Operational Amplifiers: Problems with Solution, Exercises of Electronic Circuits Design

Download Operational Amplifiers: Problems with Solution and more Exercises Electronic Circuits Design in PDF only on Docsity! Chapter Four: Operational Amplifiers  Chapter Four: Operational Amplifiers 263 4.3 An op-amp base amplifier has supply voltages of £5 V and a gain of 20. (a) Sketch the input waveform from the output waveform in Fig, P4.3. {b) Double the amplitude of your results in (a) and sketch the new output waveform. 2 6075 100128 (ms) Figure P4.3 SOLUTION: x) a} am) DISY ony dn \ 2 A anew teeee SNL Ens sop \ 4 Jt, tin) 1 2 $e IW we jes ty gb Tee i ae | \ ~. \ j bo , | ~o-2d Oasy rok \ cee ey ee 264 irwin, Basic Engineering Circuit Analysis, 8/E 44 For an ideal op-amp, the voltage gain and input resistance are infinite while the output resistance is zero. What are the consequences for (a) the op-amp’s input voltage? (6) the op-amp’s input currents? (c) the op-amp’s output current? SOLUTION: a) Since gain ig iatathe | am inavir voltage sf Zero g . ‘ 1 2 y Can Prodan @& Finike dub put vweltane. Uiy ro by Since Rig = , me Fapuk corcéet Plows. Lin? © 4 j - & cy Since Cour oO, Hu oubpel curreuk is Linthad. oly 4 excrernak cincuches yy eehernat cieethey ops. ty Pf ije De ! a i Do orand f- *& 4 Reg EQ a Chapter Four: Operational Amplifiers 265 i 4.3 Revisit your answers in Problem 4.4 under the following nonideal scenarios. | (a) R, = 00, Roy = 0, A, # 00. (1) Ri =O, Rout > G, A, =m OO. (c) R,, # 00, Ry, = 0, A, = 00. SOLUTION: 8) Since Ba =eo » tinee n . : _ . . Stee Lora , our 2 Vour FR. 4 ae Soace Wf ww vs, HO ae | ° ' iw vo a 7 e +) =O oO , tovr 6 Hina ted by balk : Pour € Ey , i | ©) Sine Aa = oe , %m eo Since be Ty FV S/S Rin [ine 0 oubg because. Vin 20 J Sena fy > O Laue Ltnthed ony by Ry g Chapter Four: Operational Amplifiers 267 4,7 Revisit the exact analysis of the inverting amplifier in Section 4.3. {a) Find an expression for the voltage gain if 2,, # 00, Rou = 0, Ay F oo. (6) For R, = 27 kO and R, = 3 kQ, plot the ratio of the actual gain to the ideal gain for A, = 1000 and 1kO =R, = 100k0. (c) From your plot, does the ratio approach unity as R,, increases or decreases? (d) From your plot in (b), what is the minimum value of R,, if the gain ratio is to be at least 0.98? SOLUTION: 2 Be Ue Ma cbeak 5 Vs Fre By ee i RD P: Gebead = ~ 2 . A « . f Bese fXy >) ates Pentet 2 oF gain ratio 0.97 | 996 Ce 6 20 40 60 80 100 i Rin (kohms) OS Crates, 268 Irwin, Basic Engineering Circuit Analysis, 8/E 4.8 An op-amp based amplifier has -18 V supplies and a gain of —80. Over what input range is the arnplifier linear? SOLUTION: of Ye . TO” nay © parehion a2 -~ Ba Wa Dye b owt pad dian : Vast & jay Linge Céad limihed b i ‘ ey Vin = Ve | lus 9.226 ~ Bo k - | Chapter Four: Operational Amplifiers 271 4.47 Using the ideal op-amp assumptions, determine the values of VY, and J, in Fig. P4.11. Figure P4.11 SOLUTION: ' a . . .- asic RHew-inverb Ay cod, oye takin. , Ve. ib Be 2m ve WM eo WE: | a Simte Pil = owt Cin eo 10 oa] 272 Irwin, Basic Engineering Circuit Analysis, 8/E 4.72 Using the ideal op-amip assumptions, determine f, h, and J, in Fig. P4.12. 