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# ELECTRONIC INSTRUMENTATION ENGR 4300

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This 35 page Class Notes was uploaded by Bruce Lowe II on Monday October 19, 2015. The Class Notes belongs to ENGR 4300 at Rensselaer Polytechnic Institute taught by Staff in Fall. Since its upload, it has received 36 views. For similar materials see /class/224821/engr-4300-rensselaer-polytechnic-institute in Engineering and Tech at Rensselaer Polytechnic Institute.

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Electronic Instrumentation ENGR 4300 Spring 2000 Section Experiment 3 Introduction to AC Circuits Purpose In the following exercises we will investigate the properties of the other two passive circuit components inductors and capacitors Equipment Required Oscilloscope HP 54603B 2 Channel 60 MHZ Oscilloscope Function Generator HP 33120A 15 MHZ FunctionArbitrary Waveform Generator Instrumented Beam Battery Each group should bring a battery to class Parts Kits Background on the Simple Pendulum An Example of an Harmonic Oscillator Before we address capacitance and inductance we will review some of the properties of the simple pendulum The instrumented beam is a very good example of a simple pendulum even though it looks more like a small diving board Let us assume that the end of the beam moves in the xdirection Obviously this is a simplification since it really travels along the arc of a circle When the beam is stationary we will assume that it is horizontal and at x0 Again this is an approximation because the beam must bend downward slightly due to its own weight When the beam is bent it experiences a restoring force like a spring F kX where k is the spring constant From Newton s Law F ma m dVdt m dzxdt2 which when combined with the previous expression gives us 2 2 7 m d Xdt kx 7 0 the harmonic oscillator equation In standard form the harmonic oscillator equation is dzxdt2 032 x 0 where OJ is the frequency of oscillation Thus the beam will oscillate at n ldm12 21tf The solution to this equation is x x0 c0snt to where x0 is the initial de ection of the beam and 10 is the initial phase For simplicity there is no need to include 10 We have also not included any damping in this model Thus we get an oscillation that will go on forever rather than decaying slowly away If you do not recall that this is the solution plug the expression for x into the differential equation and you will see that it works K A Connor 1 Revised 7162010 Rensselaer Polytechnic Institute Troy New Y ark USA Electronic Instrumentation ENGR 4300 Spring 2000 Section The kinetic energy of the beam is KE 12 m v2 while the potential energy is PE 12 k x2 Upon initial de ection the energy of the beam is w PE 12 k x02 Since we have assumed no dissipation no friction or other damping force this total energy will be conserved and w 12 m v2 12 k x2 constant To summarize the pendulum or any harmonic oscillator works by exchanging energy between two different forms Not all forms of energy can be easily converted to another state and then back again but we know this is trivial with the kinetic and potential energy of a mass We can start at this expression of energy conservation to determine the equations of motion of the beam or any other simple pendulum Since the total energy is a constant we can take the time derivative of the entire expression and set it equal to zero dWdt 12 m 2 V dVdt 12 k 2 x dxdt 0 Since v dxdt we can write m dVdt V k x V 0 m dVdt k x which is the original equation of motion Thus once we have a conservation law we can use it even to find out how things change with time Part A The Instrumented Beam as a Harmonic Oscillator In the last experiment you should have measured the oscillation frequency of the unloaded beam using the strain gauge and the bridge circuit This time we will use the output from the coil at the moveable end of the beam as the magnet clamped to the end of the beam moves through it This signal is quite a bit larger than that from the bridge However it is not proportional to the position of the beam rather it is sensitive to the beam velocity Since the velocity and the position oscillate at the same frequency we can use either signal to find the frequency Measure the frequency again since you probably do not have the same beam Then measure the frequency two more times using additional masses of your choice Make sure that the mass is at least 50 grams More than 100 grams is even better There is a scale available in the studio to measure any mass you choose to add to the beam Write down the masses you used the corresponding frequencies and the location of the masses it is difficult to place the masses at exactly the same location in the table below The location should be measured from the pivot point a you check the frequency a couple of times since you should notice that there will be a range of values for the frequency K A Connor 2 Revised 7162010 Rensselaer Polytechnic Institute Troy New Y ark USA Electronic Instrumentation ENGR 4300 Spring 2000 Section primarily because of noise and the somewhat nonideal nature of the sinusoidal voltage For now the mass of the beam mbeam and its effective location leff are not determined so you will not put numbers in those two locations in the table Now we will use the information you have just obtained to calculate the spring constant and the equivalent mass of the beam With the three measurements you have made of frequency you should have more than enoug information to determine these two constants of the simple pendulum equation However the masses you have placed on the beam were not located at exactly the same location as the center of mass of the beam Since the spring constant is very sensitive to the length you will have to take it into account in your analysis From a variety Ewt 3 413 of references it is shown that k where E is Young39s Modulus w is the width tis the thickness and l is C Ewt3 the length of the beam We can write this expression generically as 161 T where C T is a constant for the beam Again the length l is the equivalent position of the mass To determine the spring constant the equivalent mass of the beam and the location of the equivalent mass rewrite the formula for the frequency as 161 2 7 21f mTutal where mTutal mbeam mzxtra Given this information we can write out an expression for each of the three frequencies you have measured k C3 mmmr 217 C k1 F mbeam mi2 f12 1 C k2 3 mbeam m22 f22 2 From the data you should have in the table you know the values of f0 f1 f2 ll 12 m1 and mi Use these numbers to determine the values of k0 leff and mbmn Finally measure the dimensions of the beam and decide what it is made of You will probably need the width thickness length and Young s modulus of the beam to figure out what material is used Report and Conclusions What are the values for k0 leff and mbeam that you obtained Do your results seem plausible Why What is the beam made of Write a possible form for the equation of motion of the beam