Power Divider, Combiner and Coupler

8/12/2019 Power Divider, Combiner and Coupler

1/60

Power divider, combiner and

coupler

ByProfessor Syed Idris Syed Hassan

Sch of Elect. & Electron Eng

Engineering Campus USM

Nibong Tebal 14300SPS Penang


2/60

Power divider and combiner/coupler

divider combinerP1

P2= nP1

P3=(1-n)P1

P1

P2

P3=P1+P2

Divide into 4 output

Basic


3/60

S-parameter for power divider/coupler

333231

232221

131211

SSS

SSSSSS

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


4/60

Continue

If all ports are matched properly , then Sii= 0

0

00

2313

2312

1312

SS

SSSS

S

For Reciprocal

networkFor lossless network, must satisfy unitary

condition

1

2

13

2

12 SS

12

232

12 SS

12

232

13 SS

012*

23SS

023*13 SS

013

*

12 SS

Two of (S12, S13, S23) must be zero but it is not consistent. If S12=S13= 0, then

S23should equal to 1 and the first equation will not equal to 1. This is invalid.


5/60

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

232

12 SS

12

332

232

13 SSS 013*3312

*23 SSSS

023

*

13

SS033

*2313

*12 SSSS

The two equations show

that |S13|=|S23|thus S13=S23=0

and |S12|=|S33|=1

These have satisfied all


6/60

Reciprocal lossless network of two matched

S21

=ej

S12=ej

S33

=ej

1

3

2

j

j

j

e

ee

S

00

00

00


7/60

For lossless network, must satisfy unitary

condition

12

132

12 SS

12

23

2

21

SS

12

322

31 SS

032*

31SS

023

*

21 SS

013*

12SS

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


8/60

Circulator (nonreciprocal network)

010

001

100

S

001

100

010

S

1

2

3

1

2

3


9/60

Four port network

44434241

34

24

14

333231

232221

131211

SSSS

SS

S

SSSSSS

SSS

SGenerally

For reciprocal and lossless network

jiforSS

N

kkjki 01*

1

1

*

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


10/60

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

*

1413 SSSS

Say all ports are matched and symmetrical network, then

*

**

@@@

#

##


11/60

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

342

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.


12/60

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


13/60

Continue

00

0

00

00

0

j

j

j

j

S

Simplified by choosing S12= S34= ; S13=ej and S24= e

j

Where += p+ 2np

00

0

0000

0

S

1. Symmetry Coupler == p/2

2. Antisymmetry Coupler =0 ,=p

2 cases

Both satisfy 2 +2 =1


14/60

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

P14

4

1 log20

I = D + C dB

1

4 3

2

Input Through

CoupledIsolated

For ideal case

|S14

|=0


15/60

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


16/60

T-junction power divider

E-plane TH-plane T

Microstrip T


17/60

T-model

jB

Z1

Z2

Vo

Yin

21

11

ZZjBYin

21

11

ZZYin

Lossy line

Lossless line

If Zo= 50,then for equally

divided power, Z1= Z2=100


18/60

Example

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

and Z2

ino P

ZVP

31

21

1

21

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


19/60

Resistive divider

V2

V3

V1

Zo

Zo

P1

P2

P3

Zo V

oo Z

ZZ

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

o

PZ

VPP

4

12/1

2

1 2132


20/60

Wilkinson Power Divider

50

50

50

100

70.7

70.7

/4

Zo

/2 Zo

/2 Zo

2Zo

Zo

Zo

/4

2

2Te ZZ

in

oT ZZ 2

For even mode

Therefore

For Zin=Zo=50

7.70502T

Z

And shunt resistor R =2 Zo= 100


21/60

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= Vg3and

for odd mode Vg2= -Vg3. Since

the circuit is symmetrical , we

can treat separately twobisection 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. Theexcitation is effectively Vg2=4V

and Vg3= 0V.For simplicity all values are

normalized to line characteristic

impedance , I.e Zo =50 .


22/60

Even modeVg2=Vg3= 2V

Looking at port 2

Zine= Z2/2

Therefore 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

+V2e

r/2+V1e

O.CO.C outinZZZ

2

Note:

2ZIf

To determine V2e, using transmission line equation V(x) = V+ (e-jx+Ge+jx) , thus

VjVVVe G )1()4

(2

1

1)1()0(

1 GG

G jVjVVVe

Reflection at port 1, refer to is

22

22

G

2Z

Then 21

jVVe


23/60

Odd modeVg2= - Vg3= 2V

2

1

Port 1

Port 2

2V

Z

4

+V2o

r/2+V1o

At port 2, V1o =0 (short) ,

/4 transformer will belooking as open circuit ,

thus Zino= r/2 . We choose

r =2 for matching. Hence

V2o= 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)


