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# El ECE 2210

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This 28 page Class Notes was uploaded by Shyanne Lubowitz on Monday October 26, 2015. The Class Notes belongs to ECE 2210 at University of Utah taught by A. Stolp in Fall. Since its upload, it has received 36 views. For similar materials see /class/230002/ece-2210-university-of-utah in ELECTRICAL AND COMPUTER ENGINEERING at University of Utah.

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ECE 2210 00 Capacitor Lecture Notes J71 C A Stolp Now that we have voltages and currents which can be I 21703 Flurd Model rev 91806 functions of time it39s time to introduce the capacitor and the inductor Capacitor 1 C Electrical 1 equivalent 1 VC 5 permittIVIty 39 insulation C E 9 d V dV t t initial voltage Basic equations Q 1 1 d youshouldknow C V VC E 1Cdt E O 1CdtVCO 1C Cditvc Units farad 3 11 amp vial10 6 39farad mid10 12 39farad volt volt Capacitor voltage cannot change instantaneously Ener stored in electric field39 W lCV 2 series C 1 9y C 2 C 39 eq 39 1 parallel C eq C1C2C3 m CI C2 C3 GEET Example 2 S C14 uF I J Z Adt0539V C ltus Blus it rerdtO5V O539V C 4 us Sinusoids iCt Ip cosum t 1 1 1 1 1 VCt C L 1Cdt E E Ip smm t E E Ip cosm te90 deg Voltage quotlagsquot current 7 7 V p V p makes sense current has to flow in first to charge capacitor Steadystate or Final conditions If a circuit has been connected for quota long timequot then it has reached a steady state condition that means the currents and voltages are no longer changing d d long time 7V 0 1 C iV 0 dt C C dt C Vs R1 t i v R2 J r a Se R1R2 no current means it looks like an open ECE 2210 00 Capacitor Lecture Notes ECE 2210 00 Inductor Lecture Notes Eiectricai Fiuid Modei equwa em 1L VL L uoNZK p is the permeability of the inductor core K is a constant which depends on the inductor geometry N is the number of turns of wire Voltsec 3 Units henry 7 mH 1039 H pH 10 5H amp t t initiai current Basic equations d 1 1 you should know VL Ldj L lL ii VLdt EJ VLdtHLlt0gt rod 0 Inductor current cannot change instantaneoust Energy storedin eiectricfieid WL lLIL2 2 1 parallel L series LSq L1L2L3 W L L2 L3 L4 Example 2 us L4mH ZOV dt539mA L 1115 8115 1 rIOV dt5mA539mA L 4115 Sinusoids 1L0 Ipcosugtt vLm LgtiL Lmltinsinugttgt LmIpcosugt39t 90day 7 V 7 7V 7 Voltage quotIeadsquot current makes P P sense voitage has to present to make current change so voitage comes first Steadystate of Final conditions If a circuit has been connected for quota long timequot then it has reached a steady state Vs condition that means the currents and voltages are no longer changing VS 3 L l 0gt E quotlong timequot d d 71 0 v L71 0 dtL L dtL no voltage means it looks like a short ECE 2210 00 Inductor Lecture Notes t y r r Laplace transforms d3 operatlon can be replaced wlth s and J dt can be replaced by l I S ECE 2210 Lecture 2 notes p1 ECE 2210 Lecture 2 notes p2 Recall from your Ordinary Differential Equations class the Laplace transform method of solving differential equations The Laplace transform allowed you to change timedomain functions to frequencydomain functions 1 Transform your signals into the frequency domain with the Laplace transform Fs J4 fte39s39t dt Unilateral Laplace transform 0 2 Solve your differential equations with plain old algebra where 3 operation can be replaced with s and J I dt can be replaced by l S 3 Transform your result back to the time domain with the inverse Laplace transform 0 m 1 5 t 2 J4 RS 6 ds OK truth be told we neveractually use the inverse C T J Laplace transform We use tables instead So the first step is to transform the signals into the frequency domain with the Laplace transform Maybe we ought to talk a little about signals first Signals For us A timevarying voltage or current that carriers information l l Audio video position temperature digital data etc In some unpredictable fashion DC is not a signal Neither is a pure sine wave If you can predict it what information is it providing Neither DC nor pure sine wave have any quotbandwidthquot Recall Fourier series Any periodic waveform can be represented by a series of sinewaves of different frequencies Laplace transforms Let39 s evaluate some of these and see if we can make a table EX 1 ft 5t Thelmpulse or quotDiracquot function notaverylikely signal in real life Fs J 5te dt but 5tgt 5tg0 so 0 any function J 5te s39 dt J 5t1dt 1 O O ECE 3510 Lecture 2 notes p2 ECE 3510 Lecture 2 notes p3 EX 2 ft ut The unitstep function a constant value DC signal Fs J ume39 it but ut 1 from 0 to m 0 0 1 m 1 1 I 1 1 1 1e dt 7e es ie s39o 0 i1 7 0 is 0 is 7 s 7 s s if s gt 0 Im splane Re timedomain DC quotpolequot is at 0 time EX 3 ft utequot Fs J eme39s39t dt J e a SM dt 1 6Hquot m 0 0 a i S O O 1 1 39eaisw 7 1 39eais0 07 1 1gt 1 3 3 3 3 3 3 3 3 quotpolequot is at a if s gt a for positive a values for negative a values Im Im Im splane a Re a Re a Re pole is at a Unbounded Bounded signals signal eat eat That39 s not good faster response time time time timedomain This