Experiment 5 Op-Amp Circuits

Experiment 5 Op-Amp Circuits Purpse: In this experiment, yu will learn abut peratinal amplifiers (r p-amps). Simple circuits cntaining peratinal amplifiers can be used t perfrm mathematical peratins, such as additin, subtractin, and multiplicatin, n signals. They can als be used t take derivatives and integrals. Anther imprtant applicatin f an p-amp circuit is the vltage fllwer, which serves as an islatr between tw parts f a circuit. Equipment Required: HP 34401A Digital Multimeter HP 33120A 15 MHz Functin / Arbitrary Wavefrm Generatr HP E3631A Pwer Supply Prtbard Sme Resistrs 741 p-amp r 1458 dual p-amp Helpful links fr this experiment can be fund n the links page fr this curse: http://hibp.ecse.rpi.edu/~cnnr/educatin/eilinks.html#exp5 Part A Intrductin t Op-Amp Circuits Backgrund Elements f an p-amp circuit: Figure A-1 belw is a schematic f a typical circuit built with an p-amp. Figure A-1 The circuit perfrms a mathematical peratin n an input signal. This particular p-amp circuit will invert the input signal, V1, and make the amplitude 10 times larger. This is equivalent t multiplying the input by -10. Nte that there are tw DC vltage surces in additin t the input. These tw DC vltages pwer the p-amp. The circuit needs additinal pwer because the utput is bigger than the input. Op-amps always need this additinal pair f pwer surces. The tw resistrs R3 and R2 determine hw much the K.A. Cnnr and Susan Bnner 1 f 18 Revised: 1/16/2006

p-amp will amplify the utput. If we change the magnitude f these resistrs, we d nt change the fact that the circuit multiplies by a negative cnstant; we nly change the factr that it multiplies by. [These cmpnents behave much like the parameters f a subrutine.] The lad resistr RL is nt part f the amplifier. It represents the resistance f the lad n the amplifier. Pwering the p-amp: The tw DC surces, (± V CC ), that prvide pwer t the p-amp are typically set t have an equal magnitude but ppsite sign with respect t the grund f the circuit. This enables the circuit t handle an input signal which scillates arund 0 vlts, like mst f the signals we use in this curse. (Nte the sign n V4 and V5 in the circuit abve.) The schematic in figure A-2 shws a standard ± V CC cnfiguratin fr p-amps. The schematic symbls fr a battery are used in this schematic t remind us that these supplies need t be a cnstant DC vltage. They are nt signal surces. Figure A-2 The HP E3631A pwer supply prvides tw variable supplies with a cmmn grund (fr ± V CC ) and a variable lw vltage supply. The pwer supply jack labeled "COM" between the V CC supplies shuld be cnnected t circuit grund. When yu supply pwer t the p-amp, adjust the tw vltage levels s that +V CC and - V CC are equal, but ppsite in sign, at 15 V. Nte that in PSpice, there are tw ways t represent a surce with a negative sign. Figure A-3 shws the tw ptins: yu can either set the vltage surce t a negative value, r yu can reverse the plarity f the surce. V1-15V = V2 15V 0 Figure A-3 0 The p-amp chip: Study the chip layut f the 741 p-amp shwn in figure A-4. The standard prcedure n DIP (dual in-line package) "chips" is t identify pin 1 with a ntch in the end f the chip package. The ntch always separates pin 1 frm the last pin n the chip. In the case f the 741, the ntch is between pins 1 and 8. Pin 2 is the inverting input, V N. Pin 3 is the nn-inverting input, V P, and the amplifier utput, V O, is at pin 6. These three pins are the three terminals that nrmally appear in an p-amp circuit schematic K.A. Cnnr and Susan Bnner 2 f 18 Revised: 1/16/2006

