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# PATTN COMP REL RG ST 145

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This 149 page Class Notes was uploaded by Kaya Conroy on Thursday October 22, 2015. The Class Notes belongs to RG ST 145 at University of California Santa Barbara taught by Staff in Fall. Since its upload, it has received 51 views. For similar materials see /class/227180/rg-st-145-university-of-california-santa-barbara in Religious Studies at University of California Santa Barbara.

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Date Created: 10/22/15

ECE145A218A Course Notes Last note set Introduction to transmission lines 1 Transmission lines are a linear system superposition can be used 2 Wave equation permits forward and reverse wave propagation on lines VxtVt xv V txv IxtZiVt x v V t x v IDC o v L hase velocit LC P V ZO characteristic impedance L C per unit length 3 Re ections occur at discontinuities in impedance A re ection coefficient F can be defined RL 1 V Z 0 V RL i0 1 4 Transmission line can be represented by an equivalent circuit LT2 LT2 LT W CTZ CTZ LT Z01 CT TZo T lengthv time of ight Goals Understand frequency domain analysis of transmission lines phasor notation position dependence phase constant 5 re ections movement of reference plane impedance and F variation with position Smith Chart ECE145A218A Course Notes Transmission Lines 2 Transmission Lines in the Frequency Domain and the Smith Chart Time domain analysis is intellectually clearer the picture being forward and reverse waves propagating re ecting and rere ecting This analysis becomes intractable as soon as we introduce reactive impedances as multiple convolutions will be required for timedomain re ection analysis So we will analyze in the frequency domain instead Frequency domain analysis of transmission lines is a classical approach to this problem gtX ZL FIG 1 Transmission line with impedance Z0 connects the source with Thevenin equivalent generator impedance Zs to a load ZL V Re Vaejwt V0 V0ei u S0 2 vs t IVUI cosmt gt On a transmission line at position x waves travel in time as x i Vt where velocity v is the phase velocity Equivalently at a time t waves vary with position x according to It XV Thus we can represent voltage on the line vXt as c0sat 0 gtc0sat i xv gig c0sat 0 i xa v c0sat t7 i x 2 112007 Prof Steve Long ECE145A218A Course Notes Transmission Lines 2 where 5 is the phase or propagation constant 5 oav 2117 5 converts distance to radians Here 7 is the wavelength The sketches below illustrate the concept Here the function cos oat 5X is plotted 1 Let s suppose that t to so that the wave appears frozen in time on the X aXis If distance X2 X1 7 as shown the corresponding phase change over this distance is 211 l7bgt X1 3 X2 3 X FIG 2A 2 Now if we plot this same cosine function as a function of 5X we see that the wavelength is equal to a phase 5X 211 since 7 21t5 gtl27 lt WW 3 Also we can set X 0 and observe the wave function in the time domain for FIG 2B increasing time In this case one period requires a time interval T 21tn lf as shown M T I4 I W0 t FIG 2C 4 Finally you might ask why the cos oat 5X wave is the forward to the right travelling wave direction To see why track a point of constant phase with position on the X aXis as time progresses from t1 to t2 3 112007 Prof Steve Long ECE145A218A Course Notes Transmission Lines 2 t t2 gtX X2 FIG 3 Plot of wave propagation in the forward direction at times t1 and t2 We can see that t gt t1 and X2 gt X1 for a forward direction of travel and that the cosine function will have the same value at points of constant phase Therefore cosat1 6x1cos17t2 x2 1 Thus wtl mtz Bx1 xz lt 0 2 from the drawing A forward wave must have a positive phase velocity 0 v gt 0 3 3 From eq 2 v gt0 4 t2 t1 X2 must be greater than X1 therefore the wave is travelling in the forward direction for cos oat x Of course these waves can also be described by complex exponentials IVOIejwt em eij c sine wave phase position along line 4 112007 Prof Steve Long ECE145A218A Course Notes Transmission Lines 2 The ejmt time dependence is always taken as implicit and is frequently omitted when using phasor notation IVOI ej u eij x Voltage and Current on Transmission Lines Now we can use this notation to describe the voltage and current on a transmission line at any location on the line Vx Vx Vx V0ej x V0 af x Iltxgt ZiVltxgt V ltxgt 0 Zi0V0e f V390 af x 5 112007 Prof Steve Long ECE145A218A Course Notes Transmission Lines 2 Re ections in the frequency domain 0 VQb39 If IZL Zo r FIG 4 Re ection from a mismatched load impedance ZL The forward wave is travelling in the positive X direction and re ects from the load ZL We set X0 at the load end of the transmission line as our reference plane In frequency domain analysis we assume that the wave amplitudes are steady state values From the de nition of re ection coefficient V7 0 V0FL z 1 FL L zL 1 where ZL is the normalized load impedance ZLZO FL is in general complex 6 112007 Prof Steve Long ECE145A218A Course Notes Transmission Lines 2 Movement of reference plane x l 0 I Zo VQb39 too IZL r Now we must determine voltage and current on the line as a function of position This is often referred to as moving the reference plane Here we move from X 0 at the load to x Z at the left end of the drawing Vx V x V x V x1 I39x where is the positiondependent re ection coefficient Substituting for V X and VX V 0ef V0emc no I39x and VxV0e x1 1quot0e2j x From this we can see that the re ection coef cient at position Z from the load is given W 1quot l1quot0e2 l 7 112007 Prof Steve Long ECE145A218A Course Notes Transmission Lines 2 and the re ection coefficient goes through a phase shift of minus 21 200 radians minus 2B1 radians minus 360 2lZ degrees The re ection coefficient changes phase with position Some observations 1 FX is periodic ZJ39KZ jx Fx r0e A 41tX7t 211 whenever X n M 2 where n is an integer 2 VX is also periodic Vx V0e f x 1 FOe2j x 1 every n 1 every nk 2 1 every nkZ 1 every nk4 Example Let F0 1 short circuit X l l The result Standing wave pattern in voltage 0 Voltage maxima and minima are separated by M 4 o Successive maxima are separated by M2 8 112007 Prof Steve Long ECE145A218A Course Notes Transmission Lines 2 Voltage Standing Wave Ratio QYSWRg We have seen that in general there are two waves travelling in opposite directions on a transmission line We also saw that the variation of voltage along the line at position X due to the sum of these two waves is given by VxV0e fquotx1 1quot0e2 The magnitude is given by Vx V0 1F0ezj xl So we can easily determine the minimum and maximum voltage magnitude that will be found along the transmission line at some position X Vx lmaX IV 0 1 H0 I Vx lmin IV 0 1 H0 I Taking the ratio of max t0 min gives us the VSWR VSWR 2W 1F0 Vxmin I 1 I r0 An open or short circuited line will give us an in nite VSWR because the minimum voltage on the line is zero F0 l for both cases VX v1r 0 l l l x M2 M4 0 Why is VSWR important It is often used as a speci cation for an ampli er 9 112007 Prof Steve Long ECE145A218A Course Notes Transmission Lines 2 Impedance vs Position Refer to the same picture The impedance at any point on the line can be found from the current and voltage or equivalently from the re ection coefficient 2xwz w 1x Vx Vx 1 Fx 0 The normalized impedance at any point is easily found 1 zxZx 20 iti So impedance depends on the position along the transmission line While this is conceptually simple there is a lot of math involved This can become tedious So we could benefit from a graphical representation this is called the Smith Chart after Philip Smith who invented this convenient graph of transmission line parameters back in the 1930 s The relationship for the normalized impedance ZX is the key to the Smith Chart The Smith Chart is just a polar plot of re ection coefficient Impedance is determined by It is a onetoone mapping between complex numbers F and z and is in fact an analytic function and a conformal transformation You can read about this in math books on complex analysis 10 112007 Prof Steve Long ECE145A218A Course Notes Transmission Lines 2 In the twodimensional plane of F the F plane a re ection coef cient F is represented by a point 1 j0 Im F 0j1 0 1j0 Re Ojl 112007 Prof Steve Long ECE145A218A Course Notes Transmission Lines 2 As we move away from the load by a distance Z on the transmission line F