第01章-习题解答

=1023m-3T300KT=300K1631.510inm−≈×1723331010AiNncmm==233010APNm∴≈=293002.2510innpm×==1.1.2T=300Kni1.51016m-3(1)300KNA=21020m-3(2)573Kni31021m-3(a)1633001.510iTKnm−=≈×∵203210AiNnm×=2030210APNm−∴==×2123001.12510innmp−==×(b)213203573310210iATKnmNm−−==×=×∵in∴AN21300310iPnnm−≈≈=×1.2.1PNT=290K1IS9020.05V31A0.10.20.3V290TK=25TVmV≈(1)250.9(1)(1)TqVVVmVsssiIIeIe=−=−=−25ln(1(0.9))0.0576vVmV∴=×+−=−(2)10.05250.0525i(1)17.389i1(1)qVKTmVsqVmVVsIeeeIe−−−−===−−−(3)1sIAμ=(1)53.61(1)2.981(1)162.750.16vmVvmVvmViAeAiAemAiAemAAμμμμ=−==−==−==1.2.2290KIS1nA0.5pA1.2.21mA1mA1()DGe2()DSi1.2.21ImA=29025TTKVmV=≈25912(1)(1)25ln(1)1:25ln(1)0.345110125ln(1)0.518110TqVVVmVsssGeSiiIeIeiVmVImAVmVVmAVmVV−−=−=−∴=×+=×+=×=×+=×∵1.2.3)()1(102012AeITVV−×=−VT26mV11.5V210100(1)1.51214262010(1)2.2710VmVIeA−=×−=×PN1.5V122010IA−=−×(2)10Ω×10IΩ×100Ω×100IΩ×IDr100Ω×1.2.41.2.4V1=18VR=1kRL=2KDZVZ10V1VOIOIIZ2RL−OVIVRLR+−ZDIOIZI1.2.4(1)2181212LIZLRVVVRR×=×=++∵ZD10V105mA218108mA1853mAoZooLIoZoVVVIRVVIRIII∴=====−−====−=−=(2)LIZLRVVRR×+DZIVRZVLR101LLRR×+1.25LRkΩ1.2.51.2.5DZVZ=8VR=3kvi15sintVvo(t)R3kΩZD+−Ov+−Iv1.2.51Z0V(8V)v=ZDOIvv=1ZVv≥ZDO8Vv=10.7V0v−ZDOIvv=10.7Vv≥−ZDO0.7Vv=−Ov1.2.5==ΩD26261r==ΩbD0.65R3252==ΩD26132r==Ω1.3.21.3.2DVPaDVD=0.3VDbc(a)10V−R10kΩ1D2DAB0.2V3.3VPVPD1V15V2V12VR3kΩ+−PVP(b)(c)1D1V6V2V12V2D3kΩ+−PVP1.3.2=bDPV12V=−c1D2DPV6V=−1.3.31.3.3Dt1:1+−iv1D2D3D4DLR+−ov1.3.31D3D2D4D2D4D1D3D1.3.3L1utωtωtωiu1Lu2Lu1.3.311Diu2e1D3DLR1/21.3.3L2u=iu1D2D2D1D2D1D31Diu2D2D1.3.4vs(t)=9sintV1.3.4vot(a)R+−Sv1D3V6V2D+−Ov(b)R+−Sv1D3V2D+−Ov1.3.4(a)3sUv−1D2D3oUv=−6sUv−1D2D6oUv=−63svUv−−1D2D()osUUt=oUsUb3sUv1D2D3oUv=0sUv1D2D0oUv=03svUv1D2D()osUUt=(1.3.51.3.515Vvot1D2D10kΩ10kΩ2V10V+−()Svt+−()Ovt1.3.52D1D1021062oUv−=−=6sUvoU10sUv1D2D10oUv=610svUv1D2D()osUUt=oUsU=200vO+−R1kΩ1D2Div+−Ov1.3.6(1)i00.7Vv1D2Doivv=i0.7Vv≥1D2Do0.7Vv=i0.7Vv≤−2D1Do0.7Vv=−oviv1.3.6a(2)1.3.6bthV0.5V=D200r=Ωi00.5Vv1D2Doivv=i0.5Vv≥1D2Di0.5Vv≤−2D1Di0.5Vv≥ithODthDVVRvvrr−=++om(6-0.5)VV2000.5V1.42V(1000+200)=×Ω+≈Ωi0.5Vv≤−ivOv1.3.6ctωtωiVvoVvivovDr40Ω0.6VDR=1kΩtωtωiVvoVvabc1.3.6(a)vO(t)12(a)(b)+−D5V50Ω200Ω()Ivt+−()Ovt010/tms()/IvtV301.3.7DD'I()vt'III50()()0.2()(20050)vtvtvtΩ==+Ω1'I()0.2vt=I()5VvtI()25VvtDOI0.2()vvt='I()5Vvt≥I()25Vvt≥DO5Vv=Ov1.3.7a21.3.7b'I()5.7VvtI()28.5VvtD'OII()()0.2()vtvtvt=='I()5.7Vvt≥I()28.5Vvt≥DO()5.7Vvt=O()vt1.3.7co()vt50Ω5VD200ΩI()vtmstI()Vvtab1.3.71.3.81.3.8VT26mVVth=0.7VCvi(t)=15sint(mV)1vi2id+−R2V500ΩCivDiD1.3.8=−DDVuiR=−1.3.8DDU0.7V,I2.6mA≈≈TdDU2610I2.6r⎛⎞≈=Ω=Ω⎜⎟⎝⎠(t)=15sintmVivωdd()15sint()mA1.5sintmA10ivtitrωω⎛⎞===⎜⎟⎝⎠msiVuQ1.3.8