Lnear Amplfers and OpAmps eferences: Barbw (pp 7-80), Hayes & Hrwtz (pp 63-40), zzn (Chapter ) Amplfers are tw-prt netwrks n whch the utput ltage r current s drectly prprtnal t ether nput ltage r current. Fur dfferent knd f amplfers exts: ltage amplfer: Current amplfer: Transcnductance amplfer: Transresstance amplfer: A = / = cnstant A = I /I = cnsant G m = I / = cnstant m = /I = cnstant Our fcus n ths curse s n ltage amplfers (we als see a transcnductance amplfer). ltage amplfers can be accurately mdeled wth three crcut elements as shwn belw. These crcut elements are related t transfer functns f tw-prt netwrks dscussed befre. and are, respectely, nput and utput resstances. Gan, A 0, s the alue f H(jω) f the un-termnated crcut. (Fr ltage amplfers the cnentn s t use A nstead f H (jω).) A gd ltage amplfer has a large nput resstance,, and a small utput resstance,. An deal ltage amplfer has, and 0. s Zs I A 0 I Z L Practcal Amplfers ltage Amplfer Mdel eal ltage amplfers dffer frm the deal amplfers. Nt nly, the nput resstance s nt nfnte and the utput resstance s nt zer, but the amplfer wrks prperly nly n certan cndtns. One shuld always be aware f the range where the crcut acts as a lnear (deal) amplfer,.e., the utput s prprtnal t the nput wth the rat f A = / = cnstant (exactly the same waefrm). Amplfer Saturatn: Amplfers d nt create pwer. ather, they act as a ale adjustng the pwer flw frm the pwer supply nt the lad accrdng t the nput sgnal. As such, the utput ltage amplfer cannt exceed the pwer supply ltage (t s usually lwer because f ltage drp acrss sme acte elements). The fact that the utput ltage f a practcal amplfer cannt exceed certan threshld alue s called saturatn. A ltage amplfer behaes lnearly,.e., / = A = cnstant as lng as the utput ltage remans ECE60L Lecture Ntes, Sprng 004 34
belw the saturatn ltage, sat < < sat Nte that the saturatn ltage, n general, s nt symmetrc,.e., sat, < < sat,. Fr an amplfer wth a gen gan, A, the abe range f translate nt a certan range fr sat < < sat sat < A < sat sat A < < sat A.e., any amplfer wll enter ts saturatn regn f s rased abe certan lmt. The fgure shws hw the amplfer utput clps when amplfer s nt n the lnear regn. Amplfer Bandwdth: Typcally, amplfers wrk n a certan range f frequences. Ther gan, A = / drps utsde ths range. The ltage transfer functn (gan) f an amplfer s pltted smlar t that f flters (Bde plts). It lks smlar t thse f a band-pass flter r a lw-pass flter. Cut-ff frequences and bandwdth f the amplfer s defned smlar t thse f flters (3 db drp frm the maxmum alue). In terms f frequency respnse, ltage amplfers are dded nt tw categres. () AC amplfers whch nly amplfy AC sgnals. Ther Bde plts lk lke a band-pass flter. () DC amplfers whch amplfy bth DC sgnals and AC sgnals up t a certan frequency. Ther Bde plts lk lke a lw-pass flter. Fr ths class amplfers, the bandwdth s equal t the cut-ff frequency (lwer cut-ff frequency s set t zer!). se Tme: In an deal amplfer, f the nput ltage s a unt step functn, the utput ltage wll als be a unt step functn as shwn. A practcal amplfer cannt change ts utput nstantaneusly f the nput changes suddenly. It takes sme tme (a shrt but fnte tme) fr the amplfer utput ltage t reach ts nmnal leel. The maxmum rate f change n the utput ltage s called the rse tme. Ideal Amp. Practcal Amp. t ECE60L Lecture Ntes, Sprng 004 35
