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PCB 线宽与电流(PCB line width and current) G) h( |3 E% t/ ~2 I V
The withstand current of 1 square copper wire is 5 -- 8A, 380V can
) P* S: f! L" v8 i% [be brought with maximum 4KW, 220V can be maximum 2KW.2 g" t+ `. R5 K5 d8 A& W5 C
2 square? Three square meters can be pushed like that
0 B b& r" W" QPCB line width and current relationship. Z" g: b }1 g7 P
I. calculation method is as follows:
1 l. g* b, T% ?First, the section area of Track is calculated. The copper foil
; B. a/ I2 M3 t- i5 Zthickness of most PCB is 35um (the PCB manufacturer can be asked if it
0 N, l' Q# f2 ^4 k3 ?" sis not certain). The width of the PCB is the2 a, k9 c& x6 F9 a
cross-sectional area. There is an experience value of current- r; |2 o* x! U: l! A. l0 j
density for 15 ~ 25 amperes per square millimeter. We can just call it
- \& U- w* v$ Uthe area where we get the volume.3 ~- O& \0 T4 Z8 z
I = KT0.44 A0.75 (K is the correction factor, which is 0.024 in the0 k; e4 V m8 c+ g1 |0 y% Q
inner layer of the copper line, and 0.048 in the outer layer)
2 R* {/ ?3 [: ?2 b3 Q3 k6 GT to the maximum temperature rise, the unit for degrees Celsius
3 P `! Y4 k+ h0 v7 I(copper melting point is 1060 ?)
# ]/ S* o6 i. M) a. W# nA to cover the area of copper, the unit is square MIL (not mm mm,# K4 F0 L5 p5 P% Y0 e
the attention is square MIL.)
! p2 `2 b4 [5 ^, R% K* V, d8 ]I is the maximum allowable current, the unit is ampere (amp). R0 r% s# q" B% w# w* G
The average 10mil = 0.010 inch = 0.254 May be 1A, 250MIL = 6.35mm,6 W! |6 g9 q; z* E6 U! H! |
which is 8.3 A
, Z9 i3 S* @3 `$ @5 ?3 V6 i& }+ K
' J9 t/ Q( v1 c
/ A, Q, f1 f e/ h
1 ~) a+ \& u! _1 O4 e$ u2 M |
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发表于 2020-2-17 15:30
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