ECE 391 supplemental notes - #11. Adding a Lumped Series Element

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1 ECE 391 supplemental notes - #11 Adding a umped Series Element Consider the following T-line circuit: Z R,1! Z,2! Z z in,1 = r in,1 + jx in,1 Z in,1 = z in,1 Z,1 z = Z Z,2 zin,2 = r in,2 + jx in,2 z,1 = r in,2 Z,2 Z,1 + R Z,1 + jx in,2 Z,2 Z,1 renormalization è All calculations can be done on Smith chart 168 1

2 Smith Chart as Admittance Chart For circuits involving parallel (shunt) connections of circuit elements, it is more convenient to work with admittances. How can the Smith chart be used with admittances? è How is Y=1/Z = G + jb related to reflection coefficient Γ? Consider normalized admittance y y = Y Y = G + jb Y = g + jb = Y Z = Z Z = 1 z = 1 r + jx 169 Smith Chart as Admittance Chart Reflection coefficient Γ Γ = z 1 z +1 = 1 y 1 1 y +1 = y 1 y +1 Smith chart can directly be used with y=g+jb after rotating reflection coefficient by 18. y 1 jπ g = Γ = Γ e y +1 = const b = const circles 17 2

3 Impedance vs. Admittance Coordinates Impedance Chart Im(Γ) Admittance Chart Im(-Γ) inductive r + jx capacitive g + jb SC capacitive r - jx OC Re(Γ) OC inductive g - jb SC Re(-Γ) 171 Example 4 Given the normalized load admittance y =.5 + j2. Determine the normalized admittance at distance d = λ/16 =.625λ Solution: use Smith chart with admittance coordinates è see web applet (example 4) 172 3

4 Example 5 Given a 5Ω T-line that is terminated in Z = Z T = 25Ω and has a shunt resistance R sh = 5Ω connected at distance d = 1m from the termination. The wavelength on the line is λ = 3m. Determine the input impedance at the location of the shunt resistance. Solution: see web applet, example Additional Example Z = ( 25+ j25)ω Z = 5Ω z =.5 + j.5 l =.125λ Results SWR 2.6 Γ y z in.45 θ =1 j j

5 Impedance Matching Implementation Issues Types of Networks - lumped elements - distributed elements - mixed - reactive vs. resistive Design Criteria - bandwidth - lossless/lossy - complexity - parasitics 175 Impedance Matching è Matching network needs to have at least two degrees of freedom (two knobs) to match a complex load impedance 176 5

6 Transmission-ine Matching Idea: Use impedance transformation property of transmission line load z (or y ) 2βz transformation circle 177 Transmission-ine Matching Transformation to real z or y load z (or y ) 2βz real z (or y) real z (or y) transformation circle 178 6

7 Transmission-ine Matching Transformation to r=1 (or g=1) circle 2βz load z (or y ) z=1+jx (y=1+jb) r=1 or g=1 transformation circle z=1-jx (y=1-jb) 179 Single Stub Matching Stub tuners are useful for matching complex load impedances. The two design parameters in single stub matching networks are distance from load at which stub is connected length of stub Other design considerations are open or shorted stub series or shunt connected stub 18 7

8 Basic Design Principle Y in = Y Objective: Match load admittance by 1. transforming along T-line to Y in = Y +jb 2. tuning out remaining jb with a parallel shunt element of admittance jb 3. shunt element can be realized with a T-line stub 181 Design Steps 1. Determine distance d/λ from the load Y (or Z ) at which the input admittance is Y +jb (for shunt stub matching) input impedance is Z +jx (for series stub matching) 2. Realize the stub with a length l/λ of open- or shortcircuited transmission line (stub)

9 Example 6 Using a short-circuited stub, match the load impedance Z =(8 + j7)ω to a transmission line with Z = 5Ω. Z Z d Solution: use Smith chart with admittance coordinates è see web applet (example 6) 183 Additional Example R C =125Ω = 2.54pF f =1GHz = 5Ω Z z =. 5 j y in_1,2 1± j1.58 Need shunt element with Y sh_1,2 = j.58 / Z = j31.6ms

10 Quarter-Wave Transformer λ / 4 at design frequency f Z1 = ZZ increased mismatch NOTE: for complex load, insert λ/4 transformer where Z in =R 185 Example: Quarter-Wave Transformer R C =125Ω = 2.54pF z f =1GHz = 5Ω Z =. 5 j z in,1.24 From Z, T Z in,2 Z in,1 12Ω Z = Z = Z 24. 5Ω 2, T in,

11 Bandwidth Performance at f =1GHz 187 umped Element Matching Reactive -Section Matching Networks R > Z R < Z B = X ± R Z 2 R + R 2 2 X + X 2 R Z B = ± ( Z R ) Z R 1 B X Z R X = + Z B R X = ± R ( Z R ) X different solutions may result in different bandwidths

12 Example Design criteria bandwidth realizability of components parameter sensitivity cost? 189 umped-distributed and Distributed Matching Networks Series Reactance Matching Shunt Reactance Matching 19 12

13 Examples 191 Double-Stub Tuner Double-stub (and multi-stub) tuners use fixed stub locations (typical spacing: λ/8) Advantages: - easier to accommodate different loads - better bandwidth performance

14 Smith Chart with ossy T-ines Recall for a lossless T-line z in (z') = Z in(z') Z = 1+ Γ e jφ ( z' ) 1 Γ e φ(z') = θ jφ ( z' ) 2βz' For low-loss T-lines, Z real and γ = α + jβ Then z in (z') = Z in (z') Z ( ) = 1+ Γ e 2γ z' = 1+ Γ e 2αz' e jφ z' 1 Γ e 2γ z' 1 Γ e 2αz' e jφ ( z' ) 193 Smith Chart with ossy T-ines ( ) = Γ e 2αz' Γ z' Γ im As z' z in 1 (Z in Z ) Γ re

15 Approach Γ im 1. Assume lossless line and rotate by -2βz z 2. Change Γ by factor e 2αz' Γ re Γ lossy in = Γ lossless in e 2αz' Γ lossless in = Γ e jφ 195 Example 7 Given a lossy T-line with characteristic impedance Z = 1Ω, length l = 1.5m (l < λ/2), and Z SC (z =l) = 4 j28 Ω. (a) Determine α and β. (b) Determine Z in for Z = 5 + j5 Ω an l = 1.5m. Solution using Smith chart è see web applet (example 7)