diagram. The ±V CC cnnectins (7 and 4) MUST be cmpleted fr the p-amp t wrk, althugh they usually are mitted frm simple circuit schematics t imprve clarity. Figure A-4 The null ffset pins (1 and 5) prvide a way t eliminate any ffset in the utput vltage f the amplifier. The ffset vltage (usually dented by Vs) is an artifact f the integrated circuit. The ffset vltage is additive with V O (pin 6 in this case). It can be either psitive r negative and is nrmally less than 10 mv. Because the ffset vltage is s small, in mst cases we can ignre the cntributin V OS makes t V O and we leave the null ffset pins pen. Pin 8, labeled "NC", has n cnnectin t the internal circuitry f the 741, and is nt used. Op-amp limitatins: Just like all real circuit elements, p-amps have certain limitatins which prevent them frm perfrming ptimally under all cnditins. The ne yu are mst likely t encunter in this class is called saturatin. An p amp becmes saturated if it tries t put ut a vltage level beynd the range f the pwer surce vltages, ± V CC, Fr example, if the gain tries t drive the utput abve 15 vlts, the p-amp is nt supplied with enugh pwer t get it that high and the utput will cut ff at the mst it can prduce. This is never quite as high as 15 vlts because f the lsses inside the p-amp. Anther cmmn limitatin is amunt f current an p-amp can supply. Large demands fr current by a small lad can interfere with the amunt f current available fr feedback, and result in less than ideal behavir. Als, because f the demands f the internal circuitry f the device, there is nly s much current that can pass thrugh the pamp befre it starts t verheat. A third limitatin is the delay in the feedback lp f the p-amp circuit. It takes a while fr the p-amp t re-stabilize itself if there is a fundamental change in the input. This delay is called the slew rate. Delays caused by the slew rate can prevent the p-amp circuit frm displaying the expected utput instantaneusly after the input changes. The final cautin we have abut p-amps is that the equatins fr p amps are derived using the assumptin that an p-amp has infinite intrinsic (internal) gain, infinite input impedance, zer current at the inputs, and zer utput impedance. Naturally these assumptins cannt be true, hwever, the design f real p-amps is clse enugh t the assumptins that circuit behavir is clse t ideal ver a large range. K.A. Cnnr and Susan Bnner 3 f 18 Revised: 1/16/2006

The Inverting Amplifier: Figure A-5 shws an inverting amplifier. Its behavir is gverned by the fllwing equatin: Figure A-5 V ut Rf V Rin =. The negative sign indicates that the circuit will invert the signal. (When yu invert a signal, yu switch its sign. This is equivalent t an180 degree phase shift f a sinusidal signal.) The circuit will als amplify the input by Rf/Rin. Therefre, the ttal gain fr this circuit is (Rf/Rin). Nte that mst p-amp circuits invert the input signal because pamps stabilize when the feedback is negative. Als nte that even thugh the cnnectins t V+ and V- (± V CC ) are nt shwn, they must be made in rder fr the circuit t functin in bth PSpice and n yur prtbard. The Nn-inverting Amplifier: Figure A-6 shws a nn-inverting amplifier. Its behavir is gverned by the fllwing equatin: V ut Rf = 1 + Vin. Rin in Figure A-6 This circuit multiplies the input by 1+(Rf/Rin) and, unlike mst p-amp circuits, its utput is nt an inversin f the input. The verall gain fr this circuit is, therefre, 1+(Rf/Rin). The inverting amplifier is mre cmmnly used than the nn-inverting amplifier. That is why the smewhat dd term nninverting is used t describe an amplifier that des nt invert the input. If yu lk at the circuits, yu will see that in the inverting p-amp, the chip is cnnected t grund, while in the nn-inverting amplifier it is nt. This generally makes the inverting amplifier behave better. When used as a DC amplifier, the inverting amp can be a pr chice, since its utput vltage will be negative. Hwever, fr AC applicatins, inversin des nt matter since sines and csines are psitive half the time and negative half the time anyway. K.A. Cnnr and Susan Bnner 4 f 18 Revised: 1/16/2006

Experiment The Inverting Amplifier In this part f the experiment, we will wire a very simple p-amp circuit using PSpice and lk at its behavir. Wire the circuit shwn in figure A-7 belw in PSpice. Figure A-7 The input shuld have 100mV amplitude, 1k hertz and n DC ffset. The p-amp is called ua741 and is lcated in the EVAL library. Be careful t make sure that the + and inputs are nt switched and that the tw DC vltage supplies have ppsite signs. Nte the lcatin f the input vltage marker. The input t any p-amp circuit ges at Vin which will always be t the left f the input cmpnent(s). In this case, R2 is the input resistr, Rin, s the marker ges t its left. Run a transient simulatin f this circuit that displays three cycles. What des the equatin fr this type f circuit predict fr its behavir? Use the cursrs t mark the amplitudes f the input and utput f the circuit. Calculate the actual gain n the circuit. Is this clse t the gain predicted by the equatin? Print ut this plt and include it with yur reprt. Run a transient f the circuit with a much higher input amplitude. Change the amplitude f the surce t 5V and rerun the simulatin. What des the equatin predict fr the behavir this time? Des the circuit display the utput as expected? What happened? Use the cursrs t mark the maximum value f the input and utput f the circuit. What is the magnitude f the utput f the circuit at saturatin? Print ut this plt and include it with yur reprt. Build an Inverting Amplifier In this part f the circuit, yu will build an inverting amplifier. Build the inverting p-amp circuit in figure A-7 n yur prtbard. K.A. Cnnr and Susan Bnner 5 f 18 Revised: 1/16/2006