is the single mostimportant Laplace transform case In fact we really don39 t need any others Ex1 can be thought of as this case with a m Ex 2 can be thought of as a 0 And finally all sinusoids can be made from exponentials if you let the poles a be complex Remember Euler39 s equations jun 01H ij Euler39 sequations e cosmtjsinmt elt eMcosmtjsinmt Pole Locations correspond to the type of signal ECE 3510 Lecture 2 notes p3 ECE 3510 Lecture 2 notes p4 Euler39 sequations 003mt sinmt 2 2j Ex4 ft utcosut m m Jon rjwt jwisit 760349 F3 cosute s tdt 6 6 es tdt Ldt O 2 2 O O e mshdt e G39mHMdt 2 0 2 0 1 39eris so 1 39ee owrsyt so 2 jmis O 2 j39m rS 0 1 1 1 1 71 71 077 Hoff71 7 2 j 3 2 7jx3 e2jmi23 2jmi 23 1 1 Ii043 IiMM 2ju 23 207 23 2ju 232ju7 23 743 743 73 3 2jm 232jm 23 412027 432 7027 S2 024r S2 Im a Re 70 Bounded signal Doesn39 t timedomain converge time time What if the poles have a real component ft uteM31n ot Im Im o X X o a Re a Re 7 X Unbounded signal X 7 I Doesn t Bounded signal converge 39 time time converges ECE 3510 Lecture 2 notes p4 ECE 3510 Lecture 2 notes p5 Ex5 ft utteat Fs J teaters39tdt J te sgtquotdt O 0 Remember integration by parts 391 d ftgtigtgtdt ftgtgtgt7 glttgtiftgtdt dt dt choose ft t fromwhich gm 1 t 375M and digm ems fromwhich gt ea squotdt 6 dt ais d ftgtgtgt glttgtiftgt dt dt m 375M m 375M 375M m 375M Fs J tea sgt dt te e 1dt te e 0 ales 0 O ales ates 0 ais2 0 0 0 0 1 217s2 1 1 2 2 The easy way a S 3 a Use the quotmultiplication by timequot property 5 on p8 of the textbook txt ltgt fix ds mat ilt 1 i dipswli L 1 PM 1 13 sea 13 713202 13 3202 3202 Im dbl a Re dbl in Unbounded signal Doesn39 t converge ti me ECE 3510 Lecture 2 notes p5 Operational Amplifiers A Stolp 42201 rev 12505 An operational amplifieris basically a complete highgain voltage amplifier in a small package Opamps were originally developed to perform mathematical operations in analog computers hence the odd name They are now made using integrated circuit technology so they come in the typical multipin IC packages With the proper external components the operational amplifier can perform a wide variety of operations on the input voltage It can multiply the input voltage by nearly any constant factor positive or negative it can add the inputvoltage to other input voltages and it can integrate or differentiate the input voltage The respective circuits are called amplifiers summers integrators and differentiators Opamps are also used to make active frequency filters currenttovoltage converters voltagetocurrent converters current amplifiers voltage comparators etc etc These little parts are so versatile useful handy and cheap that they re kind of like electronic Lego blocks although somewhat drably colored Op amp characteristics Operational amplifiers have several very important characteristics that make them so Va An opamp has two inputs and it amplifies the voltage difference between those two inputs These two inputs are known as the noninverting input labeled and the inverting input labeled as shown in Vb Fig 1 The outputvoltage is a function of the noninverting inputvoltage minus the inverting input Figurei Opamp Symbol That is I Gwaer Where G voltage gain of the opamp a N The opamp must be connected to external sources of power not shown on the drawing above The output voltage v0 cannot be more positive than the positive ower source or more negative than the negative power source The gain G is very high typically more than 100000 Together that means that if the output VB is in the active range somewhere between its physical limits often called rails then va vb z 0 and v8 vb This is a very important point If you don t see this look back at the equation above v0 is limited G is very big so va vb must be very small If the output is The inputs must be In active range v8 v rail ltvO lt rai If the inputs are The output must be va gt vb rail va lt vb rail 3 In fact va vb must be so small that it s very difficult to make va amp vb close enough p1 ECE 2210 Opamp Notes without using some negative feedback Negative feedback makes the opamp maintain va z vb for itself With the proper negative feedback the opamp keeps va z vb so close that you can assume that va vb Without this negative feedback the opamp output will almost certainly be at one of its limits either high or low ie NOT in its active or linear range Incidentally circuits without negative feedback are also useful but then the output is either high or low digital and not linearly related to the input These types of circuits are called nonlinear circuits 9 Opamps amplify DC as well as AC 01 The input currents are almost zero In more technical terms the opamp has very high input impedance As long as you use reasonable resistor values in your circuits say 1 M0 you can neglect the input currents Simple isn39 tit OK so it doesn t sound so simple yet but the application of these characteristics really isn t hard Let s look at some circuits Linear Circuits Linear circuits employ negative feedback to keep va z vb If a circuit has a connection from the