Impedance Matching with Transmission Lines

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Impedance Matching with Transmission Lines /4 W Z L useful functions and identities Units Constants Table of Contents I. Introduction II. Inputs III. Transmission Line Synthesis Function IV.

1 Impedance Matching with Transmission Lines /4 W Z L useful functions and identities Units Constants Table of Contents I. Introduction II. Inputs III. Transmission Line Synthesis Function IV. Single Shunt Stub Transmission Line Matching V. Single Series Stub Transmission Line Matching VI. /4 Wave Transformer Matching VII. /4 Wave Binomial Transformer Matching VIII. /4 Wave Chebyshev Transformer Matching IX. Exponential Taper Matching Transformer X. Triangular Taper Matching Transformer XI. Klopfenstein Taper Matching Transformer XII. Copyright and Trademark Notice Introduction Transmission lines have some special properties that can make them advantageous for impedance matching. First they can provide an open certain frequencies, and a short at others, making them useful for biasing, where an impedance needs to be a short at DC, but an open at a desired frequency. This property also allows them to short out harmonics of a signal, without attenuation of the desired signal. At very high frequencies (>GHz), the physical dimensions of inductors and capacitor make them difficult to manufacture. At these frequencies, inductors and capacitors are replaced with transmission line, made cheap and small out of PC board. Inputs Z L Ω 5Ω S max db f c GHz f.3ghz ε r 4. h 6mil Load Impedance Source Impedance Maximum Tolerable S Center Frequency Bandwidth Needed Relative Permittivity of LTCC (Low Temperature Co-fired Ceramic) PC Board Height for LTCC

2 Transmission Line Synthesis Function A W ε r, h, Z Z 6ohm 377ohm π B Z ε r ε r + + ε r ε r ε r W W W if W 8e A h e A h π B W <, W, if h ln( B ) + ε r Single Shunt Stub Transmission Line Matching ε r W >, W, m h ln( B ) ε r d l Z L Here a stub and a transmission line are used to match a transmission line input, with characteristic impedance of, to a complex load of impedance Z L. A open circuit stub, of length l, is placed at the input with a transmission line connecting it to the load with a length, d. Both transmission line segments have a characteristic impedance of. You little (choice ot two implementions) control over the bandwidth for single stub tuners. W val W ε r, h, ω c πf c πc ω c R L Re Z L W Fig. : Single open shunt stub matching W val = 3.6 mm ω c rad = sec = 9.9 cm R L = Ω Width of Transmission Lines Center Frequency Wavelength in Free Space Real Part of Load Impedance X L Im Z L X L = Ω Imag part of Load Impedance When designing transmission line transformers, there are multiple solutions to the matching problem. For an example, an additional wavelength of transmission line can be added to yield the same matching, but with a different frequency response. This is usually not done to keep the area small. Here multiple solutions are presented. t X L + R L R L Z S + X L R L t =.44 Coefficient for First Solution t X L R L R L Z S + X L R L t =.44 Coefficient for Second Solution

3 d atan( t ) d cm π d π + atan( t ) d.44 cm π B s R L t X L t + R L + X L + t ( > ) ( d > ) ( d ) ( > ) ( d > ) ( > ) ( d > ) OK l o > > OK OK 3 l s OK 3 OK 4 l s OK 4 W val ( X L t ) Area l o + d Area.488 cm Area ( l o + d ) W val Area 4. cm Area 3 ( l s + d ) W val Area 3.74 cm Area 4 ( l s + d ) W val Area cm = Connecting Transmission L (Solution #&3) = Connecting Transmission L (Solution #&4) Susceptance of Stub (Solution #& ( X L t ) = 7.7 ohm B s B s R L t X L t + R L + X L + t B s = 7.7 ohm l o atan( B s ) π l o 9.89 mm Susceptance of Stub (Solution #& = Length of Open Circuit Stub (Solution #) l s atan l s = mm Length of Short Circuit Stub π B s (Solution #3) l o atan( B s ) l o = 9.89 mm Length of Open Circuit Stub π (Solution #) l s atan l s = mm Length of Short Circuit Stub π B s (Solution #4) Some solutions are not valid (negative lengths), so they are dropped: Is Solution # OK to use? OK l o OK = = Is Solution # OK to use? = Is Solution #3 OK to use? = Is Solution #4 OK to use? = Area for Solution # = Area for Solution # = Area for Solution # = Area for Solution # l o d 3.35 cm = Areas if white space is considered l o d = 3.47 cm l s d =.667 cm l s d = cm 3

