电路分析与电子线路综合训练题(DOC43页)

1、2018年电路分析与电子线路综合训练题注意：2018年电路分析与电子线路综合训练组队共同讨论完成综 合训练题目3-4题，以书面形式撰写报告，需要详细的讨论过程和仿真的结果。请同学们自学“数值分析软件（如MATLAB?）及电路仿真软件（如Multisim 或Pspice等）”。提交书面报告时间：6月20日前。Project 1Po wer Supp liesObjective: This p roject will show some of the basic principles of power supp lies using fullwave rectifier, Zener diode,

2、 and fixed-voltage regulator circuits.Components: Bridge Rectifier (50 PIV, 1 A), Zener diode (10 V at 500 mW), 7805 regulatorIn troducti on:Most of the direct curre nt (DC) po wer used in electro nic devices is derived by converting 60 Hz, 115 V alter nati ng curre nt (AC) po wer to direct curre nt

3、 po wer. This AC to DC con versi on usually invo Ives a ste p-dow n tran sformer,rectifier, filter, and a regulator.The ste p-dow ntran sformer is used to decrease the AC line voltage from 115 VRMSto anRMS value n ear the DC voltage n eeded. The out put of the ste p-dow n tran sformer is the n fed i

4、n to a diode rectifier circuit that on ly out puts po sitive halves of the input sinu soid. A filter is the n used to smooth the rectifier out put to achieve a n early con sta nt DC voltage level. A regulator can be added after the filter to e nsure a con sta nt out put voltage in sp ite of cha nges

5、 in load curre nt and input voltages.Twodifferenttypes of voltage regulators will be used in this project.The first invo Ives a Zener diode circuit and the sec ond invo Ives a voltage regulator circuit. A Zener diode canbe used as a voltage regulator whe nZK). For is nearly coursethe diode is revers

6、e biased and op erated in the breakdow n regi on. To mai ntai n voltage regulati on, the Zener diode must be op erated in the breakdow n regi on at a curre nt greater tha n the kn ee curre nt (I currents greater than I zk, the Zener diode characteristic curve verticaland the voltage across the diode

7、 changes very little. Ofthere is a maximum current the diode can tolerate, so good regulation is pro vided whe n the diode is reverse biased with curre nts betwee n Izk andI ZMAX Zener diodes are available with a wide variety of breakdown voltages. Ano ther type of voltage regulator is available wit

8、h the 7800 series regulators. This series of fixed-voltage regulators is nu mbered 78xx, where xx corres ponds to the value of the out put voltage. Output voltages from 5 to 24 volts are available. These regulators are easy to use and work very well.Desig n:1. Find approximations for the DCvoltage l

9、evel and ACpeak to peak ripple voltage for the bridge rectifier and filter circuit of Figure 1-1.2. For the Zener diode regulator circuit of Figure 1-2 assume that the Zener diode will regulate at 10 V over a curre nt range of 5 mA to 25 mA. Assu ming that the curre nt flow ing through R is always b

10、etwee n 5 mA and 25 mA and the Zener diode is regulati ng at 10 V, find the minimum values of R and R l required. You may assume the forward diode drop for the two diodes is 去 1 V.Figlire 1-1: Bridge RecUfer and raterTransronnerFigure 1-2: Zener Diode RegulatorFigure 1 - 3; 7805 RegulatorPin 1 = Inp

11、utPin 2 = GroukdPill M = Output12 3Figure 1-4: 7805 Piii ConiiguratioiiLab P rocedure:1. Con struct the bridge rectifier circuit of Figure 1-1 without the cap acitor. Use the Variac with the ste p-dow n tran sformer for the input voltage to the bridge rectifier. With the tran sformer pl ugged into t

12、he Variac, adjust the Variac un til the sec on dary voltage from the tran sformer equals 12 V rms BE CAREFUL not to short the sec on dary termi nals! Observe the sec on dary waveform on the oscillosc ope. Put the oscillosc ope on DC coupling and observe the load voltage waveform V_. Remembethat both

13、 the input source and the load cannot share a com mon ground term in al.2. Removepower from the circuit.Insert the capacitor as shown in Figure1-1 being sure to observe the correct polarity. Energize the circuit. With the oscillosc ope on DC coup li ngobserve VL. Measure the DCvoltage level using th

14、e digital voltmeter. With the oscillosc ope on AC coup li ng observe the ripple voltage VR. Comparethese measured values with the calculated values.3. Observe the effect of load ing on the circuit by cha nging the load resistor from 1 k Qto 500 Q. Measure the DCvoltage level with the digital voltmet