24/60

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

4The quarterwave-transformer length is


25/60

Wilkinson splitter/combiner

application

/4

100

70.7

50

matching

networks

/4

100 50

70.7

70.7

70.7

Splittercombiner

Power Amplifier


26/60

Unequal power Wilkinson

Divider

3

2

031

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


27/60

Parad and Moynihan power divider

4/1

201 1

KKZZ o

2

3

2

32

portatPower

portatPower

P

PK

K

KZR o1

4/124/302 1 KKZZ o

4/5

4/12

031

K

KZZ o

KZZ o04 KZZ o05

Zo

Zo

Zo

Z05

Zo4Zo2

Zo3

Zo1

R1

2

3


28/60

Cohn power divider

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

/f1

) = 2

1

2

3


29/60

Couplers

/4

/4

Yo

Yo

Yo

Yo

Yse

Ysh 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


30/60

Couplers

input

isolate

Output

3dB

Output

3dB 90oout of phase

3 dB Branch line coupler

/4

/4

Zo

Zo

Zo

Zo

2/Zo

2/Zo

Zo Zo

32 EE

1Ysh

2Y1Y 22se sh

1.414Yse

50oZ

50shZ

5.35seZ


31/60

Couplers9 dB Branch line coupler

355.010 209

1

3 E

E

22

1

2 355.01

E

E

935.0355.01 21

2

E

E

38.0935.0

355.0

2

3

E

E

8.0shYLet say we choose

38.0

8.01

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


32/60

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

Te

To

Ge

Go


33/60

AnalysisThe amplitude of scattered wave

oeB GG2

1

2

11

oe TTB2

1

2

14

oe TTB2

1

2

12

oeB GG2

1

2

13


34/60

Couple lines analysis

Planar Stacked

Coupled microstrip

b

w wsw

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 effective

capacitances between lines and velocity of propagation.


35/60

Equivalent circuits

+V +V

H-wall

+V-V

E-wall

C11

C22C

11C

22

2C122C12

Even mode Odd mode

C11and C22are the capacitances between conductors and the groundrespectively. For symmetrical coupled line C11=C22. C12is 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

C12is effectively open-circuited.


36/60

ContinueEven mode

The resulting capacitance Ce= C11= C22

ee

e

eoe

CC

LC

C

LZ

1Therefore, the line characteristic impedance

Odd mode

The resulting capacitance Co= C11+ 2 C12 = C22+ 2 C12

Therefore, the line characteristic impedanceo

ooC

Z

1


37/60

Planar coupled stripline

Refer to Fig 7.29 in Pozar , Microwave Engineering


38/60

Stacked coupled stripline

mFsb

bsbsb

C oWroWroWr /42/2/ 22

11

w >> s and w >> b

mFs

C oWr /12

mF

sb

bCC oWre /

4

2211

mFssb

bwCCC oro /

1222

221211

oor 1

ro

eoe

bwsbZ

CZ

41

22

ssbbwZ

CZ

r

oo

oo/1/22

11

22


39/60

Coupled microstripline

Refer to Fig 7.30 in Pozar , Microwave Engineering


40/60

Design of Coupled line Couplers

inputoutput

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

I4

I3

I2Schematic circuit

Layout


41/60

Even and odd modes analysis

3 4

1 2

Zo

Zo

Zo

Zo

Zoo

V

+V3o

+V2o

+V4o

+V1o

I1o

I4oI

3o

I2o

V

_

+

+

_

3 4

1 2

Zo

Zo

Zo

Zo

Zoe

V

+V3e

+V2e

+V4e

+V1e

I1e

I4eI

3e

I2e

V_+

+

_

I1

e= I3

e

I4e= I2

e

Sameexcitation

voltage

V1e= V3

e

V4e= V2e

Even

I1o= -I3

o

I4o=- I2

o

V1o= -V3

o

V4o= -V2

o

Odd

Reverse

excitationvoltage

(100)

(99)


42/60

Analysis

ooin

oino

ZZ

ZVV

1

tan

tan

ooe

oeo

oe

e

in jZZ

jZZ

ZZ

oe

oe

in II

VV

I

V

Z11

11

1

1

Zo= load for transmission line

= electrical length of the line

Zoeor 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)


43/60

continue

Substituting eqs. (104) - (107) into eq. (101) yeilds

ooin

ein

oein

oin

o

ooin

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

2ooeoo

ein

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)


44/60

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

oooe

ZZjZ

ZZjVV

(112)

i


45/60

continue

We define coupling as

oooe

oooe

ZZ

ZZC

Then V3/ V , from ( 112) will become

oooe

o

ZZ

ZC

2

1 2

tan1

tan

tan2

tan

23

jC

jCV

ZZ

ZZj

ZZ

Z

ZZ

ZZj

VV

oooe

oooe

oooe

o

oooe

oooe

and

sincos1

1

2

2

222jC

CVVVV oe

022444

oeoe VVVVVSimilarly

V1=V


46/60

Practical couple line coupler

V3is maximum when = p/2 , 3p/2, ...

