电子技术习题ALL

++RDEuoui(a)++RDEuoui(b)++RDEuoui(c)++RDEuoui(d)ouit10-10ouototouotouot(a)(c)(d)(b)1.1在图示各电路中，E=5V，二极管的正向压降可忽略不计，试分别画出输出电压uo的波形。55555iU2tou2t1.1：5iuuiiuuDuVu0,0,5截止，若VuDuVui5,0,50导通，若RiuouDEutiu5解:将D开路，求其阳、阴两极开路电压u，若u0，则D导通；反之D截止。(a)iU2tou2t1.1：Riu0uDEutiu5(c)iuu5VuDuVui5,0,50导通，若iiuuDuVu0,0,5截止，若1.2试求输出端Y的电位VY及各元件（R、DA、DB）中通过的电流：（1）VA=VB=0V；（2）VA=+3V，VB=0V；（3）VA=VB=3V。AB+12VDADBR3.9kΩYIAIRIB（1）DA通，DB通，VY=0V；IR=2IA=2IB=12/3.9=3.08mA（2）DB通，DA止，VY=0V；IR=IB=12/3.9=3.08mA（3）DA通，DB通，VY=3V；IR=2IA=2IB=9/3.9=2.3mAiuouIOR1.3图示电路中，若稳压管的稳定电压UZ都是6V，正向导通压降均为0.6V，试画出输出电压uo的波形。10sinViutouit10-10ouot6.66.6-6.6-6.6CCUCCUCCUCCU2.3试判断图示各个电路能不能放大交流信号？为什么？可以，偏置正确可以，偏置正确不可以，无法稳定Q点不可以，输入端短路80604020)(VuCE)(mAiCQ交流负载线264810124351CCCUCRLR0u2CBRTiu125tanCR因此RC=2.5KΏ（1）从直流负载线可以看出，UCC=10V2.41CCCUCRLR0u2CBRTiu80604020)(VuCE)(mAiCQ交流负载线264810124351040CCBECCBBBBUUUIARRR250BRkQ点对应的ICQ=2mA,IBQ=40µA，UCEQ=5V1CCCUCRLR0u2CBRTiu80604020)(VuCE)(mAiCQ交流负载线2648101243532582//1CLRR交流负载线的斜率kRL75.31CCCUCRLR0u2CBRTiu80604020)(VuCE)(mAiCQ交流负载线26481012435（2）由交流负载线可知输出电压的最大幅值为（8-5）VuuomimAAUU3t)(VuCEOutCQI)(VuCEQ)(mAiCCEQU1Q2Q交流负载线],min[CEQCEQomUUUUU21uomimAUU1U2UkmAImVrEbe96.0)()(26)1(300-7896.05.150beLCiour//RRUUAmVAAUUuuomim383mVrIriUbeBQbebmim38iut)(AiBBQIBQI)(VuBEtBEQU)(AiB)(VuBEQ1Q2QiuCRLR0uCEBRBbibiberSRSeiibBQBiIiBQbmIiibbeUir(3)输出首先出现什么失真？80604020)(VuCE)(mAiCQ交流负载线26481012435t)(VuCEOutCQI)(VuCEQ)(mAiCCEQU1Q2Q交流负载线],min[CEQCEQomUUUUU21uomimAUU1U2U(4)增大RL，对直流负载线无影响，但使交流负载线的斜率绝对值减小80604020)(VuCE)(mAiCQ交流负载线26481012435'1'LRtg交流负载线的(5)减小一半ARRURUUIBBCBBECB40101CCCUCRLR0u2CBRTiu降为原来的一半CBCIII变大CCCCCERIUU)()()()()(mAImVmAImVrBEbe26300261300减小一半beLCiour//RRUUA1CCCUCRLR0u2CBRTiu不变berCCUCRBRiU.ER01.U02.U+_2.5CCUBRCRERCIBIEICEUBEU直流通路CCUBRCRERCIBIEICEUBEUARRUUIEBBECCB301)(mAIIBE5.1)1(kmAImVrEbe18.1)()(26)1(300BEBEBBCCIRUIRU)(1CCUCRBRiU.ER01.U02.U+_BRCR01UberbIbIcIERiU+11=0.97(1)oCuibeEURAUrREbbebiRIrIU)1(CboRIU1BRCR02UberbIbIcIERiU+_CCUCRBRiU.ER01.U02.U+_BRCR02UberbIbIcIERiU+_22(1)=0.99(1)oEuibeEURAUrREbbebiRIrIU)1(EboRIU)1(2ERERUIBRUberbIbIcIERREIIEIbeErURU)(1I21beorUrIbebrIUbebrUI求输出电阻IUro2REbREEIIIII)(11oCrRCCUCRBRiU.ER01.U02.U+_求输入电阻BRCRberbIbIcIiIERiriUir//..iBiiirRIUrEbebiiRrIUr)1(而2.6在图示的两级放大电路中，试问：（1）哪些电路或元件构成直流负反馈？