小白學安規: IEC 60950 Clause 2.10.3 and 2.10.4電氣間隙和爬電距離

market

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產品: 電源Rating 100-240V,
% p; }; v. U8 h0 Y5 L" LClause 2.10.3 and 2.10.4
' A6 _/ @  P5 A按表2K查證:
  N3 q, |8 c" k- g% G8 g( RUpeak=420V, Utransient=2500, R=4.0(3.0);
4 H2 F2 P, w0 V, \7 i1 H% bUpeak=840V, Utransient=2500, R=6.4(6.0);
8 A. N( z9 n, Y5 Y7 Z" h現測得某器件初次級管腳之間Upeak=612V,
$ F! l7 d! t! {/ e  t4 o- U
& @* H% M1 f2 b考慮兩种方法:
5 O, I2 o( L7 z( _1. 查表2L 一次電路的附加電氣間隙 Upeak=640(652)V, Utransient=2500, R=0.6; 結果是4.6;
2. 查表2K 用內插法求值, Upeak=612V, Utransient=2500, R=5.1(4.4);

  ]$ V1 Z8 d( [% |, ^( v& D請教按正規來說,此處的加強絕緣到底是4.6還是5.1才是最正確呢?
請教前輩!

茶馆

For an AC MAINS SUPPLY not exceeding 300 V r.m.s. (420 V peak):
a) if the PEAK WORKING VOLTAGE does not exceed the peak value of the AC MAINS SUPPLY
voltage, minimum CLEARANCES are determined from Table 2K;
b) if the PEAK WORKING VOLTAGE exceeds the peak value of the AC MAINS SUPPLY voltage,
the minimum CLEARANCE is the sum of the following two values:
/ L; C  f4 r2 B) R0 [• the minimum CLEARANCE from Table 2K; and
& Q2 t! i. n7 D# r( K: Z  }' r• the appropriate additional CLEARANCE from Table 2L.

, B0 f6 E8 j. ]3 H0 i  _3 VNow the working voltage between primary and seconday is Upeak=612V ，we should refer to item b：
From Table 2K,it is 6.4 comply with R. From Table 2L ,it is 0.6. so the min cl is 7.
6 `5 x, T3 }9 |2 f
" u# ]" Z0 i+ Q' g: Q: D: b+ p7 kIf  consider to apply linear interpolation , it is 5.1+0.6=5.7

xadzkjdxhj

the same question, and i can quite understand the mean of the above description. is there any superior would give a direction?

茶馆

引用第2楼xadzkjdxhj于2011-07-28 15:30发表的  :
the same question, and i can quite understand the mean of the above description. is there any superior would give a direction?

2 ]5 m/ F7 g: rwhat 's your concern?

market

謝謝2樓兄弟的解答,我忽略了2.10.3.3的要求.
+ |  ^5 f" h* K5 ?# p1 G) U 但是從這樣來說, 比如說該點Upeak值不是612V,而是750V的話:
1. 查表2L值是R=4.0+1.0(附加間隙)=5.0;
* O* [* a7 g' N/ F& ]% N2. 查表2K依線性內插法R=5.9;
. T1 z' B  S0 p
但結合2.10.3.3這句話的要求,那其它條件相同但是在750Vpeak下,所需的R=5.9+1.0=6.9;
而如果依查得表2K: Upeak=840V, Utransient=2500, R=6.4(6.0);
實際值Upeak=750V << Upeak=840V; 但是R(750)=6.9 反而>> R(840)=6.4;
5 m% F1 w  X- q  m這樣的判定大家覺得合理不?疑惑中.

chenlj0832

I think it should be 5.7.

market

這個加分怎麼加啊?我想給討論的人加分怎麼操作?

茶馆

引用第4楼market于2011-07-28 16:08发表的  :
謝謝2樓兄弟的解答,我忽略了2.10.3.3的要求.
但是從這樣來說, 比如說該點Upeak值不是612V,而是750V的話:
# _( p  ^7 ?6 ~$ m% I6 f# k1. 查表2L值是R=4.0+1.0(附加間隙)=5.0;
2. 查表2K依線性內插法R=5.9;

.......

If working voltage is Upeak=750V
# G' ~! x$ N2 GForm Table 2K, value is 5.9 comply with linear interpolation;
From Table 2L, value is 1.0.
so Cl for Upeak 705V is 5.9+1.0=6.9
2 y4 e$ O0 Q% J; T+ L+ Z5 _! Z. [
4 r$ W$ u( Z6 v2 ?If woking voltage is Upeak=840V
+ L. u; o, V  I& C* L9 U' {9 V/ S  oForm Table 2K, value is 6.4
From Table 2L, value is 1.2
So cl for Upeak 840V is 6.4+1.2=7.6.
1 m; r1 X6 w" e7 Z( g: b* `" l
are you clear now?

chenlj0832

R(840)的情况下也必须要加上表2L的值，在峰值电压超过供电的峰值电压时。

anpup2010

正确认案件 4.6mm, 且现在所有的认证单位都接受这种算法(TUV, UL, NEMKE, ITS都接受的)，并且从标准出发，也是有根据，是正确的。
重点： 请不要忘记表格2K, “PEAK WORKING VOLTAGE a”， 在峰值电压有备注一个小“a”, 对应到这个表格下面的备注，就是“a If the PEAK WORKING VOLTAGE exceeds the peak value of the AC MAINS SUPPLY voltage, see 2.10.3.3 b) regarding additional CLEARANCES.
! R# t! p1 _$ u
# Y+ h. Z- H: u# `# P+ d借上面兄弟的标准描述：
For an AC MAINS SUPPLY not exceeding 300 V r.m.s. (420 V peak):
* R$ {9 y: J* Y# h, W0 ^a) if the PEAK WORKING VOLTAGE does not exceed the peak value of the AC MAINS SUPPLY
voltage, minimum CLEARANCES are determined from Table 2K;
b) if the PEAK WORKING VOLTAGE exceeds the peak value of the AC MAINS SUPPLY voltage,
the minimum CLEARANCE is the sum of the following two values:
• the minimum CLEARANCE from Table 2K; and
• the appropriate additional CLEARANCE from Table 2L.
8 [/ }# j4 V" J& f% A
5 t: X4 g4 \& v0 `故我们通过确认 Upeak=612V， 根据表格2K (420V) ，要求4.0 mm, 然后再根据2L(612V), 再增加0.6 mm.
# ]6 \- z/ c9 i& s# Y  TPEAK WORKING VOLTAGE a，请不要忘记表格2K, 在峰值电压有备注一个小“a”, 对应到这个表格下面的备注，就是“a If the PEAK WORKING VOLTAGE exceeds the peak value of the AC MAINS SUPPLY voltage, see 2.10.3.3 b) regarding additional CLEARANCES.”，所以说，最终还是用表格2K加上2L的距离，就等于最终的距离要求。故距离要求应该是 4.6 mm.

2 E8 K7 ?) a% A% {* X$ \! N4 t问题已经明了，但有一个疑问是标准表格2K中提到的线性插入法，说是可以用，但基本目前的认证单位不会去用，且在标准第一版中是“3) For WORKING VOLTAGES between 2 800 V peak or d.c. and 42 000 V peak or d.c., linear interpolation is permitted between the nearest two points, the calculated spacing being rounded up to the next higher 0,1 mm increment.”，有一点不一样。