Relationship between PCB width and current

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• 10 C
• 20 C
• 30 C
• Copper
• Trace width
• Maximum current amps
• Inch
• Mm

Relationship between PCB width and current Reprinted from: http://www.dianyuan.com/article/207334, calculation method is as follows:

First, calculate the cross-sectional area of the track. The thickness of most PCB copper foil is 35um (if you are not sure, you can ask the PCB manufacturer). The top width is the cross-sectional area. There is an experience value of current density, which is 15 ~ 25 AMPS/mm. The cross-sectional area is called to obtain the flow capacity.

I = kt0.44a0.75(K is the correction coefficient. Generally, 0.024 is used for Copper Coating on the inner layer and 0.048 is used for outer layer.
T indicates the maximum temperature rise, measured in degrees Celsius (the melting point of copper is 1060 ℃)
A is the cross-sectional area of copper clad, in the unit of square mil (not mm, note that it is square mil .)
I is the maximum allowable current, in ampere (AMP)
Generally, 10mil = 0.010 inch = 0.254 can be 1A, 250mil = 6.35mm, and 8.3a

Relationship between PCB cabling width and current

The current load of Copper Foil with different thicknesses and widths is shown in the following table:
Copper thickness 35um copper thickness 50um copper thickness 70um copper T = 10 copper T = 10 copper T = 10

Current a width mm current a width mm
6.00 2.50 5.10 2.50 4.50 2.50
5.10 2.00 4.30 2.00 4.00 2.00
4.20 1.50 3.50 1.50 3.20 1.50
3.60 1.20 3.00 1.20 1.20
3.20 1.00 2.60 1.00 2.30 1.00
2.80 0.80 2.40 0.80 2.00 0.80
2.30 0.60 1.90 0.60 1.60 0.60
2.00 0.50 1.70 0.50 1.35 0.50
1.70 0.40 1.35 0.40 1.10 0.40
1.30 0.30 1.10 0.30 0.80 0.30
0.90 0.20 0.70 0.20 0.55 0.20
0.70 0.15 0.50 0.15 0.20 0.15

Note 1: when using copper as a wire through large current, the carrying capacity of copper foil width should be taken into consideration by referring to the numerical value reduction 50% in the table, the empirical formula of the first version of Mao Nannan sun Jun 96.1 is extracted from the following Original article:
"Due to the limited thickness of copper foil, the carrying capacity of copper foil should be considered in the Strip copper foil that needs to flow through a large current. taking the typical 0.03mm thickness as an example, if the copper foil is used as a strip wire with a width of w (mm) and a length of L (mm), its resistance is 0.0005 * L/W ohm. in addition, the load of copper foil is related to the type, quantity and heat dissipation conditions of the components installed on the printed circuit board. considering the safety, we can calculate the carrying capacity of copper foil based on the empirical formula of 0.15 x W (.

PS-Ef | grep wcz ps-E | grep allegro

(2) what I saw at the e-engineering album Forum

Calculation Method of the copper width and the amount of electrical flow of the PCB circuit board:

Generally, when the Copper Foil Thickness of PCB is 35um and the line width is 1mm, the area of the horizontal section of the line is 0.035 mm², and the current density 30a/square millimeter is usually used, the line width per millimeter can flow through 1A current.

The formula for calculating PC275-A is as follows: temperature rise, Copper Foil Thickness,.

I = 0.0150 (DT 0.5453) (a 0.7349) for IPC-D-275 internal traces
I = 0.0647 (DT 0.4281) (a 0.6732) for IPC-D-275 external traces

Ii. Data:

The computation of PCB streaming capability has always lacked authoritative technical methods and formulas. Experienced CAD engineers rely on their personal experience to make more accurate judgments. However, it is difficult for new users to use CAD.
The current Loading Capacity of PCB depends on the following factors: line width, line thickness (copper foil thickness), and allowable temperature rise. As we all know, the wider the PCB strip, the larger the streaming capacity. Here, let me know: If 10 mil strip can withstand 1A under the same conditions, how much high current can the 50 mil strip bear, is it 5A? Naturally, the answer is no. See the following data from international authorities:

The unit of the line width is: inch (inch inches = 25.4 millimetres millimeters) 1 oz. copper = 35 microns thick, 2 oz. = 70 microns thick, 1 oz = 0.035mm 1mil. = 10-3 inch.

Temp rise		10 C			20 C			30 C
Copper		1/2 oz.	1 oz.	2 oz.	1/2 oz.	1 oz.	2 oz.	1/2 oz.	1 oz.	2 oz.
Trace width		Maximum current amps
Inch	Mm
. 010	0.254	. 5	1.0	1.4	0.6	1.2	1.6	. 7	1.5	2.2
. 015	0.381	. 7	1.2	1.6	0.8	1.3	2.4	1.0	1.6	3.0
. 020	0.508	. 7	1.3	2.1	1.0	1.7	3.0	1.2	2.4	3.6
. 025	0.635	. 9	1.7	2.5	1.2	2.2	3.3	1.5	2.8	4.0
. 030	0.762	1.1	1.9	3.0	1.4	2.5	4.0	1.7	3.2	5.0
. 050	1.27	1.5	2.6	4.0	2.0	3.6	6.0	2.6	4.4	7.3
. 075	1.905	2.0	3.5	5.7	2.8	4.5	7.8	3.5	6.0	10.0
. 100	2.54	2.6	4.2	6.9	3.5	6.0	9.9	4.3	7.5	12.5
. 200	5.08	4.2	7.0	11.5	6.0	10.0	11.0	7.5	13.0	20.5
. 250	6.35	5.0	8.3	12.3	7.2	12.3	20.0	9.0	15.0	24.5

Trace Carrying Capacity
Per mil std 275

3. Experiment:

The pressure drop caused by the wire resistance generated by the wire length must be considered in the experiment. The tin used in process welding is only used to increase the current capacity, but it is difficult to control the volume of tin. 1 oz copper, 1mm wide, generally used as a 1-3 a current meter, depending on your line length, pressure drop requirements.

The maximum current value is the maximum allowable value under the temperature rise limit, and the fusing value is the value that the temperature rise reaches the melting point of copper. Eg. 50mil 1 oz temperature rise of 1060 degrees (copper melting point), current is 22.8a.

AWG :( American wire gauge) American wire Specification