# Band Pass Filter Equivalent Components

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):

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:

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: 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: -3dB from that peak is not at the correct cutoff frequency:

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.

• 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. – a concerned citizen May 18 '18 at 15:08