III. Operational Amplifiers

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1 III. Operatnal Amplfers Amplfers are tw-prt netwrks n whch the utput vltage r current s drectly prprtnal t ether nput vltage r current. Fur dfferent knds f amplfers ext: ltage amplfer: Current amplfer: Transcnductance amplfer: Transresstance amplfer: A v = / = cnstant A = I /I = cnsant G m = I / = cnstant m = /I = cnstant Our fcus n ths curse s n vltage amplfers (we als see a transcnductance amplfer). ltage amplfers can be accurately mdeled wth I I three crcut elements as shwn belw. These crcut elements are related t transfer functns f tw-prt netwrks dscussed befre. and are, respectvely, A 0 nput and utput resstances. Gan, A 0, s the penlp transfer functn, H v (jω). (Fr amplfers the cnventn s t use A nstead f H(jω).) ltage Amplfer Mdel A gd vltage amplfer has a large nput resstance,, and a small utput resstance,. An deal vltage amplfer has, and Feedback Nt nly a gd amplfer shuld have suffcent gan, ts perfrmance shuld be nsenstve t envrnmental and manufacturng cndtns, shuld have a large, a small, a suffcently large bandwdth, etc. It s easy t make an amplfer wth a very large gan. A typcal transstr crcut can easly have a gan f 00 r mre. A three-stage transstr amplfer can easly get gans f 0 6. Other characterstcs f a gd amplfer are hard t acheve. Fr example, the β f a BJT changes wth peratng temperature makng the gan f the three-stage amplfer vary wdely. The crcut can be made t be nsenstve t envrnmental and manufacturng cndtns by the use f feedback. Prncple f feedback: The nput t the crcut s mdfed by feedng a sgnal prprtnal t the utput value back t the nput. There are tw types f feedback (remember the example f a car n the freeway dscussed n the class):. Negatve feedback: As the utput s ncreased, the nput sgnal s decreased and vce versa. Negatve feedback stablzes the utput t the desred level. Lnear system emplys negatve feedback.. Pstve feedback: As the utput s ncreased, the nput sgnal s ncreased and vce versa. The utput f an amplfer wth a pstve feedback s always at ts lmt (saturatn vltages). (Pstve feedback has ts uses!) ECE65 Lecture Ntes (F. Najmabad), Sprng

2 Fr feedback t wrk well, the amplfer gan shuld be large. In ths lmt, the respnse f the verall crcut s set by the feedback lp (and nt by the gan f amplfer). As such, a vltage amplfer can be made nt many ther useful crcuts such as actve flter, ntegratr, etc. We wll frst explre the cncept f feedback thrugh peratnal amplfer crcuts. 3. Operatnal Amplfers Operatnal amplfers (OpAmps) are general purpse vltage amplfers emplyed n a varety f crcuts. OpAmps are DC amplfers wth a very large gan, A 0 (0 5 t 0 6 ), hgh nput mpedance (> 0 00 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 and termnals f the OpAmp are called, respectvely, nnnvertng and nvertng termnals. s and s are pwer supply attachments. An OpAmp chp shuld be pwered fr t t wrk,.e., pwer supply attachments are necessary. These cnnectns, hwever, are nt usually shwn n the crcut dagram. p d n s 3.. OpAmp Mdels Lnear Mdel Ideal Mdel I p p p d A d 0 d A 0 d n n I n Because s very large and s very small, deal mdel f the OpAmp assumes and 0. Ideal mdel s usually a very gd mdel fr OpAmp crcuts. ery large nput resstance als means that the nput current nt an OpAmp s very small: Frst Glden ule f OpAmps: I p I n 0 (Als called rtual Open Prncple ) As dscussed befre, cncepts f very large and very small emplyed abve requre a frame f reference. A rule f thumb fr gnrng and and emplyng the deal mdel fr the OpAmp (and usng the frst glden rule) s t ensure that all mpedances cnnected t the OpAmp crcut are much smaller than and much larger than (fr a typcal OpAmp, all mpedances shuld between kω and MΩ) ECE65 Lecture Ntes (F. Najmabad), Sprng

3 Anther mprtant feature f OpAmp s that the OpAmp wll be n saturatn wthut negatve feedback because ts gan s very hgh. Fr example, take an OpAmp wth a gan f 0 5 and sat = 5. Then, fr OpAmp t be n lnear regn, = 50 µ (a very small value). As such, OpAmps are rarely used by themselves. They are always part f a crcut whch emply ether negatve feedback (e.g., lnear amplfers, actve flters) r pstve feedback (e.g., cmparatrs). Examples belw shws several OpAmp crcuts wth negatve feedback. 3.. Invertng Amplfer I L d p n A 0 d L The frst step n slvng OpAmp crcuts s t replace the OpAmp wth ts crcut mdel (deal mdel s usually very gd). p = 0, = A 0 d = A 0 ( p n ) = A 0 n Usng nde-vltage methd and ntng I n 0: n n = 0 Substtutng fr n = /A 0 and multplyng the equatn by, we have: [ = 0 ] = A 0 A 0 A 0 A 0 / = ( ) A 0 If and are chsen such that ther rat s nt very large, / A 0, then the vltage transfer functn f the OpAmp s ECE65 Lecture Ntes (F. Najmabad), Sprng

