11 directional couplers

8/20/2019 11 Directional Couplers

1/29

Directional Coupler

4-portNetwork

1 2

3 4

A directional coupler is a 4-port network exhibiting:

• All ports matched on the reference load (i.e. S11=S22=S33=S44=0)

• Two pair of ports uncoupled (i.e. the corresponding Si,j parameters are zero).Typically the uncoupled ports are (1,3) and (2,4) or (1,4) and (2,3)


2/29

Characteristic Parameters

It is here assumed that the uncoupled ports are (1-4) and (2-3). The S

parameters the ideal coupler must exhibit are:

S11=S22=S33=S44=0, S14=S41=S23=S32=0.

In this case the ports (1-2), (1-3), (3-4), (2-4) are coupled.

Let define C (Coupling) as:

C=|S13|2, CdB=-20 log(|S13|)

If the network is assumed lossless and reciprocal, the unitary conditionof the S matrix determines the following relationships:

2 2 2 2 2

11 12 13 14 12 12

2 2 2 2 2 2 2 2

21 22 23 24 12 24 24 132 2 2 2 2

31 32 33 34 34 34 12

1 1 1

1 1

1 1 1

S S S S S C S C

S S S S S S S S C

S S S S C S S S C

Then, exciting the network at port 1 (2), the output power is divided between ports

3 and 2 (4 and 1) according the factors C and 1-C. No power comes out of port 4(3)


3/29

Further implications of lossless

condition 13 3412 24* *

12 24 13 34 12 24 13 34

12 24 13 34

0 j j

S S S S S S e S S e

Assigning 12=34=0 e 13=±/2, will result 24=±/2, then

S12=S34, S13=S24 The network is symmetric

It can be demonstrated that it is sufficient to impose the matching

condition to the four ports of a lossless and reciprocal network to get a

directional coupler


4/29

Parameters of a real directional coupler In a real coupler the matching at the ports is not zeros at all the frequencies. It

is then specified the minimum Return Loss in the operating bandwidth.

The coupling parameter C, it is in general referred to the port with the lowest

coupling.It must be then considered that to the uncoupled port actually arrive a not zero

power. To characterize this unwanted effect the isolation I is introduced:

I=Power to the coupled port/Power to the uncoupled port

For the coupler considered in the previous slides (assuming the port with the

lowest coupling is the 3):

2 213 14 I S S

Note that I for the ideal coupler is infinite


5/29

Use of the directional coupler (1)

Measure of the reflection coefficient

CVi Vr

LLine, Zc

13 24,i L L r L L

r L L L

i L L

V V S j C V V V S j C V

V j C V V

V V j C V

121, 1 1C S C

V+

V-

LV

LV

LV V

1 2

3 4


6/29


Power divider (C=3 dB)

C=3dB

Pin 1 2

3 4

Pout= Pin /2

Pout= Pin /2

Ports closed on the

reference load

2 1 12 1 2 1 3 1 13 1 3 1

1 1 1 1,

2 2 2 2V V S V P P V V S V P P


7/29


Power combiner (C=3 dB)

1 2

3 4

Pin/2

Pin/2

Pin

3 2

1 2 12 3 13

2 2

2 3 2

1

2 2

1 1 1

2 2 2 2 2 2 2

in in

in

V V

V V S V S j

V V P PV P


8/29

C=3dB

Vout1

1 2

3 4

V A

VB


Sum and difference of voltages (C=3 dB)

1 1 12 13 1 ,2

out A B A BV V V S V S V V

12=13= 24=0

34=

0°

0° 0°

°Vout2

2 4 24 34

1

2out A B A B

V V V S V S V V

Ports closed on the

reference load


9/29


Balanced Amplifier

A

A

0°

90°

C=3 dB

90°

0°

Vin

Vout

Vin/2

jVin/2

2

AV in

2

A jV in

1a 2a

3a 4b

2b

3b

C=3 dB

Gain:

2 3

2 34

2 , 2

2 2

b in b in

b b out out b in

in

V A V V j A V V V P

V V j j A V AP

Reflection:

in

in

1 2 2 3

3 1 3 2

1

1

, 2 , 2 , 2 ,

12 , 2 2

2

0

a in a in a in in a in

a in in a a a in in in in

ain

a

V V V A V V A V V j A V

V j A V V jV V A V A V

V V


10/29

Coupled TEM lines as directional coupler

0 , ,= Zc p c d Z Z

Zcp, Zcd

L

1 2

3 4

We have previously seen that, for having all the ports matched, the following

condition must be satisfied:

This condition also implies that port 4 is uncoupled :

1 4

2 3

0

0

S S

S S

14 1 2 3 41

0

4

S S S S S

The maximum value |S13|2 defines the coupling: C=(|S13|

2)maxIt is obtained for L=/2 :

22 2

2

13 1 2 3 4 1 2max max max

1 1

4 2

cp cd

cp cd

Z Z S S S S S S S

Z Z


11/29

Variation of frequency

Matching and Isolation are frequency independent and equal to zero and

infinity respectively (ideal lossless TEM line).

