11 Directional Couplers

Directional Coupler 1

2

4-port Network 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)

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 condition of the S matrix determines the following relationships:

S11  S12  S13  S14  1  S12  C  1  2

2

2

2

2

S12  1  C

S 21  S 22  S 23  S 24  1  S12  S 24  1  2

2

2

2

2

2

S31  S32  S33  S34  1  C  S34  1  2

2

2

2

2

S24  S13  C 2

2

S34  S12  1  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)

Further implications of lossless condition * * S12 S24  S13 S34  0  S12 S 24 e

j  12  24 

 S13 S34 e

j  13 34 

12   24    13   34 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

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):

I  S13

2

S14

2

Note that I for the ideal coupler is infinite

Use of the directional coupler (1) Measure of the reflection coefficient

Line, Zc

V+

VL

1

2

VL

V-

C  1,

S12  1  C  1  L

V  V



Vi

Vi  VL S13  j C  VL ,

3

C

Vr  VL S24  j C  VL

Vr j C  VL VL     L  Vi j C  VL VL

L

4

Vr

Use of the directional coupler (2) Power divider (C=3 dB)

Pin

1

2

Pout= Pin /2

Ports closed on the reference load Pout= Pin /2

V2  V1S12 

4

3

C=3dB

1 1  V1  P2  P1 , 2 2

V3  V1S13 

1 1  V1  P3  P1 2 2

Use of the directional coupler (3) Power combiner (C=3 dB) Pin

Pin/2

1

3

2

Pin/2

4

 V3 V2  V1  V2 S12  V3 S13    j  2  2 1 2 V1 2

2 2  V2  1  Pin Pin  1 V3          Pin 2  2 2  2  2 2 

Use of the directional coupler (4) Sum and difference of voltages (C=3 dB) 0° Vout1 1

12=13= 24=0  34=

2

VB

Ports closed on the reference load

0°

0° 3

VA

°

4 V out2

C=3dB 1 Vout1  V1  VA S12  VB S13   VA  VB  , 2 Vout 2  V4  VA S24  VB S34 

1  VA  VB  2

Use of the directional coupler (5) Balanced Amplifier Vin

1a

2a

0°

Vin/2

in

90°

C=3 dB

Gain:

A

3a jVin/2

V2b  A Vin Vout  V4b  j

A

in

Vin

jVin

A 2 A 2

2b

90° 3b

2 , V3b  j A  Vin V2b 2



V3b 2

 j A  Vin

Reflection:

V1a  Vin , V2a  A Vin V3a  jin A Vin

2 , V2a  in A Vin

2 , V1a 

C=3 dB

0°

4b

Vout

2 

Pout A Pin

2 , V3a  j A Vin

2,

1 jV3a  V2a   in A Vin 2  in A Vin 2  2 V1a in    0 V1a

Coupled TEM lines as directional coupler L 2

1 Zcp, Zcd 3

4

We have previously seen that, for having all the ports matched, the following condition must be satisfied:

 S1  S 4   0  S 2  S 3   0

Z 0 = Zc , p  Z c , d

This condition also implies that port 4 is uncoupled : S14 

1  S1  S 2  S 3  S 4   0 4

The maximum value |S13|2 defines the coupling: C=(|S13|2)max It is obtained for L=/2 : 2

S13 max  2

2

Z cp  Z cd

1 1  S1  S 2  S 3  S 4    S1  S 2   Z cp  Z cd 4 2 max max

2

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: C  

Cmax , 1  (1  Cmax )  cot 2   

Cmax 

Z cp  Z cd Z cp  Z cd

1 0.9

For Cmax<0.1 the variation of C è practically independent on Cmax. Note that the band for C/Cmax< 0.5 dB is about 0.44 f0

B0.5 dB

0.8 0.7

Cmax=0.01-0.1

C/Cmax

0.6 0.5 0.4 0.3 0.2 0.1 0

0

0.2

0.4

0.6

0.8

1 f/f0

1.2

1.4

1.6

1.8

f0 is the frequency for which L=/2

2

Practical Restrictions They concern mainly the maximum value of Cmax. In fact, increasing Cmax, the lines 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 the Z  Z cd following figures reporting the values of Cmax  cp , Z 0  Z cp  Z cd Z cp  Z cd as a function of S and W (lines separation and width). Frequency: 1 GHz Cmax

