2.1 The Inverting Configuration

Transcription

1 /3/0 secton _ The nertng confguraton /. The Inertng Confguraton eadng Assgnment: pp One use of amps s to make amplfers! Ths seems rather obous, but remember an amp by tself has too much gan to be practcal! Thus, the amp s but one element n our amplfer desgn. The resultng amplfer wll be ery dfferent from the amp tself do not confuse the amp wth the amplfer! In ths secton, we wll consder the nertng amplfer an amplfer constructed wth resstors and one amp. HO: ANALYSIS OF THE INVETING AMPLIFIE The nertng amplfer uses feedback we close a lo! HO: CLOSEDLOOP AND OPENLOOP GAIN The result of ths feedback s the rtual short. HO: THE VITUAL SHOT Let s determne the nput and put resstances of the nertng amp! Jm Stles The Un. of Kansas Dept. of EECS

2 /3/0 secton _ The nertng confguraton / HO: IN AND OUT OF THE INVETING AMP Make sure that your feedback s negate! HO: FEEDBACK STABILITY Another mportant applcaton of the nertng confguraton s the weghted summer. HO: THE WEIGHTED SUMME Jm Stles The Un. of Kansas Dept. of EECS

3 /3/0 Analyss of the Inertng Amplfer lecture / Consder an nertng amplfer: Analyss of the Inertng Amplfer n d 0 deal Note that we use here the new notaton and. Jm Stles The Un. of Kansas Dept. of EECS

4 /3/0 Analyss of the Inertng Amplfer lecture / Pay attenton to your TA! Now what s the encrcut oltage gan of ths nertng amplfer? Let s start the analyss by wrtng down all that we know. Frst, the amp equaton: ( ) A Snce the nonnertng termnal s grounded (.e., 0): A n 0 deal Jm Stles The Un. of Kansas Dept. of EECS

5 /3/0 Analyss of the Inertng Amplfer lecture 3/ Frst some KCL Now let s apply our crcut knowledge to the remander of the amplfer crcut. For example, we can use KCL to determne that: Howeer, we know that the nput current of an deal amp s zero, as the nput resstance s nfntely large. Thus, we reach the concluson that: n 0 deal Jm Stles The Un. of Kansas Dept. of EECS

6 /3/0 Analyss of the Inertng Amplfer lecture 4/ And then some Ohm s law Lkewse, we know from Ohm s Law: and also that: And so combnng: n 0 deal Jm Stles The Un. of Kansas Dept. of EECS

7 /3/0 Analyss of the Inertng Amplfer lecture 5/ Fnally, from KCL we can conclude: Followed by KVL n n In other words, we start at a potental of n olts (wth respect to ground), we dr a potental of olts, and now we are at a potental of olts (wth respect to ground). n 0 deal Jm Stles The Un. of Kansas Dept. of EECS

8 /3/0 Analyss of the Inertng Amplfer lecture 6/ And yet another KVL Lkewse, we start at a potental of of olts (wth respect to ground), we dr a potental of olts, and now we are at a potental of olts (wth respect to ground). n 0 deal Jm Stles The Un. of Kansas Dept. of EECS

9 /3/0 Analyss of the Inertng Amplfer lecture 7/ The feedback equaton Combnng these last three equatons, we fnd: n Now rearrangng, we get what s known as the feedback equaton: n Note the feedback equaton relates n terms of put. n 0 deal Jm Stles The Un. of Kansas Dept. of EECS

10 /3/0 Analyss of the Inertng Amplfer lecture 8/ The feedforward equaton We can combne ths feedback equaton wth the amp equaton: A Ths amp equaton s lkewse referred to as the feedforward equaton. Note ths equaton relates the put n terms of. We can combne the feedback and feedforward equatons to determne an expresson nolng only nput oltage n and put oltage : A n Jm Stles The Un. of Kansas Dept. of EECS

11 /3/0 Analyss of the Inertng Amplfer lecture 9/ and the encrcut oltage gan appears! earrangng ths expresson, we can determne the put oltage n terms of nput oltage n : A n ( ) A and thus the encrcut oltage gan of the nertng amplfer s: A Ao n ( ) A ecall that the oltage gan A of an deal amp s ery large approachng nfnty. Thus the encrcut oltage gan of the nertng amplfer s: A Ao lm A ( ) A Jm Stles The Un. of Kansas Dept. of EECS

12 /3/0 Analyss of the Inertng Amplfer lecture 0/ Summarzng Summarzng, we fnd that for the nertng amplfer: A o n n 0 deal Jm Stles The Un. of Kansas Dept. of EECS

