Notes 20

ECE 5317-6351 Microwave Engineering

Fall 2019

Power Dividers and Circulators

Prof. David R. Jackson Dept. of ECE

1

Adapted from notes by Prof. Jeffery T. Williams

Power Dividers and Couplers

2

A power divider is used to split a signal.

A coupler is used to combine a signal.

These are examples of a three-port network.

Divider

Coupler(Combiner)

Three Port Networks

[ ]11 12 13

21 22 23

31 32 33

S S SS S S S

S S S

=

General 3-port network:

3

If all three ports are matched, and the device is reciprocal and lossless, we have:

[ ]12 13

12 23

13 23

00

0

S SS S S

S S

=

Such a device is not physically possible! (Please see the next slide.)

4

(There are three distinct values.)

Three Port Networks (cont.)

(The S matrix is unitary.)

Conclusion: If we want a match at all ports, we have to have a lossy device.

[ ]12 13

12 23

13 23

00

0

S SS S S

S S

=

*13 23

*12 23

*12 13

0

0

0

S SS SS S

=

=

=

2 212 13

2 212 23

2 213 23

1

1

1

S S

S S

S S

+ =

+ =

+ =

Lossless ⇒ [S] is unitary

Lossless, reciprocal, and matched at all ports is not physically possible.

These cannot all be satisfied.

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

5

Power Dividers and Couplers (cont.)

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

Hence:

Only one of them can be nonzero.

Power Dividers T-Junction: A lossless divider

1inY

02Z

03Z3inY

01Z

1

1

3

3 2

02 03

02 03

02 03

0 02 0

2

3

in in0

01 in 02 03in

02 03 02 03in

1 02 02 02 03 03

0

02

3

1 ||

1 1 1 2 1

1 1

2

Z ZZ Z

Z ZZ Z

Y

Y

Z Z

Y

Z Z Z Z

Z Z Z ZYZ

Z

Z

Z

Z Z Z

+

+

= = = =

≠

+ += + = + = =

≠

To match :

We then have :

SimilarlT y,hus,

6

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

1in02 03

1 1YZ Z

= +

Note: Matching at ports 2

and 3 would be helpful when output lines 2

and 3 are not matched to their loads.

02Z

03Z

01Z 1V+

−1

2

31

2 1

3 1

21

in01

21 01

out in02 02

21 01

out in03 03

12

12

12

VP

Z

V ZP PZ Z

V ZP PZ Z

=

= =

= =

02 0301

02 03

Z ZZZ Z

=+

Assuming port 1 matched:

7

Power Dividers (cont.)

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

out2 03

out3 02

P ZP Z

=

( )

2 32 3

1 3 1 2

1 01 111

11 01 00

02 03 011 01

1 01 02 03 01

3222 33

2 30 0

01 03 02 01 02 03

01 03 02 01 02 03

// a aa a

in

in

a a a a

V Z VSVV Z

Z Z ZZ ZZ Z Z Z Z

VVS SV V

Z Z Z Z Z ZZ Z Z Z Z Z

− −

++= == =

−−

+ += = = =

= =

−−= =

+ +

= =

− −= =

+ +

zero if port 1 is matched

Examine the reflection at each port (Sii):

8

Power Dividers (cont.)

+

-

+

-

Note: A match on port 1 requires:

01 02 01 03,Z Z Z Z< <(The two output lines combine in parallel.)

22 330, 0S S⇒ ≠ ≠The first term in the numerators must be less than the second term.

2 3

2

02 01221

1 1 02

01 0a a

VZ ZVS

V V ZZ

−

−

+ +

= =

= =

9

Also, we have:

Power Dividers (cont.)

