“We usually need to quickly estimate the resistance value of a trace or a plane on a printed circuit board instead of performing tedious calculations. Although there are now available printed circuit board layout and signal integrity calculation programs that can accurately calculate the resistance of the traces, in the design process, we sometimes hope to take a quick and rough estimate.
We usually need to quickly estimate the resistance value of a trace or a plane on a printed circuit board instead of performing tedious calculations. Although there are now available printed circuit board layout and signal integrity calculation programs that can accurately calculate the resistance of the traces, in the design process, we sometimes hope to take a quick and rough estimate.
There is a way to accomplish this task easily, called “square statistics”. Using this method, the resistance value of any geometric trace can be accurately estimated in a few seconds (with an accuracy of about 10%). Once this method is mastered, the area of the printed circuit board that needs to be estimated can be divided into several squares. After counting the number of all squares, the resistance value of the entire trace or plane can be estimated.
The key concept of square statistics is that the resistance value of any square printed circuit board trace (with a certain thickness) is the same as that of other squares. The resistance value of a square depends only on the resistivity of the conductive material and its thickness. This concept can be applied to any type of conductive material. Table 1 shows some common semiconductor materials and their bulk resistivity.
For printed circuit boards, the most important material is copper, which is the raw material for most circuit boards.
Let’s start with the copper square in Figure 1. The length of the copper block is L, the width is also L (because it is a square), the thickness is t, and the cross-sectional area of the copper foil area through which the current passes is A. The resistance of the copper block can be simply expressed as R=ρL/A, where ρ is the resistivity of copper (this is an inherent characteristic of the material, which is 0.67μΩ/in. at 25°C).
Note, however, that the section A is the product of the length L and the thickness t (A=Lt). The L in the denominator and the L in the numerator cancel each other out, leaving only R=ρ/t. Therefore, the resistance of the copper block has nothing to do with the size of the block, it only depends on the resistivity and thickness of the material. If we know the resistance value of any size copper block and can decompose the entire trace that needs to be estimated into multiple blocks, we can add up (statistics) the number of blocks to get the total resistance of the trace.
To realize this technique, we only need a table, which shows the function relationship between the resistance value of a square on the printed circuit board trace and the thickness of the copper foil. The thickness of the copper foil is generally specified by the weight of the copper foil. For example, 1oz. copper means 1oz. per square foot.
Table 2 shows the weight of the four most commonly used copper foils and their resistivities at 25°C and 100°C. Please note that due to the positive temperature coefficient of the material, the resistance of copper will increase with increasing temperature.
For example, we now know that the resistance of a 0.5oz. heavy square copper foil is about 1mΩ, which has nothing to do with the size of the square. If we can decompose the printed circuit board traces that need to be measured into multiple virtual squares, and then add these squares together, we get the resistance of the traces.
A simple example
Let’s take a simple example. Figure 2 is a rectangular copper trace, its weight is about 0.5oz. at 25°C, the trace width is 1 inch, and the length is 12 inches. We can decompose the trace into a series of squares, each of which has a side length of 1 inch. In this way, there are a total of 12 squares. According to Table 2, the resistance of each 0.5oz. heavy copper foil square is 1mΩ. There are now 12 squares, so the total resistance of the trace is 12mΩ.
How to count a turn?
In order to facilitate understanding, the previous article lists a very simple example, let’s take a look at the more complicated situation.
The first thing to know is that in the previous example, we assumed that the current flows in a straight line along one side of the square, from one end to the other (as shown in Figure 3a). However, if the current has to make a right-angle turn (as the square right-angle in Figure 3b), then the situation is a little different.
In the previous example, we assumed that the current flows in a straight line along one side of the square, from one end to the other (as shown in Figure 3a). If the current has to turn at a right angle (as the square right angle in Figure 3b), we will find that the current path in the lower left part of the block is shorter than the upper right part.
When the current flows through the corners, the current density is higher, which means that the resistance of a corner square can only be calculated by 0.56 squares.
Now we see that the current path in the lower left part of the square is shorter than the upper right part. Therefore, the current will be crowded in the lower left area where the resistance is lower. Therefore, the current density in this area will be higher than in the upper right area. The distance between the arrows indicates the difference in current density. As a result, the resistance of a corner square is only equivalent to 0.56 squares (Figure 4).