that includes friction The solution to this equation should be a damped sinusoid That is our solution for x should be multiplied by e39m where on is the damping constant Determine the value of the damping constant from the scope trace you obtained for the unloaded beam K A Connor 3 Revised 7162010 Rensselaer Polytechnic Institute Troy New Y ark USA Electronic Instrumentation ENGR 4300 Spring 2000 Section Background on Harmonic Oscillators Made from Inductors and Capacitors There are many important lessons we can learn from the harmonic oscillator but perhaps one of the most useful is the value of conservation laws It is fair to say that the most powerful problem solving technique is to first decide which conservation laws hold Once the conservation laws are identified they can be used to determine a great deal of information about any system In passive electrical systems there are three kinds of circuit elements resistors capacitors and inductors Resistors turn electrical energy into heat When a current I flows through a resistor there will be a voltage drop V across the resistor The power dissipated by the resistor is equal to the product of I times V Since resistors produce heat it should be no surprise that they play the same role as friction in a mechanical system The ideal pendulum will oscillate forever a real pendulum will oscillate until all its stored energy is converted to heat through friction Thus if we wish to create a circuit analogous to the ideal harmonic oscillator it can have no resistors in it Rather we will combine only inductors and capacitors A typical inductor consists of a coil of wire If we pass a current through the coil a magnetic field will be created Many of us have made simple electromagnets at some time in our lives by wrapping wire around some magnetic material like a nail When a battery is connected to the wire it is possible to attract small pieces of iron to the nail The field created by the coil the magnetic field can do work and thus contains energy The energy stored in an inductor is given by the expression WM 12LI2 where we have used the subscript M to indicate that the energy is stored in the magnetic field and L is the inductance in units of Henries Joseph Henry was honored by using his name for this unit because of his early work in developing practical electromagnets He began his work here in New York s capital district when he was teaching at the Albany Academy There is a statue of this great scientist outside the original location of the Academy across the street from the Albany City Hall Henry was a contemporary of Amos Eaton the intellectual force behind the founding of RPI He eventually left Albany for Princeton and the Smithsonian A typical capacitor consists of two metal plates of large area separated by some insulating material such as teflon or some other plastic When a voltage source is connected to the plates charge flows from the source to the plates with positive charge deposited on the plate with the highest voltage and negative charge deposited on the plate with the lowest voltage Since these charges are opposite in sign and since unlike charges attract one another there is a force between the two plates Again the existence of this force tells us that we have to do work to charge up the plates and that there is energy stored In this case the energy is stored in the electric field created by the charges The energy stored by a capacitor is given by the expression wE 12C v2 where we have used the subscript E to indicate that the energy is stored in the electric field and C is the capacitance in units of Farads Michael Faraday also worked on electromagnets and being British gained much more fame for his work since America was a scientific backwater at the time Henry showed him how to make better magnets but Faraday s work was much more far reaching Henry also showed Morse how to build a telegraph Aside It is somewhat interesting to note that neitherHenries nor Farads turns out to be much ofa common practical unit One Henry is a huge inductor rarely seen in practice One Farad is also rare now occasionally seen in highly lteredpowersuppliesfor computers We will need to use the pre xes milli micro nano pico etc a lot when dealing with these components We also do not see one Ohm all thatmuch but require the other kind of pre xes kilo mega etc Consider the simplest possible configuration of a single capacitor and a single inductor connected as shown below Note that since there are only two components one can describe this connection as either in parallel or in series Also assume that the capacitor has been charged up to some voltage V at time t 0 at which time it is connected to K A Connor 4 Revised 7162010 Rensselaer Polytechnic Institute Troy New Y ork USA Electr uni Innnnmentztinn ENGRVASOO 39 Secdnn m mductax n hm m um m Wm camaan The current awmg Lhmugm the monster m then chug the capsular back up and the Th2 mm quotqu gvenby W 12 c V2 12 L 12 constant dWdt 12 c 2 V dVdt 12 L 2 I dIdt 0 L 011011 1 c dVdt V 0 whchwe an up mm L dILdtVL and c chdt 1c vL m11cu v xsmzwlugeatthempafthz cucutandhsthz Eunent awmg 1fwenateLhAtV 7v Amund the cumquot and deurmmzd empmca y Please see Chapter 2 sechansl ma amegxuh and the electmmc magnum an m cums webpage We can cambme the m cunen39nmltag xelahanshlpsfm the capsular andmducmx afaur mp1 mum ablam ddt c chdt ddt 1c dILdt VL erL PVCdc Lcquot Vc 0 we saw thh m pmamum Thus on LCm mm 5 Revsad 7152010 K A Comor Riv52110 Polytechnic 1mm Troy New 1m USA Electr uni Innnnmentztinn Spring 2000 ENGRVASOO Secdnn W W W 01 1mH U n rm 3 RC and RLC Camus iExxmples quc Camus can cansst af my cambmman uf R L Dr c as ascussaamchapm 2 ameyxch We wan begn wah a cambmauan we have seen befaxe an RC mun u u a am a wnaga dwmer cambxmum mLhe same manner as am can gmed wahwa xesxsmxs anm m Pmbe pm m aim math capac m39 vR v 5 Ta determe v1 place a wnaga m4 marker niche lap gum vauaga mm saute valtage Prim uul a phi ufyuur quotsuns alfmh af qese pm 39 these mm czses mm K A Connor 5 Revised 7152010 mth rotymm 1mm Troy New York USA Electronic Instrumentation ENGR 4300 Spring 2000 Section these voltages with respect to ground Modify your plot so that you are displaying the magnitude of the capacitor voltage divided by the magnitude of the source voltage Produce a plot your results This plot is the frequency response for this filter Now reverse the order of the resistor and capacitor in the circuit and take the output voltage across the resistor Perform the same AC sweep and determine What kind of a filter you get when you take the output voltage across the resistor Produce a plot of the frequency response for the second type of lter That is plot the magnitude of the output voltage divided by the magnitude of the input voltage Set up the low pass RC filter on a protoboard Connect the function generator as the input AC source Monitor the input and output voltages with the scope Measure the relative amplitude i Vom i i Vin i at the three frequencies 100 Hz 1 kHz and 10 kHz You cannot