Maxmum Output Current: A ltage amplfer mdel ncludes a cntrlled ltage surce n ts utput crcut. Ths means that fr a gen nput sgnal, ths cntrlled surce wll hae a fxed ltage ndependent f the current drawn frm t. If we attach a lad t the crcut and start reducng the lad resstance, the utput ltage remans a cnstant and lad current wll ncrease. Fllwng ths mdel, ne culd reduce the lad resstance t a ery small alue and draw a ery large current frm the amplfer. In realty, ths des nt happen. Each ltage amplfer has a lmted capablty n prdng utput current. Ths maxmum utput current lmt s als called the Shrt-Crcut Output Current, I SC. If ne tres t exceed ths current lmt, the ltage amplfer wll nt behae lnearly; the utput current wll nt ncrease, rather the current stays cnstant and the utput ltage and the gan wll decrease. I SC I I SC If the a fxed lad resstance, L, s cnnected t the amplfer, the maxmum utput current means that the utput ltage cannt exceed the L I SC : L I SC L I SC In ths case, the maxmum utput current lmt manfests tself n a frm smlar t amplfer saturatn. The utput ltage waefrm wll be clpped at alue f L I SC. Tw ptns are aalable t ad the maxmum current lmt: () Increasng L as s seen frm the abe equatn, () decreasng by decreasng the nput ampltude. There are ther lmtatns f a practcal amplfer such as sgnal-t-nse rat, nnlnear dstrtn, etc. whch are ut f the scpe f ths curse. Hw t measure A,, Measurng A s straght frward. Apply a snusdal nput sgnal wth ampltude t the nput, measure the ampltude f the nput and utput sgnals wth a ery large lad (scpe nput resstance). Make sure that the amplfer s nt n saturatn and sgnal s nt clpped), and cmpute A = /. Nte that A s n general depends n frequency. There may als been sme phase dfference between and smlar t flters. Measurng nput and utput resstances (r mpedances) s nt tral. Part f the dffculty s due t the fact that an Am-meter s lmted t 60 Hz (gd ne can measure up t a few khz) and we shuld use scpes that nly measure ltages. In rder t measure a current wth scpe, we need t fnd the ltage acrss a resstr whse alue s accurately knwn. ECE60L Lecture Ntes, Sprng 004 36
Care shuld be taken as addtn f such a resstr t the crcut may mdfy ts behar. The general technque fr measurng nput and utput resstance f any tw-prt netwrk s gen belw: Measurng nput resstance: Add a resstance (measured accurately) n seres wth the nput. Measure ltages and A. Then: = A = A Zs s A Surce I Twprt Netwrk Nte that shuld be chsen such that a) s nt t small t be measured accurately (large cmpared t ), and b) A and are nt t clse (small cmpared t ). T see the secnd pnt, assume we measure A and t be and 0.98 respectely. Als, = kω. If we use these alues n the abe frmula, we wll fnd = 49 kω. Let us nw cnsder the errr n ur measurements. The measurements are perfrmed wth a scpe wth a relate accuracy f %. In ths case, the real alue f A s n the range f ± 0.0 r between 0.98 t.0. Smlarly wuld be n the range f 0.96 t. The term A whch appears n the frmula fr / abe wuld then be n the range f 0.0 t 0.06. Snce negate alues f A s unphyscal, we shuld hae A 0.06. The abe frmula fr then ges 6 kω. It can be seen clearly that the frst calculatn leadng t = 49 kω s ncrrect. can be as large as seeral MΩ r as small as 6 kω. Ths prblem always ccurs when we are nterested n the dfference between tw measurements that are clse t each ther. T ad the abe prblem, experment usually s repeated fr dfferent alues f untl / A s between 0. t 0.8 ( / A = 0.5 crrespnds t = ). Measurng utput resstance: T measure, cnsder the amplfer mdel f page 34 wth Z L replaced wth L. Frm the crcut: A 0 = L L If we measure fr tw dfferent alues f L (.e., L and L wth and, respectely), the abe frmula ges: A 0 = L L ECE60L Lecture Ntes, Sprng 004 37
