Electrical – Second Order Bandpass Filter vs Second Order High Pass and Low Pass in Series

This is an example of a Sallen-Key op amp. Most second order amplifier designs require 2 op amps. The Sallen-Key's clever design accomplishes this using only 1 op amp.

A mathematical discussion of the low pass Sallen-Key op amp can be found here.

For a simpler "thought" explanation of how the 2 capacitors work listen to this video starting at about 1 minute 15 seconds.

If you can not view the video, the video essentially states that C2 (in your diagram) appears as a short for all high frequencies to ground. Because of this, the output of the op amp will be very close to ground at high frequencies. Consequently, C1 (in your diagram) also shorts all high frequencies to what appears very close to ground at the output of the op amp.

Also, Sallen-Key amplifiers are usually used in unity gain mode which would not use RA nor RB (in your diagram). Instead, a unity gain amplifier would simply connect the output of the op amp to the negative input. Adding RA and RB allows for gain adjustment. But also opens the door to other types of op amps designs as the 1 less component count of the Sallen-Key design is no longer an advantage after adding the 2 extra resistors.

The roll off (what you are calling attenuation) calculations of a first order filter can be found here. Basically it is 20dB/decade. A second order filter is double that or 40dB/decade. (Note, the linked article is using 6dB/octave which is about 20dB/decade.)

Wc1 = 1 / (2piRC)

Wc2 = 1 / (2piRC)

You haven't designed a bandpass filter; you've designed a low-pass filter and a high-pass filter and connected one to the output of the other.

The equation for the low-pass only gives the cut-off frequency when the load on the output has a very high impedance. Your 2nd stage (the high-pass part) is loading the first stage (low-pass part) and changing its cut-off frequency.

Similarly the high-pass equation assumes the source feeding it has very low impedance, which is again not valid once you connect the low-pass filter as the input source.

I expect there to be some attenuation, but my "passed" frequencies for some of the filters are at best -6dB.

The best case scenario at the midpoint of the passband is that C1 looks like an open circuit while C2 looks like a short circuit. In that case you'd have an equivalent circuit given by the resistor divider formed by C1 and C2. For your "bass" filter, for example, this would be about -3 dB because the two resistor values are roughly equal.

But things won't even be that good, because you're using equal capacitor values for the two filter stages, so the capacitors will have equal impedance at any given frequency. I won't try to work out the effect mathematically since you've already run the simulation and seen what it is.