do this directly since this is not an option with the scope However you can measure the input and output voltages and then divide the latter by the former If you have time also do the measurement at 10 Hz This is such a low frequency that the display on the scope is somewhat harder to work with Add these experimentally measured points to the PSpice plots Repeat for the high pass RC filter If you are having any trouble figuring out Which configuration is which check the figures in Chapter 3 of Gingrich Where he discusses filters We will now consider an RLC circuit with all three kinds of passive components Here we have used two switches to change the input voltage from zero to 10 volts at time t 0 The switches are in the parts list Use PSpice to simulate the transient response of this circuit for a total time of lmsec Describe the voltages plotted by Probe What features of the voltages reminds you of the instrumented beam 0 R1 L1 1 X 2 AAA W J1 v WV 50 10mH 2 U2 C1 0 1 068uF Report and Conclusions I Write out the mathematical expressions for the three voltages for the first RC circuit case you considered I Determine the range of frequencies for Which the input voltage V1 is Within 5 of the output voltage either VR or Vc for the two configurations This tells you the range of frequencies that pass through the filter more orless unchanged Signals at other frequencies are attenuated by the filter I Determine the resonant frequency of the RLC circuit you analyzed with PSpice K A Connor 7 Revised 7162010 Rensselaer Polytechnic Institute Troy New Y ark USA E Iecmmic Instmmentztiun Spring 2000 ENGRVASOO s ecu un Pan c Eqmwlent 0mm er Sundard a auenes later un batta39y Please du nut The plan 15 m see er Rb R2 Hare we have eemee39ed a m n uhm resxslur armss me mammals e fa 9 van battery When yuu du um plase my a egn by wm all be m pamllel wme dawn the we vuuages here cheek wnhyuur webpage Ifyuur mBasuremEnt dues nut ages wnh what uthers have fuund yuu shuuld dB 0 War qun and Cnnclusinns Discuss me accuracy ufyuur Rb masurEmEnt Whatkmd ufbauary mdyeu study and what wine eer am yuu ub39am7 frequmcy measuremBnW Can yuu Lhmk Ufa bells Way m delermme Rm KA Connov a Riwstzd 7162010 RzmstzlawPolynzchnc mtmm My New Yovk USA Electronic Instrumentation ENGR 4300 Spring 2000 Section Experiment 3 Please list the names of all group members A TA or instructor will initial a participation box each class day you attend and participate in this experiment When you have completed all of the experimental and simulation activities have a TA or instructor initial under completed They should also look over what you have done to be sure that your results are useful If you are unable to attend class for any reason you can make up the work during an open shop time The maximum participation grade is 5 points Please answer any questions asked above under the Report and Conclusions sections Also attach any plots requested On each plot describe what is being displayed and why the results make sense Include a handdrawn or computer drawn circuit diagram for any PSpice output or plots of measurements indicating where and how the measurements were made Summarize the key points of this experiment Discuss any problems you encountered or mistakes you made and how you addressed them Names 1 2 3 4 Grade Out of 25 K A Connor 9 Revised 7162010 Rensselaer Polytechnic Institute Troy New Y ork USA Electronic Instrumentation ENGRV4300 Fall 2004 Section Experiment 6 Introduction to Diodes Voltage Limitation and Regulation Purpose We wrll by AC voltages on tlre load A Thls We Such lrrnrtauon wrll ofVoltages is usually applred to proteet erreurt eornponents Equrprnent Required HP 34401A Drgrtal Multrrneter HP 33120A 15 MHz Punetron Axbltxary Waveform Generator HP 546033 Oserlloseope HP E3631A Tnple Output DC Power Supply Prot Resrstors Capaertors Drodes Zener Drodes Benchllnk and OrCAD Capture and PSpree Background Please readpages 150 158 rn Essenee ofAnalog Eleetronres Pages 160190 snow praeueal applreauons oftlre drode that we wrll beusrng rn elass The parueular drode we wrll be usrng rn tlrrs expenrnentrs tlre N4148 Td ll l 1n praetree posruve voltage aeross tlrern reaenes a srnall tlrresnold They also have a srnall saturation eurrent and a breakdown 1n srnall srgnal dr de mA 1 V 100 V andtne t eurrent L l nA Power 15 V or so Tnereverse breakdown voltage ofpower drodes rnayrange from as low as 50V upto 1000 V or even rnuen rnore xLl anode P eatlrode Crreurt syrnbol Reversee t F d breakbown Saturauon eurrent1 orwar r l Reverseebras has re ron 7 g region region K A Comm 1 R2vlsed109ZOO4 Rensselaer Polytechnic Instlmte Troy New York USA Elem uni Innnnmentztinn Fall 2004 ENGRVASOO Sectinn and 2 me am Hi cuve butluves nut same details Ink reverse bxukdawn Junaum unnemnee etc PznALV Charactensucs hnewhase slap 15 equaltaR Thsxsxncantxastta39heLV chuactenmcafa ade Taseeune m xeastms as Shawn belaw ob serve the w Gunmenme Cme my 2 Ram39slur Draw une emu snan use v DC m an sauce m c spam v 1 W Runune munan Select AddTnce m nmune cunentthmughxesxsmx RI mu Recs um yvu mmt ennase Lhs cuxent mm me 1151 Wu see when yvu use Tncequot and men Add quot Change une mm afyaur pm asfa aws Rxght chck DnLhz pm and enaase Semng Tnen chck an me x axxs Tab Cluck an an Axxs v unable bunan at an banam Enter van 1 value dependent upan une sauce vultagz whnuh 15 menng rm 4 m 1 Wm Tne pm pmducedwx Shaw me Hi charac39znshc ranenw RI Yum PROEE pm snnmaxaak sanemngnke une ane mane fallvwmgpage K A Connor 2 Revued 792004 Rimmlnev Fobtacky lwmmm Troy New York USA Sectinn Electr uni Innnnmentztinn ENGRVASOO Fall 2004 xiv Charactensuc ufasnn Ohm Ream Observe the Channerislic Cum m 2 mm Ma fyyamPprceschemahcbyxeplacnglwnhDLADlNAIAdede Yauwx fmdums dude mtthans 1st Itxsxnthe EVAL PSPICEhbnry mm m samewayywu ammij Ch ethexraxxsafwurPROEEplataswudldmpmltaVDl 0170312 T xraxxstabethz wing acmssdmdeDl Theplalpmducedwx Shaw theer char m m prewaus secum his sets yum acunshc curve 39 yuurp m Uang m cusaxs mark at least 5 paints m uf Lhs pm Yau wm be Hang mm 5 paints m Ex 21 m help Wu pm the charactensuc clxve mm dude Chums yam um accuatelyrepusennhe fumes afyaur curve Assth an mm Cilcuil m nunquotth hm arm nd symbnl mm unmade Nhenywu Wu the mEmL make sue yum dude is puns sa mm unmade facestawazd gamd as Shawninth gue abvve Suppxymc sweep valtagetawur mm We camatacma ychaase c Sweep butaample a mmm wutputaf39hefuncum generalm39 Vm bycannzcung camectxtasthevnltagz In euuanmmenpmcw aflEIEIHz mamquotp amrlUV m pm 0 xltawur scape Selectaslwwrep mv K A Connor 3 Revued 1092004 Rimselnev Foly zchmc lwmmm Troy New York USA Electronic Instrumentation ENGR 4300 Fall 2004 Section Using the scope measure Vom the voltage at the end of the diode connected to the resistor R2 and the input voltage Take a sample in Excel of the input and output voltages Eliminate all points except for one upward ramp of the input voltage this represents a single DC sweep from 10V to HOV You can now use this data to plot the IV characteristic of the diode You need to plot