The first thing to note is that the ounce (OZ) itself is a unit of weight. The conversion formula for ounces and grams (g) is: 1OZ ≈28.35g.
In the PCB industry, 1OZ means the thickness of copper with a weight of 1OZ evenly spread on an area of ​​1 square foot (FT2). It uses the weight per unit area to express the average thickness of the copper foil. Expressed by the formula, 1OZ=28.35g/FT2 (FT2 is square feet, 1 square foot=0.09290304 square meters).
Specifically, it can be said that the conversion method of thickness and length is as follows:
First of all, we know that the density constant of copper and the conversion formula of related units are as follows:
The density of copper Ï=8.9g/cm3
1 centimeter (cm) = 10 millimeters (mm); 1 millimeter (mm) = 1000 microns (um)
1mil≈25.4um
1 FT2≈929.0304cm2
1mil≈25.4um
According to the mass calculation formula m=Ï×V (volume)=Ï×S (area)×t (thickness), knowing that the weight of the copper foil divided by the density and area of ​​the copper is the thickness of the copper foil!
As we know from the previous article, 1OZ=t×929.0304cm2×8.9g/ cm3=28.35g
Therefore, t=28.35÷929.0304÷8.9cm≈0.0034287cm=34.287um≈34.287÷25.4mil≈1.35mil
It can be seen that the thickness of 1OZ copper foil is about 35um or 1.35mil.
Copper thickness 1.OZ (0.035mm) Copper thickness 1.5OZ (0.05mm) Copper thickness 2.OZ (0.07mm)
The relationship between PCB line width and current
Calculate the cross-sectional area of ​​the track first. The copper foil thickness of most PCBs is 35um (if you are not sure, you can ask the PCB manufacturer). The cross-sectional area is multiplied by the line width. Note that it is converted to square millimeters. There is an empirical value of current density, which is 15-25 A/mm2. Call it the upper cross-sectional area to get the flow capacity.
2. Data:
The calculation of PCB current-carrying capacity has always lacked authoritative technical methods and formulas, and experienced CAD engineers can make more accurate judgments relying on personal experience. But for CAD novices, it cannot be said that they have encountered a problem.
The current carrying capacity of the PCB depends on the following factors: line width, line thickness (copper foil thickness), and allowable temperature rise. Everyone knows that the wider the PCB trace, the greater the current-carrying capacity. Here, please tell me: Assuming that under the same conditions, a 10MIL trace can withstand 1A, how much current can a 50MIL trace withstand, is it 5A? The answer is naturally no. Please see the following data from international authorities:
The unit of line width is: Inch (inch = 25.4 millimetres) 1 oz. Copper = 35 microns thick, 2 oz. = 70 microns thick, 1 OZ =0.035mm 1mil.=10-3inch.
In the experiment, the voltage drop caused by the wire resistance caused by the wire length must also be considered. The tin on the process welding is only to increase the current capacity, but it is difficult to control the volume of the tin. 1OZ copper, 1mm wide, generally used as a 1-3 A ammeter, depending on your cable length and voltage drop requirements.
The maximum current value should refer to the maximum allowable value under the temperature rise limit, and the fuse value is the value at which the temperature rise reaches the melting point of copper. Eg. 50mil 1oz temperature rise 1060 degrees (ie copper melting point), current is 22.8A
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