1. Power divider, combiner and
coupler
By
Professor Syed Idris Syed Hassan
Sch of Elect. & Electron Eng
Engineering Campus USM
Nibong Tebal 14300
SPS Penang
2. Power divider and combiner/coupler
divider combiner
P1
P2= nP1
P3=(1-n)P1
P1
P2
P3=P1+P2
Divide into 4 output
Basic
3. S-parameter for power divider/coupler
33
32
31
23
22
21
13
12
11
S
S
S
S
S
S
S
S
S
S
Generally
For reciprocal and lossless network
j
i
for
S
S
N
k
kj
ki
0
1
*
1
1
*
N
k
ki ki
S
S
1
13
12
11
S
S
S
1
23
22
21
S
S
S
1
33
32
31
S
S
S
0
*
23
13
*
22
12
*
21
11
S
S
S
S
S
S
0
*
33
23
*
32
22
*
31
21
S
S
S
S
S
S
0
*
33
13
*
32
12
*
31
11
S
S
S
S
S
S
Row 1x row 2
Row 2x row 3
Row 1x row 3
4. Continue
If all ports are matched properly , then Sii= 0
0
0
0
23
13
23
12
13
12
S
S
S
S
S
S
S
For Reciprocal
network
For lossless network, must satisfy unitary
condition
1
2
13
2
12
S
S
1
2
23
2
12
S
S
1
2
23
2
13
S
S
0
12
*
23
S
S
0
23
*
13
S
S
0
13
*
12
S
S
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.
5. Another alternative for reciprocal network
33
23
13
23
12
13
12
0
0
S
S
S
S
S
S
S
S
Only two ports are matched , then for reciprocal network
For lossless network, must satisfy unitary
condition
1
2
13
2
12
S
S
1
2
23
2
12
S
S
1
2
33
2
23
2
13
S
S
S 0
13
*
33
12
*
23
S
S
S
S
0
23
*
13
S
S
0
33
*
23
13
*
12
S
S
S
S
The two equations show
that |S13|=|S23|
thus S13=S23=0
and |S12|=|S33|=1
These have satisfied all
6. Reciprocal lossless network of two matched
S21
=ej
S12=ej
S33
=ej
1
3
2
j
j
j
e
e
e
S
0
0
0
0
0
0
7. For lossless network, must satisfy unitary
condition
1
2
13
2
12
S
S
1
2
23
2
21
S
S
1
2
32
2
31
S
S
0
32
*
31
S
S
0
23
*
21
S
S
0
13
*
12
S
S
Nonreciprocal network (apply for circulator)
0
0
0
32
31
23
21
13
12
S
S
S
S
S
S
S
0
31
23
12
S
S
S
0
13
32
21
S
S
S
1
13
32
21
S
S
S
1
31
23
12
S
S
S
The above equations must satisfy the following either
or
9. Four port network
44
43
42
41
34
24
14
33
32
31
23
22
21
13
12
11
S
S
S
S
S
S
S
S
S
S
S
S
S
S
S
S
S
Generally
For reciprocal and lossless network
j
i
for
S
S
N
k
kj
ki
0
1
*
1
1
*
N
k
ki ki
S
S
1
14
13
12
11
S
S
S
S
1
24
23
22
21
S
S
S
S
1
34
33
32
31
S
S
S
S
0
*
24
14
*
23
13
*
22
12
*
21
11
S
S
S
S
S
S
S
S
0
*
44
24
*
43
23
*
42
22
*
41
21
S
S
S
S
S
S
S
S
0
*
34
14
*
33
13
*
32
12
*
31
11
S
S
S
S
S
S
S
S
R 1x R 2
R 2x R3
R1x R4
1
44
43
42
41
S
S
S
S
0
*
44
14
*
43
13
*
42
12
*
41
11
S
S
S
S
S
S
S
S
0
*
34
24
*
33
23
*
32
22
*
31
21
S
S
S
S
S
S
S
S
0
*
44
34
*
43
33
*
42
32
*
41
31
S
S
S
S
S
S
S
S
R1x R3
R2x R4
R3x R4
10. Matched Four port network
0
0
0
0
34
24
14
34
24
14
23
13
23
12
13
12
S
S
S
S
S
S
S
S
S
S
S
S
S
The unitarity condition become
1
14
13
12
S
S
S
1
24
23
12
S
S
S
1
34
23
13
S
S
S
0
*
24
14
*
23
13
S
S
S
S
0
*
34
23
*
14
12
S
S
S
S
0
*
34
14
*
23
12
S
S
S
S
1
34
24
14
S
S
S
0
*
34
13
*
24
12
S
S
S
S
0
*
34
24
*
13
12
S
S
S
S
0
*
24
23
*
14
13
S
S
S
S
Say all ports are matched and symmetrical network, then
*
**
@
@@
#
##
11. To check validity
Multiply eq. * by S24
* and eq. ## by S13
* , and substract to obtain
0
2
14
2
13
*
14
S
S
S
Multiply eq. # by S34 and eq. @@ by S13 , and substract to obtain
0
2
34
2
12
23
S
S
S
%
$
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.