4 Ths crcut s called an nvertng amplfer because the vltage transfer functn s negatve. (A negatve snusdal functn lks nverted.) The negatve sgn actually means that there s 80 phase shft between nput and utput sgnals. Nte that the vltage transfer functn f the crcut s ndependent f the OpAmp gan, A 0, and s set by the values f the resstrs and. Whle A 0 s qute senstve t envrnmental and manufacturng cndtns (can very by a factr f 0 t 00), the resstr values are qute nsenstve and, thus, the gan f the system s qute stable. Ths stablty s acheved by a negatve feedback (see fgure). The utput vltage s sampled va and s appled n parallel t the nput sgnal (va ) t the nput termnals f the OpAmp (called parallelparallel feedback). If ncreases, ths resstr frces n t ncrease, reducng d = p n and = A 0 d, and stablzes the OpAmp utput. Feedback Crcut Ths s a negatve feedback as s cnnected t the nvertng termnal f OpAmp. Als, t s bvus that s needed, therwse feedback wuld nt wrk as n = s a fxed value and the nput t OpAmp chp wuld nt change when changes. Feedback can be appled n a dfferent manner. The utput vltage s sampled and s appled n seres wth t the nput termnals f the OpAmp (called parallel-seres feedback,.e., parallel t the utput, n seres wth n the nput). We explre ths feedback cnfguratn belw Nn-nvertng Amplfer I I p d n p A 0 d I Feedback Crcut The left crcut abve s a nn-nvertng amplfer. It emplys a parallel-seres feedback (see fgure t the rght abve) as the utput s sampled thrugh a cmbnatn f and resstrs and s appled n seres t t the nput termnals f the OpAmp. Furthermre, as the utput s cnnected t the nvertng termnal, t s a negatve feedback. ECE65 Lecture Ntes (F. Najmabad), Sprng

5 eplacng the OpAmp wth ts deal mdel (mddle crcut abve) and ntng that I n 0 and I p 0, we get: p = ltage dvder: n = d = p n = Substtutng fr d = /A 0 n the last equatn abve, we get: ( = A 0 A 0 ( = A 0 = = A 0 ) = ) A 0 ( ) If and are chsen such that ther rat s nt very large, / A 0, then the vltage transfer functn f the OpAmp s = Ths crcut s called a nn-nvertng amplfer because ts vltage transfer functn s pstve. (as ppsed t the nvertng amplfer we a negatve vltage transfer functn.) Nte that the vltage transfer functn s ndependent f the OpAmp gan, A 0, and s nly set by the values f the resstrs and OpAmp Crcuts wth Negatve Feedback An mprtant feature f OpAmp crcuts wth negatve feedback s that because the OpAmp s NOT saturated, d = /A 0 s very small (because A 0 s very large). As a result, Negatve Feedback d 0 n p Secnd Glden ule f OpAmps: Fr OpAmps crcuts wth negatve feedback, the OpAmp adjusts ts utput vltage such that d 0 r n p (als called rtual Shrt Prncple ). Ths rule s derved by assumng A. Thus, cannt be fund frm ECE65 Lecture Ntes (F. Najmabad), Sprng

6 = A 0 d = 0 = ndefnte value. The vrtual shrt prncple replace = A 0 v d expressn wth d 0. The secnd glden rule f OpAmps allws us t slve OpAmp crcuts n a much smpler manner fllwng the recpe belw. ecpe fr slvng OpAmp crcuts: ) eplace the OpAmp wth ts crcut mdel. ) Check fr negatve feedback, f s, p n. 3) Slve. Best methd s usually nde-vltage methd. Yu can slve smple crcuts wth KL and KCLs. D nt use mesh-current methd. 4) Fnd prblem unknwns n term f nde vltages. Fr example, fr the nvertng amplfer we wll have: Negatve Feedback n p 0 n n = 0 = 0 A v = = Nte that yu shuld nt wrte a nde equatn at OpAmp utput as ts a nde attached t a vltage surce. The value f s A d and s ndefnte. Instead f usng ths equatn, we used d 0. Input and Output resstances f nvertng amplfer cnfguratn can nw be fund. Frm the crcut, I = 0 = I = The nput mpedance f the nvertng amplfer crcut s (althugh nput mpedance f OpAmp s very large). The utput mpedance f the crcut s zer because s ndependent f L ( des nt change when L s changed). Smlarly, fr the nn-nvertng amplfer crcut abve, we wll have: p = I I L Negatve Feedback: n 0 n = 0 n p = ECE65 Lecture Ntes (F. Najmabad), Sprng

7 Nte that agan we d nt wrte a nde equatn at OpAmp utput as ts a nde attached t a vltage surce. Substtutng fr n =, we get = 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 abve crcut wll have nput and utput resstances equal t that f the OpAmp tself. 3.3 OpAmps as lnear amplfers 3.3. ltage Fllwer In sme cases, we have tw-termnal netwrks whch d nt match well,.e, the nput mpedance f the later stage s nt very large, r the utput mpedance f precedng stage s nt lw enugh. A buffer crcut s usually used n between these tw crcuts t slve the matchng prblem. A buffer crcuts has a gan f but has a very large nput mpedance and a very small utput mpedance. Because the gan f buffer s, t s als called a vltage fllwer. The nn-nvertng amplfer abve has and = 0 and, therefre, can be turned nt a vltage fllwer (buffer) by adjustng and such that the gan s. = = = 0 S by settng = 0, we have = r a gan f unty. We nte that ths expressn s vald fr any value f. As we want t mnmze the number f cmpnents n a crcut as a rule (cheaper crcuts!) we set = (pen crcut) and remve frm the crcut. ECE65 Lecture Ntes (F. Najmabad), Sprng 007 5

8 3.3. Nn-Invertng Summer Negatve Feedback: n 0 s n p n f = 0 = ( ) f n s p ( p = 0 p ) = p n s f Substtutng fr n n the st equatn frm the secnd (ntng p = n ): = ( f/ s ) / / S, ths crcut als sgnal adds (sums) tw sgnals. It des nt, hwever, nverts the sgnals Invertng Summer Negatve Feedback: p = 0 n = 0 n p n n n = 0 f n p f = f f S, ths crcut adds (sums) tw sgnals and nvert them Dfference Amplfer Negatve Feedback: n p p p 0 3 = 0 n p = n n f = 0 Substtutng fr n n the nd equatn, ne can get: 3 3 f p n s = ( f ) ( f 3 3 ) ECE65 Lecture Ntes (F. Najmabad), Sprng 007 5