+

-

+

-

( ) 0232 23 22

03

1 ZS S SZ

= = +

Similarly:

2 1 2 11 1/ //V VV V V V+ +− + = =

( ) ( )1 1 11 1 1 111 1/VV V S SV+ += + ⇒ = +

Also

( ) 0121 11 12

02

1 ZS S SZ

= + =

Hence ( ) 01

31 11 1303

1 ZS S SZ

= + =

( )

( )

01 03 02 01 02 0311 22 33

01 03 02 01 02 03

01 01 0321 12 11

02 02 02 03

01 01 0231 13 11

03 03 02 03

0 ; ;

1

1

Z Z Z Z Z ZS S S

Z Z Z Z Z Z

Z Z ZS S SZ Z Z Z

Z Z ZS S SZ Z Z Z

− −⇒ = = =

+ +

= = + = =+

= = + = =+

[ ]12 13

12 22 23

13 23 33

0 S SS S S S

S S S

=

If port 1 is matched:

The output ports 2 and 3 are not isolated. 10

Power Dividers (cont.) 02 03

0102 03

Z ZZZ Z

=+

11 0S =Only( ) 02

32 23 2203

1 ZS S SZ

= = +From last slide:

Summary

11

Power Dividers (cont.)

out3 02

out2 03

P ZP Z

=

( )

01 03 02 01 02 0311 22 33

01 03 02 01 02 03

0321 12

02 03

0231 13

02 03

0232 23 22

03

0 ; ;

1

Z Z Z Z Z ZS S S

Z Z Z Z Z Z

ZS SZ Z

ZS SZ Z

ZS S SZ

− −= = =

+ +

= =+

= =+

= = +

02 0301

02 03

Z ZZZ Z

=+

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

The output ports are not isolated.

Waves reflected from devices on ports 2 and 3 with cause

interference with the other devices.

Example: Microstrip T-junction power divider

12

Power Dividers (cont.)

11

22 33

21 12

31 13

32 23

012

1212

12

S

S S

S S

S S

S S

=

= = −

= =

= =

= =

01 50 [ ]Z = Ω

02 100 [ ]Z = Ω

03 100 [ ]Z = Ω

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

Incident

Note: Quarter-wave transformers could be put on the output lines to

bring the final output lines back to 50 [Ω].

The matched power divider also works as a matched power combiner.

13

Power Dividers (cont.)

01 50 [ ]Z = Ω

02 100 [ ]Z = Ω

03 100 [ ]Z = Ω

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

outgoing

Equal waves are incident from ports 2 and 3 (a2 = a3).

There is no reflection if equal waves are incident on ports 2 and 3, and port 1 is matched.

11

22 33

21 12

31 13

32 23

012

1212

12

S

S S

S S

S S

S S

=

= = −

= =

= =

= =

( )3 1 31 2 32 3 33

3 33 32

31 12 2

0

b a S a S a Sa S S

a

= + +

= +

= − +

=

[ ]0 1 11 0 0

21 0 0

jS

− =

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

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

Output ports are isolated (S23 = S32 = 0)

Obviously not unitary 14

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

2 221 31

12

S S= =

3

2

1

(The Wilkenson can also be designed to have an unequal split.)

The derivation is in the appendix.

[ ]0 1 11 0 0

21 0 0

jS

− =

15

Wilkinson Power Divider (cont.)

0Z

0Z

0Z0

2 Z

02 Z

02Z

/ 4gλ

/ 4gλ

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

11 22 33 0S S S= = =

32 23 0S S= =

Microstrip layout

[ ]0 1 11 0 0

21 0 0

jS

− =

16

Wilkinson Power Divider (cont.)

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 (the two output ports are isolated from each other).

21 31 2jS S −

= =

12 13 2jS S −

= =0Z

0Z

0Z0

2 Z

02 Z

02Z

/ 4gλ

/ 4gλ

Example: Microstrip Wilkinson power divider

17

Wilkinson Power Divider (cont.)

01 50 [ ]Z = Ω

03 50 [ ]Z = Ω

02 50 [ ]Z = Ω

0 70.7 [ ]TZ = Ω

100 [ ]R = Ω/ 4gλ

0 70.7 [ ]TZ = Ω

18

Wilkinson 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.

19

Wilkinson Power Divider (cont.)

Figure 7.12 of Pozar

Frequency response of an equal-split Wilkinson power divider. Port 1 is the input port; ports 2 and 3 are the output ports.