14. Physical interpretation
|S13 | 2 = coupling factor = 2
|S12 | 2 = power deliver to port 2= 2 =1- 2
Characterization of coupler
Directivity= D= 10 log
dB
P
P
log
20
3
1
Coupling= C= 10 log
dB
S
P
P
14
4
3 log
20
Isolation = I= 10 log dB
S
P
P
14
4
1 log
20
I = D + C dB
1
4 3
2
Input Through
Coupled
Isolated
For ideal case
|S14|=0
18. Example
• If source impedance equal to 50 ohm and the
power to be divided into 2:1 ratio. Determine Z1
and Z2
in
o P
Z
V
P
3
1
2
1
1
2
1
in
o P
Z
V
P
3
2
2
1
2
2
2
75
2
3
2
o
Z
Z
150
3
1 o
Z
Z
o
o
in
Z
V
P
2
2
1
50
// 2
1 Z
Z
Zo
19. Resistive divider
V2
V3
V1
Zo
Zo
P1
P2
P3
Zo V
o
o Z
Z
Z
3
Zo/3
Zo/3
Zo/3
o
o
o
in Z
Z
Z
Z
3
2
3
V
V
Z
Z
Z
V
o
o
o
3
2
3
/
2
3
/
3
/
2
1
V
V
V
Z
Z
Z
V
V
o
o
o
2
1
4
3
3
/
3
2
o
in
Z
V
P
2
1
2
1
in
o
P
Z
V
P
P
4
1
2
/
1
2
1
2
1
3
2
20. Wilkinson Power Divider
50
50
50
100
70.7
70.7
/4
Zo
/2 Zo
/2 Zo
2Zo
Zo
Zo
/4
2
2
T
e Z
Zin
o
T Z
Z 2
For even mode
Therefore
For Zin =Zo=50
7
.
70
50
2
T
Z
And shunt resistor R =2 Zo = 100
21. 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 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 .
22. Even mode
Vg2=Vg3= 2V
Looking at port 2
Zin
e= Z2/2
Therefore for matching
2
Z
then V2
e= V since Zin
e=1 (the circuit acting like voltage divider)
2
1
Port 1
Port 2
2V
Z
4
+V2
e
r/2
+V1
e
O.C
O.C out
inZ
Z
Z
2
Note:
2
Z
If
To determine V2
e , using transmission line equation V(x) = V+ (e-jx + Ge+jx) , thus
V
jV
V
V e
G
)
1
(
)
4
(
2
1
1
)
1
(
)
0
(
1
G
G
G
jV
jV
V
V e
Reflection at port 1, refer to is
2
2
2
2
G
2
Z
Then 2
1 jV
V e
23. Odd mode
Vg2= - Vg3= 2V
2
1
Port 1
Port 2
2V
Z
4
+V2
o
r/2
+V1
o
At port 2, V1
o =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
/
2
2
1
1 j
V
V
V
V
o
e
o
e
S13 = S31 = 2
/
j
S23 = S32 = 0 ( short or open at bisection , I.e no
coupling)
24. Example
Design an equal-split Wilkinson power divider for a 50 W system
impedance at frequency fo
The quarterwave-transformer characteristic is
7
.
