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| DSH 387
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7 a1 t' c* |* `' M0 R. r% p7 P | Hazardous Voltage
8 m$ Y+ W( J4 t# e2 A. R | 1.2.8.3 / 12.8.4+ v& z5 }, m3 t D0 b
| 60950(ed.2) & 60950(ed.3)
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Standard:
1 n: q) S& Y4 Y/ O: iIEC60950, 2nd (3rd) Edition+ t. j+ g S; V' _
Sub clause:
O. u7 r0 a& e( C2 h1.2.8.3(1.2.8.4)" o! o) _2 S2 h7 M& ^' \2 A5 b
Sheet n. 3874 \. }& H0 L) H$ K( \) e; v; f
Page 1 (1)
, N- i& c0 x6 n- w( F6 S% K# W( c# iSubject:! G5 }- b7 L' K/ ] W; Z& h! ?! P
Hazardous Voltage4 t/ `! c0 G5 h3 `7 O% G$ d- k. O
Key words:: t; D7 t$ g" ]3 w
-Hazardous Voltage; h1 v$ g# ^. ?7 c! i+ \
Decision confirmed4 O2 P" `# d: m2 A" y W/ y
at CTL 38th meeting' M& @3 p0 ^) T! [# \
Question:2 l! G9 A1 |+ `
Which of the criteria 42,4Vpeak or 60Vdc do you find should be applied as the compliance
" {7 A# }) Z5 gcriteria, in order to determine whether the voltage shown in the graph is hazardous or not?& Q2 M7 k* o1 |
Condition: The circuit that generates this voltage is not a limited current circuit.
% ?3 U" e9 ]2 K2 h' _( zDecision:
% a. S* Y+ W& W {5 [This voltage is a hazardous voltage because it exceeds the 42,4Vpeak as written in 1.2.8.35 _# v1 R, E' Y: ]( [1 R
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