3dB−

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

[ ]12 13

21 23

31 32

00

0

S SS S S

S S

⇒ =

*31 32

*21 23

*12 13

0

0

0

S SS SS S

=

=

=

2 221 31

2 212 32

2 213 23

1

1

1

S S

S S

S S

⇒ + =

+ =

+ =

Lossless

These equations will be satisfied if:

“Circulator”

12 23 31

21 32 13

01

S S SS S S

= = =

= = =

21 32 13

12 23 31

01

S S SS S S

= = =

= = =

20

(There are six distinct values.)

or

1

2

Circulators

ij jiS S≠

(unitary)

[ ]0 0 11 0 00 1 0

S =

Clockwise (LH) circulator

Circulators (cont.)

Counter-clockwise (RH) circulator

21

1

2

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

[ ]0 1 00 0 11 0 0

S =

1

2

3

2

1

3A wave goes in one port and comes out from the

adjacent port!

Circulators (cont.)

22

Frequency: 0.698 – 0.96 GHz VSWR: <1.3 Isolation : >18 dB Power: 1000 W Insertion loss: <0.35 dB Length: 1.75 in Width: 2 in Height: 1.02 in Temperature range: -20 – 70 deg C Price: $439.94

An example of a circulator:

Circulators can be made using biased ferrite materials.

Circulators (cont.)

23

Application: Wireless system

Antenna

Transmitter

Receiver

1

2

3

TX

TX

RX

RX

0

The same antenna can be used for transmit and receive. The transmit and receive frequencies can even be the same.

(matched receiver)

Note: A circulator used this way is often called a duplexer.

Duplexer

24

A duplexer:

Diplexer

25

A diplexer) is a type of filter that combines or splits two different frequencies (f1 and f2).

Diplexer

Transmission line

Diplexer

1f

2f

1f

2f

1 2f f+

Triplexer: three frequencies Multiplexer: multiple frequencies

Diplexer (cont.)

26

An example of a diplexer:

Diplexer (cont.)

27

Note: A diplexer can be used to transmit and receive two different channels with the same antenna (as with a duplexer), if the frequencies are separated enough.

Transmitter

Receiver

Diplexer

Antenna

Tf

Rf

Tf

Rf

Note: This requires a high isolation between the transmit and receive ports,

to avoid interference.

Isolator

28

An isolator is a two-port device that is nonreciprocal:

[ ] 0 01 0

S =

A wave coming in on port 1 goes through to port 2. A wave coming in on port 2 gets absorbed (does not go to port 1).

Frequency: 0.8 - 1 GHz VSWR: 1.25 Isolation : 20 dB Power: 2 W Insertion loss: <0.4 dB Length: 1.25 in Width: 2 in Height: 0.75 in Temperature range: -20 – 60 deg C Price: $279.82

Circulator as Isolator

29

A circulator can also be used to make an isolator:

A signal from the input port goes to the output port. A signal from the output port does not get to the input port.

In

Matched load

1

2

3

0

Out

30

A waveguide-based circulator with a matched load at port 3, acting as an isolator

https://en.wikipedia.org/wiki/Circulator

Circulator as Isolator (cont.)

31

Appendix

The analysis of the Wilkinson power divider is given here.

Even/odd mode analysis is used to analyze the Wilkinson power divider.

This requires two separate analyzes (even and odd).

Each analysis involves only a two-port device instead of a three-port device.

Even and odd analysis is used to analyze the structure when port 2 is excited.

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

22 32,S S⇒ To determine

11 21,S S⇒ To determine

32

Appendix (cont.)

The other five components of the S matrix can be found by using physical symmetry and reciprocity (the symmetry of the S matrix).

Top view

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

33

Appendix (cont.)

A microstrip realization is shown.

Plane of symmetry

0 022

h VZ ZI

⇒ = =

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

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

For each half 34

Appendix (cont.)

0Z/ 2I

POS

/ 2I

Z0 microstrip line

Top view

(magnetic wall)

Plane of symmetry

35

Appendix (cont.)

Ports 2 and 3 are excited in phase.

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

3 2e eV V=Note :

22 32,e eS S⇒

Port 2 Excitation “even” problem

Overview

22

ee VS

V

−

+≡

32 22

ee eVS S

V

−

+≡ =

1 3 20,o o oV V V= = −Note :36

Appendix (cont.)

