标题: 关于家用电器内,电子线路设计,熔断丝的参数选择问题 [打印本页] 作者: E_zhou 时间: 2009-7-15 15:18 标题: 关于家用电器内,电子线路设计,熔断丝的参数选择问题 安规的工作,我个人觉得,更重要的是利用自己对于安规的理解,实际贯彻到产品的设计过程中去。此贴是我之前做到一个分析报告。用于设计工程师的培训。但是,感觉还是有很多遗漏,例如关于快速熔断和慢熔断,延时熔断到选择依据等等。发上来供大家完善评阅。 1 {" j; U# _& s- `" m如下为引用标准GB4706.1—1998的内容,当熔断器的设置,是为了当模拟电子线路非正常测试时起保护作用的情况,熔断器应当按照如下进行考核,判断的条件为 “电路被充分保护”' a$ K9 @. t+ G8 |
If the safety of the appliance under any of the fault conditions depends on the operation of a 5 w2 T5 T5 S1 Y3 Ominiature fuse-link complying with IEC 60127, the test of 19.12 is carried out. 0 Q+ f% ]. m, ~" t6 r! h19.12 If safety of the appliance depends upon the operation of a miniature fuse-link - G2 p& ]8 W, O, a3 U/ y5 ?complying with IEC 60127 during any of the fault conditions specified in 19.11.2, the test is ' }9 S. L4 b7 y# l9 e8 A" H/ |repeated but with the miniature fuse-link replaced by an ammeter. If the current measured 4 e6 e" {5 ~! f, P r* k– does not exceed 2,1 times the rated current of the fuse-link, the circuit is not considered to) s5 a' X- x% A2 r- }- Q1 v" _) h) v
be adequately protected and the test is carried out with the fuse-link short-circuited; ; z8 Y1 R! ~: d2 o9 u% V– is at least 2,75 times the rated current of the fuse-link, the circuit is considered to be. v- o) m. n9 ^, k" {/ l
adequately protected; $ ~7 u2 W: k7 D3 z– is between 2,1 times and 2,75 times the rated current of the fuse-link, the fuse link is( c1 ~& d5 B; m) F
short-circuited and the test is carried out ! h0 A* m1 s% o0 P: j• for the relevant period or for 30 min, whichever is the shorter, for quick acting fuselinks; / N# h: t: B5 X( `. g• for the relevant period or for 2 min, whichever is the shorter, for time lag fuse-links. . ?8 R( [) i4 r5 ?: gNOTE 1 In case of doubt, the maximum resistance of the fuse-link has to be taken into account when determining* `6 f; U- v) w! l- E
the current.: P, R7 L" K- y+ ^
NOTE 2 The verification whether the fuse-link acts as a protective device is based on the fusing characteristics ! t9 h! F6 S$ H; [6 sspecified in IEC 60127, which also gives the information necessary to calculate the maximum resistance of the 5 L1 H' `# X7 n: Vfuse-link. 5 C! f, W8 A6 \( @8 E2 C' w* PNOTE 3 Other fuses are considered to be intentionally weak parts in accordance with 19.1.; ~' } p B, i6 l. s1 g
分析方法如下: 7 j: l+ v9 w' F* a3 q& J+ v: Y" X1、 明确熔断器的设置,是用于在何种电子线路故障时起保护作用; 2 z4 V9 h) {4 \8 C% R2、 计算或者测试,在单一故障发生的情况下,通过熔断器的电流; 0 G- h) o8 }' s0 h3、 测试和选择方法:+ {- J! I3 Z# T4 c
1)、用电流表代替熔断器的两端,逐个模拟单一故障,记录电流表的测试值; - v+ o8 C; X8 _; M: Y2)、电流表代替时,应充分考虑电流表与熔断器电阻的差异,尽量做到两者内阻相等; ; {! ^ ]! J$ t8 N6 h+ I) L+ |: }3)、用电流表测得的电流值(选取模拟故障时,需要熔断器动作来保护的最小电流值),除以2.75,所得出的值为熔断器的最大额定电流;' A$ C7 ]' `/ C$ {9 @' ~! }* }
4)、熔断器的额定电流,必须大于电子线路正常工作时通过熔断器位置的电流。(考虑正常电压波动,测量正常工作电流时,输入电压为AC240V,50Hz)。 7 e- ^; s7 N7 V$ X! m4、当按照测试值采用了熔断器之后,须进行验证。若模拟电子线路故障是,熔断器立即动作,则在同样条件下重复测试一次,作为薄弱位置的可靠保护确认。 1 A. S2 C% }, X" q% A5 j抛砖引玉,请高手完善。作者: 阿鸡米得 时间: 2009-7-15 16:31
为什么要除以2.75了? 4 |5 p& T4 h u& F. S" c9 K4 T g2 J
我在网上找的资料说是选取FUSE为正常电流的1.5-2倍。 2 t. B. a! z" ~# R8 E, q7 y" Z$ C- v. [' h$ t7 t/ e( m! q7 y. K
有标准要求异常测试时,熔断器电流和额定电流比为小于等于4A的 要大于2。1倍) d9 a- d6 E9 o6 p) G9 S/ b
/ J% i2 a% f8 ]7 g; K4到等于10A的要大于1.9 o, N' ^+ p# @' M0 U
& }2 J: H4 `) I4 ~3 p3 }1 @' P" f, ?
10到25A的要大于1.75作者: okjun 时间: 2009-7-16 09:09
楼主,如果保险丝仅仅是用来防止L、N之间短路(比如LN之间有一个电容,要做短路测试),这时候FUSE的值怎么定?! x I- B0 I0 s% W3 U/ E. Y1 u r
LN之间的短路电流一般是多少?这个我没测过,因为怕电流表烧坏,而且也没有时间读数(瞬间就会跳闸)。作者: E_zhou 时间: 2009-7-16 09:16