A 0 = L L If the tw measurements are dne fr the same alue, ddng the tw equatns ge: = L L L L whch can be sled t fnd. Typcally, we chse L t be ery large, L (e.g., nternal resstance f scpe). Call = c (pen crcut ltage). Then: L = c L = c L We als nte: = A = c A = c S, ths measurement wll ge bth gan and utput resstance. Smlar t measurng the nput resstance, we shuld chse L such that s suffcently dfferent frm c fr the measurement t be accurate. As n the case f measurng nput resstance, seeral alues f L are used untl / c s typcally between 0. t 0.8 ( / c = 0.5 crrespnds t L = ). In sme cases, we cannt reduce L t a lw enugh leel such that c / s suffcently dfferent frm. Ths s due t the maxmum utput current lmtatn (we wll rest ths fr OpAmps). In such cases, we cannt fnd the alue f the utput resstance. Hweer, we can stll fnd a bund fr (smlar t errr analyss fr nput resstance that gae a bund fr the nput resstance). Suppse L s the smallest resstance that we can use and at ths alue f L and t accuracy f ur scpe (ɛ = %), c. If the relate measurement errr s ɛ, then: measured ( ɛ) real measured ( ɛ) c measured ( ɛ) c real c measured ( ɛ) ɛ c ɛ real L ɛ ɛ L ECE60L Lecture Ntes, Sprng 004 38
S, fr example, f the accuracy f the scpe s % and the mnmum L we culd use was kω, the bund fr 0.0 000 = 40 Ω. Feedback Nt nly a gd amplfer shuld hae suffcent gan, ts perfrmance shuld be nsenste t enrnmental and manufacturng cndtns, shuld hae a large, a small, a suffcently large bandwdth, etc. It s easy t make an amplfer wth a ery large gan. A typcal transstr crcut can easly hae a gan f 00 r mre. A three-stage transstr amplfer can easly get gans f 0 6. Other characterstcs f a gd amplfer s hard t achee. Fr example, the β f a BJT changes wth peratng temperature makng the gan f the threestage amplfer ary wdely. The system can be made t be nsenste t enrnmental and manufacturng cndtns by the use f feedback. Feedback als helps n ther regards. Prncple f feedback: The nput t the crcut s mdfed by feedng a sgnal prprtnal t the utput alue back t the nput. There are tw types f feedback (remember the example f a car n the freeway dscussed n the class):. Negate feedback: As the utput s ncreased, the nput sgnal s decreased and ce ersa. Negate feedback stablzes the utput t the desred leel. Lnear system emplys negate feedback.. Pste feedback: As the utput s ncreased, the nput sgnal s ncreased and ce ersa. Pste feedback leads t nstablty. (But, t has ts uses!) We wll explre the cncept f feedback n the cntext f peratnal amplfers. ECE60L Lecture Ntes, Sprng 004 39
OpAmps as lnear amplfers Operatnal amplfers (OpAmps) are general purpse ltage amplfers emplyed n a arety f crcuts. OpAmps are DC amplfers wth a ery large gan, A 0 (0 5 t 0 6 ), hgh nput mpedance (> 0 MΩ), and lw utput resstance (< 00 Ω). They are cnstructed as a dfference amplfer,.e., the utput sgnal s prprtnal t the dfference between the tw nput sgnals. = A 0 d = A 0 ( p n ) s s and s are pwer supply attachments. They set the saturatn ltages fr the OpAmp crcut (wthn 0. ). Pwer supply grund shuld als be cnnected t the OpAmp grund. and termnals f the OpAmp are called, respectely, nn-nertng and nertng termnals. p - - n d _ - s - OpAmp Mdels p n d A 0 d Lnear Mdel p n I p d I n A 0 d Ideal Mdel Because s ery large and s ery small, deal mdel f the OpAmp assumes and 0. Ideal mdel s usually a ery gd mdel fr OpAmp crcuts. ery large nput resstance als means that the nput current nt an OpAmp s ery small: Frst Glden ule f OpAmps: I p I n 0 (Als called rtual Open Prncple ) An mprtant feature f OpAmp s that because the gan s ery hgh, the OpAmp wll be n the saturatn regn wthut negate feedback. Fr example, take an OpAmp wth a gan f 0 5 and sat = 5. Then, fr OpAmp t be n lnear regn, 5 0 5 = 50 µ (a ery small alue). OpAmps are neer used by themseles. They are always part f a crcut whch emply ether negate feedback (e.g., lnear amplfers, acte flters) r pste feedback (e.g., cmparatrs). Examples belw shws seeral OpAmp crcuts wth negate feedback. ECE60L Lecture Ntes, Sprng 004 40
Inertng Amplfer I L I I n d p n A 0 d L The frst step n slng OpAmp crcuts s t replace the OpAmp wth ts crcut mdel (deal mdel s usually ery gd). p = 0, = A 0 d = A 0 ( p n ) = A 0 n Usng nde-ltage methd and ntng I n 0: n n = 0 Substtutng fr n = /A 0 and multplyng the equatn by, we hae: [ = 0 ] = A 0 A 0 A 0 A 0 = A 0 A 0 Snce OpAmp gan s ery large, /A 0. Als f and are chsen such that ther rat s nt ery large, / A 0 r /(A 0 ), then the ltage transfer functn f the OpAmp s = The crcut s called an nertng amplfer because the ltage transfer functn s negate. (A negate snusdal functn lks nerted.) The negate sgn means that there s 80 phase shft between nput and utput sgnals. ECE60L Lecture Ntes, Sprng 004 4
Nte that the ltage transfer functn s ndependent f the OpAmp gan, A 0, and s nly set by the alues f the resstrs and. Whle A 0 s qute senste t enrnmental and manufacturng cndtns (can ery by a factr f 0 t 00), the resstr alues are qute nsenste and, thus, the gan f the system s qute stable. Ths stablty s acheed by negate feedback, the utput f the OpAmp s cnnected a t the nertng termnal f OpAmp. If ncreases, ths resstr frces n t ncrease, reducng d = p n and = A 0 d, and stablzes the OpAmp utput. An mprtant feature f OpAmp crcuts wth negate feedback s that because the OpAmp s NOT saturated, d = /A 0 s ery small (because A 0 s ery large). As a result, Negate Feedback d 0 n p Secnd Glden ule f OpAmps: Fr OpAmps crcuts wth negate feedback, the OpAmp adjusts ts utput ltage such that d 0 r n p (als called rtual Shrt Prncple ). Ths rule s dered by assumng A. Thus, cannt be fund frm = A 0 d = 0 = ndefnte alue. The rtual shrt prncple replace = A 0 d expressn wth d 0. The abe rule smplfes slutn t OpAmp crcuts dramatcally. Fr example, fr the nertng amplfer crcut abe, we wll hae: Negate Feedback n p 0 n n = 0 = 0 = Input and Output resstances f nertng amplfer cnfguratn: Frm the crcut, I = 0 = I = The nput mpedance f the nertng amplfer crcut s (althugh nput mpedance f OpAmp s nfnte). The utput mpedance f the crcut s zer because s ndependent f L ( des nt change when L s changed). Amplfer Bandwdth: The ltage gan fr the nertng amplfer s / and s ndependent f frequency same gan fr a DC sgnal (ω 0) as fr hgh frequences. Hweer, ths ltage gan has been fund usng an deal OpAmp mdel (deal OpAmp parameters are ndependent f frequency). ECE60L Lecture Ntes, Sprng 004 4
