标题: 60335 8.1.4 保护阻抗 [打印本页] 作者: addwin 时间: 2010-1-21 21:20 标题: 60335 8.1.4 保护阻抗 60335 8.1.4 里面的保护阻抗的漏电电流应该怎么测量,因标准上说有直流值和峰值。?! m1 E$ K P" K/ a8 j+ r$ B1 ]/ o1 W% A
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if protective impedance is used, the current between the part and the supply source shall not exceeed 2mA for d.c.,its peak value shall not exceed 0.7mA for a.c. and ) C0 R% |0 e' W
-for voltages having a peak value over 42.4 V up to and including 450V,the capacitance shall not exceed 0.1uF,( l6 E( X, c5 _- X* b2 S5 ]3 U
- for voltages having a peak value over 450V up to and including 15KV, the discharge shall not exceed 45uC.作者: wbfsy 时间: 2010-1-21 21:39
一般都是计算而得。 8 R2 E! `* }6 G4 V% D+ Z 根据 22章的要求,阻抗保护需要有两个,在一个实效的情况下同时需要满足8。1。4的要求 / o6 D. z% f: H/ ~+ o2 }* [: M+ u 注意交流电的峰值是 根号2的倍数关系,正玄波的关系作者: addwin 时间: 2010-1-21 21:44
但具体用什么仪器测量。因为是电流。用示波器只能测出电压。作者: wbfsy 时间: 2010-1-21 21:49
计算而得,所有测试设备都有内阻,一旦你联上电流表数值就不等同了作者: addwin 时间: 2010-1-21 21:51
if protective impedance is used, the current between the part and the supply source shall not exceeed 2mA for d.c.,its peak value shall not exceed 0.7mA for a.c1 y2 O/ `5 E/ r# E/ J5 F, I