Power divider, combiner and coupler.ppt

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

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

3. S-parameter for power divider/coupler              33 32 31 23 22 21 13 12 11 S S S S S S S S S S Generally For reciprocal and lossless network j i for S S N k kj ki     0 1 * 1 1 *    N k ki ki S S 1 13 12 11    S S S 1 23 22 21    S S S 1 33 32 31    S S S 0 * 23 13 * 22 12 * 21 11    S S S S S S 0 * 33 23 * 32 22 * 31 21    S S S S S S 0 * 33 13 * 32 12 * 31 11    S S S S S S Row 1x row 2 Row 2x row 3 Row 1x row 3

4. Continue If all ports are matched properly , then Sii= 0              0 0 0 23 13 23 12 13 12 S S S S S S S For Reciprocal network For lossless network, must satisfy unitary condition 1 2 13 2 12   S S 1 2 23 2 12   S S 1 2 23 2 13   S S 0 12 * 23  S S 0 23 * 13  S S 0 13 * 12  S S 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              33 23 13 23 12 13 12 0 0 S S S S S S S S Only two ports are matched , then for reciprocal network For lossless network, must satisfy unitary condition 1 2 13 2 12   S S 1 2 23 2 12   S S 1 2 33 2 23 2 13    S S S 0 13 * 33 12 * 23   S S S S 0 23 * 13  S S 0 33 * 23 13 * 12   S S S S 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 S21 =ej  S12=ej  S33 =ej  1 3 2                   j j j e e e S 0 0 0 0 0 0

7. For lossless network, must satisfy unitary condition 1 2 13 2 12   S S 1 2 23 2 21   S S 1 2 32 2 31   S S 0 32 * 31  S S 0 23 * 21  S S 0 13 * 12  S S Nonreciprocal network (apply for circulator)              0 0 0 32 31 23 21 13 12 S S S S S S S 0 31 23 12    S S S 0 13 32 21    S S S 1 13 32 21    S S S 1 31 23 12    S S S The above equations must satisfy the following either or

8. Circulator (nonreciprocal network)              0 1 0 0 0 1 1 0 0 S              0 0 1 1 0 0 0 1 0 S 1 2 3 1 2 3

9. Four port network                  44 43 42 41 34 24 14 33 32 31 23 22 21 13 12 11 S S S S S S S S S S S S S S S S S Generally For reciprocal and lossless network j i for S S N k kj ki     0 1 * 1 1 *    N k ki ki S S 1 14 13 12 11     S S S S 1 24 23 22 21     S S S S 1 34 33 32 31     S S S S 0 * 24 14 * 23 13 * 22 12 * 21 11     S S S S S S S S 0 * 44 24 * 43 23 * 42 22 * 41 21     S S S S S S S S 0 * 34 14 * 33 13 * 32 12 * 31 11     S S S S S S S S R 1x R 2 R 2x R3 R1x R4 1 44 43 42 41     S S S S 0 * 44 14 * 43 13 * 42 12 * 41 11     S S S S S S S S 0 * 34 24 * 33 23 * 32 22 * 31 21     S S S S S S S S 0 * 44 34 * 43 33 * 42 32 * 41 31     S S S S S S S S R1x R3 R2x R4 R3x R4

10. Matched Four port network                  0 0 0 0 34 24 14 34 24 14 23 13 23 12 13 12 S S S S S S S S S S S S S The unitarity condition become 1 14 13 12    S S S 1 24 23 12    S S S 1 34 23 13    S S S 0 * 24 14 * 23 13   S S S S 0 * 34 23 * 14 12   S S S S 0 * 34 14 * 23 12   S S S S 1 34 24 14    S S S 0 * 34 13 * 24 12   S S S S 0 * 34 24 * 13 12   S S S S 0 * 24 23 * 14 13   S S S S Say all ports are matched and symmetrical network, then * ** @ @@ # ##

11. To check validity Multiply eq. * by S24 * and eq. ## by S13 * , and substract to obtain 0 2 14 2 13 * 14         S S S Multiply eq. # by S34 and eq. @@ by S13 , and substract to obtain 0 2 34 2 12 23         S S S % $ 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                  0 0 0 0 0 0 0 0 34 24 34 24 13 12 13 12 S S S S S S S S S If all ports matched , symmetry and S14=S23=0 to be satisfied The equations reduce to 6 equations 1 13 12   S S 1 24 12   S S 1 34 13   S S 1 34 24   S S 0 * 34 13 * 24 12   S S S S 0 * 34 24 * 13 12   S S S S 24 13 S S  By comparing these equations yield * * ** ** By comparing equations * and ** yield 34 12 S S 