output to the inverting input then it has negative feedback V0 V0 Voltage follower The voltage follower shown in Fig 2 is probably the simplest linear opamp circuit Notice the feedback from the output to the inverting input If we were to Figure 2 Voltage follower hook the circuit input Va up to some voltage source say 2 volts DC what would happen If the output was lower than 2 V then the input voltage difference va vb would be positive and the huge gain of the opamp would drive the output higher If the output was higher than 2 V then va vb would be negative the output would go down Very quickly the output voltage v0 would change until va vb becomes very small Or basically until v0 2 V This is the concept of negative feedback A fraction in this case all of the output voltage is quotfed backquot to the input in order to control the gain of the opamp The opamp works very hard to maintain a very small difference between the voltages on its inputs This circuit is known as a voltage follower because the output follows the input Negative feedback is an important concept It is used in almost all systems including all natural systems A very simple example is the heating system in your house If the air temperature is too lowthe thermostat detects a difference between its setting and the air temperature and turns on the heater When the air temperature reaches the set temperature the thermostat turns off the heater negative feedback The servo system that you ve seen in lab is another example of negative feedback When the motor position sensor senses a different position than the input position sensor the circuit makes the motor turn in such a way that the difference is minimized and the positions line up p2 ECE 2210 Opamp Notes Noninverting amplifier Now suppose we feed back only a fraction of the output voltage rather than all of it The method used for this is shown in Fig 3 R1 and RT constitute a voltage divider Remember the current flowing into an op amp input is virtually nil so we can neglect its effect on the voltage divider This is one of the very nice features of an op amp In this circuit as in the voltage follower the op amp works very hard to keep va vb very small Only now vb is a fraction of V0 and the op amp has to make vO that much larger Figure 3 Noninverting amplifier F1 VinVaVb Vo 39 R1Rf For all practlcal R1Rf Rf purposes VDV 1 1 Notice that by adjusting the ratio of Rf and R we can make the gain of the op amp circuit almost anything we want sn tthat neat The circuit in Fig 3 is called a noninverting amplifier because the output voltage is in phase with the input voltage that is it is not inverted When the input voltage increases the output voltage will also increase and vice versa Yes noninverting is a double negative and kind of a dumb name Inverting amplifier Before going on observe that I ve swapped the positions of the two inputs amp on my op amp symbol Either way of drawing the op amp is OK whatever makes the whole schematic look better The noninverting input is on the bottom in this case because it s hooked to ground Draw the op amp so that the surrounding circuitry is clear Figure 4 Inverting amplifier The op amp output is still connected to the inverting input so again we have negative feedback If vb gt v3 then the output will go down taking vb with it until vb va f vb lt v3 then the output will go up until vb va Negative feedback makes the op amp do its best to equalize its inputs In this circuit v3 O which means that the op amp will try to keep vb O as well The current into the op amp is zero so iIn and if must be the same if im Using these two ideas together ijn V quot 00 V0 vi v R Rf D R In in in The minus sign means that vO will be inverted with respect to vm hence the name of this amplifier When vIn is positive V0 is negative and when vm is negative V0 is positive The gain of the inverting amplifier like that of the noninverting amplifier is completely dependent on our choices of RT and Rm p3 ECE 2210 Op amp Notes Summer The inverting amplifier can also be used as a summing amplifier that is it can be made to add the effects of several input voltages together Look at the circuit in Fig 5 if i1i2i3 I v0 v1 v2 v3 7 7 7 f R R R2 R3 Figure5 Summer V0 7iv1 7 ivz 7 iv3 The summer can be expanded to any number R1 R2 R3 of inputs See This is getting easier Differentiator Rf The differentiator looks an awful lot like the inverting amplifier and is analyzed in a very similar V C ll l dv v 1070 Ie dt R CRfdV V 0 dt lntegr ator Another useful op amp circuit is the integrator shown in Fig 7 For this circuit 1 7 7 dt V0 CRiann Figure 7 Integrator Unfortunately The simple integrator does have one little practical problem Notice that if the input voltage has any dc component the output voltage will soon try to run off to infinity Actually it will stop when the op amp reaches one of its output limits either negative or positive A resistor is usually placed in parallel with the capacitor to eliminate this rather annoying effect The circuit in Fig 8 has such a resistor This is a