1ma(f Figure P4.12 SOLUTION: y mAy 1a, * ofl ideal eR eg ) By ke oe 4 Se a E { im A Chapter Four: Operational Amplifiers 273 4.13 In a useful application, the amplifier drives a load. The circuit in Fig. P4.13 models this scenario. fés} (a) Sketch the gain V,/V; for 100 = R, Soo. (b) Sketch J, for 10O = R, Soo if = O1V. (c) Repeat (b) if Vy = 1.0 V. . (d) What is the minimum value of R,, if |J,| must be less than 100 mA for |V6] < 0.5 V? (@) What is the current Js if R, is 100 0? Repeat for R, = 10k. SOLUTION: . ae . Ops grka Q Skan. 5 wey {oe ; ° eA, Pay 2. ye be 2 yo cette Vs : 43 2. aon os VA ay ‘ a b) Vs FO9V Uys iV \ gains re ie —~ Eye Ty +0 = Yo 4 M Bomo? By Lanne enn Ry Lue} Hy Gm by if > Oe u  i i i ) Chapter Four: Operational Amplifiers 275 4,18 The op-ainp in the amplifier in Fig. P4.15 operates with +15 V supplies and can output no more than 200 mA. What is the maximum gain allowable for the amplifier if the maximum value of ¥; is 1 V? Ry + Ry = 10 kQ Figure P4.15 SOLUTION: - i @ asa Wan in derting, configu show Ve . ye Re By ePe lof 6 d Ue RB pT Fe ss Cy ' Iwel = Wel f te \ cas Sites yyslV, wt ev, ae] } ky Aloo, Tys Vo 4 ve 2 dog toh. sf eer} 4 Loom Rt, fo % FoR, E,4 8445 Chea gen Und: Final Ans way  276 Irwin, Basic Engineering Circuit Analysis, 8/E 4.7 Por the amplifier in Fig. P4.16, the maximum value of Vy is 2 V and the op-amp can deliver no more than 100 mA. (a) If+410 V supplies are used, what is the maximum allowable value of R,? (b) Repeat for +3 V supplies. (c) Discuss the impact of the supplies on the maximum allowable gain. ok V5 Op Ry = 10 kQ “ov Vo Ry, = 100 kQ ow Ry i Figure P4.16 SOLUTION: a) Basie Ren-inarting can} rncohon Vp = Vs [* ¢ Tos Ye I , RAP) : For Veg=2N, Yon 2 (ete) srov : ee Ry Rg deol | Check To Veib: te 2 4 2 \ eee LOOM * . lot # (109+ 4x8 ) a b) Vy=2(tebea\esv |e, e sobn Ios 3. em wAY VE) L a= Fens ae G.3 %S Xin” ¢) For a. Firtn Vs tatur , Aw spans is Athtank, reletil & Svgely “nla wAL Ey emit be toss oh icsue , } Chapter Four: Operational Amplifiers 279 4,79 The network in Fig. P4.19 is a current-to-voltage converter or transconductance amplifier. Find v,/ is for this network. Figure P4.19 SOLUTION: KCL gt v- inpet : le = 67 Vo Yo. Lk Vo. | ts as 280 Irwin, Basic Engineering Circuit Analysis, 8/E 4,20 Caiculate the transfer function i,/ uv, for the network shown in Fig. P4.20. SOLUTION: Kel ak Yo. ta tee vy ea . f [a vee —. ae tae No GY pio Eke Be [ Yo Ey Chapter Four: Operational Amplifiers 281 4.2) Determine the relationship between v, and i, in the circuit shown in Fig. P4,21. Rp i —— LOW lg Vo Figure P4,21 SOLUTION: oo. Beste Tetetng coh outahinn: J 7g 284 Irwin, Basic Engineering Circuit Analysis, 8/E 4.24 Show that the output of the circuit in Fig. P4.24 is v=li+ Bly oe Ry 1 Roy R, Figure P4.24 SOLUTION: ek We i wpact: “ VyoNx 20> Viz Vx 3 Vo = Vx ( | rte’ ~Ve (Pep) “L Chapter Four: Operational Amplifiers 285 4.25 Find V, in the network in Fig, P4.25. 