The following summarizes eight formulas, tables, and calculation formulas for the relationship between current and line width. Although they are different (generally similar), everyone can consider the size of the PCB board comprehensively in actual PCB board design, and choose an appropriate line width based on current.

1、 PCB current and line width

The calculation of PCB current carrying capacity has always lacked authoritative technical methods and formulas, and experienced CAD engineers can make more accurate judgments based on personal experience. But for CAD beginners, it can be said that they encounter a difficult 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. Assuming that under 

the same conditions, a 10 MIL wiring can withstand 1A, how much current can a 50 MIL wiring withstand? Is it 5A?

The answer is naturally negative. Please refer to the following data provided by international authoritative institutions:

The unit of line width is: Inch (1 inch=2.54cm=25.4mm)

Data source: MIL-STD-275 Printed Wiring for Electronic Equipment

2、 The Relationship between Copper Platinum Thickness, Line Width, and Current in PCB Design

Before understanding the relationship between copper platinum thickness, line width, and current in PCB design, let's first understand the conversion between the unit ounce, inch, and millimeter of PCB copper thickness: "In many data tables, the unit ounce of PCB copper thickness is often used, and its conversion relationship with inches and millimeters is as follows:"

1 ounce=0.0014 inches=0.0356 millimeters (mm)

2 ounces=0.0028 inches=0.0712 millimeters (mm)

Ounce is a unit of weight, and it can be converted to millimeters because the copper coating thickness of PCB is ounces per square inch.

An empirical formula can also be used to calculate: 0.15 x line width (W)=A

The above data are all the line current carrying values at a temperature of 25 ℃.

Wire impedance: 0.0005 × L/W (line length/width)

In addition, there is a direct relationship between the current carrying value of the wire and the number of through holes in the wire (currently, there is no calculation formula for the impact of through holes and the number of pads per square millimeter on the carrying value of the line. Interested friends can find it themselves.) Here, we will only briefly introduce some of the main factors that affect the current carrying value of the line.

1. The carrying capacity values listed in the table data are the maximum current carrying capacity values that can be withstood at room temperature of 25 degrees. Therefore, in actual design, various factors such as environment, manufacturing process, sheet metal process, and sheet metal quality should also be considered. So the table is only provided as a reference value.

2. In practical design, each wire is also affected by solder pads and through holes. For example, for wire segments with too many solder pads, the current carrying value of that segment of the solder pad will greatly increase after soldering. Many people may have seen some high current boards where a certain section of the circuit between the solder pad and the solder pad is burned. The reason for this is simple: after soldering, the solder pad has increased the current carrying value of that segment of the wire due to the presence of component pins and solder, and the maximum current carrying value of the solder pad between the solder pads is also the maximum current carrying value allowed by the wire width.

Therefore, when the circuit fluctuates instantaneously, it is easy to burn out the section of the circuit between the solder pads. The solution is to increase the wire width. If the board cannot allow an increase in wire width, add a layer of Solder layer to the wire (usually a wire with a left to right Solder layer of 0.6 can be added on a 1mm wire, and of course, you can also add a 1mm Solder layer wire). After passing through tin, this 1mm wire can be regarded as a 1.5mm~2mm wire (depending on the uniformity and amount of tin when the wire passes through tin), as shown in the following figure:

This type of processing method is not unfamiliar to friends engaged in small home appliance PCB layout. Therefore, if the amount of tin passing is uniform and sufficient, this 1mm wire can be seen as more than just a 2mm wire. And this is particularly important in single-sided high current boards.

3. The processing method around the solder pads in the figure is also to increase the uniformity of the current carrying capacity between the wires and the solder pads. This is particularly important in boards with large current and coarse pins (pins greater than 1.2 and solder pads greater than 3).

Because if the solder pads are 3mm or more and the pins are 1.2 or more, the current of this spot welding pad will increase several tens of times after being tinned. If there is a large fluctuation in the current instantly during high current, the current carrying capacity of the entire circuit will be very uneven (especially when there are many solder pads), which still easily leads to the possibility of burning the circuit between the solder pads. The processing in the figure can effectively disperse the uniformity of the current carrying value between a single solder pad and the surrounding circuit.

Finally, it should be noted that the current carrying capacity data table is only an absolute reference value. Without conducting high current design, adding an additional 10% of the data provided in the table can definitely meet the design requirements.

In general single panel design, with a copper thickness of 35um, it can be designed in a 1:1 ratio, which means that the current of 1A can be designed with 1mm wires, which can meet the requirements (calculated at a temperature of 105 degrees).

3、 The relationship between copper foil thickness, wire width, and current in PCB design

The current intensity of the signal. When the average current of the signal is high, the current that the wiring width can carry should be considered. The wiring width can refer to the following data:

The relationship between copper foil thickness, wire width, and current during PCB design, and the current carrying capacity of copper foils with different thicknesses and widths are shown in the table below:

Note:

i. When using copper foil as a wire to pass a large current, the current carrying capacity of the copper foil width should be selected and considered by reducing the value by 50% according to the table.

