电力电子技术王兆安第五版课后习题答案..doc

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 PDF created with pdfFactory Pro trial version  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：uAK0 且 uGK0。 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 Id1、Id2、Id3 与电流有效值 I1、I2、I3。 0 2 0 4 4  5 2 0 2 4 2 a) b) c) 图 1-43 晶闸管导电波形  解：a) Id1= 1 p I m p I m sinwtd (wt ) = ( 2 1 ) 0.2717 I 2π 4 2π 2  I1=  1 p 2 p ( I m sin wt)  d (wt) = I m 3 1  0.4767 Im 2p 4 1 p 2 4 2p I m 2 b) Id2 = π p I m sinwtd (wt ) = 4 ( 1 ) 0.5434 Im 2  I2 =  1 p 2 p ( I m sin wt)  d (wt ) =  0.6741I p 4 1 p 1 c) Id3= 2 I m d (wt ) = Im 2π 0 4 2 1 I3 = 2 I m d (wt) = Im 2p 0 2 4. 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 1 PDF created with pdfFactory Pro trial version 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 b) Im2  0.4767 I 0.6741 329.35， Id1 0.2717 Im1 89.48 232.90, Id2 0.5434 Im2 126.56 1 c) Im3=2 I = 314, Id3= 4 Im3=78.5  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