciprocal network)
010
001
100
S
001
100
010
S
1
2
3
1
2
3
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Four port network
44434241
34
24
14
333231
232221
131211
SSSS
SS
S
SSSSSS
SSS
SGenerally
For reciprocal and lossless network
ji for SS
N
k kjki 0
1
*
11
* N
k ki ki
SS
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
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Matched Four port network
0
0
00
342414
34
24
14
2313
2312
1312
SSS
S
SS
SS
SSSS
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
*
1413SSSS
Say all ports are matched and symmetrical network, then
*
**
@@@
###
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To check validityMultiply eq. * by S
24
* and eq. ## by S13
* , and substract to obtain
0214
213
*14
SSS
Multiply eq. # by S 34 and eq. @@ by S 13 , and substract to obtain
0234
21223
SSS
%
$
Both equations % and $ will be satisfy if S 14 = S23 = 0 . This meansthat no coupling between port 1 and 4 , and between port 2 and 3 ashappening in most directional couplers.
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Directional coupler
00
0
00
000
3424
34
24
13
12
1312
SS
S
S
S
SSS
S
If all ports matched , symmetry and S 14=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 SSBy comparing these equations yield
*
*
**
**
By comparing equations * and ** yield 3412 SS
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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 log20
3
1Coupling= C= 10 log
dBSP
P
144
3 log20
Isolation = I= 10 log dBSPP
14
4
1 log20
I = D + C dB
1
4 3
2Input Through
CoupledIsolated
For ideal case|S
14|=0
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Practical coupler
Hybrid 3 dB couplers
Magic -T and Rat-race couplers
= = p /2
010
1
0
00
001
10
2
1
j
j
j
j
S
0110
1
1
0
001
001
110
2
1S
=0 , =p
= = 1 / 2
= = 1 / 2
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T-junction power divider
E-plane TH-plane T
Microstrip T
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T-model
jB
Z1
Z2
Vo
Yin
21
11 Z Z
jBY in
21
11 Z Z
Y in
Lossy line
Lossless line
If Zo = 50,then for equallydivided power, Z 1 = Z2=100
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Example• If source impedance equal to 50 ohm and the
power to be divided into 2:1 ratio. Determine Z 1 and Z 2
ino P
Z V P
31
21
1
21
ino P
Z
V P
3
2
2
1
2
2
2 752
32
o Z Z
15031 o Z Z
o
oin Z
V P
2
21 50 // 21 Z Z Z o
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Resistive divider
V2
V3
V1
Zo
Zo
P 1
P 2
P 3
Zo V
oo Z
Z Z
3
Zo /3Zo /3
Zo /3
ooo
in Z Z Z
Z 3
23
V V Z Z
Z V
oo
o
32
3 / 23 / 3 / 2
1
V V V Z Z
Z V V
oo
o
2
1
4
3
3 / 32
oin Z
V P
21
21
in
oP
Z
V PP
412 / 1
21 21
32
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Wilkinson Power Divider
50
50
50
10070.7
70.7
/4
Zo
/2 Zo
/2 Z o
2Z o
Zo
Zo
/4
2
2T e Z
Z in
oT Z Z 2
For even mode
Therefore
For Z in =Zo=50
7.70502T
Z
And shunt resistor R =2 Z o = 100
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Analysis (even and odd mode)
2
2
1
1
Port 1
Port 2
Port 3
Vg2
Vg3
Z
Z
4
+V2
+V3
r/2
r/2
4
For even mode V g2 = Vg3 andfor odd mode V g2 = -V g3. Sincethe circuit is symmetrical , we
can treat separately twobisection circuit for even andodd modes as shown in the nextslide. By superposition of thesetwo modes , we can find S -
parameter of the circuit. Theexcitation is effectively V g2=4Vand V g3= 0V.
For simplicity all values arenormalized to line characteristicimpedance , I.e Z o = 50 .
