1. Power divider, combiner and couplerBy
Professor Syed Idris Syed Hassan
Sch of Elect. & Electron Eng
Engineering Campus USM
Nibong Tebal 14300
SPS Penang

2. Power divider and combiner/coupler
Basic
P2= nP1 P1 P1
divider
combiner
P3=P1+P2
P3=(1-n)P1 P2
Divide into 4 output

3. S-parameter for power divider/coupler
Generally
For reciprocal and lossless network
Row 1x row 2
Row 2x row 3
Row 1x row 3

4. Continue If all ports are matched properly , then Sii= 0
For Reciprocal
network
For lossless network, must satisfy unitary condition
Two of (S12, S13, S23) must be zero but it is not consistent. If S12=S13= 0, then S23 should equal to 1 and the first equation will not equal to 1. This is invalid.

5. Another alternative for reciprocal network
Only two ports are matched , then for reciprocal network
For lossless network, must satisfy unitary condition
The two equations show
that |S13|=|S23|
thus S13=S23=0
and |S12|=|S33|=1
These have satisfied all

6. Reciprocal lossless network of two matched

7. Nonreciprocal network (apply for circulator)
For lossless network, must satisfy unitary condition
The above equations must satisfy the following either or

8. Circulator (nonreciprocal network)

9. Four port network Generally For reciprocal and lossless network
R 1x R 2
R1x R3
R1x R4
R 2x R3
R2x R4
R3x R4

10. Matched Four port network
Say all ports are matched and symmetrical network, then
The unitarity condition become * ** @ # ##

11. To check validity
Multiply eq. * by S24* and eq. ## by S13* , and substract to obtain %
Multiply eq. # by S34 and eq. by S13 , and substract to obtain $
Both equations % and $ will be satisfy if S14 = S23 = 0 . This means that no coupling between port 1 and 4 , and between port 2 and 3 as happening in most directional couplers.

12. Directional coupler ** * ** *
If all ports matched , symmetry and S14=S23=0 to be satisfied
The equations reduce to 6 equations ** *
By comparing these equations yield **
By comparing equations * and ** yield *

13. Continue 2 cases Both satisfy a2 +b2 =1
Simplified by choosing S12= S34=a ; S13=be j q and S24= b e j j
Where q + j = p + 2np
2 cases
1. Symmetry Coupler q = j = p/2
2. Antisymmetry Coupler q =0 ,j =p
Both satisfy a2 +b2 =1

14. Physical interpretation
|S12 | 2 = power deliver to port 2= a2 =1- b2
|S13 | 2 = coupling factor = b2
Characterization of coupler
Coupling= C= 10 log
I = D + C dB
Directivity= D= 10 log
For ideal case |S14|=0
Isolation = I= 10 log

15. Practical coupler Hybrid 3 dB couplers a= b = 1 /
q = j = p/2
a= b = 1 /
Magic -T and Rat-race couplers
q =0 ,j =p
a= b = 1 /

16. T-junction power divider
H-plane T
E-plane T
Microstrip T

17. T-model Lossy line Lossless line
If Zo = 50,then for equally divided power, Z1 = Z2=100

18. Example
If source impedance equal to 50 ohm and the power to be divided into 2:1 ratio. Determine Z1 and Z2

19. Resistive divider
Zo/3
Zo/3
Zo/3

20. Wilkinson Power Divider
For even mode
For Zin =Zo=50 W
Therefore
And shunt resistor R =2 Zo = 100W

21. Analysis (even and odd mode)
For even mode Vg2 = Vg3 and for odd mode Vg2 = -Vg3. Since the circuit is symmetrical , we can treat separately two bisection circuit for even and odd modes as shown in the next slide. By superposition of these two modes , we can find S -parameter of the circuit. The excitation is effectively Vg2=4V and Vg3= 0V.
For simplicity all values are normalized to line characteristic impedance , I.e Zo = 50 W.

22. Therefore for matching
Looking at port 2
Zine= Z2/2
Therefore for matching
Even mode
Vg2=Vg3= 2V
Note: If
then V2e= V since Zine=1 (the circuit acting like voltage divider)
To determine V2e , using transmission line equation V(x) = V+ (e-jbx + Ge+jbx) , thus
Reflection at port 1, refer to is
Then

23. Odd mode
Vg2= - Vg3= 2V
At port 2, V1o =0 (short) , l/4 transformer will be looking as open circuit , thus Zino = r/2 . We choose r =2 for matching. Hence V2o= 1V (looking as a voltage divider)
S-parameters
S11= 0 (matched Zin=1 at port 1)
S22 = S33 = 0 (matched at ports 2 and 3 both even and odd modes)
S12 = S21 =
S13 = S31 =
S23 = S32 = 0 ( short or open at bisection , I.e no coupling)

24. Example
Design an equal-split Wilkinson power divider for a 50 W system impedance at frequency fo
The quarterwave-transformer characteristic is
The quarterwave-transformer length is

25. Wilkinson splitter/combiner application
Power Amplifier

26. Unequal power Wilkinson Divider2 1 3

27. Parad and Moynihan power divider2 1 3

28. Cohn power divider 2 1 3 VSWR at port 1 = 1.106
VSWR at port 2 and port = 1.021
Isolation between port 2 and 3 = 27.3 dB
Center frequency fo = (f1 + f2)/2
Frequency range (f2/f1) = 2

29. Couplers
Branch line coupler E2 E1
x dB coupling E3 or

30. Couplers 3 dB Branch line coupler input Output 3dB Output isolate
3dB 90o out of phase
isolate

31. Couplers 9 dB Branch line coupler Let say we choose
Note: Practically upto 9dB coupling

32. Couplers Hybrid-ring coupler 4 isolated Output in-phase 3 1 Input
Can be used as splitter , 1 as input and 2 and 3 as two output. Port is match with 50 ohm.
Can be used as combiner , 2 and 3 as input and 1 as output.Port 4 is matched with 50 ohm. 2
Output in-phase

33. Analysis
The amplitude of scattered wave

34. Couple lines analysis
The coupled lines are usually assumed to operate in TEM mode. The electrical characteristics can be determined from effective capacitances between lines and velocity of propagation.

