Power dividers and couplers part 1x

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Transcript Power dividers and couplers part 1x

ECE 5317-6351 Microwave Engineering

Fall 2011

Prof. David R. Jackson Dept. of ECE

Notes 19

Power Dividers and Couplers Part 1 1

Power Dividers and Directional Couplers Three-port networks

P

1 Divider

P

3

P

2  

P

1   

P

1

P

1 

P

2 

P

3 Coupler (Combiner)

P

2

P

3 General 3-port:   

S

11

S

21

S

31

S

12

S

22

S

32

S

13

S

23  

S

33 2

Power Dividers and Directional Couplers (cont.) For all ports matched, reciprocal, and lossless:     0

S

12

S

13

S

0 12

S

23

S

0 13

S

23   (There are three distinct values.) Not physically possible Lossless  [

S

] is unitary 

S

12 2 

S

13 2  1

S

12 2 

S

23 2  1

S

13 2 

S

23 2  1 

S

13 

S

12 

S

12

S

23  0

S

23  0

S

13  0 These cannot all be satisfied.

(If only one is nonzero, we cannot satisfy all three.)  At least 2 of

S

13 ,

S

12 ,

S

23 must be zero.

(If only one is zero (or none is zero), we cannot satisfy all three.) 3

Power Dividers and Directional Couplers (cont.) Now consider a 3-port network that is non-reciprocal , all ports matched, and lossless:    0

S

21

S

0 12

S

13

S

23   “Circulator”

S

31

S

32 0 These equations will be satisfied if: (There are six distinct values.) Lossless 

S

21 2 

S

31 2  1

S

12 2 

S

32 2  1

S

13 2 

S

23 2  1 

S

31 

S

21 

S

12

S

32

S

23

S

13   0 0  0 1 2

S

12

S

21  

S

23

S

32 

S

 31 

S

13 0  1 or Note that

S ij

S ji

.

S

21

S

12  

S

32

S

23  

S

13 

S

31 0  1 4

1 Circulators   0 1 0 0 0 1 1 0   0 Note: We have assumed here that the phases of all the

S

parameters are zero.

Clockwise (LH) circulator

S

21 2

S

32 1 3

S

13 2    0 0 1 1 0 0 0 1   0 Circulators can be made using biased ferrite materials.

S

12 1 2

S

23

S

31 3 Counter clockwise (RH) circulator 5

Power Dividers T-Junction: lossless divider

Y in

1

Z

02

Z

01

Y i n

1  1

Z

02  1

Z

03

Y in

3

Z

03 To mat ch 

Y i n

1 Note, however, 

Y in

3 1

Z

01  

Z

01 

Z

02 | |

Z

03 1

Z

01  1

Z

02 

Z

02 

Z

02 03 

Z

03 

Z

03 03  1

Z

02 

Z

02  2

Z

03 03  1

Z

03  

Z

02 

Z

02 2

Z

03   Th us ,

Y in

3  1

Z

0 3

Y in

2  1

Z

0 2 If we match at port 1, we cannot match at the other ports!

6

Power Dividers (cont.) Assuming port 1 matched:

Z

01 

Z

02 03 

Z

03

P in

1  1 2

V

1 2

Z

0 1

P out

2  1 2

V

1 2

Z

02

P ou t

3  1 2

V

1 2

Z

0 3 

Z

01

Z

02

P in

1 

Z

01

Z

03

P i n

1 1

Z

01 

V

1  

Z

02

Z

03 

Z

0 3   

P i n

1 

K P in

1 

Z

0 2

Z

02 

Z

0 3  

P i n

1 

K

P i n

1

Z

02 2

Z

03 3 We can design the splitter to control the powers into the two output lines.

7

Power Dividers (cont.) For each port we have:

Z

01

S

11 

V

1 

V

1 

a

2 0 

Z

02

Z

02

Z

03

Z

03 

Z

01 

Z

01  zero if port 1 is matched 

V

1 + +

V

2 -

Z

02

Z

03

S

22 

V

2 

V

2 

a

1 0 

Z

01

Z

01

Z

03 

Z

03 

Z

02

Z

02

S

33 

V

3 

V

3 

a

1

a

2 0 

Z

01

Z

01

Z

02

Z

02 

Z

03 

Z

03 8

Power Dividers (cont.) Also, we have

Z

02

S

21 

V

2 

Z

02

V

1 

Z

01

a

2 0

Z

01 +

V

1 +

V

2 -

V

1 

V

1   1  

S

21 

S

11 

S

11  ;

