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应该从两个方面来判断：! F7 W& j4 `8 w/ A. H
1. 若是申请60950,Fuse两端的距离则要求功能绝缘,标准上有明确指出满足功能绝缘只要符合以下条款中的任何一种即可：(即距离,耐压,短路任何一种方式符合要求即可),
7 t' M. U8 `; g& k! k( k/ V5.3.4 Functional insulation
: ]- N- p; z! v4 Z$ X( zFor FUNCTIONAL INSULATION,CLEARANCES and CREEPAGE DISTANCES shall satisfy one ofthe following requirements a) or b) or c).
) y; a3 e* }+ d) @' g; h2 O4 ta)They meet the CLEARANCE and CREEPAGEDISTANCE requirements for FUNCTIONAL INSULATION in 2.10(or AnnexG).2 P# {5 x5 Q5 s3 a# q8 v
b)They with stand the electricstrength tests for FUNCTIONAL INSULATION in 5.2.2.
& P# J2 S, Z5 ic)They are short-circuited where a short-circuit could cause* T5 ?6 m2 f. {3 T( ^' i8 ^
•overheating of any material creating a risk offire,unless the material that could be overheated is of V-1 CLASSMATERIAL,or2 Z9 I4 g! G) L+ ~- z
• thermal damage to BASIC INSULATION, SUPPLEMENTARY INSULATION or REINFORCED INSULATION, there by creating a risk of electric shock.+ p: h$ D: S% V. }; R
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Compliance criteria for 5.3.4 c) are in 5.3.9.$ S6 l' \. ~3 K, K, d
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不过,依目前与认证机构的配合来看,美规是接受以上方式的任何一种来满足保险丝的距离,但欧规一般不接受短路,Hit-pot这两种方式,一定要距离满足才行.3 g" j8 L% g* w5 x: D! m' o b
2. 若是申请60065则没有功能绝缘这个概念,即保险丝两端的距离要求为2.0mm,2.5mm(基本绝缘),则要求距离及Hit-pot都满足才行. |
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发表于 2010-11-16 20:58
，FUSE不应该有spacing的要求
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