Impedance Matching and Tuning 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. 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”.

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.

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

Solution ( Use Smith chart) 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.

-0.7

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

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

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

Matching Bandwidth Series-to-Parallel Transformation

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

Define QL and Qin for RLC resonator

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

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. 

=S11 =S22 BW=34%

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

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

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

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

Splitting into two “L-shape” matching networks

Four solutions for -shape matching networks

BW=4%

Case B: T- shape matching

Splitting into two “L-shape” matching networks

Four solutions for T-shape matching networks

BW=4%

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

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

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

Splitting into two “L-shape” matching networks 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

Four solutions for Cascaded L-shape matching networks

BW=61%

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.

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

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.

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.

Analytic Solution for Shunt Stub

Problem 1: Repeat example 5.5 using analytic solution.

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.

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

Analytic Solution for Series Stub

Problem 2: Repeat example 5.6 using analytic solution.

Shunt stubs are easier to implement in practice than series stubs. 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

Disadvantage is the double-stub tuner cannot match all load impedances 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.

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.

4. The susceptance of the second stubs should be 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.

Analytic Solution for Double Stub

Problem 3: Repeat example 5.7 using analytic solution.

Quarter-Wave transformer It can only match a real load impedance. The length l= /4 at design frequency f0. The important characteristics

Example5.8: Design a quarter-wave matching transformer to match a 10 load to a 50 line? Determine the percent bandwidth for SWR1.5? Solution 

Binomial Multi-section Matching The passband response of a binomial matching transformer is optimum to have as flat as possible near the design frequency, and is known as maximally flat. The important characteristics

Binomial Transformer Design If ZL<Z0, the results should be reversed with Z1 starting at the end.

Example5.9: Design a three-section binomial transformer to match a 50 load to a 100 line? And calculate the bandwidth for m=0.05? Solution 

Using table design for N=3 and ZL/Z0=2(reverse) can find coefficient as 1.8337, 1.4142, and 1.0907.

Chebyshev Multi-section Matching The Chebyshev transformer is optimum bandwidth to allow ripple within the passband response, and is known as equally ripple. Larger bandwidth than that of binomial matching. The Chebyshev characteristics

Chebyshev Transformer Design

Example5.10: Design a three-section Chebyshev transformer to match a 100 load to a 50 line, with m=0.05? Solution 

Using table design for N=3 and ZL/Z0=2 can find coefficient as 1 Using table design for N=3 and ZL/Z0=2 can find coefficient as 1.1475, 1.4142, and 1.7429. So Z1=57.37, Z2=70.71, and Z3=87.15.

Tapered Lines Matching The line can be continuously tapered instead of discrete multiple sections to achieve broadband matching. Changing the type of line can obtain different passband characteristics. Relation between characteristic impedance and reflection coefficient Three type of tapered line will be considered here 1) Exponential 2)Triangular 3) Klopfenstein

Exponential Taper The length (L)of line should be greater than /2(l>) to minimize the mismatch at low frequency.

Triangular Taper The peaks of the triangular taper are lower than the corresponding peaks of the exponential case. First zero occurs at l=2

Klopfenstein Taper For a maximum reflection coefficient specification in the passband, the Klopfenstein taper yields the shortest matching section (optimum sense). The impedance taper has steps at z=0 and L, and so does not smoothly join the source and load impedances.

Example5.11: Design a triangular, exponential, and Klopfenstein tapers to match a 50 load to a 100 line? Solution Triangular taper  Exponential taper

Klopfenstein taper

Bode-Fano Criterion The criterion gives a theoretical limit on the minimum reflection magnitude (or optimum result) for an arbitrary matching network The criterion provide the upper limit of performance to tradeoff among reflection coefficient, bandwidth, and network complexity. For example, if the response ( as the left hand side of next page) is needed to be synthesized, its function is given by applied the criterion of parallel RC For a given load, broader bandwidth , higher m. m  0 unless =o. Thus a perfect match can be achieved only at a finite number of frequencies. As R and/or C increases, the quality of the match ( and/or m) must decrease. Thus higher-Q circuits are intrinsically harder to match than are lower-Q circuits.