Power divider combiner and coupler By Professor Syed

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

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

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

Continue If all ports are matched properly , then Sii= 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.

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

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 1 x R 2 R 1 x R 3 R 1 x R 4 R 2 x R 3 R 2 x R 4 R 3 x R 4

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

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.

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

Continue Simplified by choosing S 12= S 34=a ; S 13=be j q and S 24= b e j j Where q + j = p + 2 np 2 cases 1. Symmetry Coupler q = j = p/2 2. Antisymmetry Coupler q =0 , j =p Both satisfy a 2 +b 2 =1

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

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

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

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

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

Resistive divider Zo/3

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

Analysis (even and odd mode) For simplicity all values are normalized to line characteristic impedance , I. e Zo = 50 W. For even mode Vg 2 = Vg 3 and for odd mode Vg 2 = -Vg 3. 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 Vg 2=4 V and Vg 3= 0 V.

Even mode Vg 2=Vg 3= 2 V Looking at port 2 Zine= Z 2/2 Therefore for matching Note: If then V 2 e= V since Zine=1 (the circuit acting like voltage divider) To determine V 2 e , using transmission line equation V(x) = V+ (e-jbx + Ge+jbx) , thus Reflection at port 1, refer to Then is

Odd mode Vg 2= - Vg 3= 2 V At port 2, V 1 o =0 (short) , l/4 transformer will be looking as open circuit , thus Zino = r/2. We choose r =2 for matching. Hence V 2 o= 1 V (looking as a voltage divider) S-parameters S 11= 0 (matched Zin=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)

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

Parad and Moynihan power divider 2 1 3

Cohn power divider 2 1 VSWR at port 2 and port 3 Isolation between port 2 and 3 Center frequency fo Frequency range (f 2/f 1) = 1. 106 = 1. 021 = 27. 3 d. B = (f 1 + f 2)/2 =2 3

Couplers Branch line coupler E 1 E 2 x d. B coupling E 3 or

Couplers 3 d. B Branch line coupler input Output 3 d. B isolate Output 3 d. B 90 o out of phase

Couplers 9 d. B Branch line coupler Let say we choose Note: Practically upto 9 d. B 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 2 as two output. Port is match with 50 ohm. Output in-phase • Can be used as combiner , 2 and 3 as input and 1 as output. Port 4 is matched with 50 ohm.

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 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.

Even mode Continue The resulting capacitance Ce = C 11 = C 22 Therefore, the line characteristic impedance Odd mode The resulting capacitance Co = C 11 + 2 C 12 = C 22 + 2 C 12 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 Coupling Isolated (can be matched) Layout output input Schematic circuit

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

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

continue Substituting eqs. (104) - (107) into eq. (101) yeilds (108) Let Therefore eqs. (102) and (103) become (109) (110) 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 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)

continue We define coupling as and Then V 3 / V , from ( 112) will become Similarly V 1=V

Practical couple line coupler V 3 is maximum when q = p/2 , 3 p/2, . . . Thus for quarterwave length coupler q = p/2 , the eqs V 2 and V 3 reduce to V 1=V

Example Design a 20 d. B single-section coupled line coupler in stripline with a 0. 158 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 Characteristic impedance of even and odd mode are Then multiplied by From fig 7. 29 , we have w/b=0. 72 , s/b =0. 34. These give us w=0. 72 b=0. 114 cm s= 0. 34 b = 0. 054 cm

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

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

Example Design a three-section 20 d. B 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 d. B at q =p/2. Thus C= 10 -20/20=0. 1 From (202), we C= C 2 - 2 C 1= 0. 1 Solving © and © © gives us ©© C 1= C 3 = 0. 0125 (symmetry) and C 2 = 0. 125

continue Using even and odd mode analysis, we have

continue Let say , er = 10 and d =0. 7878 mm Plot points on graph Fig. 7. 30 We have , w/d = 1. 0 and s/d = 2. 5 , thus w = d = 0. 7878 mm and s = 2. 5 d = 1. 9695 mm For section 1 and 3 Similarly we plot points We have , w/d = 0. 95 and s/d = 1. 1 , thus w = 0. 95 d = 0. 748 mm and s =1. 1 d = 0. 8666 mm For section 2

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

Analysis Simplified circuit where Equivalent circuit

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)