1
\$\begingroup\$

I am designing a very simple first-order RC band pass filter with cutoff frequencies \$f_l = 0.5 MHz\$ and \$f_h = 4.5MHz\$. I have it pictured below (I'm using 5Spice):enter image description here

This filter acts the way I want it to. The half power points are around the appropriate cutoff frequencies. For the HPF side of the filter, \$f_h = 1/(2*\pi*100nF*3\Omega)= 0.53MHz\$ and on the LPF side, \$f_l = 1/(2*\pi*1nF*35\Omega)= 4.55MHz \$. Below is the frequency domain plot: enter image description here

Now, when I change the capacitor and resistance values to something that should yield the same cutoff frequencies, the frequency domain plot looks very different. Here's the modified circuit: enter image description here And the cutoff frequencies should be \$f_h = 1/(2*\pi*31.8pF*10K\Omega)= 0.50MHz\$ and on the LPF side, \$f_l = 1/(2*\pi*3.54pF*10K\Omega)= 4.50MHz \$. However, not only does the peak of the signal exist at less than -20 dB: enter image description here -3dB from that peak is not at the correct cutoff frequency: enter image description here

How can these two configurations yield such different results? They both calculate the correct cutoff values, so why are they different in simulation?

Also, in researching this subject I realize that this is a very poor filter and that I should be using a passive RLC band-pass filter or an active one. I'm new to 5Spice, so please suggest any edits to these pictures and I will make them to make this question more readable.

\$\endgroup\$
  • 1
    \$\begingroup\$ The output impedance of the first filter affects the input impedance of the second. For example, make the 1st with 470 Ohms and 680p, then the 2nd with 4.7k and 7.5p. But, yes, RLC is the way to go for this. \$\endgroup\$ – a concerned citizen May 18 '18 at 15:08
1
\$\begingroup\$

In your first circuit, the high-pass filter (analysed on its own) has an output impedance not greater than 3 ohm (R1 dictates this). Your low-pass stage has an input impedance not less than 35 ohm (R2 dictates this) and, when looked-at together you can see that the low-pass filter stage is applying very little loading to the high-pass stage and you roughly get what you expected.

In your 2nd circuit with R1 = R2 = 10 kohm the 2nd stage is much more heavily loading the first stage and this screws up the nice response as seen in the first circuit.

If you want to use a simple RLC pass band filter you need to describe what bandwidth you need - it might be that what you want in terms of bandwidth might not be realizable from a simple RLC circuit and you may need to cascade two - then you are back to loading problems.

The best choice is made when you decide what you want in terms of bandwidth and out-of-band rejection characteristics.

\$\endgroup\$

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.