running average or Miller integrator Figure 8 Practical integrator p4 ECE 2210 Op amp Notes Active Filters If you replace the resistors in the inverting and noninverting amplifiers with frequency dependant impedances capacitors andor inductors you can make all sorts of frequency dependant circuits including filters In fact the differentiator and integrator circuits can be thought of as filters One of the main advantages of active filters is that you don t need to use inductors Real inductors are far from ideal as you ve no doubt observed in lab Real capacitors are much closer to ideal capacitors and they re cheaper than inductors Entire books are devoted to these active filters and we won t cover them any further here Differential amplifier This circuit amplifies only the difference R between the two inputs In this circuit Va f V2 Vb f V1V0 V0 there are two resistors labeled R which RinTRf RinTRf means that their values are equal Same goes for the two Rf s F F R 3 f V2 f 1V0 m fvo anTR anTR RinRf RVz RVr 39 Riva T Rinvo T Riva RVz RVr T Rinvo V2 V1 0 9 3 Figure 10 Differential amplifier Don t confuse the differential amplifier with the differentiator The differential amplifier amplifies the difference of two inputs while the differentiator amplifies the slope of an input Vi Instrumentation Amplifier The differential amplifier isn t really very practical The current that flows into the top input depends on the voltage applied to the bottom input This may not seem that bad but it is It means that the input characteristics of this circuit are not constant One way to get around this would be to place a voltage follower on each input as shown here Figure 9 Buffered differential amplifier p5 ECE 2210 Opamp Notes Now this is a perfectly good circuit If the two Rs are closely matched and the two Rms are also closely matched then this circuit will amplify differential voltages very well and reject common voltages a voltage that is common to both inputs should subtract out of the equation In EE terms it has a good CommonModeRejection Ration CMRR But what if you want to change the gain You d have to change two resistors at the same time By adding two more matched resistors and variable resistor we ll get the instrumentation amplifier shown at right The equation for this circuit is V0 2H H v 1 2 4 v a v o H1H32 1 This is an important circuit and you will probably see it again many times For instance if you had to amplify the output of a wheatstone bridge of stain gages this would be the amp for the job Figure 11 Instrumentation amplifier V Opamp with extra current amplification Most opamps cannot supply much current to the load They are often limited to 10 or 20 mA about enough to light an LED but not much more That can br very limiting The circuit at right shows a quick and dirty way to use two transistors to greatly increase the load current at a small cost in output voltage swing Notice that the feedback is taken from the output of the transistors so they sortof become part of the opamp and the opamp will do a pretty good job of eliminating the crossover deadzone that occurs as one transistor turns off and the other turn on Vin I V Figure 12 How to get more current This particular circuit is a simple voltage follower You can adapt this same current amplification to most of the other opamp circuits that we have discussed A few words of warning however The extra delay in the feedback can result in instabilities Try it with the parts you intend to use before you depend on this design Also if you use a low quality op amp with a slow slew rate you can get significant crossover distortion p6 ECE 2210 Opamp Notes Nonlinear Circuits In all cases so far the feedback signal voltage has been applied to the inverting input of the opamp This means that the feedback is negative Negative feedback tends to reduce the difference between the va and vb voltages and make linear circuits Without negative feedback the opamp cannot minimize the difference between va and vb and the very high sensitivity of the opamp results in switching or nonlinear circuits Comparator Now look at Fig 14 This circuit will not work as a linear circuit lf va gt 0 the output will be as high as the opamp can make it usually a volt or two below the positive power supply lf va lt 0 the output will be as low as the opamp can make it usually a volt or two above the negative power supply The output is no longer linearly related to the input it s more like a digital signal f high or low depending on how vm compares to ground 0 V Figure 13 Comparator The comparator is a nonlinear circuit Figure 14 Other comparators All the circuits above are also comparators In the first circuit the input is again compared to ground but this time the output goes low when the input goes high and viseversa In the remaining circuits the input is compared not to 0 V but to some voltage set by the voltage divider of Fl1 and R2 Schmitt trigger The Schmitt trigger is a variation of the simple comparator which has hysteresis that is it has a toggle action When the