Chapter 4 Impedance Matching

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1 Chapter 4 Impedance Matching Quarter-wave transformer, series section transformer Stub matching, lumped element networks, feed point location 3 Gamma match 4 Delta- and T-match, Baluns

2 -port network Smith Chart EVISION

3 -Port Network epresentation I I Two-Port V Network V Six ways to represent a two-port network in terms of V and I at each port: -matrix open-circuit impedance Y-matrix short-circuit admittance ABCD-matrix chain or transmission parameters B-matrix inverse transmission parameters H-matrix hybrid parameters G-matrix inverse hybrid parameters 3

4 -Port Network epresentation I I V Two-Port Network V 7th way to represent a two-port network in terms of waves entering and leaving each port: S-matrix scattering parameters In general, there could be n ports in a network: n-port network 3 n 4

5 [S] for Two-Port Network a a ~ b o Two-Port Network b o Definition of normalized voltage waves: Vi Incident wave: a = = Pi eflected wave: o V b = = r r o P V a = = i i o V b = = P r r o P Scattering parameters are then defined as: b + = Sa Sa b + = Sa Sa or outputs b b S = S S S a a or inputs [ b ] = [ S][ a] 5

6 Sfor a One-Port Network Measured S of an 800 MHz square microstrip patch antenna on a.6 mm F4 substrate Antenna Bandwidth ( =3 db) 6

7 S for a One-Port Network Measured S of a 54 MHz hair-pin bandpass filter on a.6 mm F4 substrate 7

8 Smith Chart Impedance or Admittance 8

9 Smith Chart: Exercises. Use as a -chart, locate s/c, o/c, o, + j. Use as a Y-chart, locate s/c, o/c, Yo, + j 3. Move towards generator l/l 4. Move towards load l/l 5. ead VSW, G, Vmax, Vmin 6. ead attenuation 9

10 Introduction Importance of IMPEDANCE MATCHING 0

11 INTODUCTION The operation of an antenna system over a frequency range is not completely dependent upon the frequency response of the antenna element itself but rather on the frequency characteristics of the transmission line-antenna element combination. In practice, the characteristic impedance of the transmission line is usually real whereas that of the antenna element is complex. Also the variation of each as a function of frequency is not the same. The efficient coupling-matching networks must be designed which attempt to couple-match the characteristics of the two devices over the desired frequency range.

12 Feed-Point Impedance: a a = antenna impedance at its feed-point. a = + a a jx a a is complex generally.

13 Importance of Impedance Matching Increased power throughput (e.g. maximum power transfer) Increased power handling capability in a transmission line (due to reduced VSW) educed effects on impedance matching sensitive circuits (e.g. frequency pulling effect on the signal source) With "controlled mismatch", an F amplifier can operate with minimum noise generation (i.e. minimum noise figure) 3

14 Concept of maximum power In lump circuit o transfer P V i I V o Power deliver at is P = V I = I = Vi + o Power maximum whence = o 4

15 continue In transmission line The important parameter is reflection coefficient ρ = + o o No reflection whence = o, hence ρ = 0 The load can be matched as long as not equal to zero (shortcircuit) or infinity (open-circuit) 5

16 Features of Complexity Networks use the simplest --> cheap, low loss Bandwidth perfect match usually at a single frequency wider bandwidth increases complexity Implementation select matching components to suit an application: lumped-element (, C, ) distributed-element (transmission line, waveguides) Adjustability for loads with variable impedance 6

17 calculations smith chart MATCHING METHODS 7

18 Matching with lumped elements The simplest matching network is an -section using two reactive element jx Configuration Whence > o o jb = +jx jx Configuration Whence < o o jb 8

19 continue If the load impedance (normalized) lies in unity circle, configuration is used.otherwise configuration is used. The reactive elements are either inductors or capacitors. So there are 8 possibilities for matching circuit for various load impedances. Matching by lumped elements are possible for frequency below GHz or for higher frequency in integrated circuit(mic, MEM). Configuration Configuration 9

20 Impedances for serial lumped elements Serial circuit eactance relationship values +ve X=πf =X/(πf) -ve X=/(πfC) C=/(πfX) C 0

21 Impedances for parallel lumped elements Parallel circuit Susceptance relationship values +ve B=πfC C=B/(πf) -ve B=/(πf) =/(πfb) C

22 umped elements for microwave integrated circuit ossy film Planar resistor Chip resistor oop inductor Spiral inductor Dielectric ε r ε r Interdigital gap capacitor Metal-insulatormetal capacitor Chip capacitor