the voltage across the diode against the current through the diode The voltage across the diode is equal to VinV0 VinVent on the scope as the difference between the triangle waves The current through the diode is VentR2 Now that you have a set of points corresponding to the behavior of the diode you can determine the equation which governs it see Part A question 1 Use the equation on the top of page 2 You already know the value of the constant VT and you have a data set for iD and VI You can pick a good guess for Is from your PSpice plot You will then have to guess values ofn between 1 and 2 and possibly modify Is slightly until the theoretical curve fits your data Include a copy of this plot in your writeup Part B Rectifiers D1 N VI D1N4148 AAA v v JIF In the circuit shown above the DC voltage source we used before is replaced by a sinusoidal source VSIN Such a circuit is called a rectifier Use PSpice to simulate the circuit above in which a diode is used as a rectifier with no smoothing I Draw the circuit shown Use VSIN for the source Use the following parameters for VSIN VAMPL 5 10V peak to peak VOFF 0 DC 0 and FREQ 60Hz Perform a transient analysis in increments of 200us up to 80ms Obtain a graph of the input and output voltages Vin V1 and Vom voltage across R vs time Print this plot Add a capacitor across the resistor Use PSpice to simulate the circuit below D1 N VI D1N4148 AAA v 0quot I Modify your PSpice circuit by adding capacitor C1 in parallel with R1 I Perform a transient analysis as you did in the previous part K A Connor 4 Revised 1092004 RensselaerPalytechm39c Institute Troy New Y ark USA Electronic Instrumentation ENGR 4300 Fall 2004 Section I Obtain a graph of the input and output voltages vs time Print this plot I Modify the frequency of the source to 1k Hz the time to lms and the step size to 01ms Generate a plot of the Vin and Vom Note the time at which the output voltage reaches 4V I Keeping the time and step size the same modify the frequency of the source to lMeg and rerun the simulation It may take a bit longer Note the time at which the output reaches 4V You will need to delete the other trace to see it What is happening as the frequency increases Hardware Implementation I Assemble the rectifier circuit without the capacitor Use the function generator to supply an input sinusoidal signal of 60Hz 10 volts peak to peak Reminder 7 Always connect the function generator to one channel of the scope so that you can be sure that you have the signal you need Also as a rule you should always connect your input signal to one channel of the scope and the output signal to the other channel To see the effect of a circuit it is essential to observe both the input and the output simultaneously I Observe both the input and the output voltages with the oscilloscope Be sure that your scope is set to DC coupling I Repeat the procedure above with a 47uF capacitor in parallel with the resistor R Note that the sine voltage from the generator will become distorted when you add the large capacitor This is due to the finite nonzero source impedance of the function generator which we have been ignoring so far in this experiment and the large variation in load seen during alternate halves of the cycle In a more critical circuit one might for example add a 50 ohm load resistor to the function generator For our purposes here we simply note the distortion 0 Measure approximately 5 points during one cycle and add the experimentally measured points to the plot generated with PSpice I Vary the frequency of the function generator and observe both the input and output voltage of the rectifier with the capacitor Try connecting and disconnecting the capacitor as you vary the frequency Part C PN Junction Voltage Limitation When you performed the triangular wave sweep of the 1N4l48 diode you should have observed that the voltage across the diode remained near 06 volts when it was on We can take advantage of this effect in the following circuit which permits small voltages to pass Without distortion but clips any voltage outside the range of about 06 to 06 volts gt4 gt pm rlt an lt 9 52 D1 2 D2 D1N4148 D1N4148 Dill Observe the diodes functioning as limiters or clippers 0 Draw the circuit shown Use VSIN for the source VAMPL 10 FREQ 1k everything else 0 Perform a transient analysis in increments of 50us up to 5ms Use PROBE to plot the input K A Connor 5 Revised 1092004 RensselaerPalytechnic Institute Troy New Y ark USA Elecmmic Instrumema nn ENGRasoo Fall 20m i Andm pmvnhxges Th mphnsm vuhxge smlrce whd 1h ampm mm acmss 39h am pm as Shawn mm u mom ph39 Change 39h mm huh smmahxmpu vm m n mu Rzpanhz tmmzm analysis mm m moan ph39 Anth I aim mm mm an ynulpxulzrhuud Us 39h mnangemmmxm supplyal kHz lEIVsmIm1da mpm Us 39h scwp ms mama vwnh 39h ascxllnscap m wm med m g 15 mm ymlrgmnrdsmanlymlz an Eecm dm sc p h 39h mph Rgpa h mummhcsmmumnmhmmmpm Dun39n39axgnmaddm m rpm palms m ymr Psm m Hulgnuld mama Ifyml recs h gure hhm shnwmg h w chmtznsnc Ufa am yml WI 522 um hrh am 5 smmzndymexse based h w Wham h m xevng mm Furthznnme h dud WM w nmmnafpmxmtzlycam am wldz range a cumms This pmpnyls hmwnasbxeakdnwn 1h bnakdnwnwlhge 5 mud 7v wlme 1h hm z mdlcahes that m 5 aim kmwn 5 h 2 WM um am an mm m wmk h h bnakdnwn m are I5uykmwn 1h h semlcandmmx mag charge Camus hm aha electmns m cammu ybemg Lhannallygemmzd whmh mm m h shun wing mdzpndzm reverse mam cnmm whzn h dud 5 reverse bused g c rs xpmme thanquotng Ifthls emxgyls 1mg emugh 39h caumaharh chm WA hanmn w gamma h hm 1w elzcrmn pk Th 21mm rumngxrchh aJsa pl elastmrs fmm u mhs am n beams large enmgh Bathpxmesses Increase 39h nnmhexafclmge cmmhmhhs Increase 39h hhhh anh am m carry a beaks cummmthz xevug anecch Eyappmprmz duping ms puss ulz m dzslgnahmxdln that dawn m hhywhm rmmh wwhs m h fewhnndredwhs ham hm ma 1 A m4 ulan Pm n Zena mag Valhge Rzgulmax KA Owner 5 mm my217m Ranssalaavl olymhnn mam Troy NM Yetg um Electronic Instrumentation ENGR 4300 Fall 2004 Section R1 A AAA B if D1 25 D1 N750 In the circuit shown above D1 is a Zener diode Be sure you have its orientation correct Obtain Zener diode characteristic 0 Draw the circuit shown Perform a DC sweep analysis from 10 to 20 volts in increments of 01 volts Be sure that you enter V1 for the name of the device that is swept Plot the current through the Zener diode IDl vs the voltage across the diode VCDl 2 to obtain the lV characteristic of the Zener diode in the same manner as we addressed the standard diode in the last experiment ie make the Xaxis variable VCDl 2 Note that your plot should look like a typical Zener diode characteristic If for some reason the current or the voltage look upside down or backwards reverse one or both of the signs until the plot looks correct Print this plot 0 Look up the typical Zener voltage in the spec sheet for the lN750A diode You can find a link to this spec sheet on the links page Draw a vertical line in the reverse bias region on your output plot corresponding to the rated Zener voltage Note that this diode will keep its reverse