A majr cncern fr any amplfer crcut s ts stablty (whch yu wll study n depth n junr and senr curses). Bascally, any amplfer crcut prduce a phase-shft n the utput ltage. Once the phase shft becmes smaller than -80 (mre negate), negate feedback becmes pste feedback. The amplfer gan shuld be less than at these frequences fr stable peratn. In a practcal OpAmp, the pen lp gan, A 0, s usually ery large, rangng 0 5 t 0 6. T reduce the gan at hgh frequences and ad nstablty, the ltage gan (r ltage transfer functn) f a practcal OpAmp lks lke a lw-pass flter as shwn (marked by pen lp meanng n feedback). Ths s acheed by an addng f a relately large capactr n the OpAmp crcut chp (nternally cmpensated OpAmps) r by prdng fr cnnectn f such a capactr utsde the chp (uncmpensated OpAmp). In rder fr the OpAmp gan t becme smaller than at hgh frequences, the pen-lp bandwdth, f 0 (whch s the same as cut-ff frequency) s usually small (0 t 00 Hz, typcally). ecall the nertng amplfer crcut dscussed abe. In that crcut, the ltage transfer functn was ndependent f A 0 as lng as / A 0. But / s the ertng amplfer crcut gan (call t A = / = / ). Thus, the negate feedback wrked as lng as A A 0. Snce A 0 decreases at hgher frequences, ne wll encuntered a frequency, abe whch A = / A 0 s lated (See fgure fr clsed-lp). Belw ths frequency, the amplfer gan s ndependent f frequency whle abe that frequency, we reert back t the pen-lp gan cure. One can shw that: A 0 f 0 = A f = f u = cnstant 0 lg A 0 0 lg A 0 lg A A (db) Open lp f f f 0 Clsed lp f (lg scale) Therefre, the prduct f the gan and bandwdth (whch the same as the cut-ff frequency) f the amplfer crcut s a cnstant and s equal t the prduct f pen-lp gan and pen-lp bandwdth. Ths prduct s gen n manufacturer spec sheet fr each OpAmp and smetmes s dented as unty gan bandwdth, f u. Nte that gan n the expressn abe s NOT n db, rather t s the alue f /. Hw t sle OpAmp crcuts: ) eplace the OpAmp wth ts crcut mdel. ) Check fr negate feedback, f s, wrte dwn p n. 3) Sle. Best methd s usually nde-ltage methd. Yu can sle smple crcuts wth KL and KCLs. D nt use mesh-current methd. ECE60L Lecture Ntes, Sprng 004 43
Physcal Lmtatns f OpAmps OpAmps, lke ther ltage amplfers, behae lnearly nly under certan cndtns (see page 34-35). Seeral lmtatns f OpAmp crcuts are dscussed befre.. ltage-supply lmt r Saturatn: s < < s whch lmts the maxmum utput ltage (r a fr a gen gan, lmts the maxmum nput ltage). If an amplfer s saturated, ne can recer the lnear regme by reducng the nput ampltude.. Frequency espnse lmt: A practcal amplfer has a fnte bandwdth. Fr OpAmps crcuts wth negate feedback, ne can trade gan wth bandwdth: A 0 ω 0 = Aω r A 0 f 0 = Af = f u Nte that gan n the expressn abe s NOT n db, rather t s the alue f /. 3. Maxmum utput current lmt: A practal ltage amplfer has a lmted capablty n prdng utput current. Ths maxmum utput current lmt s als called the Shrt- Crcut Output Current, I SC. Thus, I I SC. (see dscussn f page 36). Nte: The maxmum utput current f the OpAmp amplfer cnfguratn s NOT the same as the maxmum utput current f the OpAmp tself. Fr example n the nertng amplfer cnfguratn, the maxmum utput current f the amplfer s smaller than I SC f OpAmp because OpAmp has t supply current t bth the lad (maxmum utput current f amplfer cnfguratn tself) and the feedback resstr. 4. Slew rate (se Tme): If nput changes suddenly, the OpAmp cannt change ts utput nstantaneusly. The maxmum rate f change f the utput f an OpAmp s called the slew rate (gen usually n the unts f /µs): S 0 d dt Max Slpe f S 0 Slew rate affects all sgnals (nt lmted t square waes). Fr example, at hgh enugh frequency and/r at hgh enugh ampltude, a snusdal nput turns nt a trangular utput sgnal. As an example, cnsder an nertng amplfer wth a gan f A = 0, buld wth an OpAmp wth a slew rate f S 0 = /µs. The nput s a snusdal wae wth an ampltude f = and frequency f ω. = cs(ωt) = A cs(ωt) ECE60L Lecture Ntes, Sprng 004 44 t
