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 Feb 13 '19 at 14:42
  • \$\begingroup\$ Try the bandpass filter designs and plotting tools here. sim.okawa-denshi.jp/en/Fkeisan.htm \$\endgroup\$ – CrossRoads Feb 13 '19 at 14:55
  • \$\begingroup\$ Given the Q is much less than one, this is not a standard LC bandpass design. \$\endgroup\$ – analogsystemsrf Feb 13 '19 at 17:07

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.

| improve this answer | |
  • \$\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\$ – Sam lemonts Feb 13 '19 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 Feb 13 '19 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 Feb 13 '19 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 Feb 13 '19 at 16:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.