bias voltage quite close to the Zener voltage for a wide range of currents The smallest current for which the bias voltage is about equal to the Zener voltage is called the knee current What is the minimum current or knee current for which the reverse bias voltage is no more than 01 volts less than the rated zener voltage Mark the knee current on the plot Hardware Implementation Assemble the circuit on your protoboard Connect the DVM to measure the current through the diode Unlike volToge rheos uremenTs currenT quot1805 uremenTs are made in SERI E5 wiTh The circ uiT Use The D MM inpuTs labeled WITh The IeTTer I quot and seT it To meos ure current DC I mm The shift buTTon Adjust the DC source voltage until the current reads lmA 3mA 5mA Measure the voltage across the diode for this condition Repeat this for a current of lmA 3mA 5mA If you have questions about measuring the current PLEASE check with a TA Mark these six points on your PSpice output of the characteristic curve of the zener diode Report and Conclusions The following should be included in your written report Everything should be clearly labeled and easy to find Partial credit will be deducted for poor labeling and unclear presentation Part A Include the following plots 1 lV Characteristic curve PSpice Plot with 5 points marked 1 pt 2 Excel sample of characteristic curve taken using agilent software This should include the data points and a line found using the diode characteristic equation 1 pt Answer the following questions K A Connor 7 Revised 1092004 RensselaerPolytechnic Institute Troy New York USA Electronic Instrumentation ENGR 4300 Fall 2004 Section 1 Use the data you took for the IV characteristic of the lN4l48 diode to determine the mathematical representation of the IV curve What values did you find for Is and n 2 pts 2 Why do you know that the current through the diode is VentR2 1 pt 3 What differences if any did you notice between the IV characteristic given by PSpice and the one you measured experimentally 1 pt Part B Include the following plots 1 PSpice plot of rectifier 1 pt 2 PSpice plot of rectifier with smoothing with 5 experimentally obtained points marked 1 pt Answer the following questions 1 What is the function of a rectifier 1 pt 2 Explain why Vom changes when you add the capacitor in parallel with R Explain why this circuit would be better for use as a DC source than the circuit without the capacitor 1 pt 3 Did the circuit with the capacitor work better more like a DC source at high or low frequencies 1 pt 4 Based on your knowledge of the theoretical behavior of the capacitor at very low open and very high short frequencies how has the diode changed the expected behavior Why lpt Part C Include the following plots 1 PSpice plot of voltage limiter at 10V with 5 experimental points marked 1 pt 2 PSpice plot of voltage limiter at 01V with 5 experimental points marked 1 pt Answer the following questions 1 Why do the two plots look the way they do 1 pt 2 Why is this circuit called a limiter 1 pt 3 When and why might this circuit be useful lpt 4 Comment on the similarities and differences between the Pspice and experimental results 1 pt Part D Include the following plots 1 Zener diode characteristic curve with vertical line and knee current marked 1 pt Answer the following questions 1 What are the Zener voltage and knee current for the diode you simulated 1 pt 2 Shown below is the IV characteristic of the lN4l48 nonZener diode we looked at in part A but obtained over a much wider voltage range Compare this plot with the one you obtained for the Zener diode 1 pt K A Connor 8 Revised 1092004 RensselaerPalytechm39c Institute Troy New Y ark USA Electronic Instrumentation ENGR 4300 Fall 2004 Section n 1031 VDll 7 vDi2 3 We have seen that the voltage across the Zener diode will remain equal to the Zener voltage as long as we provide enough voltage and current from the source However the circuit configuration we have studied does not include a load A load resistor would be added in parallel to the Zener diode Discuss how this circuit will perform for a load that is much smaller than 1k ohm eg 50 ohms equal to 1k ohm and much larger than 1 k ohm eg 1 megohm That is under what conditions will it produce the desired regulated voltage Support your discussion with calculations simulations or experimental results 2 pts Summarize Key Points 1 pt Mistakes and Problems 05 pt Member Responsibilities 05 pt K A Connor 9 Revised 1092004 RensselaerPalytechm39c Institute Troy New Y ark USA Electr uni Innnnmentztinn ring 2000 See ENGReesoo Sp Exp ErimEnt 8 01 Amp Cnn gurztinns e Imegmnr Differ Entiztnr antzge Fnllnwer Puxpuse 1h lms expeumehe we wxll laak hum huhe mnde up mp ehh guehhhe that gve m Lb my hand mm m an amph ex and me change his perfmmance by hang a vahege rhuhweh am m e usedm the page represent dynamxc systems m whans called an arming camputex There are same very A Weh Mmeum hemp Huserwww sfsu eduNhlmmm hum Alsa review secums 5 2 1 and 5 2 2 an hammemhg and mvemng emph ehe aspecuvzly The hm Lb cvursewebpage Same huhe l nghhgmtsafthzse pegeeheveheehexumeahehw Flensewotefhnt both V m E we used to represent volmge We Wm he ehhseaehhg several hm up mp can gmmms The xsus the vahege humeh out m mph terminal must he the same and ha when can eh39eh ax leave anther terman Thus the mph and e Ema gt m y vallag rhuhweh aheehhnhea dawn the same The secmdxs Anxnvexhng emphnex symbolor youwi The mveneh wah wnege gem amph es the 519151 and changzs his g Em 5mm Rx Th m apexmhex emph ehe n 5 emy Shawn that the when vamge wx he K A Connor Revised 70212010 Rimming Polytecth 1mm Troy New York USA Electr uni Innnnmentztinn ring 2000 See ENGRVASOO Sp E 7R Z 4 usesme same valuest meRn m39he feedback paaum quhz mvenex mm Equipment Requued HP 3312mm MHz Funman Arbmary Wavefaxm G enemz HP mam Pawn swpxy HP SA EIZE 2 Channzl 6D MHz Oscdlascape Prambaud Same Resxsmxs andCspacnm39s 741 opeAmp Dr 1452 Dual opeAmp OxCAD Capture andPprce K A Connor Revised 7232010 mmmm mommy 1mm Troy New York USA Electronic Instrumentation ENGR4300 Spring 2000 See mmm u First r u a h thaluad resistur ubsememussmeluam RS an vs 3 RL mu Lmd Functinn Canaan In ths expmmenL we Wm be cunnecung the 1m m the funan generatur bum anecuy We Wm rst upman cme mun mure independent cme 1nd by addmg a vultage fulluwer vulume cuntxul Gama OprAmp Cunngmu39m K A 0mm 3 mm 72172010 Remxelaer Palyxedmx mum Troy New York USA E Iecmmic Instmmentztiun ENGRASw Spring2w0 anmge Funnwer PanA TheVul39zge Fulluwer a egn by emmaung me Circuit cunsxsung erme suurce me has up amp and me luad We Will add me vultage fulluwa nee Usea suumevultag en veu d2 39equency uflkHz DuanAC Sweep mm 1 Hz e e an m mu kHz Ob39amaplutuftheuulpulshuwxngth vultagesattheuutputuflhefuncnun gmemlur mm m n we quota n m a um luad resismr In um msethelatlerlvm veuages are cunneded duele meanm Tu seems an KA Connov 4 Riwstzd 7202010 RzmstzlawPolynzchnc mtmm My New 1 on USA Electronic Instrumentation ENGR 4300 Spring 2000 Sec working more or less as you would expect Now change the