15、er. Observe the ripple voltage with the oscilloscope set on ACcoupling. Comparethese values with the p reviously recorded values.4. Record the Zener diode characteristic curve from the digital curvetracer. Note the value of the breakdown voltage in the breakdown region. Also n ote the value of the k

16、n ee curre nt Izk.5. After verifying your designed values for R and RL with the instructor, con struct the Zener diode regulator circuit of Figure 1-2. Measure the DC voltage level with the digital voltmeter for the mi nimum value of R along with several values above and below the minimum value. Be

17、careful not to overload the Zener diode. Comme nt on the circuits op erati on for these differe nt load resista nces.6. Con struct the 7805 regulator circuit of Figure 1-3 being careful toobserve the correct pin configurationof the regulator. Measure the loadvoltage for R l equal to 300 Q, 200 Q, an

18、d 100 Q. Calculate the current for each of these cases. Does the value of the load resistor affect theout put voltage?7. Using R l equal to 200 Q, record the 7805 regulator input voltage (pin1) and out put voltage (pin 3). Decrease the regulator input voltage by decreas ing the sett ing of the Varia

19、c. For each decrease in amp litude, record the regulator input and out put voltages. Con ti nue decreas ing the amp litude un til the out put of the regulator drops a measurable amount below 5 V. What is the minimuminput voltage needed for the 7805 regulator to p roduce a 5 V out put?Questi ons:1. W

20、hycant the input source and load have a commonground in the bridge rectifier circuit?2. Can the Zener diode be used as a conven tio nal diode? Explain your an swer and verify with a curve from the curve tracer.3. Would the value of the out put filter cap acitor have to in crease, decrease, or rema i

21、n the same to maintain the same ripple voltage if the bridge rectifier were cha nged to a half-wave rectifier? Exp lai n your an swer.4. Howwould increasing the frequency of the input source affect the ripple voltage assu ming all components rema ined the same?Project 2An alog App licati ons of the

22、Op erati onal Amp lifierObjective: This p roject will dem on strate someof the an alog app licati ons of an op erati onal amp lifier through a sum ming circuit and a bandp ass filter circuit.Componen ts: 741 op-ampIn troducti on:Figure 2-1 shows a weighted summer circuit in the in vert ing con figur

23、ati on. This circuit can be used to sum in dividual input sig nals with a variable gain for each signal.The virtual ground at the invertinginput term inal of the op-amp kee ps the input sig nals isolated from each other. This isolati on makes it po ssiblefor each input to be summed witha differe nt

24、gain.The bandp ass filtershow n in Figure 2-2 uses an op-amp in comb in ati onwith resistors and cap acitors. Since the op-amp can in crease the gai n of the filter, the filter is classified as an active filter. This bandp ass filter circuit is extremely useful because the cen ter freque ncy can be

25、cha nged by vary ing a resistor in stead of cha nging the values of the cap acitors. The cen ter freque ncy is give n by:The cen ter freque ncycan be cha nged by vary ing the variable resistorR3. In creas ing R 3 decreases the cen ter freque ncy while decreas ing R33in creases the cen ter freque ncy

26、. The ban dwidth is give n by:Notice that the ban dwidth is independent of the variable resistor R so the cen ter freque ncy may be varied without cha nging the value of the bandwidth. The gain at the center frequency of the bandpass filter is given by:Desig n:1. Find the relatio nship betwee n the

27、out put and inputs for the weighted summer circuit of Figure 2-1.2. Desig n a bandp ass filter with a cen ter freque ncy of 2.0 kHz and a ban dwidth of 200 Hz. Let the voltage gain at the cen ter freque ncy be 20.Check your design with PSP ICE ?. Use 土 15 V suppl ies for the op-am p. UseRl = 2.4 k Q

28、.Figure 2 - 1: Weighted SummerDFigure 2 - 2: Bandp ass FilterLab P rocedure:1. Con struct the sum mingamp lifier of Figure 2-1. Desig n for the tran sfer function to be V O= -2 V in1 - V in2. Use 15 V supplies for the op -amp. Use RL = 2.4 k Q.2. Let VIN1 be a 1 V peak sine wave at 1 kHz and VIN2 eq

29、ual to 5 V DC. Verify the amp lifiers op erati on by mon itori ng the out put waveform on the oscillosc ope.3. Con struct the bandp ass filter of Figure 2-2. Use the desig ned valuesfor the resistorsand capacitors.Use 土 15 V supplies for the op-amp. UseRL = 2.4 k Q.4. Record and plot the freque ncy