4Q iy Ee O Figure P4.25 SOLUTION: ' aA gy Von 4 as By Be Met ak WL imped: 5 \ YQ = OV Loe 286 Irwin, Basic Engineering Circuit Analysis, 8/E 4.26 Find the voltage gain of the op-amp circuit shown in Fig. P4.26, 20 kD Ve i AK a Vy @) . ¥ a * ~) ¢, 1 KOS 0 Figure P4.26 SOLUTION: Two Step solutim: Pf) Fred Vo / 24 Find Vo f¥s, . Vy 2) [+ Fe. 2s ey ; : . Ny ! Dealt Jaane ws Vory, = (os 4 teAp ) \ Chapter Four: Operational Amplifiers 289 4,29 In the network in Fig. P4.29 ey the expression for v, in terms of the inputs v, and vp. fes} o_ |. Vy os : EE [" Figure P4.29 SOLUTION: LL ak Vy inact xy xe te ( Bes ey Bz +b.) Boe Uf fe. Bey, 290 Irwin, Basic Engineering Circuit Analysis, 8/E i t i 4.30 Find V, in the circuit in Fig. P4.30. 40 kQ BV ‘) Figure P4,30 SOLUTION: Koh ot Ve iepak | gee \ bytde No fytte i | Rye] Ry | Chapter Four: Operational Amplifiers 291 oo 4.31 Find V, in the circuit in Fig. P4.31. 10 ko bs VN I ng £, 100 kO 10V : V, —O Figure P4,31 SOLUTION: EL abe jag a 4 &- Ye if Ry "EE woe A é woe Me, feb Sexto* Axtot in fp S In 294 Inwin, Basic Engineering Circuit Analysis, 8/E 4.34 Find the input/output relationship for the current amplifier shown in Fig, P4.34. Figure P4.34 SOLUTION: kéL at Ve fagud of 0? op ang, a “pet of UO" op ep iat O-te Up = - Re by Ep wy Ade . - fi : Ze bye ttop in chasare rnvsety wy cee fgeretrn g Co . ? a . - 4 Ver Up fit Ey t Tae Vee Ue 1 fy / Bk Pehigy te Chapter Four: Operational Amplifiers 295 4.35 Find V, in the circuit in Fig. P4.35, 80 kD AN: ee bs $—Vv €, oy ip 10 kO, e) 5V b Figure P4,35 SOLUTION: web ak UO. ef PP op aos Sl Ye Lg wm Wwe “Fae. bs y } eee & * ie bs ane % o 2” Ker ak Wy oh et op dep | Voy > te F fy + Esp é \ ty ° ny “ey Put ih momboes E hind nate, Ux , po 296 Irwin, Basic Engineering Circuit Analysis, 8/E 4.3% Find v, in the circuit in Fig, P4.36. Ro Figure P4,36 SOLUTION: {= Op aap baw. an- me erkina, ten ney vet em :  Chapter Four: Operational Amplifiers 299 4.38 Find the output voltage, u,, in the circuit in Fig, P4.39, ann, dg Vo i Ape UE oky Figure P4.39 SOLUTION: KEL aU. of ampli Ay WAR en tif i). ter, ez gy 4 Kee ak AC 2: yet. Vet = (ee) Bey 4 te te e > "t V" &, Zz"! Ket ok Ue ol oy aug 3: Ay-Ue | We wy rige ¥ (e: i “Ts Re o\ Bre) Kee a ve | op tag? Yee Verte ag ep Vee, [14 P46) ty by & none y an enh yp Use Ry fiz BeN\ fay ops a E(t) tee 300 Irwin, Basic Engineering Circuit Analysis, 8/E 4.49 The electronic ammeter in Example 4.9 has been modified and is shown in Fig. P4.40. The selector switch allows the user to change the range of the meter. Using values for R, and R, from Example 4.9, find the values of Ry and R, that will yield a 10-V output when the current being measured is 100 mA and 10 mA, respectively. oO . aks Oth f | Unknown R, @Rn =R 1kO current A 2Ry @ Re= oO Ry =1kQ & Selector 7 switch & 0 Figure P4.40 SOLUTION: de ame in non ~ineaTing, torbraace bin § é Vyas x by F Vy = 10% © fo (of Rela Ba te Vas 2G, = o. ° = iw Vays pe OV Chapter Four: Cperational Amplifiers 301 4.47 Given a box of 10-kQ resistors and an op-amp, design a circuit that will have an output