1. Calculation method is as follows
First, calculate the cross-sectional area of the track. Most FPCs have a copper foil thickness of 35um (if you are unsure, you can ask the FPC manufacturer; 1 ounce is equal to 35um, although it can be less than 35um in practice). Multiply the thickness by the width of the line to obtain the cross-sectional area. Make sure to convert it into square millimeters.
There is an empirical current density value of 15 to 25 amps per square millimeter. Let’s refer to it as the upper cross section to determine the flow capacity. The formula is I=KT0.44A0.75, where K is the correction factor. For general copper-coated wires, the value of K is 0.024 for the inner layer and 0.048T for the outer layer, representing the maximum temperature rise in Celsius (the melting point of copper is 1060℃). A represents the cross-sectional area of the copper-coated wire, measured in square mil (not mm). Please note that 1 mil is equal to 0.001 inch or 0.0254 mm. I represents the maximum allowable current in amperes (amps). As a general guideline, 10 mils can carry approximately 1A, and 250 mils (6.35mm) can carry 8.3A.
2.Data
The calculation of FPC current carrying capacity has lacked authoritative technical methods and formulas. Experienced CAD engineers can make more accurate judgments based on their personal experience. However, this may be challenging for CAD novices. The current carrying capacity of the FPC depends on the following factors: line width, line thickness (copper foil thickness), and allowable temperature rise. As we all know, wider FPC lines have a greater current carrying capacity. Let me give you an example: assuming that under the same conditions, a 10-mil line can withstand 1A, can a 50-mil line withstand 5A?
The answer is no. Please refer to the following data provided by international authorities: The unit for line width is inch (1 inch = 25.4 millimeters). 1 ounce of copper is 35 microns thick, 2 ounces is 70 microns thick, and 1 ounce is equal to 0.035 mm. Additionally, 1 mil is equal to 0.001 inch or 0.0254 mm. The trace carrying capacity can be determined using mil std 275.
3.experiment
In the experiment, it is important to consider the voltage drop caused by the line resistance, which is determined by the length of the wire. The soldering process only increases the current capacity but controlling the amount of solder can be challenging. For a 1 oz copper with a width of 1mm, it is generally suitable for a current range of 1-3A, depending on the length of the line and the desired voltage drop.
When determining the maximum current value, it is necessary to refer to the maximum allowable value within the temperature rise limit. The melting point of copper is reached at the fusing value, which indicates the temperature rise at which the copper melts. For example, for a 50 mil (0.050-inch) trace with 1 oz copper, at a temperature rise of 1060 degrees Celsius (which is the melting point of copper), the current carrying capacity is approximately 22.8A.
4.FPC Design: Copper Thickness, Line Width, and Current Relationship
To understand the relationship between FPC design parameters such as copper thickness, line width, and current, let’s first clarify the conversion between the units of ounce, inch, and millimeter for FPC copper thickness.
In many data tables, FPC copper thickness is often specified in ounces, and the conversion relationships with inches and millimeters are as follows:
1 ounce = 0.0014 inches = 0.0356 mm ≈ 36 microns (um)
2 ounces = 0.0028 inches = 0.0712 mm ≈ 72 microns (um)
The ounce is a unit of weight, and it can be converted to millimeters because the thickness of the copper deposited on the FPC is measured in ounces per square inch.
