标题: CL.8.1.4 OF IEC 60335-1 电量的计算 [打印本页] 作者: plmijb 时间: 2016-8-15 16:43 标题: CL.8.1.4 OF IEC 60335-1 电量的计算
0 u/ A& e' W; f- {# \5 ^ 求解：标准中有说电量的计算是通过2000 Ω电阻，测出电压/时间曲线，求出电路二端的放电电量。好久没用高数了，对于此电压/时间曲线，如何计算？积分公式是怎样的？, v. Y* J9 j9 V- ]% X' h. h' W
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If protective impedance is used, the current between the part and the supply source shall & x" k( q/ O2 inot exceed 2 mA for d.c., its peak value shall not exceed 0,7 mA for a.c. and & V, O; f% X7 k0 J: T– for voltages having a peak value over 42,4 V up to and including 450 V, the capacitance * Z4 y r" \8 C! |shall not exceed 0,1 μF;0 I: _1 u2 i$ I4 D8 H. x
– for voltages having a peak value over 450 V up to and including 15 kV, the discharge shall0 L+ a' L$ e" r
not exceed 45 μC;; n5 n* ~# Y3 s/ G s0 Y: U5 v
– for voltages having a peak value over 15 kV, the energy in the discharge shall not exceed 9 p' S* A6 H. p" i1 B, {: z350 mJ.9 i z/ W; M& i3 d8 _, ]& q2 j
Compliance is checked by measurement, the appliance being supplied at rated voltage.; j' I1 h" \) U) [; w
Voltages and currents are measured between the relevant parts and each pole of the supply; H+ k" R0 Z3 ~, R3 t6 V
source. Discharges are measured immediately after the interruption of the supply. The" U }/ _5 X( q5 ?. b* V2 ?; w
quantity of electricity and energy in the discharge is measured using a resistor having a5 F: D9 r8 q' m1 p0 m
nominal non-inductive resistance of 2 000 Ω . " B- X8 l! v, W x$ {4 b$ k0 ? . U; H3 p+ {" j* [* r$ L作者: ID长就牛b 时间: 2016-8-15 17:03
有的示波器好像是带这个功能的，作者: vcgtrf 时间: 2016-8-16 13:36 本帖最后由 vcgtrf 于 2016-8-16 13:39 编辑 + V' C5 L' }! Q- m8 A; e6 V ' k2 ^! a* C7 m% c直接用示波器的都是，积分出来的值再除以2000就是要的值了作者: plmijb 时间: 2016-8-16 16:29
查DLM 示波器，只有此项功能：通过计算焦耳积分(I2t) 测量浪涌电流，是用它除以2000？依据是什么？作者: zyhua1985 时间: 2016-8-17 09:36
直接用示波器作者: vcgtrf 时间: 2016-8-17 10:20 本帖最后由 vcgtrf 于 2016-8-17 10:23 编辑 & g% [# i* R' e0 L( M3 X3 g4 P1 d
: ,看来坛子里IT 人不多
:):):):)
