### The relationship between PCB line width and current, look-up table and calculation!

“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, 1 ounce is 35um, in fact, it is less than 35um) multiplied by the line width is the cross-sectional area, pay attention to the conversion to square millimeters.

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1. The calculation method is as follows

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, 1 ounce is 35um, in fact, it is less than 35um) multiplied by the line width is the cross-sectional area, pay attention to the conversion to square millimeters.

There is an empirical value of current density, which is 15~25 amperes/mm2. Call it the upper cross-sectional area to get the flow capacity. I=KT0.44A0.75 (K is the correction factor, generally 0.024 for the inner layer of the copper clad wire, and 0.048T for the outer layer as the maximum temperature rise, in degrees Celsius (the melting point of copper is 1060°C) A is the copper clad Cross-sectional area, the unit is square MIL (not millimeter mm, note that it is square mil.) I is the maximum allowable current, the unit is ampere (amp) generally 10mil=0.010inch=0.254 can be 1A, 250MIL=6.35mm, it is 8.3A

2. Data

The calculation of PCB current-carrying capacity has always lacked authoritative technical methods and formulas. Experienced CAD engineers can make more accurate judgments relying on personal experience. But for CAD novices, it can not 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 authoritative organizations: The unit of line width is: Inch (inch inch = 25.4 millimetres) 1 oz. Copper = 35 microns thick, 2 oz. = 70 microns thick, 1 OZ = 0.035mm 1mil. =10-3inch. Trace Carrying Capacity per mil std 275

Three, experiment

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. 1 OZ 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 (that is, the melting point of copper), the current is 22.8A.

Fourth, PCB design copper platinum thickness, line width and current relationship

Before understanding the relationship between PCB design copper platinum thickness, line width and current, let us first understand the conversion of PCB copper thickness in units of ounces, inches and millimeters: “In many data sheets, PCB copper thickness is often used in ounces. As a unit, its conversion relationship with inches and millimeters is as follows:

1 ounce = 0.0014 inch = 0.0356 millimeter (mm)

2 ounces = 0.0028 inches = 0.0712 millimeters (mm)

Ounce is a unit of weight, and the reason why it can be converted to millimeters is because the copper thickness of the pcb is ounces/square inch” PCB design copper platinum thickness, line width and current relationship table

There is a direct relationship between the current carrying value of the wire and the number of vias on the wire. The calculation formula for the influence of the pad and the hole diameter per square millimeter on the load value of the line has not been found. Friends who are interested can find it by themselves. The individual is not too clear, so I won’t explain.) Here are just a few simple main factors that affect the current carrying value of the line.

1. The load-bearing value listed in the table data is the maximum current load-bearing value at a normal temperature of 25 degrees. Therefore, various environments, manufacturing processes, board processes, board quality, etc. must be considered in the actual design. Kinds of factors. Therefore, the table is provided only as a reference value.

2. In the actual design, each wire will also be affected by the pads and vias, such as the line segment with a lot of pads, after tinning, the current carrying value of the pad section will greatly increase, maybe Many people have seen that a certain section of the wire between the pad and the pad in some high-current boards is burned. The reason is simple. The pad has component feet and solder after the tin is completed, and the current of that section of the wire is increased. The maximum current carrying value of the pad between the pad and the pad is the maximum current carrying value allowed by the wire width.

Therefore, when the circuit fluctuates momentarily, it is easy to burn the section of the line between the pad and the pad. The solution: increase the width of the wire. A wire of about 0.6 Solder layer can be added to the wire, of course, you also add a 1mm Solder layer wire) So after tinning, this 1mm wire can be regarded as a 1.5mm~2mm wire (depending on the wire) The uniformity and amount of tin when tin is passed), as shown in the figure below:

This kind of processing method is no stranger to those who are engaged in PCB Layout of small home appliances, so if the amount of tin is even and the amount of tin is enough, this 1mm wire can be regarded as more than a 2mm wire. This is very important in single-sided high-current boards.

3. The processing method around the pad in the figure is also to increase the uniformity of the current carrying capacity of the wire and the pad. This is especially true for the board with large current and thick pins (the pins are greater than 1.2 and the pads are more than 3). Very important.

Because if the pad is above 3mm and the pin is above 1.2, the current of the pad will increase dozens of times after tinning. If there is a big fluctuation in the moment of large current, the current of the entire line The carrying capacity will be very uneven (especially when there are many pads), and it is still easy to cause the possibility of the circuit between the pads and the pads to burn. The processing in the figure can effectively disperse the uniformity of the current carrying value of a single pad and surrounding lines.

Finally, I will explain: the current carrying value data table is only an absolute reference value. When not doing large current design, an additional 10% of the data provided in the table can definitely meet the design requirements. In the general single-panel design, the copper thickness is 35um, which can basically be designed at a ratio of 1:1, that is, 1A current can be designed with a 1mm wire, which can meet the requirements (calculated at a temperature of 105 degrees) .