rotates by an angle A9 where A6 432 360 One whole rotation is required in the F plane for each halfwavelength movement on the line 1quot0 Note the following 1 g is de ned as quotelectrical lengthquot in degrees 2 The degree scale on the edge of the Smith Chart represents Z f the angle of the re ection coefficient Note that this is TWICE the electrical length L Note that phase delay corresponds to a clockwise movement in angle 12 112007 Prof Steve Long ECE145A218A Course Notes Transmission Lines 2 Lines 01 L r quoton L 11quot z l I39 Thesewill L Sect 2 L line across the center of the chart as shown below Constant reactants circle centers on vertical ax s zrjx m Constant resistance circles on honz axis Note that the unis ofr are normalized to 20 in this case 50 ohms So the circle labeled r 05 corresponds to the 25 ohm resistance circle in this case But 20 can be any convenient real Value IfZO were 100 ohms F05 would represent 50 ohms Similarly the reactances x can be represented by circles These have their centers on the Vertical 39 4 the chart quot L quot Positive x 112007 Prof Steve Long ECE145A218A Course Notes Transmission Lines 2 corresponds to inductive reactance and is above the center line of the chart Negative X represents capacitive reactance Given this mapping onto the F plane we can associate any re ection coef cient a point on the plane with an impedance simply by reading the z coordinates of the point We can also associate the change of impedance with position with a rotation on the chart Just rotate the F vector clockwise around the chart at the rate of one rotation for every half wavelength of movement on the line Then read off the impedance directly from the chart Fx FOe 2j x Note that all points on this chart represent series equivalent impedances 14 112007 Prof Steve Long ECEIAsAzlsA CnurseNntes Transmissinn Linesz So for example 2 1 11 represents a senes RL network 21 71115 asenes RC 2 1 15 apoan ngm m are center ofthe chart Normahzauon Consxderthe pomtz 1 11 IfZU 5011nen z 50 150 when he 1rnpedanee1s denomahzed IfZU were 1000 ohms we would havez 1000 11000 for are same pomt In 11an way the enan ean be used over an arbnrary range ofxmpedance 15 112007 Prof SteveLong ECEIAsAzlsA CnurseNntes Transmide Linesz on ths ehmwe see ammpedanee 1 11 comespondmgto me senes RL network Ifwe 1 e e resxstance lme r 1 to the center The reactancehas been cancelled 16 1120U7 Prof SteveLong ECEIAsAzlsA CnurseNntes Transmide Linesz 1mm example we have stanedthh a senes RC network 2 111 By addmg 1nduct1ve reactants 11 m senes we can agam move along the r1cmle up Lethe center othe ehan 17 11211177 Prof SteveLong ECEIAsAzlsA CnurseNntes Transmide Linesl Senes resxstance The chm can also represent the addmon ofresxstance Thxs W111 correspond to movement along a mrcle of constant reactants For Example1fwe begm vmhz111agam and addanormahzedr 1 to thatweamve atz211 18 11211177 Prof SteveLong ECE145A218A Course Notes Transmission Lines 2 The Complete Smith Chart Black Magic Design RADIALLY SCALE PARAMETERS TOWARD LOAD gt lt TOWARD GENERATOR 3 25 2 1 16 14 12 11 15 0 7 5 4 3 z 1 1 5 4 3 2 l 1 12 13 14 16 18 2 3 4 5 5 a a 910 12 14 20 30210 0 02 04 06 08 1 5 2 3 4 5 6 04 03 02 01 005 001 00 11 12 13 14 15 16171819 2 25 a 4 05 05 04 03 02 01 0 1 099 095 09 03 07 06 05 04 03 CENTER 00 01 02 03 04 05 011 07 011 09 1 11 12 13 14 15 15 17 18 19 2 19 112007 Prof Steve Long ECE145A218A Course Notes Transmission Lines 2 Let s look at some interesting examples Quarter wavelength transmission line How will impedances be transformed by this line x 74 0 2 gt39 39 Fx FOe2 x The re ection coef cient retains the same magnitude but changes phase with position The variable X is de ned as shown in the above diagram Let X M4 The propagation constant can be used to calculate the electrical length x 22 7r so x 7Z 2 2 Thus r x4 r0 e 139 In other words the angle of the re ection coef cient is rotated clockwise 1800 for a quarterwave line negative angle clockwise rotation 20 112007 Prof Steve Long ECEIAsAzlsA CnurseNntes Transmide Linesl 5nuuuu What does thxs do tothexmpedanee7 2144 1 1quot44 1r1quot44 1 170 4 1801r 170 4 180 1 We rotate 180 on the Smxth chart eloekwxse for the quarter wave lme Ths transforms areal Z mto areal Zm m the example above Ex Z ZLZa 20050 4 Rotating 180quot takes us to Zm 0 25 Th151512 5 ohms when denormahzed 2 Onusmg the above equamon fornormahzedxmpedanee and 11 11 1quot0zL r 1zL 1 35 2 1 05 4 180 1 705 4 180 025 21 11211177 Prof SteveLong ECE USA218A Cnurse Nate Transmide Lines 2 3 You can also use the formula below which is speeme forthe impedance unnormahzedxmpedances 2m zj ZL Clearlyd12 Smith Chart is the easiest m Ilse 4 The Smnh Chan also can be used with complexloads 111490 ZL 0 j1 Pure inductor ZL0j1 Pure capacitor 1120U7 Prof SteveLong E CE USA218A Course Notes Transmission Lines 2 5 What about an open circuit at the load end ZN 0 PIN 017180 OPEN Conversely a shorted quanerwave line produces an Open circuit Ly L L a mm transformations will be exact only at the design frequency whalemm Aquot A 39 39 m a 21 lines ofother lengths L L L L l l 2Bx Iadmdnn 39 quot quot 39 39 39 from the load toward the generator 23 112007 Prof Steve Long ECE145A218A Course Notes Transmission Lines 2 Exercise Use the Smith Chart to verify that the transmission line combination below will transform 50 Q into 25 Q 509 M12 259 M12 259 I 509 24 112007 Prof Steve Long ECE145A218A Course Notes Transmission Lines 2 Admittance Chart We can also use the Smith Chart for admittances Impedance Z RjX Normalized Impedance Z r jX Admittance Y lZ G jB Conductance j Susceptance Normalized Admittance Y g j b Y YYO We can plot the values of both r and X and g and b on the F plane by rotating the impedance coordinates by 180 r and X then become constant conductance circles and constant susceptance circles Since Y lZ taking lZ is equivalent to a 1800 rotation in angle This is the impedance 7 admittance chart shown below The admittance coordinates are the bold red lines You can see from this that it is possible to use the chart to quickly translate between impedance and admittance Every point on the chart can be interpreted in both ways 25 112007 Prof Steve Long ECEIAsAzlsA Cnnrse Nate Transmissinn Linesl br l m senes The admlttance charm useful for components m parallel EXAMPLES 26 112007 Prof Steve Long ECE145A218A Course Notes Transmission Lines 2 Example Use the YZ Smith Chart to determine Zin of the circuit below X j7so X j7so B1j7so RHQ l Normalize the reactances and admittances to Z0 and Y0 What would be a good choice for Z0 in this case 2 Starting from the resistor add capacitive reactance 7 j l to the impedance 3 Using the admittance chart add susceptance 7 j l 4 Going back to impedance add another 7 jl The result is Zin 75 j0 27 112007 Prof Steve Long E CEIAsAZLEA C muse Nate Transmissinn Lines 2 An important observation through the cxrcult load to source or source toload x711 21111 From our prevxous example we proceeded from A to 1m by Starling W1 A 1 7 J1 28 112007 Prof Steve Long ECElASAZISA CnurseNntes Transmide Linesl capacitor What happens when we go the other way7 1 Start atthe center 2 Go downm reactants 11 2 Go up m suseeptanee 11 3 We are now all ms is just agraphmal version othe statement Match the load withxts complex eomugate impedance or admittance 29 1120U7 Prof SteveLong ECE 145A218A Course Notes Transmission Lines a review and explanation An apology 1 We must quickly learn some foundational material on transmission lines It is described in the book and in much of the literature in a highly mathematical way DON39T GET LOST IN THE MATH We want to use the Smith chart to cut through the boring math but must understand it first to know how to use the chart 2 We are hoping to generate insight and interest not pages of equations Goals 1Recognize various transmission line structures 2Define re ection and transmission coefficients and calculate propagation of voltages and currents on ideal transmission lines 3Learn to use S parameters and the Smith chart to analyze circuits 4Learn to design impedance matching networks that employ both lumped and transmission line distributed elements 5Able to model Agilent ADS and measure network analyzer nonideal lumped components and transmission line networks at high frequencies What is a transmission line ECE 145A218A Course Notes Transmission Lines 1 Qoaxial SIG GND quotunbalancedquot Microstrip