13 /3/0 Analyss of the Inertng Amplfer lecture / The nonnertng termnal s at ground potental One last thng. Let s use ths fnal result to determne the alue of, the oltage at the nertng termnal of the amp. ecall: Insertng the equaton: we fnd: n n n 0 n n The oltage at the nertng termnal of the amp s zero! Jm Stles The Un. of Kansas Dept. of EECS

14 /3/0 Analyss of the Inertng Amplfer lecture / The logc behnd the rtual short Thus, snce the nonnertng termnal s grounded ( 0), we fnd that: and 0 ecall that ths should not surprse us. We know that f amp gan A s nfntely large, ts put wll also be nfntely large (can you say saturaton?), unless s nfntely small. We fnd that the actual alue of to be: n A A a number whch approaches zero as A! Jm Stles The Un. of Kansas Dept. of EECS

15 /3/0 Closed and Open Lo Gan lecture /5 ClosedLo and OpenLo Gan Consder the nertng amplfer a feedback amplfer constructed wth an amp: n deal The encrcut oltage gan of ths amplfer: A o s also referred to by engneers the closed lo gan of the feedback amplfer. Jm Stles The Un. of Kansas Dept. of EECS

16 /3/0 Closed and Open Lo Gan lecture /5 A closed lo Q: Closed lo? What does that mean? A: The term closed lo refers to lo formed by the feedforward path and the feedback (.e., feedback) path of the amplfer. In ths case, the feedforward path s formed by the amp, whle the feedback path s formed by the feedback resstor. Feedback Path n ClosedLo deal Feedforward Path Jm Stles The Un. of Kansas Dept. of EECS

17 /3/0 Closed and Open Lo Gan lecture 3/5 An en lo If the lo s broken, then we say the lo s en. The gan ( o ) for the en lo case s referred to as the enlo gan. n OpenLo deal Jm Stles The Un. of Kansas Dept. of EECS

18 /3/0 Closed and Open Lo Gan lecture 4/5 Open and closed lo gans For example, n the crcut we know that: 0 n 0 ( ) A 0 n deal Combnng, we fnd the enlo gan of ths amplfer to be: A en A n Once we close the lo, we hae an amplfer wth a closedlo gan: A closed n whch of course s the encrcut oltage gan of ths nertng amplfer. Jm Stles The Un. of Kansas Dept. of EECS

19 /3/0 Closed and Open Lo Gan lecture 5/5 Feedback s a wonderful thng Note that the closedlo gan ( ) does not explctly nole the amp gan A. * The closedlo gan s determned by two resstor alues, whch typcally are selected to prode sgnfcant gan ( A o > ), albet not so large that the amplfer s easly saturated. * Conersely, the enlo gan (A ) obously does nole the amp gan. Moreoer, as n ths case, the enlo gan of a feedback amplfer often only noles the amp gan! * As a result, the amp gan s often alternately referred to as the enlo gan. Note that closng the feedback lo turns a generally useless amplfer (the gan s too hgh!) nto a ery useful one (the gan s just rght)! Jm Stles The Un. of Kansas Dept. of EECS

20 /3/0 The Vrtual Short lecture /6 The Vrtual Short For feedback amplfers constructed wth amps, we hae found (and wll contnue to fnd) that the two amp termnals wll always be approxmately equal ( ). Of course, ths must be true n order to aod saturaton, as the gan of an amp s deally nfntely large. Snce for feedback applcatons, t appears that the two amp termnals are shorted together! Vrtual Short o Jm Stles The Un. of Kansas Dept. of EECS

21 /3/0 The Vrtual Short lecture /6 There s no short nsde the amp! The condton n amp feedback amplfers where s known as the rtual short. emember, although the two nput termnal appear to be shorted together, they are most certanly not! If a true short were present, then current could flow from one termnal to the other (.e., ). Howeer, we know that the nput resstance of an amp s deally nfnte, and thus we know that the nput current nto an amp s zero. 0 en crcut o 0 Therefore, t s not the amp that enforces the condton, t s the feedback that makes ths so! Jm Stles The Un. of Kansas Dept. of EECS

22 /3/0 The Vrtual Short lecture 3/6 The rtual short: your new BFF Applyng the concept of a rtual short can greatly smplfy the analyss of an amp feedback amplfer. For example, consder agan the nertng amplfer: n 0 deal Jm Stles The Un. of Kansas Dept. of EECS

23 /3/0 The Vrtual Short lecture 4/6 The rtual ground Ths tme, we begn the analyss by applyng the rtual short condton: Snce the nonnertng termnal s grounded ( 0), the rtual short means that the nonnertng termnal s lkewse at zero potental ( 0)! We refer to ths condton as a rtual ground. n 0 deal 0 Jm Stles The Un. of Kansas Dept. of EECS

