Low Pass Filter

After some deliberation, I have decided to not use an RLC circuit, and to go with an active, two pole, low pass filter.

I think it would be simplest to use Nodal Analysis, with two unknowns, and two equations. The use of conductances can make nodal analysis much easier, so I think I will use that. The conductance is simply the reciprocal of the resistance.

The first nodal equation:

The second nodal equation:

Solving the second nodal equation for V1:

Substituting this into the first nodal equation:

 

Simplifying

Solving for V2, which is the output:

Combining like terms:

Where omega is the cut off frequency, and Q is the gain of the amplifier at omega. For the best possible filter performance, while remaining maximally flat, Q should be between 0.707 and 1. I do not want the gain to roll off too quickly, nor do I want too high of a peak at the cut off frequency.

R1 and R2 should be on the same order as the output resistance of the microphone, C1 could equal C2. If that is the case, the equations simplify.

Unfortunately, this configuration can only produce a Q of 0.5, and only with equal resistances. To simplify the number of components needed, both resistances will be equal.

C1 must be between two and four times C2 to achieve a high enough Q. The resistance is 1000 Ohms.

With C1 two time C2, and R equal to 1000 Ohms

C2 is 100 nF, C1 is 200 nF.

Looking at the simulation, in log log scale, I see that it does not roll off at a low enough frequency. I choose C2 to be 200 nF, and C1 to be 400 nF. Now the drop off starts at 200 Hz, instead of at around 700 Hz.

In conclusion, this is the operation of the active, two pole, low pass filter. Unlike the RLC, higher resistance smaller capacitances can achieve smaller cut off frequencies. Care must be taken to make sure operational amplifier non-linearities are properly accounted for, but this is a valid way of implementing a low pass amplifier.

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Electrical Engineer, Minneapolis, Minnesota