The coupling varies with the frequency according to the following expression:

maxmax2

max

,

1 (1 ) cot

cp cd

cp cd

Z Z C C C

C Z Z

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

f/f0

C / C m a x

Cmax=0.01-0.1

B0.5 dB For Cmax


12/29

Practical Restrictions

They concern mainly the maximum value of Cmax. In fact, increasing Cmax, thelines become closer and closer until the practical implementation is no more

possible with sufficient accuracy. Typically the maximum value of Cmax must be

lower than 0.1 (CdB=10)

Example: Design a stripline coupler with C=0.1 with Z0=50 using thefollowing figures reporting the values of

as a function of S and W (lines separation and width). Frequency: 1 GHz

max 0,

cp cd

cp cd

cp cd

Z Z C Z Z Z

Z Z

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

S (mm)

Cmax

0

0.025

0.05

0.075

0.1

0.125

0.15

0.175

0.2

Re(Eqn(Cmax))

Output Equations

W=9.5

W=12

W=11.18

S=0.195

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

S (mm)

Z0

45

50

55

60

65

Re(Eqn(Z0))Output Equations W=9.5

10

10.5

11

11.5

12

W=11.18

S=0.195


13/29

Solution:

We draw the line C=0.1 on the first graph and the line Z0=50 on the second graph. A point on each of these lines has to be then found, which must characterized by

the same pair of values (S, W) .

10 mm11.18 mm 11.18 mm

0.19565 mmr =1

Note: If a coupler dimensioned for the requested C but with a different valueof Z0 is available, impedance transforming networks can be used in place of

redesigning a new coupler.

0 2

75 mm

L Lc

L

C=0.1

Z’0

Z0

Z0

Z0

Z0

Z’0

Z’0

Z’0

Z’0


14/29

Couplers with quasi-TEM Lines

If quasi-TEM coupled lines are considered (Microstrip), the phase velocity of theeven and odd modes are not exactly coincident. Strictly speaking, that would

not allow to apply the model here assumed for the characterization of the

directional coupler.

In the practice, until the difference between the two velocities is not too large,

the same phase velocity can be assumed for both modes (equal to the averageof the actual values), assuming the lines as TEM. There are however some

differences comparing the performances with the ideal TEM coupler (a perfect

matching is no more possible)

800 850 900 950 1000 1050 1100 1150 1200

Frequency (MHz)

Microstrip Coupler

-40

-35

-30

-25

-20

-15

-10

-5

0

1000.3 MHz-35.83 dB

999.18 MHz

-26.38 dB

1000.1 MHz-10.03 dB

S21

S31

S41

S11

1.575 mm

2 mm 2 mm

0.4037 mm r =2.33

0

, ,

0, ,

0.1 0.4037 mm

74.12

1.97, 1.71

1.84,2

55.24 mm

cp cd

eff p eff d

eff medio eff medio

m S

Z Z Z

L L

c

L


15/29

Coupler with lumped couplingsTo realize couplers with a large coupling (C


16/29


17/29

Branch-line coupler with C=0.5 (3 dB)

The couplers with C=3 dB are identified with the word Hybrid.To realize an hybrid of branch-line type the characteristic impedances of the

lines must be:

0 0 00.5

1 0.5 35.35 , 50 500.5c c

Z Z Z Z Z

Practical restrictions

One can easily verify that, with C tending to 0: Z’cZ0 and Z’’c∞. In the

practice, even with con C=0.1 (10 dB) the corresponding value of Z’’c is

very difficult to realize (Z’’c =3Z0). Usually, C must be between 3 and 6 dB.

Frequency dependence

For this device, both matching and isolation vary with frequency (the

nominal value is obtained at the frequency where the length of the lines is

0/4). Also the coupling depends on f (the max is again at f 0).The bandwidth for a given value of maximum coupling increases with Cmax;

Usually, the frequency variation of matching and isolation is more

pronounced than that of the coupling.


18/29

Branch Line: a summary

11 33 22 44 13 24

2 2

14 23

2 2

12 34

0, 0

1

S S S S S S

S S C

S S C

Design equations:

S parameters obtained:

14 23 12 34, 1S S C S S j C

For C=0.5 (3dB), it has (assuming Z0=1/Y0=50):

Conditions to be imposed:

0 0

1,

11c c

C Y Y Y Y

C C

0 0 00.5

1 0.5 35.35 , 50 500.5c c Z Z Z Z Z

=0180°

-90°cY

cY

cY cY

11 22

33 44


19/29

800 850 900 950 1000 1050 1100 1150 1200

Frequency (MHz)

Branch-line C=3dB

-50

-45

-40

-35

-30

-25

-20

-15

-10

-5

0

DB(|S(1,1)|)