70
2 o
Z
Z
100
2 o
Z
R
r
o
4
The quarterwave-transformer length is
26. Unequal power Wilkinson
Divider
3
2
03
1
K
K
Z
Z o
)
1
( 2
03
2
02 K
K
Z
Z
K
Z o
K
K
Z
R o
1
R2
=Zo
/K
R
R3=Zo/K
Z02
Z03
Zo
2
3
2
3
2
port
at
Power
port
at
Power
P
P
K
1
2
3
27. Parad and Moynihan power divider
4
/
1
2
01
1
K
K
Z
Z o
2
3
2
3
2
port
at
Power
port
at
Power
P
P
K
K
K
Z
R o
1
4
/
1
2
4
/
3
02 1 K
K
Z
Z o
4
/
5
4
/
1
2
03
1
K
K
Z
Z o
K
Z
Z o
04 K
Z
Z o
05
Zo
Zo
Zo
Z05
Zo4
Zo2
Zo3
Zo1
R
1
2
3
28. Cohn power divider
VSWR at port 1 = 1.106
VSWR at port 2 and port 3 = 1.021
Isolation between port 2 and 3 = 27.3 dB
Center frequency fo = (f1 + f2)/2
Frequency range (f2/f1) = 2
1
2
3
29. Couplers
/4
/4
Yo Yo
Yo
Yo
Yse
Ysh Ysh
Branch line coupler 2
sh
2
se Y
1
Y
2
se
2
sh
sh
2
3
Y
Y
1
2Y
E
E
20
1
3 10
E
E x
x dB coupling
2
3
2
2
2
1 E
E
E
2
1
3
2
1
2
E
E
E
E
1
or
E1
E2
E3
30. Couplers
input
isolate
Output
3dB
Output
3dB 90o out of phase
3 dB Branch line coupler
/4
/4
Zo
Zo
Zo
Zo
2
/
Zo
2
/
Zo
Zo Zo
3
2 E
E
1
Ysh
2
Y
1
Y 2
2
se
sh
1.414
Yse
50
o
Z
50
sh
Z
5
.
35
se
Z
31. Couplers
9 dB Branch line coupler
355
.
0
10 20
9
1
3
E
E
2
2
1
2 355
.
0
1
E
E
935
.
0
355
.
0
1 2
1
2
E
E
38
.
0
935
.
0
355
.
0
2
3
E
E
8
.
0
sh
Y
Let say we choose
38
.
0
8
.
0
1
8
.
0
2
1
2
2
2
2
2
se
se
sh
sh
Y
Y
Y
Y
962
.
1
36
.
0
38
.
0
6
.
1
se
Y
50
0
Z
5
.
62
8
.
0
/
50
sh
Z
5
.
25
962
.
1
/
50
se
Z
Note: Practically upto 9dB coupling
32. 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
2
1
2
OC
1/2
1/2
2
2
2
2
2
2
/8
/8
/4
/4
/8
/8
Te
To
Ge
Go
33. Analysis
The amplitude of scattered wave
o
e
B G
G
2
1
2
1
1
o
e T
T
B
2
1
2
1
4
o
e T
T
B
2
1
2
1
2
o
e
B G
G
2
1
2
1
3
34. Couple lines analysis
Planar Stacked
Coupled microstrip
b
w w
s
w
s
w w
s
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.
35. Equivalent circuits
+V +V
H-wall
+V -V
E-wall
C11
C22
C11
C22
2C12
2C12
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.
