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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 SPS Penang

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Power divider and combiner/coupler divider combiner P1P1 P 2 = nP 1 P 3 =(1-n)P 1 P1P1 P2P2 P 3 =P 1 +P 2 Divide into 4 output Basic

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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 S ii = 0 For Reciprocal network For lossless network, must satisfy unitary condition Two of (S 12, S 13, S 23 ) must be zero but it is not consistent. If S 12 =S 13 = 0, then S 23 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 |S 13 |=|S 23 | thus S 13 =S 23 =0 and |S 12 |=|S 33 |=1 These have satisfied all

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Reciprocal lossless network of two matched

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For lossless network, must satisfy unitary condition Nonreciprocal network (apply for circulator) 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 R 2x R3 R1x R4 R1x R3 R2x R4 R3x R4

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

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To check validity Multiply eq. * by S 24 * and eq. ## by S 13 *, and substract to obtain Multiply eq. # by S 34 and eq. by S 13, and substract to obtain % $ Both equations % and $ will be satisfy if S 14 = S 23 = 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 S 14 =S 23 =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 Simplified by choosing S 12 = S 34 = ; S 13 = e j and S 24 = e j Where + = + 2n 1. Symmetry Coupler = = /2 2. Antisymmetry Coupler =0, = 2 cases Both satisfy 2 + 2 =1

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Physical interpretation |S 13 | 2 = coupling factor = 2 |S 12 | 2 = power deliver to port 2= 2 =1- 2 Characterization of coupler Directivity= D= 10 log Coupling= C= 10 log Isolation = I= 10 log I = D + C dB For ideal case |S 14 |=0

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Practical coupler Hybrid 3 dB couplers Magic -T and Rat-race couplers = = /2 =0, = = = 1 /

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

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T-model Lossy line Lossless line If Z o = 50,then for equally divided power, Z 1 = Z 2 =100

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

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Resistive divider Z o /3

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Wilkinson Power Divider For even mode Therefore For Z in =Z o =50 And shunt resistor R =2 Z o = 100

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Analysis (even and odd mode) For even mode V g2 = V g3 and for odd mode V g2 = -V g3. 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 V g2 =4V and V g3 = 0V. For simplicity all values are normalized to line characteristic impedance, I.e Z o = 50

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Even mode V g2 =V g3 = 2V Looking at port 2 Z in e = Z 2 /2 Therefore for matching then V 2 e = V since Z in e =1 (the circuit acting like voltage divider) Note: If To determine V 2 e, using transmission line equation V(x) = V+ (e -j x + e +j x ), thus Reflection at port 1, refer to is Then

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Odd mode V g2 = - V g3 = 2V At port 2, V 1 o =0 (short), /4 transformer will be looking as open circuit, thus Z in o = r/2. We choose r =2 for matching. Hence V 2 o = 1V (looking as a voltage divider) S-parameters S 11 = 0(matched Z in =1 at port 1) S 22 = S 33 = 0 (matched at ports 2 and 3 both even and odd modes) S 12 = S 21 = S 13 = S 31 = S 23 = S 32 = 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 1 2 3

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Parad and Moynihan power divider 1 2 3

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Cohn power divider VSWR at port 1 = VSWR at port 2 and port 3 = Isolation between port 2 and 3 = 27.3 dB Center frequency f o = (f 1 + f 2 )/2 Frequency range (f 2 /f 1 ) =

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Couplers Branch line coupler x dB coupling or E1E1 E2E2 E3E3

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

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

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Couplers Input Output in-phase isolated 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. Hybrid-ring coupler

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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 Even mode Odd mode C 11 and C 22 are the capacitances between conductors and the ground respectively. For symmetrical coupled line C 11 =C 22. C 12 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 C 12 is effectively open-circuited.

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Continue Even mode The resulting capacitance C e = C 11 = C 22 Therefore, the line characteristic impedance Odd mode The resulting capacitance C o = C C 12 = C C 12 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 input output Isolated (can be matched) Coupling Schematic circuit Layout

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Even and odd modes analysis I 1 e = I 3 e I 4 e = I 2 e Same excitation voltage V 1 e = V 3 e V 4 e = V 2 e Even I 1 o = -I 3 o I 4 o =- I 2 o V 1 o = -V 3 o V 4 o = -V 2 o Odd Reverse excitation voltage (100) (99)

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Analysis Z o = load for transmission line = electrical length of the line Z oe or Z oo = characteristic impedance of the line By voltage division From transmission line equation, we have where (101) (102) (103) (104) (105) (106) (107)

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

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continue Since Z in = Z o, then by voltage division V 1 = 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 We define coupling as Then V 3 / V, from ( 112) will become and Similarly V 1 =V

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Practical couple line coupler V 3 is maximum when = /2, 3 /2,... Thus for quarterwave length coupler = /2, the eqs V 2 and V 3 reduce to V 1 =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 , and a center frequency of 3 GHz. Coupling factor is C = /20 = 0.1 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 Then multiplied by

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Multisection Coupled line coupler (broadband) For single section, whence C<<1, then V 4 =0 and For = / 2 then V 3 /V 1 = C and V 2 /V 1 = -j

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

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

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Continue Apply (201) Midband C o = 20 dB at = /2. Thus C= /20 =0.1 From (202), we C= C 2 - 2C 1 = 0.1 © © © Solving © and © © gives us C 1 = C 3 = (symmetry) and C 2 = 0.125

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continue Using even and odd mode analysis, we have

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

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

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Analysis Simplified circuit Equivalent circuit where

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