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|  DSH 3877 T G, W3 i: Y
7 i [& W _ S- O) ^) m; u9 ~4 ~! a
| Hazardous Voltage
, T+ d/ X; e2 u, [# n | 1.2.8.3 / 12.8.4
. n; S* t, f9 U" j" [; g; e. C5 O3 C | 60950(ed.2) & 60950(ed.3)/ _; X% p6 D# g6 [
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8 v! Z( W! a% J. e7 v* i0 V) D8 `, N& I
Standard:
8 O" G& l. n4 [IEC60950, 2nd (3rd) Edition
8 d4 h4 x# g6 z: [Sub clause:6 R- E3 [2 t$ ?# G/ H* W% |
1.2.8.3(1.2.8.4)
7 l/ X% Y; C5 t6 M# S L8 YSheet n. 387
1 P, V% n: }' W) a. z# B E$ iPage 1 (1)
( C5 ^, A4 U7 V. a( @Subject:
1 D& c& _) K! l; _7 A/ b! \: DHazardous Voltage
' w3 y" u7 ^) W1 j9 U# eKey words:
2 Z3 n6 p' Y6 s/ ^' b. Q6 V+ a6 _-Hazardous Voltage* F- t% a2 j& J0 M8 M
Decision confirmed
% ?; \5 e& @; x! c" |' ^% Uat CTL 38th meeting
9 i" K7 U2 `% B8 e [Question:
- C5 p/ A( G2 _( rWhich of the criteria 42,4Vpeak or 60Vdc do you find should be applied as the compliance7 n7 \& ^# _% w4 |- b r
criteria, in order to determine whether the voltage shown in the graph is hazardous or not?
$ O( w6 z* T* n V/ GCondition: The circuit that generates this voltage is not a limited current circuit.8 ~! `) U6 ]9 v* G1 k
Decision:4 s6 U! A+ l* w9 ^
This voltage is a hazardous voltage because it exceeds the 42,4Vpeak as written in 1.2.8.3
0 q- ~0 j9 G6 @2 O t9 {0 L7 S* N `) n, U0 L) _+ ]
! h) i3 N1 D& p% Y+ ?
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发表于 2012-11-6 17:16
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