Super finishing! The relationship between current and line width in PCB design

Regarding the empirical formulas of PCB line width and current, there are many relationship tables and software on the Internet. This article organizes the online ones to provide convenience for the majority of engineers when designing PCB boards.

    The following summarizes the eight kinds of current and line width relationship formulas, tables and calculation formulas, although they are different (generally similar), but you can comprehensively consider the size of the PCB board in the actual PCB board design, through the current, choose one Appropriate line width.

    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 relying on personal experience. But for CAD novices, it cannot 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. 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 look at the following data from international authorities:

    Data provided:

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

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



    2. 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 between ounces, inches and millimeters of PCB copper thickness: "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 inches = 0.0356 millimeters (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 thickness of the copper coating of the PCB is ounces/square inch"

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



    You can also use the empirical formula to calculate: 0.15 × line width (W) = A

    The above data are all line current carrying values at a temperature of 25°C.

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

    In addition, the relationship between the current carrying value of the wire and the number of vias pads of the wire

    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 via hole diameter on the load value of the line per square millimeter 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 current load-bearing value that can be withstood at a normal temperature of 25 degrees. Therefore, various environments, manufacturing processes, plate processes, plate quality, etc. must be considered in the actual design. factor. 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 line between the pad and the pad in some high-current boards is burned. The reason is very 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 carrying value, and the current carrying value of the pad between the pad and the pad is also the 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. If the board cannot allow the width of the wire to be increased, add a layer of Solder layer on the wire (usually 1 mm). You can add a wire of about 0.6 Solder layer 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 in the board with large current and thick pins (the pin is greater than 1.2, and the pad is above 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 large 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.

    Again: the current carrying value data table is only a reference value. When not doing large current design, add 10% to the data provided in the table to 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) .

    Third, 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:

    i. When copper is used 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 for selection consideration.

    ii. In PCB design and processing, OZ (ounce) 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, which corresponds to a physical thickness of 35um; 2OZ copper thickness It is 70um.

    Fourth, how to determine the line width of a large current wire





    Five use PCB temperature impedance calculation software to calculate (calculate line width, current, impedance, etc.) PCBTEMP

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



    It can be seen that the results of the same method are similar (20 degrees Celsius, 10mil line width, which is 0.010inch line width, copper foil thickness is 1 Oz)

    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 0.048 for the outer layer

    T is the 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 allowable current in ampere (amp)

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

    7. The calculation method provided by a netizen 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). Multiplying 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 A/mm2. Call it the upper cross-sectional area to get the flow capacity.

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

    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) for small current signals. For some electromechanical control systems, sometimes the instantaneous current flowing in the wire can reach more than 100A, so the thinner wire will definitely have a problem.

    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 reverse 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)

    The general PCB drawing software often has several options when laying copper on the via pads of the device pins: right-angle spokes, 45-degree spokes, and direct laying. What is the difference between them? Novices often don't care too 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 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-hole 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 mention this? Because I have been working on a motor driver for a while, the H-bridge components in this driver are always burned out, and 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 circuit was copper-layed with right-angle spokes (and because of the poor copper drawing, only two spokes actually appeared). This greatly reduces the overcurrent capability of the entire power loop. Although the product has no 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 a big landmine was detonated with a strand of hair.

    So, why are only two spokes used on the pads in the power loop? Why not let the copper foil go straight over it? Because the staff in the production department said that this pin is too difficult to solder!