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# BRASS QUINTET MUS A 145

UCSB

GPA 3.96

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This 28 page Class Notes was uploaded by Cecelia Emmerich on Thursday October 22, 2015. The Class Notes belongs to MUS A 145 at University of California Santa Barbara taught by Staff in Fall. Since its upload, it has received 54 views. For similar materials see /class/226994/mus-a-145-university-of-california-santa-barbara in Music at University of California Santa Barbara.

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Date Created: 10/22/15

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

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