24 /3/0 The Vrtual Short lecture 5/6 Isn t ths smpler? Analyzng the remander of the crcut, we fnd: 0 n 0 deal n n 0 Combnng, we fnd: n earrangng, we agan fnd the encrcut, closedlo oltage gan: A o n Jm Stles The Un. of Kansas Dept. of EECS

25 /3/0 The Vrtual Short lecture 6/6 Your TA s een smarter than ths guy! Note ths s exactly the result we found before, yet n ths case we neer consdered the amp equaton: ( ) A The rtual short equaton ( ) replaced the amp equaton n our analyss of ths feedback amplfer. Eeffectely, the two equatons say the same thng (proded A s nfntely large, and we hae negate feedback n the crcut). Jm Stles The Un. of Kansas Dept. of EECS

26 /3/0 and o of the Inertng Amplfer lecture / n and of the Inertng Amplfer ecall that the nput resstance of an amplfer s: n n n For the nertng amplfer, t s edent that the nput current n s equal to : n n Jm Stles The Un. of Kansas Dept. of EECS

27 /3/0 and o of the Inertng Amplfer lecture / Its nput resstance From Ohm s Law, we know that ths current s: The nonnertng termnal s connected to rtual ground: n n 0 and thus the nput current s: n n n n We now can determne the nput resstance: n n n n n The nput resstance of ths nertng amplfer s therefore! n Jm Stles The Un. of Kansas Dept. of EECS

28 /3/0 and o of the Inertng Amplfer lecture 3/ Output resstance s harder Now, let s attempt to determne the put resstance. ecall that we need to determne two alues: the shortcrcut put current sc ( ) and the encrcut put oltage( ). To accomplsh ths, we must replace the amp n the crcut wth ts lnear crcut model: n A ( ) sc 0 Jm Stles The Un. of Kansas Dept. of EECS

29 /3/0 and o of the Inertng Amplfer lecture 4/ Frst, the short crcut put current From KCL, we fnd that: sc where: A A o o n A ( ) 0 sc and: Therefore, the shortcrcut put current s: sc A A A Jm Stles The Un. of Kansas Dept. of EECS

30 /3/0 and o of the Inertng Amplfer lecture 5/ Now, the en crcut put oltage The encrcut put oltage can lkewse be determned n terms of A and. n A ( ) 0 Here, t s edent that snce 0: where we fnd from Ohm s Law: ( A) A Jm Stles The Un. of Kansas Dept. of EECS

31 /3/0 and o of the Inertng Amplfer lecture 6/ The encrcut put oltage Now from KVL: Insertng the expresson for : A o A A n ( A ) A ( ) 0 Jm Stles The Un. of Kansas Dept. of EECS

32 /3/0 and o of the Inertng Amplfer lecture 7/ Now we fnd the put resstance Now, we can fnd the put resstance of ths amplfer: sc A A o o o o o In other words, the nertng amplfer put resstance s smply equal to the alue of the feedback resstor n parallel wth amp put resstance. Jm Stles The Un. of Kansas Dept. of EECS

33 /3/0 and o of the Inertng Amplfer lecture 8/ Ths s zero f the amp s deal Ideally, of course, the amp put resstance s zero, so that the put resstance of the nertng amplfer s lkewse zero: 0 0 Note for ths case where the put resstance s zero the put oltage wll be the same, regardless of what load s attached at the put (e.g., regardless )! of n 0 A ( ) n Jm Stles The Un. of Kansas Dept. of EECS

34 /3/0 and o of the Inertng Amplfer lecture 9/ For real amps the put resstance s small Thus, f 0, then the put oltage s equal to the encrcut put oltage een when the put s not en crcuted: for all n!! ecall that t s ths prerty that made 0 an deal amplfer characterstc. We wll fnd that real (.e., nondeal!) amps typcally hae an put resstance that s ery small ( ), so that the nertng amplfer put resstance s approxmately equal to the amp put resstance: Jm Stles The Un. of Kansas Dept. of EECS

35 /3/0 and o of the Inertng Amplfer lecture 0/ A summary Summarzng, we hae found that for the nertng amplfer: n (deally zero) Thus, ths nertng amplfer n( t ) n ( t ) ( t ) ( t ) Jm Stles The Un. of Kansas Dept. of EECS