Sig 1 S ig 2 quotbalancedquot Sig 1 Sig 2 Coglanar waveguide ECE 145A218A Course Notes Transmission Lines 1 Common features 39 a pair of conductors 39 geometry doesn39t change with distance A guided wave will propagate on these lines An unbalanced line is characterized by 1 Has a signal conductor and ground 2 Ground is at zero potential relative to distant objects True unbalanced line Coaxial line Nearly unbalanced Coplanar waveguide microstrip Balanced Lines 1 Have two symmetric conductors 2 Conductor potentials are symmetric with respect to distant objects twin lead twisted pair coplanar strips ECE 145A218A Course Notes Transmission Lines 1 Compare microstrip to striQIine Stripline 8r ground planes 4 signal This is made by sandwiching 2 pc boards together All fields are within material of Sr Hence velocity cJg In microstrips the fields are partly in air a1r 8r l J s i me lt So for wide lines the fields are almost all in board while narrower lines will have proportionally more field energy in air 8r 8 board Wide lines v c J ECE 145A218A Course Notes Transmission Lines 1 For lines of non infinite width w H v at cJ This leads to the idea of an effective dielectric constant 8r e CZVZ so that v c Igr e The text gives for example that 1 8rl gr M h 112 K w se r 2 2 very roughly W the line width and h substrate thickness Advice Ignore above formula and use LINECALC part of ADS For higher accuracy there is a free version of Sonnet an easy to use EampM silnulation tool Some typical material parameters Line parameters are for microstrip lines with Wh Dielectric 8 v 22 075 c 943 Q 48 054 c 68 Q 98 039 c 49 Q 13 034 c 43 Q 38 058 73 Q Duroid is available in a variety of dielectric constants Sapphire crystalline A1203 is sometiInes used but is anisotropic SiOz is used both in its crystalline form Quartz and in amorphous form fused silica ECE 145A218A Course Notes Transmission Lines 1 Voltage and Current on Transmission Lines Transmission lines can be described from EM Viewpoint or Circuit We will use Circuit point of View since it will tie into the application better AV de Vx V05 dx l I gt gt Cdx 1v 1x 39vl Ix dx Ill A 31 ideal lossless line Vx alx I Vx I 0 KVL I C capacitance length L inductance length g LE 1 C telegrapher s equatlons ac dt Ixt Ixdxt C0 KCL U 32V 32V 2 LC 2 wave equatlon dc a has solutions of form Vxt V t V39 t xv superposition Ixt lVt V tJZO IDS ECE 145A218A Course Notes Transmission Lines 1 1 V velocit of r0 a ation JL C y P P g Z0 L C characteristic impedance Voltage at any point use superposition Vxt Vx t V xt forward reverse voltage waves Vxt V xt Zo pUQU V Zo mc c V I x t current ECE 145A218A Course Notes Transmission Lines 1 Reflection parameters What happens if we connect an arbitrary resistive load to the line X I 0 Z 0 i N R L v ratio of voltage current Z0 on line must be equal at X 0 KL at load K R V I L Z E72 x 0 V V 0 if we define the re ection coefficient as FL x0 then R 1rZQ 1 PL and RV Z R Z F 0 L 0 We see that F varies between 1 and 1 for an R Z 0 If RltO then IFI gt 1 This is a negative resistance still can be described by F ECE 145A218A Course Notes Transmission Lines 1 3 examgles 1 OEen circuit RL 00 FL 1 VJr V7 re ected voltage wave is same ampl as forward wave Vx0zV V 2V voltage is doubled at open circuit 1x 2 OJ 2 H makes sense Z 0 2 Short circuit RL 0 FL 1 V7 V re ected wave is negative of forward wave Vx 0t 0 short circuit Ix0t 2VZ0 current is doubled 3 Matched load RL ZO FL 0 no re ected voltage or current ECE 145A218A Course Notes Transmission Lines 1 Return Loss Re ection coefficients can also be represented as return loss Return Loss 20 log F Notice that this does not provide information about phase A matched load has a RL oo no re ected power A short or open has RL 0 dB everything re ected What about the source end RS V0t gt w Z0 I V0 Vs I0Rs Vt V 10 Vt VZo Solve for Vz V0t rSV 0t VStTS Zo ZORS TS source transmission coefficient voltage divider if line is terminated with RL ZO V7 0 and so V is the total voltage But if V7 is finite must account for source re ection coefficient At some t n2T the re ected wave will return to x 0 and the voltage at that node will change accordingly To find F5 use the Z0 of the line that the incident wave is traveling and the R9 of the generator ECE 145A218A Course Notes Transmission Lines 1 V0t rs V390t RsZO 1 S RS20 ECE 145A218A Course Notes Transmission Lines 1 Transmission Parameters Transmission through an interface or device Vincident Vtrans mitt DUT V Transmission Coefficient T rm mm T 4 lVlL leVll Insertion Loss IL 613 20 10g10 ITI Transmission Gain GaindB 2010g10 ITI Also at an interface between two transmission lines T 1 F Vim nsmilled V VT V 1 1 F21 12 4 201 I I Z02 Z02 Z01 Note that the sign of F changes for waves traveling from Z02 to Z01 ECE 145A218A Course Notes Transmission Lines 1 Transmission line junction OK Let s consider a junction between two transmission lines with different characteristic ilnpedances Z01 and Z02 V2 T12 V1 F21 Vz Where T12 transmission coefficient between Z01 and Z02 1 H 1 Z02Z01 1Z02Z01 1 2Z02Z01 Z02 F21 F12 V2quot V2 FL depends on the termination at the far end of line 2 FLZZLZ0 ZLZ0 ECE 145A218A Course Notes Transmission Lines 1 Time Domain Given the relationships that exist between forward and re ected waves on transmission lines it is relatively siInple to sketch the progression of a step function on a line in the tiIne domain Example A drawing of position of the wave vs tiIne can be used to graphically show the progression of the traveling wave on the line Here is an example of a very silnple case with a T line whose iInpedance matches the source and load The re ection coefficients at both source and load are zero The step propagates from source to load without any re ections The final voltage at the load is reached at tilne T1 Rs Zo 20 T1 Vs ECE 145A218A Course Notes Transmission Lines 1 Example Here is a more complicated example with re ections at both source and load Plot the source voltage V1 and load voltage VL as a function of tiIne for the network below After many round trip re ections the final voltage must be given by the resistor division ratio 125 50 125 07143 V We see the solution approaching the steady state value after two round trips 20 759 T 100ps R8 509 V8 RL 1259 v 06 T VLV1rL075 V39 FLV 015 2T v391r8072 FSFLWDOIOE gt 3T 4T FLFS FL V 00075 07125 0708 I ECE 145A218A Course Notes Transmission Lines 1 Transmission line laws 1 Source and load iInpedances should be equal to the characteristic i1npedance of the line if re ections are to be avoided 2 Think about the voltages on transmission line conductors before connecting them 3 Think about the currents on transmission line conductors before connecting them Example T junction in coplanar waveguide Circuit diagram Incorrect iInplementation T junction interrupts the current in the ground plane ECE 145A218A Course Notes Correct implementation Transmission Lines 1 The air bridge connection between ground restores the current path Example Junction between balancec and unbalanced lines V2 I V2 Balanced line Coplanar strips Unbalanced line Coplanar waveguide Air Bridge This is a problem because the balanced line has potentials V 2 and V 2 on both conductors while the coplanar waveguide has zero potential on the ground conductors A balun is needed to make this junction more on this later ECE 145A218A Course Notes Transmission Lines 1 Let39s go back to equiv circuit for transmission line g 3 g Z 0 4L C characteristic iInpendance v L hase veloci LC P W if line is of length Z Z Z LT total inductance 0 0T v but I T the propagation tune delay on the line v and Z T CT Z 0v Z 0 So a given line of length 6 can be modeled by a T or H section or a series of these sections LTZ LTZ LT O C ICZICZ 6T T 9T Physical length is not relevant Impedance and delay tilne describe electrical behavior ECE145AECE 218A Notes Set 5 Impedance Matching Why do we impedance match gt Power transfer is reduced when we have a mismatch Example Suppose we have a 1V source with 100 ohms source resistance Rs The available power is the largest power that can be extracted from the source and this is only possible when matched RL RS 2 V Pm 125 mW SRS If we were to attach a 10009 load PLaad R6VL1L VL Vgen 10001100 IL Vgen1100 PLOAD 041 mW Alternatively we could calculate the re ection coefficient 12 1 FL 00818 RL 1 0 Z PL Pm1 FL2 Pm033 04lmW So if the source and load impedances are not