Component Tolerance

Accuracy Unmasked

We take for granted sometimes that the tolerances of our electronic components fit the needs of our designs. In this article, Robert takes a deep look into the subject of tolerances, using the simple resistor as an example. He goes through the math to help you better understand accuracy and drift along with other factors.

By Robert Lacoste

One of the last projects I worked on with my colleagues was a kind of high-precision current meter. It turned out to be far more difficult than anticipated, even with our combined experience totaling almost 100 years. Maybe this has happened with your projects too: You discover that, even when you’re not looking for top performance out of your electronic components, the accuracy and stability of those components can be pernicious. My topic this month is examining component tolerances. And, for simplicity, I will focus on the simplest possible electronic device: a resistor.

FIGURE 1 A very simple voltage divider. With these values, Uout will be 1 V with Uin=100 V

Let’s start with a basic application. Imagine that you have to design a voltage divider with a ratio of 1/100 (Figure 1). I will assume that the source impedance is very low and that the load connected on the output draws no current at all. With those parameters the calculations are very easy. You just need to know Ohm’s Law. Because the resistors are in series, the current circulating through the two resistors is:

Similarly, the output voltage is:

Given that the current I is the same in both equations, we get:

This circuit is indeed a voltage divider, with a ratio of R2/(R1+R2). We want a ratio of 1/100, so one resistor could be fixed arbitrarily and the second easily calculated. For example: R1=9,900 Ω and R2=100 Ω will do the job as:

Of course, you can easily simulate such a circuit with any SPICE-based circuit simulator if you wish. I personally used Proteus from Labcenter to draw and simulate the small schematic provided on Figure 1, and the output voltage is 1 V with 100 V applied on the input, as expected. As usual, I encourage you to reproduce these small examples with your preferred simulator: for example the free LT-Spice.

Now let’s talk about accuracy. You want your divider to be as precise as possible and therefore you want to buy reasonably accurate resistors. But what if your budget is constrained? Will you use a high accuracy resistor for R1 (9,900 Ω)? Or for R2 (100 Ω)? Or for both? The good answer is both. In that case, a 1% error on either R1 or R2 gives close to a 1% error of the output voltage, as shown in Figure 2. Even if R1 has a stranger value than R2—9,900 Ω vs. 100 Ω—their accuracy is just as critical.

Figure 2
A 1% error either on the top or bottom resistors will induce a roughly 1% error on the output. That would not be the case for other division ratios.

Maybe you think this is too obvious? In that case I will give you another exercise: What happens with a divide-by-2 circuit using two resistors of the same value? Do the calculation or simulate it and you will find that both resistors have still the same impact on accuracy. But now a 1% error on one of the resistors has only a 0.5% impact on the output voltage. That means you could buy slightly less expensive resistors for the same overall precision! In fact, the higher the division ratio, the higher is the impact of each resistor on the overall accuracy.

E Series Resistors

Let’s go back to the 1/100 divider example. If you want to build it and look for a
9,900-Ω resistor, you will have some difficulties because nobody sells them.. …

Read the full article in the April 333 issue of Circuit Cellar

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