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标题: 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 [打印本页]

作者: 信思哲    时间: 2019-4-25 16:13
标题: 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
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/ }2 _& L* _" v) v/ i5 d
1.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
$ h8 e/ \4 Z6 G" ~) i3 w, C' c2 K$ ?7 ?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?
/ ]( F& M' D$ K. R* _, Z3 x0 v3.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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. {4 h) R: m' m8 n) G0 G8.7.2 Component failure test
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2 V1 e# M- \, @3 _# _6 w8.7.2.1 A unit having components such as resistors, semiconductor devices, capacitors, and the like shall( S" h% E! x; d2 c: n
not exhibit a risk of fire or electric shock when a simulated short circuit or open circuit is imposed. In* ?+ y: O  Z/ t# o, f
preparation for component failure tests, the equipment, circuit diagrams, and component specifications
0 c2 A. r9 Z9 K+ iare examined to determine those fault conditions that might reasonably be expected to occur. Examples: Z  N  j; V' j$ x
include: short-circuits and open circuits of semiconductor devices and capacitors, faults causing open; y7 ?3 L4 B4 _; @& E
circuits of resistors and internal faults in integrated circuits.7 u3 V7 l' G" Q. T

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9 A+ Z7 W0 G' G* m! x+ C) uException No. 1: Circuits in which maximum power levels have been determined to not exceed 50 W need
' K& _: d! d3 p2 _* |not be evaluated for component failure.
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% V7 k3 E5 B1 e  G$ F1 E1 Z7 mException No. 2: Devices supplied by a source operating within the limits for risk of fire and electric shock
6 H8 _6 V5 [4 R5 }) P' x; q, Zneed not be subject to this test.
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, d/ C$ _9 O' I& }/ \9 J3 o# D8.8 Circuit power limit measurement test2 C$ _: M9 N: }" V
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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,
6 h& M$ H# B% W( a5 p# `5 Fincluding short circuit, measured after one minute of operation exceeds a defined limit. For the purposes
3 F: |" B% O: ~# u8 Z) aof this test, the limit (for example, 15W or 50W) is referred to as PLIMIT. This test is applied to a circuit
$ \8 l7 L% v* u; j8 r: ?5 G) R% @(under evaluation) to determine if it has a power draw of PLIMIT or less.( K. C# K$ k, t8 h
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