35. Equivalent circuits Odd mode Even mode
C11 and C22 are the capacitances between conductors and the ground respectively. For symmetrical coupled line C11=C22 . C12 is the capacitance between two strip of conductors in the absence of ground. In even mode , there is no current flows between two strip conductors , thus C12 is effectively open-circuited.

36. Continue Even mode The resulting capacitance Ce = C11 = C22
Therefore, the line characteristic impedance
Odd mode
The resulting capacitance Co = C C12 = C C12
Therefore, the line characteristic impedance

37. Planar coupled stripline
Refer to Fig 7.29 in Pozar , Microwave Engineering

38. Stacked coupled stripline
w >> s and w >> b

39. Coupled microstripline
Refer to Fig 7.30 in Pozar , Microwave Engineering

40. Design of Coupled line Couplers
Isolated
(can be matched)
Coupling
Layout
output
input
Schematic circuit

41. Even and odd modes analysis
I1e = I3e
Same excitation voltage
I4e = I2e
V1e = V3e
(99)
V4e = V2e
Odd
I1o = -I3o
Reverse excitation voltage
I4o =- I2o
V1o = -V3o
(100)
V4o = -V2o

42. Analysis (101) By voltage division
From transmission line equation , we have
(104)
(102)
(105)
(103)
(106)
where
Zo = load for transmission line
q = electrical length of the line
Zoe or Zoo = characteristic impedance of the line
(107)

43. continue
Substituting eqs. (104) - (107) into eq. (101) yeilds
(108)
Let
Therefore eqs. (102) and (103) become
(110)
(109)
For matching we may consider the second term of eq. (108) will be zero , I.e or
and (108) reduces to Zin=Zo

44. continue
Since Zin = Zo , then by voltage division V1 = V. The voltage at port 3, by substitute (99), (100) , (104) and (105) is then
(111)
Substitute (109) and (110) into (111)
Then (111) reduces to
(112)

45. continue V1=V We define coupling as and
Then V3 / V , from ( 112) will become
Similarly
V1=V

46. Practical couple line coupler
V3 is maximum when q = p/2 , 3p/2, ...
Thus for quarterwave length coupler q = p/2 , the eqs V2 and V3 reduce to
V1=V

47. Example
Design a 20 dB single-section coupled line coupler in stripline with a cm ground plane spacing , dielectric constant of 2. 56, a characteristic impedance of 50 W , and a center frequency of 3 GHz.
Coupling factor is C = 10-20/20 = 0.1
Then multiplied by
Characteristic impedance of even and odd mode are
From fig 7.29 , we have w/b=0.72 , s/b =0.34. These give us
w=0.72b=0.114cm
s= 0.34b = 0.054cm

48. Multisection Coupled line coupler (broadband)
For single section , whence C<<1 , then
V4=0
For q = p / 2 then V3/V1= C and V2/V1 = -j
and

49. Analysis
Result for cascading the couplers to form a multi section coupler is
For symmetry C1=CN , C2= CN-1 , etc
(200)
At center frequency
Where M= (N+1)/2

50. Example
Design a three-section 20 dB coupler with binomial response (maximally flat), a system impedance 50 W , and a center frequency of 3 GHz .
Solution
For maximally flat response for three section (N=3) coupler, we require
(201)
From eq (200) and M= (N+1)/2 =( 3+1)/2=2 , we have
(202)

51. Continue
Apply (201)
Midband Co= 20 dB at q =p/2. Thus C= 10-20/20=0.1
From (202), we C= C2 - 2C1= © ©
Solving © and © © gives us C1= C3 = (symmetry) and C2 = 0.125

52. continue
Using even and odd mode analysis, we have

53. continue Let say , er = 10 and d =0.7878mm
Plot points on graph Fig. 7.30
We have , w/d = 1.0 and s/d = 2.5 , thus
w = d = mm and s = 2.5d = mm
For section 1 and 3
Similarly we plot points
We have , w/d = 0.95 and s/d = 1.1 , thus
For section 2
w = 0.95d = 0.748mm and s =1.1d = mm

54. Couplers Lange Coupler Evolution of Lange coupler 1= input 2=output
3=coupling
4=isolated

55. Analysis
Equivalent circuit
Simplified circuit
where

56. Continue/ 4 wire coupler
Even mode
All Cm capacitance will be at same potential, thus the total capacitance is
(300)
Odd mode
All Cm capacitance will be considered, thus the total capacitance is
(301)
Even and Odd mode characteristic impedance
(302)

57. continue
Now consider isolated pairs. It’s equivalent circuit is same as two wire line , thus it’s even and odd mode capacitance is
Substitute these into (300) and (301) , we have
And in terms of impedance refer to (302)

58. continue Characteristic impedance of the line is Coupling
The desired characteristic impedance in terms of coupling is

59. VHF/UHF Hybrid power splitter

60. Guanella power divider (VHF/UHF)