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Chapter 3 Matching and Tuning By Professor Syed Idris Syed Hassan Sch of Elect. & Electron Eng Engineering Campus USM Nibong Tebal 14300 SPS Penang © Prof.

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Presentation on theme: "Chapter 3 Matching and Tuning By Professor Syed Idris Syed Hassan Sch of Elect. & Electron Eng Engineering Campus USM Nibong Tebal 14300 SPS Penang © Prof."— Presentation transcript:

1 Chapter 3 Matching and Tuning By Professor Syed Idris Syed Hassan Sch of Elect. & Electron Eng Engineering Campus USM Nibong Tebal 14300 SPS Penang © Prof Syed Idris Syed Hassan

2 Why matching or tuning is important? To maximize power delivery and minimize power loss. To improve signal to noise ratio as in sensitive receiver components such as LNA, antenna, etc. To reduce amplitude and phase error as in distributed network such as antenna array. Basic idea of impedance matching

3 Concept of maximum power transfer Power maximum whence Z L = Z o Power deliver at Z L is In lump circuit

4 continue In transmission line No reflection whence Z L = Z o, hence The load Z L can be matched as long as Z L not equal to zero (short-circuit) or infinity (open-circuit) The important parameter is reflection coefficient

5 Factors in selecting matching network Complexity: simpler, cheaper, more reliable and low loss circuit is preferred. Bandwidth: match over a desirable bandwidth. Implementation: depend on types of transmission line either cable, stripline, microstripline, waveguide, lump circuit etc. Adjustability:some network may need adjustment to match a variable load.

6 Matching with lumped elements The simplest matching network is an L-section using two reactive elements Configuration 1 Whence R L >Z o Configuration 2 Whence R L { "@context": "http://schema.org", "@type": "ImageObject", "contentUrl": "http://images.slideplayer.com/1577269/5/slides/slide_5.jpg", "name": "Matching with lumped elements The simplest matching network is an L-section using two reactive elements Configuration 1 Whence R L >Z o Configuration 2 Whence R L Z o Configuration 2 Whence R L

7 continue Configuration 1Configuration 2 If the load impedance (normalized) lies in unity circle, configuration 1 is used.Otherwise configuration 2 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 1 GHz or for higher frequency in integrated circuit(MIC, MEM).

8 Impedances for serial lumped elements Serial circuit Reactancerelationshipvalues +veX=2 fLL=X/(2 f) -veX=1/(2 fC)C=1/(2 fX)

9 Impedances for parallel lumped elements Parallel circuit Susceptancerelationshipvalues +veB=2 fCC=B/(2 f) -veB=1/(2 fL)L=1/(2 fB)

10 Lumped elements for microwave integrated circuit Spiral inductor Loop inductor Interdigital gap capacitor Planar resistor Chip resistor Metal-insulator- metal capacitor Chip capacitor

11 Matching by calculation for configuration 1 For matching, the total impedance of L-section plus Z L should equal to Z o,thus Rearranging and separating into real and imaginary parts gives us * **

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

13 Matching by calculation for configuration 2 For matching, the total impedance of L-section plus Z L should equal to 1/Z o,thus Rearranging and separating into real and imaginary parts gives us * **

14 continue Solving for X and B from simultaneous equations (*) and (**), we obtain Since R L { "@context": "http://schema.org", "@type": "ImageObject", "contentUrl": "http://images.slideplayer.com/1577269/5/slides/slide_13.jpg", "name": "continue Solving for X and B from simultaneous equations (*) and (**), we obtain Since R L

15 Matching using Smith chart (-ve) (+ve)

16 10 50 Serial C Parallel L Serial L Parallel C Matching using lumped components

17 Example Design an L-section matching network to match a series RC load with an impedance Z L =200-j100, to a 100 line, at a frequency of 500 MHz. Solution Normalized Z L we have : Z L = 2-j1 Z L = 2-j1 Parallel C (+j0.3) Serial L (j1.2) Serial C (-j1.2) Parallel L (-j0.7) Solution 1 Solution 2

18 continue Solution 2 seems to be better matched at higher frequency

19 Single stub-matching Short-stub matching Open-stub matching Parallel configuration

20 Example Smith Chart Design two single–stub shunt tuning networks to match a load Z L =15+j10 to 50 at 2GHz. The load consists of a resistor and inductor in series. Plot the reflection magnitude from 1 GHz to 3 GHz for each solution. Solution Normalized the load z L =0.3+j0.2 Construct SWR circle and convert load to admittance Then move from load to generator to meet a circle (1+jb) to obtain two points i.e y 1 =1-j1.33 and y 2 =1+j1.33. The distance d from the load to stub is obtained either of these two points i.e. d 1 = 0.044 and d 2 =0.387. To improve bandwidth, the stub is chosen as close as possible to the load. The tuning requires a stub of j1.33 for y 1 and –j1.33 for y 2. For open circuit-stub 1 =0.147 and 2 =0.353.

21 Continue Convert j0.2 to inductance value, thus Use this value to calculate reflection coefficient

22 Formulas for calculation Let Z L =R L +jX L, then the impedance at distance d from the load is Admittance at this point is Thus, by substituting Z d and separating real and imaginary, we have

23 continue Equating G = Y o = 1/Z o,and, thus we have Solving

24 continue The two principle solution are To find the stub length,

25 Single stub-matching Short-stub matching Open-stub matching Serial configuration

26 Example Smith Chart Match a load impedance of Z L =100+j80 to a 50 W line using a single series open-stub.Assuming the load consists of resistor and inductor in series at 2GHz. Plot the reflection coefficient from 1 GHz to 3 GHz. Solution Normalized the load z L =2+j1.6 Construct SWR circle Then move from load to generator to obtain two points on unity circle(1+jx) z 1 =1-j1.33 and z 2 =1+j1.33. The distance d from the load to stub is obtained either of these two points i.e. d 1 = 0.120 and d 2 =0.463. To improve bandwidth, the stub is chosen as close as possible to the load. The tuning requires a stub of j1.33 for z 1 and –j1.33 for z 2. For open circuit-stub 1 =0.397 and 2 =0.103.