V

2  

V

2 

V

1

Z

01

Z

02 

S

12 Similarly,

S

31 

S

13 

S

11  

Z

01

Z

0 3 and

S

32 

S

23

V

2  /

V

1  

V V

1 / 1   1 

S

11 

S

22 

Z

02

Z

0 3

Z

03 9

Power Dividers (cont.) If port 1 is matched:

Z

01 

Z

02 03 

Z

03 

S

11  0 ;

S

22 

Z

01

Z

01

Z

03

Z

03 

Z

02 

Z

02 ;

S

33 

Z

01

Z

01

Z

02

Z

02 

Z

03 

Z

03

S

21 

S

12

S

13 

S

31

S

32 

S

23   

S

11

S

11  

S

22 

Z

01

Z

02 

Z

01

Z

03 

Z

01

Z

02 

Z

01

Z

03 

Z

02

Z

03   1 

S

22 

Z

02

Z

03 

Z

03

Z

02

Z

02 

Z

03

Z

02

Z

03    0

S

21

S

31

S

21

S

22

S

32

S

31

S

32  

S

33 The output ports are not isolated.

10

Powers: Power Dividers (cont.)

P

2

P

1 

S

2 1 2 

Z

02

Z

 03

Z

03  

P

3

P

1 

S

31 2 

Z

0 2

Z

 02

Z

03   Hence

P

3

P

2 

Z

02

Z

03 Check :

P

1 

P

2 

P

3   

Z

02

Z

03 

Z

0 3  

P

1   

Z

0 2

Z

 02

Z

03  

P

1 

P

1 11

Power Dividers (cont.) Summary

Z

02

Z

01 

Z

02 03 

Z

03 1

Z

01 2

Z

03 3

S

11  0 ;

S

22 

Z

01

Z

01

Z

03

Z

03 

Z

02 

Z

02 ;

S

33 

Z

01

Z

01

Z

02

Z

02 

Z

03 

Z

03

P

3

P

2 

Z

02

Z

03

S

21 

S

12 

S

13 

S

31 

S

32 

S

23 

Z

02

Z

03 

Z

03

Z

02

Z

02 

Z

03

S

22 

Z

02

Z

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

 The output ports are not isolated.

12

Power Dividers (cont.)

S

11  0

S

22 

S

33   1 2

S

21 

S

12 

S

31 

S

13  1 2 1 2

S

32 

S

23  1 2 Example: Microstrip T-junction power divider

Z

02 

Z

01 

Z

01

Z

02 

Z

01

Z

03  50 100 

Z

03  13

Resistive Power Divider

Z in

1 

Z

0 3  4

Z

0 3 4

Z

0 3 

Z

0 3  2

Z

0 3 

Z

0 Same for

Z in

1 and

Z in

2  All ports are matched.

S

11 

S

22 

S

33  0

Z

0

Z

0

Z in

1

V

  1

Z

0 3 port1

Z

0 3 

V Z

0 3 

V

3 port3 port2

Z

0 14

Resistive Power Divider (cont.)

S

21 

V

2 

Z

0

V

1 

Z

0

a

2 0

V

1 

V

1   1 

S

11  

V

1 

V

2  

V

2 

V

1 

V

1   

Z

3 0 2 3

Z

0  2 3

Z

0 2 3 3      1 2

V

1     

Z

0

Z

0 

Z

0 3    

Z

0

Z in

1

V

  1 port1

Z

0 3

Z

0 3 

V Z

0 3 

V

3 port3 port2 By reciprocity and symmetry

Z

0

Z

0 

S

21  1 2 

S

12 

S

3 1 

S

13 

S

32 

S

23 15

Resistive Power Divider (cont.)

P

2 Hence we have

P

1

Z in

1

Z

0 3

Z

0 3 

V

2  0 1 1 1 0 1 1 1   0

Z

0

V

  1

Z

0 3 

V

3 port1 port3

P

1 

P in

 1 2

V

1 2

Z

0  1 2

a

1 2

P

2 

P

3  1 2

b

2 2  1 2 21 2  21 2 

P

1 1 2  2

P in

4 All ports are matched, but 1/2

P in

output ports are not isolated.

is dissipated by resistors, and the

Z

0 port2

P

3

Z

0 16

Even-Odd Mode Analysis (This is needed for analyzing the Wilkenson.) Example: We want to solve for

V

.