9 If ne chse the resstrs such that 3 = f, then = f ( ) Transcnductance Amplfer r Current Surce Negatve Feedback: I = I L = n = n p = I L L I = = G m = cnst S, ths crcut s a transcnductance amplfer: the utput current s prprtnal t the nput vltage (see page 45). Alternatvely, f a fxed value f s appled t the crcut, the current I L s ndependent f value f L and utput vltage. As such, ths crcut s als an ndependent current surce. Nte that when the utput f a crcut s the current, a lad s necessary fr the crcut t functn prperly. Ths crcut s smlar t the nn-nvertng amplfer crcut. It s an example f crcuts whch perfrm dfferent functns dependng n the lcatn f the utput termnals. Fr ths crcut, f the utput termnals are between the utput f the OpAmp chp and the grund, t s a nn-nvertng amplfer whle f the utput termnals are taken acrss L t wuld be a current-surce. Anther example f ths type f crcuts s C frst-rder flters that we examned befre (see pages 7 and 3). If the utput s taken acrss the resstr, t wuld be a hgh-pass flter whle f the utput s taken acrss the capactr, t wuld be a lw-pass flter Grunded Current Surce The prblem wth the abve current surce s that the lad s nt grunded. Ths may nt be desrable n sme cases. Ths crcut s als a current surce wth a grunded lad f f / = 3 /. Exercse: Cmpute I L and shw that t s ndependent f L. f 3 I L L ECE65 Lecture Ntes (F. Najmabad), Sprng

10 3.4 Actve Flters, Integratrs & Dfferentatrs Cnsder the crcut shwn. Ths s an nvertng amplfer wth mpedances nstead f resstrs. Fllwng the nvertng amplfer slutn, we fnd: H(jω) = = Z Z Z Z arus flter crcuts can be made wth dfferent chces fr Z and Z st Order Lw-Pass Flter: Z = Z = C = jωc C } Z H(jω) = Z Z = / jω C Cmparng the abve wth the generalzed transfer functn f a st rder lw-pass flter: H(jω) = K jω/ω c We fnd that the abve 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 abve amplfer has a phase shft f 80 at lw frequences and 70 at hgh frequences (r alternatvely 80 at lw frequences and 90 at hgh frequences as we can add 360 t the phase angle). Anther dfference wth passve 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 actve flters. Input esstance: =. Output esstance: = 0 (OpAmp utput resstance). ECE65 Lecture Ntes (F. Najmabad), Sprng

11 3.4. st Order Hgh-Pass Flter: Z = ( ) = j jωc ω C Z = H(jω) = Z = / Z j ω C C } Z Cmparng the abve wth the generalzed transfer functn f a st rder hgh-pass flter, H(jω) = K jω c /ω We fnd that the abve 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 abve 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) 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 vltage transfer functn lks lke a hgh-pass and a lw-pass flter put tgether (a wde-band, band-pass flter). ECE65 Lecture Ntes (F. Najmabad), Sprng

12 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 abve 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 abve expressn wth the general frm fr nd rder band-pass flters t fnd K, Q, ω 0. Nte that H(jω) abve s very smlar t H(jω) f the wde-band bandpass flter f page 36 (wth the exceptn f / n the nmnatr. Fllwng the same prcedure as n page 36, 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 values abve (see page 36-38). As dscussed n pages 37-38, 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 Questn: Cmputer and f the abve flter Integratr In the ntegratr and dfferentatr crcuts belw, the nput s assumed t be an arbtrary functn f tme (t des nt have t be a snusdal functn). ltages and currents are wrtten n lwer case t remnd yu f ths fact. ECE65 Lecture Ntes (F. Najmabad), Sprng

13 Negatve feedback v p = v n = 0 dv c = C dt = C d(v v n ) dv = C dt dt Als, = v n v = v Thus: v dv = C dt v (t) v (0) = C t 0 v (t )dt dv dt = C v 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 wave frm a square wave (as s dne n the functn generatr), charge amplfers, and analg cmputers. The prblem wth ths crcut s that t s t gd! It ntegrates everythng ncludng nse. Lw frequency nse s a cnsderable prblem because even small DC nputs are ntegrated rapdly, ncreasng v, and saturatng the OpAmp. Ths can be seen by examnng the frequency respnse f ths crcut: H(jω) = = /(jωc ) = j ω C whch becmes nfnte fr ω = 0. Slutn s t add a resstr parallel t C that dscharges the capactr n lng tmes, gettng rd f ntegrated DC nse. The value f the resstr s chsen such that the tme cnstant f ths C crcut τ = C s abut 0-00 tmes the perd f the lwest frequency sgnal f nterest. Nte that the addtn f makes the crcut lk lke a lw-pass flter. Ths s an example f a crcut that behaves dfferently dependng n the frequency range f nterest. If we are nterested n ths crcut as an ntegratr, we shuld set the cut-ff frequency f the flter t be much smaller than the smallest frequency f nterest (.e., sgnals that we want t be ntegrated). C H (j ω) Wthut H (j ω) Integratr Wth Lwpass flter ωc ω ωc ω ECE65 Lecture Ntes (F. Najmabad), Sprng

