The cut-off frequency of a RC filter is obtained from the well known equation, $$1\over(2 \pi RC)$$

This is one equation with two variables. For example, R=100, C=10 has the same result as R=10, C=100.

Based on what should I prefer one over the other?

  • 4
    \$\begingroup\$ Aug, you've asked several questions along this line and, in each case, you've received a "hint" to be more specific. Is it too much to ask for you to include in your question the following: best according to which additional constraint?. You have two degrees of freedom and one specification - the time constant RC. If you're interested in what additional constraints exist then ask that question instead. \$\endgroup\$ Oct 10, 2013 at 22:57

2 Answers 2


It's a compromise.

With R at 1000 ohm and C at 100nF (cut-off frequency = 1.59kHz), the driving voltage at the input may be required to drive signals with frequencies well above 1.59kHz into what is getting close to a 1000 ohm load. Consider what the impedances are at 1.59kHz - R of course is 1000 ohm and C's impedance also has a magnitude of 1000 ohms whereas, at 10 kHz, C's impedance has a magnitude of only 100 ohms.

In other words, at 10kHz, the signal feeding into the RC low pass filter "sees" an impedance of about 1000 ohms. This is due to the following formula: -

Z = \$\sqrt{R^2 + X_C^2}\$ = \$\sqrt{1,000,000 + 10,000} = 1005\space \Omega\$

If the signal feeding the RC network has an output resistance of 100 ohms then this adds an error to the "R" part of the equation and distorts the "true" spectral shape of the filter.

On the other hand....

The benefit in having a low R and a high C means that the output impedance is affected less by the circuit its output connects to. In the example above, even at DC the output impedance of the network is 1000 ohms. If R was (say) 10k ohms and C was 10nF, the output impedance at DC is 10k ohms and may be affected by some loads.

So, you have to consider what your driving impedance is and what your RC network may have to "drive" into. There are many examples where the output will connect to an op-amp which will usually have a DC input resistance in the Gohm range but, it may have an input capacitance of 10pF. This input capacitance offsets the output capacitance by a small amount and, in the example above, would make the 100nF capacitor into 100.01nF - hardly a big deal of course but if you are designing a filter that has a cut-off at 50kHz, it's starting to become a potential source of error.

Cascading RC low pass filters (or any filter types) is also a serious issue. Say you want to passively connect two RC low pass filters - if you picked both resistors to be 1000 ohm and both capacitors to be 100nF you are not going to get the same filter response should you have connected them via a high impedance buffer amp.

A partial solution is to make the first network low impedance and the 2nd network high impedance. To give you an idea make the first RC network from a 1,000 ohm and 100nF and the connecting network from 10,000 ohms and 10nF - there will still be a little bit of interaction but it is far less than when both are the same impedance.


Two degrees of freedom and one specification means that you can only fix the product \$RC\$.

As has been pointed out in answers and comments to similar questions, the answer is: it depends - it depends on some other constraint. Seriously, a question like this is almost unanswerable without additional context.

Here are some considerations that might go into a another constraint.

  1. since resistor and capacitor values are not continuous, one must find a combination of standard values that give a time constant that is "close enough" to the desired one.

  2. common capacitor values are far coarser than common resistor values.

  3. in general, it's far cheaper to find a large value of R than a large value of C

  4. capacitors with relatively large values of capacitance are often far from ideal at high frequencies

  5. capacitors with very stable capacitance over time, temperature, etc. may constrain the range of capacitance available


There's plenty more and the list above is not, in any case, meant to be exhaustive but, instead, meant to give you some idea of the context in which such a evaluation as best must be made.


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