I'm trying to create a passive RLC band-stop filter for my Sony EX71SL earphones, which has a nasty little spike in one treble frequency area.

I found a similar such filter on the likwitz website: http://www.linkwitzlab.com/reference_earphones.htm

But my output impedance is much lower and my frequency center point higher. I'm having trouble calculating the values I need for the filter to work with my devices/amps. I've played with online calculators, such as http://sim.okawa-denshi.jp/en/RLCbekeisan.htm but I still can't seem to generate a similar notch depth AND Q value at the same time.

Here are my devices and their output impedances:

Apogee Duet: 32ohms - iPhone 6: 3.18 ohms - C5 amp: 2.2 ohms - Sansa Fuze: ~1 ohm

Center frequency: 6,000hz

Cut: 7.8db

I was thinking I can use the 32 ohm for my calculations and simply add a 30 ohm adapter to my other two devices, thus essentially giving me roughly 32 ohms for each device when used.

He lists a Q value of 17, but I'm using EQ apps with lower values. I'm using a 7.2 Q value or 0.2 BW (bandwidth) if you want to convert it to that.

These result in a perfect reduction of the 6,000hz treble peak as found with a sine sweep.

Using the values on his site result in a different amount of reduction and whatnot if I plug them into the online calculator.

Can someone tell me what values I would need to use to achieve a 7.8 or rounded to 8.0db reduction at 6,000hz with a 7.2 Q if my output impedance is 32 ohms, and would it matter if the output is off by 1 ohm on each device? How much would that effect the shape?

I'm starting to understand some of the equations, but I can't seem to isolate the exact relationships and what I need to do to arrive at these values even just putting in values on the online calculator and trying to tweak them.

Any help would be greatly appreciated.

Amplification shouldn't be an issue either. The C5 seems to easily drive these earphones well past iphone levels and much louder than I'd ever use.

I'd love to understand this all, so any detailed responses would be great.


1 Answer 1


All this work you have done is very good but the load impedance of your headphones needs to be modeled - it will significantly alter the Q of your circuit if you do not take this into account. The Okawa site is brilliant for most things but the model you have chosen does not take the headphone load into account.

There is another type of band reject filter worth trying - this consists of a parallel L and C placed in series between audio output and headphones. At resonance this circuit produces total signal rejection so, a resistor needs to be placed in parallel with the L and C to allow a certain amount of the 6kHz through but the amount depends on the headphone impedance at that frequency.

It can be assumed that the headphone impedance is 32 ohms of course and this may give decent results: -

enter image description here

The above picture was taken from a simulation in microcap. C is 6.8uF, L is 103uH and R is 56 ohms.

This will work better because it takes into account the impedance of the headphones but, as I mentioned earlier the actual headphone impedance may be different at 6kHz and if you don't get the results you expect this may need investigating.

Regards output impedance of the amplifier and how it might affect things, here's a plot around the main area of interest as output impedance increases from 0R to 1, 2, 5, 10 and 20 ohms: -

enter image description here

If you feel that Q needs to be bigger then make L smaller and C bigger but keep the product of L and C identical i.e doubling C means reducing L by 2.

The best way to do this would be to make a headphone buffer amp and incorporate filters into that.

  • \$\begingroup\$ the earphone impedance is 15 ohms... gotta run, will read your full post in a bit. thanks!!! \$\endgroup\$
    – luisdent
    Commented Jan 20, 2015 at 14:03
  • \$\begingroup\$ So I've read through your post, and your response graph looks perfect. How does the earphones being 15ohms (rated) effect the response? Also, regarding your graph on the differences of the output impedances... it doesn't seem to change much other than the overall plot amplitude being higher or lower, or am I mis-reading the graph? Lastly, can we discuss how the components relate to the response/output? Either here or is some private manner? \$\endgroup\$
    – luisdent
    Commented Jan 20, 2015 at 18:15
  • \$\begingroup\$ ...I'd like to be able to formulate different scenarios for other earphones, but don't know the formulas/relationships for this design. \$\endgroup\$
    – luisdent
    Commented Jan 20, 2015 at 18:32
  • \$\begingroup\$ @luisdent sorry dude I don't have the time to help you much further but may I suggest you get hold of LTSpice (a free simulator from Linear technology) and do exactly what I have done. Sure there's a learning curve but if you are interested in electronics, the sooner you start down this path the better. Hey you may not thank me for saying that but I'm 100% right!! With different source impedances the depth of the cut gets smaller btw. \$\endgroup\$
    – Andy aka
    Commented Jan 20, 2015 at 21:31
  • \$\begingroup\$ There are formulas that can be used to simulate this in mathlab or in excel but they are a little bit long winded to prove - maybe ask a new question about how to solve the equation for this? \$\endgroup\$
    – Andy aka
    Commented Jan 20, 2015 at 21:33

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