14 3.4.5 Dfferentatr Negatve feedback v p = v n = 0 dv c = C dt = C d(v v n ) dt = C dv dt C Als, = v n v = v Thus: v (t) = C dv dt S the utput f ths crcut s prprtnal t the dervatve f the nput. In practce, ths crcut des nt wrk as advertsed. The prblem can be seen by examnng the transfer functn fr the crcut n the frequency dman: H(jω) = = /(jωc ) = jω C As can be seen when ω becmes large, crcut gan becmes nfnte. S, practcally, ths crcut s nt a dfferentatr, rather t s a hgh-frequency-nse amplfer. 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. C Nte that the addtn f makes the crcut lk lke a hgh-pass flter. Ths s anther example f a crcut that behaves dfferently dependng n the frequency range f nterest. If we are nterested n ths crcut as a dfferentatr, we shuld set the cut-ff frequency f the flter t be much larger than the largest frequency f nterest (.e., sgnals that we want t be dfferentated). H (j ω) Wthut H (j ω) Dfferentatr Wth Hghpass flter ω ω ωc ωc ECE65 Lecture Ntes (F. Najmabad), Sprng

15 3.5 Practcal Amplfers and OpAmp Lmtatns eal vltage 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 v = / = cnstant (exactly the same wavefrm) ltage-supply lmt r Saturatn: Amplfers d nt create pwer. ather, they act as a valve adjustng the pwer flw frm the pwer supply nt the lad accrdng t the nput sgnal. As such, the utput vltage amplfer cannt exceed the pwer supply vltage (t s usually lwer because f vltage drp acrss sme actve elements). The fact that the utput vltage f a practcal amplfer cannt exceed certan threshld value s called saturatn. A vltage amplfer behaves lnearly,.e., / = A v = cnstant as lng as the utput vltage remans belw the saturatn vltage, sat < v < sat Nte that the saturatn vltage, n general, s nt symmetrc,.e., sat sat. Fr an amplfer wth a gven gan, A v, the abve range f v translate nt a certan range fr v sat < v < sat sat < A vv < sat sat sat A v A v < v <.e., any amplfer wll enter ts saturatn regn f s rased abve certan lmt. The fgure shws hw the amplfer utput clps when amplfer s nt n the lnear regn. Alternatvely, f an amplfer s saturated, ne can recver the lnear respnse by reducng the nput ampltude. Fr OpAmps, the saturatn vltages are clse t pwer supply vltages that pwer the OpAmp chp, r s < v < s ECE65 Lecture Ntes (F. Najmabad), Sprng

16 3.5. Maxmum Output Current: A vltage amplfer mdel ncludes a cntrlled vltage surce n ts utput crcut. Ths means that fr a gven nput sgnal, ths cntrlled surce wll have a fxed vltage ndependent f the current drawn frm t. If we attach a lad t the crcut and start reducng the lad resstance, the utput vltage remans a cnstant and lad current wll ncrease. Fllwng ths mdel, ne culd reduce the lad resstance t a very small value and draw a very large current frm the amplfer. In realty, ths des nt happen. Each vltage amplfer has a lmted capablty n prvdng utput current. Ths maxmum utput current lmt s called the Shrt-Crcut Output Current, I SC. 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 vltage cannt exceed the L I SC : L I SC v L I SC In ths case, the maxmum utput current lmt manfests tself n a frm smlar t amplfer saturatn. The utput vltage wavefrm wll be clpped at value f L I SC. Smlar lmtatn als apples t OpAmp chps and crcuts. Hwever, ne shuld nte that the maxmum utput current f the OpAmp amplfer cnfguratn s NOT the same as the maxmum utput current f the OpAmp chp tself. Fr example n the nvertng 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 Frequency espnse lmt r Amplfer Bandwdth: Typcally, amplfers wrk n a certan range f frequences. Ther gan, A v = / drps utsde ths range. The vltage 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 value). In terms f frequency respnse, vltage amplfers are dvded 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 ECE65 Lecture Ntes (F. Najmabad), Sprng

17 sgnals. 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!). A majr cncern fr any amplfer crcut s ts stablty (whch yu wll study n depth n junr and senr curses). A requrement fr stablty s that the crcut gan shuld be less than unty at hgh frequences fr stable peratn. T reduce the gan at hgh frequences and avd nstablty, the vltage gan f a practcal OpAmp lks lke a lw-pass flter as shwn (marked by pen lp meanng n feedback). Ths s acheved by an addng f a relatvely large capactr n the OpAmp crcut chp (nternally cmpensated OpAmps) r by prvdng 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 cut-ff frequency f the OpAmp chp tself, f 0, s usually small (0 t 00 Hz, typcally). The pen-lp gan f the OpAmp chp, A, s gven by: A = jω/ω 0 A = A 0 A 0 (ω/ω 0 ) 0 lg A 0 0 lg A 0 lg A A (db) v Open lp f f f 0 Clsed lp f (lg scale) Where ω 0 s the cut-ff frequency f the chp. Fr frequences much larger than ω 0 = πf 0 (slped lne n the fgure), ω ω 0, the pen-lp gan f the chp scales as A A 0 (ω/ω 0 ) = A 0ω 0 ω A 0 ω 0 = A ω r A 0 f 0 = A f = f u Therefre, the prduct f the pen-lp gan and bandwdth (whch the same as the cut-ff frequency) f the OpAmp chp s a cnstant. Ths prduct s gven n manufacturer spec sheet fr each OpAmp and smetmes s dented as unty gan bandwdth, f u. Nte that gan n the expressn abve s NOT n db. ecall that we fund the vltage gan fr the nvertng amplfer t be /. Ths gan was ndependent f frequency: same gan fr a DC sgnal (ω 0) as fr hgh frequences. Hwever, ths vltage gan was fund usng an deal OpAmp mdel (deal OpAmp parameters are ndependent f frequency) whle n a practcal OpAmp chp, the pen-lp vltage gan s reduced sgnfcantly at hgh frequences. T examne ths effect, recall the nvertng amplfer crcut (pages 47) that we examned n detal. In that crcut, the vltage transfer functn was ndependent f A 0 as lng as ECE65 Lecture Ntes (F. Najmabad), Sprng 007 6