matched we can lose lots of power In this example we have delivered only 33 of the available power to the load Therefore if we want to deliver the available power into a load with a nonzero re ection coefficient a matching network is necessary ECE145AECE218A Impedance Matching Notes set 5 Page 2 L Matching Networks 8 possibilities for single frequency narrowband lumped element matching networks C 01 E C 39 50m q C l ZLOAD CI ZtOAD ZLOAD AOAD ZLOAD Figure 242 Matching networks Figure is from G Gonzalez Microwave Transistor Amplifiers Analysis and Design Second Ed Prentice Hall 1997 These networks are used to cancel the reactive component of the load and transform the real part so that the full available power is delivered into the real part of the load impedance 1 Absorb or resonate imaginary part of Z S and Z L 2 Transform real part as needed to obtain maximum power transfer Rev January 22 2007 Prof 8 Long ECE UCSB ECE145A ECE218A Impedance Matching Notes set 5 Page 3 How to proceed Recall the Series Parallel transformations that you derived in homework 1 RP RSQ2 1 Remember that these relationships between the series circuit and parallel circuit elements are valid only at one frequency And Q is the unloaded Q as defined in lecture 1 RS RP CP C S l 1 Here of course X P and X S wCP wCS Design a matching network We want to match Rp to Rs and cancel reactances with a conjugate match Rev January 22 2007 Prof 8 Long ECE UCSB ECE145A ECE218A Impedance Matching Notes set 5 Page 4 Matching Network 39X RS J S LOAD L For this configuration of L network Rp must be greater than Rs T SOURCE l we know Rs and RP given Use the first Series Parallel transforming equation to determine the Q such that Rp will be transformed into Rs We can know Q because Q2 1 2 R P or Q amp 1 Rs Rs 2 Now using the de nition of unloaded Q for the series and parallel branches compute XS 2 Q Rs Xp RP Q 3 Then determine their values L Xsw C lpr Note that these reactive elements must be of opposite types Now to show that it works convert the parallel RP 39jXp into its series equivalent We started by determining the Q based on the relationship between Rs and RP so we know that R31 R32 Rev January 22 2007 Prof 8 Long ECE UCSB ECE145A ECE218A Impedance Matching Notes set 5 Page 5 Series equivalent 39X 39X J 81 J 32 LOAD RSI E RszR31 T SOURCE Then 2 Q J QRP QRs1Xs1 X 2X S2 PQ21 Q21 So we see that X32 X31 and we have cancelled the reactance as well as transforming the real part 39X 39X J 81 J 32 LOAD Z1N Rs RszR31 The input impedance is simply Rs Rev January 22 2007 Prof 8 Long ECE UCSB ECE145A ECE218A Impedance Matching Notes set 5 Page 6 Matching Network Z1N Rs sz LOAD L Same process applies with high pass form Same XS Xp but different C L values are required Let s complete our matching network design Suppose f 1590 MHZ 0 1 x 1010 radsec RpSOOQ Rs50 2 500 1 3 Q 50 Xs3 Rs150 2 Xp RpQ 5003 167Q Then evaluate at 0 C 06 pF L 15 nH Rev January 22 2007 Prof 8 Long ECE UCSB ECE145A ECE218A Impedance Matching Notes set 5 Page 7 Of course we can also do this quite nicely on the Smith Chart 0 9 Normalize to 50 Q Then rp 10 on real axis Move on constant conductance circle down 03 to the r l circle capacitive susceptance So bp 03 Denormalize Bp 0350 0006 03C Xp lBp 167 Q C 06 pF Next the series branch Move on constant resistance circle from 1 j3 to center inductive reactance Denormalize Xs 30 X 50 150 Q 03L L 15 nH Rev January 22 2007 Prof 8 Long ECE UCSB ECEMSA ECEleA Impedance Matchmg Notes set 5 Page 8 Also note chat d1 Q canbe read offche Smich Chart Q xr 3010 bg 0301 3 Rev January 222007 Prof 5 Lung ECE ucsa ECE145AECE218A Impedance Matching Notes set 5 Page 9 134 Matching Networks and Signal Flow Graphs Chap 2 Figure 2416 Constant Qn contours for Qquot 15 and 10 Figure is from G Gonzalez Microwave Transistor Ampli ers Analysis and Design Second Ed Prentice Hall 1997 Rev January 22 2007 Prof 8 Long ECE UCSB ECE145A ECE218A Impedance Matching Notes set 5 Page 10 Why choose one form highpass vs lowpass over the other 1 Absorb load reactance into matching network Ex L kag BJT I I Rs r 7 V CC Cp needed forL network L kag Ls needed forL network 2 Resonate load reactance necessary if C7 gt Cp Matching Network 39X J S LOAD WV ll 3 Harmonic suppression lowpass 39 We can use the Smith chart and get the answer directly 39 We can calculate the Qquot of the network X S X p can be determined from Rs and RP Rev January 22 2007 Prof 8 Long ECE UCSB ECE145A ECE218A Impedance Matching Notes set 5 Page 11 Example Suppose C7 1 pF r7 5009 This could be the base ofa bipolar transistor Matching Network 39X J S LOAD Mi L Pr 01 We know from the example above that j Xp j 167 Q Convert to susceptance BP 1 XP 0006 S This is the total susceptance required in the parallel branch But we have already from C7 Bp m1x1012 0018 This is more than we need So we must subtract BL 0004 S by putting an inductor in parallel as shown in the figure above L 1oaBL25nH Then add the required series Xs to bring to 50 ohms Check the result on a Smith Chart Also note that there are other solutions possible Rev January 22 2007 Prof 8 Long ECE UCSB ECE145A ECE218A Impedance Matching Notes set 5 Page 12 Matching with Distributed Elements There are cases where transmission line elements are more effective than lumped elements in the design of matching networks at higher frequencies when parasitics of lumped elements cannot be controlled when very small capacitors or inductors are required Suppose we have designed a lumped ilnpedance matching network This example has shunt and series inductors and a shunt capacitor Think for a moment as to why no series capacitor has been chosen L2 C3 14 We may not have m and T available to us only of iInpedances over the range Zmin to Zmax Rev January 22 2007 Prof 8 Long ECE UCSB ECE145A ECE218A Impedance Matching Notes set 5 Page 13 Basis for distributed matching using transmission line segments the equivalent circuit model of a short transmission line L2 L2 IL I CZI ECZ LTZO CT ZO rz vp Let s approximate a shunt inductor with a transmission line section Ll Z1151 012 tlZl L1 thl E So we obtained the inductor L1 we desire together with a C1 2 which we do not want C1 does vary as 1 Z1 and L1 as Z1 so using a high impedance line greatly helps to reduce C1 relative to L1 To make a good inductor we need to keep C1 small Rev January 22 2007 Prof 8 Long ECE UCSB ECE145A ECE218A Impedance Matching Notes set 5 Page 14 Series inductor 12 z2 12 12 mm l WY L gt gt C2 2I E C2 2 Again L2 Z2 3952 C2 t2 Z2 SO Z2 should be high Shunt Cagacitor I L32 L32 C3 Z3 13 WYIVYY g gt C E 3 C3 13 Z3 L3 3953 Z3 SO Z3 should be kept low to minilnize L3 We started with this Circuit C3 14 And approxilnated it with transmission lines Rev January 22 2007 Prof 8 Long ECE UCSB ECE145A ECE218A Impedance Matching Notes set 5 Page 15 Z212 Z1 quot1 Z313 Which has an equivalent circuit approximately like this 12 C3 Z32 2 C3 Z32 2 J J J L1 f V I L12Z12 12 2Z22 lf Z1 and Z2 are sufficiently high and Z3 sufficiently low this will approximate the desired network Rev January 22 2007 Prof 8 Long ECE UCSB ECE145A ECE218A Impedance Matching Notes set 5 Page 16 It is helpful to think of transmission lines in both their equivalent circuit form and in a distributed form 39 39 39 WWW l SCI icl CIZO 36 36 36 U W20 L If we merge all of these sections together oI Zo we have ordinary t line with wide bandwidth neglecting loss What would happen new if we add extra capacitance to the line 12 139 139 12 CX g CX CX We have changed Z 0 of the composite line Rev January 22 2007 Prof 8 Long ECE UCSB ECE145A ECE218A Impedance Matching Notes set 5 Page 17 CXC CXC L 23 E W L39 E quotLC CX X per section We also have now a frequency limitation on the transmission line The Bragg cutoff frequency a 4 E uations above limited to a ltlt a C LC CX q C39 This occurs when you construct an artificial line with discrete L and C L 120 C r Z0 Shorter line sections small T lead to higher wc Why do we care Nice trick for broadband designs 1 Distributed or traveling wave amElifier 121 rZl LZI Cgs of FETs is absorbed into transmission line Ler L 0 CCgs Rev January 22 2007 Prof 8 Long ECE UCSB ECE145A ECE218A Impedance Matching Notes set 5 Page 18 make Z1 high to get mainly inductance and keep sections short Rev January 22 2007 Prof 8 Long ECE UCSB ECE145A ECE218A Impedance Matching Notes set 5 Page 19 2 Wideband ingut match Rev January 22 2007 Prof 8 Long ECE UCSB