# precision of components [closed]

Especially passives like resistors and capacitors.

It looks like the E12 series is still the most used for resistor values, despite that it only covers the full range for 10% tolerance. It looks like 10% is hardly used anymore.

What precision for resistors do you normally use, and why? Do you always use the same tolerance, or for which applications would it be different? Same for ceramic capacitors (I realise that electrolytics are a completely different story).

• Not a good question, IMO. Anyone has his requirements, and will choose the tolerance accordingly. Sometimes is easier to buy a batch of 5% or 1% resistors instead of going for the cheapest ones. – clabacchio May 30 '12 at 8:45
• @clabacchio: I'm trying to learn from you guys. If you use a 1% resistor, why is that? Does the application need it. Is it cheaper? Better available? – Federico Russo May 30 '12 at 8:48
• Like everything, if you need to be accurate you need accurate things, which will likely be more expensive. This keeping power apart. If you have a problem to solve, you can get advice, but this is like asking: "How fast my car should be?" – clabacchio May 30 '12 at 8:50
• @clabacchio - It will depend on the application, but I think the question has some value. For instance, do you use 5% or 1% resistors for the setting of an LM317? Will that depend on application? Not always. Is it possible that you use 5% out of habit, when you should know that it even worsens the already 5% tolerance on the reference voltage?tolerance – stevenvh May 30 '12 at 8:54
• @stevenvh but to me the questions lacks of scope, it's too broad and subjective (in the sense that everyone has different applications, and different needs). A better scoped one might be better. – clabacchio May 30 '12 at 8:56

I will use mostly 1%. They're not much more expensive than 5%, and you have theE96 range to choose from. Otherwise you calculate a resistor at 20k$\Omega$, and have to do the calculation again because only 18k$\Omega$ and 22k$\Omega$ is available.

If you take the equation for the output voltage of the LM317:

$V_O = 1.25V \cdot \left(1 + \dfrac{R2}{R1}\right) + R2 \cdot I_{ADJ}$

The 1.25V is the reference, but that can vary from 1.2V to 1.3V. ${I_{ADJ}}$ is typically 50$\mu$A, but can be maximum 100$\mu$A.
Many will just calculate R2/R1 = 3 to get a 5V output. But if you use 5% resistors you won't get a 3:1 ratio in the E12 series. So you pick 680$\Omega$ and 220$\Omega$. That will give you nominal 5.15V, that's a 3% error, that's OK. But worst case you'll get

$V_O = 1.3V \cdot \left(1 + \dfrac{680\Omega \cdot 1.05}{220\Omega \cdot 0.95}\right) + (680\Omega \cdot 1.05) \cdot 100\mu A = 5.74V$

That's a 15% error. I didn't play any dirty tricks, just used standard tolerances.

With E96 values you can pick 715$\Omega$ and 237$\Omega$ 1% resistors, and then you get 5.06V nominal, a 1% error. Worst case it will be 5.37V, a 7% error.

Use 1% resistors. The only drawback is that you may have to store more different values, though you still can use a 10k$\Omega$, only it will be 1% instead of 5% or 10%.

• Well, in case you want to accuse me of using dirty tricks by using E12 for the 5% resistors: you show me an engineer who uses E24 and I'll buy you a beer (Belgian!). – stevenvh May 30 '12 at 9:41
• Well, if you use E12 resistors you will store more values buying less stocks, just because they will be spread :) then you just have to measure them one by one and hope for luck ;) – clabacchio May 30 '12 at 9:47

I mainly use 0805 5% resistors for my generic needs. In case I am dealing with current sense application (various drivers), I use 1% for sensing resistor. I never really need higher precision as most of my designs are digital electronics. For few drivers that I make, I think 1% is enough and resistance will be influenced by temperature anyway and driver circuits tend to get hot...