load resistor to 10 ohms and do the transient analysis again What do you observe now Can you explain it 02 E o n R E mu Finally it was noted above that the input impedance of the voltage follower should be very large Determine the input impedance by finding the ratio of the input voltage to the input current for the follower That is find the ratio of the voltage to the current for the noninverting input of this device Since PSpice tries to be a realistic as possible you should get a large but not infinite number You can obtain the voltage we need by placing a voltage marker near the noninverting input However it is not as easy to identify current Fortunately we can use what we know about basic circuits to figure it out The current we need is the difference between the current through R4 and R5 From your AC sweep results plot VU2lR4IR5 Note that your voltage divider resistors might have different names if you placed them on the schematic in a different order You will note that we are getting into the noise levels for the simulation so that the plot will be a bit ragged Results and Discussion I Compare the AC sweep outputs with and without the buffer circuit in place What is the function of the buffer circuit I Why is the follower unable to work properly with a small load resistor What is the typical value of the input impedance of the voltage follower when it is working properly Part B Integrator Now remove Cl from your circuit and replace it with a wire Also go back to the original 100 ohm load Repeat the transient analysis of this circuit and obtain a plot of your results You should observe that this circuit does work approximately as an integrator What is there about the transient response that tells you this If you have trouble answering this question you might want to do all the tasks in part B and then come back to it Now repeat the AC Sweep In addition to the voltages that you have been plotting also plot the phase of the voltage across the load What should the value of the phase be approximately if the circuit is to be working more or less like an integrator Hint 7 look at Gingrich You should observe that this circuit will work better if you lower the characteristic RC frequency by increasing C2 Try C2luF Repeat the simulation Does it now work better Why Print a copy of the transient output and AC Sweep Indicate on these plots why you think the integrator is now integrating correctly In addition to looking at the phase of the output we can also check its magnitude to see when this circuit acts best as an integrator Assume that the input voltage is sinusoidal Vin Vo sinth Using the equations on page 112 of Gingrich show that the magnitude of Vom VinOJRC where R R1 and C C2 In our PSpice simulation Vin is the output voltage of the function generator voltage between R3 and R1 while K A Connor 5 Revised 7202010 Rensselaer Polytechnic Institute Troy New Y ark USA Electronic Instrumentation ENGR 4300 Spring 2000 Sec Vom is the voltage at pin 6 of the opamp Change the plot for the AC sweep to show just Vom and Vin oJRC which should show when they are approximately equal and thus when the circuit is acting like an integrator Note that you need to input the frequency OJ as 2piFrequency in your PSpice plot 02 l u M R E mu 39 0 A more direct way of demonstrating that integration can be accomplished with this circuit is to replace the source with a DC source and a switch Note that the switch is set to close at time t001 sec Use a voltage of 01 volts to avoid saturation problems Do the transient analysis for times from 0 to 50ms with a step of lOus Rather than plotting the output voltage voltage across the load resistor plot the negative of the output voltage You should see that this circuit does seem to integrate reasonably well Indicate on your plot how close it has come to doing what it is designed to do Does it do its job for the entire simulation time Evaluate the slope of the output sign 1 and use this information in your discussion tClDSe 1 l I 392 U3 R6 mu To show what our problem was before when it did not produce an output proportional to the integral decrease C2 to its original value of 001uF and repeat the simulation Decreasing the capacitance decreases K A Connor 6 Revised 7202010 Rensselaer Polytechnic Institute Troy New Y ark USA Electronic Instrumentation ENGR 4300 Spring 2000 Sec the RC corner frequency See section 332 of Gingrich for the general discussion of approximate integrators Print your output and describe the performance of the circuit That is does it integrate even approximately for any period of time Increase the capacitance to lOuF and plot your results Use the information in your plot to show that it does indeed work better Can you think of any reason why we might prefer to use the luF capacitor in the feedback loop even though the circuit does not integrate quite as we Please note that the ideal integrator does not have a feedback resistor We have been looking at the configuration with such a resistor because it is more practical Set the feedback capacitor back to its value 0 luF Remove the resistor from the feedback loop and run your transient analysis again You should see that the circuit no longer works Plot your results and indicate what is wrong with the output The problem with this circuit is that there is no DC feedback The circuit will operate on both the AC and DC inputs There will always be a small DC offset voltage at the inputs This voltage will be amplified by the full intrinsic gain of the opamp and thus the output will be saturated Hardware Implementation Using either a 1458 or 741 op amp set up the integratorlow pass filter circuit in the most effective configuration analyzed above Use the sine wave from the function generator for the voltage source set the amplitude to 01 V remember to use the scope to set the amplitude correctly Obtain measurements of the input and output voltages at frequencies of lOOHz lkHz and ZkHz Add your experimental points to your PSpice plot Try playing around with a square wave input vary the frequency and amplitude to see if you can make the output look like an integrated version of the input Describe what you did to make this happen and produce a plot showing your input and output Results and Discussion Derive the relationship between Vom and Vin for the integrator circuit as described above I What are the features of the AC sweep and transient analysis of an integrator that shows it is working moreorless correctly I Why is the integrator also called a lowpass filter In the hardware implementation you should have used a squarewave input to demonstrate that the integrator was working approximately correctly If it was a perfect integrator what would the output waveform look like Make a sketch of the input and output voltages If we had built a differentiator instead what would the output waveform look like for the squarewave input What would the differentiator circuit output look like for a triangular wave input Again make a sketch of the input and output voltages I