output is high positive feedback makes the switching level higherthan it is when the output is low A little positive feedback makes a comparator with better noise immunity Increase the positive feedback and the Schmitt trigger can be used in other switching applications Look at the Schmitt ttrigger circuit shown at right Notice i that va Fl1Fl1 Rfv0 it depends on the output Lets say Figure 15 Schmitt trigger p7 ECE 2210 Opamp Notes the output ts tow and the tnput ts decreastng When v lt va the output goes htgh and suddenty va goes a ttttte btt htgher wtth tt That makes the dtfference between vb and v8 even btgger To make the ctrcutt swttch agatn v has to go back up beyond the ortgtnat swttchtng tevet It has to reach the new V8 before the output wttt swttch tow n thts ctrcutt the two swttchtng tevets are above and betow ground by the same amount untess you have nonsymmetrtc power suppttes Figure 16 Other Schmttt trtggers The ctrcutts above are vartattons of the Schmttt trtgger In the ftrst ctrcutt the tnput ts agatn compared to tevets above and betow ground but thts tt me the output goes htgh when the tnput goes htgh and vtseversa In the rematntng ctrcutts the swttchtng tevets are not symmetrtc about 0 V but about some vottage set by the vottage dtvtder of Ft1 and R2 Multivibrator square wave generator R The heart of the mutttvtbrator ts a Schmttt trtgger wtth tots of postttve feedback Usuatty Ft2 R3 whtch set the swttchtng tevets at about V2 V and V2 V When the i output ts htgh the capacttor charges through Ft1 unttt tt reaches the V2 V swttchtng tevet the output swttches tow and the capacttor dtscharges to zero and then charges up down unttt tt reaches the V2 V swttchtng tevet That makes the output swttch htgh and the process repeats th Figure 17 Mutttvtbrator Concluston n att of these ctrcutts wtth etther negattve or postttve feedback the output vottage vu cannot tncrease wtthout bounds t ts bounded tn the postttve dtrectton by V the opamp postttve power suppty vottage and ts bounded tn the negattve dtrectton by V the opamp negattve suppty vottage If the output vottage ts wtthtn these bounds va vb must be very smatt If ve vb were not very smatt vu woutd soon be forced to one of tts ttmtts Ltnear ct rcutts use negattve feedback to keep thts dtfference smatt thhout negattve feedback you can reasonabty assume that the ctrcutt ts some ktnd of swttchtng ctrcutt and that the output ts atways at one or the other of tts ttmtts Thts onty scratches the surface of what you can do wtth opamps Get a copy of The Op amp Cookbook for tots more tdeas presented tn a nononsense way p8 ECE 2210 Opamp Notes Translslor Notes BJT ECE 2210 Astop 32500 to t tztoe Imagtne tt you wttt a hyorauttc oevtce where the ow tn a smatt ptpe controts avatve tn a targerptpe ttt ut t t ttt tt tarae ptpe owrcorttroHed artoff vatve swttch Or oepenotng on the ows and pressures tnvotveo t couto be used as an ampmet hat ts tt couto be used to make some hvorauttc stanat taroer and more powertut A st nat ts a ow or pressure whtch convevs tntormatton and an amptttter ts a oevtce whtch tncreases the power of astgnat The etectrtcat equtvatent of thts owrcorttroHed vatve ts a tranststor Spectttcattythe NPN btpotartunctton tranststor EMT There are other tvpes The symbot for a tranststor ts shown betow Nottce that 5 a A t ttt u tt t out of the bottom the emttter current The vatve orawn betow ts a more accurate anatogytor the etectrtcat tranststor CoHector quottransistor Em A Ya STSLOY has three termtnatsw the base the coHector and the emttter The current ow from the cottector to the emttter through the tranststor ts controtteo by the current ow from the base to the emt er cott ctot cutt ttt by a stmpte tactor catteo beta 5 For a gtven base current the tranststor wttt attowp ttmes as much K M TH Lt tttut cottectorcurrent The ey u to w t ut u tt ttttwt t u Eta power 100 rut BTS usuaHy between 100 and 400 10005 Atran t tut quot quot k 39 5 off We an open swttch When there ts a base current tt s on If somethtng outstoe of the tranststor ts ttmtttng the cot ctot ttt tuttt on as much as tt can We actoseo swttch A tranststor that ts off ts operattng tn tts quotcutoffquot regton A tt tt t tot p tttts tut ttott t utott tt tt t tot p acttve controt of t cott ctot ctt regton Note the vatve anatogy has a probtem wtth the quotopenquot and quotdosedquot terms ttt t t ttt tt p ttrartststors rwr except that aH the currents and M vottaoe tnsteao of current and come tn many vartettes In thts ctass we H onty work wtth NPN tranststors Transistor Notes am p1 Silicon diodes are made of two layers of doped silicon a P layer is the anode and an N layer is the catho e A PN junction isadiode Anode D Cathode Bipolar junction transistors BJTs consist of three layers of doped silicon The NPN transistor has a thin layer of Pdoped silicon sandwiched between two a ers of Ndoped silicon Each PN junction can act like a diode C In fact this is a fairly good way to check atransistor with an ohmmeter set to the diode setting B The baseemitter junction always acts like