（2）哪些构成交流负反馈，并说明其类型；（3）如果RF不接在T2的集电极，而接在C2与RL之间，两者有何不同？（4）如果RF的另一端不是接在T1的发射极，而是接在它的基极，则有何不同？是否会变成正反馈？CCU1ERLR+_SESR+1CRB1RB2RC1RC2+2T1T+2EC2ERFR+_oU.CCU1ERLR+_SESR+1CRB1RB2RC1RC2+2T1T+2EC2ERFR+_oU.直流负反馈：RE1、RE2RF温度IC2UBE1IB1VC2VE1IC1VC1IB2IC2CCU1ERLR+_SESR+1CRB1RB2RC1RC2+2T1T+2EC2ERFR+_oU.(+)(-)(-)(+)+RF：电压串联负反馈RE1：电流串联负反馈V20'1ERLR+_iU.+1CRB1RB2RC1+2T1T2ER+_oU.''1ER+2.7（1）画直流通路V20'1ERRB1RB2RC12T1T2ER''1ER第一级：VRRRVBBBB4202121mARUVIIEBEBEC139.36.0411111AIICB25111VRRRIUUEECCCCCE6.64.1320)(''1'1111V20'1ERRB1RB2RC12T1T2ER''1ERkIrEbe34.126)1(300111第二级：VRIVVCCCB1020112mARUVIIEBEBEC84.11.56.01022222AIICB46222VRIUUECCCCE6.10222V20'1ERRB1RB2RC12T1T2ER''1ERkIrEbe879.026)1(300222V20'1ERLR+_iU.+1CRB1RB2RC1+2T1T2ER+_oU.''1ER+（2）求交流参数1BR2BR1CRiU1ber1bI1bI1cI+__+2ber2bI22bILR.OU''1ER2ER)1(//''1121111EbeiCiouRrrRUUA+_1.OU1BR2BR1CRiU1ber1bI1bI1cI+__+2ber2bI22bILR.OU''1ER2ER2ir''1122211)1(]//)1(//[EbeLEbeCRrRRrR+_1.OU1BR2BR1CRiU1ber1bI1bI1cI+__+2ber2bI22bILR.OU''1ER2ERLEbeLEOouRRrRRUUA//)1(//)1(222221221uuiouAAUUA求输入电阻irir////21..iBBiiirRRIUr''E11)1(RrIUrbebii而+_1.OU1BR2BR1CRiU1ber1bI1bI1cI+__+2ber2bI22bILR.OU''1ER2ER+_1.OU1BR2BR1CRiU1ber1bI1bI1cI+_2ber2bI22bI''1ER2ERro求输出电阻2122121)(beCEbeCEorRRrRRr）（1CR2ber2bI22bI2ERroERCCUBRRC1TRC2TBR+uoui1ui2_uo1uo2-6V2.9(1)求静态值CCUBRRC1TERICIBIE2IE+_UBE-6V62EEBEBBRIURImARRUEIEBBEEB01.0)1(2CCUBRRC1TERICIBIE2IE+_UBE-6VmARRUEIEBBEEB01.0)1(2EBCImAII5.0VRIVBBB1.0VUVVBEBE7.0VRIVCCB45.36(2)信号分解ERCCUBRRC1TRC2TBR+uoui1ui2_uo1uo2-6Vui1=5+2mVui2=5-2mVmV222121iiididuuuumV522121iiicicuuuuBRRC1T++uic1__2RE1ocuERCCUBRRC1TRC2TBR+uoui1ui2_uo1uo2-6V(3)求单端共模输出EbeBCcRrRRA)1(2mVuRRuAuicECiccoc5.22111BRRC1T++uic1__2RE1ocumVuucoc5.21o2BRRC1T++uid1__1oduERCCUBRRC1TRC2TBR+uoui1ui2_uo1uo2-6V(4)求单端差模输出BRRC1T++uid1__1odumVurRRuAuidbeBCiddod401111mVuudod401o2ERCCUBRRC1TRC2TBR+uoui1ui2_uo1uo2-6V(5)求单端总输出mVuuuodoco5.42405.2111mVuuuodoco5.37405.2222(6)求双端总输出mVuuuooo80212(1)icBbeErRrRBRRC1T++uic1__2RE1ocu(7)求共模输入电阻及差模输入电阻2idBberRr（）ERCCUBRRC1TRC2TBR+uoui1ui2_uo1uo2-6VuiuoT3T5T4T1T22.9（1）图示是什么电路？T4和T5是如何联接的，起什么作用？（2）在静态时，VA=0，这时T3的集电极电位VC3应调到多少？设各管的发射结电压为0.6V。（3）忽略UCES，求最大输出功率Po22maxmax15()225/2212.59CESooLLUUPRRouot