Why is the thickness of PCB copper foil in ounces? After reading this, I finally understand!

First of all, we need to make a clarification is that the ounce (OZ) itself is a product weight as well as the unit. Ounces and grams (g) conversion relationship formula: 1OZ ≈ 28.35g.

In the printed circuit board industry, 1OZ means that the thickness of 1OZ of copper is evenly distributed over an area of 1 FT2. The weight per unit area indicates the average thickness of the copper foil. Using the formula,pcblink 1oz vs 2oz Copper 1 oz = 28.35 g/FT2 (FT2 is square feet, 1 square foot = 0.09290304 square meters).

Specifically, it converts to length or thickness as follows.

First, we know the density constant for copper and the associated unit conversion formula as follows.

Density of copper ρ = 8.9 g/cm3

1 centimeter (cm) = 10 millimeters (mm); 1 millimeter (mm) = 1000 microns (um)

1mil≈25.4um

1 FT2≈929.0304cm2

1mil ≈ 25.4um

Based on the formula of mass m = ρ x V (volume) = ρ x S (area) x T (thickness), we know that the weight of the copper foil divided by the density and area of the copper is the thickness of the foil!

As mentioned earlier, 1 ounce = T × 929.0304 cm2 × 8.9 g/cm3 = 28.35 g

So, t = 28.35 ÷ 929.0304 ÷ 8.9 cm cm ≈ 0.0034287 cm = 34.287 microns ≈ 34.287 25.4 mils ≈ 1.35 mils

From this we can see that the thickness of 1OZ copper foil is typically about 35um or 1.35 mil.

Copper thickness 1.OZ (0.035mm) Copper thickness 1.5OZ (0.05mm) Copper thickness 2.OZ (0.07mm)

PCB line width and current relationship

First, calculate the cross-sectional area of the track. Most PCBs have a copper foil thickness of 35um (ask the PCB manufacturer if you're not sure). That's the cross-sectional area multiplied by the line width, note the conversion to square millimeters. There is an empirical value for current density of 15 to 25 amps/square millimeter. Called cross-sectional area, get the flow capacity.

Second, the data.

PCB current carrying capacity calculation has been the lack of authoritative technical methods, formulas, experienced CAD engineers rely on personal experience to make more accurate judgments. But for new CAD users, there is a problem.

The current carrying capacity of a PCB depends on several factors: line width, line width (copper foil thickness) and allowable temperature rise. It is well known that the wider the PCB alignment, the higher the current carrying capacity. Here, please tell me:Suppose under the same conditions, 10 mils of wire can withstand 1A, 50 mils of wire can withstand how much current, is 5A? Naturally, the answer is no. Please see the following data provided by international authorities.

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

The voltage drop due to wire resistance caused by wire length must also be considered in the experiment. The soldering process can only increase the current capacity, but it is difficult to control the volume of tin. 1 OZ copper, 1 mm wide, is typically used as a 1-3a meter, and depending on the length of the line, a voltage drop is required.

The maximum current value should refer to the maximum allowable value at the limit of temperature rise, and the fusing value is the value when the temperature rise reaches the melting point of the copper. For example, for a 50 mil 1oz temperature rise of 1060 degrees (i.e. the melting point of copper), how to test pcb board with a multimeter ?the current is 22.8A