30、res ponse (you may want to use computer con trol for the swee p and data collect ion). Find the cen ter freque ncy, corner freque ncies, ban dwidth, and cen ter freque ncy voltage gain to verify that the sp ecificati ons have bee n met.5. ChangeR3 to lower the center frequency from 2.0 kHz to 1.0 kH

31、z. Repeat part 4 for the newfrequency response. Verify that the newcenter frequency is 1.0 kHz. What is the new ban dwidth? What is the new cen ter freque ncy voltage gain? Compare with the measureme nts of P rocedure 4.Questio ns:1. Could the summer circuit be used with the inputs conn ected to the

32、 nonin vert ingtermi nal and p roduce the sameaffect without the inv ersi on?Exp la in.2. What is/are the ben efit(s) of using an op-amp circuit to p roduce a bandp ass filter over using an RLC circuit with a nonin vert ing op-amp at the out put of the RLC circuit?Project 3An alog Compu ter App lica

33、ti ons using the Op erati onal Amp lifierObjective: This project will focus on the use of the operational amplifier in p erform ing the mathematical op erati ons of in tegrati on anddifferentiation.The design of a simple circuit (analog computer) to solvea differe ntial equatio n will also be in clu

34、ded.Componen ts: 741 op-ampIn troducti on:Figures 3-1 and 3-2 illustrate two op-ampbased circuits desig ned toperform differentiation and integrationrespectively.The operations arep erformed real-time and can be helpful in observ ing both in itial tran sie nts and steady state res pon se. The an aly

35、sis of the circuits is based on the idealop-amp assumptions and performed in the time domain.The resistor R I show n in the two circuits is in cluded to help with stability and for gen eral circuit p rotectio n. The value for RI isnomin ally set equal to the feedback resistor (Figure 3-1) or the inp

36、ut resistor (Figure 3-2). The purp ose of the op ti onal resistor is left for stude nt inv estigati on in conjunction with the summary questi ons.The differe ntiator and in tegrator circuits may be comb ined withsta ndard in vert ing and non-i nverting op-amp circuits to pro vide the build ing block

37、s for an alog compu ters. The resulta nt an alog compu ter circuits are designed to solve differentialand/or integral/differentialequations in a real time environment. The ability to easily include,andcha nge, in itialcon diti onsand forcing fun cti ons are additi onal ben efitsof the an alog compu

38、ters. Figure 3-3 illustrates a circuit desig ned to solve the second order differentialequation KY - Y = 0 with the initialcon dition Y(0) = - V x and K = R1R2C1C2. The in itial con dition is set by using the mome ntary con tact switch to force the out put to equal the app lied voltage at t = 0 (the

39、 time the switch is closed).While the major adva nces in digital compu ters and digital sig nal p rocess ing have reduced the use of these three circuits, they are stilla fast and relatively inexpen sive method for p rocess con trol and stability/operationanalysis for systems that can be represented

40、 in termsof differe ntial equati ons.Desig n:1. Derive the exp ressi ons relat ing the input and out put sig nals for the circuits show n in Figures 3-1 and 3-2.哼心釜叫h y(0) = 2.2. Design an analog computer to solveSolve the differe ntial PSp ice?.Figure 3 - 1: Differe ntiatorV.inFigure 3 - 2: In tegr

41、atorequation whenf(t)= 0 and verify your results usingFigure 3 - 3: An alog Compu ter (li near, sec ond order, homoge nous differe ntial equati on)Lab Procedure:1. Con struct the circuit show n in Figure 3- the op-amp and a load resista nee of1. Use 土 15 V supplies for 龟 2.4 k Q.using a 500 mVpeak,

42、50 Hz sin ewave2. Verify the operation of the circuitas the input sig nal. Be sure to desig n the ga in such that the out put does not saturate.3. Rep eat ste p 2 with a sin ewave freque ncy of 500 Hz. Does the circuit still op erate correctly? What cha nges n eed to be made to p reve nt out put sat

43、urati on?4. Rep eat ste ps 2 and 3 using a tria ngle wave and the n using a square wave with the same magn itudes and freque ncies as used in ste ps 2 and 3.5. Con struct the circuitshow n in Figure 3-2. Aga in, use 15 V supp liesfor the op-amp and a load resista nee of比 2.4 k Q.6. Rep eat ste ps 2