adiode but because the base is very thin it makes the other junction act E like a controlled valve you probably don39 twant to know the details so call it magic A bipolar junction transistor contains Translsmr Symbols FNF two diode junctions 0 v Collector V Emitter c c m s C E 55 E E reverse reverse tonyard VB Bu e t d t d TEE VPE Base YEC E E mmjunoc ti big E VE VBE VD C H t NPN PNP Emitter a or NPN PNP Notice the subscripts Fleplace vBE With vEB and VBE vBrvE VCEWlthVECer equationsbelow V VEC VCE VC VE Modes or regions of operation vBE and vCE are approXimate Cutoff off Active partially on Saturation fully on vBE lt 07V vBE 07V vBE 07V lE0 lEgt0 lEgt0 l in utfrom vCE 2 02V vCE 02V 1 09 CW iC 0 i0 iiiE onE or 1 i0 lt BiB limited bysomething l emiroired wireirarrseior outside ol he VWSW l l Motor Driver 7 The Transistor as a switch One of the most common uses of a transistor is as a currentcontrolled switch Transistor switches are the basisfor all digital circuits but that39 s probably not where you39 ll use the transistor More likely you39 ll want to control a highcurrent device like a motor with arintegratedcircuit output from acomputer or logic circuit The small integrated circuit won39 the able to sup l enou h current to run e motor so you39 ll use a transistor to switch the larger current that flows through the motor The input is hooked to the base of the transistor Often through a current limiting resistor since VR will only e 07V when the transistor is on A small IB can switch on the much larger l0 and VCE can be as low as 02V Vcc The terminal marked V99 above isjust a circuit terminal hooked to a power supply drawn in dotted lines here but usually not shown at all Power supply wires like ground wires are often not shown explicitly on schematics It makes the schematics a little less cluttered and easier to read Diode If you39 re switching an inductive load like a motor you should add adiode so that you39 re not trying to switch off the motor current instantly The diode called a flyback diode when used like this 39 hed off provides a path for the current still flowing through the motor when the transistor is swr c Transistor Notes BJT p2 Hbridge or course tt you want to make the motor turn tn both otrecttons you tt need a more comptexctrcutt Look at the ctrcutt at rtght Q tt s has the shape ot an hence the name It tranststors t an A are on then the current ttows as shown ieftrtorrtght through the motor It A G1 are art tt u the motor and the motor wttt turn tnthe oppostte otrectton The motor here ts a permanentrmagnet DC motor In my ctrcutt the top two tranststors are PNPs whtch makes the ctrcutt more etttctent The I I Hrbrtdqe couto atso be made wtth att NPNs or wtth power MOSFET vths Vtho tranststors An Hrbrtdge reoutrestour tnputs att operated tn concert To turn on 3 and 3 T as shown Vtm wouto have to be tow and VW wouto have to be htgh At the same ttme the other two tranststors wouto have to be off so sz wouto have to be htgh and VWa wouto have to be tow It the controt ctrcutt makes a mtstake and turns on at and G1 or 37 and Jr at thesamettmevou U have atoaster tnsteao ot a motor ortver at teast tor ashort w t e The ctrcutt at tett reoutres onty two tnputs Tranststors o and 3 work as mverfers when thetr tnputs are htgh thetr outputs are tow and ytc a ersa The reststors are known as puHIF reststors The Hrbrtdge Shouid aiso tnciude yback dtodes Linear Amplifiers 7 The obTecttve of a Mnear amphfter ts to output a fatthfui reproducttort of an mput stanai omv bta er A vottaae amptttter makes the startai vottaae btaaer A current amptttter makes the startat current btgger Marty amptttters do both Att amptttters shouto make the stgnat power btgger oepenos wer generatty DC power trom a battery or power suppty The stgnats are usuatty Ac Unttketranststorswttches whtch operate tn cutott ano saturatton ttnear amptttters must operate tnthe acme ream Imponam relations acttve regtort E 07v VCE yceyE gt 07V muzvnsaturateo C quotHE E Bias Outstoe ot A t ttn tt current mytttt t n n t tn Thatmeans that the tranststor must be turned on part way even when there s no stgnat at at Look back at the yatye anatogytt m tt out tttttt t to m H Itettherttow ever stops the hortzontat ptpe ow OE ts no tonger tn controt Att vottages and currents can be shown th T w nt e acttye regtort tEat td tC must be postttye tor att threedttteventweys yatues ot the AC stgnats ha and t m b sea to some postttye DC yatue We use capttat tetters IP and In tor these CAPW 343 DC btas yatues and tower case tetters tr ano trifortheAC c stgnats that wttt appear as ttuctuattons ot these DC vatues Wm Vb t A stgha sync yE 1C Do and AcTogether Transistor Notes am pa The objective of bias then is to partially turn on the transistor to turn it sortof halfway on Now if I twiddle i3 i0 will show a similar but bigger twiddle that39 s the whole idea The transistor should never go into cutoff for any expected input signal otherwise you39 ll gecipping at the output Clipping is a form of distortion where the output no longer looks like