模拟电子电路 练习卷 3

习题1.402.2110.220log1120.8ovivAVVvdB或dBVVmAiRviiAiLoioi8.26log22202211002.2或dBWWivRvPPAiiLoiop8.23log242102422222或mVmAPdc1202032mWPPPPodcidis8.95%2.20%100dcoPP(a)当VI=2V时，Vomax=L+=10V当VI=Vo+2时，Vomin=L-=1.317V此时VI=3.317V（b）偏置点Vo=5V,此时VI=3V(c)习题1.44VVvvvAVvIVviovii1021033tVtVVtVvdiiiio2cos5.2cos105.252cos3510)(222VVmVVVVoiii4.04001.0105.22习题1.46VVAvo100dBWWRvRvAdBVVvvvvAiiLopiiiov17.791026.81.399.9010100100100722或或WRVPVVVVVVmAVmAiLopeakoipeakRMSopeakipeakopeako5.02078.029.90109.9010100100,1002max则若习题1.51VvvVVRRRRRRRRRRRRvvvvvvvvvvVVmVVvvsooLLoiioiisiiioiiiisisoso066.26.2066.206101010200102323312211323121实际：理论：单极AVO=10，则至少需要3级放大器。2.2、VMkkvvI10010.4111210010.4)(212AAvvvAvo2.13、512RRvvIo又R1+R2=120kΩ。，kRkR10020212.19、。，，则阻为。设需要并联的分流电。，，其中有限时：k84.5R49.12//RRk49.1Rk100R200A50/11/aaa121212kkARRRRvvAio2.37。，，，时，当k10RRmA1.0ii1,212121Vvv。，则选，，又kRkR120k40R,1031RR4RR23123132.602114234314dcm10220(1)0.009ACMRR20log60.8dBAoindidcmvRARRkvRRRRARRRR，。。。2.72ABCD。。。令)(sin08.8)(),(sin02.25.12)(sin04.43),(sin04.43)(sin08.02i)(sin04.03),(sin04.0334434222111221VtvvRRvVtvRRRvvvVtviRvVtiRvvmAtRvvVtvVtvABoBBDcIBIAIIII2.81。。时，，时，MHzfAfkHzfAffAAkHzfHzfffjAAbotbobobo44.637108.76103.8,8.761068.7,103.8)/(1103.8A10101.5A100,133432332.99）。，（有用的输入电压范围为。则对应的则）、（。。有用工作频率范围为而。则）、若（。则不失真的）、（率。为输出不失真的最大频，则、V1V1-110,10,108.312,5dkHz200~020010/13.3181022,50795.010,95.72,20b8.312SRV,5,5.0)(maxmaxmaxmax'123maxomaxVvvVvVVfSRvkHzfkHzfRRffkHzvSRVSRfmVvcVvvVfSRvkHzfkHzVSRfVVvVvaoioottdBioMioiooooi2.106。，又。，又。取F0.16C102CR1F18.3C1002CR1502199100,2001,100R22311111232kRRRkR2.115。时，加上）、而当（。时，当。时，则当）、（。，时，，又当、)11000(cos2100011000cos21000sin21)(t1000sin234)2(2t20002ms2t0,12nF10k1k1001Ck1RC11C1Tk100R)1(000inttRCtdtRCtvvVmsvvmsttRCvdtvRCvRRttoioootio3.9、(a)1212D1D2V050I1,50(5)0.5100.5mAkImAkIIImA假设、导通，则。。。假设成立。(b)、1212D1D2V050I0.5,100(5)15II0.50,D1D2I05(5)5551.671053mAkImAkImAVkVVkk假设、导通，则。。假设不成立。则截止，导通，。。3.19。则AieeeeiiVVeIieIieIiTTTTTVVTVSVSvvS6220/7.0/5.012/5.023/7.01/10335.0,1111,025.0),1(,10)1(),1(283.44,,0.98(0.65),0.650,0.65IDoIDoDIIIoovvvRvvrRvvVvvV3.69Iz=20mA时，VZ=6.8V，rZ=5Ω，则VZ=VZO+IZrZ，VZO=6.7V。①电源电流不受限（反向击穿）则VO=6.8V，IZ=20mA，R=(9-6.8)/20mA=110Ω。