模电作业题答案 黄丽亚_杨恒新_机械工业出版社(汇总)

1模电作业题答案电子科学与工程学院2+6VDR100Ω图1.3(1)流过二极管的直流电流1.3D(on)D60.7A=53mA100EUIR(2)二极管的直流电阻D(on)D3D0.7V13.25310AURI二极管的交流电阻3D3D2610V0.495310ATUrI3+10VDRL100Ω图1.41.4(1)设二极管为理想二极管，流过负载LR的电流10A100mA100LEIRE(2)设二极管为恒压降模型，流过负载LR的电流D(on)L100.7A93mA100EUIR4+10VDRL100Ω图1.41.4E(3)设二极管为折线模型，流过负载LR的电流D(on)LD100.777.5mA10020EUIRr(4)将电源电压反接时，流过负载电阻的电流S0III或5(5)增加E，其他参数不变，+10VDRL100Ω图1.41.4EEDIDDTUrI故二极管的交流电阻下降61.6(a)设V1、V2截止，则D13VuD25VuV2正偏电压高，优先导通。再设V1截止，则假设不成立；D12Vu假设成立。o5VU71.6(b)设V1、V2截止，则D13VuD23VuV1导通。再设V2截止，则假设不成立；D26Vu假设成立。o0VU81.6(c)设V1、V2截止，则D16VuD20VuV1优先导通。再设V2截止，则假设不成立；D20Vu假设不成立。4o1234622V(//)4(1//0.051)2ERURRRR91.7(a)设V截止，则Di20.7Vuu即o2Vui2.7Vu当i2.7Vu时，V导通，则oi0.7Vuu此时101.7111.7(b)设V截止，则Di20.7Vuu即oiuui1.3Vu当i1.3Vu时，V导通，则o1.3Vu此时121.7131.9u1、u2中至少有一个为3V，则uo＝3-0.7＝2.3V。u1、u2均为0时，uo＝-0.7V。141.9151.10当ui≥5V时，Vz击穿，uo＝5V。当ui≤-0.7V时，Vz正向导通，uo＝-0.7V。当-0.7V＜ui＜5V时，Vz截止，uo＝ui。161.11(1)max3maxmin22104003010iZZLUUII3181080045780051010RILIZDZRLUiUo171.11(2)250LR1040mA250LImaxmax()(3040)0.11017ViZLZUIIRUminmin()(540)0.11014.5ViZLZUIIRURILIZDZRLUiUo1.11(3)RILIZDZRLUiUoLRkRL1o2o1100010.2410.12V121000LZLRUUrR10.1210.240.12VoUiZ()ZZUUrRI310(12)2010R488Ro1ZiZ12()101010.24V12488ZZLrUUUUrR19NPNPNPEBUUBCUUEBCUUUEBUUBCUUEBCUUU2.10.7V8V0V(a)-0.7V-8V0V(b)4V1V3.7V(c)4V4.3V9V(d)题图2.1202.10.7V8V0V(a)-0.7V-8V0V(b)4V1V3.7V(c)4V4.3V9V(d)题图2.1NPNPNPCBEUUUCBEUUU先确定基极，再确定发射极，最后确定集电极根据发射结电压确定材料212.10.7V8V0V(a)-0.7V-8V0V(b)4V1V3.7V(c)4V4.3V9V(d)题图2.18V0V0.7V硅NPN-8V0V-0.7V硅PNP1V4V3.7V锗PNP锗NPN9V4.3V4V题图（a）题图（b）题图（c）题图（d）222.3-5V-0.3V0V3AX81(a)12V3V0V(d)3DG86.6V6.3V7V3CG21(c)8V2.5V3V(b)3BX1题图2.3(b)e结反偏，c结反偏，截止状态(c)e结正偏，c结正偏，饱和状态(d)e结开路，晶体管损坏(a)e结正偏，c结反偏，放大状态EESTeuiIU14E10Ai0.7VuE4.9mAi3Vu36E1.310Ai232.61215Q50,Q00.3点：点：。(BR)CEOCM40V,1524320mWUP242.712V-2V(a)1k20kC1BE0.7VU(a)设管子截止，则有故假设成立CB0U管子处于截止状态252.7TRB470kRC2kRE1k(b)15V(b)BE0.7VU设管子截止，因故发射结导通，假设管子处于放大状态BQBE150.725μA(1)IRRCQBQ2.5mAIICEQ152.537.5V0.7VU故假设成立，管子处于放大状态。262.7TRB100kRC2kRE1k(c)15V(c)BE0.7VU设管子截止，故发射结导通，假设管子处于放大状态因BQ71μAICQ7.1mAICEQ157.1(21)6.3V0.7VU故假设不成立，管子处于饱和状态。