Dn t neglect t wire the DC pwer vltages at pins 4 and 7. D nt cnnect pin 4 and 7 t grund. They g thrugh the pwer supply t grund. D nt frget t set bth the psitive and negative values n the DC pwer supply. One des nt autmatically set when yu set the ther. D nt frget t attach the cmmn grund fr the pwer supply vltages t the grund fr the circuit as a whle. Examine the behavir f yur circuit. Take a picture with the Agilent sftware f the input and utput f the circuit at 1K hertz and 100mV amplitude and include it in yur reprt. What was the gain f yur circuit at this amplitude and frequency? [Use the signals t calculate the gain, nt the values f the resistrs.] Increase the amplitude until the p-amp starts t saturate. At abut what input amplitude des this happen? What is the magnitude f the utput f the circuit at saturatin? Hw des this cmpare with the saturatin vltage fund using PSpice? Summary As lng as ne remains aware f sme f their limitatins, p-amp circuits can be used t perfrm many different mathematical peratins. That is why cllectins f p amp circuits have been used in the past t represent dynamic systems in what is called an analg cmputer. There are sme very gd pictures f analg cmputers and ther cmputers thrugh the ages at H. A. Layer s Mind Machine Web Museum. The link is lcated n the curse links page. Part B Vltage Fllwers Backgrund The Vltage Fllwer: The p-amp cnfiguratin in figure B-1 is called a vltage fllwer r buffer. Nte that the circuit abve has n resistance in the feedback path. Its behavir is gverned by the equatin: V =. ut V in Figure B-1 If ne cnsiders nly the equatin V ut = Vin, this circuit wuld appear t d nthing at all. In circuit design, hwever, vltage fllwers are very imprtant and extremely useful. What they allw yu t d is cmpletely separate the influence f ne part f a circuit frm anther part. The circuit supplying Vin will see the buffer as a very high impedance, and (as lng as the impedance f the input circuit is nt very, very high), the buffer will nt lad dwn the input. (This is similar t the minimal effect that measuring with the scpe has n a circuit.) On the utput side, the circuit sees the buffer as an ideal surce with n internal resistance. The magnitude and frequency f this surce is equal t Vin, but the pwer is supplied by ± V CC. The vltage fllwer is a cnfiguratin that can serve as an impedance matching device. Fr an ideal p- K.A. Cnnr and Susan Bnner 6 f 18 Revised: 1/16/2006

amp, the vltages at the tw input terminals must be the same and n current can enter r leave either terminal. Thus, the input and utput vltages are the same and Z in = V in /I in. In practice Z in is very large which means that the vltage fllwer des nt lad dwn the surce. Experiment A Vltage Fllwer Applicatin In this part, we will investigate the usefulness f a vltage fllwer using PSpice Begin by creating the circuit pictured in figure B-2 belw in PSpice. Figure B-2 The surce has amplitude f 100mV and a frequency f 1K Hz. R1 is the impedance f the functin generatr R2 and R3 are a vltage divider and R4 is the lad n the vltage divider. Run a simulatin that displays three cycles f the input. Run the simulatin, mark the amplitude f the vltages shwn, and print the plt fr yur reprt. If we cmbine R3 and R4 in parallel, we can demnstrate that the amplitude f the utput is crrect fr this circuit. What if ur intentin when we built this circuit was t have the input t the 100 hm resistr be the utput f the vltage divider? Ie. We want the vltage acrss the lad (R4) t be ½ f the input vltage. Clearly the relatinship between the magnitudes f the 100 hm resistr and the 1K hm resistr in the vltage divider will nt let this ccur. A vltage fllwer is needed. Mdify the circuit yu created by adding an p-amp vltage fllwer between R3 and R4, as shwn in figure B-3 n the next page: K.A. Cnnr and Susan Bnner 7 f 18 Revised: 1/16/2006