Thus for quarterwave length coupler = p/2 , the eqs V2and V3 reduce to

V1=V

04V

VC

jCjCV

jCjCV

jCjCVV

2223

11)(

2/tan12/tanp

p

22

2

2

2 11

2/sin2/cos1

1CjV

j

CV

jC

CVV

pp C

CZZ ooe

1

1

C

CZZ ooo

1

1


47/60

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

1.01

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=0.114cm

s= 0.34b = 0.054cm

Then multiplied by r


48/60

Multisection Coupled line coupler (broadband)

V1

V3 V4

V2

input Through

IsolatedCoupled

C1

CN-2

C3

C2 CN

CN-1....

jejCj

jC

jC

jCVV

sintan1

tan

tan1

tan21

3

je

jC

C

V

V

sincos1

1

2

2

1

2

For single section , whence C


49/60

AnalysisResult for cascading the couplers to form a multi section coupler is

)1(21

212113

sin...

sinsin

NjjN

jjj

eVejC

eVejCVejCV

)1(

)2(222

)1(2113

...

1sin

NjM

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 frequency2/1

3

p

V

VCo

(200)


50/60

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

d

n

n

p

From eq (200) and M= (N+1)/2 =( 3+1)/2=2 , we have

21

1

3

2

12cossin2 CC

V

VC

sin)(3sinsinsin3sin 12121 CCCCC

(201)

(202)


51/60

ContinueApply (201)

0cos)(3cos32/

121 p

CCCd

dC

010sin)(3sin9 212/1212

2

CCCCCd

Cd

p

Midband Co= 20 dB at =p/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


52/60

continue

Using even and odd mode analysis, we have

63.500125.01

0125.0150

1

131

C

CZZZ ooeoe

38.49

0125.010125.01

31 ooooo ZZZ

69.56

125.01

125.0150

1

12

C

CZZ ooe

1.44125.01

125.012 ooo ZZ


53/60

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


54/60

Couplers

Lange Coupler

Evolution of Lange

coupler

1= input

2=output

3=coupling

4=isolatedw

w

w

w

w

s

s

s

s

1

4 3

2

1

34

2

1

2

3

4

2

4

1

3


55/60

Analysis

1

4 3

2

1

34

2C C

90o

Ze4 Zo4

Zo4Ze4

1

4321

2C

m

Cex

Cex

C

Cex Cex

CinCin

CmC

mC

m

Simplified circuit Equivalent circuit

mex

mexexin

CC

CCCC

where


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Continue/ 4 wire couplerEven mode

All Cmcapacitance will be at same potential, thus the total capacitance is

inexe CCC 4

minexo CCCC 64

Odd modeAll Cmcapacitance will be considered, thus the total capacitance is

Even and Odd mode characteristic impedance

44

1

ee

CZ

44

1

oo

CZ

lineontransmissiinvelocity

(300)

(301)

(302)


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continueNow consider isolated pairs. Its equivalent circuit is same as two wire line ,

thus its even and odd mode capacitance is

exe CC

mexo

CCC 2

Substitute these into (300) and (301) ,

we have

oe

oee

e CC

CCCC

3

4

mex

mexexin

CC

CCCC

oe

eooo

CC

CCCC

34

And in terms of impedance refer

to (302)

oeoeoo

oeoo

e Z

ZZ

ZZZ

34

oooooe

oeooo Z

ZZ

ZZZ

34

i


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continue

oooeoeoo

oeoooooeoeo

ZZZZ

ZZZZZZZ

33

2

44

Characteristic impedance of the line is

oooeoooeoooe

oe

oe

ZZZZ

ZZZZZZC

23

322

22

44

44

Coupling

The desired characteristic impedance in terms of coupling is

ooe Z

CCCCCZ

1/128934

2

ooo Z

CCC

CCZ

1/12

8934 2


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VHF/UHF Hybrid power splitter

50input

50output

50output

100C

T1

T21

5

6

7

8

2

3

4


60/60

Guanella power divider

(VHF/UHF)

RL

V2

I2

I1

V1

Rg

Vg I

1

V2

I2

Documents

Power Divider, Combiner and Coupler