13. Continue                0 0 0 0 0 0 0 0         j j j j S Simplified by choosing S12= S34= ; S13=e j  and S24=  e j  Where  +  = p + 2np                  0 0 0 0 0 0 0 0         S 1. Symmetry Coupler  =  = p/2 2. Antisymmetry Coupler  =0 , =p 2 cases Both satisfy 2 +2 =1

14. Physical interpretation |S13 | 2 = coupling factor = 2 |S12 | 2 = power deliver to port 2= 2 =1- 2 Characterization of coupler Directivity= D= 10 log dB P P  log 20 3 1   Coupling= C= 10 log dB S P P 14 4 3 log 20    Isolation = I= 10 log dB S P P 14 4 1 log 20   I = D + C dB 1 4 3 2 Input Through Coupled Isolated For ideal case |S14|=0

15. Practical coupler Hybrid 3 dB couplers Magic -T and Rat-race couplers  =  = p/2                0 1 0 1 0 0 0 0 0 1 1 0 2 1 j j j j S                  0 1 1 0 1 1 0 0 0 1 0 0 1 1 1 0 2 1 S  =0 , =p =  = 1 / 2 =  = 1 / 2

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

17. T-model jB Z1 Z2 Vo Yin 2 1 1 1 Z Z jB Yin    2 1 1 1 Z Z Yin   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 in o P Z V P 3 1 2 1 1 2 1   in o P Z V P 3 2 2 1 2 2 2      75 2 3 2 o Z Z    150 3 1 o Z Z o o in Z V P 2 2 1     50 // 2 1 Z Z Zo

19. Resistive divider V2 V3 V1 Zo Zo P1 P2 P3 Zo V o o Z Z Z   3 Zo/3 Zo/3 Zo/3 o o o in Z Z Z Z    3 2 3 V V Z Z Z V o o o 3 2 3 / 2 3 / 3 / 2 1    V V V Z Z Z V V o o o 2 1 4 3 3 / 3 2      o in Z V P 2 1 2 1    in o P Z V P P 4 1 2 / 1 2 1 2 1 3 2   

20. Wilkinson Power Divider 50 50 50 100 70.7  70.7  /4 Zo /2 Zo /2 Zo 2Zo Zo Zo /4 2 2 T e Z Zin  o T Z Z 2  For even mode Therefore For Zin =Zo=50     7 . 70 50 2 T Z And shunt resistor R =2 Zo = 100

21. Analysis (even and odd mode) 2 2 1 1 Port 1 Port 2 Port 3 Vg2 Vg3 Z Z 4 +V2 +V3 r/2 r/2 4 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 .

22. Even mode Vg2=Vg3= 2V Looking at port 2 Zin e= Z2/2 Therefore for matching 2  Z then V2 e= V since Zin e=1 (the circuit acting like voltage divider) 2 1 Port 1 Port 2 2V Z 4 +V2 e r/2 +V1 e O.C O.C out inZ Z Z  2 Note: 2  Z If To determine V2 e , using transmission line equation V(x) = V+ (e-jx + Ge+jx) , thus V jV V V e  G     ) 1 ( ) 4 ( 2  1 1 ) 1 ( ) 0 ( 1  G  G  G     jV jV V V e Reflection at port 1, refer to is 2 2 2 2    G 2  Z Then 2 1 jV V e  

23. Odd mode Vg2= - Vg3= 2V 2 1 Port 1 Port 2 2V Z 4 +V2 o r/2 +V1 o At port 2, V1 o =0 (short) , /4 transformer will be looking as open circuit , thus Zin o = r/2 . We choose r =2 for matching. Hence V2 o= 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 = 2 / 2 2 1 1 j V V V V o e o e     S13 = S31 = 2 / j  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    7 . 70 2 o Z Z    100 2 o Z R r o   4   The quarterwave-transformer length is

25. Wilkinson splitter/combiner application /4 100 70.7 50 matching networks /4 100 50 70.7 70.7 70.7 Splitter combiner Power Amplifier