Extra Credit Who was the N ller of the Miller Integrator What else is he known for The answer to the rst question might be quite hard to nd butyou might nd some information on the second question K A Connor 7 Revised 7202010 Rensselaer Polytechnic Institute Troy New Y ork USA Electronic Instrumentation ENGR 4300 Spring 2000 Sec Experiment 8 Please list the names of all group members A TA or instructor will initial a participation box each class day you attend and participate in this experiment When you have completed all of the experimental and simulation activities have a TA or instructor initial under completed They should also look over What you have done to be sure that your results are useful If you are unable to attend class for any reason you can make up the work during an open shop time The maximum participation grade is 5 points Please answer any questions asked above under the Report and Conclusions sections Also attach any plots requested On each plot describe what is being displayed and why the results make sense Include a handdrawn or computer drawn circuit diagram for any PSpice output or plots of measurements indicating where and how the measurements were made Summarize the key points of this experiment Discuss any problems you encountered or mistakes you made and how you addressed them Names Grade Out of 25 K A Connor 8 Revised 7202010 Rensselaer Polytechnic Institute Troy New Y ork USA Electr uni Innnnmentztinn ENGReuoo Fall 2004 See Exp ErimEnt 8 01 Amp Cnn gurztinns e Integxztnr Differ Entiztnr antzge Fnllnwer Puxpuse 1n thus exbeumem we WI 1m bum quhe mm by my cm gunuansthat 9v usLhe my land aflaadm an amph ex andnat change xtsperfmmance by ebmg avaluge fullwwex v WI eee memes quiz eeeyebpeerbm addihms subkachans denvauves mdmugals Thanswhy mung ebmpmee Thu be same very gbbapeme br malag ebmpmeee and mm ebmpumemmuge me Fma y check numb webmeshsudm me 1121pr mfa secum bub cums web page sum babe Voltage Upmp Con gmnnom We WI be canadenng several bm up mp can guahans The mm Lb wxuge rbnbwex out m mputtemunal mustbe me same mam emem can enter be luv usz terminal Tm me mm and Mm m mege bubwee abeemum dawn me saute Tb seems an memng embmee symbolor youwi Tb mvenex wah Wing gm amphfxes me 519151 and changzs 115 g Em EAR A Tb m apexahanal amph exs nus eaaly sbbwn um me buqbuwmuge wxll be K A Connor 1 Revued mam 04 Rimmlnev rotymbm lwmmm Troy New York USA Elem uni Innnnmentztinn ENGRVASOO Fall 2004 E 7R Z 4 usesme same valuest meRn Am recs chatthue 5 an apeemp cm gunuanwhlchwdl nahnventhe mpm Thsxs called anan mvemng amph ex and was mtmducedta yaum expenmem A belaw m me feedback waan uf me mvenex emu K A Connor 2 Revued weran Rimmlnev Foly zchmc lwmmm Troy New York USA Electronic Instrumentation ENGR 4300 Fall 2004 Sec Basic Circuit Block this experiment we will be connecting a voltage divider to the function generator but not directly We will first modify 39 using a 39 39 L quot 39 4 L 39 quot quot affect the function ofthe voltage divider Then we will add a voltage follower and demonstrate that we can 39 39 39 f L 39 39 39 correctly In order to do this we need a circuit with 39 in our 39 quot parasi 39c 39 39 39 between wires The circuit we have chosen is a combination of an integrator and a differentiator It performs neither of these functions AL A L L L L L t 4 a voltage follower quotquot basic 39 39 39 39 L39 r 39 e function generator the circuit with impedance the voltage follower optional the voltage divider and a load Rather than drawing the quot 39 quot 39 39 39 39 that each function canbe identified First there is the function generator that we will represent as a Thevenin equivalent followed by the voltage divider with a load at the center pin know that RL will affect the voltage at Vout depending upon its size Calculate what Vout should be when RL is equal to 1 Meg 1K and 100 ohms Function Generator Load Next is our circuit with impedance This is the basic op amp configuration with enough componenm to be either an integrator or differentiator 01 R H tuF t k RU 1x RLZ w D General OpiAmp Con guration Finally the voltage follower 3 Ikvised 10302004 Remselaer Polytechnic Imtitute Way New York USA E lecmmic Instrumentatiun ENGRASOO F2112004 anngFnllnwer PanA TheVul39zge Fulluwer m n a a and39heluad lkHz HzmmnkHz mm m n mquot vultage lde quota m vultageatpm cunslslenlwnhyuurexpemzuuns rum mu uhmluadresmu bur u A dd me n m r sweep umpum suurce 0mm 1mm me yuurresults can yuu Explam whalthalmans Tu see Lhs du KA Connov 4 Rivsad 10502004 zmstzlaav Polytechmc mtmm My NW 1 on USA Electronic Instrumentation ENGR4300 Fall 2004 See analysxs agam what an yuu ubserve huwv Can yuu Explamx Refer te the spec sheet fur the 741 up amp many fulluwer That xs Recall that I Hu ever Furtuhately te gurext eut The emehtweheea is the differencebetween the eunehtthmugh R4 and R5 me yuur AC sweep results pletvwz IR471R5 Nutethatyuurvultage atvtaeheststehs mghthave impedance 7 Is ths a large but hut m mtequot humhehv Include this plat Part B Integratur In this seetaeh we will ehsemethe uperatmn efah mtegratur Mudxfy yuur ehemt nmpanA Remuve R m c1 and change the values ule and cz tn 1K and luFrespecnvely Yuur euemt shuuldnuwluukhke K A Comm 5 Reused 10517200 Remxelaer Patyteshms Inmtm e Troy New Yulc USA Electronic Instrumentation ENGR 4300 Fall 2004 Sec Repeat the transient analysis of this circuit and obtain a plot of your results Just like in mathematical integration integrators can add a DC offset to the result Adjust your input so that it is centered around zero by adding a trace that adds or subtracts the appropriate DC value Print this plot The equation that governs the behavior of this integrator at high frequencies is given by 1 1 then m r m I v 0dr R2 C2 R1 C2 Recall that the integration of sinth 1oJcosth Therefore the circuit attenuates the integration of the input by a constant equal to lOJR1C2 The negative sign means that the output should also be inverted What is there about the transient response that tells you that the circuit is working correctly Is the phase as expected The amplitude Above what frequencies should we expect this kind of behavior if a25gtgt Now we can look at the behavior of the circuit for all frequencies Repeat the AC sweep you did in part A In addition to the voltage plot the phase of the voltage at Vout What should the value of the phase be approximately if the circuit is to be working more or less like an integrator Print this plot Mark the frequency fCl21R2C2 Also mark the frequency above which the circuit is within 5 degrees of the desired phase shift We can also use PSpice to check the magnitude to see when this circuit acts best as an integrator Using the equation above we know that at frequencies above fc Vom VnOJRC where R R1 C C2 and OJ21f In our PSpice simulation Vin is the output voltage of the function generator voltage between R3 and R1 while Vout is the voltage at pin 6 of the opamp Change the plot for the AC sweep to show just Vom and V OJRC Note that you need to input the frequency OJ as 2piFrequency