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# Ounces are weight units. Why is it used to indicate thickness in PCB?

Ounces are weight units. Why is it used to indICate thickness in PCB?

First of all, the ounce (OZ) itself is a unit of weight. The conversion formula of ounce and gram (g) is: 1OZ ≈ 28.35g.

In the PCB industry, 1OZ means the thickness of 1OZ copper evenly spread on an area of 1 square foot (FT2). It is the average thickness of copper foil expressed by the weight per unit area. The formula is 1OZ=28.35g/FT2 (FT2 is square foot, 1 square foot=0.09290304 square meters). To be specific, the conversion method between length and thickness is as follows:

First of all, we know that the density constant of copper and the conversion formula of relevant units are as follows:

Density of copper ρ= 8.9g/cm3

1 cm=10 mm; 1 mm=1000 μ m

1mil≈25.4um

1 FT2≈929.0304cm2

1mil≈25.4um

According to the calculation formula of mass, m= ρ× V (volume)= ρ× S (area) × T (thickness), know that the weight of copper foil divided by the density and area of copper is the thickness of copper foil!

As we know, 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 1.5 OZ (0.05mm) Copper thickness 2. OZ (0.07mm)

Relationship between PCB line width and current:

First, calculate the sectional area of the Track. The copper foil thickness of most PCBs is 35um (if you are uncertain, you can ask the PCB manufacturer). It is the sectional area when multiplied by the line width. Note that it is converted into square millimeters. There is an empirical value of current density, which is 15~25A/mm2. Call it the upper sectional area to get the flow capacity.

The calculation of PCB current carrying capacity has been lacking of authoritative technical methods and formulas, and experienced CAD engineers can make more accurate judgments based on their personal experience. But for CAD novices, it can be said that there is a problem.

The current carrying capacity of a 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 wiring, the greater the current carrying capacity. Here, please tell me: suppose that under the same conditions, the 10 MIL cable can withstand 1A, then how much current can the 50 MIL cable withstand? Is it 5A? The answer, of course, is no. Please see the following data provided by international authorities:

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

In the experiment, the voltage drop caused by the line resistance caused by the length of the wire must also be considered. The tin on the process soldering station is only used to increase the current capacity, but it is difficult to control the volume of tin. 1OZ copper, 1mm wide, generally used as 1 - 3A galvanometer, depending on your wire length and voltage drop requirements.

The maximum current value shall refer to the maximum allowable value under the temperature rise limit, and the fusing value is the value when the temperature rise reaches the copper melting point. Eg. Temperature rise of 50mil 1oz is 1060 ℃ (i.e. copper melting point), and the current is 22.8A

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