the input Furthermore the transistor must not saturate That will also cause clipping at the output Because 5 can vary widely from transistor to transistor of the same part number and V5 changes with temperature achieving a stable bias can be a bit of a problem Usually an emitter resistor RE is needed to stabilize the bias DC Analysis in the active region DC analysis applies to both switching and bias although the circuits we39 ll look at here will include an RE and we39 ll be working in the active region meaning they are bias circuits The key to DC analysis with an RE is usually finding VB The circuit at right shows a typical bias arrangement The equations below are for that circuit adapt them as necessary to fit your actual circuit If you can neglect IB Often in duick anddirty analysis you can neglect the base current IE In that case V E R B2 VB VCci VE VB707V 1E 71C VC VCC71CRC RB1 R B2 R E This assumption is OK if RB1 ll RB2 ltlt BR E Quick check RB1lt 10RE andor RB2 lt 10R E Should result in lt10 error if 3 100 Vcc I If you can t neglect IB 0 RC Then you need to make a Thevenin equivalent of the base bias resistors R 1 B2 R 7 V BB V cc iR R BB 1 1 B1 t B2 77 RB1 RB2 From the base39 5 pointofview the emitter resistor will look 1 times bigger than it really is This is because 3 1 times as much current flows through REthan into the base We can ignore the fact that the current is bigger if we pretend that the resistor is bigger That leads to the simplified circuit Usuall we use 3 as the factor ratherthan 13 1 after all Blust isn39 tthat well known anyway VBB707V 1B R7 1C B1B11E VE IERE ICRE VB VEO7V BBTB RE Vc VCC IC RC VE OR VB IBBREQ7V VE VB707V 1E TE 1 IC VC VCC71CRC Transistor Notes BJT p4 m 5 N5 mg E 3 322m E x E x J m 97 x m m gt x00gt m gt m a m 5506 H n H 00 93me Zn mgt gtZm gtH mgt lt53 0H 33n0m 0m NE qb nmm 55995 gm gt27 gtTm0gt mgtx0gtu m0gt 535 am Em Nmm m mgt am am Em lt57 0 lt53 N5 gt0 0gt gt3 mgt gtH 00gt ca 352 522 20 gtH00gt Em 0em0gt 0m isme m H NE gt 354 Sagozsm Ewian gtwoA gtEwnm0gt mgtx0gtum0gt qmomn0m L0m mm P H m z 53 H mgt gtmnmgt gt Tmgtu mgt 937 mm E E m m gtTmgt 00gt mgt 5 5335 an Em ng 8gt am 0 mgt mgt E 937 mm 9 Em 9 Em gt3 0gt gt2 00gt m 5 2 20 BEETS 0em0gtu 0m Sagozsm Em xo gt27 gt dnm0gt mgtx0gtn m0gt gt4 gt 0m0700 gt mm m 0H u 530an H m mgt Sum gtonmgtumgt 5535 0n Em 8gt 0gt o E 9 mm 30H0m gton00gt gee Hy 8 a 395 3 wwo amg 3 m gt A 520 gt3 00gt F mmgtmlt WEE on mmEmem AC Analysis of Common emitter CE amplifier th an RE any AC swgna apphed m we base SEWquot vans g a a o a S a 1 mm Recang ma meswgna a me emmer TS abam mesame as meswgna a we base RC basemcauecmer gem e 7 z 7 Vb RE dmereme gm T vEmHaws VE jus 0 7V DC Tower n a T T a a a gaeswayup W u Tms TS caHed me smaTTswgnaT emmer resws ance r C To nd We gems when We mpm has a Source res s ance and We ampm TS cannec ed 0 a Toad res s ar We ca cu auans became a We more campTeX YOU DON T NEED TO KNOW THE FOLLOWING MATERIAL R E s me DC resws ancetmm emmerm ground R 6 TS me AC swgna resws ance Tram emmer m ground may be zero npm mpedance R e R E1 HR E2 HB39rcRcgt Ompm mpedance R0 Rc Hru gmermsgbced AC caHec arresws ance re RC R L We b TS a charac ensnc av me ransws ar and TS onen negTec ed v AC Signal Example 0 n me vs swgna were apphed a we base an AC swgna waqu aTsa appear a me caHecmr How much Targerwau d n be7 Vanage gem R basemcauecmer gam e 433 umesbwgger E Transistor Noles BJT p6 t lsrrtthere 39 39 39 mulllpllcalion and dlvlslon Laplace transforms i operatlon can be replaced wrths and I dt can be replaced by 1 d1 5 ECE 2210 Lecture 2 notes p1 ECE 2210 Lecture 2 notes p2 Recall from your Ordinary Differential Equations class the Laplace transform method of solving differential equations The Laplace transform allowed you to change tim edomain functions to frequencydomain functions 1 Transform your sig nal s into the freq uency domain with the Laplace transform Fs J fte s39tdt Unilateral Laplace transform 0 2 Solve your differential equationswith plain old algebra where Ed operation can be replaced with s and J I dt can be replaced by l t s 3 Transform your result backto the time domain with the inverse Laplace transform 1 cj m Fs 215 2 OK truth be told we never actually use the inverse ch Laplace transform We use tables instead So the first step is to transform the signals into the frequency domain with the Laplace transform Maybe we ought to talk a little about signals first Signals For us A tim evarying voltage or current that carriers information Audio video position tem perature dig ital data etc In some unpredictable fashion DC is not a signal Neither is a pure sine wave If you can predict it what information is it providing Neither DC nor pure sine wave have any quotbandwidthquot Recall Fourier series Any periodic waveform can be represented by a series of sinewaves of different frequencies Laplace transforms Let s evaluate some of these and see if we can make a table EX 1 ft 5t The Impulse or quotDiracquot function not a very likely signal in real life Fs j 5te squotdt but 5tgt 5tg0 so 0 