18 ( / ) A 0. eplacng A 0 wth A = A 0 /( jω/ω 0 ) n the vltage transfer functn f the nvertng amplfer n page 47 frm the frmula abve, we get: A = / / = ( ) = ( jω/ω 0 ) ( ) A A 0 / / = ( ) j ω ( ) j ω ( ) A 0 ω 0 A 0 ω 0 A 0 whch s smlar t the transfer functn f a st-rder lw-pass flter, H = K/( jω/ω c ). Therefre, the nvertng amplfer has a cut-ff frequency gven by: ω c = A 0 ω 0 / ( A)ω c = A 0 ω 0 ( A)f c = A 0 f 0 = f u n whch we have replaced / = A, the gan f the nvertng amplfer. As can be seen, the feedback has caused the cut-ff frequency f the amplfer crcut t ncrease drastcally frm the value f f 0 t [A 0 /( A)]f 0. Als nte that f A, we recver the gan-bandwdth frmulas f the OpAmp chp tself: Af = cnst. A smlar analyss can be frmed fr the nn-nvertng amplfer f page 49. We wll fnd: A = ( = ) j ω ( ω 0 A 0 Therefre, the nn-nvertng amplfer has a cut-ff frequency gven by: ω c = A 0 ω 0 / Aω c = A 0 ω 0 Af c = A 0 f 0 = f u In general, the bandwdth (r the cut-ff frequency) f OpAmp crcuts shuld be calculated ndvdually. Hwever, as mst crcut cntanng ne OpAmp chp s cnfgured n wther nvertng r nn-nvertng cnfguratn, smple estmates f the crcut bandwdth can be derved: ) Invertng Amplfer: ( A)f c = A 0 f 0 = f u Nn-nvertng Amplfer: Af c = A 0 f 0 = f u Invertng and Nn-nvertng Summers: There s a cut-ff frequency asscated wth each nput and s gven by the abve frmulas fr the nvertng r nn-nvertng amplfer. The smallest f these cut-ff frequences sets the band-wdth f the summer. ECE65 Lecture Ntes (F. Najmabad), Sprng 007 6

19 Dfference Amplfer: s a cmbnatn f an nvertng and nn-nvertng amplfers. Fnd the cut-ff frequency fr each nput frm the abve frmulas fr the nvertng and nn-nvertng amplfer. The smallest f these cut-ff frequences sets the band-wdth f the dfference amplfer. Actve flters: The three actve flters we wrked n are all n nvertng amplfer cnfguratn. In rder t ensure prper peratn, we shuld set the effectve cut-ff frequency set by the chp (.e., cut-ff frequency based n the gan f the flter) t be at least 5 tmes larger than the upper cut-ff frequency f the flter se Tme In an deal amplfer, f the nput vltage s a unt step functn, the utput vltage 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 vltage t reach ts nmnal level. The maxmum rate f change n the utput vltage s called the rse tme. v Ideal Amp. v v Practcal Amp. The maxmum rate f change f the utput vltage f an OpAmp s called the slew rate (gven usually n the unts f /µs): S 0 dv dt Max Slew rate (r rse tme) affects all sgnals (nt lmted t square waves). Fr example, at a hgh enugh frequency and/r at a hgh enugh ampltude, a snusdal nput turns nt a trangular utput sgnal. As an example, cnsder an nvertng amplfer wth a gan f A, buld wth an OpAmp wth a slew rate f S 0. The nput s a snusdal wave wth an ampltude f and frequency f ω. v = cs(ωt) v = A cs(ωt) dv dt = A ω sn(ωt) dv = A dt ω S 0 Max Fr example, fr =, A = 0, and S 0 = /µs, we have dv dt = 0ω 0 6 ω 0 5 Max ECE65 Lecture Ntes (F. Najmabad), Sprng t

20 Whch means that at frequences abve 0 5 rad/s, the utput wll depart frm a snusdal sgnal due t the slew rate lmt. Because fr the snusdal wave, the slew rate lmt s n the frm A ω S 0, ne can avd ths nnlnear behavr by ether decreasng the frequency, r by lwerng the amplfer gan, r reducng the nput sgnal ampltude. Other lmts: OpAmps have ther physcal lmtatns such as Input ffset vltage, Input bas current, and Cmmn-mde reject rat (CMM) that yu wll see n junr and senr curses. 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 nvertng amplfer, the maxmum peratng frequency may be lmted by the frequency respnse lmt r by the slew rate. Alternatvely, the utput vltage f an amplfer may be lmted by saturatn and/r maxmum current and/r slew rate lmts. In general, we shuld cnsder all relevant lmts and use the mst restrctve as the peratng cndtn fr the crcut. ECE65 Lecture Ntes (F. Najmabad), Sprng

21 3.6 Exercse Prblems Use the fllwng nfrmatn n desgnng crcuts: ) OpAmps have a unty-gan bandwdth f 0 6 Hz, a maxmum utput current lmt f 00 ma, and a slew rate f /µs. OpAmps are pwered by ±5 pwer supples (pwer supples nt shwn). ) In crcut desgn, use cmmercal resstr and capactr values f,.,.,.3,.5,.6,.8,,.,.4,.7, 3., 3.3, 3.6, 3.9, 4.3, 4.7, 5., 5.6, 6., 6.8, 7.5, 8., 9. ( 0 n where n s an nteger). Yu can als use 5 mh nductrs. Prblem. Fnd f = = and 3 = 4 =. (Assume OpAmp s deal). Prblem. What s the transfer functn f OpAmp crcut belw. What s ts functn? (Assume OpAmp s deal). 4 3 L C Prblem Prblem Prblem 3. Fnd the gan f the amplfer crcut shwn belw. (Assume OpAmp s deal and = = 3 = 3 and 4 = ). Prblem 4. Fnd v /v. (Assume OpAmp s deal). 00k 00k 0k 3 4 v 0k v Prblem 3 Prblem 4 ECE65 Lecture Ntes (F. Najmabad), Sprng

