1、习题6.1R0.83CCBEEFVVImAR11212(1)REFCBBCIIIII10.81421REFOCIIImA习题6.21001故忽略三极管基极分流,可得11212151BECRRVIIImARR22330.5RRIRImAR30.5ORIImAor 习题6.3由VT1和R1可得方程:111GSDDDVIRV2111()2DnoxGStWICVVL12.8470.82GSVVV或(舍去)1107.65DIA2211(/)21.53(/)DDWLIIAWL习题6.4该电路作为恒流源电路设计,基本要求VT2要工作在放大区,因此有2()0.3ECEC satVVV由电路可得22222222
2、220.30.30.314.7ECECEEECEEOCEEOEEEEOOVVVVIRVVI RVVI RVVVVVRkII121214.7RRRRk又习题6.550.70.7(5)18.6REFVImAk 足够大,故所有三极管基极电流为0,则所有BJT均匹配,故I1=IREF=1mAI2=3IREF=3mAI3=2IREF=2mA习题6.6假设所有晶体管的其他参数均匹配,且都工作在饱和区,则VT1和VT2、VT3分别构成两个电流源,可得 221(/)10(/)REFWLIIAWL3341(/)20(/)REFWLIIAIWL又VT4和VT5构成一个电流源,可得5544(/)80(/)WLIIA
3、WL习题6.7VT1、R1和VT2构成参考支路,可得12110(10)1.86EBBEREFVVImAR 0,BCEIII VT1和VT3、VT4构成电流源,可得3451.86REFIIImAIVT5和VT6构成电流源,可得651.86IImAVT10和VT11构成电流源,可得VT2和VT7、VT8、VT9构成电流源,可得723.72REFIImA9101.86REFIImAIIREFI3I4=I5I6I7I9=I10I1110111.86IImA求各支路电流IREFI3I4=I5I6I7I9=I10I11习题6.713410EEEVVVV求各节点电压11349.3CBBBVVVVV3323.
4、72CVI RV45560.7CCBBVVVVV560EEVV66351.86CVI RV227899.3CBBBBVVVVVV 278910EEEEVVVVV 78743.72CCVVI RV 91010114.3CCBBVVVVV10115EEVVV111151.86CVI RV习题6.8由题意可得该电路的共模增益为0.1/ocmivAV Vv 则共模抑制比为15020lg|20lg|20lg150063.520.1dcmACMRRdBA习题6.9i120.005sin(V)vti20.50.005sin(V)vt121.50.01sin()idiivvvt V121.252iiicmvv
5、vV习题6.10直流分析:12/20.25DDIIImA小信号参数:1212(/)2.5/mmmnoxDgggCWL ImA V12140AoooDVrrrkI(/)4.762/2LdmoDRAgrRV V(1)差分输出时的差模增益(2)单端输出时的差模增益为11(/)3.125/2dmoDLAgrRRV V 共模增益为1/0.0033/2DLvcmSSRRAV VR 共模抑制比为11|937.5dvcmACMRRA习题6.1112/21CCIIImA小信号参数:12/40/mmmCTgggIVmA V121100AoooCVrrrkI(/)363.636/dmoCAgrRV V(1)差分输出
6、时的差模增益(2)单端输出时的差模增益为11(/)181.818/2dmoCAgrRV V 共模增益为1|0.025/2CvcmEERAV VR 共模抑制比为11|7272.72dvcmACMRRA习题6.1211212Q1Q21(/)(|)22,22Q2Q120.630.63GSGSGSidGSOVidpttOVtStStOVtOVidIkWLvVvVVvVVVvvVVVVVvVVvV 。当所有电流流经管,截止。此时即当所有电流流经管,管截止时,同理可推得,0.8tpVV 2/3.5/pk WLmA V122121211210.630.7,0,0.8,2.520.71.1,2.520.63,
7、0,0.7,0.630.81.432.5,1.1idDDStpDDidDDSidGSDDvVimA iVVV vkmVvVvV iimAVvvVvV vV 、当,。、当习题6.12习题6.13(1)直流分析:假设所有BJT都工作在放大区VT330.9784.7ZBEEVVImAk1233/2/20.485CCCCEIIIIImA1230.7EECVVVV 129.5265CCCCCCVVVI RV37.4EEEZBEVVVVV 显然三个BJT都工作在放大区(2)小信号参数:98.94/dmCAg RV V 差模电压增益/19.4/mCTgIVmA V(3)1216,016,8iiidicmvm
8、V vvmV vmV习题6.14(1)静态工作点的求解02.75022715BBEEEIkVIIk(1)EBII0.264EImA0.2621CEIImA2BECIIIA2.75.4BBVIkV 6.1EBBEVVVV 155.568CCCVI RV(2)小信号参数(1)9.564TEVrkI习题6.14(2)差模电压增益(/)241.585/(1)2LCdPBRRAV VRRr 差模输入电阻(1)17.3142PidBRRRrk差模输出电阻272odCRRk习题6.15(1)VT2和VT3构成一对电流源,为VT1提供直流偏置,并作为VT1的有源负载312.3BEVImA111.25TVrkI
9、140/CmTIgmA VV(2)112(/)1000/vmooAgrrV V 12|50AooVrrkI11.25iRrk12/25oooRrrk习题6.16VT2和VT3组成恒流源电路,作为放大管VT1的有源负载,则2123(/)100(/)DDREFWLIIIAWL1112(/)0.632/mnoxDgCWL ImA V11200AnoDVrkI22|100ApoDVrkI112(/)42.164/vmooAgrrV V iR 12/66.667oooRrrk习题6.1721212()2ovvvmmrAA Ag gAnAp|VV且偏置电流相同1212oooVTVTrrr和的输出电阻为又偏
10、置电流源的输出电阻与VT1和VT2的输出电阻相等 112(/)2ovmirAgR 222ovmrAg 2iR 又习题6.181CImA40/CmTIgmA VV2.5TCVrkI2.5iRrk100AoCVrkI4000/vom oAg rV V 100ooRrk(1)(2)调整电流源I使得Ri变成原来的4倍,则/40.25IImA10/CmTIgmA VV400AoCVrkI4000/vomoAg rV V 400ooRrk10iRk(3)如果放大器由一个内阻为Rsig=5k的信号源提供信号,且接有一个100k的负载电阻 第一种情况下2.510040002.55100 100666.67/i
11、LvvoisigLoRRAARRRRV V 第二种情况下101004000105100400533.33/iLvvoisigLoRRAARRRRV V 习题6.18习题6.1924(/)80odmooidvAgrrv1234/2DDDDIIIII2(/)mnDgk WL I24|AooDVrrI0.2ImA(*)习题6.20当I=100A时1234/250DDDDIIIIIA122(/)0.141/mmmnDgggk WL ImA V24/200oooRrrk24|400AooDVrrkI28.2/vdmoAg RV V习题6.20当I=400A时1234/2200DDDDIIIIIA122(/)0.282/mmmnDgggk WL ImA V24/50oooRrrk24|100AooDVrrkI14.1/vdmoAg RV V习题6.211234/21CCCCCIIIIIImA1240/CmmmTIgggmA VV24|100AoooCVrrrkI122.5mrrkg24(/)1600/vdmooLAgrrRV V25idRrk24/50oooRrrk