模拟电子技术基础学习指导与习题解答(谢红主编)第三章 思考题与习题

经典word整理文档，仅参考，双击此处可删除页眉页脚。本资料属于网络整理，如有侵权，请联系删除，谢谢！1))交)QQuu解(。３-２如图题3－2所示两级阻容耦合放大电路中,三极管的均为5.3kΩr6kΩbe2R20kΩR1.5MΩR7.5kΩ，，e1r,,,be1SbR30kΩ,RΩ,RΩ,R12kΩ,CC10μF,C30μF，b21b22e2c2132C50μF,V。eCC()求和r;rioR3.6kΩAR和。LLuS,V、V12解(1)2（IU,b1iVUI,U与U极1ib1o1iI的方向受UUIVb2o1o1b22I由ce)Ic111b2I=I)e)，输出电压U,cc12b222oRR//rR//R//R//rb21e1e1i2e1b22be27.5//30//91//62.9kΩ则rRr)RΩ3ibbe11e1Rc2rR12kΩoc2UU求A。oSSRR12kΩc2当RLL1)RAu1r)R1RAu2r26AAA0.98(200)６uu1u2r则AA(iRrSuSi3.6kΩRR//R12//3.62.77kΩ当RLLc2LRLr26Au2A0.98(46)45u249A454220249SAuSA与RR3.6k,AuSLLuS33)３3r、r、、。be1be212h;UA并指出U与UioiUuoNP,V即RR。1c2e1解U求A。oiUuUrIUbe1b1ioR(II),RR//R求I:b1UoLe1c2Le1LI(1)Ie11b1rRIR即c2c1be2e2(II)RIr)R(c1b2c1b2be22e2III将I:c11b1b2b11RIIRr)R21RIII2Rr)R22将I、I代回U得e1c2o1RUR(1)I2c1Rr)RoL1c1be22e2UIRR)R12LRr)R1L2将I代入U得b1irU1URRio)R12L1LRr)R2RR)R112LRr)RUULAoibe22RRur)R12LRr)R1L2所以U与UoirorR,U,R被UR看V管:oLibie12RRVr。c2e12be1r将rbe1Re111rrR1o1、C、C,C12335)hUUUAu2rA，；oUu1ii）分析U与UoiV1管VR2bR与RC。C这F1F222RRR//RF1F2c1F2解(h()),UoUUUAu2r求A、、。oUu1iiR//R//r(1)RAc1F2be2r2eu11be1RAu2r)R22erR//riF1be1(U与Uoi1UUBE17，，23BE2U02且当U,UBE3SoIB(满足U０时U的RSoe2用解从II:C3C312VI1mAC312kΩII1mAE3C3U12９VE3(对地)IIA3URV:视Rc2BE3e3CCIRIRUc2c2E3e3BE3IRU13IIIRc2UUVIR120.328ＶC1(对地)C2(对地)CCC1c1U0.7VE1,2求Re2RR7.66kΩ2I2(1))(２)当输入信号频率为f和f时，放大倍数的幅值约下降为中频时的LH(３3)解３(=50,VCμF,CμCC12ΩΩΩΩΩΩR，R=2０，Rb2，R,R＝２．5,R=７.５Sb1ecLCf,1L)rrf解I。EQbeiLRUVVRRBUUIBR1e与N26r30050)be0.9kΩ2.2rRRr)R0.91Ωib1b2beef:L11fL2RrC2336Si1ΩΩ3r,r，4gbebbmC。pFπA；uSf及f；LH(3-rrΩ解rbebebbrR//r377//1kΩibbegr0.04mbeUURRcri6//31ASoLrRr111SSif及f。LH11f2RrC2π236Si111f2RRC2π536cL2取fL1fL2Cr//(rRR)πbebS10.64MHz2π50010900//(10037710//110-