1.The relationship between the thickness of copper and platinum, line width and current of PCB board design
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.
1.2 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 segment will greatly increase, and there may be many
Everyone has seen that a certain section of the circuit 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,which enhances the current carrying capacity of that section
of the wire. 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 line between the pad and the pad. The solution: increase the wire
width. If the board cannot allow the wire width to be increased, add a
Solder layer (generally 1 mm 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 of tin and the amount of tin when
tin is passed),as shown in the figure below: the relationship between
the line width and the thickness of copper and platinum in the PCB board
design and the current. Unfamiliar, so if the amount of tin is even and
the amount of tin is enough,this 1mm wire can be more than a 2mm wire.This is very important in single-sided high-current boards.
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.
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.