44、through 4 for this circuit. Be sure to adjust your ga in as n ecessary to mai nta in an out put sig nal within the saturati on limits of 12 V.7. Con struct the circuit desig ned to solve the differe ntial equatio n in part 2 of the desig n secti on. Verify the op erati on of the desig n using three

45、differe nt input waves (sine, tria ngle, and square). Determ ine the op erati on for at least three differe ntfreque ncies - 10 Hz, 1 kHz, and 100 kHz. Explain any differe nces inoperationof the circuit.the result?What affect does the initialconditionhave on1. Howcould you use the rate for the op-am

46、p?differentiatorto obtain an estimate of the slew2. Why should you in clude a resistor in p arallel with the cap acitor in the in tegrator?3. What is the purpose of the resistor in series with in the differe ntiator?the input capacitor4. Is it po ssibleto desig n a circuit to p erform thein tegratio

47、 n functions using the non-i nverting input?differe ntiati on and Explain your answer.Project 4Com mon Emitter Amp lifier(desig ned for two lab p eriods)Objective: This p roject will show how the h-p arameters for a BJT can be measured and used in an equivale nt circuit model for the BJT. A CE small

48、 signal amplifierwill be biased and designed to specificationsalong withboth low and high freque ncy res ponse and adjustme nt. Series-series feedback will also be used to con trol the ban dwidth and input imp eda nee of the CE amp lifier.Components: 2N2222 BJTIn troducti on:In order for circuits in

49、v olvi ng tran sistors to be an alyzed, the terminal behavior of the transistormust be characterized by a model. Twoof the models ofte n used for a BJT are the hybrid-兀 and the h-p arametermodels. The complete hybrid-兀 circuit model for the BJT is shown in Figure 9-1. This model in cludes the intern

50、alcap acita nces and out put resista neeof the BJT. I nclusio n of the in ternal tran sistor cap acita nces makes the hybrid-兀 model valid throughout the entire frequency range of theQuesti ons:transistor. Typical data sheet values of C兀 and C k are 13 pF and 8 pFresp ectively. These values are so s

51、mall that C兀 and Cp may be con sideredopen circuits for midba nd freque ncies. The resista nee rx typ ically hasa value in the tens of ohms and can be con sidered a short circuit while rPand ro are usually extremely large in value and can be considered open circuits.The h-p arameter small sig nal mo

52、del for the BJT is characterized by the four h-p arameters and is show nin Figure 9-2. Un like the hybrid-兀 model, the h-p arameter model does not ordin arily in clude freque ncy related effects and components and is therefore gen erally valid only at midba nd frequencies and below . However the h-p

53、arameter model is very useful since the h-p arameters can be easily measured for a BJT. The value of hre isusually on the order of 10-4 and can be con sidered a short circuit. Thevalue of h oe is usually on the order of 10-5 S making 1/h oe effectivelyan open circuit for most circuit con figurati on

54、s and biases. Making the same assu mp ti ons, the hybrid- 兀 and h-p arameter models are equivale nt at midba nd freque ncies.beFor a tran sistor to op erate as an amp lifier, it must have a stable bias in the active regi on. To bias a tran sistor, a con sta nt DC curre nt must be established in the

55、collector and emitter. This current should as insensitive as possible to variations in temperature andP (or h fe).The voltage across the base-emitter junction decreases about 2 mVfor each 1 C ris e in temperature, therefore it is important to stabilize Vbe toensure that the transistordoes not overhe

56、at. The circuitshown in Figure9-3 is the biasing schememost often used for discrete transistor circuits. For this circuit, the base is supp lied with a fracti on of the supply voltage V CCthrough the voltage divider RB1, R B2. For ease of circuitan alysis, the Theve nin equivale nt circuitshow n in

57、Figure 9-4 can rep lacethe voltage divider n etwork. To en sure that the emitter curre nt is insensitive to variations inP and V be, v bb should be much greater thanVBe and R bb should be much less thanPRE. R bb is usually 20-30% of thePtheproduct PfE. The voltage across R e is also usually 2-3 volt

58、s for good stabilizati on.This same bias ing scheme can be used for all three ofBJT amp lifier con figuratio ns (CB, CC, CE).The BJT CE amp lifier is show n in Figure 9-5. The sig nal source and resistive load are cap acitively coupled to the amp lifier. The coup li ng cap acitors C1 and C2, emitter byp ass cap acitor Ce, and in ternaltran sistorcap acita nces sha pe the freque ncy res ponse of the amp lifier. A typi cal amp lifier