电力电子技术王兆安第五版课后习题答案

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··································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I

、I、I与

电流有效值I、I、I。

a)

b)

c)

图1-43 晶闸管导电波形

21πIm

Isinωtd(ωt)=(+1)≈0.2717 Im 解：a) Id1=πm

2π42π2

I1=

12π

π

π4

(Imsinωt)2d(ωt)=

Im231+≈0.4767 Im 42π

1πIb) Id2 =πImsinωtd(ωt)=m(+1)≈0.5434 Im

π4π2

I2 =

1π2Im2

(Isinωt)d(ωt)=πm

2π431

+≈0.6741I42π

多少？这时，相应的电流最大值Im1、Im2、Im3各为多少?

解：额定电流I T(AV) =100A的晶闸管，允许的电流有效值I =157A，由上题计算结果知

I

a) Im1≈≈329.35， Id1≈0.2717 Im1≈89.48

0.4767I

b) Im2≈≈232.90, Id2≈0.5434 Im2≈126.56

0.6741

1

c) Im3=2 I = 314, Id3= Im3=78.5

4

5. 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α1和α2，由普通晶闸管的分析可得，α1+α2=1是器件临界导通的条件。α1+α2＞1，两个等效晶体管过饱和而导通；α1+α2＜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：