求助：易触及部件放电量的测量方法

jack1295254

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IEC60335试验指甲

在60335。 60065中都有关于易触及部件放电量的测量，对峰值在450-15KV的部件，标准上说是用无感电阻测试其放电电压和时间的曲线得出其放电量，

4 ~" T) ^( k. j5 m我觉得测量的时候是将无感电阻接在易触及部件和电源的每一级之间，然后测量无感电阻两端的电压放电曲线。
* f3 W% j, X, ]$ K! P- T3 m; k另外有个担心就是怕两端有高压，会不会把示波器给烧掉呢？

3 ]/ A3 H7 [* G7 @不知道这样是不是正确的，希望大家指点下。。。

* u. a$ B! ~5 m
还有个问题就是，前阵子遇到有人问到干电池灭蚊拍的放电量测量，其格栅的电压在450-15KV的范围内， 要测试其放电量又应该怎么测呢，请注意这是干电池式的，不是充电式，标准中也未提到，所以想问下大家的意见！
4 m/ E1 V, Z/ o/ e
  o$ t' ?8 T% j- g# k# k; ?& A' H希望大家不惜赐教。谢谢~~！~！

jack1295254

顶一下,今天上班了,大家帮帮忙,指点下

fasten

8.1.4 An accessible part is not considered to be live if
– the part is supplied at safety extra-low voltage, provided that
• for a.c., the peak value of the voltage does not exceed 42,4 V,
• for d.c., the voltage does not exceed 42,4 V,
or
– the part is separated from live parts by protective impedance.
8 J8 [) z8 w/ c4 E8 UIf protective impedance is used, the current between the part and the supply source shall
not exceed 2 mA for d.c., its peak value shall not exceed 0,7 mA for a.c. and
– for voltages having a peak value over 42,4 V up to and including 450 V, the capacitance
% s7 R! h. q( O$ f+ k$ mshall not exceed 0,1F,
– for voltages having a peak value over 450 V up to and including 15 kV, the discharge shall
4 z, o! L# _# C9 Y& x5 o- R) lnot exceed 45 C.
Compliance is checked by measurement, the appliance being supplied at rated voltage.
1 A$ I; f- |% y. j% Y+ g: C5 dVoltages and currents are measured between the relevant parts and each pole of the supply
6 E7 A1 s  u( @3 k8 F( xsource. Discharges are measured immediately after the interruption of the supply.
NOTE Details of a suitable circuit for measuring the current are given in figure 4 of IEC 60990.

你的問題似乎比較的偏, 不過測試方法應該沒有問題.

8 J5 U% a& h' ?7 D1 }0 a
我觉得测量的时候是将无感电阻接在易触及部件和电源的每一级之间，然后测量无感电阻两端的电压放电曲线。
另外有个担心就是怕两端有高压，会不会把示波器给烧掉呢？
* v) y/ l& S! I+ w# }( r
4 }/ R0 m2 R5 L; `$ S根據電路圖, 略去Cs不計, 在最大峰值電壓15kV下, 測量腳間電壓為:15000*[Rb/(Rs+Rb)]=15000*[500/(1500+500)]=3750V, 你查查示波器說明書, 估計此電壓範圍應該可以測量不致燒毀.
4 Z( G5 P, s+ v" |3 C

还有个问题就是，前阵子遇到有人问到干电池灭蚊拍的放电量测量，其格栅的电压在450-15KV的范围内， 要测试其放电量又应该怎么测呢，请注意这是干电池式的，不是充电式，标准中也未提到，所以想问下大家的意见！

" _, Z$ F& {- A2 e7 \: r, ~不是很熟悉示波器, 更沒做電蚊拍, 估計根據放電圖, 用微積分可以求出放電量.

jack1295254

谢谢FASTEN的解答,但我有一个疑问就是,你所给出的这个电路图是60990中关于接触电流的一种测试电路图, 似乎和2000欧无感电阻的测量电路没有连系,

5 r, A- A/ K6 a' J5 c我的理解是无感电阻直接接在易触及部件和电源的每一级之间,然后测量电阻两端的电压和时间的曲线,最后求出的面积就是放电量吗?

fasten

引用第3楼jack1295254于2008-03-31 15:00发表的 :
谢谢FASTEN的解答,但我有一个疑问就是,你所给出的这个电路图是60990中关于接触电流的一种测试电路图, 似乎和2000欧无感电阻的测量电路没有连系,
7 \# B& x- P5 B9 [
+ P0 P9 O+ I. B1 B  o) y我的理解是无感电阻直接接在易触及部件和电源的每一级之间,然后测量电阻两端的电压和时间的曲线,最后求出的面积就是放电量吗?