ECE145A ECE218A Impedance Matching Notes set 5 Page 20 Transmission Iine matching networks T line sections can also substitute for lumped matching elements in L networks Short stubs open or shorted F 1 A 0 or 180 degrees Series lines constant F Z 2 Quarter wave transformers Z1 N 7 L ZL jZO tanBZ Z 2 IN 0 ZO jZL tan z From G 1340 Shorted Stub Z N jZO tanBZ for shorted stub It 7239 SifZ Z tan 1 2 o 8 4 w and we get an inductor with X L jZO OEen Stub Z L 00 X X0 2m 0 0 0 0 ZIN jZO COtBZ Rev January 22 2007 Prof 8 Long ECE UCSB ECElALSAECEZlBA ImpedanceMatChmg Notesset5 PageZl 7L and x z E We get a shunt Capacitor thh 1X 12quot SHORT Rev January 22 2007 me 5 Lung ECE ucsa ECE145A ECE218A Impedance Matching Notes set 5 Page 22 Comment on electrical length The microwave literature will say a line is 43 long at 5 GHZ What does this mean f 72f Z Electrical length E 360 l ef Recall f t v so frzf 2ny D Z gtE 360 ref360 V L V fref E T gf360 a line which is 1 ns long has an electrical length E 360 at frzf 1 GHZ and an electrical length E 36 at Eel 100 MHZ Why not just say T 1 ns 7 you should be conversant with both terminologies Rev January 22 2007 Prof 8 Long ECE UCSB ECE145A ECE218A Impedance Matching Notes set 5 Page 23 Transmission Line L Network design examples Our goal as usual is to match the load to the source Let s start with a normalized load impedance ZL 18 j19 at a design frequency of 1 GHZ We want to match this load to a 50 Q source impedance There are many possible solutions to this design 1 The first example below uses a combination of series and shunt transmission lines all of characteristic impedance Z0 50 50 Q Series Tline of length 11 gt V ZIN 50 Q SOQ Shunt shorted Tline Stub of length 12 T zL18j19 First step Determine length of series T line l1 necessary to transform the load impedance so that it intersects the unit conductance circle Using a 50 ohm YZ Smith Chart draw a circle with radius FL around the center of the chart Moving clockwise from the load impedance negative angle 2011 since Fx l 0e2139l3X and x l1 we arrive at point A on the unit conductance circle The length in wavelengths can be determined from the outside wavelength scale around the perimeter of the Smith chart If i 211 Aref then the wavelength scale represents 1 units of wavelength kref We can later determine the physical length of the line from the frequency and phase velocity Draw a straight line from the center of the chart through ZL This intersects the wavelength scale at 0204 7 Add a series line until the unit conductance circle is reached Next draw another straight line through point A This intersects the scale at 0427 7 So the electrical length of the required series line in wavelengths is 0427 0204 02237 Converting to electrical length in degrees 0223 X 360 802 degrees Rev January 22 2007 Prof S Long ECE UCSB ECElALSAECEZlBA ImpedanCeMatchmg Notes set 5 Page 24 S11 The secund step 5 to apply shunt susceptance from the shorted stub Aeeordmg to the Chart We now have a normalized edmmenee ya 10 1153 Thus We must add b 7 153 to Cancel the susceptance We Wm men arnve at pomtB 50 ohms m e e e m e KS3 360 331 nges us the requued length m degrees Rev January222007 Prof s LungECEUCSE ECElALSAECEZlBA ImpedanCeMatchmg Notes set 5 Page 25 freq 1 OOOGHz to 1000GHz Secund step Add shunt shorted stub at end of senes lme b 7153 TLSC TL2 PARAMETERS 5 Pararr em m WED Ohm Several other examples wrll be shown m Class Rev January222007 Prof s LungECEUCSE ECE145A ECE218A Impedance Matching Notes set 5 Page 26 Appendix 3 Element matching networks Why 3 elements instead of 2 Q R2 Q of L network is determined by res1stance ratlo Q E 1 no freedom to change Q 1 If higher g 2 is desired then a 3 element network is needed Why would we want higher Q For narrow bandwidth applications We will get better suppression of out of band frequencies Also provides more opportunity for parasitic absorption in active circuits PI Network X 2 must be opposite to X 1 X 3 Can be considered as 2 back to back L networks resistance at this point lt Rs or RL Rev January 22 2007 Prof 8 Long ECE UCSB ECE145A ECE218A Impedance Matching Notes set 5 Page 27 R For 7r network use Q E H 1 where R H higher of the two resistances R P or R S R RERL Q21 See example 4 4 in Bowick p 73 Design both sides to match to R at center of network T network X2 must be opposite to X1 X3 Rs Consider again as 2 back to back L networks Rs BL V V R39 gtRS or RL in this topology R R small R z 2 1I2small Can you still use the design equations when the source and load is complex 1 Rsman least of the two resistances R P or R S Yes Just absorb the series or parallel reactance susceptance into the design Rev January 22 2007 Prof 8 Long ECE UCSB ECE145A ECE218A Impedance Matching Notes set 5 Page 28 Examgle L network R S XS XS1 XP1 X L ZL RL 1 Convert series Z L to parallel equivalent 1 1 1 Y ReY ImY L ZL L RP L XP RS XS X31 RP combine combine Q amp1 Rs Xswm Xs Xs1 QRS amp Ptotal XP XP1 Q Rev January 22 2007 Prof 8 Long ECE UCSB ECE 145a 218 a notes set 3 Impedance Matching Mark Rodwell University of California Santa Barbara rodwelleceucsbedu 8058933244 8058933262 fax ImpedanceMatching Goals Recall line BBIIBBIiOIIS c At end of line V z O FLV z 0 where FL At beglnnlng of line Vz l FSV Z l TSVgen ZLZo1 ZLZo1 ZS20 1 h F M ZS20 565145301865 notes M Rodw ll mpwgmeaznog line BBIIBGIiOIIS aml Standing Waves C r3 ZO rLD l 22 20 Retlectlons V z O FLV z O and V z l l1V z l Tygw Waves traveling along line VZ VZ Oej2 ll andViZ l VZ 0ej27dl Varying frequency V and V will vary from in phase to out of phase and will vary from constructlve to destructrve 1nterIerence 3 load voltage will vary with frequency ECE1453218Ca notes M Rodwell copyrighted 2009 Gain ripples resulting from standing waves on line Matching Will eliminate this asimula an1page 111 HEW 512 gm Mew insert Marker hlsrory Qpnons 10015 gage mndow Help DEE g k W D l l hh acl ml 2 11 a A v 4 z 9 l azlsimulatmn v 1 a E a m E dBVoutpos L E l ADSnr2 ll x lesmunlnn2 schmucus Fk39 Edn mu m bum 011mm Tm Luvwl Sammie Warm own ka Dwgucuuu m w I lsn39lenwris l ev o a may e 5 mg 1 H E D quot39l d 39 22 139 IL luN Dei companyquot R R3 R200 Ohm Vom p03 MW 1 R quot Egg TUN R1 TL1 R2200 Oh Vacpoiar20 v 250 0 Ohm m Freqfreq Ezae39o 7 F1O GHz Illllllllllllllll 8101214161820 freq GHz gt 1 6 8 0 4 6 C O T l 1 T MW page Topolog yChec kyes VReIToI1e3 IRelToI1ev3 GiveAIlWamingsyeS MaxWarningsz39l 0 AC AC1 Stanz l 0 GHz Stop200 GHz Step01 GHz Recall M Xillllllll Power Transfer Theorem 5 Maximum Power Available from the Generator pg 2 VW 2 4Reden RMS quantities a Load Power PL PAVG iff ZL Z PL lt PAVG otherwise gen 7 den E ZL ngen Rgen J E L 25145301865 notes M amen mpwgmeaznog 565145301865 notes M Rodw N mpwgmeaznog Matching For M Xillllllll Power Transter By adding a lossless matching network no resistances between the generator and the load we obtain PL PAVG gen JX R gen J gen ECEMSaWBCS mazes M Rudweu cupyngmed 2009 Matching IIII Ill I B lillll quot8 matching IIII max POWBI Transfer ISt den Zn Ziamg ZautJine ZinM ZinT 20 I 20 lenghtl J L L Zoutline Z0 f I Vgequot C Fgen TL 39 den 21quot We our line inM Zin T L 20 lenghtl quotI L L nd 2 case den gt ZO Z gen FL I 0 C Fm gtZ outJine i Z 0 VW IfZ outJine 7t Z0 then matching for zero re ection and matching for maximum power transfer are not the same ECE1ASa21BCE notes M Rodweti mpyngmed 2009 Matching each to Z 0 Z Z 20 20 Zw out 0le ZO ZO 1 Iquot ZD lenghtl l L L w 39 r 0 P503 39 gen Direct Stage Stage Matching Z Z Z Z Z gtxlt 011 17 line 011 NILinc mM Z0 lenghtzl J L L A 39V ltr 40 n0 I gen Z Z in M in T Note Z mJine Zol 1162151 1 1162151 I can be any length including zero ImpedanceMatching Using Agilent I ADS in quottunequot mode as a study tool ECE145321805 notes M Rodwell copyrighted 2009 Classic Texts present Matching with On Paper Smith Chart Exercises Today Matching easily presented graphically in CAD program First Show how to tune networks in ADS Second Illustrate matching examples ECE745527BCS notes M Rome4 copyIgmed 2009 Tuning Elements in ABS 1 ADSfurZ1Ea gaintesrbenh 15chematic1 Biz gum eiecl Mew insert Qpnons IooLs Layout Smuiale mndow Dynamicunk DesignGde eip R own w n s 2 EB 1 a 5 c a mi Lumped39Cumpanents w MO at m as x m 9 W a E m s Param Maxeam Start95 GHZ Voneam MAemaxigaWS ant Stop105 GHZ Yoptt Stew GHZ Sprcie Yopttyopt80ptPortZt CaicNmseyes GDCWC et Freq GpCH cie1gpicirce8 2 51 M some 20m ESCH39Cie C c e ZopH Zopm