anyfunction j 5te 539 dt f 5t1dt 1 0 0 ECE 3510 Lecture 2 notes p2 ECE 3510 Lecture 2 notes p3 EX 2 ft ut The unitstep function a constant value DC signal Fs J ute dt but 111 1 from o to co 0 0 1 Go 1 1 I 1 1 1 1e5tdt ie s39t co e sm 50 Off1 7 0 is 0 is is is s if sgt 0 1m splane Re tim edomain 39 DC quotpolequot is at 0 time Ex 3 11 utequot FS J eate s39tdt J e017 t dt 1 ea s39t Go 0 o a S 0 0 1 1 ea 539mea s390 071 ais ais aas S a quotpolequotisata if s gt a for positive a values for negative a values 1m 1m 1m s plane A S a a Re a Re a Re pole is at a at Unbounded Bounded signals e at at signal e e That39s faster res onse not good D 1 time time time timedomain This isthe single mostimportant Laplace transform case In fact we really don39t need anyothers EX1 can be thought of as this case with a oltgt Ex 2 can be thought of as a 0 And finally all sinusoids can be madefrom exponentials if you let the poles a be com plex Remember Euler39s equations j Euler39sequations e39m39t cos00tjsjnmt eWtHmt emcos Jot jsin 000 Pole Locations correspond to the type of signal ECE 3510 Lecture 2 notes p3 ECE 3510 Lecture 2 notes p4 ejm 640 ejmmi 640 Euler s equations cos00t sin00t 2 2j Ex4 ft utcos00t Go jmt ijcot jmist 7jmst Fs cosmte squotdt e H it d e H dt 0 2 2 0 0 1 w does 1 w 7jmst 7 e dt 7 e dt 2 0 2 0 1 1 gteltjmesgtt L 1 eiltjmsgtt 2 joois o 2 jMS 0 1 1 1 1 71 71 07 10 7 1 7 2 jDis 2 7jDs 723300725 2j0072s 1 71 21007 2s72j002s 2joo2s 200725 2j002s2j072s 745 745 is s 2jm2s2jm2s 4J20027452 7002 S2 m2s Im 1m s plane m As B 2 2 Re s a 7 0 Doesn39t converge time time Bounded signal tim edomain What ifthe poles have a real component ft ute quotsinat Im a x As B 2 2 a 5 at a Re Re Unbounded signal DOGS t Bounded signal converge time time converg es ECE 3510 Lecture 2 notes p4 ECE 3510 Lecture 2 notes p5 Ex 5 Multiply by time property ft umte Fs tea39te s39tdt J te squotdt o 0 Remember integration by parts hlttgtiglttgtdt lamguy glttgtihlttgtdt dt dt choose ht t from which ihm 1 t d eaist and igt e014quot fromwhioh gt ea s39tdt dt ais d htgt gthtdt dt co aist Go Go aist aist aist Fs t a mdt te e 1dt te e 0 ais 0 ais ais 0 as2 0 0 0 0 of Hf 1 2 2 The easyway a S 5 a Use the quotmultiplication by timequot property 5 on p8 of the Bodson textbook txt ltgt eiXs ds teat ilt 1 gt dipHyli L 1 26 1 1 ds sia ds 71Sia2 ds Siaf 57a2 Anything that works for exponentials also works for sines and cosines 1m AsB dbl 0 And quotDCquot too 2 52w2gt 1 Re 7 S2 dbl Re dbl foo Unbounded signal Doesn39t tsmmtut converge Unbounded sIgnaI me Doesn39t tum converge time ECE 3510 Lecture 2 amp 3 notes p5 Signal Type Boundedness and Convergence can be predicted from the poles Poles in the OpenLeftdePlane OLHP Real part olpole is negative Reltsp lt o A Im Im AsB 1m alt0 2 2 K 1 arm 5 a a a s7 a 1 Re Re alt0 a Re urm aster response Bounded signals Converge to zero Single Poles on Imaginary Axis Real part olpole iszero Respgt 0 Im 1m ASB 1m 2 m A szm 5 Re m Re Re 39pole39is alorig39n rm Bounded signal Convergesto DC value Bounded signals Don39t Converge Double Poles on Imaginary Axis or 1 A HE am a m h 1m 7 i lt2 2r 2 s m s dbl Re R L D Lsinm L ut andorcosine line in the OpenRightHalfPlane ORHP I ma 1 Im Re f R 7h x 1 time ECE 3510 Lecture 2 amp 3 notes p6 Unbounded signals Don39t Converge DC Notes Th venin equivalent Norton equivalent To calculate a ctrcutt s Thevehth equtvalertt To calculate a ctrcutt s Nortoh equtvalertt r 1 n Vt t t J t a t I voltage where the load used to be Thts ts the calculate the shortctrcutt curreht th thts Wtre Thevehth voltage Wm Thts ts the Norton curreht N Remove the short 2 t t a t w m t t replace ttthh art opert replace tt Wltl t art open a r a t t termthals DO NOT trtcludethe load ththts reststahce Thts ts termthals DO NOT trtclude the load ththts reststahce Thts ts he Thevertthsource RT the Nortohsource quot1 reststahce Flm 39 1 reststahce FlN g 4 Draw the t v my Exactly the same as RN Rf equtvaleht ctrcutt and add nquot sgt theThe vehtrt source 3 your values aquot reststahce Flm J 4 Draw the Norton y at vatue OR the more common wa l FtrtdtheThe vertth equtvaleht ctrcutt 2 N Flm ahd Nodal Analysis IN mem 1 tt A mt one node as grourtd zero voltage lithe grourtd can be deitrted as one stde of a voltage source that thl make the to thrtg steps easter 2 Label uhkrtowrt hode voltages as Va Vb and label the current th each reststor as I IQ 3 Wrtte Ktrchoits curreht equattorts for each SUPerP 5m n urtkrtowrt hode For Will s Wlm W 4 Fleplace the currertts th your KCL equattorts thh 1 zero a W me soume T0 19m VO tage expressions ttke the one below source replace tt thh a short To zero a current source replace tt thh art opert V R1 V 2 Compute your wahted voltage or current due to Lm b I the remathtrtgsource Careful some may be 5 1 hegattve I1 3 Flepeatthe itrst two steps for all thesources 4 Sum all the cohtrtbuttohs from all the sources to 5 Solve the multtple equattorts for the multtple ithd the actual voltage or current Watch your urtkrtowrt voltages

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