38659404 Power Divider Combiner and Coupler

8/2/2019 38659404 Power Divider Combiner and Coupler

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Power divider, combiner andcoupler

ByProfessor Syed Idris Syed HassanSch of Elect. & Electron EngEngineering Campus USM

Nibong Tebal 14300SPS Penang



Power divider and combiner/coupler

divider combinerP1

P2= nP1

P3=(1-n)P 1

P1

P2

P3=P1+P2

Divide into 4 output

Basic



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



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.



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



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



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



Circulator (nonreciprocal network)

010

001

100

S

001

100

010

S

1

2

3

1

2

3



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



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

*

**

@@@

###



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.



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





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



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



T-junction power divider

E-plane TH-plane T

Microstrip T



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



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



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



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



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 .



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



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)



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



Wilkinson splitter/combiner

application

/4

100

70.7

50

matchingnetworks

/4

100 50

70.7

70.7

70.7

Splittercombiner

Power Amplifier



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



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



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



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



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



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



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



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



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.



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.



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



Planar coupled stripline

Refer to Fig 7.29 in Pozar , Microwave Engineering



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



Coupled microstripline

Refer to Fig 7.30 in Pozar , Microwave Engineering



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



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)



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)



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)



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 )



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



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



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



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



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)



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)



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



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



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



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



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



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)



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



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



VHF/UHF Hybrid power splitter

50input

50output

50output

100C

T1

T21

5

67

8

23

4



Guanella power divider

(VHF/UHF)

RL

V2

I2

I1

V1

Rg

Vg I1

V2

I2

38659404 Power Divider Combiner and Coupler

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