Figure B-3 The p-amp is called ua741 and is lcated in the EVAL library. Be careful t make sure that the + and inputs are nt switched and that the tw DC vltage supplies have ppsite signs. Rerun the simulatin Place vltage markers at the three lcatins shwn. Rerun the simulatin, mark the amplitude f the vltages shwn, and print the plt fr yur reprt. What is the vltage acrss the 100 hm lad nw? Have we slved ur prblem? The vltage fllwer has islated the vltage divider electrically frm the lad, while transferring the vltage at the center f the vltage divider t the lad. Because every piece f a real circuit tends t influence every ther piece, vltage fllwers can be very handy fr eliminating these interactins when they adversely affect the intended behavir f ur circuits. It is said that the vltage fllwer is used t islate a signal surce frm a lad. Frm yur results, can yu explain what that means? Vltage fllwers are nt perfect. They are nt able t wrk prperly under all cnditins. T see this, change R4 t 1 hm. Rerun the simulatin, mark the amplitude f the vltages shwn, and print the plt fr yur reprt. What d yu bserve nw? Can yu explain it? Refer t the spec sheet fr the 741 p amp n the links page. Hw have we changed the current thrugh the chip by adding a smaller lad resistance? Finally, it was nted abve that the input impedance f the vltage fllwer shuld be very large. Determine the input impedance by finding the rati f the input vltage t the input current fr the fllwer. Return the value f R4 back t the riginal 100 hms. Recall that R=V/I. We can btain the vltage we need by placing a vltage marker at the nninverting input (U1:+) f the p-amp. PSpice will nt allw us it place a current marker at the psitive p amp input. We can find the current anyway by finding the difference between the current thrugh R2 and R3. Place a current marker n R2 and anther n R3. Set up an AC sweep fr the circuit frm 1 t 100k Hz. K.A. Cnnr and Susan Bnner 8 f 18 Revised: 1/16/2006

Frm yur AC sweep results, add a trace f V(U1:+)/(I(R2)-I(R3)). (Nte that yur vltage divider resistrs might have different names if yu placed them n the schematic in a different rder.) Include this plt in yur reprt. What is the input impedance f the p-amp in the vltage fllwer at lw frequencies? (Since PSpice tries t be as realistic as pssible, yu shuld get a large but nt infinite number.) Run the sweep again frm 100K Hz t 100 Meg Hz. Is the input impedance still high at very high frequencies? (Nte M is mega and m is milli in PSPice vltage displays.) Summary The vltage fllwer is ne f the mst useful applicatins f an p-amp. It allws us t islate a part f a circuit frm the rest f the circuit. Circuits are typically designed as a series f blcks, each with a different functin. The utput f ne blck becmes the input t the next blck. Smetimes the influence f ther blcks in a circuit prevents ne blck frm perating in the way we intended. Adding a buffer can alleviate this prblem. Part C Integratrs and Differentiatrs Backgrund Ideal Differentiatr: Figure C-1 shws an ideal differentiatr. Its behavir is gverned by the fllwing dvin equatin: V ut = RfCin. dt Figure C-1 The utput f this circuit is the derivative f the input INVERTED and amplified by Rf Cin. Fr a sinusidal input, the magnitude f the gain fr this circuit depends n the values f the cmpnents and als the input frequency. It is equal t (ω Rf Cin). The circuit will als cause a phase shift f -90 degrees. It is imprtant t remember that there is an inversin in this circuit. Fr instance, if the input is sin(t), then yu wuld expect the utput f a differentiatr t be +cs(t) (a +90 degree phase shift). Hwever, because f the inversin, the utput phase f this circuit is -90 degrees (+90-180). Als nte that, because ne cannt build a circuit with n input resistance, there is n such thing as an ideal differentiatr. A real differentiatr differentiates nly at certain frequencies. This distinctin is discussed in the pwer pint ntes fr the curse. Ideal Integratr: The circuit shwn belw in figure C-2 is an ideal integrating amplifier. Its behavir is 1 gverned by the fllwing equatin: V ut = Vin dt. Rin Cf K.A. Cnnr and Susan Bnner 9 f 18 Revised: 1/16/2006

Figure C-2 The utput f this circuit is the integral f the input INVERTED and amplified by 1/(Rin Cf). Fr a sinusidal input, the magnitude f the gain fr this circuit depends n the values f the cmpnents and als the input frequency. It is equal t 1/ (ω Rin Cf). The circuit will als cause a phase shift f +90 degrees. It is imprtant t remember that there is an inversin in this circuit. Fr instance, if the input is sin(t), then yu wuld expect the utput f an integratr t be -cs(t), a -90 degree phase shift. Hwever, because f the inversin, the utput phase shift f this circuit is +90 degrees (-90+180). Als, because the integratin f a cnstant DC ffset is a ramp signal and there is n such thing as a real circuit with n DC ffset (n matter hw small), wiring an ideal integratr will result in an essentially useless circuit. A Miller integratr is an ideal integratr with an additinal resistr added in parallel with Cf. It will integrate nly at certain frequencies. This distinctin is discussed in the pwer pint ntes fr the curse. Experiment Using an p-amp circuit t integrate an AC signal In this sectin, we will bserve the peratin f a Miller integratr n a sinusid. Yu will examine the way in which the prperties f the integratr change bth the amplitude and the phase f the input. Build the integratin circuit shwn belw in figure C-3. V1 shuld have a 100mV amplitude and 1K Hz frequency. Figure C-3 K.A. Cnnr and Susan Bnner 10 f 18 Revised: 1/16/2006

