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

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.