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

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

22 32,o oS S⇒

Port 2 Excitation “odd” problem

Overview

22

oo VS

V

−

+≡

32 22

oo oVS S

V

−

+

−≡ = −

( )2

0in2 0

0

in2 022

in2 0

2

2

0

e

ee

e

ZZ Z

ZZ ZSZ Z

= =

−⇒ = =

+

37

2T

inL

ZZZ

=

Recall:

(quarter-wave transformer)

Appendix (cont.)

Port 2 Excitation “even” problem

Analysis

33 22 0e eS S= =Also, by physical symmetry:

32 22 32 0e e eS S S= ⇒ =Also, in the even case:

02Z

0Z

02 Z

OC

/ 4gλ

V +

eV −

2eV

+

1eV+

2 22,e einZ S

Port 2

in,2eZ

in2 0 0

in2 022

in2 0

0

o

oo

o

Z Z Z

Z ZSZ Z

= ∞ =

−⇒ = =

+

38

Appendix (cont.)

Port 2 Excitation “odd” problem

Analysis

33 22 0o oS S= =Also, by physical symmetry:

32 22 32 0o o oS S S= − ⇒ =Also, in the odd case:

1 0oV =

Port 2

Short

in,2eZ

( ) ( )

( ) ( )

1 3

1 3

222 22 22

2 0

33

332 32 32

2 0

23

1 1 0 0 02 2 2

0

1 1 0 0 02 2 2

0

e o e oe o

a a

e o e oe o

a a

V V V V VS S SV V V V

S

V V V V VS S SV V V V

S

− − − − −

+ + + += =

− − − − −

+ + + += =

+ += = = = + = + =

+

⇒ =

− −= = = = − = − =

+

⇒ =

(by symmetry)

(by reciprocity)

39

22

33

32 23

00

0

SSS S

=== =

In summary, for port 2 excitation, we have:

We now 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.

Appendix (cont.)

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

40

Port 1 Excitation “even” problem

Overview

Port 1

Appendix (cont.)

11 21,e eS S⇒

41

Appendix (cont.)

1 1

1 1

/ 2e

e

I IV V

=

=

Top view

Z0 microstrip line #1

1eI

1I

PMC

0Z

0Z

0Z0

2 Z

02 Z

02Z

/ 4gλ

/ 4gλ

1V +

1V −

02 Z

0Z

02ZOC

/ 4gλ1V +

1V −

O.C. symmetry

plane

Port 1 1eV −

1eV +

1 1 1 1,e eV V V V+ + − −= =

Port 1 Excitation “even” problem

Analysis

( )

2 3 2

2

0in1 0

0

in1 011

in1 0

1 111 11

1 10 0

22

2 02

0

e

ee

e

ee

ea a a

ZZ Z

ZZ ZSZ Z

V VS SV V

− −

+ += = =

= =

−= =

+

= = = =

42

11 0S =

Hence

Appendix (cont.)

2T

inL

ZZZ

=

Recall:

(quarter-wave transformer)

1

2

Port 1

in,2eZ

Port 1 Excitation “even” problem Analysis (cont.)

2 3 2 3

2 221 21

1 10 0

ee

ea a a a

V VS SV V

− −

+ += = = =

= = =

( )2 2

1 1 11 11

e e

e e e e

V V

V V S V

−

+ +

=

= + =

(symmetry of S matrix)

( )( )

2 221

1 1

1 21 2 2

e ee

e eV VS j jV V

−

+

+ Γ⇒ = = = − = −

− Γ

43

21 122jS S−

= =

( ) ( )20 1

e j z j zTV z V e e

z

β β+ − += + Γ

= distance from port 2

0 0

0 0

2 1 22 1 2

Z ZZ Z

− −Γ = =

+ +

Along λg/4 wave transformer:

Appendix (cont.)

Port 1

1eV −

1eV +

2eV −

( ) ( )( ) ( )

2 0

1 0

0 1

/ 4 1

e eT

e eT g

V V V

V V V jλ

+

+

= = + Γ

= − = − Γ

2 2 21 11 2 1 2

+ Γ = − Γ =+ +

Port 1 Excitation “even” problem Analysis (cont.)

31 21 2jS S −

= =By physical symmetry:

44

13 31 2jS S −

= =By reciprocity (symmetry of S matrix):

For the other two components:

Appendix (cont.)

[ ]0 1 11 0 0

21 0 0

jS

− =

We then have the final S matrix: