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Chapter 5

Chapter 5. Impedance Matching and Tuning. Why need Impedance Matching. Maximum power is delivered and power loss is minimum. Impedance matching sensitive receiver components improves the signal-to-noise ratio of the system.

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Chapter 5

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  1. Chapter 5 Impedance Matching and Tuning

  2. Why need Impedance Matching • Maximum power is delivered and power loss is minimum. • Impedance matching sensitive receiver components improves the signal-to-noise ratio of the system. • Impedance matching in a power distribution network will reduce amplitude and phase errors. • Basic Idea The matching network is ideally lossless and is placed between a load and a transmission line, to avoid unnecessary loss of power, and is usually designed so that the impedance seen looking into the matching network is Z0. ( Multiple reflections will exist between the matching network and the load) • The matching procedure is also referred to as “tuning”.

  3. Design Considerations of Matching Network • As long as the load impedance has non-zero real part (i.e. Lossy term), a matching network can always be found. • Factors for selecting a matching network: 1) Complexity: a simpler matching network is more preferable because it is cheaper, more reliable, and less lossy. 2) Bandwidth: any type of matching network can ideally give a perfect match at a single frequency. However, some complicated design can provide matching over a range of frequencies. 3) Implementation: one type of matching network may be preferable compared to other methods. 4) Adjustability: adjustment may be required to match a variable load impedance.

  4. Lumped Elements Matching • L-Shape (Two-Element) Matching • Case 1: ZL inside the 1+jx circle (RL>Z0) • Use impedance identity method

  5. Example5.1: Design an L-section matching network to match a series RC load with an impedance ZL= 200-j100, to a 100 line, at a frequency of 500 MHz? Solution ( Use Smith chart) 1. Because the normalized load impedance ZL= 2-j1 inside the 1+jx circle, so case 1 network is select. 2. jB close to ZL, so ZL YL. 3. Move YL to 1+jx admittance circle, jB=j 0.3, where YL 0.4+j 0.5. 4. Then YLZL, ZL 1+j 1.2. So jX=j 1.2. 5. Impedance identity method derives jB=j 0.29 and jX=j 1.22.  6. Solution 2 uses jB=-j 0.7, where YL 0.4-j 0.5. 7. Then YLZL, ZL 1-j 1.2. So jX=-j 1.2. 8. Impedance identity method derives jB=-j 0.69 and jX=-j 1.22.

  6. -0.7

  7. Use resonator method (Case 1: RS<1/GL) Goal: Zin=Rs =S11=(Zin-Rs)/(Zin+Rs)=0

  8. Case 2: ZL outside the 1+jx circle (RL<Z0) • Use admittance identity method

  9. Use resonator method (Case 2: RS>RL) Goal: Yin=1/Rs =S11= =0

  10. Matching Bandwidth Series-to-Parallel Transformation

  11. Case 1: ZL inside the 1+jx circle (RL>Z0)

  12. Define QL and Qin for RLC resonator

  13. Similarly, for case 2: ZL inside the 1+jx circle (RL>Z0). Summary

  14. Example5.2: Design an L-section matching network to match load impedance RL= 2000, to a RS= 50, at a frequency of 100 MHz? Solution Because RS <1/ GL, so case 1 network is select. 

  15. =S11 =S22 BW=34%

  16. Three Elements Matching (High-Q Matching) • Use resonator method for complex load impedance. • Splitting into two “L-shape” matching networks. • Case A: - shape matching • Case B: T- shape matching Goal: Zin(=0)=RS ,(=0)=0 Goal: Zin(=0)=RS ,(=0)=0

  17. Case A: - shape matching Conditions: RV<1/GL , RV<RS P.S. RV : Virtual resistance

  18. Case B: T- shape matching Conditions: RV>RL ,RV>RS P.S. RV : Virtual resistance

  19. Example5.3: Design a three elements matching network to match load impedance RL= 2000, to a RS= 50, at a frequency of 100 MHz, and to have BW<5%? Solution Case A: - shape matching

  20. Splitting into two “L-shape” matching networks

  21. Four solutions for -shape matching networks

  22. BW=4%

  23. Case B: T- shape matching

  24. Splitting into two “L-shape” matching networks

  25. Four solutions for T-shape matching networks

  26. BW=4%

  27. Cascaded L-Shape Matching (Low-Q Matching) • Use resonator method for complex load impedance. • Splitting into two “L-shape” matching networks. • Low Q value but large bandwidth. • Case A: • Case B: Conditions: RL<RV<RS Goal: Zin(=0)=RS ,(=0)=0 Conditions: 1/GL>RV>RS Goal: Zin(=0)=RS ,(=0)=0 P.S. RV : Virtual resistance