36 /3/0 and o of the Inertng Amplfer lecture / The nertng amp equalent crcut has the equalent crcut: n( t ) ( t ) n( t ) ( ) n( t) ( t ) Note the nput resstance and encrcut oltage gan of the nertng amplfer s VEY dfferent from that of the amp tself! Jm Stles The Un. of Kansas Dept. of EECS

37 3/0/0 Feedback Stablty lecture.d /8 Feedback Stablty ecall that for the nertng amp: n deal we hae the feedforward equaton: A and the feedback equaton: n Jm Stles The Un. of Kansas Dept. of EECS

38 3/0/0 Feedback Stablty lecture.d /8 Nothng more fun than calculus! Takng derates of these equatons, we fnd that: ( A ) A and: n These derates are ery mportant n determnng the stablty of the feedback amplfer. Jm Stles The Un. of Kansas Dept. of EECS

39 3/0/0 Feedback Stablty lecture.d 3/8 Feedforward To see ths, consder what happens when, for some reason, changes some small alue Δ from ts nomnal alue of 0. The put oltage wll then lkewse change by a alue Δ : Δ Δ A Δ Note f Δ s poste, then decrease n. Δ wll be negate an ncrease n leads to a Ths descrbes the feedforward porton of the lo. Jm Stles The Un. of Kansas Dept. of EECS

40 3/0/0 Feedback Stablty lecture.d 4/8 Feedback The feedback equaton states that a small change n the put oltage (.e., Δ ), wll lkewse result n a small change n : Δ Δ Δ Note n ths case, a decreasng put oltage wll result n a decreasng nertng termnal oltage. Thus, f the nertng termnal oltage tres to ncrease from ts correct alue of 0, the control lo wll react by decreasng the oltage essentally counteractng the ntal change! Jm Stles The Un. of Kansas Dept. of EECS

41 3/0/0 Feedback Stablty lecture.d 5/8 Note that the lo product: Negate feedbackn ths case t s a good thng! A s a negate alue; we refer to ths case as negate feedback. Negate feedback keeps the nertng oltage n place (.e., 0) t enforces the concept of the rtual ground! Jm Stles The Un. of Kansas Dept. of EECS

42 3/0/0 Feedback Stablty lecture.d 6/8 Let s try some poste feedback Contrast ths behaor wth that of the followng crcut: n deal Q: Isn t ths precsely the same crcut as before? A: NO! Note that the feedback resstor s now connected to the nonnertng termnal, and the nertng termnal s now grounded. Jm Stles The Un. of Kansas Dept. of EECS

43 3/0/0 Feedback Stablty lecture.d 7/8 Poste derates! The feedforward equatons for ths crcut are thus: And so: A ( A ) A n whle the feedback equatons are: deal and: n n Note n ths case, both derates are poste. Jm Stles The Un. of Kansas Dept. of EECS

44 3/0/0 Feedback Stablty lecture.d 8/8 Poste feedbackn ths case t s a bad thng! Ths means that an ncrease n wll lead to an ncrease n. The problem s that the feedback wll react by ncreasng een more the error s not corrected, t s nstead renforced! The result s that the put oltage wll be sent to the amplfer wll saturate). Note that the lo product for ths case s poste: or (.e., A Thus, we refer to ths case as poste feedback. Poste feedback typcally leads to amplfer nstablty! As a result, we fnd that the feedback porton of an amp crcut almost always s connected to ts nertng () termnal! Jm Stles The Un. of Kansas Dept. of EECS

45 /6/0 The Weghted Summer lecture /3 The Weghted Summer Consder an nertng amplfer wth multple nputs! f deal From KCL, we can conclude that the currents are related as: 3 and because of rtual ground ( 0), we can conclude from Ohm s Law: Jm Stles The Un. of Kansas Dept. of EECS

46 /6/0 The Weghted Summer lecture /3 Lkewse: The put oltage f Insertng these results nto the ntal KCL expresson: f f deal 3 f 3 Now, we sprnkle on some algebrac pxe dust, and fnd the put oltage: f f f 3 3 The put s thus a weghted summaton of each of the nput sgnals! We therefore refer to ths crcut as the weghted summer. Jm Stles The Un. of Kansas Dept. of EECS

47 /6/0 The Weghted Summer lecture 3/3 How to combne sgnals Note that f 3, the put s an unweghted summer: f For example, f: ( ) ( t) ( t) ( t) ( t) 3 ( ) ( ) t.0 cos πt π ( ) ( πt ) t.0 cos π 3 ( ) ( πt 4) 3 t.5cos π then: ( ).0 cos( π π).0 cos( π π 3).5cos( π π 4) t t t t The summer s a method for combnng seeral sgnals! Jm Stles The Un. of Kansas Dept. of EECS