Run a transient analysis. We want t set up the transient t shw five cycles, but we als want t display the utput starting after the circuit has reached its steady state. Set the run time t 15ms, the start time t 10ms, and the step size t 5us. Obtain a plt f yur results. Just like in mathematical integratin, integratrs can add a DC ffset t the result. Adjust yur utput s that it is centered arund zer by adding a trace that adds r subtracts the apprpriate DC value. After yu have dne this, mark the amplitude f yur input and utput with the cursrs. Print this plt and include it in yur reprt. Use the equatins fr the ideal integratr t verify that the circuit is behaving crrectly. The equatin that gverns the behavir f this integratr at high frequencies is given by: 1 1 if ω c >> then vut ( t) vin ( t) dt R C R C 2 2 Recall that the integratin f sin(ωt) = (-1/ω)cs(ωt). Therefre, the circuit attenuates the integratin f the input by a cnstant equal t -1/ωR1C2. The negative sign means that the utput shuld als be inverted. What is there abut the transient respnse that tells yu that the circuit is wrking crrectly? Is the phase as expected? The amplitude? Abve what frequencies shuld we expect this kind f behavir? Nw we can lk at the behavir f the circuit fr all frequencies. D an AC sweep frm 100m t 100K Hz. Add a secnd plt and plt the phase f the vltage at Vut. (Either add a p(vut) trace r add a phase marker at the utput f the circuit.) What shuld the value f the phase be (apprximately) if the circuit is wrking mre-r-less like an integratr. Mark the regin n the plt where the phase is within +/- 2 degrees f the expected value. Print this plt Yu will mark this sweep with the data frm the circuit that yu build. We can als use PSpice t check the magnitude t see when this circuit acts best as an integratr. Rerun the sweep. D nt add the phase this time. Using the equatin abve, we knw that at frequencies abve fc, V ut = -V in / (ωrc), where R = R1, C = C2, and ω=2πf. [We plt the negatin f the input because the equatin fr the transfer functin f the circuit has an inversin. In a sweep, nly the amplitude matters, s the sign is nt imprtant.] Change the plt fr the AC sweep f the vltage t shw just V ut and -V in / (ωr1c2). Nte that yu need t input the frequency ω as 2*pi*Frequency in yur PSpice plt. (Pspice recgnizes the wrd pi as the value f π and the wrd Frequency as the current input frequency t the circuit. Als nte that yu must enter numbers fr R1 and C2..) When are these tw signals apprximately equal? It is at these frequencies that the circuit is acting like an integratr. Mark the pint at which the tw traces are within 100mV f each ther. Calculate fc=1/(2πr2c2). Hw clse are the amplitudes f the tw signals at that frequency? At a frequency much greater than fc, the circuit shuld start behaving like an integratr. Mark the crner frequency n yur plt. Print this plt. 1 2 Using an p-amp integratr t integrate a DC signal Anther way t demnstrate that integratin can be accmplished with this circuit is t replace the AC surce with a DC surce and a switch. K.A. Cnnr and Susan Bnner 11 f 18 Revised: 1/16/2006

