I am working on designing a precision analog filter for a university electronics project.

The resistance values I am using (to meet the filter specifications) are not E96 values, so I am combining smaller value resistors in series to make the non-standard resistor. Is there any guidance for choosing these resistor values?

For example, to create an 11.9 kΩ resistor, I could use a 100 Ω resistor and an 11.8 kΩ resistor; or I could use a 7.15 kΩ resistor in series with a 4.75 kΩ resistor.

Intuitively I feel having the resistors a similar size may be better, but I am not sure if that is correct, or why it may be the case.

Also, would this change significantly based on the precision of my resistors? For example if I moved from 1% tolerance resistors to 0.5% or 0.1% values?

  • \$\begingroup\$ qsl.net/in3otd/parallr.html \$\endgroup\$ – Jeroen3 May 19 '20 at 13:05
  • \$\begingroup\$ I was actually using that site @Jeroen3 However, using the example of 11.9, the site provides 6 combinations which could make up a 11.9 value. Is there any way of choosing which combination is best? \$\endgroup\$ – JAS May 19 '20 at 13:09
  • \$\begingroup\$ While this may be of some theoretical interest, it has no practical value. The tolerance of your resistor combination is going to be no better than that of the individual resistors, so you might as well just pick the closest standard value to begin with. \$\endgroup\$ – Dave Tweed May 19 '20 at 13:11
  • \$\begingroup\$ Besides which, if you're designing filters, can you get 1% (or 0.5% or 0.1%) capacitors? If you need filters that are THAT precise, you really should be doing DSP. \$\endgroup\$ – Dave Tweed May 19 '20 at 13:17
  • \$\begingroup\$ @DaveTweed I was planning to use 1% capacitors, but appreciate that may not be feasible. Appreciate the comment \$\endgroup\$ – JAS May 19 '20 at 13:33

If you are making these combinations based only on the marked (nominal) value of the resistors then you may not going like the results.

For example, an 11.8k\$\Omega\$ resistor with a 1% tolerance can actually have a resistance that is +/- 118\$\Omega\$ different from the marked value. Adding a 100\$\Omega\$, 1% resistor in series will give you a nominal resistor of 11.9k\$\Omega\$ +/-119\$\Omega\$. So, the actual resistance value could be as low as 11.781k\$\Omega\$.

If you really need 11.900k\$\Omega\$ you will need to buy resistors with a much smaller tolerance or measure each one. Oh, and be sure to calculate the accuracy of your ohmmeter.

  • \$\begingroup\$ So essentially regardless of whether its 11.8kΩ + 1 Ω or, 7.15KΩ + 4.75Ω, there will be no difference, it will still be 11.9kΩ +/-119Ω? I.e., the only way to improve the accuracy of the circuit would be to use lower tolerance resistors? \$\endgroup\$ – JAS May 19 '20 at 13:17
  • \$\begingroup\$ Yes, or measure the actual resistor values instead of relying on the nominal values. \$\endgroup\$ – Elliot Alderson May 19 '20 at 13:21
  • \$\begingroup\$ Thanks for clarifying \$\endgroup\$ – JAS May 19 '20 at 13:29
  • \$\begingroup\$ while the potential values will be wider due to the tolerance stackup, there is increase probability around the nominal as the peak of the normal curve is flattened. NOTE: i am not advocating "on average" as I would always do the tolerance stackup, just adding a note about the impact of tolerance stacking \$\endgroup\$ – JonRB May 19 '20 at 13:30
  • \$\begingroup\$ @JonRB You are making an assumption that is not supported by data. You assume that the distribution of values for both resistors has a mean equal to its nominal (marked) value and that there is a normal distribution of values. \$\endgroup\$ – Elliot Alderson May 19 '20 at 14:00

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