I've been trying to design a band-stop filter for my headphones to cut back on excessive highs and to provide a more flat frequency response.

With a software EQ and lots of tries, I managed to arrive at a reasonable one: \$f_0 = 4800Hz\$, \$gain = -10dB\$, \$1/3 < Q < 0.4\$ (or in octave bandwidth, \$3.447 > BW > 3\$). See the raw frequency response and the filtered one.

I would like to turn this into a parallel RLC filter in series with the headphones. Unfortunately I have no formal education in electronics, so I have to rely mainly on various websites and tutorials.

However all of them describe RLC circuits without a load attached. Adding a load resistor completely changes the gain and the \$Q\$ factor, and I am unaware of the exact relationship. Is it related to the voltage divider rule or something similar?

I managed to get an appropriate \$R\$ by trial and error, from \$Rload\$ and gain. I also managed to calculate \$LC\$ from the relationship \$w_0 = \frac{1}{\sqrt{LC}}\$. However, the equation \$Q = R \sqrt{\frac{C}{L}}\$ that is appropriate for a parallel RLC without a load, is plain wrong in this case.

My question would be is to how to determine \$L\$ and \$C\$ from these parameters:

f0 = 4800Hz
gain = -10dB
Q between 1/3 and 0.4
impedance of headphones: 32 ohm nominally, can get around 35 ohm at 4800Hz
impedance of amplifier: 0.09 ohm

The circuit so far (with imprecise \$L\$ and \$C\$ values):

CircuitLab Schematic u8shfw

  • \$\begingroup\$ You say you are trying to reduce "excessive highs" and, traditonally for this, you need to use a low pass filter, not a BPF - can you explain your reasoning? Can you also write down, one term at a time what you reckon you needed - it's the 1/3<Q<0.4 or 3.447>BW>3 that is bugging me - are you saying Q can be x OR BW can be y as an alternative? \$\endgroup\$
    – Andy aka
    Apr 14, 2013 at 14:32
  • \$\begingroup\$ Well, the excessive highs are centered around 4800Hz with peaks at 4800Hz, ~7300Hz and somewhere around ~11300Hz. Band-stop filter seems more appropriate than a simple low pass filter. See raw and filtered \$\endgroup\$
    – Frigo
    Apr 14, 2013 at 14:34
  • \$\begingroup\$ @Andyaka: He said band-stop, not band-pass. \$\endgroup\$
    – Dave Tweed
    Apr 14, 2013 at 14:35
  • \$\begingroup\$ BW is in octave bandwidth \$\endgroup\$
    – Frigo
    Apr 14, 2013 at 14:36
  • \$\begingroup\$ I think you need a better spec of what you are trying to achieve - are you saying ~7300Hz because there are some signals in this area that are bothering you? Ditto ~11300Hz \$\endgroup\$
    – Andy aka
    Apr 14, 2013 at 14:37

2 Answers 2


In general, designing passive filters to work at the low, highly frequency dependent (i.e., reactive) and poorly-controlled impedances associated with things like headphones is not the best approach.

It would be better to attenuate the headphone signal as needed to run it into an active (opamp-based) filter with the desired characteristics, and then boost it back up to headphone levels at the output. The results will be much more stable and easier to fine-tune for specific applications.

In fact, you can purchase such units — both parametric and graphic headphone equalizers — off-the-shelf.

  • \$\begingroup\$ What side effects can it have? I have seen a hybrid filter addressing the 7300Hz peak, containing both a series RLC in shunt and a parallel RLC in series with the load. Never understood the need for both, perhaps is it built to compensate for unstable impedance? \$\endgroup\$
    – Frigo
    Apr 14, 2013 at 14:52
  • \$\begingroup\$ Also here is the impedance graph \$\endgroup\$
    – Frigo
    Apr 14, 2013 at 15:01
  • \$\begingroup\$ Can it have other side effects apart from unstable gain and Q? Because I can live with that. And I'm not exactly familiar with opamps, so it would add additional difficulty for me. \$\endgroup\$
    – Frigo
    Apr 15, 2013 at 8:22
  • \$\begingroup\$ The impedance graph is rather useless with the vertical scale compressed like that. It hides the variations with respect to frequency, which are clues to the reactive components of the load. These components can create unwanted resonances -- new peaks and dips in the frequency response that are different from what you intended. \$\endgroup\$
    – Dave Tweed
    Apr 15, 2013 at 11:13

Assuming you have noted Dave Tweed's comments about getting a circuit to work with headphones but have decided to go-ahead with a directly connected RLC, I would choose LC like this: -

The "R" factor is in series with the inductor - that's where your confusion arose possibly

If you know what Q you want then convert this to Damping ratio, ζ: -

\$ \zeta = \dfrac{1}{2Q}\$

Then equating: -

\$\zeta = \dfrac{R}{2}\sqrt{\dfrac{C}{L}}\$

Using your values of 6.8mH and 150nF, with a Q of 0.4 gives an R value of 530 ohms. This is in series with the inductor. Make sure your inductor has a self-resonant frequency well-above 20kHz - aim for 100kHz.

  • \$\begingroup\$ @jippie - thanks for the cool edits - I which I could do that - is there some crib-sheet so that I can easily remember. Maybe I need to go on a training course? \$\endgroup\$
    – Andy aka
    Apr 14, 2013 at 16:52
  • \$\begingroup\$ meta.electronics.stackexchange.com/questions/434/… \$\endgroup\$
    – jippie
    Apr 14, 2013 at 17:46
  • \$\begingroup\$ "The R factor is in series with the inductor" - Is it otherwise unchanged, like this? It seems to be a highpass filter instead of a band-stop one, unless I screwed up something. \$\endgroup\$
    – Frigo
    Apr 14, 2013 at 21:38
  • \$\begingroup\$ @Frigo it's a band-stop filter definitely but maybe you estimated Q incorrectly. I used Q=0.4 (as per your instructions) which, to me seems intuitively a little low. \$\endgroup\$
    – Andy aka
    Apr 14, 2013 at 21:50
  • 1
    \$\begingroup\$ @Frigo The most common determination of Q for a parallel tuned circuit uses the formula with R in series with the inductor because inductors are made from wire and they have losses that can be lumped together as series elements. Capacitors, on the other hand, have losses that would be naturally seen as a parallel resistor but these losses can be ignored compared to the major loss contributor, the inductor. \$\endgroup\$
    – Andy aka
    Apr 17, 2013 at 19:26

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