Notes 19 ECE 5317-6351 Microwave Engineering Fall 2011 Power Dividers and Couplers Part 1 Prof....

1

Notes 19

ECE 5317-6351 Microwave Engineering

Fall 2011

Power Dividers and Couplers Part 1

Prof. David R. JacksonDept. of ECE

2

Power Dividers and Directional Couplers

Divider1P 2 1P P

3 11P P

1 2 3P P P 2P

3P

Coupler(Combiner)

Three-port networks

11 12 13

21 22 23

31 32 33

S S S

S S S S

S S S

General 3-port:

3

For all ports matched, reciprocal, and lossless:

12 13

12 23

13 23

0

0

0

S S

S S S

S S

13 23

12 23

12 13

0

0

0

S S

S S

S S

2 2

12 13

2 2

12 23

2 2

13 23

1

1

1

S S

S S

S S

Lossless [S] is unitary

Not physically possible

These cannot all be satisfied.

At least 2 of S13, S12, S23 must be zero.

Power Dividers and Directional Couplers (cont.)

(There are three distinct values.)

(If only one is zero (or none is zero), we cannot satisfy all three.)

(If only one is nonzero, we cannot satisfy all three.)

4

Now consider a 3-port network that is non-reciprocal, all ports matched, and lossless:

12 13

21 23

31 32

0

0

0

S S

S S S

S S

31 32

21 23

12 13

0

0

0

S S

S S

S S

2 2

21 31

2 2

12 32

2 2

13 23

1

1

1

S S

S S

S S

Lossless

These equations will be satisfied if:

“Circulator”

12 23 31

21 32 13

0

1

S S S

S S S

21 32 13

12 23 31

0

1

S S S

S S S

Power Dividers and Directional Couplers (cont.)

(There are six distinct values.)

or

1

2Note that Sij Sji.

5

21S

32S

13S

1

2

3

12S

23S

31S

2

1

3

0 0 1

1 0 0

0 1 0

S

Clockwise (LH) circulator

Circulators

Counter clockwise (RH) circulator

1

2

Circulators can be made using biased ferrite materials.

Note: We have assumed here that the phases of all the S parameters are zero.

0 1 0

0 0 1

1 0 0

S

6

Power DividersT-Junction: lossless divider

1inY

02Z

03Z3inY

01Z

1

23

1

3

02 03 02 03 02 03

01 02 02

02 03

02 0301 02 03

01 02 03

03 02 02 03 03 02

0 03 2

2 21 1

1 1

1|

1

1

|

1

1

in

in

i

n

in

n

i

Z Z Z Z Z ZY

Z Z Z Z Z Z Z Z Z

YZ

YZ Z

Z ZY Z Z Z

Z Z Z

YZ

Note,however,

Th

To mat

us ls, A o,

ch

If we match at port 1, we cannot match at the other ports!

7

02Z

03Z

01Z 1V

1

2

3

2 1

3

1

1

11

1 1

01 03

02 02 0

2

1

02

2

3

01 021

03 0

2

1

30 0

1

3 2

0

1

2

1

2

1

2

1

in iout in

i

in

ni nt

n

n iou

VP

Z

V

Z ZP P

Z Z Z

Z ZP P

ZP

ZK

K

P

P

ZP

V

Z

Z

02 0301

02 03

Z ZZ

Z Z

Assuming port 1 matched:

Power Dividers (cont.)

We can design the splitter to control the powers into the two output lines.

8

02Z

03Z

01Z 2V1V+

-

+

-

2 3

1 3 1 2

02 03 01

02 03 01

01 03 02 01 02 03

111

1 0

2 322 33

2

01 03 02 01 02 03

30 0

a a

a a a a

Z Z Z

Z Z Z

Z Z Z Z Z Z

Z

VS

V

V VS

Z Z

V

Z

SV

Z Z

zero if port 1 is matched

For each port we have:


9

2 3

0121 11 12

02

01 0231 13 11 32 23 22

0

1 1 11 2 2 1 2

2

0221

1

01 0

1 11

0

1 1

3 3

1 ; /

1

1 1

/ 1

a a

V V S V V V V V V V

ZS S S

Z

Z ZS S S S S S

Z Z

Z

V

S

V

S

Z

Similarly,

and


02Z

03Z

01Z 2V1V+

-

+

-

Also, we have

10

01 03 02 01 02 0311 22 33

01 03 02 01 02 03

01 0121 12 11

02 02

01 0113 31

03

02 03

0211

03 03 02 03

0222

03

0232 23 22

03

0

1

; ;

1

1

1

Z

Z Z

Z

Z Z

ZS

Z Z Z Z Z ZS S S

Z Z Z Z Z Z

Z ZS S S

Z Z

Z ZS S S

Z Z

ZS S S

Z Z

21 31

21 22 32

31 32 33

0 S S

S S S S

S S S

If port 1 is matched:

The output ports are not isolated.