Five, the relationship between copper foil thickness, trace width and current in PCB design

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

The relationship between copper foil thickness, trace width and current in PCB design

The current carrying capacity of copper foil of different thickness and width is shown in the following table:

Note:

・ When using copper as a conductor to pass large currents, the current carrying capacity of the copper foil width should be derated by 50% with reference to the value in the table.

・ In PCB design and processing, OZ (ounces) is commonly used as the unit of copper thickness. 1 OZ copper thickness is defined as the weight of copper foil in 1 square foot area is one ounce, and the corresponding physical thickness is 35um; 2OZ copper thickness is 70um.

Extracted from: Huawei PCB Wiring Specification Internal Information P10

Six empirical formula

I=KT0.44A0.75

(K is the correction factor, generally 0.024 for the inner layer of the copper clad wire, and 0.048T for the outer layer as the maximum temperature rise, in degrees Celsius (the melting point of copper is 1060°C)

A is the cross-sectional area of the copper clad, and the unit is square MIL (not millimeter mm, note that it is square mil.)

I is the maximum allowable current in ampere (amp)

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

7. The calculation method provided by a netizen is as follows

First calculate the cross-sectional area of the track, most of the PCB copper foil thickness is 35um (if you are not sure, you can ask the PCB manufacturer), multiplying the line width is the cross-sectional area, pay attention to convert it to square millimeters. There is an empirical value of current density, which is 15~25 amperes/mm2. Call it the upper cross-sectional area to get the flow capacity.

8. A little experience about line width and copper paving of vias

We generally have a common sense when drawing PCBs, that is, use thick wires (such as 50 mils or more) where large currents are used, and thin wires (such as 10 mils) can be used for small current signals. For some electromechanical control systems, sometimes the instantaneous current flowing in the wire can reach more than 100A. In this case, the thinner wire will definitely cause problems.

A basic empirical value is: 10A/mm2, that is, the current value that a wire with a cross-sectional area of 1mm2 can pass safely is 10A. If the line width is too thin, the line will be burnt when a large current passes. Of course, the current burned traces also need to follow the energy formula: Q=I*I*t, for example, for a trace with 10A current, a 100A current burr suddenly appears and the duration is us level, then the 30mil wire is Definitely can bear it.

(At this time, there will be another problem-the stray inductance of the wire. This burr will generate a strong back electromotive force under the action of this inductance, which may damage other devices. The thinner the longer the wire is stray The greater the inductance, so the actual length of the wire must be considered)

General PCB drawing software often has several options when coppering the via pads of device pins: right-angle spokes, 45-degree spokes, and straight-laying. What’s the difference between them? Novices often don’t care much, just choose one at random and just look good. actually not. There are two main considerations: one is to consider not cooling too fast, and the other is to consider the over-current capability.

The characteristic of using the direct laying method is that the overcurrent capability of the pad is very strong, and this method must be used for the device pins on the high-power loop. At the same time, its thermal conductivity is also very strong. Although it is good for heat dissipation of the device when it works, it is a problem for the circuit board soldering personnel. Because the heat dissipation of the pad is too fast and it is not easy to hang the tin, it is often necessary to use a larger wattage soldering iron and Higher welding temperature reduces production efficiency.

The use of right-angle spokes and 45-angle spokes will reduce the contact area between the pins and the copper foil, and the heat dissipation is slow, and the soldering is much easier. Therefore, the choice of copper connection method for via pads should be based on the application, and the overall overcurrent capability and heat dissipation capability should be considered together. Do not use direct routing for low-power signal lines, and for pads that pass large currents, they must be straight. shop. As for the right angle or the 45 degree angle, it looks good.

Why did you bring this up? Because I have been working on a motor driver a while ago, and the H-bridge components in this driver are always burnt out. I can’t find the reason for four or five years. After a lot of hard work, I finally found out: It turned out that the pad of a device in the power loop was copper-layed with right-angle spokes (and because the copper was not painted well, only two spokes actually appeared). This greatly reduces the overcurrent capability of the entire power loop.

Although the product does not have any problems during normal use, it is completely normal under the condition of 10A current. However, when the H bridge is short-circuited, a current of about 100A will appear on the loop, and the two spokes will be burnt instantaneously (uS level).

Then, the power circuit becomes an open circuit, and the energy stored in the motor is emitted through all possible means without a discharge channel. This energy will burn the current-measuring resistor and related operational amplifier devices, and destroy the bridge control chip. And infiltrate into the signal and power supply of the digital circuit part, causing serious damage to the entire device. The whole process was as thrilling as the explosion of a large landmine with a strand of hair.

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