27 Continue Convert j1.6 to inductance value, thus Use this value to calculate reflection coefficient

28 Formulas for calculation Let Y L =G L +jB L, then the load admittance at distance d from the load is Impedance at this point is Thus, by substituting Y d and separating real and imaginary, we have Where t = tan d

29 continue Equating R = Z o = 1/Y o, thus we have Solving

30 continue The two principle solution are To find the stub length,

31 Double-stub matching The advantage of this technique is the position of stubs ( d and x) are fixed. The matching are done by changing the length of stubs. The disadvantage of this technique is not all impedances can be matched. Open or short stubs

32 A= load admittance A=admittance of A at stub 2 B= Adjust of stub S2 to bring to the S2 circle B=Admittance of B at S1 Then by adjusting stub 1 will bring the admittance to the center of the chart A A B B

33 Example Design a double-stub shunt tuner to match a load Z L =60-j80 to a 50 line. The stubs are to be short circuited stubs, and are spaced /8 apart. The load consists of a series resistor and capacitor that match at 2 GHz. Plot the reflection coefficient magnitude versus frequency from 1 GHz to 3GHz. Solution Plot normalized load z L =1.2 –j 1.6 and convert to admittance we have y L = 0.3 +j0.4. Construct a rotated 1+ j b circle which is a distance d= /8 from a 1+jb circle. Get two possible points on the rotated 1+jb circle, y 1 and y 1 by adding susceptance of the first stubs. In this case we take x=0(match section immediate after the load. I.e b 1 =1.314 and b 1 =-0.114.The length of stub will be 1 =0.396 and i 1 =0.232. Now transform both point onto 1+jb circle along SWR circles.This bring two solution y 2 =1-j3.38 and y 2 =1+j1.38. Then the second stub should be b 2 = 3.38 and b 2 =-1.38. The length of stub will be 2 =0.454 and i 2 =0.100.

34 continue yLyL y1y1 y 1 b1b1 b 1 y 2 y2y2 b 2 b2b2 Rotated 1+jb circle

35 continue The shorter the stubs the wider will be the bandwidth

36 Formulas for calculation Let Y L =G L +jB L, then the load admittance at distance x from the load is Admittance at first stub x is distance between stub and load Where t = tan d After transforming to stub 2, we have

37 continue Solving At this point the ral part of Y 2 =Y o,thus Since G L is real, the quantity in square root must be nonnegative, thus Simplified to

38 continue So susceptances of stubs are and To find the stub length, B either B 1 or B 2

39 continue The two principle solution are To find the stub length,

40 Plot Z L normalised to 50 ohm Find bisector and point intersect the chart axis (I.e A) Draw circle at center A touching the center of the chart Obtain R1 and calculated new characteristic impedance of the line Renormalised the Z L to Z T, thus we have Z L2. Then plot on the chart. Obtain x for the length of the matching section by moving towards generator. Graphical method Suitable for load impedance laying inside the unity circle.

41 Graphical method

42 Transmission line transformer Quarter-wave transformer Z1Z1 Matching at one frequency.For other frequency the impedance at the input of matching section will be Where t = tan d At matched frequency tan d = tan ( /2)

43 continue Reflection coefficient Since Z o 2 =Z 1 Z 2, this reduces to 0 m m

44 continue Fractional bandwidth is given by Reflection coefficient with different ratio of Z 1 :Z 2

45 Example Design a single section quarter-wave matching transformer to match a 10 load to a 50 line at f o =3GHz. Determine the percentage bandwidth for which the SWR < 1.5 For a pure resistive load we can just calculate directly and the length of transformer is /4 Then determine reflection coefficient for SWR=1.5 Hence fractional bandwidth is or 29%

46 Multisection transformer Binomial multisection Chebyshev multisection Two types of transformer Reflection coefficient for multisection can be written as Or its magnitude At center frequency (i.e = /4)

47 Binomial transformer Solving forto obtain value of A or For binomial expansion the reflection coefficient can be written as where Note C n N =C N-n N, C 0 N =1 and C 1 N =N=C N-1 N Let then Hence

48 continue For bandwidth Therefore The fractional bandwidth, thus

49 Example Design a three section binomial transformer to match 50 load to a 100 line Solution N=3, Z L =50, Z 0 =100 then But Z 1 =91.7 Z 2 =70.7 Z 3 =54.5

50 Chebyshev transformer Useful forms of Chebyshev polinomial are Chebyshev polinomial in general

51 continue Solving forto obtain value of A or For maximum reflection coefficient magnitude Therefore For N section,Chebyshev expansion the reflection coefficient can be written as or

52 Example Design a three section Chebyshev transformer to match a 100 load to a 50 line. Taking m =0.05. Solution * ** Equating * and ** and A= m =0.05 cos

53 continue cos But Then Start n=0 n=1 For symmetrical and n=2

54 Simple form of multisection transformers 2-quarter-wave transformer 3-quarter-wave transformer

55 Tapered Transmission line transformer Exponential Taper transformer Triangular Taper transformer Klofenstein Taper transformer where Impedance distributionReflection

56 Transmission line transformer Nonsynchronous transformer E.g matching Z 1 =75 to Z 2 =50 T = 29.3 o

57

58 Complex to Complex Impedance Matching


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