4 Single Series Stub Transmission Line Matching d l Z L l s atan X s l s 9.89 mm π l o atan l o mm π X s l s atan X s l s 9.89 mm π OK l o > > OK ( d ) ( > ) ( d > ) ( > ) ( d > ) ( > ) ( d > ) ( + ) W val ( + ) W val ( + ) W val ( + ) W val W Fig. : Single series stub matching B L Im G L Re Z Z L L Load Conductance and Suseptance Y S Source Admittance G L Y Y S G L S t G L Y S G L Y Y S G L S t G L Y S d atan( t ) π d.99 cm B L + + B L Coefficient for First Solution B L + B L Coefficient for Second Solution d π + atan( t ) d cm π ( B L t Y S ) G L t Y S t B L + X s Y S G L + ( B L + t Y S ) X s Ω ( B L t Y S ) G L t Y S t B L + X s Y S G L + ( B L + t Y S ) X s Ω l o atan l o mm π X s OK l o OK OK 3 l s OK 3 OK 4 l s OK 4 Area l o d Area.488 cm Area l o d Area 4. cm Area 3 l s d Area cm Area 4 l s d Area cm 4 = Connecting Transmission Line (Solution #&3) = Connecting Transmission Line (Solution #&4) = Stub Reactance (Solution #) = Stub Reactance (Solution #) = Length of Open Circuit Stub (Solution #) = Length of Short Circuit Stub (Solution #3) = Length of Open Circuit Stub (Solution #) = Length of Short Circuit Stub (Solution #4) = Is Solution # OK to use? = Is Solution # OK to use? = Is Solution #3 OK to use? = Is Solution #4 OK to use? = Area for Solution # = Area for Solution # = Area for Solution # = Area for Solution #

5 /4 Wave Transformer Matching /4 W Z L No control over bandwidth achieved π β ε r β m - len len cm 4 Wval W ε r, h, Z _4 Wval.6 mm Area lenwval Area.3 cm Outputs Z _4 Fig. : Single shunt stub matching f f_f c f_f c 3 % f c S max Γ max Γ max. Z _4 Z L Z _4 7.7 Ω 4 f_f cmax π acos = Fractional Bandwidth Needed = Reflection Coefficient Needed = Impedance of Quarter Wave Transm Γ max Z L Actual Bandwidth Z L Γ max f_f cmax = 36.7 % OK f_f cmax > f_f c OK = OK to use this circuit? = Phase Constant (Wave Number) = Length of Transformer = Width of Transformer = Area of Transformer = 7.7 Ω Impedance of Line OK = OK to use this circuit? Area =.3 cm Area of Transformer 5

6 /4 Wave Binomial Transformer Matching /4 /4 W ZL W N Z L Γ DC Z L + ln ln cos π Γ max Γ DC f_f c =.88 Fig. : /4 wave binomial transformer matching Γ DC =.333 DC Reflection Coefficient Fractional Order Needed N ceil A( N) N Γ DC 4 f_f c ( N) π acos ln ln cos π Γ max Γ DC f_f c Γ max A( N) N N = A( N) =.67 f_f c ( N) = % Order Estimation Binomial Coefficient Actual Fractional Bandwidth f_f c ( Nval) % i.. ( N ) N! C( n, N) ( N n)! n! C( i, N) Nval = Index Vector Binomial Coefficients Γ( n, N) A( N) C( n, N) Γ( i, N).67 = Reflection Coefficients 6

7 j.. N Z Zval Z S Wval W ε j r, h, Z j max( Wval) =.643 mm len Area 4 N Maximum Width Outputs for i.. ( N ) Γ( i, N) + Zval Zval i+ i Γ( i, N) Zval lenmax( Wval) Wval =.643 mm len = cm Area =.8 cm Z = j 7 Ω Wval = len = cm.643 mm Area =.8 cm T-Line Impedances Widths of Transmission Line Segme Length of Transformer Area of Transformer Widths of Line Segments Length of Transformer Area of Transformer 7

