Power Dividers and Directional Couplers
Three-port network : T-junction
Four-port network: Directional coupler, hybrid
Basic Properties
Three-port network:
[ S ]

If no anisotropic material :




S
11
S
21
S
31
S ij

S
12
S
22
S
32
S ji
S
13
S
23
S
33



 or [S] is symmetric.
If all ports are matched,
S ii
= 0.
If all ports are matched and the network is lossless , [S] is unitary , or
S
12
2 
S
13
2 
1 S
31
* S
32

0
S
21
2 
S
23
2
S
31
2 
S
32
2

1

1
S
21
* S
23
S
12
* S
13

0

0
At least two of ( S
12
,
S
13
, S
23
) must be zero ⇒ inconsistent!
A three-port network cannot be lossless , reciprocal, and matched at all ports.

Matched Lossless Three-port Nonreciprocal Network
Matched at all ports :
[ S ]





S
0
21
S
31
S
12
0
S
32
S
13
S
23
0




(nonreciprocal)
Lossless, [S] is unitary:
S
12
2 
S
13
2 
1 S
13
* S
23

0
S
12
2 
S
23
2 
1 S
23
* S
12

0
S
13 or
2 
S
23
S
12
S
21


2
S
23
S
32

1

S
31

0
S
12
&

S
13

0
&
*
S
12
S
13
S
21



S
23
0
S
32


S
13
S
31
The results show that
S ij



1
1
S ji
, for i ≠ j, which implies that the device must be nonreciprocal. The two possible solutions are the following two circulators.
The two circulators:
[ S ]





0
1
0
0
0
1
1


0
0


[ S ]



0


0
1
1
0
0
0


1
0

 clockwise circulator counterclockwise circulator

If only two ports are matched, a lossless and reciprocal three-port network can be realized.
(1) Port 1 & 2 are matched ,
[ S ]





0
S
21
S
31
S
12
0
S
32
S
13
S
23
S
33




(2) Lossless
S
13
* S
23

0
S
12
* S
13

S
23
S
23
* S
12

S
33
* S
33
* S
13


0
0
S
12
2 
S
13
2 
1
S
12
2 
S
23
2 
1
S
S
13
2
13


S
23
S
32
2 
S
33
2 
1

0 and S
12

S
33

1
Four-port Networks (Directional Couplers)
(1) Reciprocal and matched at all ports (2)Lossless , [S] is unitary
[ S ]







0
S
12
S
13
S
14
S
12
0
S
23
S
24
S
13
S
23
0
S
34
S
S
14
24
S
34
0






S
13

) S
13
* S
23
* S
23


S
14
S
14
* S
24
* S
24
S
14
* ( S
13
Similarly, S
14
2
* ( S
13

S
24
2
2
)

S
24



2
)
0
0
0

S
24
*

S
13
*

0

S
12
* S
13

S
24
* S
34

0 
(1)Symmetric coupler:
[ S ]





 j
0

0
 j

0
0
 j


0
0
    j
0


0






/ 2
    
2 n

(2)Anti-symmetric coupler:
[ S ]






0


0


0
0


0
0


0

0







For a hybrid coupler : c=3dB or
  j
 
1 / 2
What does the S – parameter matrix mean? e.g., symmetric coupler , quadrature hybrid ,
   
90
0

0 ,
  

Directional Couplers:
Coupling: C= 10 log(
P
1
P
3
)
 
20 log

dB
Directivity: D= 10 log(
P
3
P
4
)
 
20 log

/ S
41
dB
Isolation: I= 10 log(
P
1
P
4
)
 
20 log S
41
dB
Note that I=C+D (dB)
D →∞ and I →∞ for an ideal coupler.
Hybrid coupler: C= 3dB or
   
1 / 2

The T-junction(Three-port) Power Divider jB is used to model fringing fields & higher order modes.
Usually, B is cancelled by using a certain reactive element.
P in

V
0
2
Z
0
(if  in

0 ),
P
1

V
0
2
Z
1
,
P
2

V
0
2
Z
2
V
Z
0
2
0

V
0
2
Z
1
V
0
2
Z
1
:
V
0
2
Z
2

V
0
2
Z
2

2 : 1

Z
1
: Z
2

1 : 2
(1)Z
1
=3Zo=150Ω,
Z
2
=1.5Zo=75Ω,
Z in
=3Zo//1.5Zo=Zo.
(2)Looking into the Z1=150Ω line,
Z in1
=50//75=30Ω,
Γ in1
=(30-150)/(30+150)=-2/3.
(3)Looking into the Z
2
=75Ω line,
Z in2
=50//150=37.5Ω,
Γ in2
=(37.5-75)/(37.5+75)=-1/3.

A Resistive Divider:
Z in

Z
0
3
V

2
3
V
1

( Z
0

Z
0 ) //( Z
0
3

Z
0 )

Z
0
3
V

V
3

Z
0

Z
0
Z
0
/ 3
V

3
4
V

1
2
V
1

S
11

S
22

S
21

S
31

S
33

S
23

0

1 / 2
[ S ]

1
2


0


0
1
1
0
1
1


1
0


 Not unitary because of loss.
P in

V
1
2
2 Z
0
,
P
2

P
3

( V
1
/ 2 )
2
2 Z
0

V
1
8 Z
0

P in
4
Resistive dividers for unequal power division ratios are possible.
The Wilkinson Power

In summary, the S-parameter matrix is
[ S ]









0

2 j
2 j

0
0
2 j

0
0
2 j







(1)Matched at all ports.
(2) S
12

S
21

S
13

S
31
(3) S
23

S
32

0 , ports 2 & 3 are isolated.
(4)Note that when the divider is driven at port 1, and the output ports
(ports 2 &3) are matched, no power is dissipated in the resistors.

