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Line width and copper thickness in PCB design
Line width and copper thickness in PCB design

# Line width and copper thickness in PCB design

2022-01-07
View：218
Author：pcb

1. The relationship between the thickness of copper and platinum, line width and current of PCB boarddesign
Before understanding the relationship between PCB board design copper platinum thickness, line width and current, let us first understand the conversion between the unit ounces, inches and millimeters of the PCB copper thickness: "In many data sheets, the PCB copper thickness is often Using ounces as the unit, the conversion relationship between it and inches and millimeters is as follows: 1 ounce = 0.0014 inches = 0.0356 mm, 2 ounces = 0.0028 inches = 0.0712 millimeters, ounces is the unit of weight, and the reason why it can be converted to millimeters is because of the PCB layout. Copper thickness is ounces/square inch", the relationship table of PCB design copper platinum thickness, line width and current, the relationship between PCB design line width and copper platinum thickness and current can also be calculated using the empirical formula: 0.15&TImes ;Line width (W)=A, the above data are all line current carrying values at a temperature of 25 degree Celsius., Wire impedance: 0.0005&TImes;L/W, the current carrying value is directly related to the number of components/pads and vias on the line. In addition, the current carrying value of the wire is related to the number of vias and pads of the wire. There is a direct relationship between the current carrying value of the wire and the number of vias of the wire and the number of pads (currently, there is no calculation formula for the influence of pads and vias per square millimeter on the carrying value of the circuit. 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.

1.1 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 factors such as various environments, manufacturing processes, plate processes, and plate quality must be considered in the actual design. . Therefore, the table is provided only as a reference value.

1.3 The processing method around the pads in the figure is also to increase the uniformity of the current carrying capacity of the wires and the pads. This is especially true in the boards with large current and thick pins (the pins are greater than 1.2 and the pads are more than 3). 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. Again: the current carrying value data table is just 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) .

2. The relationship between copper foil thickness, trace width and current in PCB board 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 thickness of the copper foil in the design of the PCB board, the relationship between the trace width and the current, different thicknesses and different widths of copper The current carrying capacity of the foil is shown in the following table: the relationship between the line width and the thickness of copper and platinum and the current in the PCB board design
Note: 2.1 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 for selection consideration. 2.2 In the design and processing of PCB boards, OZ 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 is 70um.

3. How to determine the line width of large current wires
The relationship between line width and copper platinum thickness and current in PCB board design
The relationship between line width and copper platinum thickness and current in PCB board design
The relationship between line width and copper platinum thickness and current in PCB board design
The relationship between line width and copper platinum thickness and current in PCB board design

4. Use the temperature impedance calculation software of the PCB board to calculate (calculate the line width, current, impedance, etc.) PCB board TEMP fill in the Location (External/Internal) wire on the surface or inside the FR-4 board, Temp temperature (Degree C), Width line width (Mil), Thickness thickness (Oz/Mil), and then click Solve to find the passing current, and you can also know the passing current to find the line width. Very convenient. In the PCB board design, the relationship between the line width and the thickness of copper and platinum and the current, you can see 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)

5. Empirical formula
I=KT0.44A0.75 (K is the correction factor. Generally, the copper-clad wire is 0.024 for the inner layer and 0.048 for the outer layer. T is the temperature rise in degrees Celsius (the melting point of copper is 1060°C), and A is the coating Copper cross-sectional area, the unit is square MIL, I is the allowable current, the unit is ampere. Generally 10mil=0.010inch=0.254 can be 1A, 250MIL=6.35mm, it is 8.3A

6. A little experience about line width and copper paving
When drawing PCB boards, we generally have a common sense, that is, use thick lines (such as 50 mils or more) where large currents are used, and thin lines (such as 10 mils) can be used for low-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 problems. A basic empirical value is: 10A/square mm, that is, the current value that a wire with a cross-sectional area of 1 square millimeter can safely pass is 10A. If the line width is too thin, the line will be burnt when a large current passes. Of course, the current burned trace must also 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, 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 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. Using right-angle spokes and 45-angle spokes will reduce the contact area between the pins and the copper foil, heat dissipation is slow, and 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 mention this? Because I have been working on a motor driver a while ago, 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-plated with right-angle spokes (and because of the poor copper painting, 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 PCB board equipment.