22 Prblem 5. Fnd v /v. (Assume OpAmp s deal). Prblem 6. The crcut belw s a vltage-t-current cnverter wth I L = g. Fnd the value f g and shw that t s ndependent f L. (Assume OpAmp s deal). I L 3 L Prblem 5 Prblem 6 Prblem 7. Assume the OpAmp n the crcut belw s deal. Fnd the vltage transfer functn H(jω) = /. Prblem 8. Fnd v n terms f v and v. (Assume OpAmp s deal). C 4 Prblem 7 Prblem 8 Prblem 9. Cnsder the amplfer crcut belw wth v = 0.5 cs(ωt). a) what s the amplfer gan n db? b) Fr what range f frequences, des ths amplfer behave as a lnear amplfer? (Assume a unty-gan bandwth f 0 6 Hz, I SC = 00 ma, S 0 = /µs and OpAmps are pwered by ±5 pwer supples.) Prblem 0. Cnsder the nvertng amplfer belw wth L = 500 Ω. The nput sgnal s v = cs(0 6 t). We want the amplfer t behave lnearly (.e., utput sgnal t be snusdal) and / = K. What s the maxmum value f K we can chse? (Assume a unty-gan bandwth f 0 6 Hz, I SC = 0 ma, S 0 = 4 /µs and OpAmps are pwered by ±5 pwer supples.) ECE65 Lecture Ntes (F. Najmabad), Sprng

23 5kΩ K v kω v - L - Prblem 9 Prblem 0 Prblem. The nput and utput sgnals t an OpAmp crcut are gven belw. A) Draw the crcut (assume t r = 0 fr ths part nly). B) If the slew-rate f the OpAmp s /µs, calculate t r. C) Suppse that we cnsder a sgnal as a square wave as lng as t r /(T/) < 0.. What s the frequency range that ths amplfer prduce a square wave? (Assume a unty-gan bandwth f 0 6 Hz, I SC = 00 ma, S 0 = /µs and OpAmps are pwered by ±5 pwer supples.) Prblem. Cnsder the amplfer crcut belw. The nput and utput wavefrms are shwn. Explan why the utput s nt a trangular wavefrm. (Assume a unty-gan bandwth f 0 6 Hz, I SC = 00 ma, S 0 = /µs and OpAmps are pwered by ±5 pwer supples.) - T t(s) v () t ( µ s) 5 t(s) 0k Ω 00kΩ 50Ω v () 5-5 t r t ( µ s) Prblem Prblem Prblem 3. Desgn a nn-nvertng amplfer wth a gan f 0 db t drve a 0 kω lad. What s the bandwdth f the crcut yu have desgned? What s ts nput mpedance? Prblem 4. Desgn an actve wde-band flter wth f l = 0 Hz, f u = khz, a gan f 4 and an nput mpedance larger than 5 kω. Prblem 5. Desgn a flter wth the transfer functn f H(jω) = mnmum nput mpedance f 50 kω. 3 jω/5000 and a Prblem 6. We have an scllatr that makes a square wave wth a frequency f khz, a peak-t-peak ampltude f 0, and a DC-ffset f 0. Desgn a crcut that cverts ths square wave t a trangular wave wth a peak-t-peak ampltude f 0, and a DC-ffset f 0. What wuld be the frequency f the trangular wave? ECE65 Lecture Ntes (F. Najmabad), Sprng

24 3.7 Slutn t Exercse Prblems Prblem. Fnd f = = and 3 = 4 =. (Assume OpAmp s deal). We have negatve feedback: n = p =. Usng nde-vltage methd: Nde n n 0 n = 0 Nde n 0 = n 4 3 Lettng n = and usng = = and 3 = 4 =, the abve tw equatns becme: 0 = 0 = 0 = 0 = 0 = 4 = 4( ) = 6 Prblem. What s the transfer functn f OpAmp crcut belw. What s ts functn? (Assume OpAmp s deal). The crcut s n general frm f an nvertng amplfer wth mpedances. S: H(jω) = = Z = Z jωl jωc L C H(jω) = / ( j ωl ω ) C Ths s a band-pass actve flter as ts transfer functn s n the general frm f transfer functn f a band-pass flter: H(jω) = K jq ( ω ω0 ω 0 ω ) ECE65 Lecture Ntes (F. Najmabad), Sprng

25 The center frequency and bandwdth can be fund frm: K = Qω ω 0 = ωl and Qω 0 ω = ω C L Q = C and ω 0 = LC Prblem 3. Fnd the gan f the amplfer crcut shwn belw. (Assume OpAmp s deal and = = 3 = 3 and 4 = ). Negatve feedback: n p = 0 Usng Nde-vltage Methd: Nde n n n 4 = 0 3 n 4 Nde 3 n = 0 where we have utlzed the frst glden rule f OpAmp (I n = I p = 0). Substtutng fr n 0, = = 3 = 3 and 4 =, and multplyng bth equatns by 3 we get: = 0 = = 0 3 = 0 3(3 ) = 0 0 = A v = = 0. Prblem 4. Fnd v /v. (Assume OpAmp s deal). eplace OpAmp wth ts Ideal mdel. Negatve feedback: v n v p = 0 Usng nde-vltage methd: 00k v 0k 00k v n v 0, 000 v n v 00, 000 = 0 v = 0v v 0k v n v v v n 00, 000 v v 00, 000 v 0, 000 = 0 v v = 0 v = v = 0v ECE65 Lecture Ntes (F. Najmabad), Sprng