  28. Splitting into two “L-shape” matching networks for case A

  29. Splitting into two “L-shape” matching networks for case B

  30. Example5.4: Design a cascaded L-shape matching network to match load impedance RL= 2000, to a RS= 50, at a frequency of 100 MHz, and to have BW60%? Solution Select 1/GL>RV>RS Splitting into two “L-shape” matching networks

  31. Four solutions for Cascaded L-shape matching networks

  32. BW=61%

  33. Multiple L-Shape Matching • Lower Q value but larger bandwidth follows the number of L-section increased. • Conclusions: • Lossless matching networks consist of inductances and capacitances but not resistances to avoid power loss. • Four kinds of matching techniques including L-shape, -shape, T-shape, and cascaded L-shape networks can be adopted. Generally larger Q value will lead to lower bandwidth. • A large range of frequencies (> 1GHz) and circuit size may not be realizable.

  34. Transmission-Line Elements Matching • Single-Stub Matching • Easy fabrication in microstrip or stripline form, where open-circuit stub is preferable. While short-circuit stub is preferable for coax or waveguide. • Lumped elements are not required. • Two adjustable parameters are the distance d and the value of susceptance or reactance provided by the shunt or series stub. • Shunt Stub • Series Stub

  35. Example5.5: Design two single-stub (short circuit) shunt tuning networks to match this load ZL= 60-j 80 to a 50 line, at a frequency of 2 GHz? Solution 1. The normalized load impedance ZL= 1.2-j1.6. 2. SWR circle intersects the 1+jb circle at both points y1 = 1.0+j1.47 y2 = 1.0-j1.47. Reading WTG can obtain: d1= 0.176-0.065=0.11 d2= 0.325-0.065=0.26. 3. The stub length for tuning y1 to 1 requires l1 = 0.095, and for tuning y1 to 1 needs l2 = 0.405.

  36. 1. ZL= 60-j 80 at 2 GHz can find R= 60,C=0.995pF. 2. Solution 1 is better than solution 2; this is because both d1 and l1 are shorter for solution, which reduces the frequency variation of the match.

  37. Analytic Solution for Shunt Stub

  38. Problem 1: Repeat example 5.5 using analytic solution.

  39. Example5.6: Design two single-stub (open circuit) series tuning networks to match this load ZL= 100+j 80 to a 50 line, at a frequency of 2 GHz? Solution 1. The normalized load impedance ZL= 2-j1.6. 2. SWR circle intersects the 1+jx circle at both points z1 = 1.0-j1.33 z2 = 1.0+j1.33. Reading WTG can obtain: d1= 0.328-0.208=0.12 d2= 0.672-0.208=0.463. 3. The stub length for tuning z1 to 1 requires l1 = 0.397, and for tuning z1 to 1 needs l2 = 0.103.

  40. 1. ZL= 100+j 80 at 2 GHz can find R= 100,L=6.37nH.

  41. Analytic Solution for Series Stub

  42. Problem 2: Repeat example 5.6 using analytic solution.

  43. Double-Stub Matching adjustable tuning • Variable length of length d between load and stub to have adjustable tuning between load and the first stub. • Shunt stubs are easier to implement in practice than series stubs. • In practice, stub spacing is chosen as /8 or 3/8 and far away 0 or /2 to reduce frequency sensitive. • Original circuit • Equivalent circuit

  44. Disadvantage is the double-stub tuner cannot match all load impedances. The shaded region forms a forbidden range of load admittances. • Two possible solutions b1,b2 and b1’,b2’ with the same distance d.

  45. Example5.7: Design a double-stub (open circuit) shunt tuning networks to match this load ZL= 60-j 80 to a 50 line, at a frequency of 2 GHz? Solution 1. The normalized load impedance YL= 0.3+j0.4 (ZL= 1.2-j1.6). 2. Rotating /8 toward the load (WTL) to construct 1+jb circle can find two values of first stub b1 = 1.314 b’1 = -0.114. 3. Rotating /8 toward the generator (WTG) can obtain y2= 1-j3.38 y’2= 1+j1.38.

  46. 4. The susceptance of the second stubs should be b2 = 3.38 b’2 = -1.38. 5. The lengyh of the open-circuited stubs are found as l1 = 0.146, l2 = 0.482, or l1 = 0.204, l2 = 0.350. 6.ZL= 60-j 80 at 2 GHz can find R= 60, C=0.995pF.

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