in your PSpice plot Pspice recognizes the word pi as the value of 1 and the word Frequency as the current input frequency to the circuit When are these two signals approximately equal It is at these frequencies that the circuit is acting like an integrator Print out this plot and indicate fCl21R2C2 How close are the amplitudes of the two signals at that frequency Also indicate on the plot the approximate frequency above which the two signals are equal What will happen if we change the value of the capacitor Try increasing C2 to 10uF Run the transient analysis at 1k Hz What happens to the phase and the amplitude Print this plot Now decrease C2 to 001 HF Run the transient analysis again What happens to the phase and the amplitude this time Print this plot Note that the opamp is saturating for part of this plot What feature of the output signal tells us this As indicated below a more direct way of demonstrating that integration can be accomplished with this circuit is to replace the source with a DC source and a switch Note that the switch is set to close at time t001 sec Use a voltage of 01 volts to avoid saturation problems TCLOSE n m R 1 2 K A Connor 6 Revised 10302004 RensselaerPalytechnt39c Institute Troy New Y ark USA Eledronic Instrumenta on ENGR74300 Fall 2004 Sec step oflOus maul l r r llllll at ll all lll 4 L deslgned to do Calculate the approxlmate slope othe output erte lt On your outputplot Also wnte the lyolts be7 Does lt do ltsjob for the ehtlre slrrlulatlorl lee7 To show llal ll r Prlrlt your output That ls A l l l W for k a llll lllul l l l u a lllalll aya lll qulte as well7 Please note thatthe ldeal lrltegrator does rlothaye a feedback reslstor We have been looklng at the ll 4 Mlller Integrator In thS course we also refer to lt as a Real Integratorquot Now we W111 look atwhy although an eal llll gal l la lll ul l luF lull ul ll Negatethe output boLhtheAC and DC lrlputs d good your o r t th r Therefore eohtlrluously m addm n h r Thls lrrlrrledlately saturates the Opramp Part c Hardware Inplementau39un Uslnga741 op amp r lul aya sll ll l Dohot uw lelllelll l l u ul obtalrl measurements othe lrlput arld outputyoltages at frequmcles of mm mom and lkHz Add your R othese Slgnals wlth the Agllehtlhtulhhk soltware K A Connor 7 Revoed 10302004 Remxelaer Polytecth Instlture Troy New York USA Electronic Instrumentation ENGR 4300 Fall 2004 Sec Set the function generator to a frequency which gives a reasonable signal amplitude and integrates fairly well This is somewhat subjective we just want you to see the shapes of the outputs for different input wave shapes You can use this to verify that the circuit does indeed integtrate Set the function generator to the following types of inputs 1 sine wave 2 triangular wave 3 square wave Take a picture of each situation with the Agilent software Now we will create a differentiator Remove the feedback capacitor C2 Replace R1 with an input capacitor C11pF Replace the 100K feedback resistor with a 1K resistor Your circuit should now look like this R3 AAA W n W n VOFF u 5 VAMPLEI FREQHlt r 0 Set the function generator to a frequency which gives a reasonable signal amplitude and differentiates fairly well Take a picture of the output and the input when the function generator is adjusted to the following types of inputs 1 sine wave 2 triangular wave square wave Take a picture of each situation with the Agilent software Results and Conclusions The following should be included in your report Everything should be labeled and easy to find Partial credit will be deducted for poor labeling or unclear presentation Part A Include the following plots 1 PSpice AC sweep without voltage follower 1 pt 2 PSpice AC sweep with voltage follower 1 pt 3 PSPice plot of voltage follower input impedance 1 pt Answer the following questions 1 Compare the AC sweep outputs with and without the buffer circuit in place What is the function of the buffer circuit 1 pt 2 Why is the follower unable to work properly with a small load resistor 1 pt 3 What is the typical value of the input impedance of the voltage follower when it is working properly 1 pt Part B Include the following plots 1 PSpice transient plot of integrator C21uF 1 pt K A Connor 8 Revised 10302004 RensselaerPalytechm39c Institute Troy New Y ark USA Electronic Instrumentation ENGR 4300 Fall 2004 Sec 2 AC sweep plot of integrator voltage with three points marked and phase with three points marked Also the loacation of fc and the place where the phase becomes close to ideal should be indicated 1 pt 3 PSpice plot of Vout vs VinoJRC On this plot fc and the frequency where the output and input become equal should be marked1 pt 4 PSpice transient plot of integrator C210uF 05 pt 5 PSpice transient plot of integrator C2001uF 05 pt 6 PSpice plot of integrator with DC source C21uF with slope and theoretical slope if any indicated on plot 05 pt 7 PSpice plot of integrator with DC source C2001uF with slope and theoretical slope if any indicated on plot 05 pt 8 PSpice plot of ideal integrator without feedback resistor 1 pt Answer the following questions 1 Derive the relationship between Vout and Vin for the integrator circuit as described above 1 pt 2 What are the features of the AC sweep and transient analysis of an integrator that shows it is working moreorless correctly Consider the phase shift and the change in amplitude of the output in relation to the input 1 pt 3 Why would we prefer to use the luF capacitor in the feedback loop even though the circuit does not integrate quite as well 1 pt 4 Why is the integrator also called a lowpass filter Take the limits of the transfer function at high and low frequencies to demonstrate this 1 pt Part C Include the following plots 1 Agilent lntuilink pictures of your circuit trace input vs output at 100 Hz 1K Hz and 2K Hz 3 plots 1 pt 2 Agilent lntuilink pictures of your integrator output with sine wave triangular wave and square wave inputs input vs output 3 plots 1pt 3 Agilent lntuilink picture of your differentiatoroutput with sine wave triangular wave and square wave inputs input vs output 3 plots 1pt Answer the following questions 1 1n the hardware implementation you should have used a squarewave input to demonstrate that the integrator was working approximately correctly If it was a perfect integrator what would the output waveform look like 1 pt 2 When we built the differntiator what did the ouptput waveform look like for the squarewave input What did the differentiator circuit output look like for a triangular wave input If it was a perfect differentiator what would the output waveform look like 1 pt 3 Does a differentiator need an additional resistance to be added in parallel with the capacitor in order to function effectively Why or why not Hint Find Hjoa at low and high frequencies 2 pt Summarize key points 1 pt Discuss mistakes and problems 05 pts List member responsibilities 05 pts K A Connor 9 Revised 10302004 RensselaerPalytechm39c Institute Troy New Y ark USA

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