d dt = A ω sn(ωt) d = A dt ω Max The slew rate lmt means that d = A dt ω S 0 Max Fr the example abe, =, A = 0, and S 0 = /µs, we hae d = 0ω 0 6 ω 0 5 dt Max Whch means that at frequences abe 0 5 rad/s, the utput wll depart frm a snusdal sgnal due t the slew rate lmt. Because fr the snusdal wae, the slew rate lmt s n the frm A ω S 0, ne can ad ths nnlnear behar by ether decreasng the frequency, r by lwerng the amplfer gan, r reducng the nput sgnal ampltude. 5. Other lmts: OpAmps hae ther physcal lmtatns such as Input ffset ltage, Input bas current, and Cmmn-mde reject rat (CMM). Cnsult zzn fr a descrptn f these lmtatns. Nte: In rder t ensure that an OpAmp crcut wuld perate prperly, we shuld cnsder all f the physcal lmtatns f OpAmps. Fr example, fr an nertng amplfer, the maxmum peratng frequency may be lmted by the frequency respnse lmt r by the slew rate. The mst restrcte f these lmts wuld then s used. Alternately, the utput ltage f an amplfer may be lmted by saturatn and/r maxmum current and/r slew rate lmts. In general, we shuld cnsder all releant lmts and use the mst restrcte as the peratng cndtn fr the crcut. ECE60L Lecture Ntes, Sprng 004 45
Nn-nertng Amplfer I - p = Negate Feedback n p = n 0 n = 0 Nte that yu shuld nt wrte a nde equatn at OpAmp utput as ts a nde attached t a ltage surce. The alue f s A d and s ndefnte. Instead f usng ths equatn, we use d 0 as was dscussed n page 4. Substtutng fr n =, we get I p d n p A 0 d = 0 = Input esstance: I = I p = 0. Therefre,. Output esstance: s ndependent f L, s = 0. Nte that and = 0 shuld be taken n the cntext that we are usng an deal OpAmp mdel. In realty, the abe crcut wll hae nput and utput resstances equal t that f the OpAmp tself. ltage Fllwer In sme cases, we hae tw-termnal netwrks whch d nt match well,.e, the nput mpedance f the later stage s nt ery large, r the utput mpedance f precedng stage s nt lw enugh. A buffer crcut s usually used n between these tw crcuts t sle the matchng prblem. These buffer crcut typcally hae a gan f but hae a ery large nput mpedance and a ery small utput mpedance. Because ther gans are, they are als called ltage fllwers. ECE60L Lecture Ntes, Sprng 004 46
The nn-nertng amplfer abe has and = 0 and, therefre, can be turned nt a ltage fllwer (buffer) by adjustng and such that the gan s. = = = 0 S by settng = 0, we hae = r a gan f unty. We nte that ths expressn s ald fr any alue f. As we want t mnmze the number f cmpnents n a crcut as a rule (cheaper crcuts!) we set = (pen crcut) and reme frm the crcut. Inertng Summer p = 0 Negate Feedback: n p = 0 n n n = 0 f = f f S, ths crcut adds (sums) tw sgnals. An example f the use f ths crcut s t add a DC ffset t a snusdal sgnal. n p f A 0 d Nn-Inertng Summer p Negate Feedback: n p p p = 0 n 0 s p ( ) = n f ( f s = = 0 ) n n s A 0 d f ECE60L Lecture Ntes, Sprng 004 47