36. Continue
Even mode
The resulting capacitance Ce = C11 = C22
e
e
e
e
oe
C
C
LC
C
L
Z
1
Therefore, the line characteristic impedance
Odd mode
The resulting capacitance Co = C11 + 2 C12 = C22 + 2 C12
Therefore, the line characteristic impedance
o
oo
C
Z
1
38. Stacked coupled stripline
m
F
s
b
b
s
b
s
b
C oW
r
oW
r
oW
r /
4
2
/
2
/ 2
2
11
w >> s and w >> b
m
F
s
C oW
r /
12
m
F
s
b
b
C
C oW
r
e /
4
2
2
11
m
F
s
s
b
b
w
C
C
C o
r
o /
1
2
2
2
2
2
12
11
o
o
r
1
r
o
e
oe
bw
s
b
Z
C
Z
4
1 2
2
s
s
b
b
w
Z
C
Z
r
o
o
oo
/
1
/
2
2
1
1
2
2
40. 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
Zoo
Zoe
2V
+V3
+V2
+V4
+V1
I1
I4
I3
I2
Schematic circuit
Layout
41. Even and odd modes analysis
3 4
1 2
Zo
Zo Zo
Zo
Zoo
V
+V3
o
+V2
o
+V4
o
+V1
o
I1
o
I4
o
I3
o
I2
o
V
_
+
+
_
3 4
1 2
Zo
Zo Zo
Zo
Zoe
V
+V3
e
+V2
e
+V4
e
+V1
e
I1
e
I4
e
I3
e
I2
e
V
_
+
+
_
I1
e = I3
e
I4
e = I2
e
Same
excitation
voltage
V1
e = V3
e
V4
e = V2
e
Even
I1
o = -I3
o
I4
o =- I2
o
V1
o = -V3
o
V4
o = -V2
o
Odd
Reverse
excitation
voltage
(100)
(99)
43. continue
Substituting eqs. (104) - (107) into eq. (101) yeilds
o
o
in
e
in
o
e
in
o
in
o
o
o
in
e
in
o
o
in
e
in
o
e
in
o
in
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
0
2
o
e
in
o
in Z
Z
Z or 2
o
oe
oo
e
in
o
in Z
Z
Z
Z
Z
(108)
Let oe
oo
o Z
Z
Z
Therefore eqs. (102) and (103) become
tan
tan
oo
oe
oe
oo
oe
e
in
Z
j
Z
Z
j
Z
Z
Z
tan
tan
oe
oo
oo
oe
oo
o
in
Z
j
Z
Z
j
Z
Z
Z
and (108) reduces to Zin=Zo
(110)
(109)
44. continue
Since Zin = Zo , then by voltage division V1 = V. The voltage at port 3, by
substitute (99), (100) , (104) and (105) is then
o
o
in
o
in
o
e
in
e
in
o
e
o
e
Z
Z
Z
Z
Z
Z
V
V
V
V
V
V 1
1
3
3
3 (111)
Substitute (109) and (110) into (111)
tan
2
tan
oo
oe
o
oo
o
o
o
in
o
in
Z
Z
j
Z
jZ
Z
Z
Z
Z
tan
2
tan
oo
oe
o
oe
o
o
e
in
e
in
Z
Z
j
Z
jZ
Z
Z
Z
Z
Then (111) reduces to
tan
2
tan
3
oo
oe
o
oo
oe
Z
Z
j
Z
Z
Z
j
V
V
(112)
45. continue
We define coupling as
oo
oe
oo
oe
Z
Z
Z
Z
C
Then V3 / V , from ( 112) will become
oo
oe
o
Z
Z
Z
C
2
1 2
tan
1
tan
tan
2
tan
2
3
j
C
jC
V
Z
Z
Z
Z
j
Z
Z
Z
Z
Z
Z
Z
j
V
V
oo
oe
oo
oe
oo
oe
o
oo
oe
oo
oe
and
sin
cos
1
1
2
2
2
2
2
j
C
C
V
V
V
V o
e
0
2
2
4
4
4
o
e
o
e V
V
V
V
V
Similarly
V1=V
46. Practical couple line coupler
V3 is maximum when = p/2 , 3p/2, ...
Thus for quarterwave length coupler = p/2 , the eqs V2 and V3 reduce to
V1=V
0
4
V
VC
j
C
jC
V
j
C
jC
V
j
C
jC
V
V
2
2
2
3
1
1
)
(
2
/
tan
1
2
/
tan
p
p
2
2
2
2
2 1
1
2
/
sin
2
/
cos
1
1
C
jV
j
C
V
j
C
C
V
V
p
p C
C
Z
Z o
oe
1
1
C
C
Z
Z o
oo
1
1
47. Example
Design 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
.
55
1
.
0
1
1
.
0
1
50
oe
Z
23
.
45
1
.
0
1
1
.
0
1
50
oo
Z
4
.
88
oe
r Z
4
.
