Power Divider and Combiner

7/28/2019 Power Divider and Combiner

1/28

Power divider and combiner/coupler

divider combinerP1

P2= nP1

P3=(1-n)P1

P1

P2

P3=P1+P2

Divide into 4 output

Basic


2/28

S-parameter for power divider/coupler

[ ]

=

333231

232221

131211

SSS

SSSSSS

SGenerally

For reciprocal and lossless network

jiforSSN

k

kjki ==

0

1

*1

1

*=

=

N

k

ki 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


3/28

ContinueIf all ports are matched properly , then Sii= 0

[ ]

=

0

0

0

2313

2312

1312

SS

SS

SS

S

For Reciprocal

networkFor lossless network, must satisfy unitary

condition

1

2

13

2

12 =+ SS

12

23

2

12 =+ SS

12

23

2

13 =+ 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.


4/28

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

13

2

12 =+ SS

12

23

2

12 =+ SS

12

33

2

23

2

13 =++ 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


5/28

Reciprocal lossless network of two matched

S21=ej

S12=ej

S33=ej

1

3

2

[ ]

=

j

j

j

e

ee

S

00

00

00


6/28

For lossless network, must satisfy unitary

condition

12

13

2

12 =+ SS

12

23

2

21 =+ SS

12

32

2

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


7/28

Circulator (nonreciprocal network)

[ ]

=

010

001

100

S

[ ]

=

001

100

010

S

1

2

3

1

2

3


8/28

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 =

=

01

*

1

1

*=

=

N

k

ki 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


9/28

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

*

**

@@@

#

##


10/28

To check validity

Multiply eq. * by S24

* and eq. ## by S13

* , and substract to obtain

02

142

13*14 =

SSS

Multiply eq. # by S34 and eq. @@ by S13 , and substract to obtain

02

34

2

1223 =

SSS

%

$

Both equations % and $ will be satisfy if S14 = S23 = 0 . This meansthat no coupling between port 1 and 4 , and between port 2 and 3 as

happening in most directional couplers.


11/28

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 =


12/28

Continue

[ ]

=

00

0

00

00

0

j

j

j

j

S

Simplified by choosing S12= S34= ; S13=ej 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


13/28

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

Plog20

3

1=Coupling= C= 10 log

dBSP

P

144

3 log20

=

Isolation = I= 10 log dBSP

P

14

4

1 log20=

I = D + C dB

1

4 3

2

Input Through

CoupledIsolated

For ideal case |S

14

|=0


14/28

Practical coupler

Hybrid 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


15/28

T-junction power divider

E-plane TH-plane T

Microstrip T


16/28

T-model

jB

Z1

Z2

Vo

Yin

21

11

ZZjBYin ++=

21

11

ZZYin +=

Lossy line

Lossless line

If Zo = 50,then for equallydivided power, Z1 = Z2=100


17/28

Example

If source impedance equal to 50 ohm and thepower to be divided into 2:1 ratio. Determine Z1

and Z2

ino

PZ

VP

3

1

2

1

1

2

1 ==

ino

P

Z

VP

3

2

2

1

2

2

2 == == 752

3

2

oZZ

== 15031 oZZ

o

oin

Z

VP

2

2

1=

== 50// 21 ZZZo


18/28

Resistive divider

V2

V3

V1

Zo

Zo

P1

P2

P3

Zo 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 =

+

=

VVV

ZZ

ZVV

oo

o

2

1

4

3

3/32==

+

==

o

inZ

VP

2

1

2

1=

( )in

o

PZ

VPP

4

12/1

2

1 21

32 ===


19/28

Wilkinson Power Divider

50

50

50

100

70.7

70.7

/4

Zo

/2 Zo

/2 Zo

2Zo

Zo

Zo

/4

2

2

Te ZZ

in=

oT ZZ 2=

For even mode

Therefore

For Zin =Zo=50== 7.70502TZ

And shunt resistor R =2 Zo = 100


20/28

Example

Design an equal-split Wilkinson power divider for a 50 W systemimpedance at frequency fo

The quarterwave-transformer characteristic is

== 7.702 oZZ

== 1002 oZR

r

o

4

=The quarterwave-transformer length is


21/28

Wilkinson splitter/combiner

application

/4

100

70.7

50

matching

networks

/4

100 50

70.7

70.7

70.7

Splitter combiner

Power Amplifier


22/28

Unequal power Wilkinson

Divider

3

2

03

1

K

KZZ o

+

=

)1( 2032

02 KKZZKZ o +==

+=

KKZR o

1

R2=Z

o/K

R

R3=Z

o/K

Z02

Z03

Zo

2

3

2

32

==

portatPower

portatPower

P

PK

1

2

3


23/28

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

03

1

K

KZZ o

+

=

KZZ o=04 K

ZZ

o=05

Zo

Zo

Zo

Z05

Zo4Zo2

Zo3

Zo1

R1

2

3


24/28

Cohn power divider

/4

98 241 50

82 61

82 61

50

50 /4

VSWR at port 1 = 1.106VSWR 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/f

1) = 2

1

2

3


25/28

Couplers

/4

/4

Yo

Yo

Yo

Yo

Yse

Ysh Y

sh

Branch line coupler 2sh

2se Y1Y +=

2

se

2

sh

sh

2

3

YY1

2Y

E

E

+

=

( )20

1

310

E

E x=

x dB coupling

23

22

21 EEE +=

2

1

3

2

1

2

E

E

E

E1

+

=or

E1E2

E3


26/28

Couplers

input

isolate

Output3dB

Output3dB 90o out of phase

3 dB Branch line coupler

/4

/4

Zo

Zo

Zo

Zo

2/Zo

2/Zo

Zo Zo

32 EE =

1Ysh=

2Y1Y22

se =+= sh

1.414Yse =

= 50oZ

= 50sh

Z

= 5.35seZ


27/28

Couplers9 dB Branch line coupler

( )355.010

209

1

3==

E

E

( )22

1

2355.01 +

=

E

E

( ) 935.0355.01 2

1

2==

E

E

38.0935.0

355.0

2

3==

E

E

8.0=shYLet say we choose

38.0

8.01

8.02

1

2

2222=

+

=

+sesesh

sh

YYY

Y

962.136.038.0

6.1==seY

= 500Z

== 5.628.0/50shZ

== 5.25962.1/50seZ

Note: Practically upto 9dB coupling


28/28

Couplers

/4

/4

/4

3/4

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

3/8

3/8

Te

To

e

o

Documents

Power Divider and Combiner