Substtutng fr n n the secnd equatn frm the frst (ntng p = n ): = ( f/ s ) / / S, ths crcut als sgnal adds (sums) tw sgnals. It des nt, hweer, nerts the sgnals. Dfference Amplfer Negate Feedback: n p p p 0 3 = 0 n n p = 3 3 n f = 0 3 n p f A 0 d Substtutng fr n n the nd equatn, ne can get: = ( f ) ( f 3 3 ) If ne chse the resstrs such that 3 = f, then = f ( ) Current Surce Negate Feedback: L = n = s = cnstant n p = s Fr a fxed alue f s, the current I L s ndependent f alue f L and utput ltage. As such, ths crcut s an ndependent current surce. s p n L I L The alue f the current can be adjusted by changng s (fr a fxed L. Therefre, ths crcut s als a ltage t current cnerter r a transcnductance amplfer. ECE60L Lecture Ntes, Sprng 004 48
Grunded Current Surce The prblem wth the abe current surce s that the lad s nt grunded. Ths may nt be desrable n sme cases. Ths crcut here s als a current surce wth a grunded lad f f / = 3 /. Questn: Cmpute I L and shw that t s ndependent f L. s n p f 3 I L L Acte Flters, Integratrs & Dfferentatrs Cnsder the crcut shwn. Ths s an nertng amplfer wth mpedances nstead f resstrs. Fllwng the nertng amplfer slutn, we fnd: H(jω) = = Z Z Z arus flter crcuts can be made wth dfferent chces fr Z and Z : Z st Order Lw-Pass Flter: C } Z Z = Z = C = jωc H(jω) = Z Z = / jωc Cmpare the abe ltage transfer functn wth the general expressn fr a st rder lw-pass flter: H(jω) = K jω/ω c ECE60L Lecture Ntes, Sprng 004 49
We fnd that the abe crcut s a lw pass flter wth K = and ω c = C The mnus sgn n frnt f K ndcates an addtnal 80 phase shft. A lw-pass C r L flter has a phase shft f 0 at lw frequences and 90 at hgh frequences. The abe amplfer has a phase shft f 80 at lw frequences and 70 at hgh frequences (r alternately 80 at lw frequences and 90 at hgh frequences as we can add 360 t the phase angle). Anther dfference wth passe C r L flters s that the gan, K = / can be set t be larger than ne (.e., amplfy the sgnals n the pass band). As such ths knd f flters are called acte flters. Input esstance: =. Output esstance: = 0 (OpAmp utput resstance). st Order Hgh-Pass Flter: Z = ( ) = j jωc ωc Z = H(jω) = Z = / Z j ωc C } Z Cmparng the abe ltage transfer functn wth the general expressn fr a st rder hgh-pass flter, H(jω) = K jω c /ω We fnd that the abe crcut s a hgh-pass flter wth K = and ω c = C Agan, the mnus sgn n K ndcates an addtnal 80 phase shft. A hgh-pass C r L flter has a 90 at lw frequences and 0 at hgh frequences. The abe amplfer has a phase shft f 90 at lw frequences and 80 at hgh frequences. Input esstance: = Z and mn =. Output esstance: = 0 (OpAmp utput resstance). ECE60L Lecture Ntes, Sprng 004 50
nd Order Band-Pass Flter: Z = ( ) = j jωc ωc Z = C = jωc Defnng ω c =, ω c = C C H(jω) = Z = Z jω c /ω jω/ω c C C } Z C } Z C As can be seen, the ltage transfer functn lks lke a hgh-pass and a lw-pass flter put tgether (a wde-band, band-pass flter). T fnd the cut-ff frequences, bandwdth, etc. f ths flter, t s smplest t wrte H(jω) n the frm smlar t the general frm fr nd rder band-pass flters: H(jω) = K ( ω jq ω0 ω ) 0 ω T d s, we rearrange the terms n the expressn fr H(jω) f the abe flter t get: H(jω) = / ( jω/ω c )( jω c /ω) = ( ω c /ω c ) j(ω/ω c ω c /ω) We cmpare the abe expressn wth the general frm fr nd rder band-pass flters t fnd K, Q, ω 0. Nte that H(jω) abe s ery smlar t H(jω) f the wde-band bandpass flter f page 30 (wth the exceptn f / n the nmnatr. Fllwng the same prcedure as n page 30, we get: K = / ω c /ω c Q = ω c /ω c ω c /ω c ω 0 = ω c ω c alues f band-wdth, B, and upper and lwer cut-ff frequences can then be calculated frm ω 0 and Q alues abe (see pages 6 and 7). ECE60L Lecture Ntes, Sprng 004 5