线性稳压性：。VmVRrrVVzzSO/5.43②电源电流受限（未击穿）IZ=0.25mA，VZO=VZK=6.7V，rZ=750Ω，R=(9-6.7)/0.25=9.2kΩ。线性稳压性：。VmVRrrVVzzSO/4.753.72VVVVVVVSSSO7.0,07.0,7.0a)、c)、Vs=15sinθ=-0.7,θ=π+0.0467rad。导通终止时间为2π-0.0467rad0467.020467.0.429.4)7.0sin15(21argVdvOd)、峰值电流出现在VS=VS.peak=-15V时，iD.peak=14.3/15k=9.53mA。e)、二极管的PIV=VS=15V。3.81。反向击穿电压为考虑安全因子，二极管。。：线圈匝数比为V55PIV5.16.367.01.121222%,10212V15%,10224%,102120.S.peakVVVPIVDOpeakS3.99(图略)4.11当rDS足够小时，iD=k‘n（W/L)(VGS-Vt)VDS,。，具有两倍大时，当。”’100r21r500r21rW5.3,)1)((1200,5001)(,)15.1)((11,))((1DSDSDSDS''''VVVLWkLWkLWkkVVLWkiVrGSGSnnntGSnDDSDS4.19。，，斜率相等：028.0V1V36VV424-82-2.2,20)4822.2(,)(AAA11kmmrVirODSDO4.26。、饱和区：、饱和区：。、饱和区：、变阻区：变阻区：饱和区：AiVVVVAiVVVVAiVVVVAiVVVVVVVVVLWkiVVVLWkiDDSDDDSDDDSDDDSDGSDSDStGSpDDStGSpD588,10,54;539,5,033.524,5.3,5.12;8.499,1,415],21))[(()1())((212'2'4.48。。（）、（。（）、（VVVAVVLWkiVVVVAiAiAiiiiLWbVVVVVLWkiVVLWaGStGSnDDDDDDDGStGSnD35.1,120)()(212.0,8.1,80,1202005.1,20)5.1LW32.1,1.32VV,))((21,1k40100-5V,20)LW321'112212121213GS2'21214.58。，舍去），解得：）（时，）（当。舍去），解得：’’’’’’%13II13.1I(mA53.3II-25.1I3LWk1mAI4mA(I,045)2())((21,44DDD2DDnDD22'2mAIIIVVLWkIIRIVDDDDtGSnDDSDGS4.60。。，的峰峰值为。。）（，舍去），或k11IVRV1-V2V220,k7I10--3-RV3-V(V1V3,))((21DDDDDSSGS2'DDDtGDtGDtGStGSnDVVVVvvVvVVVVVLWkI4.732()()1()()21.720GStmnoxDGStnoxWgCVVLWICVVLVGSVWm4.75Goo2()RM)22*5000.91/,100k20.9()2.456,DAmoOVDDGSGStDAIVAgmAVrVIIVVVVIGDDimLiD由偏置电路结构可知，该电路工作在饱和模式且V=V=2V,I=I=500直流忽略沟长效应又=10，其上交流分流很小，视为开路v/v=-g(ro||R而v/v=-8.27V/V若电流增加到1mA,则I=1mA,即o2.45621.285/50DGDAmoDOVVVVIVgmAVrkIVi，v/v=-10.7V/V4.77(a)直流分析(b)求器件小信号交流参数(c)画交流小信号等效电路(d)求解放大器交流参数(a)直流分析'25155510*3(1)1()()22(1)(2)21/3115*7.57.5GGGSDDnGStGSGSDDDMVVMMVVIkWIkVVLVVVVImAVIkV直流通路如图所示。栅极电压又假设该管工作在饱和区，忽略沟长效应，则有（）由可得或（舍去）首先满足饱和区工作条件，其次工作点与器件参数值一致。(b)求器件小信号交流参数22/,100DAmoOVDIVgmAVrkVI(c)画交流小信号等效电路1005||103.3337.510sigGDLRkRMMMRkRk3.333/0.97//(||||)8.22//(/)(/)7.97/inGingssiginsigogsmoDLRRMRvvVVRRvvgrRRVVvovsigvovgsvgsvsigVV(d)求解放大器交流参数4.79(||||)11.2/GmoDLGsigRGvgrRRRRVV4.872DDminDminmaxD()10,1()21.50.5()1.5(5)35,011V0VVR50kI20.4/,4GDnGStGSSSSDGDtDGtDDDDAmoOVDaRMWIkVVLVVVVVVRkIMOSvVvvVVvVVIVgmAVrVI忽略沟长效应或舍去若要求在饱和区工作，则有。的最大信号幅度为1，。。(b)00k()(||||)7.66/GmoDLGsigcRGvgrRRVVRR共源放大器()||||(