opusOptPonzw w i 2 GsCircie2gsicircie8 2 51 9 StabMeaS Sta StabMeast tabimeamS GaCirciet mom GaCH cie1gaicircie zm NSCH39de NSCH39ciet NSQrcietnsicircean Nme SoptRnw t StabFact MeasEqH measi UreaiZZ 1rreaiZ1 2 Z magZ2 1rmagZ1 2 24reaiZ1 1reaiZ2 2rreaiZ1 2reaiZ2 1 h21Y2 1Y 11 ZtZ2 i 50Z2 250 tabFacM StabFacMStabjactS no 5 quotMatchquot here is a circuit having a MOSFET and an input matching network Tuning Elements in ABS 2 ADSfnr2183 match Schematic3 Ele gar 5212 MW lnsart gpuons Iools Layout Emulate mndow Dynamchmk Dwgnemde elp k MN m s x saga s 1 c a El rLumped39CumpnnEnts simplehybridpi X3 lv sjupmammmrkswneMo 4 51 m as m e mm a E E CE 1 45521 805 notes M Rodwell copyrighted 2009 Use series L shunt C network to match Z1n to Z0 Very simple MOSFET small signal model c my mum payment m w wuulungvamp ql 39GENE R R1 R185 Ohm Port CZ P1 C335 fF Num1 HM VCCS SRC1 a G1OO m8 R11e100 Ohm R2200 Ohm ECE145521803 notes M Rodwell copyn39gMed 2009 I j FEFEVEelEFEHWMndE Inm ceh me nameltstar s npgt sum ard u M then press TuneOptStat Tumng Status C ear L m mm 391 h a mm mm 3 t TuneUptStatDDE Setup sp ay parameter nn schemauc cm Cnmpunent Dptmns L Inductance L 3 mm 0K cancel Help msemp Esaups Elnducmrs Tumng Upamlzauon 5mm gt Tumng 39 Upmmzmmn 5mm L Wimm yinwmm 3 WW Slam Ow Tum Status Enah 7 M Ifflame Name namalt5zrtstnpgt Standard v Hm I then set mm max 7 v DH M Mammum Value TunEpr slmDDE Setup ste values p 39 39 39 m Step vaqu w 1 Cw Logarrthmzc E msplay parameter an schemanc t m Cancel HE p Component Optmns Est E L Inductance I Hem mammes notes M Rodwel mpyngmed 2009 Setting III Element Tuning ill Ans L1 is now a tunable element ignplejybridmi L1 Do the same for C1 P1 L0 pH 1 Setting III Element Tuning ill Ans mammes notes M Rodwel mpyngmed 2009 G0 to the main testbench and select tuning Then manipulate Windows until you can see both the quot Tune Parametersquot Window and the Plot Window ADSfor115a 1 gainJestbench Schematic gpnons loos Layout Simulate mndow Dynamlcunk Desxgnauida e p gag m tumpezgnmme n M0 Stmutamn sew snap and ebase swmamr W Inning 3 Smart smutamn wizard SPARAM ETERS en Mthth node 4 MM l Cf mm W SP1 3 State5 0H2 u Stop105 GHZ e U DCA Stew GHz 3 U nquot J M D gamso 11E65I CatcNomeye5 Word Snapshot Freq GpCH C Etzgpi Update Qpnmlzauon wines e generate Budget Pam I Immumm hMMgHvaah1mm wowquot u req 95mm m 105DGHZ L umme A mumM J Agt4A Huueur wv A mum Setting III Element Tuning in M18 Usually easiest to make the plot update after every tuning change Need to update Schematic after tuning otherwise you will lose the changes made ECE745a2786a notes M Rome4 copyIgmed 2009 21 Tune Parameters WMe Shder anes ncude upt P m Enaptemsame D Snap Shdertu step Traces and lahtes Trace v Resetvames Update Schemam match a W Va ue El Max pun L Mm u step 5 same Lm v 4 Tune Parameters S mu ale WME Shder Maves v D Snap Shderto step Traces and Vames Mm n step 5 sate tm t ImpedanceMatching MethOdSItxamples E CE 1 45521 805 notes M Rodwell copyrighted 2009 Trajectories l l adding series I Slllllll I and I Adding Series L or C Adding Shunt Parallel L or C Z Yum ML L L TBIIIUIL39II 7 7 a l l igaimpln S Jiramstars0 iigainJulot SJarameterslm le Edit mew Lnsert Marker story mens Incl gage mnde elp le gdk mew Ensert Marker mow Qpnons Iools Bags l l Vcs S11 quotquot I S11 YIILM llt l N gain clrdas v pmmeters z gammzters sparameters gures nf meritu mulch l llt l Ill gain circles v Qarameters 2 parameters SJmmexersl innures of merlt nolse figure s 11 at l jYmrm l mammes notes M Rodwel mpyngmed 2009 IS lllmll ll 0 Matching Network Network To 010W S11 before matching at 100 GHz simplehybridpi X3 Port P2 Num2 LEE 3 2 1freq1oo0GHz S11O 711 89 646 impedance ZO quot O 330 j0 950 e freq 95006Hz Io 1050GHz mam275 notes M Rodwel copyrighted 2009 1Slllllllllell 0 Matching Network Increase L until YM Y0 10 jB Reached when L1 112 pH lwa39hlsvul t I E um luvm D2 muggy 7 as t w w t t r u u cr 7 I X 12 i 1 w 1m 1 J m2 We have moved on a constant r circle towards values of higher reactance jx crease C until Zm Z0 210 jO Reached when C1 44 fF x E151 our innmu Quill Eml SM n su n We have moved on a constant g circle towards values of higher susceptance jb ECEMEaZf ca mazes M Rumell cupwgmed 2009 1S llllllllell 0 Matching NBIWIII k X25 Port 1 Port Final Values P1 P2 Num1 C L412 9 t Num2 01 044 fF t MPH 39 Performance vs Frequency DC 200 GHZ frequency sweep marker at 100 GHZ DB39 blwuwlsJ my v Em v w um r If i gt mm 20 40 60 80100120140160180 20 freq GHz mammes notes M Rodwel mpyngmed 2009 2nd lumnetl 9 Matching Network Network To olou 1 S11 before matching at 100 GHz mam275 notes M Rodwel copyrighted 2009 ZIIII llllllllell 0 Matching Network Ym Y0 10 jB Reached when L1 38 pH Increase L1 until 7 W W W D H HTTETIS L V n a x L 13 l 3 y if a 1 f 73 l y a m2 Q2 quoti2 quot 12 i We have moved on a constant r circle towards values of higher reactance jx 56pH Reached when L2 Increase L2 from 00 until Zm Z0 10 jO d w y my 2 N EE 139quot x CL A v 7 m2 l J m2 m2 quot alues of higher susceptance jb ECE145521805 notes M Rodwell copyrighted2009 2nd lumnetl 9 Matching Newark simplehybridpi X3 Final Values Performance vs Frequency DC 200 GHZ frequency sweep marker at 100 GHZ 18 I I anAm JarMum E w w W gala um 95mm 30m gm Hymn1 M dBS 1 i i i i i 20 40 60 80 100 120 140 160 180 20 freq GHz More direct matching trajectory gt broader bandwidth mam278 notes M Rodwel copyrighted 2009 3rd lllmll ll 0 Matching Network simpehybridpi X3 Matching network with values 0 Port Port C L P1 C1 L2 P2 Num1 C22 fF t L568 pH 1 Num2 S11 matching trajectory at 100 GHz A original Z in C after adding L2 in parallel D after adding C1 in series 1 4 5 S11 ECE745527BCS notes M Romet copyngmed 2009 3rd llllllll ll H3 Matching Network simplehybridpi X3 Final Values Qr 39 O Pon Port C L P1 C1 L2 P2 Num1 C22 fF t L5518 pH 1 Num2 Performance vs Frequency DC 200 GHz frequency sweep marker at 100 GHZ Dz Illumw stmnlm p m m m mm 121va gm m Eng MW m dBS 1 x x x x 20 40 60 80 100 120 140 160 180 20 freq GHz Long matching trajectory gt narrower bandwidth ECE74Sa27BCa notes M Rodwel copyrighted 2009 4th lumnell 0 Matching Network simplehybridpi X3 L Port L2 P2 L1696 pH t Num2 Matching network with values S11 matching trajectory at 100 GHz A original Zin B after adding L2 in parallel 1 4 5 S11 D after adding L1 in series ECEMSaMBCS mazes M Rumell cnpynghted 2009 4th lumnell 0 Matching Network simple hybrid pi X3 Final Values Performance vs Frequency DC 200 GHZ frequency sweep marker at 100 GHZ 1563 v u in mm mm M mm m 9 ram W dBS I l ttt ttttr stttt r39o o v LI t t t t 1 t t t t t 20 40 60 80 100 120 140 160 180 2 0 freq GHz short matching trajectory gt wider bandwidth MumSection 0 Matching Newark Tuning with E C multiple series shunt elements L L I L 6 6 I gainplot 5paramet2rs0 Q Elle gal Mew insert Marker lsmry gowns Eels Rage Mndow Help in nite of possible matching networks u A 51 1 aws2180a Hales M Rumell cnpyrrghled 2009 lines of constant 0 Q maximum energy storedenergy dissipated per radian X Rseries 2 H B H Gmm for simple 2 element impedances EEK I gainplot Sparameterso le Edit Mew Insert Marker lstory Mons Eols gage Mndow Help series series curves of constant Q look roughly like this we S11 Matching networks passing through high Q points will have narrow bandwidth M llt l 3 gain Ircles Y garameters Z parameters SJaramemrs gures 0f merit HDISe gure 5 11 db aws2180a mazes M Rumell cupwgmed 2009 Narrowhaml VS Willellallll Matching BIWIII kS 4 element wideband 2 element narrowband humus ummuqm 1153 NEEDr ym um Lum lm39y W 1m 1m mam m V 7 51 5a Maw mammes notes M Rodwel copyrighted 2009 limits to Matching Network Bandwidth LowQ load ZL nghQ load ZL I quotcum stmwrnzo Ania I gnuth Synmvslto a 53 if M 139 WM Lnrw MW W E03 EBB HMN m El id 1m M Barkr aw W 305 EWE EmirV 53910 7 1 it Ti 2 t t I Starting point ZL is low Q Starting point ZL is high Q gt match can be made wide or narrow gt match cannot be made wide ShuntStub Matching Networks ECEMSaMBCS mazes M Rumell cnpyngmed 2009 Adding Series TRX line IzainJlorlsiarzmnl a V qumi HR at my mm xer JIW J h 5X5 ijndml PM 7 mme i39 increasing the line length rotates the ex 4 load re ection coef cient Fm 126721 1 LgIWLEFLESJLL J U AA JJ4 serles llne Zline Zsystemistandard ZO if the standard impedance is 509 the line is 502 E CE 1 45521 805 notes M Rodwell copyrighted 2009 Recall Traiectorv l l adding SIIIIIII SIISGGIIHIIIBB Adding Shunt Parallel Susceptance Susceptance