0.2

Z0

65

Re(Eqn(Z0)) Output Equations

Re(Eqn(Cmax)) Output Equations

0.175

W=11.18 S=0.195

0.15 0.125

60

W=9.5 10

W=11.18 S=0.195

10.5 11

W=9.5

55

0.1

11.5 12

0.075

W=12

50

0.05 0.025

45

0 0.1

0.2

0.3

0.4

0.5 0.6 S (mm)

0.7

0.8

0.9

1

0.1

0.2

0.3

0.4

0.5 0.6 S (mm)

0.7

0.8

0.9

1

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) . 11.18 mm

r=1

11.18 mm

L 

10 mm

0.19565 mm

0 L

c L  75 mm



 2

Note: If a coupler dimensioned for the requested C but with a different value of Z0 is available, impedance transforming networks can be used in place of redesigning a new coupler. Z’0

Z’0 Z0

Z0

C=0.1 Z’0

Z0 Z’0

Z0 Z’0

Couplers with quasi-TEM Lines If quasi-TEM coupled lines are considered (Microstrip), the phase velocity of the even 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 average of 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) 2 mm

Microstrip Coupler

0

S21

-5 -10

r=2.33

0.4037 mm

1.575 mm

1000.1 MHz -10.03 dB

m  0.1  S  0.4037 mm

S31

-15

2 mm

Z 0  Z cp  Z cd  74.12

999.18 MHz -26.38 dB

-20

 eff , p  1.97,  eff ,d  1.71

-25

S41

-30

1000.3 MHz -35.83 dB

-35

 eff ,medio  1.84,  L 

S11

-40 800

850

900

950 1000 1050 Frequency (MHz)

1100

1150

1200

L  55.24 mm

0 L c

 eff ,medio 

 2

Coupler with lumped couplings To realize couplers with a large coupling (C<10 dB) structure with lumped couplings are used. In planar technology two kinds of such couplers are employed, the branch-line and the rat-race. Branch-line Eigenvalues of Y and S: 1

Y’c, 

2

Y1  j Yc  Yc , S1  Y0  jBs  Y0  jBs 

3



 

0 

Y 3  j Yc  Yc , S 3  Y0  jBd  Y0  jBd 

Y’’c, 

Y’’c, 

Y 2   j Yc  Yc , S 2  Y0  jBd  Y0  jBd  Y 4   j Yc  Yc , S 4  Y0  jBs  Y0  jBs 

L=  2  4

Y’c, 

4

In this case, to obtain S11=0 and S13=0 (ports 1-3 uncoupled), we must impose:

S1  S 2  0, S 3  S 4  0

This condition is actually feasible, giving in fact:

Bs  Bd  1  Yc2  Yc2  Y02 with Bs  Yc  Yc , Bd  Yc  Yc 2 Y0

For S12 e S14 we have: S12 

2 jbs 1 1  S1  S 2  S 3  S 4    S1  S 4   4 2 1  bs2

1  bs2 1 1 , S14   S1  S 2  S 3  S 4    S1  S 4   4 2 1  bs2

bs 

Bs Y0

Moreover, the unitary condition of S implies the following relation between the phases of S12 and S14:  12  14   2 Then, being 12=-/2  14=. The parameter bs must be then >1 for having S14 negative. By imposing now the requested coupling (|S14|2=C), we can obtain the necessary value of bs:

bs2  1  S14   C, 2 1  bs

 bs 

1  C Yc  Yc  Y0 1 C

Taking into account the first found condition ( Yc  Yc  Y0 finally obtain the expressions of Y’c e Y’’c: 2

Yc  Y0

1 C , Yc  Y0 1 C 1 C

2

2

) we

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:

Z c  Z 0 1  0.5  35.35 , Z c  Z 0

0.5  50  0.5

 Z 0  50  

Practical restrictions One can easily verify that, with C tending to 0: Z’cZ0 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 f0). 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.