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Even modeVg2=Vg3= 2V
Looking at port 2Zin
e= Z2 /2Therefore for matching
2 Z
then V 2e= V since Z in
e=1 (the circuit acting like voltage divider)
2
1
Port 1
Port 2
2V
Z 4
+V2e
r/2+V1e
O.CO.Cout in Z Z Z
2
Note:
2 Z If
To determine V 2e , using transmission line equation V(x) = V+ (e -j x + Ge+j x) , thus
V jV V V e G)1()4(2
11
)1()0(1 GGG jV jV V V e
Reflection at port 1, refer to is
22
22G
2 Z
Then 21
jV V e
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Odd modeVg2= - Vg3= 2V
2
1
Port 1
Port 2
2V
Z 4
+V2o
r/2+V1o
At port 2, V 1o =0 (short) ,
/4 transformer will belooking as open circuit ,thus Z in
o = r/2 . We chooser =2 for matching. HenceV2
o= 1V (looking as a
voltage divider)S-parameters
S11= 0 (matched Z in=1 at port 1)
S22 = S33 = 0 (matched at ports 2 and 3 both even and odd modes)
S12 = S21 = 2 / 22
11 jV V
V V oe
oe
S13 = S31 = 2 / j
S23
= S32
= 0 ( short or open at bisection , I.e nocoupling)
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Example
Design an equal-split Wilkinson power divider for a 50 W systemimpedance at frequency fo
The quarterwave-transformer characteristic is
7.702 o Z Z
1002 o Z R
r
o
4The quarterwave-transformer length is
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Wilkinson splitter/combiner
application
/4
100
70.7
50
matchingnetworks
/4
100 50
70.7
70.7
70.7
Splittercombiner
Power Amplifier
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Unequal power Wilkinson
Divider3
2
031
K
K Z Z o
)1( 203
202 K K Z Z K Z o
K K Z R o
1
R2=Zo /K
R
R3=Zo /K
Z02
Z03
Zo
23
2
32
port at Power port at Power
PP
K
1
2
3
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Parad and Moynihan power divider
4 / 1
2011
K K Z Z o
23
2
32
port at Power port at Power
PP
K
K
K Z R o1 4 / 124 / 3
02 1 K K Z Z o
4 / 5
4 / 12
031
K
K Z Z o
K Z Z o04 K Z Z o
05
Zo
Zo
ZoZ05
Zo4Zo2
Zo3
Zo1
R12
3
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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 f o = (f 1 + f 2)/2Frequency range (f
2 /f
1) = 2
1
2
3
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Couplers
/4
/4
Yo Yo
YoYo
Yse
Ysh Ysh
Branch line coupler 2sh
2se Y1Y
2se
2sh
sh
2
3
YY1
2YEE
20
1
3 10EE x
x dB coupling
23
22
21 EEE
2
1
3
2
1
2
E
E
E
E1
or
E1 E2
E3
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Couplers
input
isolate
Output3dB
Output3dB 90 o out of phase
3 dB Branch line coupler
/4
/4
Zo
Zo
Zo
Zo2 / Z o
2 / Z o
Zo Zo
32 EE
1Ysh
2Y1Y 22se sh
1.414Yse
50o Z
50sh Z
5.35se Z
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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
38.0935.0355.0
2
3
E E
8.0shY Let say we choose
38.0
8.01
8.02
1
22222
sesesh
sh
Y Y Y
Y
962.136.038.06.1
seY
500 Z
5.628.0 / 50sh Z
5.25962.1 / 50se Z
Note: Practically upto 9dB coupling
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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 3as two output. Port is match with 50 ohm.•Can be used as combiner , 2 and 3 as inputand 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
Te
To
Ge
Go
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AnalysisThe amplitude of scattered wave
oe B GG21
21
1
oe T T B21
21
4
oe T T B21
21
2
oe B GG21
21
3
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Couple lines analysis
Planar Stacked
Coupled microstrip
bw ws w
s
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 effectivecapacitances between lines and velocity of propagation.
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Equivalent circuits
+V +V
H-wall
+V-V
E-wall
C11C22C11 C22
2C 122C 12
Even mode Odd mode
C11 and C 22 are the capacitances between conductors and the groundrespectively. For symmetrical coupled line C 11=C22 . C12 is thecapacitance between two strip of conductors in the absence of ground. Ineven mode , there is no current flows between two strip conductors , thusC12 is effectively open-circuited.
