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PCB 线宽与电流(PCB line width and current)3 ^ p; [! H- I1 w+ C( D
The withstand current of 1 square copper wire is 5 -- 8A, 380V can z' J1 i! @' {+ K* U; e
be brought with maximum 4KW, 220V can be maximum 2KW.1 B3 C5 N5 @6 L* e" j
2 square? Three square meters can be pushed like that
# K5 P) u0 \5 ` UPCB line width and current relationship: L! P# |4 Y/ W4 _5 d0 z% D: J
I. calculation method is as follows:) Y+ R5 F6 C6 H& }4 _6 f
First, the section area of Track is calculated. The copper foil8 y% n, k- \9 l' I
thickness of most PCB is 35um (the PCB manufacturer can be asked if it
3 A3 L" G. B4 y+ U/ B0 y# o0 @is not certain). The width of the PCB is the* c+ l) w# u; W/ y
cross-sectional area. There is an experience value of current
; i$ d3 g2 i _. ndensity for 15 ~ 25 amperes per square millimeter. We can just call it# W7 v; @. G, |% v
the area where we get the volume.
8 Z9 p/ [$ C/ i( t3 z7 \6 B) ZI = KT0.44 A0.75 (K is the correction factor, which is 0.024 in the3 O1 Q4 x7 X% b
inner layer of the copper line, and 0.048 in the outer layer)
4 m7 Y" s: E% ]" a- UT to the maximum temperature rise, the unit for degrees Celsius0 R" N5 p% f' Z$ |
(copper melting point is 1060 ?)
% h. b8 Z5 h- iA to cover the area of copper, the unit is square MIL (not mm mm,
, x+ P9 S# R* k* Ithe attention is square MIL.)
# u( D8 |/ e7 S, Q4 z, d3 rI is the maximum allowable current, the unit is ampere (amp).) v/ }! l8 k9 [9 ?
The average 10mil = 0.010 inch = 0.254 May be 1A, 250MIL = 6.35mm,3 E# F; O# y' O5 s0 G
which is 8.3 A
) V9 p2 G& X4 u- \. T$ \! _3 y1 H: o1 p5 x9 Y
1 G9 @4 q+ W6 w U) z
" W1 |- d( c9 F: F$ S" D- R2 y7 e2 @! N, }. b- p
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