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by: Miss Alberto Prohaska
Miss Alberto Prohaska
GPA 3.94


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Class Notes
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This 50 page Class Notes was uploaded by Miss Alberto Prohaska on Thursday October 22, 2015. The Class Notes belongs to MATRL 218 at University of California Santa Barbara taught by Staff in Fall. Since its upload, it has received 24 views. For similar materials see /class/226929/matrl-218-university-of-california-santa-barbara in Material Science and Engineering at University of California Santa Barbara.




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Date Created: 10/22/15
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IT 4quot g V5 1 541 roz aa zzM M 4 a 4 42 11461 2 L535 S MAM 2 4 Mz I aa 417Q75m NLZQ w Izzy1 g 5 quot 2 404 gt rag4y 5 1 6 cvrz r Z 4 4433 47 remi LivIAOJ w 74gKnuy wig644 an zr 4 074 aai tI 3 544 u 9 7 11 4 294 4a 4 ar a max1174 W a 5A 5 Qf b u ouml 4445244 3452 7M a n 1 PC 39 sac s7 1 m I D L D 3 3 a 7 464 p a g m 9 7 7 zgtgt22e M if 7 4a 3 i W w M 39g i 714 A a x rsr 7 m AM 7 71 may B Sam mile9617 aaa Ax 6 M i sm hth klNamp nn Mo knx Sn R m n c mm3v k En k 3 xak km RR WxxhkV K39 Rd Dn u RX Kme KKK Nlh k oi Exngkwghw WMamp axme Rxmhmbxwk R39 2 amex k PQ Xk Nx l u kxnx X2n at tkbkx NNx RN Ar n nmnszK inth V vomr c 3th ESK manna 91 IKNnR N E E g 58 hm nr 0 ECE145a 218a Resistive Feeback Amplifiers Mark Rodwell University of California Santa Barbara rodwelleceucsbedu 8058933244 8058933262 fax c ass notes M Rodwel copyrighted 2009 BBSiSIWB FBBIIIIEIGK For line Termination C r5 Z TL 7 J K If s K 0 BVoutpos A I i i ii It 7 I 2 4 6 81012141618 20 freq GHZ Recall there are 2 different reasons for impedance matching 1 Maximumpower trarsfer 2 Avoidinggain ripple from twn inU waves on lines In resistivefeedback amplifiers feedback providesZm ZWt ZO in order to eliminateline re ections These are not designedfor maximumpower trarsfer hence we do not obtain the transistor maximum gain Class notes M Rodwel copyrighted 2009 V dram Z 20 0 E Z0 Z0 V V gen gate Ampli er consists of transistor and bias tees connectedto input and output transmission lines Transmission lines are external to ampli er reference planes are at line ends Analysis Z pod 2 To find S parameters 0 port 1 we compute gains and impedances Z wichgw ZloadZO Take the DC bias elementsas implicit Use for now a highly simplified deVice model Circuit toanalyze Mass notes M Rodwel mpwgmeamg Mass notes M Rodwel copnamed 2009 Analysis 0 port 1 port 2 By inspection Z ij gtSH ZinZo 1 V 2 Z m g Zm Zo 1 5252 R S ZWZo 1 gen de fzo 1 0 gs out 00 gt 22 W What is wrong with this amplifier Though gain is lower thanGmax no matchingit may meet our needs More seriousproblem large S11 S22 gt standing waves gt gain ripple cass notes M Rome4 copyIgmed 2009 Standing WEIIIBS Z Th Re ections at both line ends will cause gainphasevarationsof the form simplehybridpi X4 1 Port TLIN Port 2 39 z 7 P1 TL1 P2 1 e m FSFL Num1 2100 Ohm Num2 E90 t F10 GHZ hence gain will vary Withboth 5w cable length and with frequency 40 We need to x this 97 2 freq GHZ cass Hales M Rudweu cnpyrrghled 2009 So While we can directly f F 1i Jul 0111 L 117 match two stages directlythus l L W I IE this is only acceptableif line lengths are short and the connectionpermanent If lines are long or the componentsmodularso that connection can be changedwe must match each device to Z 0 F F011 POI10 rin0 011 I AL J Z L 0 l class notes M Rodwell copyrighted 2009 Twas oi Imnedance Matches Reactive ImpedanceMatchinglossless Re ectioncoef cierts gt O No power re ected No signal power absorbedin matchingnetworks gt Circuit Gain 2 device maximumavailable gain maybe Resistivelmpedance Matchinglossy Re ectioncoef cierts gt O No power re ected Signal power absorbedin matching networks gt Circuit Gain less than device maximumavailable gain cass mazes M Rudweu cupwgmed 2009 World39s Second Simnlest Amnlitier Z uuv Z V m R 2 V Km In the absence of transistor parasitics H lines are now terminated in Z 0 RL ZO Re ectiom are now nearly