Power Dividers (cont.)02 03

0102 03

Z ZZ

Z Z

11

03

02 03

02

0

222

2 03

11

2331

1

PS

P

P

Z

Z Z

Z

Z ZS

P

Powers:


1 203 02

1 1 102 0 0

33 2 03

Z ZP P P

ZP P P

Z Z Z

Check :

3 02

2 03

P Z

P ZHence

12

Summary


3 02

2 03

P Z

P Z

01 03 02 01 02 0311 22 33

01 03 02 01 02 03

0321 12

02 03

0213 31

02 03

0232 23 22

03

0 ; ;

1

Z Z Z Z Z ZS S S

Z Z Z Z Z Z

ZS S

Z Z

ZS S

Z Z

ZS S S

Z

02Z

03Z

01Z1

2

3

02 0301

02 03

Z ZZ

Z Z

The input port is matched, but not the output ports.

The output ports are not isolated.

13

Example: Microstrip T-junction power divider


01 50 [ ]Z

02 100 [ ]Z

03 100 [ ]Z

11

22 33

21 12

31 13

32 23

0

1

2

1

2

1

2

1

2

S

S S

S S

S S

S S

01 02 01 03 50 100 33.333 [ ]Z Z Z Z

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Resistive Power Divider

0Z

0Z

0Z

1V

2V

3V

0

3

Z0

3

Z

0

3

Z

1inZ

port1

port3

port2

0 0 01

0 00

4 4

3 3 3

2

3 3

in

Z Z ZZ

Z ZZ

11 22 33 0S S S

Same for Zin1 and Zin2

All ports are matched.

15

1 1 11 1

2

1 1

00

2 10 0

0 0

1

23

23 3

1

2

3

2 3

3 4

V

V

V

V V S

Z ZV V

Z Z

V

Z Z

2 3

2

021

1

0 0a a

V

ZS

V

Z

12 321 1 13 32 23

1

2S S S SS S

By reciprocity and symmetry

Resistive Power Divider (cont.)

0Z

0Z

0Z

1V

2V

3V

0

3

Z0

3

Z

0

3

Z

1inZ

port1

port3

port2

16

221

1 10

22 2 2

2 3 2 1 21 1 21 1

0 1 11

1 0 12

1 1

1 1

2 2

4

0

1 1 1

2 2 2

in

in

VP P a

Z

PP P b a S

S

P S P

All ports are matched, but 1/2 Pin is dissipated by resistors, and the output ports are not isolated.

0Z

0Z

0Z

1V

2V

3V

0

3

Z0

3

Z

0

3

Z

1inZ

port1

port3

port2

1P

2P

3P

Resistive Power Divider (cont.)

Hence we have

17

Even-Odd Mode Analysis1

2 1 3S

S

VV V

2 2

2 SV V

Plane of symmetry 2 2 2 2

2 eSV

oSV 0V

eV

POS POS

2 eSV

oSV

1

2e o

S S SV V V Let

e oV V V

Obviously,

Example: We want to solve for V.

using even/odd mode analysis

“odd” problem“even” problem

(This is needed for analyzing the Wilkenson.)

18

“Even” problem

2 2 2 2

2 eV

eV

2SV

2SV

4 4

0I

eSV

POS POS

eSV

Open circuit (OC) plane of symmetry

2

4 2SV eV

4

2 2 4 3e S SV V

V

Even-Odd Mode Analysis (cont.)

19

“Odd” problem

oSV

2 2 2 2

2 oSV 0oV

oV

2SV

2SV

4 4

POS POS

0oV

2

4 2SV

oV

short circuit (SC) plane of symmetry


20

2 2

2 SV V

2 2 2 2

2 2SV oV

eV

2SV

2SV

2SV 2

03

3

Se

S

oVV

VV

V V

“even” problem “odd” problem

By superposition:


21

0Z

0Z

0Z0

2Z

02Z

02Z

g/ 4

g / 4

3

2

1

0 1 1

1 0 02

1 0 0

jS

Wilkenson Power DividerEqual-split (3 dB) power divider

All ports matched (S11 = S22 = S33 = 0)

Output ports are isolated (S23 = S32 = 0)

Obviously not unitary

Note: No power is lost in going from port 1 to ports 2 and 3.

2 2

21 31

1

2S S

22

Example: Microstrip Wilkenson power divider

Wilkenson Power Divider (cont.)

01 50 [ ]Z 0 70.7 [ ]TZ

03 50 [ ]Z

02 50 [ ]Z

0 70.7 [ ]TZ

100 [ ]R

23

22 32,S S

11 21,S S

• Even and odd analysis is required to analyze structure when port 2 is excited.

• Only even analysis is needed to analyze structure when port 1 is excited.

To determine

To determine


The other components can be found by using symmetry and reciprocity.

24

Top view

0Z

0Z

0Z

0

2 Z

02 Z

02Z

/ 4g

/ 4g

1V

1V 2V

2V

3V

3V

Plane of symmetry

Split structure along plane of symmetry (POS)

Even voltage even about POS place OC along POS Odd voltage odd about POS place SC along POS


A microstrip realization is shown.