Let

V S e

V S o

 1 2

V S

using even/odd mode analysis

V S

2 Obviously,

V

V S

2 

V

 2 1 

V S

3

V S e

2 POS 2 

V e

 2 “even” problem Plane of symmetry

V S e

V S o

2 POS 2  V 0  2

V

V e

V o

“odd” problem

V S o

17

V S

2

V S e

2 Even-Odd Mode Analysis (cont.) “Even” problem POS POS 2

I

 0 2 2 

V e

 2

V S e V S

2 4 

V e

 4 Open circuit (OC) plane of symmetry 2

V S

2 4 

V e

V e

V S

2 4 2  4 

V S

3 18

V S o

Even-Odd Mode Analysis (cont.) “Odd” problem POS POS 2 

V o

 2 2

V S o V S

2 2 4 

V o

  0 4 2

V S

2 short circuit (SC) plane of symmetry 2

V S

2 4 

V o

V o

 0 19

V S

2 Even-Odd Mode Analysis (cont.) 2 2

V S

V

 2 2 

V e

 2 2

V S

2 

V S

2 2 

V o

 2 2 “even” problem By superposition:

V

V e

V o

V S

3 

V

V S

3  0 “odd” problem

V S

2 20

Wilkenson Power Divider Equal-split (3 dB) power divider

Z

0 1  g / 4 2

Z

0 2  g / 4 2

Z

0 3 2

Z

0

Z

0

Z

0  All ports matched (

S

11 =

S

22 =

S

33 = 0 )  Output ports are isolated (

S

23 =

S

32 = 0 ) Note: No power is lost in going from port 1 to ports 2 and 3.

S

21 2 

S

31 2  1 2  2  0 1 1 0 1 0 1 0    0 Obviously not unitary 21

Z

01  Wilkenson Power Divider (cont.) Example: Microstrip Wilkenson power divider

Z

02 

Z

0

T

Z

0

T

R

Z

03  22

Wilkenson Power Divider (cont.) • Even and odd analysis is required to analyze structure when port 2 is excited.

 To determine

S

22 ,

S

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

 To determine

S

11 ,

S

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

23

Wilkenson Power Divider (cont.) Top view

Z

0

V

1 

V

1  

g

/ 4 2

Z

0 

g

2

Z

/ 4 0 A microstrip realization is shown.

Z

0 2

Z

0

Z

0

V

2 

V

2 

V V

3  3  Plane of symmetry Split structure along plane of symmetry (POS) Even  Odd  voltage even about POS voltage odd about POS   place OC along POS place SC along POS 24

Wilkenson Power Divider (cont.)

Z

0

V V

1  1  

g

/ 4 

g

2

Z

0 / 4 2

Z

0

Z

0 2

Z

0

Z

0

V V

2  2  Plane of symmetry

V V

3  3  How do you split a transmission line? (This is needed for the even case.) top view

I

/ 2

Z

0 POS Voltage is the same for each half of line (

V

) Current is halved for each half of line (

I

/2 )

I

/ 2 (magnetic wall)

Z

0 microstrip line 

Z

0

h

I V

2  2

Z

0 For each half 25

Wilkenson Power Divider (cont.) “Even” Problem 2

Z

0 

V

1

e

g

/ 4 2

Z

0

V

2

e

Z

0 OC

Z

0 2

Z

0 

V

1

e

OC 2

Z

0

Z

0 

g

/ 4 

V

3

e Z

0 Note :

V

3

e

V

2

e V

V

e

Ports 2 and 3 are excited in phase.

V

V

e

Note: The 2

Z

0 resistor has been split into two

Z

0 resistors in series.

26

Wilkenson Power Divider (cont.) “Odd” problem 2

Z

0 2

Z

0 

V

1

o

g

/ 4 2

Z

0

V

2

o

Z

0

Z

0

V

V

o

Note: The 2

Z

0 resistor has been split into two resistors in series.

