I wanted to improve my understanding of active filters, beyond the very simple ones that I included in my textbook, so I decided to try implementing a bandpass filter using the multiple-feedback topology. To make the design more concrete, I decided to make one that was centered at 2kHz, but firmly rejected 1.5kHz and 2.5kHz.

Here is the design I came up with. The top part is the filter, while the bottom is just a reference voltage buffer. I’ll derive the values below.

We can simplify a bit, by replacing with , where . Note that this corresponds to replacing the amplifier, R2, and C2 with an inverting high-pass filter.

It is conventional to recenter the voltages so that is zero, which simplifies the algebra. Doing that gives us

Multiplying both sides by and rearranging gives us

More correctly, that should be , but I’ll just talk about the gain .

Defining we get

If we have , which is a common design constraint, we can simplify to

We can define and . Then we have

If we let , we get

,

whose magnitude is maximized when or , and at that maximum the gain is .

The corner frequencies, where the gain drops by the square-root of 2 occur when the real and imaginary parts of the denominator have the same magnitude, or . If we take only the positive values of , we get , for a bandwidth of , or Hz.

For C=10nF and R2=220kΩ, we get a bandwidth of 144.7Hz. With R1=6.8kΩ and R3=300Ω, Rp=287.32Ω, with time constant 2.8732 µs, giving a center frequency of 2001.8kHz. I built this filter and tested it with the Analog Discovery 2:

The overall match of the filter to the theory is pretty good.

But when we look in detail at the peak, we can see that the center frequency is off a little and the gain isn’t nearly as big as it should be. I suspect that the problem is poor matching of the very cheap 10nF capacitors.

I also tried exciting the filter with wideband noise from the Analog Discovery 2, and doing an FFT of the input and output to get a different view of the response:

The filter response looks a little larger taking the difference of FFTs (after averaging many FFTs, since each individual one has a lot of noise). Note the 60 Hz spike.

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