26. Unequal power Wilkinson Divider 3 2 03 1 K K Z Z o   ) 1 ( 2 03 2 02 K K Z Z K Z o            K K Z R o 1 R2 =Zo /K R R3=Zo/K Z02 Z03 Zo 2 3 2 3 2     port at Power port at Power P P K 1 2 3

27. Parad and Moynihan power divider 4 / 1 2 01 1         K K Z Z o 2 3 2 3 2     port at Power port at Power P P K         K K Z R o 1   4 / 1 2 4 / 3 02 1 K K Z Z o     4 / 5 4 / 1 2 03 1 K K Z Z o   K Z Z o  04 K Z Z o  05 Zo Zo Zo Z05 Zo4 Zo2 Zo3 Zo1 R 1 2 3

28. Cohn power divider            VSWR at port 1 = 1.106 VSWR at port 2 and port 3 = 1.021 Isolation between port 2 and 3 = 27.3 dB Center frequency fo = (f1 + f2)/2 Frequency range (f2/f1) = 2 1 2 3

29. Couplers /4 /4 Yo Yo Yo Yo Yse Ysh Ysh Branch line coupler 2 sh 2 se Y 1 Y   2 se 2 sh sh 2 3 Y Y 1 2Y E E      20 1 3 10 E E x   x dB coupling 2 3 2 2 2 1 E E E   2 1 3 2 1 2 E E E E 1                   or E1 E2 E3

30. Couplers input isolate Output 3dB Output 3dB 90o out of phase 3 dB Branch line coupler /4 /4 Zo Zo Zo Zo 2 / Zo 2 / Zo Zo Zo 3 2 E E  1 Ysh  2 Y 1 Y 2 2 se    sh 1.414 Yse    50 o Z   50 sh Z   5 . 35 se Z

31. Couplers 9 dB Branch line coupler   355 . 0 10 20 9 1 3    E E  2 2 1 2 355 . 0 1           E E   935 . 0 355 . 0 1 2 1 2            E E 38 . 0 935 . 0 355 . 0 2 3           E E 8 . 0  sh Y Let say we choose 38 . 0 8 . 0 1 8 . 0 2 1 2 2 2 2 2        se se sh sh Y Y Y Y 962 . 1 36 . 0 38 . 0 6 . 1    se Y   50 0 Z     5 . 62 8 . 0 / 50 sh Z     5 . 25 962 . 1 / 50 se Z Note: Practically upto 9dB coupling

32. Couplers /4 /4 /4 /4 Input Output in-phase Output in-phase isolated 1 2 3 4 •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 OC 1 2 1 2 OC 1/2 1/2 2 2 2 2 2 2 /8 /8 /4 /4 /8 /8 Te To Ge Go

33. Analysis The amplitude of scattered wave o e B G  G  2 1 2 1 1 o e T T B 2 1 2 1 4   o e T T B 2 1 2 1 2   o e B G  G  2 1 2 1 3

34. Couple lines analysis Planar Stacked Coupled microstrip b w w s w s w w s b d r r r 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 +V +V H-wall +V -V E-wall C11 C22 C11 C22 2C12 2C12 Even mode Odd 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 e e e e oe C C LC C L Z  1    Therefore, the line characteristic impedance Odd mode The resulting capacitance Co = C11 + 2 C12 = C22 + 2 C12 Therefore, the line characteristic impedance o oo C Z  1 

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

38. Stacked coupled stripline     m F s b b s b s b C oW r oW r oW r / 4 2 / 2 / 2 2 11             w >> s and w >> b m F s C oW r / 12    m F s b b C C oW r e / 4 2 2 11      m F s s b b w C C C o r o / 1 2 2 2 2 2 12 11                o o r     1  r o e oe bw s b Z C Z   4 1 2 2        s s b b w Z C Z r o o oo / 1 / 2 2 1 1 2 2      

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

40. Design of Coupled line Couplers input output Isolated (can be matched) Coupling w w s 2 3 4 1 wc /4 3 4 1 2 Zo Zo Zo Zo Zoo Zoe 2V +V3 +V2 +V4 +V1  I1 I4 I3 I2 Schematic circuit Layout