23 Matching by calculation for configuration jx o jb For matching, the total impedance of -section plus should equal to o,thus o = jx + jb + / ( + jx ) earranging and separating into real and imaginary parts gives us B ( X X o ) = o X ( BX ) = Bo X * ** 3

24 continue Solving for X from simultaneous equations (*) and (**) and substitute X in (**) for B, we obtain B = X ± / o + X + X o +ve capacitor -ve inductor Since > o, then argument of the second root is always positive, the series reactance can be found as X = B + X o o B +ve inductor -ve capacitor Note that two solution for B are possible either positive or negative 4

25 Matching by calculation for configuration jx o jb For matching, the total impedance of -section plus should equal to / o, thus = jb + + j + o ( X X ) earranging and separating into real and imaginary parts gives us B o ( X + X ) = o ( X + X ) = Bo * ** 5

26 continue Solving for X and B from simultaneous equations (*) and (**), we obtain X = ± ( o ) X +ve inductor -ve capacitor B = ± ( ) o o / +ve capacitor -ve inductor Since < o,the argument of the square roots are always positive, again two solution for X and B are possible either positive or negative 6

27 parallel capacitance (+ve) serial inductance (+ve) (-ve) parallel inductance serial capacitance (-ve) 7

28 Serial C Serial Matching using lumped components Parallel C 0Ω C 50Ω C Parallel C 50Ω C! 0Ω 8

29 Example Design an -section matching network to match a series C load with an impedance =00-j00 W, to a 00 W line, at a frequency of 500 MHz. Solution Normalized we have : = -j Parallel (-j0.7) Serial C (-j.) Serial (j.) = -j Parallel C (+j0.3) Solution Solution 9

30 continue C b = π f o = 0.9 pf 38.8nH 0.9pF 00-j00 x o = = 38. 8nH π f C = =.6pF π fx o = o = 46. nh π f b.6pf 46.nH =00-j00 eflection coefficient Solution seems to be better matched at higher frequency reflection coefficient freq (GHz) solution solution 30

31 Discrete lumped-element Transmission ine MATCHING TECHNIQUES 3

32 Impedances for serial lumped elements Serial circuit eactance relationship values +ve X=πf =X/(πf) -ve X=/(πfC) C=/(πfX) C 3

33 Impedances for parallel lumped elements Parallel circuit Susceptance relationship values +ve B=πfC C=B/(πf) -ve B=/(πf) =/(πfb) C 33

34 umped elements for microwave integrated circuit ossy film Planar resistor Chip resistor oop inductor Spiral inductor Dielectric ε r ε r Interdigital gap capacitor Metal-insulator-metal capacitor Chip capacitor 34

35 esistive -Section (Using Advantage: Disadvantage: esistors) Broadband Very lossy Matched conditions: (a) o = + ( // o ) (b) = //( + ) o o o o Solving (a) and (b) to give and : = ( ) o o o = o o o o > (a) o o (b) 35 35

36 esistive -Section (Using esistors) Advantage: Disadvantage: Broadband Very lossy Attenuation: o V V out in = o + // + ( o // o ) V in ~ (a) (b) o V out > o o 36

37 esistive -Section: Example Design a broadband resistive -section matching network to match a 75 Ω TV antenna output to a 50 Ω transmission line. Calculate the attenuation of the matching network. Solution: o = 75 Ω, o = 50 Ω Substituting these values into previous equations, we get: = 43.3 Ω, = 86.6 Ω V V out in = = 0.3 = 3.5 db Balun 75 Ω txn line 75 Ω-50 Ω matching network 50 Ω txn line 37 37

38 eactive -Section (Using & C) Advantages: Disadvantages: ow loss, simplicity in design Narrow-band, fixed Q Matched conditions: p > s X s & X p : opposite sign (a) = jx + s s ( jx p // p ) jx s (b) p = jx p //( s + jx s ) s jx p p Solving (a) and (b) to give X s and X p : Q = X s p s p = Q. s = s s X p = p Q = p s p (a) (b) X s = +ve for, -ve for C X p = -ve for C, +ve for 38 38

39 eactive C -Section: Example Design an -section matching network with and C to match a 50 Ω source to a 600 Ω load for maximum power transfer at 400 MHz. Give two solutions. Solution: s = 50 Ω, p = 600 Ω Substituting these values into previous equations, we get: Q = p s 600 = = = X s = Q. s = 66 Ω X p = p /Q = 8 Ω 39