voltage of Y= ~4% [Esl SOLUTION: SAG Sy oe YS BSS0 evaded uth Vy at, are belle ke : ve Pps aeqediue , a Simpl! cemmmuc will sofAiie . a & Vy AE pty i bs y Vg bd Po | Yor obs y - Fa gy eave av, br pee Be 4 Ke LY Hh. feitewing tire ik with alt resists = teen, wale, é loka NV vy Coy «PL t Ske I ‘ t i | 304 Irwin, Basic Engineering Circuit Analysis, 8/E 4.44 Design an op-amp circuit that has the following input/output relationship: V,= -5V, + 0.5V, SOLUTION: A aihgte op “arp mit Lo Af wt une bed + 4 ~ ines, eeu ak Ve, \ wath ‘e Vie = Vx = Yr Be Fa Oa Va Baty ae ey Kin ak vn | Z= IT Vo Me 2 Ves vy, Ez Ry You Wx (14 Be) - be vy y Voz (ve (f- uM W/E ke) * 2 De) Per, = 5 Cheer Relkt => —zp= Seo Now, ole 2 Cheat Roe let => Bae ie Ratha 2 Chapter Four: Operational Amplifiers 305 4.45 A voltage waveform with a maximum value of 200 mV must be amplified to a maximum of 10 V and inverted. However, the circuit that produces the waveform can provide no more than 100 A. Desien the required amplifier. SOLUTION: ‘ ‘ : : : Reok Youn 2: Rows inverting Tp owned. by Laue ciKa, Si Peg ate qaaw 2.4 308 Irwin, Basic Engineering Circuit Analysis, 8/E 4.48 An amplifier with a gain of 7 + 1% is needed, Using resistor values from Table 2.1, design the amplifier. Use as few resistors as possible. SOLUTION: Fo pe 3: hue gan » “se han - thy ects cop rg, < G i¢ 2 2 A i u Vin Chapter Four: Operational Amplifiers 309 4.43 Show that the circuit in Fig. P4.49 can produce the output V, = BY, — Kav, only forO0 = K, =K,+ 1. Rs V2 Figure P4.49 SOLUTION: KEL ak Vy inputs Y oe vx op Vxs “EY By Kee apuk: = vet YL Re ve PA Rs} Rs 1+ Ee \. fy = Ky - Gm Ky Rye eg ad ere rhein rameters 7 & lhe | o - 4 i 20, Kys ftPiRe ole ky L 310 Irwin, Basic Engineering Circuit Analysis, 8/E 4.50 A 170°C maximum temperature digester is used in a paper milf to process wood chips that will eventually i become paper. As shown in Fig. P4.50a, three electronic thermometers are placed along its length. Each thermometer ourputs 0 V at 0°C, and the voltage changes 25 mV/°C. We will use the average of the three thermometer voltages to find an aggregate digester temperature. Furthermore, t volt should appear at V, for every 10°C of average temperature. Design such an averaging circuit using the op-amp configuration shown in Fig. 4.50b if the final output voltage must be positive. Paper mill digester tb) Chapter Four: Operational Amplifiers 311 4.51 A 0.1-O shunt resistor is used to measure current in a fuel-cell circuit. The voltage drop across the shunt resistor is to be used to measure the current in the circuit. The maximum current is 20 A. Design the network shown in Fig. P4.51 so that a voltmeter attached to the output will read 0 volts when the current is 0. A and 20 V when the current is 20 A. Be careful not to load the shunt resistor, since loading will cause an inaccurate reading. 