26 Prblem 5. Fnd v /v. (Assume OpAmp s deal). Bth OpAmps have negatve feedback, s v = v p = v n and v p = v n. Then, by nde-vltage methd: v n 0 v n v n = 0 v n v n v n 0 v n v = 0 n - n p - Substtute fr v n = v and elmnate v n between the tw equatns t get the vltage transfer functn. v n 0 = v n v ( n = v ) ( v = v n ) v n v v = v v = ( ) 3 = v ( ( ) v n = v ( ) ) v Prblem 6. The crcut belw s a vltage-t-current cnverter wth I L = g. Fnd the value f g and shw that t s ndependent f L. (Assume OpAmp s deal). p n Bth OpAmps have negatve feedback: n p and n p Usng Nde-vltage Methd and substtutng fr p = n and p = n : n p 3 I L L Nde n n p 0 Nde p n 0 Nde p L n p = 0 n n n = 0 n = n p = 0 n n = 0 n = n 3 = 0 L = n L = n 3 ECE65 Lecture Ntes (F. Najmabad), Sprng

27 where we have utlzed the frst glden rule f OpAmp (I n = I p = 0). Elmnatng n between the frst tw equatns gve: n = n = and substtutng fr n the thrd equatn, we get: L = n 3 = n n 3 = 3 Thus g = / 3 and s ndependent f L. Prblem 7. Assume the OpAmp n the crcut belw s deal. Fnd the vltage transfer functn H(jω) = /. Ths s the same crcut as prblem f sample quz. The crcut s made f tw tw-prt netwrks: A lw-pass C flter and a nn-nvertng amplfer. As the nput mpedance f the nn-nvertng amplfer s nfnte, t wll nt lad the C flter and each tw-prt netwrk cat be analyzed ndependently. We knw: I C I - = (/jωc) (/jωc) = jωc = H(jω) = = / jωc S the crcut s a st rder lw pass flter wth a gan f K = / and a cut-ff frequency f ω c = /(C). ECE65 Lecture Ntes (F. Najmabad), Sprng 007 7

28 Prblem 8. Fnd v n terms f v and v. (Assume OpAmp s deal). 4 The OpAmp t the left s cnfgured as an nvertng amplfer. Therefre, v = v = v The OpAmp t the rght s cnfgured as an nvertng summer. Thus, v = 4 v 4 v = 4v v = 4(v ) v = 8v v Prblem 9. Cnsder the amplfer crcut belw wth v = 0.5 cs(ωt). a) what s the amplfer gan n db? b) Fr what range f frequences, des ths amplfer behave as a lnear amplfer? (Assume a unty-gan bandwth f 0 6 Hz, I SC = 00 ma, S 0 = /µs and OpAmps are pwered by ±5 pwer supples.) part a: Ths s an nvertng amplfer: A = = 5k k = 5 A db = 0 lg 0 A = 0 lg 0 (5) = 4 db Tw lmts mpact the bandwth f ths amplfer: Bandwdth f OpAmp tself: v kω 5kΩ v A 0 f 0 = f u = ( A)f f = f u A = 06 6 = = 66 khz Slew ate: A ω S ω /µs = 0 6 /s ω 06.5 = 4 05 rad/s f = ω π = 64 khz The slew rate s the mst restrctve, s ths OpAmp crcut behaves lnearly n the frequency range 0 f 64 khz. ECE65 Lecture Ntes (F. Najmabad), Sprng 007 7

29 Prblem 0. Cnsder the nvertng amplfer belw wth L = 500 Ω. The nput sgnal s v = cs(0 6 t). We want the amplfer t behave lnearly (.e., utput sgnal t be snusdal) and / = K. What s the maxmum value f K we can chse? (Assume a unty-gan bandwth f 0 6 Hz, I SC = 0 ma, S 0 = 4 /µs and OpAmps are pwered by ±5 pwer supples.) ω = 0 6 rad/s, f = 0 6 /(π) Hz =, = K = K ) ltage supply lmt (Saturatn): - K L - v s < v < v s K 5 ) Frequency respnse lmt: A 0 f 0 = f u = ( A)f = 0 7 ( K) 06 π 07 K 0π = 6 3) Maxmum utput current lmt: I L = L I SC K K 5 4) Slew ate: dv dt = A mω S 0 K K 4 Slew rate lmt s mst restrctve wth K 4. ECE65 Lecture Ntes (F. Najmabad), Sprng

30 Prblem. The nput and utput sgnals t an OpAmp crcut are gven belw. A) Draw the crcut (assume t r = 0 fr ths part nly). B) If the slew-rate f the OpAmp s /µs, calculate t r. C) Suppse that we cnsder a sgnal as a square wave as lng as t r /(T/) < 0.. What s the frequency range that ths amplfer prduce a square wave? (Assume a unty-gan bandwth f 0 6 Hz, I SC = 00 ma, S 0 = /µs and OpAmps are pwered by ±5 pwer supples.) Part A: Fr t r = 0, v = 5v. S, the OpAmp crcut s a nn-nvertng amplfer wth a gan f 5. Snce: - T t(s) A = = 5 = 4 Part B: The departure f the utput frm a square wave s due t the slew rate: 5 t(s) dv S 0 = /µs = dt = 5 5 t = 0 r t r -5 t r t r = Part C: 0 /µs = 0µs I t r T/ < 0. T > t r 0. f = T < 0.05 t r = 0.05 = 5 khz ECE65 Lecture Ntes (F. Najmabad), Sprng

