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UL 8750 故障测试中的8.7.2 中Exception No1对于不超过50W的电路可以豁免故障测试。三个问题
0 v/ A+ b) u" D0 t$ l1.是不是说只要经过8.8 Circuit power limit measurement test的评估确定这个电路是小于50W的就可以豁免故障测试$ L, t6 M+ B: q. q- n. y7 p) Q
2.这个50W的电路不论输出是高压还是低压都可以豁免吗？还是说低压输入的电路？' B3 M" A/ E# Z+ d2 B! X8 j
3.目前有一个带电池的灯具，电池输出为12V，灯具只有10W，里面有电路，这个电路需要做故障测试吗？
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8.7.2 Component failure test2 h) j, ]. U& P# E& u. i x
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8.7.2.1 A unit having components such as resistors, semiconductor devices, capacitors, and the like shall" n9 s' m3 ?/ K! ^! T
not exhibit a risk of fire or electric shock when a simulated short circuit or open circuit is imposed. In
6 t7 _5 U8 U0 Z |' W3 M2 }preparation for component failure tests, the equipment, circuit diagrams, and component specifications, O% d5 R7 i8 m O% a
are examined to determine those fault conditions that might reasonably be expected to occur. Examples
. A- u8 ~3 t9 e, S F2 {include: short-circuits and open circuits of semiconductor devices and capacitors, faults causing open- p) h7 Q6 e' N2 j) ?" ~' C9 Z
circuits of resistors and internal faults in integrated circuits.
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- p+ ]* _' g! y" {3 tException No. 1: Circuits in which maximum power levels have been determined to not exceed 50 W need k' x4 ~8 z" d L+ p
not be evaluated for component failure.. O X, q* L5 v( w8 u* U+ }. @
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Exception No. 2: Devices supplied by a source operating within the limits for risk of fire and electric shock9 h# W5 ^5 a" G$ w# X. \3 @
need not be subject to this test." e; a( `$ l$ c, d2 a* O
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1 W' a7 j) ~" i& x! ]* }8.8 Circuit power limit measurement test5 c* _" ?1 C* K
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8.8.1 This test shall be used to determine if the power available to a circuit under any loading condition,2 V: c- Y( r) ~
including short circuit, measured after one minute of operation exceeds a defined limit. For the purposes
* w% {& b: s& S) b7 kof this test, the limit (for example, 15W or 50W) is referred to as PLIMIT. This test is applied to a circuit
# T9 T8 t+ ^ M8 w$ v+ _6 o5 n F(under evaluation) to determine if it has a power draw of PLIMIT or less.
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发表于 2019-4-25 16:13
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