0.10, pO + Vs /) Voltmeter o~ Figure P4.51 SOLUTION: whe. Dp =@ody V, = (NZ = 2V¥ and Vo = Zov, Needk gai of 10 wilh « brfbbPered buput =» non -fnuerbieg conf! Ve oA ze pA TS Vo je fe 2} —v Vo Ve g te. nt  Chapter Four: Operational Amplifiers 313 4.53 An operator in a chemical plant would like to have a set of indicator lights that indicate when a certain chemical tlow is between certain specific values. The operator wants a RED light to indicate a flow of at least 10 GPM (gallons per minute), RED and YELLOW lights to indicate a flow of 60 GPM, and RED, YELLOW, and GREEN lights to indicate a flow rate of 80 GPM. The 4-20 mA flow meter instrument outputs 4 mA when the flow is zero and 20 mA when the flow rate is 100 GPM. An experienced engineer has suggested the Circuit shown in Fig. P4.53, The 4-20 mA flow meter and 250 2, resistor provide a 1-5 V signal, which serves as one input for the three comparators. The light bulbs will turn on when the negative input to a comparator is higher than the positive input. Using this network, des a circuit that will satisfy the operator’s requirements. 12 12V kL Yellow / Ry N, 4 RE Figura P4.53 SOLUTION: . . p De sive Na? Ve wher fu = Bo gta : ‘ r m (fino) +b tents te [opel ' Bowl > pa foe) tb mA SG pie Lig Pe LT Ve = (fx) wy Owe Rg Ne oy 2) 2 Pb. é Sinn reel ak Lines 6ogem, Ty Need Vez 3 4V = 12 bok Cheaag Pgs lhe byt hy Fiavily, at Flags 12 GOR, Dos Sb Mh and Py Nos b= ie he chime Gs le oe nen Clhsains Kye Bye d= wn | uid usd (Lys VRS Res 253 | | hee RSV nn eee ene nanan - 2 B.bm and Vex 3-9V a3 flys Z.$8k4, Vee 14 Ree 2 stk i 314 Irwin, Basic Engineering Circuit Analysis, 8/E 4,54 An industrial plant has a requirement for a circuit that uses as input the temperature of a vessel and outputs a voltage proportional to the vessel’s temperature. The vessel’s temperature ranges from 0°C to 500°C, and the corresponding output of the circuit should range from 0 to 12 V. A RTD (resistive thermal device), which is a linear device whose resistance changes with ternperature according to the plot in Fig. P4.54a, is available. The problem then is to use this RTD to design a circuit that employs this device as an input and produces a 0- to 12-¥ signal at the output, where 0 V corresponds to 0°C and 12 V corresponds to 500°C. An engineer familiar with this problem suggests the use of the circuit shown ' in Fig. P4.54b in which the RTD bridge circuit provides the input to a standard instrumentation amplifier. Determine the component values in this network needed to satisfy the design requirements. SOLUTION: yoy redahhs Legs yor VY, Vy 1% and T. KeL at ETD: e-Wy A ~~» 12 Reger = Uy (ty + Le ) Vie WeRery AQ, a Reno) Ry Reap eek a& uy Ves fz R/U Ry rhe) Rery (7): Ray > Kyte At Pos, Rern = Ky = loon ft Te Boot, Lar = 40052 = Jootk, (sv Kae be a fe 2 lene loo+t 316 Irwin, Basic Engineering Circuit Analysis, 8/E 4FE-7 Given the summing amplifier shown in Fig. 4PFE-1 select the values of R, that will produce an output voltage of -3 V. [cs} 4kO, Wi if 4V 12kO 2V Figure 4PFE-1 SOLUTION: For Fue en ing A egy | G i] “ge i he) 4 2 pa * [ey = -3 “ atpon / \ be oad  Chapter Four: Operational Amplifiers 317 4FE-2 Determine the output voltage V,, of the summing 1 Op-amp circuit shown in Fig, 4PFE-2. Figure 4PFE-2 SOLUTION: