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|  DSH 387
. a- Y: f1 u1 `4 {3 s" I) u/ e& p7 t7 |) i
| Hazardous Voltage2 w0 ?4 s) ~5 u. T+ k
| 1.2.8.3 / 12.8.4
; Q/ ]! D, q1 Q5 R- ?3 m | 60950(ed.2) & 60950(ed.3)( x; F, ~0 _: n* k* e
|
, b. [1 f8 n7 X5 Z
Standard:: M0 {$ y, \5 l8 `& h* K$ _
IEC60950, 2nd (3rd) Edition
) h5 K- J3 L2 k$ Q* G- v7 _Sub clause:
$ _/ J3 [- n% X& j; L" K8 X T" {1.2.8.3(1.2.8.4)7 c- v4 w/ @- j: D
Sheet n. 387* Y8 C6 Y; M3 C3 R
Page 1 (1)
* J' L M8 X* M1 A1 l, YSubject:+ g" x; A8 |2 T8 t
Hazardous Voltage
/ X) m5 ~4 r i6 g% xKey words:. z0 ~8 ~6 L5 s x* A# y
-Hazardous Voltage
/ q s( r3 A" u7 ^Decision confirmed/ S9 a! ?& i/ l
at CTL 38th meeting8 u3 z( m& ~0 _, y3 l
Question:" Q0 [+ ^3 c) y' h' d: o" ~* x
Which of the criteria 42,4Vpeak or 60Vdc do you find should be applied as the compliance
: X' g! v4 A. d+ `4 @% ?( X; J- Acriteria, in order to determine whether the voltage shown in the graph is hazardous or not?
. K" w( m5 C0 n S' G: S6 g# XCondition: The circuit that generates this voltage is not a limited current circuit.; d+ I! b5 d) ~& |% M* F
Decision:
8 t- [' V6 Z9 ^+ h; Z$ L8 U6 BThis voltage is a hazardous voltage because it exceeds the 42,4Vpeak as written in 1.2.8.38 O0 T X) s* I% }/ T+ f+ n
, E5 X N) Z. j4 e0 n5 ^0 V; D
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发表于 2012-11-6 17:16
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