2/20/2013. EE 101 Midterm 2 Review

Transcription

1 //3 EE Mderm eew

2 //3 Volage-mplfer Model The npu ressance s he equalen ressance see when lookng no he npu ermnals of he amplfer. o s he oupu ressance. I causes he oupu olage o decrease as he load ressance becomes smaller. oc s he open crcu olage gan.

3 //3 Volage-mplfer Model Theenn equalens Negae Feedback 3

4 //3 IDEL OPETIONL MPLIFIES 4

5 //3 SUMMING-POINT ONSTINT Operaonal amplfers are almos always used wh negae feedback, n whch par of he oupu sgnal s reurned o he npu n opposon o he source sgnal. In a negae feedback sysem, he deal opamp oupu olage aans he alue needed o force he dfferenal npu olage and npu curren o zero. We call hs fac he summng-pon consran. 5

6 //3 Ideal op-amp crcus are analyzed by he followng seps:. Verfy ha negae feedback s presen.. ssume ha he dfferenal npu olage and he npu curren of he op amp are forced o zero. Ths s he summng-pon consran. 3. pply sandard crcu-analyss prncples, such as Krchhoff s laws and Ohm s law, o sole for he quanes of neres. 6

7 //3 The Basc Inerer pplyng he Summng Pon onsran 7

8 //3 pplyng he Summng Pon onsran V n V = V ou n ou n ou n n ou ou Summng mplfer 8

9 //3 9 Summng mplfer B F B B F o B B F o F o F B B B F B B B V V V V V V V V Summng mplfer V = V V ou f V B B ou B B B F ou B B F ou B ou F ou F ou ou B B B

10 //3 Non nerng mplfer n o n n o o n Volage Follower o n

11 //3 d d d q n c o c o n n c n n d o n

12 //3 Dfferenaor rcu n dq d d d n o o d d n Dfferenaor rcu o d d n

13 //3 Lecure Inducance and apacance apacor Energy s sored n he elecrc feld ha exss beween he plaes when he capacor s charged. q 3

14 //3 q apacance q q d q d w L L d d d d p d L d Ld d Inducance p L L d d 4

15 //3 Lecure Frs Order Transen esponse Dscharge of a apacance hrough a essance d d KL a he op node of he crcu: q dq d q d d d d 5

16 //3 Dscharge of a apacance hrough a essance d d Ke d s d We need a funcon ha has he same form as s derae. Subsung hs n for c Kse s Ke s Dscharge of a apacance hrough a essance Solng for s: s Subsung no c : Ke Inal ondon: V Full Soluon: V e 6

17 //3 Dscharge of a apacance hrough a essance Ke To fnd he unknown consan K, we need o use he boundary condons a =. = he capacor s nally charged o a olage V and hen dscharges hrough he ressor. V V e hargng a apacance from a D Source hrough a essance 7

18 //3 hargng a apacance from a D Source hrough a essance KL a he node ha jons he ressor and he capacor d d V S urren no he capacor: d c d urren hrough he ressor: V S hargng a apacance from a D Source hrough a essance d d V S earrangng: d d V S Ths s a lnear frs order dfferenal equaon wh consan coeffcens. 8

19 //3 hargng a apacance from a D Source hrough a essance The boundary condons are gen by he fac ha he olage across he capacance canno change nsananeously: hargng a apacance from a D Source hrough a essance Try he soluon: K K e s Subsung no he dfferenal equaon: Ges: d V d s s Ke K S V S 9

20 //3 hargng a apacance from a D Source hrough a essance s s Ke K V S For equaly, he coeffcen of e s mus be zero: s s Whch ges K =V S hargng a apacance from a D Source hrough a essance Subsung n for K and s: K K e s V S K e / Ealuang a = and rememberng ha += VS Ke VS K K V s Subsung n for K ges: K K e s V S V S e /

21 //3 D Seady Sae d d In seady sae, he olage s consan, so he curren hrough he capacor s zero, so behaes as an open crcu. D Seady Sae L L d L d In seady sae, he curren s consan, so he olage across and nducor s zero, so behaes as a shor crcu.