8 /4 Wave Chebyshev Transformer Matching /4 /4 W ZL W N acosh N ceil acoshsec π Γ DC Γ max acosh N if( N >, N, ) f f c acoshsec π 4 f_f c ( N) π asec cosh N acosh Γ DC Γ max Γ DC Γ max =.878 f f c Fig. : /4 wave chebyshev transformer matching N = N = f_f c ( N) = % Fractional Order Needed Order Estimation Actual Fractional Bandwidth f_f c ( Nval) % A Γ max 5 5 A =. Nval Transformer Coefficient ln θ max asec cosh N acosh Z L Γ max θ max = deg 8

9 T( n, x) if( x <, cos( nacos( x) ), cosh( nacosh( x) )) i.. N Chebyshev Polynomial Coefficients Index Vector Γ( n) Γ( i) = A sec ( θ max) 3 if n = 3 A sec ( θ max) 3 sec( θ max ) if n = 3 A sec ( θ max) 3 sec( θ max ) if n = A sec ( θ max) 3 if n = 3 Reflection Coefficients j.. N Z Zval Z S for i.. N Zval Zval e Γ( i) i+ i Zval Wval W ε j r, h, Z j Z = j Wval = Ω.65.4 Transmission Line Impedances mm Widths of Transmission Line Segme max( Wval) =.65 mm len Area 4 N Outputs lenmax( Wval) len = 4.95 cm Area =.47 cm Wval =.65 mm Widths of Transmission Line Segments Maximum Width Length of Transformer Area of Transformer.4 len = 4.95 cm Length of Transformer Area.47 cm = Area of Transformer 9

10 Exponential Taper Matching Transformer L W st ZL W end Fig. : Exponential transformer matching π L β π L β L ln Z L z Z( z) e L = 7.95 cm L = 4.59 cm ln Z L L Z ( z) e Length of Transformer (Option #: βl=π) Length of Transformer (Option #: βl=π z Z( z) 8 Z ( z) (,, Z( cm) ) = 3.6 mm (,, Z( L) ) =.74 mm W ε r h W ε r h 5 z cm TotArea L W ε r, h, Z( cm) TotArea =. cm 5 z cm Width at Beginning of Taper Width at End of Taper Total Area of Transformer

11 Triangular Taper Matching Transformer L W st ZL W end Fig. : Triangular transformer matching π L L = 4.59 cm β z ln Z L 4 z L L Z S Z( z) if z <, e L z L ln Z L Z S, e Length of Transformer (Option #: Z( z) 8 6 (,, Z( cm) ) = 3.6 mm (,, Z( L) ) =.74 mm W ε r h W ε r h TotArea L W ε r, h, Z( cm) 5 z cm TotArea = 4.4 cm Width at Beginning of Taper Width at End of Taper Total Area of Transformer

12 Klopfenstein Taper Matching Transformer L W st ZL φ( x, A) Γ DC e Γ βl A acosh Γ DC Γ max A L β x I A y dy A y ln Z L Γ DC Z( z) e cosh( A) A φ z L, A + βl cos βl A cosh( A) W end Fig. : Klopfenstein transformer matching A =.874 L = 4.35 cm Z( cm) = Ω Z( L) = Ω Coefficient for Transformer Length of Transformer Z( z) 8 Γ( βl). 6 (,, Z( cm) ) =.49 mm (,, Z( L) ) =.956 mm W ε r h W ε r h TotArea L W ε r, h, Z( cm) 4 z cm TotArea =.84 cm 5 Width at Beginning of Taper Width at End of Taper Total Area of Transformer Copyright and Trademark Notice All software and other materials included in this document are protected by copyright, and are owned or controlled by Circuit Sage. The routines are protected by copyright as a collective work and/or compilation, pursuant to federal copyright laws, international conventions, and other copyright laws. Any reproduction, modification, publication, transmission, transfer, sale, distribution, performance, display or exploitation of any of the routines, whether in whole or in part, without the express written permission of Circuit Sage is prohibited. βl

2 nd Order PLL Design and Analysis

2 nd Order PLL Design and Analysis nd Order PLL Design and Analysis S REF Phase Detector Σ K f Loop Filter VCO K V s R C Loop Divider Fig. : nd Order PLL with Current-Mode Phase Detector useful functions and identities Units Constants Table