# Parallel RLC bandpass filter Bandwidth incorrect

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:

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:

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?

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

• 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? Commented Feb 13, 2019 at 14:56
• @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.,
– LvW
Commented Feb 13, 2019 at 15:36
• @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.
– LvW
Commented Feb 13, 2019 at 15:41
• This link shows you how the formula can be derived (see my detailed answer): electronics.stackexchange.com/questions/234752/…
– LvW
Commented Feb 13, 2019 at 16:02