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ContinueEven mode
The resulting capacitance C e = C11 = C22
ee
e
eoe C C
LC
C L
Z 1
Therefore, the line characteristic impedance
Odd mode
The resulting capacitance C o = C11 + 2 C 12 = C22 + 2 C 12
Therefore, the line characteristic impedanceo
oo C Z
1
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Planar coupled stripline
Refer to Fig 7.29 in Pozar , Microwave Engineering
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Stacked coupled stripline
mF
sbb
sbsbC oW r oW r oW r / 4
2 / 2 / 2211
w >> s and w >> b
mF s
C oW r / 12
mF sb
bC C oW r
e / 4
2211
mF ssb
bwC C C or o /
1222
221211
oor 1
r o
eoe
bwsb Z
C Z
41
22
ssbbw Z
C Z
r o
ooo
/ 1 / 22
1122
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Coupled microstripline
Refer to Fig 7.30 in Pozar , Microwave Engineering
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Design of Coupled line Couplers
input output
Isolated
(can be matched)
Coupling
w
w
s
2
3 4
1
wc
/4
3 4
1 2
Zo
Zo Zo
Zo
ZooZoe
2V
+V3
+V2
+V4
+V1
I1
I4I3
I2Schematic circuit
Layout
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Even and odd modes analysis
3 4
1 2
Zo
Zo Zo
Zo
Zoo
V
+V3o
+V2o
+V4o
+V1o
I1o
I4oI3
o
I2oV
_
++
_
3 4
1 2
Zo
Zo Zo
Zo
Zoe
V
+V3e
+V2e
+V4e
+V1e
I1e
I4eI3
e
I2e
V _ +
+
_
I1
e = I3
e
I4e = I2
e Sameexcitationvoltage
V1e = V3
e
V4e = V2e
Even
I1o = -I3
o
I4o =- I2
o
V1o = -V3
o
V4o = -V2
o
Odd
Reverse
excitationvoltage
(100)
(99)
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Analysis
ooin
oino
Z Z
Z V V 1
tan
tan
ooe
oeooe
ein jZ Z
jZ Z
Z Z
oe
oe
in I I
V V
I V
Z 11
11
1
1
Zo = load for transmission line= electrical length of the line
Zoe or Zoo = characteristic impedance of the line
tantan
ooo
ooooo
oin jZ Z
jZ Z Z Z
By voltage division
oein
eine
Z Z
Z V V 1
ooin
o
Z Z
V I 1
oein
e
Z Z
V I 1
From transmission line equation , we have
where
(101)
(102)
(103)
(104)
(105)
(106)
(107)
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continueSubstituting eqs. (104) - (107) into eq. (101) yeilds
ooin
ein
oein
oin
oo
oin
ein
ooin
eino
ein
oin
in Z Z Z
Z Z Z Z
Z Z Z
Z Z Z Z Z Z Z
2
2
2
2
For matching we may consider the second term of eq. ( 108) will be zero , I.e
02o
ein
oin Z Z Z or 2
ooeooein
oin Z Z Z Z Z
(108)
Let oeooo Z Z Z
Therefore eqs. (102) and (103) become
tan
tan
oooe
oeoooe
ein
Z j Z
Z j Z Z Z
tan
tan
oeoo
oooeoo
oin
Z j Z
Z j Z Z Z
and (108) reduces to Z in=Zo
(110) (109)
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continueSince Z in = Zo , then by voltage division V1 = V. The voltage at port 3, bysubstitute ( 99), (100) , (104) and ( 105) is then
ooin
oin
oein
einoeoe
Z Z
Z
Z Z
Z V V V V V V 11333 (111 )
Substitute (109) and (110) into (111)
tan2
tan
oooeo
ooo
ooin
oin
Z Z j Z
jZ Z
Z Z
Z
tan2
tan
oooeo
oeo
oein
ein
Z Z j Z
jZ Z
Z Z
Z
Then (111) reduces to
tan2
tan3
oooeo
oooe Z Z j Z
Z Z jV V (112 )
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continueWe define coupling as
oooe
oooe
Z Z
Z Z C
Then V 3 / V , from ( 112) will become
oooe
o Z Z
Z C
21 2
tan1
tan
tan2
tan
23
jC
jC V
Z Z Z Z
j Z Z
Z Z Z Z Z
jV V
oooe
oooe
oooe
o
oooe
oooe
and
sincos1
12
2
222 jC
C V V V V oe
022444oeoe V V V V V Similarly
V1=V
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Practical couple line couplerV3 is maximum when = p /2 , 3p /2, ...
Thus for quarterwave length coupler = p /2 , the eqs V 2 and V 3 reduce to
V1=V
04V
VC
jC jC V
jC jC V
jC jC V V
2223
11)(
2 / tan12 / tan
p p
22
2
2
2 11
2 / sin2 / cos1
1C jV
jC
V jC
C V V
p p C C
Z Z ooe 11
C C
Z Z ooo 11
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ExampleDesign a 20 dB single-section coupled line coupler in stripline with a 0.158 cmground plane spacing , dielectric constant of 2. 56, a characteristic impedanceof 50 , and a center frequency of 3 GHz.