Figure C-4 Mdify yur circuit by replacing the AC surce with a DC surce and a switch as shwn in figure C-4. Nte that the switch is set t clse at time t=0.01 sec. Use a vltage f 0.1 vlts t avid saturatin prblems. The switch is called Sw_tClse and is in the EVAL library. Analyze the circuit with PSpice. D a transient analysis fr times frm 0 t 50ms with a step f 10us. Rather than pltting the utput vltage (vltage at Vut), plt the negative f the utput vltage. Yu shuld see that this circuit des seem t integrate reasnably well. Print this plt. Hw clse is the utput f yur circuit t an integratin f the input? The integratin f a cnstant shuld be a ramp signal f slpe equal t the cnstant. The utput f an integrating p-amp circuit shuld be the inversin f the ramp signal multiplied by a cnstant equal t (1/(R1C2)). Nte that since the input is a cnstant, the utput des nt depend n the input frequency. Calculate the apprximate slpe f the utput. Write it n yur utput plt. Als write the theretical slpe n the plt. Fr what range f times des it integrate reasnably well? (This is smewhat subjective.) Mdify the feedback capacitr Decrease C2 t 0.01µF and repeat the simulatin. Only run it frm 0 t 14ms this time. Dn t frget t plt the negative f the utput vltage. Print yur utput. Mark the theretical slpe n the plt. Calculate the theretical slpe f the utput. Dn t frget that the cnstant, (1/(R1C2)), is different because C2 has changed. Des the circuit integrate -- even apprximately -- fr any perid f time? Can yu think f any reasn why we might prefer t use a smaller capacitr in the feedback lp, even thugh the circuit des nt integrate as well ver as lng a perid f time? K.A. Cnnr and Susan Bnner 12 f 18 Revised: 1/16/2006

Create an ideal integratr The circuit we have been lking at is a Miller integratr. An ideal integratr des nt have an extra resistr in the feedback path. What wuld happen if we changed ur circuit t an ideal integratr? Set the feedback capacitr back t its initial value f 1uF. Remve the resistr frm the feedback lp and run yur transient analysis again. Yu shuld see that the circuit n lnger wrks. Negate the utput vltage again. Print yur utput. What is wrng with the utput? The ideal integratr circuit will perate n bth the AC and DC inputs. In any real circuit -- n matter hw gd yur equipment is nise will create a small variable DC ffset vltage at the inputs. The prblem with this circuit is that there is n DC feedback t keep the DC ffset at the input frm being integrated. Therefre, the utput vltage will cntinuusly increase and, in additin, it will be amplified by the full intrinsic gain f the p-amp. This immediately saturates the p-amp. Building an p-amp integratr and an p-amp differentiatr In this part f the experiment, we will build an p-amp integratr and an p-amp differentiatr n the prtbard and lk at the utput fr a variety f inputs. Build the p-amp integratr circuit as shwn in figure C-3. Observe the behavir f the circuit at three representative frequencies. Use the sine wave frm the functin generatr fr the vltage surce, set the amplitude t 0.1 V. Obtain measurements f the input and utput vltages at frequencies f 10Hz, 100Hz, and 1kHz. Add yur experimental pints fr bth the amplitude and phase t yur PSpice AC sweep plt fr the abve circuit. Obtain a picture f each f these signals with the Agilent Intuilink sftware. Observe the utput f the integratr fr different types f inputs Set the functin generatr t a frequency that gives a reasnable signal amplitude and integrates fairly well. This is smewhat subjective, we just want yu t see the shapes f the utputs fr different input wave shapes. Set the functin generatr t the fllwing types f inputs: 1. sine wave 2. triangular wave 3. square wave What shuld the integratin f each f these types f inputs be? Take a picture f the utput fr each input with the Agilent sftware. Create a differentiatr. R3 1k V2 VOFF = 0 VAMPL = 0.1v FREQ = 1k V1 R1 50 C1 1u V U1 3 + 2 - ua741 7 V+ V- 4 9 OS2 OUT OS1 V3 0 5 6 1 V R4 1k 0 Figure C-5 9 K.A. Cnnr and Susan Bnner 13 f 18 Revised: 1/16/2006

Remve the feedback capacitr, C2. Replace R1 with an input capacitr, C1=1µF. Replace the 100K feedback resistr with a 1K resistr, R3. Yur circuit shuld nw lk like figure C-5. Set the functin generatr t a frequency that gives a reasnable signal amplitude and differentiates fairly well. This is smewhat subjective, we just want yu t see the shapes f the utputs fr different input wave shapes. Observe the utput f the differentiatr fr different types f inputs Set the functin generatr t the fllwing types f inputs: 1. sine wave 2. triangular wave 3. square wave What shuld the differentiatin f each f these types f inputs be? Take a picture f each situatin with the Agilent sftware. Summary Op-amp circuits can be used t d bth integratin and differentiatin. The ideal versins f bth circuits are nt realizable. Therefre, the real versins f these circuits d nt wrk well at all frequencies. Als, as bth types f circuits apprach ptimal mathematical perfrmance, the amplitude f the utput decreases. This makes designing an integratr r a differentiatr a trade-ff between the desired mathematical peratin and signal strength. Part D Using Op-Amps t Add and Subtract Signals Backgrund Op-Amp Adders: Figure D-1 belw shws an adder. Its behavir is gverned by the fllwing equatin: V1 V 2 V ut = Rf +. R1 R2 Figure D-1 The gain fr each input t the adder depends upn the rati f the feedback resistance f the circuit t the value f the resistr at that input. The adder is smetimes called a weighted adder because it prvides a means f multiplying each f the inputs by a separate cnstant befre adding them all tgether. It can be used t add any number f inputs and multiply each input by a different cnstant. This makes it useful in applicatins like audi mixers. K.A. Cnnr and Susan Bnner 14 f 18 Revised: 1/16/2006