22 //3 D Seady Sae The seps n deermnng he forced response for L crcus wh dc sources are:. eplace capacances wh open crcus.. eplace nducances wh shor crcus. 3. Sole he remanng crcu. Lecure 3 /L rcus, Tme Dependen Op mp rcus

23 //3 L rcus The seps noled n solng smple crcus conanng dc sources, ressances, and one energy-sorage elemen nducance or capacance are:. pply Krchhoff s curren and olage laws o wre he crcu equaon.. If he equaon conans negrals, dfferenae each erm n he equaon o produce a pure dfferenal equaon. 3. ssume a soluon of he form K + K e s. 3

24 //3 4. Subsue he soluon no he dfferenal equaon o deermne he alues of K and s. lernaely, we can deermne K by solng he crcu n seady sae 5. Use he nal condons o deermne he alue of K. 6. Wre he fnal soluon. L Transen nalyss Fnd and he olage = for < snce he swch s open pror o = pply KVL around he loop: V S 4

25 //3 L Transen nalyss V S d L d V S L Transen nalyss d L d V S Try K Ke s d L d K V S Try K Ke K slk e s V S s VS V K V K 5 S K slk s L 5

26 //3 6 L Transen nalyss L e K / K K e K L e / f x d dx d d L d d L and L rcus wh General Sources Frs order dfferenal equaon wh consan coeffcens Forcng funcon

27 //3 and L rcus wh General Sources The general soluon consss of wo pars. The parcular soluon also called he forced response s any expresson ha sasfes he equaon. dx x d f In order o hae a soluon ha sasfes he nal condons, we mus add he complemenary soluon o he parcular soluon. 7

28 //3 The homogeneous equaon s obaned by seng he forcng funcon o zero. dx x d The complemenary soluon also called he naural response s obaned by solng he homogeneous equaon. Sep by Sep Soluon rcus conanng a ressance, a source, and an nducance or a capacance. Wre he crcu equaon and reduce o a frs order dfferenal equaon. 8

29 //3. Fnd a parcular soluon. The deals of hs sep depend on he form of he forcng funcon. 3. Oban he complee soluon by addng he parcular soluon o he complemenary soluon x c =Ke / whch conans he arbrary consan K. 4. Use nal condons o fnd he alue of K. Lecure 4 Second Order Transen esponse 9

30 //3 Second Order rcus L d d d Dfferenang wh respec o me: s L d d d d ds d Defne: Second Order rcus d d ds d L d L L d Dampenng coeffcen L f L d L d s Undamped resonan frequency Forcng funcon d d d d f 3

31 //3 3 Soluon of he Second Order Equaon d d d d soluon omplemenary f d d d d soluon Parcular Soluon of he omplemenary Equaon : : : s s equaon haracersc Ke s s Facorng Ke ske Ke s Ke Try s s s s s d d d d

32 //3 Soluon of he omplemenary Equaon Qaudrac Equaon : ax bx c b x s b a 4ac haracersc Equaon : 4 s s Soluon of he omplemenary Equaon oos of he characersc equaon: s s Dampenng rao 3

33 //3. Oerdamped case ζ >. If ζ > or equalenly, f α > ω, he roos of he characersc equaon are real and dsnc. Then he complemenary soluon s: c s e K K e In hs case, we say ha he crcu s oerdamped. s. rcally damped case ζ =. If ζ = or equalenly, f α = ω, he roos are real and equal. Then he complemenary soluon s c s e s K K e In hs case, we say ha he crcu s crcally damped. 33

34 //3 3. Underdamped case ζ <. Fnally, f ζ < or equalenly, f α < ω, he roos are complex. By he erm complex, we mean ha he roos nole he square roo of. In oher words, he roos are of he form: nds j s j a n n whch j s he square roo of and he naural frequency s gen by: n n rcus wh Parallel L and V L d L L dq d c d d We can replace he crcu wh s Noron equalen and hen analyze he crcu by wrng KL a he op node: d d L d L n 34

35 //3 35 rcus wh Parallel L and d d L d d d d d d L d d d d dfferenang d L d d n n n L : rcus wh Parallel L and f d d d d d d f funcon Forcng L frequency resonan Undamped coeffcen Dampenng d d L d d d d n n

36 //3 rcus wh Parallel L and d d d d f Ths s he same equaon as we found for he seres L crcu wh he followng changes for : Parallel crcu Seres crcu L 36