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' i. U: Q( u( m3 @
% H% e+ N2 L) F' ^1 U# i | Figure 21, presence of resistor of 45 kO6 @+ Q- S G1 U0 C5 v
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| 60601-1(ed.2);am1;am2+ X0 S/ i W* N+ Y
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2 x& [, d4 V2 Z/ hStandard(s)- (year and edition):
3 f; [: s- k' AIEC 60601-1:1988 Ed.2/ d$ I( f ^% A9 b) S
Am1+Am2
* j( Y$ C4 e& ^; i3 CSub clause(s):
( o+ L, L% B- ~8 n7 V( O! N+ r/ I+ `19% Z' B& q- L8 p
Sheet n°: 5347 k! H5 t- I$ G+ k+ C
Subject:+ z* H' p( C1 @2 G0 f _; d" R
Figure 21, presence of resistor, P4 k# s1 `: v$ v: ^
of 45 kO3 t8 W6 q, k6 p+ D
Key words: Decision taken at the 40th
8 U. @- m r5 w7 H Cmeeting 2003% d5 z; f3 |- |/ e5 Y1 z+ x& T
Question:
/ J l) ~5 X3 v/ {/ ]Should the 45 kO resistor be used since the current is limited to 5 mA and the limit is 5 mA?! R+ s1 `" v+ p- ~
Decision:" }3 D" R8 J. \0 [4 L& x
For the 1st edition, use the method of 2nd edition (use any resistance).
3 G5 L! m$ m% H2 R$ U9 h+ NFor 2nd edition, instead of using any resistance, you may also use alteration of the voltage or fuses.0 p' X- p+ _4 ^ O1 a G
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发表于 2012-9-19 15:49
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