Introduction

We often have a need to quickly estimate the resistance of a printed wire board trace or plane without resorting to a lengthy calculation. Although printed circuit board layout and signal integrity computer programs exist that can accurately compute trace resistance, we sometimes want to be able to make a fast rough estimate as part of the design process.

A method that allows us to accomplish this with very little effort is called “counting squares.” Using this method, we can make accurate (within 10% or so) estimates of any trace geometry in just seconds. Once you understand the method, you simply divide the printed wire board area that you want to estimate into squares and then count all the squares to estimate your total trace or plane resistance.

The Basic Concept

The key concept in counting squares is that any size square of printed wire board trace (of a given thickness) has the same resistance as any other size square. The resistance of the square depends only upon the resistivity of the conducting material and the thickness.

We can use this concept on any type of conducting material. Table 1 shows a number of common conductors, along with their bulk resistivity.

Table 1: Bulk Resistivity of Common Metals Used for Electrical Interconnects

For printed circuit boards, the most important material is copper, which is used to fabricate most circuit boards. (Note that aluminum is used to metalize integrated circuit die, and these principles apply there, too.)

Let's start by looking at the square of copper represented by Figure 1.

Figure 1: A Representative Square of Copper

The copper has a length (L), and because it is a square, a width (L). It has a thickness (t), and the current flows through the cross sectional area of copper (A). The resistance of this copper square is simply:

R = ρL / A

where ρL is the resistivity of the copper (an intrinsic property of the material – 0.67µΩ-in. at 25°C). But notice that the cross section A is just the length (L) times the thickness (t). The result is that the L in the numerator is cancelled by the L in the denominator, leaving:

R = ρ / t

Hence, the resistance of the copper is independent of the square's size that we measure. It just depends upon the resistivity of the material and the thickness.

If we know the resistance of any size square of copper and if we can break up the entire trace that we want to estimate into a number of squares, then we can simply add up (count) the number of squares to find the total resistance of the trace!

Implementation

To implement this technique, we need only a table showing the resistance of a square of printed wire board trace as a function of the thickness of our copper. Copper thickness is commonly specified by copper weight (e.g., 1oz., 2oz., etc.). For instance, 1oz. copper weighs 1oz. per square foot.

The table below shows four of the most commonly used copper weights and the resistivity of each at 25°C and 100°C. Note that the copper resistance increases with increasing temperature, owing to the positive temperature coefficient of the material.

Table 2: Copper Resistance versus Copper Weight

We now know, for example, that a square of ½ oz. copper has a resistance of about 1 mΩ. This is regardless of the size of the square. If we can break up the printed wire board trace that we want to measure into imaginary squares, then adding up all the squares in series will give us the resistance of the trace.

A Simple Example

Let's take a somewhat trivial example. Figure 2 shows a rectangular trace of copper, which we will assume to be ½ oz. weight at 25°C. The trace is 1 inch wide and 12 inches long. We can divide this trace into a series of squares, each 1 inch on a side. There would be twelve squares altogether. Since each square of ½ oz. copper is 1 mΩ according to our table, and we have twelve squares in series, the total resistance of the trace is 12 mΩ.

Figure 2: A Rectangular Copper Trace

What about Corners and Connectors?

Before we look at a less trivial example to realize the power of this technique, let's looks at a few refinements.

The first thing to realize is that, in the previous case, we assumed that current in our square flowed in a straight line along the length of the square, from one end to the other, as illustrated in Figure 3a.

Figure 3a: Current Flow Straight Through a Conductor

However, if we have current taking a right angle turn – for example, in a corner square, as shown in Figure 3b – the situation is a bit different.

Figure 3b: Current Flow around a Corner

Here we see that the current has a shorter path to take in the lower left section of the square than it does in the upper right. As a result, the current tends to crowd in the lower resistance, lower left hand section. The resultant current density is higher in this section than what we see in the upper right hand section. The spacing of the arrows illustrates this disparity in current density. As a result, calculated by Jaeger [1], the resistivity of a corner square counts as only 0.56 squares. See Figure 4.

Figure 4: Corner Square Equivalent Value

Similarly, we can make corrections for connectors that are soldered onto a printed circuit board. Here we make the assumption that the resistance of the connector is negligible compared with the resistance of the copper board.

We can see that if a connector occupies a significant portion of the copper square that we are evaluating, then the resistance of that square should be commensurately lower. Some connector pin configurations and their equivalent square counts (per Jaeger [1]) are given in Figures 5, 6, and 7. The shaded regions represent the connector pin in the field of copper.

