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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

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**Power divider and combiner/coupler**

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

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**S-parameter for power divider/coupler**

Generally For reciprocal and lossless network Row 1x row 2 Row 2x row 3 Row 1x row 3

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**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.

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**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

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**Reciprocal lossless network of two matched**

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**Nonreciprocal network (apply for circulator)**

For lossless network, must satisfy unitary condition The above equations must satisfy the following either or

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**Circulator (nonreciprocal network)**

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**Four port network Generally For reciprocal and lossless network**

R 1x R 2 R1x R3 R1x R4 R 2x R3 R2x R4 R3x R4

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**Matched Four port network**

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

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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.

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**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 *

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**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

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**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

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**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 /

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**T-junction power divider**

H-plane T E-plane T Microstrip T

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**T-model Lossy line Lossless line**

If Zo = 50,then for equally divided power, Z1 = Z2=100

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

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Resistive divider Zo/3 Zo/3 Zo/3

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**Wilkinson Power Divider**

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

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**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.

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**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

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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)

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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

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**Wilkinson splitter/combiner application**

Power Amplifier

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**Unequal power Wilkinson Divider**

2 1 3

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**Parad and Moynihan power divider**

2 1 3

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**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

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Couplers Branch line coupler E2 E1 x dB coupling E3 or

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**Couplers 3 dB Branch line coupler input Output 3dB Output isolate**

3dB 90o out of phase isolate

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**Couplers 9 dB Branch line coupler Let say we choose**

Note: Practically upto 9dB coupling

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**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

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Analysis The amplitude of scattered wave

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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.

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**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.

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**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

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**Planar coupled stripline**

Refer to Fig 7.29 in Pozar , Microwave Engineering

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**Stacked coupled stripline**

w >> s and w >> b

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**Coupled microstripline**

Refer to Fig 7.30 in Pozar , Microwave Engineering

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**Design of Coupled line Couplers**

Isolated (can be matched) Coupling Layout output input Schematic circuit

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**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

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**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)

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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

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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)

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**continue V1=V We define coupling as and**

Then V3 / V , from ( 112) will become Similarly V1=V

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**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

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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

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**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

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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

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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)

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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

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continue Using even and odd mode analysis, we have

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**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

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**Couplers Lange Coupler Evolution of Lange coupler 1= input 2=output**

3=coupling 4=isolated

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Analysis Equivalent circuit Simplified circuit where

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**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)

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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)

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**continue Characteristic impedance of the line is Coupling**

The desired characteristic impedance in terms of coupling is

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**VHF/UHF Hybrid power splitter**

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**Guanella power divider (VHF/UHF)**

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