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PCB 线宽与电流(PCB line width and current)/ a# }5 f2 t- f! u
The withstand current of 1 square copper wire is 5 -- 8A, 380V can2 [3 h1 n% {3 H6 v- X2 q. @! e, J+ X
be brought with maximum 4KW, 220V can be maximum 2KW.
+ O6 v4 W9 f8 u- y+ P: G2 square? Three square meters can be pushed like that5 F, n2 W+ K+ q% K
PCB line width and current relationship
4 x( q8 U# D! o7 x/ nI. calculation method is as follows:, H7 [& c8 i) _& Z# [( V q
First, the section area of Track is calculated. The copper foil
9 Z: h3 s' H' `4 \0 cthickness of most PCB is 35um (the PCB manufacturer can be asked if it" B6 D# Q: g$ N' Q5 U2 v* o
is not certain). The width of the PCB is the
' m, W7 i/ {7 xcross-sectional area. There is an experience value of current( K$ Q" Z/ A+ j" r: ` G. J2 {
density for 15 ~ 25 amperes per square millimeter. We can just call it
- ]1 x- K7 D5 W6 cthe area where we get the volume.
/ q7 c0 k5 t. k. t: tI = KT0.44 A0.75 (K is the correction factor, which is 0.024 in the
% B4 C2 t8 p2 D2 U- }" r* }! ginner layer of the copper line, and 0.048 in the outer layer)! u! x3 t5 v) H2 c# B L
T to the maximum temperature rise, the unit for degrees Celsius# Z+ H8 v1 \ q, t% L8 k( R/ N& z
(copper melting point is 1060 ?)
; ?, w7 d5 ?/ e, SA to cover the area of copper, the unit is square MIL (not mm mm,3 T. L8 g; T9 U- Y. i" X* B5 u
the attention is square MIL.)
/ `; t/ {8 a0 u& t+ s3 w* W5 xI is the maximum allowable current, the unit is ampere (amp).
( L I1 ]) e9 oThe average 10mil = 0.010 inch = 0.254 May be 1A, 250MIL = 6.35mm,& r0 c/ A4 d1 x( \! \
which is 8.3 A
0 Z, Z0 [9 O2 x; Q
+ r3 G, [/ e& f: L3 f) C# {' X6 K$ {% { c) l/ g- `
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! Q& a9 M# b: L6 s4 @' Z |
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发表于 2020-2-17 15:30
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