eliminated V Vdrain gate Power gain is wasted signal dissipaticn in terminations cass mazes M Rudw all mpyrrgmed 2009 ll lVSiS port2 Zo Pom C1rcu1tb1as 1mp11c1t V 20 RL 20 gen over simpli eddevice model Vgen ZU port1 Z0 Cgs I ng nggS RDS RL Portz Zl C1ru1t Model J 39 Vgs We will pick Rds RL Z0 cass Hales M Rudweu cnpyrrghled 2009 Analysis As alwa s y Vgen Z pom Z C135 Icy nggS RDS Rl 10quot 2 20 v u S 2I0ut 21 V gs gen denZLZO By nodal analysis or MOTC l bls l als azs2 Where a1 Cgsz0 2 ng Z0 21 ngO 2 Z0 2 the terms a2 and 91 are relevant only at very high frequencies 521 ngO 2 The Miller effect term shouldbe obvious cassnozes M RodweL mpyngmed 2009 ll lVSiS 1 whereal Cgsz0 2 ngkz0 21 ngO 2 Z0 2 S 2 22 21 gm 0 1jwa1 f127zza1 l N 2 E 3 2 1 0 1 2 3 a m 0 g Z 2 U m 0 freq 1 OOOGHZ to 1 OOOTHZ 10 1E9 1E1U 1E11 1E12 freq HZ Note the standard Bode Plot behavior With 20 dBdecadeslope and 3 dB corner Note that the 521 trajectory is a semicirclein the complex plane cass Hales M Rudweu cnpyrrghled 2009 Illlllll Imnetlance ZO ng ng1 A V TT Note the single pole Z LIN Using the Miller Approximation Note that thisa1most follows a constant G circle on the Smith chart behaVior of Sn 7 quot o j 30 I 39 40 quot xx e quot I 9 15 10 15 11 1512 freq 1OOOGHZ to 1 000THZ qu HZ class notes M Rodwell copyrighted 2009 ll S21baseband E and lzmigh E CgsZ0 ng gt 0 SO gain prOdUCt S21baseband zigh gmz Cgs f A rough calculation only poorer bandwidthwhen other device parasitics are considered cass Hales M Rudweu cnpyrrghled 2009 Simnle ResistiveTerminated Amnlitier General Form This stage generalizes to use with any gm block Recall extrinsic tranconduetance for a transistor with emittersairce degeneration Rx is lt1gmmmnyl Hf 9 PH PE cass mazes M Rudweu cupwgmed 2009 ResistiveFeedbackused to provide mput and output impedancesof Z0 We want V out Iin AV andZin 20 We mustfind gm Rf necessaryto obtainthis If we are lucky we will also get Zom ZO Inputcurrent ImVi Z V Z n m in 0 Current in R f Rf Iin I0utRf Iin Butlef 11 Vin20 Knlt14gtRf gt Rf 201 AV cass mazes M R ndweH cupwgmed 2009 cass mazes M Rudweu cupwgmed 2009 Ill 39 Currentin R f Rf IinRin2IinZO Current in load R f out IoutZO V A Zo inv gm block output current gm ngin out Rf ImAvZo IinZO Vm1AvZo gtgm1AVZO cass mazes M Rudweu cupwgmed 2009 1171 ItestZO pom Rf port2 I W 2 current m R f current 1n gm 5 c I m 1171 Z 0 20 FM gm Vin rest 1 I Z 1 Goutz z z 0 gm Rout Itest ZO ZO Gout L i1AV ZOZO1AV Z0 Z0 1 1 G 2 out 24 14V ZO G Finn ou 20 Output is impedance matched cass mazes M Rudweu cupwgmed 2009 Broadband Feedback Amnlitier Summary We want V out IinAV 7 ZinZOZOLtt This is obtainedby setting gm and 14 ZO Rf 1AVZO Thoughnote that Z in Z0 only if Z L Z0 and Zom Z0 onlyif den Z0 Mass was M Rodwel copnamed 2009 Broadband Feedback Amnlilier WIIv do it El Why do this instead0f this gm2AvZo Answer more bandwidthless noise noise for a later discussion Mass was M Rodwel copnamed 2009 Broadband Feedback Amnlilier WIIv do it El Why do this instead0f this gm2AvZo Answer more bandwidthless noise noise for a later discussion CSSSNUE S PU Highly simplified G or B D or C device model If 2 gm 27ZCln l g Cin K17 gm m Cingm27i39 SOVE The device input capacitance C m is proportional its transcendictance gm The feedback amplifierrequiresless gm gt less device Cm cass mazes M Rudweu cupwgmed 2009 canacitance as the Price quotIf Transcontluctance Jt a C a T T T Cd Rd Cd Rd Cd Rd N 95inzgm27yr Even with resistive degeneration input capacitance is proportional to transcondictance relationships from analog desrgn review notes set cass mazes M Rudweu cupwgmed 2009 Broadband Feedback Amnlitier Bandwidth Analysis Z Z 0 0 Vgen ZO Rf ngin Z0 Cm gm27gfr butgm 1AVZO so Cm 1AV27rfTZO but f3dB 127ra1 Where a1 CmZ0 ZO CmZ0 2 2 so fwg 221 Less gm required gt less Cm gt more bandwidth cass mazes M Rudw all mpyrrgmed 2009 comnare I0 Simnle Resistive Amnliiier gen gm Vin Z0 Z V ZU ZU Cir Z 0 1 Vin Cm gm27rf butgm 2AVZ0 so Cm 2AV27zfrZO but f3dB 127Ia1 where a1CmZ0 ZO CmZ0 2 50 f3de AT V More gm required gt More C m gt less bandwidth


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