can be Yum Y L if an ideal L or C D gainplot SJarameterslw I Susceptance can be a shunt le gum yew insert Marker story gptions Iools gage Vnndow ew transmission line stub 3H Susceptance can be a shunt transmission line stub A aws2180a mazes M Rumell cupwgmed 2009 T Open Terminated Short Terminated CB match L Y0j0 IpmJIoz39 SJiramlmlw a in 39av 1m 9me NF new mm m Ewe nmw new 322 31 1 39 39 start Yinfinity ShuntStun Matching EIWIII k I gainplot Sparameterso Elle Edit Mew Insert Marker story gnuons 1MB gage Mndow Help lt w g a 1 n C I rc es V gammeters Z gE mElEI S SParamarers qures of men muse figure 5 11 db gt v aws2180a mazes M Rumell cupwgmed 2009 CB LL L Series line brings load to Y Y0 1 i 18 Shunt stub adds YstubYO Combination brings load to Y Y0 1 jO mammes notes M Rodwel mpyngmed 2009 18 SIIlllltStllll Matching NBMIIIK TLIN Z500 Ohm simpehybridpi X3 Network T01 OIOU F 100 GHz 1 TLIN P1 TL1 P2 Num1 Z500 Ohm Num2 E0 t F 1 00 GHZ i uluwr SJIVIWMI39O 1 32 S T 7quot 7 quotTEE u V a g m S11 before matching at 100 GHz m2 vfreq100 GH 0 z S110711 89646 jlmpedance 20 quot 0330 7 0 950 52 2 51 1 mam275 notes M Rodwel copyrighted 2009 1S SIIlllllSlllll Matching BIWIIIII Increase T L1 length until YM Y0 10 jB Reached when 2711 xl 68 degrees 33 39wax ummn 5 i x aw lIuuu39uJn nn nail luuu39lx mu We have moved on a constant F circle ees aw Increase T L2 length until Yin Y0 10 jO Reached when 2 m y l xp m V W iExq n quot j gl 39 We have moved on a constant g circle towards values of higher susceptance jb ECE74Sa27BCa notes M Rodwel mpyngmed 2009 18 SIIlllltStllll Matching NBMIIIK z5oro Ohm simplehybridpi x3 Final Values Performance vs Frequency DC 200 GHz frequency sweep marker at 100 GHz SEE 20 40 60 80 100 120 140 160 180 2 O freq GHz 2nd SIIlllltStllll Matching HWMK A shorter series line section TL1 The shunt stub must now be 1nduct1ve from A to B brings Z matched lemumz mammmnka M37 52 gar gm mm Mam my m 1 ma gnaw new W aws2180a Hales M Rumell cnpyrrghled 2009 3rd ShuntStun Matching EIWIII k I gainplot39 SparametersD Elle Edit Mew Insert Marker story gnuons Eels gage Mndow Help o Series line brings load to Y Y0 1 i 18 v m Combination brings load llt l I g a t n c I rrles f garameters Z QEIEMEIEISA SParamatars gures of merit muse gurz 5 11 db to Y Yo 1 jO Radial stub a wedge shaped capacitor with distributed effects accurately modelled ECE74Sa27BCa notes M Rodwel mpyngmed 2009 3rd SIIlllltSIIIII Matching NBMIIIK simplehybridpi X3 Final Values MRSTUB F100 GHz Stub1 MSUb Substquotthinfilmmicrostripquot Wi10 um MSUB L98 um 1 thinfilmmicrostrip Angle90 H4 um Performance vs Fre uenc Er27 A T025 um DC 200 GHz frequency sweep marker at 100 GHz ID i kW 1Nquot WW WW WK M 29 MW UM n 39 v 7 2 10 3 a r v quotw I 20 an 7 I I I I I I I I I I I I I I I I I I I 20 40 60 80 100 120 140 160 180 2 0 7 freq GHz ShuntStub Matching Networks with general line impedance mammes notes M Rodwel copyrighted 2009 Traiectories for Adding a Highlo Series line IAdding Series inductance I ZOI IAdding series line Z gt ZOI line IAdding series line Z line I gaithn ummrzlm Ee m ym mm mm dswv mums 1m Ema mam urn Behavior is intermediate between series inductance amp series line of impedance ZO aws2180a Hales M Rumell cnpyrrghled 2009 Matching Network Will HighZn Series 8 lowlo SIIIIIII lines I gainplot39 Sparameters0 C B Elle Edit Mew Insert Marker lsmry gnuons Eels gage Mndow Help lt 0 v 7 Series line brings load a toYY01ijB m m Shunt stub adds YstubYO l l l l l gain Ircles v garameters z gammElEmA SParametarS gures cf ment noise gure s 11 db Series line is mostly inductive shunt line is mostly capacitive ECEi4sa2180a mazes M Rumell cupwgmed 2009 l Zjiarched naiched L series Lshun2 J 3 T T shim C C V T SEVIES VEVICS shun Tshunr h series series T Z39 Shunt series Lshunt Tshum ZShum CSthl Z series Tseries series series Z shunt series If we fr 2mm gt 00 n Zshum gt 0 Z marched L Whlle hOIdlng Lseries amp Cshum constant L series 2 then Cseries Lseries Zseries 0 2 and Lshunt CshuritZshurit 0 T Cslmm mam275 notes M Rodwel copyrighted 2009 Trajectories If llllillg a OWIII Series line Addin shunt ca acitance I g p I IAdding ser1es 11ne ZW lt Zol IAdding ser1es 11ne Z W Zol m i7iiri E g f L piniu 321 Behavior is intermediate between shunt capacitance amp series line of impedance Z0 aws2180a Hales M Rumell cnpyrrghled 2009 Matching Network Will High10 Series 8 OW10 Series lines B I gainplot Sparametersl L L L Elle Edit Mew Insert Marker stury gnuons Eels gage Mndow Help Z line gtZ0 line lt20 Z Series line is mostly inductive Shunt line is 5 11 mostly capacitive Dotted lines show lumped L C trajectory l l V l l gam curdas Y garameters z gammeteus SJarametars gures of menl muse gurz 5 11 db ECEi4sa2180a miss M RUINell cupynghied 2009 Z I Zmalched Umtched LIOW Z Z L61 low Tlow high Thigh I T Clow Claw2 L Z C Thigh L Z high Thigh high high Z W 710w low high If f Z gt 00 d Z gt o LZFWW thh we orce high an 10w while holding Lhigh amp C W constant then Chigh Lhigh Zfigh gt O andLZUW C 212W gt O 10w m vvxtugiug ma twinigji 1 h guyiuwig 53 aws2180a mazes M Rumell cupwgmed 2009 llllllllell VS niSll illlllell Matching BIWIII kS Given this mask layout 3 ADSforZ1Ea 1 match Layout3 H He gm galecr mew gnsert Qp ons look Sgheman Mumantum EMgs Mnduw DwgnGulde elp pea 5 t Msan 9 39E39E Q h hiD 429 zitSu Are 9 BOA v Elm r 4 0 x5 v O 2 ill ask vs default Lumpadwm Artwork v md ARF SlmSchem 1000 V500 lt Select Emerme staan pomt u rtems 1000 0 00 your eyes should see this EWING7861 L law aws2180a mazes M Rumell cupwgmed 2009 llllllllell VS niSll illlllell Matching BIWIII kS IU DSJMVH mm m w y Given this mask layout your eyes should see this N Z marched L L LShW2 series LMW2 serzes C2C2 C Series series C series Series slmnt shim b b Quarterwave and related Matching Networks Quarterwave lmnedance Tranlormer ECEMSaMBCS mazes M Rumell cnpyngmed 200 Atload ZL 1FL line 1 line 1 L z 1 f At 1nput i F Egijxw giynmuia J Elg B m in ma Zu39m wow mm m m gnaw m N 17 A rm aw fL if I 24 aws2180a mazes M Rumell cupwgmed 2009 Insight Series lines Z R L m Lin F F D D B D Regardless of the starting point Z L Zine and ending 130th 217 the impedance trajectory contains the points A B with resistive input impedance lnlnjnf SJnrnmmni39 re 11 a1 nsm 1mg sww 415mm 2 By nds m w gt such that 2 RinA I RinB Zline aws2180a mazes M Rumell cupwgmed 2009 Quarterwave Illlllellall e Tranlormer BBSiSIWB loads 331 AR m L L 2m L D D R FL 17 77 IuanaE ISJaumnhzrsl o DE 1 at you we r V n M nmn mm m m mm W 2 Rm RL 2 Z line l B Pick line impedance to match Z m to Z0 Zline VRLZO ECE145A218A Course Notes Last note set Introduction to transmission lines 1 Transmission lines are a linear system superposition can be used 2 Wave equation permits forward and reverse wave propagation on lines VxtVt xv V txv IxtZiVt x v V t x v IDC o v L hase velocit LC P V ZO characteristic impedance L C per unit length 3 Re ections occur at discontinuities in impedance A re ection coefficient F can be defined RL 1 V Z 0 V RL i0 1 4 Transmission line can be represented by an equivalent circuit LT2 LT2 LT W CTZ CTZ LT Z01 CT TZo T lengthv time of ight Goals Understand frequency domain analysis of transmission lines phasor notation position dependence phase constant 5 re ections movement of reference plane impedance and F variation with position Smith Chart ECE145A218A Course Notes Transmission Lines 2 Transmission Lines in the Frequency Domain and the Smith Chart Time domain analysis is intellectually clearer the picture being forward and reverse waves propagating re ecting and rere ecting This analysis becomes intractable as soon as we introduce reactive impedances as multiple convolutions will be required for timedomain re ection analysis So we will analyze in the frequency domain instead Frequency domain analysis of transmission lines is a classical approach to this problem gtX ZL FIG 1 Transmission line with impedance Z0 connects the source with Thevenin equivalent generator impedance Zs to a load ZL V Re Vaejwt V0 V0ei u S0 2 vs t IVUI cosmt gt On a transmission line at position x waves travel in time as x i Vt where velocity v is the phase velocity Equivalently at a time t waves vary with position x according to It XV Thus we can represent voltage on the line vXt as c0sat 0 gtc0sat i xv gig c0sat 0 i xa v c0sat t7 i x 2 112007 Prof Steve Long ECE145A218A Course Notes Transmission Lines 2 where 5 is the phase or propagation constant 5 oav 2117 5 