40 eactive C -Section: Example (cont'd) Solution : X s = +66 Ω (inductive), X p = 8 Ω (capacitive) X = s 66 s 66 nh 6 πf = π * 400 *0 = C p = = =. 6 πf X π * 400 *0 *8 p pf Solution : X s = 66 Ω (capacitive), X p = +8 Ω (inductive) X = p 8 p 7 nh 6 πf = π * 400 *0 = C πf X = = =. 4 s 6 s π * 400 *0 *66 pf 40

41 eactive C -Section: Example (cont'd) Solution : X s = +66 Ω (inductive), X p = 8 Ω (capacitive) Solution : X s = 66 Ω (capacitive), X p = +8 Ω (inductive) 4

42 -Section for Complex Impedances If s has a reactance jx', simply introduce a series with an equal but opposite-sign reactance ( jx'). extra reactances jx' jx' jx s s jx p p The jx' reactance is then combined with jx s to give its final value. Similar treatment can be applied to p by introducing a parallel jx' and a parallel jx'. 4 4

43 T-Section Advantages: can be < or > Variable Q (higher than -section) Disadvantages: More C elements jx jx 3 jx ' jx jx " - - Method: h (a) Introduce a hypothetical resistance level h at jx, such that h > and h > ( h determines the Q of the matching network) (b) Split jx into jx ' and jx " (in parallel) (c) Treat the T-section as two -sections (- and -) NOTE: Q of - section > Q of -element -section (as h > ) 43 43

44 T-Section: Q Values Using the resistance values of the previous -element -section: 50 Ω 600 Ω (a) Q of original -section: Q = 600 = = (b) Q of T-section if h = 5050 Ω (Note: Q > Q ) - section: - section: h 5050 Q = = = h 5050 Q = = =

45 T-Section: Design Example Design a T-section C matching network to match = 50 Ω and = 600 Ω at 400 MHz, and Q associated with is Ω 600 Ω (a) For - network, select Q = 0. (b) (c) (d) h X X ' = ( Q + ) = 50(0 + ) = 5050 Ω h 5050 = = = 505Ω Q 0 = = 50 *0 = 500Ω Q ve for C (arbitrarily selected) +ve for (opposite of X ' ) 45 45

46 T-Section: Design Example 50 Ω 600 Ω (a) For - network, (b) (c) (d) (e) (f) (g) X X X h 5050 " = = = 854 Ω Q.73 = = 600 *.73 = 634 Ω 3 Q 46 h 5050 Q = = = X '. X " 505 *854 = X ' // X " = = = 694Ω X ' + X " X 500 X = = = 99nH 6 πf π * 400 *0 X C = = = pf 6 πf X π * 400 *0 * 694 X 3 C3 = = = 0. 4 pf 6 πf X π * 400 *0 * ve for (arbitrarily selected) ve for C (opposite of X '' ) ve means C 46

47 T-Section: Design Example (cont'd) C = 99 nh = pf C = pf 50 Ω 600 Ω "AUTO-TANSFOME" another possible solution 47 47

48 π-section Advantages: can be < or > Variable Q (higher than -section) Disadvantage: More C elements jx ' jx jx " jx - - jx 3 Method: h (a) Introduce a hypothetical resistance level h at jx, such that h < and h < ( h determines the Q of the matching network) (b) Split jx into jx ' and jx " (in series) (c) Treat the π-section as two -sections (- and -) NOTE: Q of - section > Q of -element -section (as h < ) 48 48

49 π-section: Design Example Design a π-section C matching network to match = 50 Ω and = 600 Ω at 400 MHz, and Q associated with is Ω 600 Ω (a) For - network, select Q = 0. (b) (c) h = 50 = = 0. Ω Q ' X = h = *0 = 4. 95Ω +ve for (arbitrarily selected) Q (d) X 50 = = = 5. 0Ω Q 0 ve for C (opposite of X ' ) 49 49

50 π-section: Design Example (cont'd) 50 Ω 600 Ω (a) For - network, (b) (c) (d) (e) (f) (g) X X X " hq 3 = = * = 7. 3Ω 600 = = = 7. 4Ω Q Q = = = h = X ' + " = =. 8Ω X X C = = = pf 6 πf X π * 400 *0 *5 X.8 X = = = 8. 83nH 6 πf π * 400 *0 X 3 C3 = = = pf 6 πf X π * 400 *0 * ve for (arbitrarily selected) ve for C (opposite of X '' ) +ve means 50