31 Prblem. Cnsder the amplfer crcut belw. The nput and utput wavefrms are shwn. Explan why the utput s nt a trangular wavefrm. (Assume a unty-gan bandwth f 0 6 Hz, I SC = 00 ma, S 0 = /µs and OpAmps are pwered by ±5 pwer supples.) The crcut s a nn-nvertng amplfer wth a gan f / =. Input s a trangular sgnal wth a peak-t-peak ampltude f. If the OpAmp was deal, we wuld expect that utput t be a trangular sgnal wth a peak-t-peak ampltude f =. 0k Ω 00kΩ 50Ω The utput sgnal, hwever, s clpped at 5. Output can be clpped by ether ) amplfer saturatn and/r ) maxmum utput current lmt. We need t examne bth lmts: v () t ( µ s) Saturatn: As the OpAmp s pwered by ±5 supples, the saturatn vltage fr the OpAmp shuld be clse t ±5. As the utput sgnal s clpped at 5, clppng s NOT due t the OpAmp saturatn. v () 5 t ( µ s) Maxmum utput current lmt: As the lad resstr (50 Ω) s much smaller than the feedback resstr (00 kω), almst all f OpAmp utput current wll flw n the lad. Then: L 00 ma v 0 = 50 L = 5 Therefre, maxmum utput current lmt wll frce the utput sgnal t be clpped at 5. Ths explans the shape f utput sgnal. ECE65 Lecture Ntes (F. Najmabad), Sprng

32 Prblem 3. Desgn a nn-nvertng amplfer wth a gan f 0 db t drve a 0 kω lad. What s the bandwdth f the crcut yu have desgned? What s ts nput mpedance? Prttype f a nn-nvertng amplfer s shwn wth A = /. 0 db gan translates t: I - 0 db = 0 lg(a) lg(a) = A = 0 ( A = ) = 0 = 9 As the utput mpedance f a nn-nvertng amplfer s zer, we can chse and arbtrarly. A reasnable chce s: = 0 kω, = 9 kω Bandwdth: A f c = f u = 0 6. Thus, f c = 0 5 Hz= 00 khz. Input mpedance f the amplfer s nfnty as I = 0. Prblem 4. Desgn an actve wde-band flter wth f l = 0 Hz, f u = khz, a gan f 4 and an nput mpedance larger than 5 kω. The prttype f ths crcut s shwn wth: ω c ω c = ω u ω c = C ω l ω c = C v C C v K Z mn = Then, frm desgn cndtn f Z 5 kω, > 5 kω. T make capactrs small (and als t make nput mpedance large) chse and t be large wth K = / = 4. A reasnable set s = 00 kω and = 390 kω (cmmercal values). Then: ω u π0 3 = C C = F ω l π0 = C C = F Cmmercal values are: C = 8 nf and C = 390 pf. We need t cnsder the mpact f the bandwdth f OpAmp chp. A f c = f u = 0 6 leads t f c = 0 6 /( 4) = 00 f u = khz. S the crcut shuld wrk prperly. ECE65 Lecture Ntes (F. Najmabad), Sprng

33 Prblem 5. Desgn a flter wth the transfer functn f H(jω) = mnmum nput mpedance f 50 kω. The transfer functn s n the general frm fr frst-rder lw-pass flters: 3 jω/5000 and a H(jω) = K jω/ω c wth K = 3 and ω c = 5000 rad/s. As K >, we need t use an actve flter. The prttype f the crcut s shwn belw wth C H(jω) = / jω/ω c ω c = C The nput mpedance f ths flter s Z mn =. Cmparng the transfer functn f ths prttype crcut wth the desred ne, we get: = 3 ω c = C = 5, 000 Z mn = 50 kω Chsng cmmercal value f = 5 kω, we get: = 3 = 53 kω 50kΩ (Cmmercal) C = 5, 000 = = nf (Cmmercal) We need t cnsder the mpact f the bandwdth f OpAmp chp. A f c = f u = 0 6 leads t f c = 0 6 /( 3) = 50 f u = 5/(π) = 0.8 khz. S the crcut shuld wrk prperly. ECE65 Lecture Ntes (F. Najmabad), Sprng

34 Prblem 6. We have an scllatr that makes a square wave wth a frequency f khz, a peak-t-peak ampltude f 0, and a DC-ffset f 0. Desgn a crcut that cverts ths square wave t a trangular wave wth a peak-t-peak ampltude f 0, and a DC-ffset f 0. What wuld be the frequency f the trangular wave? Snce a trangular wave s the ntegral f a square wave, we need t use an ntegratr crcut. The prttype, the nput sgnal and the desred utput sgnal are shwn belw. The utput wll have the same frequency as nput ( khz) and same perd ( ms). v C v 5 v 0.5 t (ms) 5 v 0.5 t (ms) 5 5 Because the nput sgnal s symmetrc (0 DC ffset), we can cnsder nly the half perd 0 < t < T/ = 0.5 ms. In ths range, v = 5 and we want v = t (fund frm v (0) = 5 and v (t = 0.5 ms) = 5 ). The utput f the ntegratr s gven by: v (t) v (0) = t v (t )dt C t 5 = t 5dt C t = C t C = esstr s needed t dscharge the capactr n lng tme scale (s that small DC nput des nt add up and saturate the OpAmp). Ths capactr als ensure that the utput has n DC ff-set. Settng: τ = C = 00T = = 0. We have tw equatns n three unknwns,, and C and we can chse value f ne arbtrarly. One has t be careful as f we dvde the tw equatns, we get: C C = = 0. = easnable chces t keep resstrs between 00 and MΩ are = kω and = 400 kω: C = C = = = 50 nf Thus, reasnable cmmercal chces are = kω, = 390 kω, and C = 0.4 µf. ECE65 Lecture Ntes (F. Najmabad), Sprng

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