Schematic 1DB(|S(2,1)|)Schematic 1

DB(|S(3,1)|)

Schematic 1DB(|S(4,1)|)Schematic 1

Adattamento

Isolamento

Accoppiamento


20/29

Rat-race coupler

cY cY

1 3

2 4

cY

cY

0 4

0 4 0 4

03 4

There is only one symmetry axis (vertical)

It is anyhow possible to still use the eigenvalues method for finding the

dimensioning equations

P1P1

P2P2

P3P3

P4P4

cY

cY

cY

c

Y


21/29

Conditions to be imposed:

11 33 22 44 23 14

2 2

13 24

2 2

12 34

0, 0

1

S S S S S S

S S C

S S C

Design equations:

0 01 ,

c cY Y C Y Y C

S parameters obtained:

13 24 12 34, , 1S j C S j C S S j C

11

22 44

33

=0

cY cY

cY

cY

-90°

-90°

11

22 44

33

=0

cY cY

cY

cY

-90°

90°

For C=0.5 (3dB), we obtain (assuming Z0=1/Y0=50):

0 0 0 070.707 , 70.707

1 0.5 0.5

c c

Z Z Z Z Z Z

C C


22/29

Parameters dependence on f

0.8 0.85 0.9 0.95 1 1.05 1.1 1.15 1.2

Frequency (GHz)

Rat-race C=3dB

-50

-45

-40

-35

-30

-25

-20

-15

-10

-5

0

DB(|S(1,2)|)RatRace

DB(|S(1,4)|)RatRace

DB(|S(1,3)|)RatRace

DB(|S(1,1)|)RatRace

Accoppiamento

Adattamento

Isolamento

• The bandwidth is larger than that of the branch-line

• With the decreasing of the coupling the bandwidth increases

• La practical feasibility limits the coupling between about 3 and 8 dB


23/29

3-port lossless networks A 3-port reciprocal lossless network cannot be matched at

the 3 ports (i.e. S11

=S22

=S33

=0 not possible).

In fact, imposing the unitary of S:

2 2

12 13

2 2 2 2 2

12 23 12 13 23

2 2

13 23

1

1 0.5

1

S S

S S S S S

S S

*

13 23

*

12 23 12 23 23

*

12 13

0

0 =0 or 0 or 0

0

S S

S S S S S

S S

These conditions are

incompatible so it is

impossible to haveS11=S22=S33=0


24/29

Is it still possible to realize a power splitter

with a 3-port?

Possible solution: a 3 dB hybrid with the uncoupled

port closed on a matched load

C=3dB

Pin

Z0

2

4 3 Pout= Pin /2

Pout= Pin /2

Circuito a 3 porte

Portadisaccoppiata

The 3-port network is lossy

due to the presence of Z0.This resistor however does

not dissipate power

because the port 4 is

uncoupled. Pin is then split

between ports 2 and 3without losses


25/29

A 3-port divider: the Wilkinson network

c Z

1

2

3

RW

0 4

0 4

c Z

0 Z

0 Z

0 Z

Due to the presence of RW the 3-

port network is lossy, so the

condition S11=S22=S33=0 can be

imposed

2-port network obtained by closing port 1 with Z0:

2 3

0 4 0 4

c Z

0 Z

c Z

RW

S22=S33 and S23 can be computed thorugh the eigenvalues of this network


26/29

Zp

0 4

02 Z

c Z Zd

0 4

2

W R

c Z

2

3

04

04

c Z

02 Z

c Z

02 Z

2

W R

2

W R

2

3

04

04

c Z c Z

02 Z

c Z c Z

02 Z

2

W R

2

W R

Even Network

2 2 2

0

2 2

0 0

2,

2 2

c c p p

c

Z Z Z Z

Z Z Z

0

0

2,

2 2

W W d p

W

R R Z Z

R Z

22

0 0

0 0, 0

2

2 , 2

p d

p d

c W

S

Z Z R Z

Also S23=0:

Per Z0=50 Zc=70.7, RW=100

Odd Network

23 02

p d S


27/29


28/29

Frequency response

0.8 0.85 0.9 0.95 1 1.05 1.1 1.15 1.2

Frequency (GHz)

Divisore di Wilkinson

-50

-45

-40

-35

-30

-25

-20

-15

-10

-5

0

DB(|S(1,1)|)Ideale

DB(|S(2,1)|)Ideale

DB(|S(3,1)|)

Ideale

Adattamento

Trasmissione

• The bandwidth for RL=20 dB is about 40% of f 0• Transmission (|S21|=|S31|) is independent on frequency

• Dissipation in RW is zero provided that the load at ports 2 and 3 is the

same


29/29

Microstrip implementation

RW

The very small size of Rw (pseudo

lumped component) must be

accounted for. The two output linesmust be enough close to allow the

connection of Rw.

On the other hand is not advisable to have

the output lines too close each other

because an unwanted coupling may arise.

So diverging lines are often used.

11 directional couplers

Documents