Z

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

V

1

o

2

Z

0

Z

0 

V

 

g

/ 4 

V

3

o Z

0 

V

o

Note :

V

1

o

 0,

V

3

o

 

V

2

o

27

Wilkenson Power Divider (cont.) Even Problem Port 2 excitation 

V

1

e

g

/ 4

Z e in

2 ,

S e

22

V

2

e

 2

Z

0 OC

Z

0

V

V

e

Port 2

e Z in

2  

e S

22   2

Z

0  2  2

Z

0

e Z in

2

Z in e

2  

Z

0

Z

0

Z

0  0 2

Z

0 Also, by symmetry,

e S

33  0 28

Wilkenson Power Divider (cont.) Odd Problem Port 2 excitation 2

Z

0

o Z in

2 ,

o S

22

V

2

o

 

g

/ 4 2

Z

0 

V

1

o V

1

o

short  0

Z

0

Z

0

V

V

o

Port 2

Z in o

2   

o S

22 

Z

0 

Z Z in o

2

Z in o

2  

Z

0

Z

0 0  0 Also, by symmetry,

o S

33  0 29

S

22 

V

2 

V

2  Wilkenson Power Divider (cont.) We add the results from the even and odd cases together: 0

S

22 

V V

e

 

V

o

V

 

V

e

V

o

2

V

  1 2 

e S

22 

o S

22   2 1  

S

33  0 (by symmetry)   0

S

32 

V

3 

V

2 

a

1

a

0

S

32 

V

e V

 

V

o

V

 

V

e

 2

V

V

o

 1 2 

e S

22 

o S

22   2 1  

S

2 3  0 (by reciprocity) Note: Since all ports have the same

Z

0 , we ignore the normalizing factor 

Z

0 in the

S

parameter definition. In summary,

S

22

S S

33 32  0  0 

S

23  0  0 30

Wilkenson Power Divider (cont.)

Z

0 Port 1 excitation Port 1

Z

0

V

1 

V

1  

g

/ 4 

g

2

Z

/ 4 0 2

Z

0

Z

0 2

Z

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

V

2 

V

3  Even Problem

Z

0

Z

0

V V

1  1  

g

/ 4 2

Z

0 

g

/ 4 2

Z

0 2

Z

0

Z

0 O.C. symmetry plane 2

Z

0

V V

1  1  

g

/ 4 2

Z

0 OC

Z

0 31

Wilkenson Power Divider (cont.) Even Problem Port 1 excitation

Z e in

1   2

Z

0  2  2

Z

0

Z

0

S e

11 

e Z in

1

Z e i n

1   2

Z

0 2

Z

0  0

S

11 

V

1 

V

1 

a

2

a

3 0 

V

1 

V

1 

e e a

2  0 

e S

11  0 2

Z

0 2

V

1 

V

1 

e e

g

/ 4 2

Z

0 1

Z e in

1 ,

e S

11 OC

Z

0 Hence

S

11  0 32

Wilkenson Power Divider (cont.)  Even Problem

V

1 

V

1  Port 1 excitation

S

21 

V

2 

V

1 

a

2 

g

/ 4 2

Z

0

z

0 2

Z

0 OC

Z

0

V

1 

V

1 

V

2 

V

2   1  

V

2 

S

11  

V

1  

S

21 

V

2 

V

1  

V

2

V

1  

j

  1 1         2

j

2 2 

S

21   2

j

S

12 (reciprocal) Along 

g

/4 wave transformer:   

V e

0  1  

e

j

2 

z

V

2

V

1

z

 distance from port 2 

V

V

  

g

V

0  / 4   1    

V

0 

j

 1     

Z

0

Z

0   2

Z

0 2

Z

0  1  1  2 2

V

2  33

Wilkenson Power Divider (cont.) For the other components: By symmetry:

S

31 

S

21  

j

2 By reciprocity:

S

13 

S

31  

j

2 34

Wilkenson Power Divider (cont.)  2  0 1 1 0 1 0 1 0    0

Z

0

Z

0 

g

/ 4 2

Z

0 

g

2

Z

0 / 4 2

Z

0

Z

0

S

11 

S

22 

S

33  0

S

32 

S

23  0 All three ports are matched, and the output ports are isolated.

35

Wilkenson Power Divider (cont.)

Z

0  2  0 1 1 0 1 0 1 0    0

Z

0 

g

/ 4 2

Z

0 

g

2

Z

0 / 4 2

Z

0

Z

0

S

21 

S

31  

j

2

S

12 

S

13  

j

2  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. 36

Wilkenson Power Divider (cont.) Figure 7.15 of Pozar Photograph 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.

37