Ii In PCB design and processing, OZ (ounces) is commonly used as the unit of copper thickness. The definition of 1 OZ copper thickness is that the weight of copper foil in 1 square foot is one ounce, corresponding to a physical thickness of 35um; The thickness of 2OZ copper is 70um.

4、 How to determine the line width of high current conductors

Wire thickness 35 μ M

Wire thickness 70 μ M

Wire thickness 105 μ M

5、 Using PCB temperature impedance calculation software to calculate

PCBTEMP (calculating line width, current, impedance, etc.)

Fill in the Location (External/Internal) wire on the surface or inside the FR-4 board, Temperature (Degree C), Width (Mil), and Thickness (Oz/Mil) in order, and then click Solve to calculate the current passing through. You can also know the current passing through and calculate the line width. Very convenient.

You can see that the results are similar to the first method (20 degrees Celsius, 10 mil line width, which is 0.010 inches line width, with a copper foil thickness of 1 Oz)

Six empirical formulas

I=KT0.44A0.75

(K is the correction factor, usually taken as 0.024 for the inner layer and 0.048 for the outer layer of copper clad wire)

T is the maximum temperature rise, in degrees Celsius (the melting point of copper is 1060 ℃)

A is the cross-sectional area of copper coating, in square MIL (not in millimeters, please note that it is square mil.)

I is the maximum allowable current, in amperes (amps)

Generally, 10mil=0.010inch=0.254 can be 1A, 250MIL=6.35mm, which is 8.3A

7、 The calculation method provided by a certain netizen is as follows

First, calculate the cross-sectional area of the track. The copper foil thickness of most PCBs is 35um (if unsure, you can ask the PCB manufacturer). Multiplying it by the line width is the cross-sectional area, please convert it to square millimeters. There is an empirical value for current density, which is 15-25 amperes per square millimeter. Call it the cross-sectional area to obtain the flow capacity.

Eight Experiences on Line Width and Copper Laying through Holes

We usually have a common sense when drawing PCBs, which is to use thick wires (such as 50mil or even more) for high current areas, and thin wires (such as 10mil) for low current signals. For some electromechanical control systems, sometimes the instantaneous current flowing through the wiring can reach over 100A, which will definitely cause problems with thinner wires.

A basic empirical value is 10A per square millimeter, which means that a wire with a cross-sectional area of 1 square millimeter can safely pass through a current value of 10A. If the line width is too thin, the wiring will burn out when a large current passes through. Of course, if the current burns out the wiring, the energy formula should also be followed: Q=I * I * t. For example, for a wiring with a 10A current, if a 100A current spike suddenly appears for a duration of us level, then a 30mil wire can definitely withstand it. (At this point, another problem arises? The stray inductance of the wire, which will generate a strong reverse electromotive force under the action of this inductance, which may damage other devices. The thinner and longer the wire, the greater the stray inductance, so in practice, the length of the wire should be considered comprehensively.).

General PCB drawing software often has several options for laying copper on through-hole pads of device pins: right angle spokes, 45 degree angle spokes, and straight laying. What is the difference between them? Novices often don't pay much attention and just choose one that looks beautiful. Actually, it's not the case. There are two main considerations: first, it is necessary to consider not dissipating heat too quickly, and second, it is necessary to consider the overcurrent capacity.

The characteristic of using the direct laying method is that the solder pads have strong overcurrent capacity, and this method must be used for device pins on high-power circuits. At the same time, its thermal conductivity is also very strong. Although it is beneficial for device heat dissipation when working, it is a challenge for circuit board soldering personnel because the heat dissipation of the solder pad is too fast and it is not easy to hang tin. It often requires the use of larger wattage soldering irons and higher welding temperatures, which reduces production efficiency.

Using right angle and 45 angle spokes will reduce the contact area between pins and copper foil, slow down heat dissipation, and make soldering much easier. So the choice of copper laying on through-hole solder pads should be based on the application situation, taking into account both overcurrent and heat dissipation capabilities. For low-power signal lines, do not use straight laying, but for solder pads that pass through high currents, it is necessary to lay them straight. As for right angles or 45 degrees, it depends on the aesthetics. Why did you mention this? Because I was studying a motor driver a while ago, and the components of the H-bridge in this driver kept burning out. It has been four or five years and I can't find the reason.

After a lot of hard work, it was finally discovered that the solder pad of a device in the power circuit used a right angle spoke copper laying method (and due to the poor drawing of the copper laying, only two spokes actually appeared). This greatly reduces the overcurrent capability of the entire power circuit. Although the product has no problems during normal use and operates perfectly at a current of 10A.

However, when there is a short circuit in the H-bridge, a current of around 100A will appear on the circuit, and these two spokes will instantly burn out (uS level). Then, the power circuit becomes open circuit, and the energy stored on the motor is released through all possible channels without a discharge channel. This energy will burn out the current measuring resistor and related operational amplifier devices, destroy the bridge control chip, and enter the signal and power supply of the digital circuit, causing serious damage to the entire equipment. The whole process was as thrilling as detonating a large landmine with a strand of hair.