最全电力电子技术西安交大练习题及答案.doc

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1、目录第1章电力电子器件·········································································1第2章整流电路··················································································4第3章直流斩波电路·······································································20第4章交流电力控制电路和交交变频电路·········

2、·························26第5章逆变电路················································································31第6章PWM控制技术·····································································35第7章软开关技术············································································40第

3、8章组合变流电路·······································································42第1章电力电子器件1.使晶闸管导通的条件是什么？答：使晶闸管导通的条件是：晶闸管承受正向阳极电压，并在门极施加触发电流（脉冲）。或：uAK>0且uGK>0。2.维持晶闸管导通的条件是什么？怎样才能使晶闸管由导通变为关断？答：维持晶闸管导通的条件是使晶闸管的电流大于能保持晶闸管导通的最小电流，即维持电流。要使晶闸管由导通变为关断，可利用外加电压和外电路的作用使流过晶闸管的电流降到接近于零的某一数值以下，即

4、降到维持电流以下，便可使导通的晶闸管关断。3.图1-43中阴影部分为晶闸管处于通态区间的电流波形，各波形的电流最大值均为Im，试计算各波形的电流平均值Id1、Id2、Id3与电流有效值I1、I2、I3。02044520242a)b)c)图1-43晶闸管导电波形解：a)Id1=1pImpImsinwtd(wt)=(21)0.2717Im2π42π2I1=1p2p(Imsinwt)d(wt)=Im310.4767Im2p41p242pIm2b)Id2=πpImsinwtd(wt)=π4(1)0.5434Im2I2=1p2p(Imsinwt)2

5、Im31242pd(wt)=0.6741Ip41p1c)Id3=2Imd(wt)=Im2π041p21I3=2Imd(wt)=Im2p024.上题中如果不考虑安全裕量,问100A的晶闸管能送出的平均电流Id1、Id2、Id3各为多少？这时，相应的电流最大值Im1、Im2、Im3各为多少?解：额定电流IT(AV)=100A的晶闸管，允许的电流有效值I=157A，由上题计算结果知Ia)Im1b)Im20.4767I0.6741329.35，Id10.2717Im189.48232.90,Id20.5434Im2126.561c)Im3=2I=314,Id3=4

6、Im3=78.55.GTO和普通晶闸管同为PNPN结构，为什么GTO能够自关断，而普通晶闸管不能？答：GTO和普通晶闸管同为PNPN结构，由P1N1P2和N1P2N2构成两个晶体管V1、V2,分别具有共基极电流增益a1和a2，由普通晶闸管的分析可得，a1+a2=1是器件临界导通的条件。a1+a2＞1，两个等效晶体管过饱和而导通；a1+a2＜1，不能维持饱和导通而关断。GTO之所以能够自行关断，而普通晶闸管不能，是因为GTO与普通晶闸管在设计和工艺方面有以下几点不同：1)GTO在设计时a2较大，这样晶体管V2控制灵敏，易于GTO关断；2)GTO导通时的a1+a2

7、更接近于1，普通晶闸管a1+a21.15，而GTO则为a1+a2条件；1.05，GTO的饱和程度不深，接近于临界饱和，这样为门极控制关断提供了有利3)多元集成结构使每个GTO元阴极面积很小，门极和阴极间的距离大为缩短，使得P2极区所谓的横向电阻很小，从而使从门极抽出较大的电流成为可能。6.如何防止电力MOSFET因静电感应应起的损坏？答：电力MOSFET的栅极绝缘层很薄弱，容易被击穿而损坏。MOSFET的输入电容是低泄漏电容，当栅极开路时极易受静电干扰而充上超过20的击穿电压，所以为防止MOSFET因静电感应而引起的损坏，应注意以下几点：①一般在不用时将其三

8、个电极短接；②装配时人体