272.9B2RB2CCB1B215(1)122.9V4715RUURRRB2CQE0.32.90.32mA1.3UIRCEQECQCCCQEC()122(1.32)5.4VUUUIRR28CCBB1E0.3120.30.066mA(1)471011.3UIRR(2)当RB1开路时，IBQ=0，管子截止，UC=0当RB2开路时，设管子处于放大状态，则有2.9CCCE(c-sat)B(c-sat)EC120.30.035mA()(1.32)100UUIRRBB(c-sat)IICCCE(c-sat)CCCE120.327.1V21.3UUURRR故管子处于饱和状态CEQECQCCCQEC()120.066100(1.32)9.78V0.3VUUUIRR故，管子处于饱和状态或ECEBUUBICI略CEBEUUC结正偏29CCBECQB1EB1120.31002mA(1)1011.3UUIRRR2.9(3)当RB2开路时，设管子处于放大状态B1454KΩR＝解得假设成立，故CEQECQCCCQEC()123(1.32)2.1V0.3VUUUIRR302.18C1RB500kRC2kC2TuoCERE1kuiUCC312468101221345610μA20μA30μA40μA50μA60μAuCE/ViCE/mA①②③④02.18322.18(1)输出特性理想化，100,r,UceACCBE(on)BQBE1220μA(1)500100UUIRRCECCCCEC()123UUIRRICECCCE0,4mA;0,12VUIIUCEQCQ(6,2mA)UVI(2)先求工作点直流负载线取两点可得直流负载线如图2.18（b）中①线，工作点Q交流负载线的斜率为可得图2.18（b）中②线（交流负载线）C112R332.18(3)此时直流负载线不变，仍如图2.14(b)中①线，而交流负载CL11//RR如图2.18(b)中③线。线的斜率为342.18omUQCEQCQ(3V,3mA)UIBRBQ30μAICCBEon63BEB123010(1)10110UURRRB300kR(4)为得最大,工作点应选在交流负载线中点。是此交流负载线之中点，即点此时，调节使,则解得将图2.18(b)中③线平移使之与直流负载线①线的交点om3VU352.20(1)要求动态范围最大，应满足CQLCEQCESCCCQCCESIRUUUIRUCQCQ(2//20)122IICQCQBQ503.1mA,0.062mA3IIICCBEBBQ120.7182KΩ0.062UURI即解得故36CCCCEQCQ1212V,4KΩ,4V,2mA3URUILCLLL2//4//1,1.3KΩ2RRRRRCQCCBEBQBBQ2120.70.04mA,283KΩ500.04IUUIRIOPPom22(64)4VUU(2)由直流负载线可知2.20372.23RSCRCVRL++UoUs+UiUCCRB+382.23ebr'cerBQ20μAICQBQ1000.022mAIICEQCEQCEQCCCCCCQCL9222UUUUIRUIRRCEQ2.5VUbeEQCQ2611001.3kΩ2TTUUrIIAceCQ10050kΩ2UrI(1)计算工作点和解得392.23(2)计算源电压放大倍数suAii=ibRCRLrceicioUo++Us+RoRiRsibeUUberbioceCLbececLisisbeSbeSbe////////uUrRRrrRRUAUUrRrRr50//2//2100690.151.3402.23ibe1.3kΩRroccec//2kΩRRrR(3)计算输入电阻和输出电阻ii=ibRCRLrceicioUo++Us+RoRiRsibeUUberbi412.27RSCVRLUs+UiIBRiRoREUCCUo422.27+RiUiUo+RoRERL+UsibRsbiRSCVRLUs+UiIBRiRoREUCCUo432.27CQBQ1000.022mAIIbeCQ261001.3kΩ2TUrIibeEL1//1.311002//2102.3kΩRrRRSbeoE0.151.3//2//kΩ14.41101RrRR+RiUiUo+RoRERL+UsibRsbi442.27ELooiissisiSi1//uRRUUURAUUURRRELSbeEL1//101(2//2)0.9861//0.15102.3RRRrRR+RiUiUo+RoRERL+UsibRsbi452.30462.30B2BBEBCCCQB1B2E151240.7(1)4V,1.65mA30152RUUUUIRRRCEQCCCQCE()121.65(23)3.75VUUIRRbebbCQ2626(2)50508381.65rrIoLibe50(3//389.50.838uURAUr）＝beiE0.838//2//16151rRRoC3KΩRR472.34V1V2RBRCRERiRoRL+UiUoUCC+48+UiRBV1V2RERiRi1Ri2RoRCRL+Uo2.34iBbeEbe//(1)(//)22.5kΩRRrRroLibe50(5//51251uURAUr）＝o5kR493.1uGS/ViD/mA12312UDS=10V(a)uGS/ViD/mA12312UDS=10V(b)34uDS/ViD/mA24651015204.5V4V3.5V3V2.5VUGS=2V(c)-2-1-3-2-1-3DSS2mAI图(a)：N沟耗尽MOSFET，GS(off)3VUD0i