Similarly, we can make some corrections to the connectors soldered on the printed circuit board. Here, we assume that the connector resistance is negligible compared to the copper foil resistance.
We can see that if the connector occupies a large part of the copper foil area to be evaluated, the resistance in this area should be reduced accordingly. Figure 5 shows the calculation of the three-terminal connector structure and its equivalent block (Reference 1). The shaded area represents the connector pins in the copper foil area.
A more complex example
Now, we use a more complicated example to illustrate how to use this technique. Figure 6a shows a more complicated shape, and calculating its resistance takes some time. In this example, we assume that the weight of the copper foil is 1oz. at 25°C, and the current direction is along the entire length of the trace, from point A to point B. Connectors are placed on both ends A and B.
Using the same technique described above, we can decompose a complex shape into a series of squares, as shown in Figure 6b. These squares can be of any suitable size, and squares of different sizes can be used to fill the entire area of interest. As long as we have a square and know the weight of the copper trace, we can know the resistance value.
We have six complete squares, two squares including connectors, and three corner squares. Since the resistance of 1oz. copper foil is 0.5mΩ/square, and the current flows linearly through six full squares, the total resistance of these squares is: 6×0.5mΩ=3mΩ.
Then, we have to add two blocks with connectors, each calculated as 0.14 blocks (Figure 5c). Therefore, two connectors count as 0.28 squares (2×0.14). For 1oz. copper foil, this adds a resistance of 0.14mΩ (0.28×0.5mΩ=0.14mΩ).
Finally, add three corner squares. Each is calculated by 0.56 squares, and the total is 3×0.56×0.5mΩ=0.84mΩ.
Therefore, the total resistance from A to B is 3.98mΩ (3mΩ+0.14mΩ+0.84mΩ).
Some friends will say: How can PCB traces be so strange? However, it is the power signal that often needs to calculate the wiring resistance, and the power signal is sometimes realized by copper coating, forming some irregular shapes.
Summarized as follows:
●Six full squares of 1 = 6 equivalent squares; two connector squares of 0.14 = 0.28 equivalent squares; three corner squares of 0.56 = 1.68 equivalent squares
●Total number of equivalent squares = 7.96 equivalent squares
●Resistance (A to B) = 7.96 squares of resistance, because each square is 0.5mΩ, so the total resistance = 3.98mΩ
This technique can be easily applied to complex geometries. Once the resistance value of a certain trace is known, it is very simple to calculate other quantities (such as voltage drop or power consumption, etc.).
How to count vias?
Printed circuit boards are usually not limited to a single layer, but are stacked in different layers. Vias are used to connect traces between different layers. The resistance of each via is limited, and the resistance of the via must be taken into account when calculating the total resistance of the trace.
Generally speaking, when a via connects two traces (or planes), it constitutes a series resistance element. Multiple vias in parallel are often used to reduce the effective resistance.
The calculation of the via resistance is based on the simplified via geometry shown in Figure 7. The current (as indicated by the arrow) along the length (L) of the via passes through a cross-sectional area (A). The thickness
After some simple algebraic transformations, the via resistance can be expressed as R=ρL/[π(Dt-t2)], Where ρ is the resistivity of copper plating (2.36μΩ/in. at 25°C). Note that the resistivity of copper plating is much higher than that of pure copper. We assume that the thickness t of the plating layer in the via is generally 1 mil, which has nothing to do with the weight of the copper foil of the circuit board. For a 10-layer board, the layer thickness is 3.5mil and the copper weight is 2oz., L is approximately 63mil.
Based on the above assumptions, Table 3 shows the common via size and its resistance. We can adjust these values according to our own special plate thickness. In addition, there are many free and easy-to-use via calculation programs on the Internet.
The above is a simple method for estimating the DC resistance of printed circuit board traces or planes. Complex geometric shapes can be broken down into multiple copper squares of different sizes to approximate the entire copper foil area. Once the weight of the copper foil is determined, the resistance value of any size square is known. In this way, the estimation process is simplified to a simple statistics of the number of copper cubes.
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