converts distance to radians Here 7 is the wavelength The sketches below illustrate the concept Here the function cos oat 5X is plotted 1 Let s suppose that t to so that the wave appears frozen in time on the X aXis If distance X2 X1 7 as shown the corresponding phase change over this distance is 211 l7bgt X1 3 X2 3 X FIG 2A 2 Now if we plot this same cosine function as a function of 5X we see that the wavelength is equal to a phase 5X 211 since 7 21t5 gtl27 lt WW 3 Also we can set X 0 and observe the wave function in the time domain for FIG 2B increasing time In this case one period requires a time interval T 21tn lf as shown M T I4 I W0 t FIG 2C 4 Finally you might ask why the cos oat 5X wave is the forward to the right travelling wave direction To see why track a point of constant phase with position on the X aXis as time progresses from t1 to t2 3 112007 Prof Steve Long ECE145A218A Course Notes Transmission Lines 2 t t2 gtX X2 FIG 3 Plot of wave propagation in the forward direction at times t1 and t2 We can see that t gt t1 and X2 gt X1 for a forward direction of travel and that the cosine function will have the same value at points of constant phase Therefore cosat1 6x1cos17t2 x2 1 Thus wtl mtz Bx1 xz lt 0 2 from the drawing A forward wave must have a positive phase velocity 0 v gt 0 3 3 From eq 2 v gt0 4 t2 t1 X2 must be greater than X1 therefore the wave is travelling in the forward direction for cos oat x Of course these waves can also be described by complex exponentials IVOIejwt em eij c sine wave phase position along line 4 112007 Prof Steve Long ECE145A218A Course Notes Transmission Lines 2 The ejmt time dependence is always taken as implicit and is frequently omitted when using phasor notation IVOI ej u eij x Voltage and Current on Transmission Lines Now we can use this notation to describe the voltage and current on a transmission line at any location on the line Vx Vx Vx V0ej x V0 af x Iltxgt ZiVltxgt V ltxgt 0 Zi0V0e f V390 af x 5 112007 Prof Steve Long ECE145A218A Course Notes Transmission Lines 2 Re ections in the frequency domain 0 VQb39 If IZL Zo r FIG 4 Re ection from a mismatched load impedance ZL The forward wave is travelling in the positive X direction and re ects from the load ZL We set X0 at the load end of the transmission line as our reference plane In frequency domain analysis we assume that the wave amplitudes are steady state values From the de nition of re ection coefficient V7 0 V0FL z 1 FL L zL 1 where ZL is the normalized load impedance ZLZO FL is in general complex 6 112007 Prof Steve Long ECE145A218A Course Notes Transmission Lines 2 Movement of reference plane x l 0 I Zo VQb39 too IZL r Now we must determine voltage and current on the line as a function of position This is often referred to as moving the reference plane Here we move from X 0 at the load to x Z at the left end of the drawing Vx V x V x V x1 I39x where is the positiondependent re ection coefficient Substituting for V X and VX V 0ef V0emc no I39x and VxV0e x1 1quot0e2j x From this we can see that the re ection coef cient at position Z from the load is given W 1quot l1quot0e2 l 7 112007 Prof Steve Long ECE145A218A Course Notes Transmission Lines 2 and the re ection coefficient goes through a phase shift of minus 21 200 radians minus 2B1 radians minus 360 2lZ degrees The re ection coefficient changes phase with position Some observations 1 FX is periodic ZJ39KZ jx Fx r0e A 41tX7t 211 whenever X n M 2 where n is an integer 2 VX is also periodic Vx V0e f x 1 FOe2j x 1 every n 1 every nk 2 1 every nkZ 1 every nk4 Example Let F0 1 short circuit X l l The result Standing wave pattern in voltage 0 Voltage maxima and minima are separated by M 4 o Successive maxima are separated by M2 8 112007 Prof Steve Long ECE145A218A Course Notes Transmission Lines 2 Voltage Standing Wave Ratio QYSWRg We have seen that in general there are two waves travelling in opposite directions on a transmission line We also saw that the variation of voltage along the line at position X due to the sum of these two waves is given by VxV0e fquotx1 1quot0e2 The magnitude is given by Vx V0 1F0ezj xl So we can easily determine the minimum and maximum voltage magnitude that will be found along the transmission line at some position X Vx lmaX IV 0 1 H0 I Vx lmin IV 0 1 H0 I Taking the ratio of max t0 min gives us the VSWR VSWR 2W 1F0 Vxmin I 1 I r0 An open or short circuited line will give us an in nite VSWR because the minimum voltage on the line is zero F0 l for both cases VX v1r 0 l l l x M2 M4 0 Why is VSWR important It is often used as a speci cation for an ampli er 9 112007 Prof Steve Long ECE145A218A Course Notes Transmission Lines 2 Impedance vs Position Refer to the same picture The impedance at any point on the line can be found from the current and voltage or equivalently from the re ection coefficient 2xwz w 1x Vx Vx 1 Fx 0 The normalized impedance at any point is easily found 1 zxZx 20 iti So impedance depends on the position along the transmission line While this is conceptually simple there is a lot of math involved This can become tedious So we could benefit from a graphical representation this is called the Smith Chart after Philip Smith who invented this convenient graph of transmission line parameters back in the 1930 s The relationship for the normalized impedance ZX is the key to the Smith Chart The Smith Chart is just a polar plot of re ection coefficient Impedance is determined by It is a onetoone mapping between complex numbers F and z and is in fact an analytic function and a conformal transformation You can read about this in math books on complex analysis 10 112007 Prof Steve Long ECE145A218A Course Notes Transmission Lines 2 In the twodimensional plane of F the F plane a re ection coef cient F is represented by a point 1 j0 Im F 0j1 0 1j0 Re Ojl 112007 Prof Steve Long ECE145A218A Course Notes Transmission Lines 2 As we move away from the load by a distance Z on the transmission line F rotates by an angle A9 where A6 432 360 One whole rotation is required in the F plane for each halfwavelength movement on the line 1quot0 Note the following 1 g is de ned as quotelectrical lengthquot in degrees 2 The degree scale on the edge of the Smith Chart represents Z f the angle of the re ection coefficient Note that this is TWICE the electrical length L Note that phase delay corresponds to a clockwise movement in angle 12 112007 Prof Steve Long ECE145A218A Course Notes Transmission Lines 2 Lines 01 L r quoton L 11quot z l I39 Thesewill L Sect 2 L line across the center of the chart as shown below Constant reactants circle centers on vertical ax s zrjx m Constant resistance circles on honz axis Note that the unis ofr are normalized to 20 in this case 50 ohms So the circle labeled r 05 corresponds to the 25 ohm resistance circle in this case But 20 can be any convenient real Value IfZO were 100 ohms F05 would represent 50 ohms Similarly the reactances x can be represented by circles These have their centers on the Vertical 39 4 the chart quot L quot Positive x 112007 Prof Steve Long ECE145A218A Course Notes Transmission Lines 2 corresponds to inductive reactance and is above the center line of the chart Negative X represents capacitive reactance Given this mapping onto the F plane we can associate any re ection coef cient a point on the plane with an impedance simply by reading the z coordinates of the point We can also associate the change of impedance with position with a rotation on the chart Just rotate the F vector clockwise around the chart at the rate of one rotation for every half wavelength of movement on the line Then read off the impedance directly from the chart Fx FOe 2j x Note that all points on this chart represent series equivalent impedances 14 112007 Prof Steve Long ECEIAsAzlsA CnurseNntes Transmissinn Linesz So for example 2 1 11 represents a senes RL network 21 71115 asenes RC 2 1 15 apoan ngm m are center ofthe chart Normahzauon Consxderthe pomtz 1 11 IfZU 5011nen z 50 150 when he 1rnpedanee1s denomahzed IfZU were 1000 ohms we would havez 1000 11000 for are same pomt In 11an way the enan ean be used over an arbnrary range ofxmpedance 15 112007 Prof SteveLong ECEIAsAzlsA CnurseNntes Transmide Linesz on ths ehmwe see ammpedanee 1 11 comespondmgto me senes RL network Ifwe 1 e e resxstance lme r 1 to the center The reactancehas been cancelled 16 1120U7 Prof SteveLong

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