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# ee321ch8 - G Riz-wni Principles and Applications of...

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Unformatted text preview: G. Riz-wni, Principles and. Applications of Electrical Engineering Problem solutions, Chapter 8 b. The input voltage of the first stage can be determined in terms of the source voltage using voltage division: Re - - ”3 - + R 7 7 A“: Ezmm=mmﬂn_s__149_9._.,;_=144_3=20mgm[144.3]=43.13d3 VS vn vs vn vs ' 0.3 +7.7 8.3 Solution: Known quantities: Figure P85. Find: What approximations are usually made about the voltages and cun-ents shown for the ideal operational ampliﬁer [op-amp} model. Analysis: Bo‘l Solution: Known quantiﬁes: Figure P8.6. Find: - . ‘ What approximations are usually made about the circuit components and parameters shown for the ideal Operational ampliﬁer [op-amp] model. Analysis: n. goo p, a: 00 re :30. —————-————w—-—-——-—-——___.___._____m__—_~___ 8.5 V5 (ehllgk = 3m. _. 51k .. ‘1: V"V3\[EE]' a (“Q . C31)“ Amp 1% :1 \JOHQQ ﬂFOl,\OW'\f gag Voljgage mm “1L” ampU‘JC 7: 13V , %\\*’l§€_ \ngo [Vouagx-Q \\ H \ CLQVTJ'SS \WA'PU'E 012 019*60MKP) \Jot‘xcﬁ-Q (Wk — W1??? (1130 =- \3V- ' . ESL \3543’ : QJQCSJA 3 \EW V0: \au+\&v':30V L241, : = {0px 1‘ L36, ﬂincﬂ ("MFA “~ 0 (\Aeqjéyﬂ'a3 Lio/L: 30"”; :- 3A \$MUL C’rCn’vult :0 GOLE‘JUVGMP) 8,10: G. Rlzzoni, Principles and Applications of Electrical Engineering Problem solutions. Chapter 3 Section 8.2: The Operational Amplifier 8.24 ' Seiuiion: . Known Quantities: The circuit shown in Figure P81 V Find: _ ' § The resistor R, that will accompiish the nominal gain requirement, and stare what the maximum and minimum values (1fo can be. Will a Standard 5 percent tolerance resistor be adequate to satisfy this requirement? Analysis: For a nun-inverting ampiiﬁer the voltage gain is given by: i =-————=-~——-—=-—+l ; =16=~——+1 => R =-—=1k.0. - i A” vb. R, 11, Am” a, ‘ 16 1 E 1 To ﬁnd the 1113inme and minimumRS we note that R, 0: —-—, so to ﬁnd the maximumRs We consider 15m ' - +—-——~—=1.02m. "m 16(1-0.02)-—1 ' 15m mm=-——.———_=9 12. 1 (161+ 0112)— 1 80 Since a. standard 5% tolerance I-kﬂ resistor has mistame 950 < R < 10502511111de resistor will [33? sufﬁce In this application. am the minimum Av: R Conversely, to ﬁnd the minimum Rs we consider the maximumAp: R 8.2K — 33k ' Lad AU: "'— -———- ‘1: “ ——__ :"270g 11% ’ 12L 3 .5 “\$312+L393uo \3 US [Ml . \‘Jk— 02mm) (A ”Wk : 331<~L33kx00 \QK + 03110 I): ’ouﬁk l 008: 301 7k 133?. =33. (cl - 225 * 8.26 Rs .. , Um}? Jr 2W ’1'“ A - ' -.- ‘ \hm: wHSgsimfr I ' 2-0 ”VA IVA ““ Vaujf 0* KC - V‘t (1e A ' - = M U L C", 0 RS \olc .. I A 3 faintly} - VA“; \Mﬁ‘) =7 \onSanch -.-. 00*” —Uou£¥ - R's . \OK \30 4' : ._.. \OV‘UD'WDS 'SmmJC) + 10+“) sSimhﬁ Re. TO Aumxnoéﬁ. TX. 3.9}; -- \bkL‘Q +10 f: O W \$0M 14w is =3 123mm -3 ‘ 8.28 2" Solution: Known quantities: 1 1393:): —(2sin 031t+4sin (92:4.