51 π-section: Design Example (cont'd) C = pf = 8. 83nH C = pf 3 50 Ω 600 Ω This is just one of the four possible solutions. 5 5

52 Common Transmission-ine Inductor and capacitor Quarter-wave transformer Single-stub tuner Double-stub tuner Triple-stub tuner Applications [A good reference for the Double-and Triple-stub tuners is:.e.collin, Foundations for Microwave Engineering, McGraw Hill] 5 5

53 Transmission ine Inductor The feed-point impedance of a terminated transmission line is: in = o o + + j j o tan( βl ) tan( βl ) If the load impedance is a short-circuit, = 0, then in = j o tan( β l ) = jω Thus, the terminated transmission line behaves like an inductor (inductance = ) if bl< p/. o tan( β l ) = ω 53 53

54 Example What is the equivalent inductance at GHzat the feedpoint of a 50 W, l/8 transmission line which is terminated with a short-circuit? π λ 50tan. o tan( βl ) 8 λ = = = 7. 06nH 9 ω π ( 0 ) l o, β in in 54

55 Transmission ine Capacitor The feed-point impedance of a terminated transmission line is: in = o o + + j j o tan( βl ) tan( βl ) If the load impedance is an open-circuit, =, then in o = = j tan( βl ) jωc Thus, the terminated transmission line behaves like a capacitor (capacitance = C) if bl< p/. C = tan( βl ω o ) 55 55

56 Example What is the equivalent capacitance at GHzat the feed-point of a 50 W, l/8 transmission line which is terminated with an opencircuit? C = tan( βl) ω o = π λ tan. λ 8 π (0 )50 9 = 3.8pF in l o, β o/c in C 56

57 Quarter-Wave Transformer The feed-point impedance of a terminated transmission line is: in = o o + + j j o tan( βl ) tan( βl ) When l = l/4, i.e. bl= p/, then in o = o in Thus, we can transform the load impedance to another value inby designing the l/4 transmission line with a suitable characteristic impedance o: = o in = = 50 x75 = 6. Ω 57 57

58 Example Match a 75 W load to a 50 W source at 300 MHz using a l/4 transformer. The physical length of the transmission line is determined by the wavelength, which depends on the velocity factor of the transmission line. l=λ/4 o, β in 58

59 STUB MATCHING 59

60 Single-Stub Tuner In a single-stub tuner, a short transmission line (called "stub") is added in parallel with the main transmission line. The stub usually has the same o as the main transmission line. By choosing the appropriate stub length d and its distance l from the load, it is possible to achieve zero reflection from the source to the point where the stub is added. Therefore, the stub is always placed close to the load. The stub is usually a short-circuit stub to minimize radiation loss (which is more likely to occur with an open-circuit stub)

61 Single-Stub Tuner Source o oad o l Shortcircuit stub d 6

62 Single-Stub Tuner (cont'd) At point A where the stub is added, it is desired to have the impedance equal to A=in=o. It will be more convenient to use the normalized impedance za=zin=. Also, as the stub is added in parallel, it is more convenient to work in admittances: y o = z = o y in = z = in The matching process involves: y = z using the distance l to transform y to +jb, and using the stub to add j b to cancel the +j b susceptance. Although the matching process can be carried out using equations, a Smith Chart is most conveniently used for this application. 6 6

63 Single-Stub Tuner Using the Smith Chart ocate z and y. (y is diagonally opposite z.) Using a compass, draw a locus towards the Generator along a constant Γ circle until r = circle is reached (Point A). The distance moved is l (in units of λ). ead the jbvalue at point A. (Two possible solutions.) The stub is required to have jb. Choose either a short-circuit or open-circuit stub. Determine the stub length dto produce the required susceptance. EXAMPE: (z = ) P A ocate z =, and y = 0.5. Move towards Generator until point P A or P B. ead y A = + j0.7 and y B = j0.7 y P B z 63

64 A Source o oad P A o l Shortcircuit stub d y z P B 64

65 Single-Stub Tuner Using the Smith Chart (cont d) EXAMPE: (z = ) ocate z =, and y =0.5. Source o A oad Move towards Generator until point P A or P B. ead on the circumference: l A = 0.5λand l B = 0.348λ. Shortcircuit stub o d l ead at point P A and P B : y A = + j0.7 y B = j0.7 l A At point P A, the stub must have a susceptance of j0.7. If it is a short-circuit stub, its length d AS = 0.40λ 0.5λ= 0.5λ. If it is an open-circuit stub, its length d AO = 0.40λ. (see next slide.) At point P B, the stub must have a susceptance of +j0.7. If it is a short-circuit stub, its length d BS = 0.098λ + 0.5λ= 0.348λ. If it is an open-circuit stub, its length d BO = 0.098λ. y stub = 0 y P A P B z l B y stub = 65 65

