# Power divider, combiner and coupler

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

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

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

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.

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

Reciprocal lossless network of two matched

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

Circulator (nonreciprocal network)

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

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

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.

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 *

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

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

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 /

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

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

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

Resistive divider Zo/3 Zo/3 Zo/3

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

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.

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

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)

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

Wilkinson splitter/combiner application
Power Amplifier

Unequal power Wilkinson Divider
2 1 3

2 1 3

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

Couplers Branch line coupler E2 E1 x dB coupling E3 or

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

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

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

Analysis The amplitude of scattered wave

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.

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.

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

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

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

Coupled microstripline
Refer to Fig 7.30 in Pozar , Microwave Engineering

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

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

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)

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

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)

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

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

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

For single section , whence C<<1 , then V4=0 For q = p / 2 then V3/V1= C and V2/V1 = -j and

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

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)

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

continue Using even and odd mode analysis, we have

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

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

Analysis Equivalent circuit Simplified circuit where

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)

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)

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

VHF/UHF Hybrid power splitter

Guanella power divider (VHF/UHF)