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|  DSH 387
6 ^, z. t' F, O, s, H7 N" j, ~1 A; O2 P: B8 Z# Y
| Hazardous Voltage
" W) g: f$ B# O3 `- Y | 1.2.8.3 / 12.8.4
6 A/ v) U3 B! U9 J A1 O5 Q | 60950(ed.2) & 60950(ed.3)6 X+ g7 U- ^8 U3 a
|
: `9 m% ?5 W4 K: C# _) ]
Standard:
1 `$ d4 i4 x" Z- b Z/ R# w. zIEC60950, 2nd (3rd) Edition8 q0 U+ j+ ]8 V
Sub clause:$ \5 X" L& F: ^" k& k
1.2.8.3(1.2.8.4)
/ @1 V J8 ]$ k" R1 y) {Sheet n. 387 C; a$ u; Z6 i5 y/ A
Page 1 (1)1 \+ l+ h, H4 W9 H% c8 @
Subject:
5 o2 v( ]5 |' p* R9 OHazardous Voltage
( y4 e6 I: F. x8 i; M7 D0 W! ~% GKey words:
: H H/ N( h3 @' G9 ^- c5 T9 C-Hazardous Voltage
/ ~/ ]7 n6 M4 q/ z K+ iDecision confirmed! N& w7 A4 g) W2 @* B% K5 T8 T& N5 a+ j
at CTL 38th meeting
* d) S8 Q: J! XQuestion:
* M$ N2 j. i- s1 x, C& t* l. j, @! AWhich of the criteria 42,4Vpeak or 60Vdc do you find should be applied as the compliance$ y4 Z- I+ Y3 w/ F3 W: `5 `
criteria, in order to determine whether the voltage shown in the graph is hazardous or not?
' q3 N, d/ [) k, @% d; o ?5 ECondition: The circuit that generates this voltage is not a limited current circuit.
! S1 w' [) G! A: i$ O& oDecision:. ]$ p+ P( o0 z& Z3 `- T) o% t6 q7 m
This voltage is a hazardous voltage because it exceeds the 42,4Vpeak as written in 1.2.8.3
$ `7 C& J5 C9 [- o& l9 I5 k, V2 i. G
( r( C5 X( W! g6 v6 x
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发表于 2012-11-6 17:16
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