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.