Coupling factor is C = 10 -20/20 = 0.1
Characteristic impedance of evenand odd mode are
28.55
1.01
1.0150oe Z
23.451.011.01
50oo Z
4.88oer Z
4.72oor Z
From fig 7.29 , we havew/b=0.72 , s/b =0.34. Thesegive usw=0.72b=0.114cms= 0.34b = 0.054cm
Then multiplied by r
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Multisection Coupled line coupler (broadband)
V1
V3 V4
V2
input Through
IsolatedCoupled
C1CN-2C3C2 CN
CN-1....
je jC j
jC jC
jC V V sin
tan1tan
tan1tan
213
je jC
C V V
sincos1
12
2
1
2
For single section , whence C<<1 , then
V4=0
and For = p / 2 then V 3 /V1= C
and V 2 /V1 = -j
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AnalysisResult for cascading the couplers to form a multi section coupler is
)1(21
212113
sin...
sinsin
N j j N
j j j
eV e jC
eV e jC V e jC V
)1(
)2(222
)1(2113
...
1sin
N j M
N j j N j j
eC
eeC eC e jV V
M
jN
C
N C N C e jV
21
...
3cos1cossin2 211
Where M= (N+1)/2
For symmetry C 1=CN , C2= CN-1 ,etc
At center frequency2 / 1
3
p V V
C o
(200)
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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 /
n for C d
d n
n
p
From eq (200) and M= (N+1)/2 =( 3+1)/2=2 , we have
211
321
2cossin2 C C V V
C
sin)(3sinsinsin3sin 12121 C C C C C
(201)
(202)
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ContinueApply (201)
0cos)(3cos32 /
121p
C C C d dC
010sin)(3sin9 212 /
1212
2
C C C C C d
C d
p
Midband C o= 20 dB at =p /2. Thus C= 10 -20/20 =0.1
From (202), we C= C 2 - 2C1= 0.1 © ©
©
Solving © and © © gives us C 1= C3 = 0.0125 (symmetry) and C 2 = 0.125
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continue
Using even and odd mode analysis, we have
63.500125.010125.01
5011
31 C C
Z Z Z ooeoe
38.490125.010125.01
31 ooooo Z Z Z
69.56125.01
125.0150
1
12
C
C Z Z ooe
1.44125.01125.01
2 ooo Z Z
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continueLet say , r = 10 and d =0.7878mm
63.5031 oeoe Z Z 38.4931 oooo Z Z
69.562oe Z 1.442oo Z
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
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CouplersLange Coupler
Evolution of Lange
coupler1= input2=output3=coupling4=isolated
w
w
w
w
ws
s
s
s
1
4 3
2
1
34
2
1
2
3
4
2
41
3
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Analysis
1
4 3
2
1
34
2 C C
90 o
Ze4 Zo4
Zo4Ze4
1
4321
2C
m
Cex
Cex
C
Cex C
exC
inCin
CmC
mC
m
Simplified circuit Equivalent circuit
mex
mexexin C C
C C C C where
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Continue/ 4 wire couplerEven mode
All C m capacitance will be at same potential, thus the total capacitance is
inexe C C C 4
minexo C C C C 64
Odd modeAll C m 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)
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continue Now consider isolated pairs. It’s equivalent circuit is same as two wire line ,
thus it’s even and odd mode capacitance is
exe C C
mexoC C C 2
Substitute these into (300) and (301) ,we have
oe
oeee C C
C C C C
3
4
mex
mexexin C C
C C C C
oe
eooo C C
C C C C
34
And in terms of impedance referto (302)
oeoeoo
oeoo
eZ
Z Z
Z Z Z
34
oooooe
oeooo Z
Z Z
Z Z Z
34
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continue
oooeoeoo
oeoooooeoeo Z Z Z Z
Z Z Z Z Z Z Z
33
244
Characteristic impedance of the line is
oooeoooe
oooe
oe
oe
Z Z Z Z Z Z
Z Z Z Z C
233
22
22
4444
Coupling
The desired characteristic impedance in terms of coupling is
ooe Z C C C
C C Z 1 / 128934
2
ooo Z C C C
C C Z
1 / 12
8934 2
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VHF/UHF Hybrid power splitter
50input
50output
50output
100C
T1
T21
5
67
8
23
4
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Guanella power divider
(VHF/UHF)
RL
V2
I2
I1
V1
Rg
Vg I1
V2
I2