Figure 5: Connector in Pad – Equivalent to 0.65 Squares

Figure 6: Connector in Pad – Equivalent to 0.35 Squares

Figure 7: Connector in Pad – Equivalent to 0.14 Squares

A More Complex Example

Let's now look a less trivial example to see how we can use this technique. Figure 8a shows a more complex shape that would require some work to calculate its resistance. Our assumption for this example is that we are using 1 oz. copper at 25°C, and current is flowing along the entire length of the trace, from point A to point B. Connectors are placed at each end, A and B.

Figure 8a: An Arbitrary Shape with Connector Pads

Using the same technique discussed above, we can break the complex shape into a series of squares, as in Figure 8b. The squares can be any size that is convenient, and different size squares can be used to fill the entire area of interest. As long as we have a square and the weight of the copper trace is known, we know the resistance.

Figure 8b: Arbitrary Shape Partitioned into Squares

We count up six full squares, two squares containing connectors, and three corner squares. Because 1oz. copper has a resistance of 0.5 mΩ per square, and the current flows linearly through six full squares, the total resistance for these squares is 6 x 0.5 mΩ = 3.0 mΩ.

Then we add the two squares that have connectors attached, which count as 0.14 squares each (see Figure 7). Therefore, the two connectors count as 0.28 squares (2 x 0.14). For our 1oz. copper, this adds 0.14 mΩ (0.28 x 0.5 mΩ = 0.14 mΩ).

Lastly, add the three corner squares. These squares count as 0.56 squares each, contributing a total of 3 x 0.56 x 0.5 mΩ, = 0.84 mΩ.

So the total resistance from A to B is 3.98 mΩ (3.0 mΩ + 0.14 mΩ + 0.84 mΩ).

To summarize, we have:

Six full squares @ 1.00 = 6.00 equivalent squares

Two connector squares @ 0.14 = 0.28 equivalent squares

Three corner squares @ 0.56 = 1.68 equivalent squares

Total equivalent squares = 7.96 equivalent squares

Resistance (A to B) = 7.96 squares @ 0.5 mΩ per square = 3.98 mΩ

The technique can be easily extended to more complex geometries.

Once the resistance of a particular trace is known, it is simple to calculate other quantities of interest, such as voltage drop or power dissipation.

What About Vias?

Often, printed wire traces or planes are not confined to a single layer, but continue on a different layer in the stack-up. Vias are used to connect traces together on different layers. Each via has a finite resistance that must be considered in the overall calculation of the trace resistance.

Generally, vias constitute series resistance elements when they connect two traces (or planes) together. Multiple vias are frequently employed in parallel to reduce their effective resistance.

The calculation of via resistance is based upon the simplified via geometry of Figure 9. Current flows along the length of the via (L), as indicated by the arrow, through a cross sectional area (A). The thickness (t) is based upon the plated thickness of copper inside the walls of the via.

Figure 9: Via Geometry

After some simple algebra, the resistance of the via is given by:

R = ρL / (π (Dt – t 2 ))

where ρ is the resistivity of plated copper (2.36µΩ-in. at 25°C). Note that the resistivity of plated copper is much higher than pure copper. We can assume that t, the thickness of the plating in the via hole, will generally be 1 mil, regardless of the copper weight of the board. For a 10-layer board, built using 3.5-mil cores and 2-oz. copper, L is about 63 mils.

Based on these assumptions, a table of commonly used via sizes, along with their resistances, is provided below. You can easily ratio the numbers up or down for your particular board thickness.

Table 3: Resistance of Common Via Sizes

Alternately, a number of free easy-to-use via calculator programs are available on the internet [2], [3].

Summary

A simple method of estimating the DC resistance of a printed wire board trace or plane was presented. Fairly complex geometries can be broken up into various different size squares of copper to approximate the entire region of interest.

Once the copper weight is determined, the resistance of any size square becomes a known quantity. The estimation process is then reduced to simply counting the squares of copper.

References

[1] Jaeger, Richard C, and Neudeck, Gerold, W. and Pierret, Robert, F., editors. Introduction to Microelectronic Fabrication, Volume V. Addison-Wesley Publishing Co, Inc. 1988.

[2] Saturn PCB Design Inc., Turnkey Electronic Engineering Solutions, http://www.saturnpcb.com/pcb_toolkit.htm

[3] The CircuitCalculator.com Blog, A Blog with Live Web Calculators, http://circuitcalculator.com/wordpress/2006/03/12/pcb-via-calculator/

About the author

Vince Spataro is a Sr. Member of Technical Staff at BAE Systems, Wayne, NJ, a leading global defense, aerospace, and security company. He has over 30 years experience in the analysis, simulation, and design of commercial/military/aerospace power supplies, as well as experience with a wide range of analog and mixed signal circuitry, from precision DC to 1GHz. Vince received a BS degree in Physics, summa cum laude, from Fairleigh Dickinson University and a MS degree in Engineering Physics from Stevens Institute of Technology. He holds two patents in the power conversion technology field.