As dscussed n pages 30 and 3, ths s a wde-band band-pass flter and wuld wrk prperly nly f ω c ω c. In ths lmt: K, Q ωc ω c, ω u ω c = C, ω l ω c = C Nte that sgn n K sgnfes an addtnal -80 phase shft. Questn: Cmputer and f the abe flter. Imprtant Nte: In the analyss f acte flters usng OpAmps, we hae assumed an deal OpAmp wth an nfnte bandwdth. In prncple, ne shuld nclude the fact that OpAmp gan wll drp at hgh frequences. Alternately, ne can assume an deal OpAmp n the analyss f the bandwdth f the OpAmp s at least ten tmes larger than the hghest frequency f nterest. The bandwdth f the OpAmp shuld be calculated based n the gan f the crcut, K. ECE60L Lecture Ntes, Sprng 004 5
Integratr In the ntegratr and dfferentatr crcuts belw, the nput s assumed t be an arbtrary functn f tme (t des nt hae t be a snusdal functn). ltages and currents are wrtten n lwer case t remnd yu f ths fact. C Negate feedback p = n = 0 d c = C dt = C d( n ) dt Als, = n = Thus: d = C dt (t) (0) = C t 0 = C d dt (t )dt d dt = C S the utput f ths crcut s prprtnal t the ntegral f the nput. Examples f use f such crcut nclude makng a trangular wae frm a square wae (as s dne n the functn generatr), charge amplfers, and analg cmputers (see zzn fr these crcuts). The prblem wth ths crcut s that t s t gd! It ntegrates eerythng ncludng nse. Lw frequency nse s a cnsderable prblem because een small DC nputs are ntegrated rapdly, ncreasng, and saturatng the OpAmp. Slutn s t add a resstr parallel t C that dscharges the capactr n lng tmes, gettng rd f ntegrated DC nse. The alue f the resstr s chsen such that the tme cnstant f ths C crcut τ = C s abut 00 tmes the perd f the lwest frequency sgnal f nterest. Addtn f makes the crcut lk lke a lw-pass flter. n p C ECE60L Lecture Ntes, Sprng 004 53
Dfferentatr Negate feedback p = n = 0 d c = C dt = C d( n ) dt Als, = n = = C d dt C Thus: (t) = C d dt S the utput f ths crcut s prprtnal t the derate f the nput. In practce, ths crcut des nt wrk as adertsed. The prblem can be seen by assumng that the nput s a snusdal wae and fndng the transfer functn fr the crcut: H(jω) = = /(jωc ) = jω C As can be seen when ω becmes large, becmes large (because f derate f cs ωt = ω sn(ωt)). S, practcally, ths crcut s nt a dfferentatr, rather t s a hgh-frequency-nse amplfer. Bde plts f H(jω) shws that the transfer functn s a lne wth a slpe f 6 db/ctae. The slutn s t attenuate the ampltude f hghfrequency sgnals by addng a resstance n seres wth C. At lw frequences, C dmnate and the crcut s a dfferentatr. At hgh frequences, dmnates and the crcut becmes a smple amplfer. H (j ω) Wth & C C Wthut & C Wth nly C ω Stablty f OpAmp crcuts wth negate feedback As nted befre, the OpAmp gan has a negate phase shft at hgh frequences. If the clsed-lp gan s greater than when OpAmp phase shft becmes smaller than 80 (becmes mre negate), negate feedback becmes pste feedback and the crcut becmes unstable. Stablty f feedback amplfers wll be dscussed next year n the breadth curses (such as ECE0). Bth acte hgh-pass flter and dfferentatr crcuts abe are specally susceptble t ths prblem because f the rase n the gan at lw frequences. If a crcut encunters ths prblem (specally fr uncmpensated OpAmps), the slutn s t add a capactr C t the crcut as shwn t make the crcut lk lke a band-pass flter attenuatng the hgh-frequency sgnals. Fr the dfferentatr crcut abe, alue f C s chsen such that the crcut gan s less than when phase shft s 80. ECE60L Lecture Ntes, Sprng 004 54