Branch Line: a summary Conditions to be imposed: S11  S 33  S 22  S 44  0, S13  S 24  0

S14  S 23  C 2

2

S12  S 34  1  C 2

-90°

Yc

1

2

2

Design equations:

Yc  Y0

1 C , Yc  Y0 1 C 1 C

Yc =0

3

Yc Yc

S parameters obtained:

S14  S 23   C , S12  S 34   j 1  C

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

Z c  Z 0 1  0.5  35.35 , Z c  Z 0

0.5  50  0.5

 Z 0  50  

180° 4

Branch-line C=3dB

0

Accoppiamento

-5 Adattamento

-10

Isolamento

-15 -20 -25 -30

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

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

-35

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

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

-40 -45 -50 800

850

900

950 1000 1050 Frequency (MHz)

1100

1150

1200

Rat-race coupler 0 4

1

Yc

P1

3

P3 Yc

0 4 Yc

2

Yc 0 4

Yc

Yc

Yc

P2

4

P4 Yc

30 4

There is only one symmetry axis (vertical) It is anyhow possible to still use the eigenvalues method for finding the dimensioning equations

-90°

Conditions to be imposed: S11  S 33  S 22  S 44  0,

1

S 23  S14  0

Yc

S13  S 24  C 2

2

4

2

=0

Yc

-90° =0 1

Yc  Y0  1  C , Yc  Y0  C

Yc

Yc

2

S parameters obtained:

Yc

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

Z c 

3

Yc

90° 4

S13   j C , S 24  j C , S12  S 34   j 1  C

Z0 Z  0  70.707, 1 C 0.5

Yc

2

Design equations:

Z c 

3

-90°

S12  S 34  1  C 2

Yc

Z0 Z  0  70.707 0.5 C

Parameters dependence on f Rat-race C=3dB

0 -5

Accoppiamento

-10 -15

Adattamento

-20 Isolamento

-25 -30

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

-35

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

-40

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

-45

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

-50 0.8

0.85

0.9

0.95 1 1.05 Frequency (GHz)

1.1

1.15

1.2

• 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

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: S12  S13  1 2

2

S12  S23  1  2

2

S12  S13  S23  0.5 2

2

2

S13  S 23  1 2

2

* S13 S 23 0 * 0  S12 S 23

S12 S13*  0

S12 =0 or S 23  0 or S 23  0

These conditions are incompatible so it is impossible to have S11=S22=S33=0

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

Pin

2

Pout= Pin /2

C=3dB Z0

4

Porta disaccoppiata

Circuito a 3 porte

3

Pout= Pin /2

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 3 without losses

A 3-port divider: the Wilkinson network 0 4

2

Zc Z0

1

Z0

RW

Zc

3

0 4

Due to the presence of RW the 3port network is lossy, so the condition S11=S22=S33=0 can be imposed

Z0

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

Zc

Zc

0 4

0 4

Z0

3

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

RW 2

2

RW 2

Zc

Zc

0 4

0 4

Even Network

2Z 0

3

Odd Network

22Z Z0

Zp

Zd

Zc

Zc

0 4

2Z 0

RW 2

Z c2 Z c2  2 Z 02 Zp  , p  2 Z c  2Z 02 2Z 0 S 22 

 p  d 2

Zd   0   p  0,  d  0 

Zc  2  Z0 , Per Z0=50  Zc=70.7, RW=100

RW  2Z 0

0 4

RW R  2Z 0 , p  W 2 RW  2 Z 0 Also S23=0:

S 23 

 p  d 2

0

Evaluation of S11 Exciting port 1 with ports 2 and 3 closed with Z0, there is no current through RW, so it can be discarded. 0 4

Z in

Zc Zc

Z0

Z c2  2Z 02 1 Z c2 , in  S11  2 Z in  2 Z0 Z c  2Z 02

Z0

0 4

Being

Z c  2  Z c also S11 vanishes.

Evaluation of S21= S31 Power entering port 1 is not reflected and there is there is no dissipation in RW. So the power is all transferred to ports 2 e 3; for the symmetry, the power exiting at each port is the half of Pin:

S 21  S31  0.5 2

2

The phase for both parameters is -90°.

Frequency response Divisore di Wilkinson

0

Trasmissione

-5 -10 -15 -20 -25

Adattamento

-30

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

-35

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

-40

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

-45 -50 0.8

0.85

0.9

0.95 1 1.05 Frequency (GHz)

1.1

1.15

1.2

• The bandwidth for RL=20 dB is about 40% of f0 • Transmission (|S21|=|S31|) is independent on frequency • Dissipation in RW is zero provided that the load at ports 2 and 3 is the same

Microstrip implementation The very small size of Rw (pseudo lumped component) must be accounted for. The two output lines must be enough close to allow the connection of Rw. 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.