Electronic Design - How to estimate the resistance value of PCB wiring skills

We usually need to quickly estimate the resistance of a single line or plane on a printed circuit board, rather than making tedious calculations. Although there are printed circuit board layout and signal integrity calculations that can be used to calculate the resistance of the exit line, there are times when we want to make quick and rough estimates in the design process.

An easy way to do this is called "cube counting." Using this method, the resistance of any geometry can be estimated in a few seconds (about 10%). Once this method is mastered, the PCB area to be estimated can be divided into squares, and the resistance of the entire line or plane can be estimated by counting the number of squares.

The basic concept

The key concept of square statistics is that a square printed circuit board of any size (thickness determined) will have the same resistance as a square of any other size. The resistance value of a positive square depends only on the resistivity of the conducting 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 important material is copper, which is the raw material for most circuit boards (note: aluminum is used to metallize the chip core of integrated circuits, and the same principles apply to aluminum)

Let's start with the copper square 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 area through which the current flows 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 property of the material, 0.67μ ω /in. At 25 degree Celsius).

But note that section A is the product of length L and thickness t (A=Lt). The L in the denominator cancels with the L in the numerator, leaving only R=ρ/t. Therefore, the resistance of the copper block is independent of the size of the block, it only depends on the resistivity and thickness of the material. If we know the resistance of copper squares of any size and can divide the entire line to be estimated into squares, the number of squares can be added to obtain the total resistance of the line.

implementation

To implement this technique, all we need is a table showing a function of the resistance value of a square on the PCB line and the thickness of the copper foil. The thickness of copper foil is usually specified by the weight of copper foil. For example, 1oz. Copper means 1oz. Per square foot.

Table 2 shows the weights of four commonly used copper foils and their electrical resistivity at 25 degree Celsius and 100 degree Celsius. Note that the copper resistance value increases with increasing temperature due to the positive temperature coefficient of the material. For example, we now know that a 0.5oz square of copper foil has a resistance of about 1 MEgohm, a value that is independent of the size of the square. If we can break down the PCB wiring we need to measure into virtual squares, and then add these squares together, we can get the resistance of the wiring.

Let's take a simple example. Figure 2 shows a rectangular copper wire weighing approximately 0.5oz. At 25 ° C, with a wire width of 1 inch and a length of 12 inches. We can break the line up into a series of squares, with each square side 1 inch long. So there are 12 beepers. According to Table 2, the resistance of each 0.5oz. Weight copper foil square is 1m ω, now there are 12 squares, so the total resistance of the wire is 12m ω.

What about turning?

This is a very simple example to understand, but let's look at something a little more complicated.

First, remember that in the previous example, we assumed that the current flowed in a straight line along one side of the square, from one end to the other (see Figure 3A). However, if the current has to make a right Angle turn (the square right Angle in Figure 3B), the situation is somewhat different.

In the previous example, we assumed that the current flowed in a straight line along one side of the square, from one end to the other (see Figure 3A). If the current takes a right Angle turn (the square right Angle in Figure 3B), we will find that the current path is shorter in the lower left part of the square than in the upper right part. When current flows through a corner, the current density is high, meaning that the resistance of a corner square can only be calculated in terms of 0.56 squares.

Now we see that the current path is shorter in the lower left part of the square than in the upper right part. As a result, the cu

rrent will crowd into the lower left region where the resistance is low. So the current density in this region is going to be higher than the current density in the upper right. The distance between the arrows shows the difference in current density. As a result, the resistance of a corner square is equal to 0.56 squares (Figure 4).

Also, we can make some modifications to the connectors that are welded 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 portion of the copper foil area to be evaluated, the resistance of that area should be reduced accordingly. Figure 5 shows the three-terminal connector structure and its equivalent square calculation (reference 1). The shaded area indicates the connector pins in the copper foil area.

A more complicated example

Here, we use a more complex example to illustrate how to use this technique. FIG. 6A is a more complex shape and it takes some work to calculate its resistance. In this example, we assume that the copper foil weighs 1oz at 25 degree Celsius, and that the current flows along the entire length of the line, from point A to point B. Connectors are placed on both A and B terminals. Using the same technique as above, we can decompose a complex shape into a series of squares, as shown in Figure 6b. The squares can be any appropriate size, and you can fill the entire area of interest with squares of different sizes. As long as we have a square, and we know the weight of the copper wire, we know the resistance.

We have six perfectly square squares, two square squares with connectors, and three corner squares. Since the resistance of 1oz. Copper foil is 0.5m ω / square, and the current flows linearly through the six squares, the total resistance of these squares is: 6 x 0.5m ω =3m ω.

Then, we add two squares with connectors, each calculated at 0.14 squares (Figure 5C). Thus, the two connectors count as 0.28 squares (2*0.14). For 1oz. Copper foil, this adds 0.14m ω of resistance (0.28*0.5m ω =0.14m ω)., plus three corner squares. Calculated by 0.56 squares each, the total is 3*0.56*0.5m ω =0.84m ω. Thus, 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 so strange shape? However, it is the power supply signal that often needs to calculate the wiring resistance, and the power supply signal is sometimes realized by covering copper, forming some irregular shapes.

The summary is as follows:

. Six full squares of 1 =6 equivalent squares; Two connector blocks of 0.14 =0.28 equivalent blocks; 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 square resistance, since each square is 0.5m ω, so the total resistance =3.98m ω this technique can be easily applied to complex geometry. Once you know the resistance of a particular wire, it is easy to calculate other things, such as voltage drop or power consumption.

How to calculate the hole?

Printed circuit boards are usually stacked in different layers rather than single layers. The through-hole is used for wiring connections between different layers. The resistance of each hole is limited. Therefore, the resistance of each hole must be taken into account when calculating the total resistance of cabling. In general, when two wires (or planes) are connected through a hole, it constitutes a series resistance element. Multiple parallel perforations are often used to reduce the effective resistance. The through-hole resistance is calculated based on the simplified through-hole geometry shown in Figure 7. The current (as indicated by the arrow) along the length of the hole (L) passes through A cross-sectional area (A). The thickness (t) depends on the thickness of the copper layer electroplated on the inner wall of the hole.

By some simple algebraic transformations, the through-hole resistance can be expressed as R=ρL/[π(dt-T2)], where ρ is the resistivity of copper plating (2.36μ ω /in. At 25 degree Celsius). Note that the resistivity of copper plating is much higher than that of pure copper. We assume that the thickness t of the coating in the hole is generally 1mil, which is independent of the weight of the copper foil on the circuit board. For a 10-layer plate with a thickness of 3.5mil and copper weight of 2Oz., L is about 63mil.

Based on the above assumptions, common hole sizes and resistances are given in Table 3. We can adjust these values for our particular plate thickness. There are also many easy-to-use through-hole computing programs available online.

This is a simple way to estimate the dc resistance of a PCB line or plane. The complex geometry can be decomposed 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 of any square of size is a known quantity. In this way, the estimation process is simplified to a simple count of copper squares.