The Differential Amplifier: The circuit in figure D-2 is a differential amplifier, als called a difference Rf amplifier. Its behavir is gverned by the fllwing equatin: V ut = ( V1 V 2). Rin Figure D-2 It amplifies the difference between the tw input vltages by Rf/Rin, which is the verall gain fr the circuit. Nte that the ability f this amplifier t effectively take the difference between tw signals depends n the fact that it uses tw pairs f identical resistances. Als nte that the signal that is subtracted ges int the negative input t the p-amp. Be careful with the term differential. In spite f its similarity t the term differentiatin, the differential amplifier des nt differentiate its input. Amplifying the utput f a bridge circuit: Yu may recall frm Experiment 4, that it was difficult t measure the AC vltage acrss the utput f the bridge circuit because bth f the utput cnnectins had a finite DC vltage. Withut a special prbe, the black leads f the scpe are always attached t grund. That meant that ne culd nt just cnnect ne f the scpe channels acrss the utput, since the scpe wuld shrt ne f the vltages t grund. The differential amplifier allws us t get by this prblem, since neither input is grunded. A very large fractin f measurement circuits use sme kind f a bridge cnfiguratin r are based n sme kind f cmparisn between tw vltages. Thus, the peratin f the differential amplifier is very imprtant t understand fr prject 2. Experiment PSpice simulatin f an adder Set up the circuit shwn in figure D-3 in PSpice. Nte the plarity f the vltage surces prviding +/- Vcc t the p-amp. Nte that the amplitude f the surce, V3, is 2V and that the amplitude f the surce, V4, is 1V. Run a simulatin that shws 3 cycles f the input. What is the gain f the adder? What shuld it d t the tw input signals? Is the adder wrking crrectly? Include this simulatin with yur reprt. Adders are ften used as mixers that give different emphasis t each input signal and then cmbine the inputs tgether int ne signal. What wuld we have t set R2 t, if we wanted twice as much f the signal frm V3 t pass thrugh the adder as the signal frm V4. [Nte: This des nt mean changing K.A. Cnnr and Susan Bnner 15 f 18 Revised: 1/16/2006

nthing because V3 already has twice the amplitude as V4. It means mixing in twice as much f the amplitude f V3 as the amplitude f V4 int the final utput signal.] Mdify resistr R2, rerun the simulatin, and verify that the utput f the signal is as expected. Include the utput f the simulatin in yur reprt. Summary Figure D-3 In this experiment we used an adder t add tw signals. Then, we mdified it s that it wuld cmbine the signals with different emphasis, as in an audi mixer. Reprt and Cnclusins The fllwing shuld be included in yur written reprt. Everything shuld be clearly labeled and easy t find. Partial credit will be deducted fr pr labeling r unclear presentatin. Part A (10 pints) Include the fllwing plts: 1. PSpice transient f inverting amplifier with input amplitude f 100mV and bth traces marked. (1 pt) 2. PSpice transient f inverting amplifier with input amplitude f 5V and bth traces marked. (1 pt) 3. Agilent picture f inverting amplifier circuit (1 pt) Answer the fllwing questins: 1. What is the theretical gain f yur inverting amplifier? What gain did yu find with PSpice when the input amplitude was 100mV? Hw clse are these? (2 pt) 2. What was the actual gain yu gt fr the inverting amplifier yu built? Hw did this cmpare t the theretical gain? t PSpice? (2 pt) 3. What value did yu get fr the saturatin vltage f the 741 p-amp in PSpice? What value did yu get fr the saturatin vltage f the real p-amp in yur circuit? Hw d they cmpare? (2 pt) 4. At what input vltage did the p-amp in the amplifier yu built n the prtbard begin t saturate? (1 pt) K.A. Cnnr and Susan Bnner 16 f 18 Revised: 1/16/2006

