HW2_1_Solutions_OpAmps

HW2_1_Solutions_OpAmps - Solution of the Homework...

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Unformatted text preview: Solution of the Homework Assignment 2 (14:332.361) Excerises: We Know +ko+ v3=H(V;—V.) and 924000. Wue‘ora: (a) v.—_- V;— X1: o-_3;3=-o.oo2v =-1rnv Io ”V‘=1rnv. v =V1_V‘-_-_ 0+1MV=pvug+1nV Id 1‘m=_2|_._(y.+vz)=_i_(o-Zrnv)=—IH\V V3 —|o =V -——=5-___—-=5.0IV (b)v' l H Iooo v —- vz-v,_—_ 5- 5.00 =.°-O| v: lOmV u" - l =_!_ 5.0; 5 = 5.005 v: 5v VI“, 3 ( vI +V1) 2-( + ) (0) v3: Fuvrv.) = looo.(o.qqs — Loans-"1v V1,=v,_.v.=o.998_n.ooa.=.+|mv mez—L-(v,+v3)=3':<l.ool+ onqs): I v -16 looo =-3.6036v (d) V1=V.+%_ =_3.6+ VI: = v,__v, = -36036 -(— 3 .6): -o. oozév =-‘s.en« VI“; J£(v.+v1)=-3.eomvz-3.ev From Tabla Ll we have: Rm: .V__ 13 16:0 .ia. odpuf is open chm-3+ 1F: neaod’ive inpud‘ +erm'moJ OF H" °PMB La.) V.‘ 3s 5 Vif+ml aroma ,+kus VI' :0 ““33 R3“... ”‘5 R4»... or: nomknal Valuzs 11:” IMP“ 'es;s+°‘"°°' °‘F ““5 invaH'B For R3 and R.“ respecl-iyely, we have: in‘I'esrafl'or is K, «Hive-Fore, R=|OKIL Ram“: 1 K-fi. Md RAM“: 19° K-(L,‘kus, Si'nce. “Ne desirté irH-earxlfion ‘Hme. Cons-‘urfl' . - —5 .5 l0 3s, we have- (LR—.1": s =#- m°:0.qq < RH < lofixl-O' C: IO 35 .__. 0 ”up 1u|.ol ‘ Ra‘ lxofil‘l mun. qg 01 <___ R4. < ,01'01 From efiua‘l'n'on (2.50) +52 #Ma'per flamenc- R3 0-; ‘Hfis 5n+e3ro~+or is: Simllar'y,R we Con Show 1450:} V00.” =.._.'.—- V;(J'u) JuC—R 98°1<_:_’- (101.03. , I For walo rod/s +I~e in+e3m+or fians‘ex Hence,-loz.ol ‘< - R ———-<— QB 02- ‘Fwnc-Hon has Maanl‘b-cle. I TRereFore, I... = ___‘___. = loo V/v and Phase. ¢=9o’. R Rh 2:. IO: '0 3 -4‘4 _fi.___=_<4g.l.__._[44 R3 RI ’23 RI For «9:! rad/s 44.2 in+earc~+°r +MHS£U In 'H‘e Nors+ case ‘Func‘Hon has moani‘l’ude %l [XL = ' = IOOOV/V and Phase ¢=¢IOY l%— V; 0 x 10'3 R <———-=?"9c,.[<o.ofl l+ 3%; “38' 01 Using egm‘Hon (1.53) +52 grec‘mC—Y 01" Noi'g +ko.+ +45: wars-i- Case. n“, MPF‘“ whzch 'Hne in+83ra~+or gain Manama: is when 51:48. 01 d 3—: 1,0 ' is w- =_.'_=_L_=Iooora.d R: an R. lo 2. and)! "‘4' CR [0'3 /5 We, du‘FFexomfi‘al 30.6.». Ad 0; 'H\e mpfl‘ger is Rd: 52", “hem-Fore, fie correspondirs Value- 0? CMRR «For +he. wors+co~s¢ Ha» CMRR: logos ‘9“ =1oQ°3 ‘02-'01 =7 0.041 CM RR = 1.0 9.03 (2550.5) 2 EBdB 2.19 V;(+) v.9) v.=v;_ Rig: o_Ri;=—Ri; Rn: 12—- = “.Ri- . . \' t‘ l.:.o ‘ =‘R :RM=_R=_IOKA R;.-.: L and V; ‘59. vir'l'ual around (v;=o), M Ri=a9-=O :R;=OJ\_ '3 550“ we are. ass-“fins 'H‘\o:}- -H\e op MP 'm His {ransres'vs-fMCe amplifier is )dea'fike gran-P has zero oufl'Putl' resis-Iunczand Mm +ke ou+Pu+ resis‘l-nnce :1? «His mares;s+mce amplifier is also we .1de us R.=o.n_. ConnOCHnj "H‘e signal Source shown in ‘3‘" 22.5 +a +k¢ inPwl' 0F +hu's MPH-FIG: we have: RsIOV-n V; isa vithn.‘ around “‘04" is v; =O,‘H1u_5 N wm+ ‘r’kwinj ‘HVWQH H‘e. IckJL “5+0" €0an bah-teen v; and around is We. TFerekre V.: V;.. RIO.5mH=O-— IOKx°.5mR= —5V 1.13 iL=i._—. Imfi . v°=V.+ 31x9Kn=I+lx9=|°v 3.5:...— = ———= IOmR,;°=.IL+iL=|ImQ ; I v V I... __ "Dru“- ‘I. o P. = Vol“. mon t 00 P1 v1.1; Ix o 1.15 R: R. VII V O V12. R (as) :' R.=R3=1K.fl., R1=Rq=1nohJL Since. Kalli-5: Ra/R. we Rowe: Rd: V0 2 EL:_J;‘;°_=.IOOV/V sz—Vxn ' 2' (b) Rza=1 R.= 1x2nn=4 Kn. Since we are “55mins H‘e. °P amp is n... Rig—PR3 R0 R4 :( ' )(l— 5?. £2. ‘+ R; R! R‘I Ra Bulfi- : R: Bi 3.1 1 R3 + We Wor5+ Case common—mod: gain 9a.. happens Wkehlacml has H’s maximum value. SH" ‘H‘e res€s+ors have I7o +olerance,we hO‘VE RMM‘ "°'°') R‘l < R‘Qnom (| 1-0-0.) R'snw (‘+°‘°')‘ R3 \ R3non(‘—O.°l) (Cont. ...
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