41. Even and odd modes analysis 3 4 1 2 Zo Zo Zo Zo Zoo V +V3 o +V2 o +V4 o +V1 o I1 o I4 o I3 o I2 o V _ + + _ 3 4 1 2 Zo Zo Zo Zo Zoe V +V3 e +V2 e +V4 e +V1 e I1 e I4 e I3 e I2 e V _ + + _ I1 e = I3 e I4 e = I2 e Same excitation voltage V1 e = V3 e V4 e = V2 e Even I1 o = -I3 o I4 o =- I2 o V1 o = -V3 o V4 o = -V2 o Odd Reverse excitation voltage (100) (99)

42. Analysis o o in o in o Z Z Z V V   1   tan tan o oe oe o oe e in jZ Z jZ Z Z Z    o e o e in I I V V I V Z 1 1 1 1 1 1     Zo = load for transmission line  = electrical length of the line Zoe or Zoo = characteristic impedance of the line   tan tan o oo oo o oo o in jZ Z jZ Z Z Z    By voltage division o e in e in e Z Z Z V V   1 o o in o Z Z V I   1 o e in e Z Z V I   1 From transmission line equation , we have where (101) (102) (103) (104) (105) (106) (107)

43. continue Substituting eqs. (104) - (107) into eq. (101) yeilds       o o in e in o e in o in o o o in e in o o in e in o e in o in in Z Z Z Z Z Z Z Z Z Z Z Z Z Z Z Z Z 2 2 2 2            For matching we may consider the second term of eq. (108) will be zero , I.e 0 2   o e in o in Z Z Z or 2 o oe oo e in o in Z Z Z Z Z   (108) Let oe oo o Z Z Z  Therefore eqs. (102) and (103) become   tan tan oo oe oe oo oe e in Z j Z Z j Z Z Z      tan tan oe oo oo oe oo o in Z j Z Z j Z Z Z    and (108) reduces to Zin=Zo (110) (109)

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                 o o in o in o e in e in o e o e Z Z Z Z Z Z V V V V V V 1 1 3 3 3 (111) Substitute (109) and (110) into (111)     tan 2 tan oo oe o oo o o o in o in Z Z j Z jZ Z Z Z Z          tan 2 tan oo oe o oe o o e in e in Z Z j Z jZ Z Z Z Z      Then (111) reduces to       tan 2 tan 3 oo oe o oo oe Z Z j Z Z Z j V V     (112)

45. continue We define coupling as oo oe oo oe Z Z Z Z C    Then V3 / V , from ( 112) will become oo oe o Z Z Z C    2 1 2               tan 1 tan tan 2 tan 2 3 j C jC V Z Z Z Z j Z Z Z Z Z Z Z j V V oo oe oo oe oo oe o oo oe oo oe           and   sin cos 1 1 2 2 2 2 2 j C C V V V V o e       0 2 2 4 4 4      o e o e V V V V V Similarly V1=V

46. Practical couple line coupler V3 is maximum when  = p/2 , 3p/2, ... Thus for quarterwave length coupler  = p/2 , the eqs V2 and V3 reduce to V1=V 0 4  V VC j C jC V j C jC V j C jC V V              2 2 2 3 1 1 ) ( 2 / tan 1 2 / tan p p 2 2 2 2 2 1 1 2 / sin 2 / cos 1 1 C jV j C V j C C V V          p p C C Z Z o oe    1 1 C C Z Z o oo    1 1

47. Example Design a 20 dB 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  , and a center frequency of 3 GHz. Coupling factor is C = 10-20/20 = 0.1 Characteristic impedance of even and odd mode are      28 . 55 1 . 0 1 1 . 0 1 50 oe Z 23 . 45 1 . 0 1 1 . 0 1 50     oo Z 4 . 88  oe r Z  4 . 72  oo r Z  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 r 

48. Multisection Coupled line coupler (broadband) V1 V3 V4 V2 input Through Isolated Coupled        C1 CN-2 C3 C2 CN CN-1 ....       j e jC j jC j C jC V V        sin tan 1 tan tan 1 tan 2 1 3    j e j C C V V       sin cos 1 1 2 2 1 2 For single section , whence C<<1 , then V4=0 and For = p/ 2 then V3/V1= C and V2/V1 = -j