Part B (10 pints) Include the fllwing plts 1. PSpice transient f vltage divider with 100 hm lad and n vltage fllwer (1 pt) 2. PSpice transient f vltage divider with 100 hm lad and vltage fllwer (1 pt) 3. PSpice transient f vltage divider with 1 hm lad and vltage fllwer (1 pt) 4. PSpice AC sweep f input impedance fr the vltage fllwer. (2 pt) Answer the fllwing questins: 1. Cmpare the transients f the utput with and withut the buffer circuit in place. What is the functin f the buffer circuit? (2 pt) 2. Why is the fllwer unable t wrk prperly with a small lad resistr? (1 pt) 3. What is the typical value f the input impedance f the vltage fllwer when it is wrking prperly at lw frequencies? (1 pt) 4. Is the magnitude f the input impedance f the vltage fllwer high enugh at high frequencies fr it t wrk effectively? (1 pt) Part C (38 pints) Include the fllwing plts: 1. PSpice transient plt f integratr. (1 pt) 2. AC sweep f amplitude (with three experimental pints marked) and phase (with three experimental pints marked.) The frequency at which the phase gets clse t ideal shuld als be marked. (3 pt) 2. AC sweep plt f integratr vltage and -Vin/ωRC with the lcatin f fc and the place where the vltage gets clse t ideal indicated. (2 pt) 3. PSpice plts f integratr with DC surce with slpe and theretical slpe (if any) indicated n plt. One shuld be when C2=1uF and the ther fr C2=0.01uF (2 plts) (2 pt) 4. PSpice plt f ideal integratr (withut feedback resistr) (1 pt) 5. Agilent Intuilink pictures f yur circuit trace (input vs. utput) at 10 Hz, 100 Hz and 1K Hz. (3 plts) (3 pt) 6. Agilent Intuilink pictures f yur integratr utput with sine wave, triangular wave and square wave inputs (input vs. utput) (3 plts) (3 pt) 7. Agilent Intuilink picture f yur differentiatr utput with sine wave, triangular wave and square wave inputs (input vs. utput) (3 plts) (3 pt) Answer the fllwing questins: 1. Using the rules fr analyzing circuits with p amps, derive the relatinship between V ut and V in fr the integratr circuit. (3 pt) 2. Why is the integratr als called a lw-pass filter? Take the limits f the transfer functin at high and lw frequencies t demnstrate this. (3 pt) 3. What are the features f the AC sweep and transient analysis f an integratr that shw it is wrking mre-r-less as expected accrding t the transfer functin? Fr abut what range f frequencies des it act like an inverting amplifier?...like an integratr? (3 pt) 4. Cnsider the phase shift and the change in amplitude f the utput in relatin t the input when the circuit is behaving like an integratr. Use the expected change in phase and amplitude (frm the ideal equatin) t demnstrate that the circuit is actually integrating. (3 pt) 5. Why wuld we prefer t use the 0.01uF capacitr in the feedback lp even thugh the circuit des nt integrate quite as well ver as large a range? (1 pt) 6. What happens when we try t use an ideal integratr? (1 pt) 7. In the hardware implementatin, yu used a square-wave input t demnstrate that the integratr was wrking apprximately crrectly. If it were a perfect integratr, what wuld the utput wavefrm lk like? Is it clse? (3 pt) 8. When we built the differentiatr, what did the utput wavefrm lk like fr the square-wave input? What did the differentiatr circuit utput lk like fr a triangular wave input? If it were a perfect differentiatr, what wuld the utput wavefrm lk like? Is it clse? (3 pt) K.A. Cnnr and Susan Bnner 17 f 18 Revised: 1/16/2006

Part D (10 pints) Include the fllwing plts: 1) Transient simulatin f the utput f the adder with bth input resistrs set t 1K. (1 pt) 2) Transient simulatin frm PSpice with R2 mdified. (1 pt) Answer the fllwing questins: 1) Demnstrate that the riginal adder circuit (plt D-1) wrks as expected. (3 pt) 2) Demnstrate that the mdified adder circuit (plt D-2) wrks as expected (3 pt). 3) Give an example f a system (electrical, mechanical, chemical r sme cmbinatin) with negative feedback and an example f a system with psitive feedback. (2 pt) Summary (12 pints) 1. Summarize key pints (8 pts) 2. Discuss mistakes and prblems (2 pts) 3. List member respnsibilities (2 pts) Ttal: 80 pints fr write up + 20 fr attendance = 100 pints Attendance: 3 classes (20 pints) 2 classes (10 pints) 1 class (0 pints) ut f 20 pssible pints Minus 5 pints fr each late. N attendance at all = N grade fr experiment. K.A. Cnnr and Susan Bnner 18 f 18 Revised: 1/16/2006