25

0 022

h VZ Z

I

0Z/ 2I

POS

/ 2I

Z0 microstrip line

How do you split a transmission line? (This is needed for the even case.)

top view

Voltage is the same for each half of line (V)Current is halved for each half of line (I/2)

For each half


0Z

0Z

0Z

0

2 Z

02 Z

02Z

/ 4g

/ 4g

1V

1V 2V

2V

3V

3V

Plane of symmetry

(magnetic wall)

26

“Even” Problem


02Z

0Z

0Z

0

2 Z

02Z

OC

/ 4g

/ 4g

V

eV

V

eV

02Z

1eV

1eV

3eV

0Z

0Z

OC

2eV

Ports 2 and 3 are excited in phase.

Note: The 2Z0 resistor has been split into two Z0 resistors in series.

3 2e eV VNote :

271 3 20,o o oV V V Note :


“Odd” problem

02Z

0Z

0Z

0

0

g

g

V

oV

V

oV

02Z

1oV

1oV

3oV

0Z

0Z

2oV

Ports 2 and 3 are excited 180o out of phase.

Note: The 2Z0 resistor has been split into two Z0 resistors in series.

28

02Z

0Z

0

OC

gV

eV

2eV

1eV

2 22,e einZ S

33

2

0

2 00

2 022

2 0

2

2

0

0

ein

ee in

ein

e

ZZ Z

Z

Z ZS

S

Z Z

Also, by symmetry,


Even ProblemPort 2 excitation

Port 2

29

02Z

0Z

02Z

/ 4g V

oV

0Z

2oV

1oV

2 22,o oinZ S

short

2 0 0

2 022

2 0

33 0

0

oin

oo in

oin

o

Z Z Z

S

Z ZS

Z Z

Also, by symmetry,

1 0oV


Port 2

Odd Problem

Port 2 excitation

30

1 3

1 3

222

2 0

33

332

22 22

2 0

2

22

32 22 22

3

1 10 0 0

2 2 2

1 10 0 0

2 2 2

0

0

e o e oe o

a a

a

e o e oo

a

e

V V V VS S S

V V V

V V V VS S S

V V V

VS

V

S

VS

V

S

(by symmetry)

(by reciprocity)


22

33

32 23

0

0

0

S

S

S S

In summary,

We add the results from the even and odd cases together:

Note: Since all ports have the same Z0, we ignore the normalizing factor Z0 in the S parameter definition.

31

0Z

0Z

0Z

0

2 Z

02 Z

02Z

/ 4g

/ 4g

1V

1V 2V

3V

0Z

0Z

0Z

0

2 Z

02 Z

02Z

/ 4g

/ 4g

1V

1V

02 Z

0Z0

2Z

OC

/ 4g

1V

1V

O.C. symmetry

plane

When port 1 is excited, the response, by symmetry, is even. (Hence, the total fields are the same as the even fields.)

Even Problem


Port 1

Port 1 excitation

32

02Z

0Z0

2Z

OC

/ 4g

1eV

1eV

1

2

1 11,e einZ S

2 3 2

1 011

1 0

1 111 11

2

0

0 0

1

1

0

1

0

20

2

0

22

ee in

ei

ein

n

ee

e

a a a

Z

Z ZS

Z Z

V VS S

V V

Z ZZ


11 0S

Hence

Even Problem

Port 1 excitation

33

02Z

0Z0

2Z

OC

/ 4g

1V

1V 2V

z

1 1 11 1

2 2 2

1V V S V

V V V

2 3

221

1 0a a

VS

V

(reciprocal)

2 221

1 1

1 2

1 2 2

V VS j j

V V


21 122

jS S

20

2 0

1 0

1

0 1

/ 4 1

j z j z

g

V z V e e

V V V

V V

z

V j

distance from port 2

0 0

0 0

2 1 2

2 1 2

Z Z

Z Z

Along g/4 wave transformer:

Even Problem

Port 1 excitation

34

31 21 2jS S By symmetry:


13 31 2jS S By reciprocity:

For the other components:

35

0 1 1

1 0 02

1 0 0

jS


0Z

0Z

0Z

0

2 Z

02 Z

02Z

/ 4g

/ 4g

All three ports are matched, and the output ports are isolated.

11 22 33 0S S S

32 23 0S S

36

0 1 1

1 0 02

1 0 0

jS


When a wave is incident from port 1, half of the total incident power gets transmitted to each output port (no loss of power).

When a wave is incident from port 2 or port 3, half of the power gets transmitted to port 1 and half gets absorbed by the resistor, but nothing gets through to the other output port.

21 312

jS S

12 132

jS S

0Z

0Z

0Z

0

2 Z

02 Z

02Z

/ 4g

/ 4g

37


Figure 7.15 of PozarPhotograph of a four-way corporate power divider network using three

microstrip Wilkinson power dividers. Note the isolation chip resistors.

Courtesy of M.D. Abouzahra, MIT Lincoln Laboratory.

Notes 19 ECE 5317-6351 Microwave Engineering Fall 2011 Power Dividers and Couplers Part 1 Prof....

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