标准并没有说到无感电阻：

5 y# R$ _# U) Y4 MNOTE Details of a suitable circuit for measuring the current are given in figure 4 of IEC 60990.

jack1295254

引用第4楼fasten于2008-03-31 20:26发表的  :
" k, x0 V& k: f1 Q: @0 N) N
$ m. B& j. T; X6 E0 c) ]# k$ z5 o
标准并没有说到无感电阻：

NOTE Details of a suitable circuit for measuring the current are given in figure 4 of IEC 60990.

% D+ D/ y- [$ C$ E我认为这个NOTE 1只是针对
; `+ ]3 F% N: [If protective impedance is used, the current between the part and the supply source shall
2 B; ]7 Q# k" o, c  F+ F4 @not exceed 2 mA for d.c., its peak value shall not exceed 0,7 mA for a.c. and
! c3 E+ a- U2 t+ W5 Z
  Q2 g; ?  L- y3 i, v/ X7 z而下面的三项要求是在其基础上还应该满足的要求, 因此我认为是NOTE1是指这里的电流要求,而且60990中其电流是U2/500得出的.
% `& W7 X, l. e7 [6 n, Q" k. R/ u
而且在FIGURE 4中并无提到2000欧的无感电阻的问题,因此我认为测放电量并不是用这个电路图进行的测量,此电路是用于测量60950接触电流及60335的泄漏电流所用的电路,
/ y) g0 _  l0 j3 Y* V" M& A4 \/ B
而关于NOTE 2
NOTE 2 The quantity of electricity is calculated from the sum of all areas recorded on the voltage/time graph without taking voltage polarity into account.
' k! L9 E. q0 i这句话4706翻译过来是  
4 m5 v" X4 i: \, |6 S2 r  t 注2：电量是通过记录在电压/时间曲线中的总面积计算得出，面积求和时不考虑电压极性。

  [; z, n/ {! ^% P, {/ v8 ]这里是电压/时间曲线, 因此我觉得是测量2000欧无感电阻两端电压跌落曲线来求出相应的放电量.


个人看法,望指点.欢迎讨论.

jack1295254

还有我发现FASTEN的不是最新的IEC 60335,在这里贴出2006年份的,里面有关于无感电阻的讲述..

* F! B2 o3 u9 j3 @( T
* M5 u& u. p: B' @8.1.4 An accessible part is not considered to be live if
, f. P% `! Z4 R$ ~$ o& S– the part is supplied at safety extra-low voltage, provided that
• for a.c., the peak value of the voltage does not exceed 42,4 V,
  t+ ^$ v% {& ~! Z8 c3 O1 T9 b• for d.c., the voltage does not exceed 42,4 V,
$ a( |( Q% |6 [' S) u4 o3 @or
– the part is separated from live parts by protective impedance.
/ p% O% M; J" [# ^& _not exceed 2 mA for d.c., its peak value shall not exceed 0,7 mA for a.c. and
) x( m7 ^& p0 Q
– for voltages having a peak value over 42,4 V up to and including 450 V, the capacitance
shall not exceed 0,1 μF,
$ g; ]; v  r0 d9 Q, a/ W– for voltages having a peak value over 450 V up to and including 15 kV, the discharge shall
not exceed 45 μC,
# Q' y0 ?9 L3 R1 ~5 ]7 A( l– for voltages having a peak value over 15 kV, the energy in the discharge shall not exceed
% w1 X7 H6 b& S2 w" p350 mJ.

Compliance is checked by measurement, the appliance being supplied at rated voltage.
  }8 H( V: ~5 R! L, I, V, DVoltages and currents are measured between the relevant parts and each pole of the supply
source. Discharges are measured immediately after the interruption of the supply. The
quantity of electricity and energy in the discharge is measured using a resistor having a
! d3 y) I, Y# L2 \8 Snominal non-inductive resistance of 2 000 Ω.

NOTE 1 Details of a suitable circuit for measuring the current are given in figure 4 of IEC 60990.
3 h/ @& C1 c/ K' W+ q5 h+ A# z6 UNOTE 2 The quantity of electricity is calculated from the sum of all areas recorded on the voltage/time graph without taking voltage polarity into account.

fasten

我查的是BSI EN60335-1：2006。你的是新版，有待查阅新版。

) v4 C* K5 H  ?' A! y( S, {# R所以讨论时讲清标准号与年号非常之重要有空再回你了

jack1295254

  [s:3]    [s:90]  [s:90]  [s:90]
, W$ ?5 q- e4 c
1 e9 ~/ }1 u9 r5 |好的,等着你有空来讨论哈,先谢了~!!

rxq20059

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