49. Analysis Result for cascading the couplers to form a multi section coupler is               ) 1 ( 2 1 2 1 2 1 1 3 sin ... sin sin          N j j N j j j e V e jC e V e jC V e jC V             ) 1 ( ) 2 ( 2 2 2 ) 1 ( 2 1 1 3 ... 1 sin              N j M N j j N j j e C e e C e C e jV V       M jN C N C N C e jV 2 1 ... 3 cos 1 cos sin 2 2 1 1           Where M= (N+1)/2 For symmetry C1=CN , C2= CN-1 , etc At center frequency 2 / 1 3 p    V V Co (200)

50. 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 2 , 1 0 ) ( 2 /    n for C d d n n p    From eq (200) and M= (N+1)/2 =( 3+1)/2=2 , we have          2 1 1 3 2 1 2 cos sin 2 C C V V C          sin ) ( 3 sin sin sin 3 sin 1 2 1 2 1 C C C C C       (201) (202)

51. Continue Apply (201)   0 cos ) ( 3 cos 3 2 / 1 2 1     p    C C C d dC   0 10 sin ) ( 3 sin 9 2 1 2 / 1 2 1 2 2         C C C C C d C d p    Midband Co= 20 dB at  =p/2. Thus C= 10-20/20=0.1 From (202), we C= C2 - 2C1= 0.1 © © © Solving © and © © gives us C1= C3 = 0.0125 (symmetry) and C2 = 0.125

52. continue Using even and odd mode analysis, we have          63 . 50 0125 . 0 1 0125 . 0 1 50 1 1 3 1 C C Z Z Z o oe oe       38 . 49 0125 . 0 1 0125 . 0 1 3 1 o oo oo Z Z Z         69 . 56 125 . 0 1 125 . 0 1 50 1 1 2 C C Z Z o oe      1 . 44 125 . 0 1 125 . 0 1 2 o oo Z Z

53. continue Let say , r = 10 and d =0.7878mm    63 . 50 3 1 oe oe Z Z    38 . 49 3 1 oo oo Z Z   69 . 56 2 oe Z   1 . 44 2 oo Z Plot points on graph Fig. 7.30 We have , w/d = 1.0 and s/d = 2.5 , thus w = d = 0.7878mm and s = 2.5d = 1.9695mm 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 = 0.8666mm For section 1 and 3 For section 2

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

55. Analysis 1 4 3 2 1 3 4 2 C C 90o Ze4 Zo4 Zo4 Ze4 1 4 3 2 1 2 Cm Cex Cex C Cex Cex Cin Cin Cm Cm Cm Simplified circuit Equivalent circuit m ex m ex ex in C C C C C C    where

56. Continue/ 4 wire coupler Even mode All Cm capacitance will be at same potential, thus the total capacitance is in ex e C C C   4 m in ex o C C C C 6 4    Odd mode All Cm capacitance will be considered, thus the total capacitance is Even and Odd mode characteristic impedance 4 4 1 e e C Z   4 4 1 o o C Z   line on transmissi in velocity   (300) (301) (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 ex e C C  m ex o C C C 2   Substitute these into (300) and (301) , we have   o e o e e e C C C C C C    3 4 m ex m ex ex in C C C C C C      o e e o o o C C C C C C    3 4 And in terms of impedance refer to (302) oe oe oo oe oo e Z Z Z Z Z Z    3 4 oo oo oe oe oo o Z Z Z Z Z Z    3 4

58. continue      oo oe oe oo oe oo oo oe o e o Z Z Z Z Z Z Z Z Z Z Z      3 3 2 4 4 Characteristic impedance of the line is     oo oe oo oe oo oe o e o e Z Z Z Z Z Z Z Z Z Z C 2 3 3 2 2 2 2 4 4 4 4        Coupling The desired characteristic impedance in terms of coupling is     o oe Z C C C C C Z       1 / 1 2 8 9 3 4 2     o oo Z C C C C C Z       1 / 1 2 8 9 3 4 2

59. VHF/UHF Hybrid power splitter 50 input 50 output 50 output 100 C T1 T2 1 5 6 7 8 2 3 4

60. Guanella power divider (VHF/UHF) RL V2 I2 I1 V1 Rg Vg I1 V2 I2

Power divider, combiner and coupler.ppt

Vous êtes en train de lire un aperçu.

Consultez-les par la suite

Power divider, combiner and coupler.ppt

Plus De Contenu Connexe

Similaire à Power divider, combiner and coupler.ppt (20)

Plus récents (20)