I have been trying to create a bandpass filter with the cut-off frequency at \$\mathrm{200Hz}\$ and \$\mathrm{4800Hz}\$ I have managed to get the centre frequency to \$\mathrm{2500Hz}\$. The bandwidth on the to be non-existent and I have somehow created a really bad high-pass filter.

Below is the circuit I have created: enter image description here

I am using a \$\mathrm{40nF}\$ capacitor, \$\mathrm{0.1H}\$ and a \$\mathrm{1718}\Omega\$ resistor in parallel which is all in series with another \$\mathrm{1718\Omega}\$ resistor.

This then creates the AC sweep: enter image description here

In order to get the values I have I did the following:

I knew that resonant frequency is $$ f_o= \frac{1}{2\pi\sqrt{LC}} $$

I assigned a value to \$L\$ of \$\mathrm{0.1H}\$ and rearranged to get the capacitance

Then for bandwidth: $$ B = R\sqrt{\frac{C}{L}} $$ I knew the values of \$C\$, \$L\$ and the bandwidth (\$\mathrm{2600Hz}\$)

*I realise now that the values of the resistors are incorrect but I have also tried \$\mathrm{1450}\Omega\$ resistors and the same issue

Can anyone please explain to me why this has happened and how to fix it?

  • 1
    \$\begingroup\$ What method and calculations have you use to get to the current schematic? Note that you can not cascade "individual" calculated filters. \$\endgroup\$
    – Oldfart
    Commented Feb 13, 2019 at 14:42
  • \$\begingroup\$ Try the bandpass filter designs and plotting tools here. sim.okawa-denshi.jp/en/Fkeisan.htm \$\endgroup\$
    – CrossRoads
    Commented Feb 13, 2019 at 14:55
  • \$\begingroup\$ Given the Q is much less than one, this is not a standard LC bandpass design. \$\endgroup\$ Commented Feb 13, 2019 at 17:07

1 Answer 1


I have managed to get the centre frequency to 2500 Hz

If you want equal amplitude cut-off frequencies of 200 Hz and 4800 Hz, the centre frequency you need is 980 Hz. This is calculated as \$\sqrt{200\times 4800}\$ = 979.8 Hz.

That is the centre frequency you need to aim for.

Also, when you are so asymmetrical with your 3 dB frequencies (relative to Fc) the bandwidth formula you used becomes inappropriate because it relies on both 3 dB points being close to each other. You would probably fair better with a double RC filter given the gulf between 200 Hz and 4800 Hz.

  • \$\begingroup\$ I see that is a very good point with respect to the large difference between the cut-off frequencies. Please, could you provide a link to how you determined the centre frequency? \$\endgroup\$ Commented Feb 13, 2019 at 14:56
  • \$\begingroup\$ @Andy aka...I think, the relations between Q, B and Fo for such a filter are always valid - independent on the width of the passband. Hence, it does not matter if the 3dB frequencies are close to each other - or not. Such a bandpass is always "unsymmetric" with respect to the center frequency., \$\endgroup\$
    – LvW
    Commented Feb 13, 2019 at 15:36
  • \$\begingroup\$ @sam lemonts...The square root formula mentioned by Andy aka applies to ALL second-order bandpass functions. The center frequency is always the geometric mean of the two edge frequencies. \$\endgroup\$
    – LvW
    Commented Feb 13, 2019 at 15:41
  • \$\begingroup\$ This link shows you how the formula can be derived (see my detailed answer): electronics.stackexchange.com/questions/234752/… \$\endgroup\$
    – LvW
    Commented Feb 13, 2019 at 16:02

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