Sol : (1)Quarter-wave transmission lines Z=70.7Ω,
(2)Shunt resistors r= 2Zo= 100Ω.
Unequal Power Division

N-way, Equal-Split Wilkinson Divider
Matched at all ports , all ports, isolation between all ports.
Crossover resistors , difficult to implement in planar form.
Quadrature( 90
0
) Hybrid
(1) Quadrature hybrids are 3dB directional couplers with a difference in the outputs of through and coupled arms.
(3) Also known as a branch-line hybrid for a microstrip realization.
Example:
90
0 phase
[ S ]

1
2





0
1
0 j 0
0
1 j 1
0
0 j
0
1
0 j






The branch-line has a symmetric structure, as any port can be used as the input port.
Analysis:
Original Hybrid = Even-mode+ odd-mode circuits
A.Even-mode :
 e
( S
11
)

T e
( S
21
)

A
A


B
B

C

C


D
D
A

B
2

C

D

0
 
( 1

2 j )

B. Odd-mode :
 o
( S
11
)

T o
( S
21
)

A
A


B
B

C

C


D
D
A

2
B

C

D

0

( 1

2 j )
B
1

1
2
 e

1
2
 o

0 (port 1 is matched) B
3

1
2
T e

1
2
T o


1
(half power)
2
B
2

1
T e

1
T o

 j
(half power) B
4

1
2
 e

1
2
 o

0 (isolated , no power)
2 2 2

S
11



[ S ]

0
,
S
21

 j
2
, S
31


1
2
,
S
41

0 (Input = port 1)
If input=port 2, then through=port 1, coupled=port 4, and isolated = port 3.
If input = port 3, then through = port 4, coupled = port 1, and isolated = port 2.

If input = port 4, then through = port 3, coupled = port 2, and isolated = port 1.
1
2

0

 j


1
0 j
0
0
1
1
0
0 j
0

1 j
0





Example Performance of a 50Ω Branch-Line Quadrature Hybr
(1) Bandwidth 10 ~ 20%.
(2) By using multiple sections in cascade, BW can be increased to a decade or more.
(3) The basic design can be modified for unequal power division and/or different characteristic impedances at the output ports.
(4) The discontinuity effects at the junction may require that the shunt arms be lengthened by 10
0
~ 20
0
.

Coupled-Line Directional Couplers
(1) Coupled-line theory
Equivalent capacitance networks:
Z oe

L e
C e

L e
C e
C e

1 v e
C e
Z oo

L o
C o

L o
C o
C o

1 v o
C o
*Small s , strong EM interaction ,
Z oe
Z oo

1
*Large w , small Z oe and Z oo
.
*e.g. s/b=0.1 , w/b=0.4 , Z oe
=140

,
Z oo
=
75

Z oo
=75 

Normalized Z oe and Z oo design data for a microstrip line .
(1) Quasi-static results , realistic microstrip is dispersive. Usable up to
5 –6 GHz.
(2) If
 r is different, data must be changed.
(3) Given Z oe and Z oo
, s/d and w/d are uniquely specified.
(4) Phase velocity data is not shown.
Design of Coupled-Line Couplers

Even-Odd mode analysis :
Let
 e
  o
 
,
  l
 and Z oe
Z oo be the modal characteristic impedances.
(1)Even-mode (2) Odd-mod
I
1 e 
I
3 e
, I
2 e 
I e
4
I
1 e 
I
3 e
, I
2 e 
I e
4
V
1 e
Z e in

V
3 e

Z oe
,
Z o
Z oe


V
2 e 
V
4 e jZ oe jZ oe tan tan


V
1 e
Z o in

V
3 e
,

Z oo
Z o
Z oo
V
2 e



V
4 e jZ oo jZ oo
V
1 e

Z e in e
Z in

Z o
V
,
I
1 e 
Z e in
V

Z o
V
1 o
Input impedance at any port:
Z in

V
1
I
1

V
1 e
I
1 e

V
1 o

I
1 o

Z o in
Z o in

Z o
V , I
1 o  o
Z in
V

Z o
Z e in e
Z
 in
Z o

Z o in o
Z
 in
Z o

 e
Z in
Z o

1

Z o

2 ( Z e
Z in
 e o in
Z in o
Z in o
Z in
1



Z o
2
)
2 Z o
Z o o
Z in
 e
Z in
Z e in

(
Z o
Z o in


Z o
Z o
 in
)
Z o

Z o

At the midband (
 l
   
/ 2 )
(1) V
2 and V
3 have a phase difference of quadrature hybrid .
(2) As long as Z o

Z oo
Z oe
90
0
, the coupler can be used as a holds , the coupler will be matched at the input and have perfect isolation, at any frequency .
(3) Coupled-line couplers are best suitable for weak couplings.
Power conservation:
P in
V
2

2 Z
0
P
2

V
2
2 Z
0
2

P
2

P
3

P
4

( 1

C 2 ) V
2
2 Z
0
P
3

V
3
2
2 Z
0
P 0
4


C
2
V
2
2 Z
0
Usually, the coupling C is given to find Z oe and
Z oo
C

Z oe
Z oe


Z oo
Z oo
  o e
Z o o

Z oe

Z o
1

C
1

C
,
Z oo

Z o
1

C
1

C