72
oo
r Z
From fig 7.29 , we have
w/b=0.72 , s/b =0.34. These
give us
w=0.72b=0.114cm
s= 0.34b = 0.054cm
Then multiplied by r
48. Multisection Coupled line coupler (broadband)
V1
V3 V4
V2
input Through
Isolated
Coupled
C1
CN-2
C3
C2 CN
CN-1
....
j
e
jC
j
jC
j
C
jC
V
V
sin
tan
1
tan
tan
1
tan
2
1
3
j
e
j
C
C
V
V
sin
cos
1
1
2
2
1
2
For single section , whence C<<1 , then
V4=0
and For = p/ 2 then V3/V1= C
and V2/V1 = -j
49. Analysis
Result for cascading the couplers to form a multi section coupler is
)
1
(
2
1
2
1
2
1
1
3
sin
...
sin
sin
N
j
j
N
j
j
j
e
V
e
jC
e
V
e
jC
V
e
jC
V
)
1
(
)
2
(
2
2
2
)
1
(
2
1
1
3
...
1
sin
N
j
M
N
j
j
N
j
j
e
C
e
e
C
e
C
e
jV
V
M
jN
C
N
C
N
C
e
jV
2
1
...
3
cos
1
cos
sin
2 2
1
1
Where M= (N+1)/2
For symmetry C1=CN , C2= CN-1 ,
etc
At center frequency
2
/
1
3
p
V
V
Co
(200)
50. Example
Design 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
,
1
0
)
(
2
/
n
for
C
d
d
n
n
p
From eq (200) and M= (N+1)/2 =( 3+1)/2=2 , we have
2
1
1
3
2
1
2
cos
sin
2 C
C
V
V
C
sin
)
(
3
sin
sin
sin
3
sin 1
2
1
2
1 C
C
C
C
C
(201)
(202)
52. continue
Using even and odd mode analysis, we have
63
.
50
0125
.
0
1
0125
.
0
1
50
1
1
3
1
C
C
Z
Z
Z o
oe
oe
38
.
49
0125
.
0
1
0125
.
0
1
3
1 o
oo
oo Z
Z
Z
69
.
56
125
.
0
1
125
.
0
1
50
1
1
2
C
C
Z
Z o
oe
1
.
44
125
.
0
1
125
.
0
1
2 o
oo Z
Z
53. continue
Let say , r = 10 and d =0.7878mm
63
.
50
3
1 oe
oe Z
Z
38
.
49
3
1 oo
oo Z
Z
69
.
56
2
oe
Z
1
.
44
2
oo
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
56. Continue/ 4 wire coupler
Even mode
All Cm capacitance will be at same potential, thus the total capacitance is
in
ex
e C
C
C
4
m
in
ex
o C
C
C
C 6
4
Odd mode
All Cm capacitance will be considered, thus the total capacitance is
Even and Odd mode characteristic impedance
4
4
1
e
e
C
Z
4
4
1
o
o
C
Z
line
on
transmissi
in
velocity
(300)
(301)
(302)
57. continue
Now consider isolated pairs. It’s equivalent circuit is same as two wire line ,
thus it’s even and odd mode capacitance is
ex
e C
C
m
ex
o C
C
C 2
Substitute these into (300) and (301) ,
we have
o
e
o
e
e
e
C
C
C
C
C
C
3
4
m
ex
m
ex
ex
in
C
C
C
C
C
C
o
e
e
o
o
o
C
C
C
C
C
C
3
4
And in terms of impedance refer
to (302)
oe
oe
oo
oe
oo
e Z
Z
Z
Z
Z
Z
3
4
oo
oo
oe
oe
oo
o Z
Z
Z
Z
Z
Z
3
4
58. continue
oo
oe
oe
oo
oe
oo
oo
oe
o
e
o
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
3
3
2
4
4
Characteristic impedance of the line is
oo
oe
oo
oe
oo
oe
o
e
o
e
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
C
2
3
3
2
2
2
2
4
4
4
4
Coupling
The desired characteristic impedance in terms of coupling is
o
oe Z
C
C
C
C
C
Z
1
/
1
2
8
9
3
4 2
o
oo Z
C
C
C
C
C
Z
1
/
1
2
8
9
3
4 2