66 Single-Stub Tuner Using the Smith Chart (cont d) For a stub with a susceptance of j0.7, if it is a short-circuit stub, its length d AS = 0.40λ 0.5λ = 0.5λ. If it is an opencircuit stub, its length d AO = 0.40λ. open-circuit y stub = 0 short-circuit y stub = 0.0λ 0.5λ 0.40λ 66 66

67 Single-Stub Tuner: Exercise The impedance of a WiFi monopole antenna at.48 GHz was 0 j5 Ω. The antenna was connected to the transmitter via a 50 Ω microstrip line with an effective dielectric constant of. Design a single-stub matching network to match the antenna to the transmission line. Use the shortest short-circuit stub. Source Shortcircuit stub o o d l oad 67 67

68 GAMMA MATCH 68

69 GAMMA MATCH Gamma Match Unbalanced transmission lines. Equivalent circuit The gamma match is equivalent to half of the T-match equires a capacitor in series with the gamma rod The input impedance is: [( ) ] + α g a in = jx c + g + ( + α ) a [3.] a is the center point free space input impedance of the antenna in the absence of gamma-match connection

70 GAMMA MATCH Design procedure Determine the current division factor αby using Eq. [3.3] Find the free space impedance (in the absence of the gamma match) of the driven element at the center point. Designate it as a Divide a by and multiply by the step-up ratio (+α). Designate the result as Determine the characteristic impedance 0 of the transmission line form by the driven element and the gamma rod using Eq. [ 3.5a] = + jx = ( + α ) a 70 [3.] 70

71 GAMMA MATCH Normalized of Eq. [3.]by 0 and designate it as z Invert z of Eq. [3.3]and obtain its equivalent admittance y =g +jb The normalized value Eq. [3.4] z = 0 z = g = j + 0 jx tan k = l' r + jx [3.3] [3.4] 7 7

72 GAMMA MATCH Equivalent admittance y g =g g +jb g Add two parallel admittances to obtain the total input admittance at the gamma feed. Invert the normalize input admittance y in to obtain the equivalent normalized input impedance y in in ( g + g ) + j( b + b ) = y + yg = g z = r in + jx in g [3.5] [3.6] 7 7

73 GAMMA MATCH Obtain the unnormalized input impedance by multiplying z in by 0 Select the capacitor Cso that its reactance is equal in magnitude to X in in C = = in + jx ω π fc in = X = in z 0 in [3.7] [3.8] 73 73

74 How they work How they are made BAUNS 74

75 What is a balun? A Balun is special type of transformer that performs two functions: Impedance transformation Balanced to unbalanced transformation The word balun is a contraction of balanced to unbalanced transformer 75

76 Why do we need a balun? Baluns are important because many types of antennas (dipoles, yagis, loops) are balanced loads, which are fed with an unbalanced transmission line (coax). Baluns are required for proper connection of parallel line to a transceiver with a 50 ohm unbalanced output The antenna s radiation pattern changes if the currents in the driven element of a balanced antenna are not equal and opposite. Baluns prevent unwanted F currents from flowing in the third conductor of a coaxial cable. 76

77 Balanced vs Unbalanced Transmission ines A balanced transmission line is one whose currents are currents are symmetric with respect to ground so that all current flows through the transmission line and the load and none through ground. Note that line balance depends on the current through the line, not the voltage across the line. 77

78 An example of a Balanced ine Here is an example of a balanced line. DC rather than AC is used to simplify the analysis: V = +6 VDC I = 5 ma 6 V 40 Ω 6 V 40 Ω V = -6 VDC I = -5 ma Notice that the currents are equal and opposite and the that the total current flowing through ground = 5mA- 5mA = 0 78

79 FAQ s Do I really need a balun? Not necessarily. If you feed a balanced antenna with unbalanced line and you don t want feed line radiation, use a balun! What kind of balun is best? There is no best balun for all applications. The choice of balun depends on the type of antenna and the frequency range. Will you make a Balun for me? No. However, I will be happy to show how to make your own. 79

RF Devices and RF Circuit Design for Digital Communication

RF Devices and RF Circuit Design for Digital Communication Agenda Fundamentals of RF Circuits Transmission ine Reflection Coefficient & Smith Chart Impedance Matching S-matrix Representation Amplifiers