We usually need to quickly estimate the resistance of a wire or a plane on a printed circuit board, rather than make a tedious calculation. Although there are available printed circuit board layout and signal integrity calculation programs, which can accurately calculate the wiring resistance, we sometimes hope to adopt a fast and rough estimation method in the design process.
There is a way to do this easily, called “block statistics.”. With this method, the resistance of any geometry can be accurately estimated in a few seconds (with an accuracy of about 10%). Once this method is mastered, the PCB area to be estimated can be divided into several blocks. After counting the number of all blocks, the resistance value of the whole wiring or plane can be estimated.
The key concept of block statistics is that the resistance value of square printed circuit board of any size (determined by thickness) is the same as that of other size blocks. The resistance value of the positive block only depends 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 block in Figure 1. The length of the copper block is l, the width is l (because it is square), the thickness is t, and the cross-sectional area of the copper foil through which the current passes is a. The resistance of the copper block can be simply expressed as R = ρ L / A, where ρ is the resistivity of copper (this is the inherent property of the material, which is 0.67 μ Ω / in.) at 25 ℃.
Note, however, that section a is the product of length L and thickness t (a = LT). The L in the denominator and the L in the molecule cancel each other, leaving only r = ρ / T. Therefore, the resistance of 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 whole route to be estimated into multiple blocks, then we can add (count) the number of blocks to get the total resistance of the wire.
In order to realize this technology, we only need a table, which gives the function relationship between the resistance value of a block on the printed circuit board trace and the thickness of copper foil. The thickness of copper foil is generally specified by the weight of copper foil. For example, 1 oz. copper means 1 oz. per square foot.
Table 2 gives the weights of the four most commonly used copper foils and their resistivity at 25 ℃ and 100 ℃. Note that since the material has a positive temperature coefficient, the copper resistance increases with increasing temperature.
For example, we now know that the resistance of a 0.5 oz. square copper foil is about 1 m Ω, which is independent of the size of the block. If we can decompose the printed circuit board wiring to be measured into several virtual blocks, and then add these blocks together, we can get the wiring resistance.
A simple example
Let’s take a simple example. Figure 2 shows a rectangular copper wire with a weight of about 0.5oz at 25 ℃, with a width of 1 inch and a length of 12 inches. We can decompose the routing into a series of squares, each of which is an inch long. So, there are a total of 12 squares. According to table 2, the resistance of each 0.5oz. Heavy copper foil block is 1m Ω. Now there are 12 blocks, so the total resistance of the wiring is 12m Ω.
How about turning?
To facilitate understanding, the previous article listed a very simple example, let’s take a look at the situation of complex points.
First of all, in the previous example, we assume that the current flows in a straight line along one side of the block, from one end to the other (as shown in Figure 3a). However, if the current is to be bent at a right angle (as in the square in Figure 3B), the situation is somewhat different.
In the previous example, we assume that the current flows in a straight line along one side of the block, from one end to the other (as shown in Figure 3a). If the current is to take a right angle bend (such 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 that in the upper right part.
When the current flows through the corner, the current density is high, which means that the resistance of a corner square can only be calculated as 0.56 squares.
Now we see that the current path in the lower left part of the square is shorter than that in the upper right part. As a result, the current will crowd in the lower left area where the resistance is low. So, the current density in this area will be higher than that 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 0.56 square (Figure 4).
Similarly, we can make some modifications to the connectors soldered to 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 of that area should be reduced accordingly. Figure 5 shows the structure of the three terminal connector and the calculation of its equivalent block (ref. 1). The shaded area represents the connector pins in the copper foil area.
A more complex example
In, we use a more complex example to illustrate how to use this technology. Fig. 6a is a complicated shape, and it takes a lot of time to calculate its resistance. In this example, we assume that the weight of copper foil is 1oz at 25 ℃, and the current direction is along the whole length of the line from point a to point B. There are connectors on both a-end and b-end.
Using the same technique described above, we can decompose a complex shape into a series of squares, as shown in Figure 6B. These blocks can be of any suitable size and can be filled with blocks of different sizes to fill the whole area of interest. As long as we have a square and know the weight of the copper wire, we can know the resistance value.
We have six full squares, two of which contain connectors, and three corner squares. Because the resistance of 1 oz. copper foil is 0.5 m Ω / square, and the current flows through six blocks linearly, the total resistance of these blocks is 6 × 0.5 m Ω = 3 m Ω.
Then, we will add two blocks with connectors, each calculated as 0.14 blocks (Figure 5C). Therefore, two connectors count 0.28 squares (2 × 0.14). For 1 oz. copper foil, this increases the resistance by 0.14 m Ω (0.28 × 0.5 m Ω = 0.14 m Ω).
Finally, add three corner squares. According to 0.56 blocks per block, the total is 3 × 0.56 × 0.5m Ω = 0.84M Ω.
Therefore, the total resistance from a to B is 3.98M Ω (3m Ω + 0.14m Ω + 0.84M Ω).
Some friends will say: how can PCB wiring be so strange? However, it is the power signal that needs to calculate the wiring resistance. Sometimes the power signal is realized by copper coating, forming some irregular shapes.
The summary is 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 equivalent blocks = 7.96 equivalent blocks
Resistance (a to b) = 7.96 pieces of resistance, because each block is 0.5m Ω, so the total resistance = 3.98M Ω
This technique can be easily applied to complex geometry. Once the resistance value of a wire is known, it is easy to calculate other quantities (such as voltage drop or power consumption).
What about vias?
Printed circuit boards are usually not limited to single layers, but stacked in different layers. Vias are used for wiring connections between different layers. The resistance of each vias is limited, and the resistance of vias must be taken into account in the calculation of total wiring resistance.
Generally speaking, when a via connects two wires (or planes), it forms a series resistance element. Multiple parallel vias are often used to reduce the effective resistance.
The calculation of via resistance is based on the simplified via geometry shown in Figure 7. The current along the length (L) of the via (as indicated by the arrow) passes through a cross-sectional area (a). The thickness (T) depends on the thickness of copper plating on the inner wall of the via hole.
Through some simple algebraic transformations, the through hole resistance can be expressed as R = ρ L / [π (dt-t2)], where ρ is the resistivity of copper plating (2.36 μ Ω / in.) at 25 ℃. Note that the resistivity of copper plating is much higher than that of pure copper. We assume that the thickness of the plating layer in the via hole is generally 1mil, which is independent of the weight of the copper foil of the circuit board. For a 10 ply plate, l is about 63 mil when the thickness is 3.5 mil and the weight of copper is 2 oz.
Based on the above assumptions, table 3 shows the common via sizes and their resistances. We can adjust these values according to our own special plate thickness. In addition, there are many free and easy to use vias calculation programs on the Internet.
The above is a simple method to estimate PCB wiring or Planar DC resistance. The complex geometry can be decomposed into several copper blocks 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 block is known. In this way, the estimation process is simplified as a simple statistics of the number of copper blocks.