- Ssinw3r+ lﬁsinmg) V, RF =5 k9. - 1 Design an inverting summing ampliﬁer to obtain vm (t) and demine the require some resistam. ‘! , Analysis: ' ' i The inverting summing ampliﬁer is shown in the folluwing ﬁgme. : I E a ) E i i By superposition and by sclecnng R5: = 3%, vs, a 51!! (qt, I =1...4. the output voltagc is vm=—£——ivg= —22‘vg= -;2'sinm,t that coincides with the dashed output voltage. So. the required some msistors are . ‘ 1 14: R Iii R,.=-§-=2.,5mxﬂ=f=1.2sm, Rﬁ= -R 7‘ =625£2,R 5 _ 3 8 “=11; =3.129 W . Wald: sin”; “Um‘ki ““039 R‘ -—__ 0‘10”“ 83411 ‘ VORcm‘e 01m \‘—h.in\)u'\' '3 V“) I%N\¢Q \Aeqn'a \Juwcwxe, (LONﬁS Dyvoxmp \n’PA i% ¢o Luau?— LID, Lainey; thug MAO gram-930).. .- 1h: 3 A3 \bhg (19; “\$50 (Lowe/VJ“ an“ Row - Améue, 40 W Jt-A’Muqlx L‘d) «AWMQOIQ “QmeoXJOi‘ax M “\W, \Q \JgLO ‘ LUWQNJQ wmﬁi Hm up ~\5\\voug\\ ”0/7, %T M W341 Jr0 .. L61) M “Me/12w: Mé-Q- lawmug‘k (KM . \‘mmaghovg ask mi“ wk \ E ; ‘; E 80%2 L V E. r Sanding: Known quantiﬁes: ' The circuit shown in Figurc P821 Find: ' ‘ Show that the voltage VM is proportional to the current gcnmtad by the Cats solar cell. Show that the transimpedancc of the circuit Van/I, is —R. Analysis: 011 Assuming an ideal op~ampz 11"" .=. v" = 0 21> Vm = —RIS V i i i i . s E E 1 a 1' , J 1 i i E , -R . The transimpbdance is gchn by R. .___. ii S All Isqmgwbg Is ‘19- ‘ n—ﬁor 8.16 8 .L‘Ll G. Rizzoni. Principlm and Applications of Electrical Engineering Problem solutions. Chapter 8 v0 =_-.R£ +§L 5'. R52 R51 1 g § 1 1‘ 1 a . i ‘i 1 1 , g i : i I . | : ! l 3 BS1 G. Rizzoni, Principles and Applications of Electrical Enginaering Problem solutions. Chapter 8 In fact. the output voltage is R31 V a) =———-—-——-—-—V ' 0U ) R 1+ .iszR 5U”) . Vo(jq£} R2 1 13. H0 E0 - =-—-———-—- 0) var» ) R 1+jmcirzz c. The gain in decibel is obtaiiwd by evaluating 1H,, (ice)! at (0:0. is. [H,(j0)Lm = 20Log%= 201033221 =7'_55 dB. The cutoﬂ’ {raquency is SST—““Hﬁ Solution: Known quantities: . Forthe circuit shown in Figure P8.40: C = 0.47 11F R = 2.2169 R2 = .6816} Rt. 21m ‘ Find: a. An expression in standard form fur the vultage ﬁequency response function. b. Ther gain in dwibeis' m the passbmd,tl1at'1s, at the frequmcies being passed by tho ﬁlter. and mi: cutoff frequency. Analysis: 3. From“ rigure 3. 21 itresults that the ampliﬁer in Figure P8. #0' Is aiompass ﬁlter Th: voltage frequency response function is . V (1w) R 1' HAW) V,(jw) R, 1+ jay CR2 ' b. The gain in decibel is obtained by evaluating [H11 jag] iii «i=0, Le. . R2 68 - H o = Lo———=20Lo ——-=29.3dB. I ”(1)!” 20' th g2.2 The cumﬁ'fmquﬂncy is . mo=L=———l-_—=31289iadjs CR2 0.47 10‘9 68 103 8.13 FY‘OM U} '. EH ...
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