How do I calculate the cutoff frequency with diff values...?

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What if my low-pass filter circuit has different R (ohms) and C (capacitance) values, and I'm using a 3rd order low pass filter? Do I add every resistor (R) and capacitance value (C)? Then I'd apply the formula?

1/[2*pi*(R1+R2+R3) * (c1+c2+c3) ]

Would that be right? ...

If I had another scenario with same values for R and C, do I need to add them as well?

This is the schematic I'm using: http://img810.imageshack.us/img8...

What's the cutoff frequency?

Thanks in advance.

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Looks like there should be an input resistor on the left side that got cutoff somehow. I assume you want a 3rd order Butterworth filter. The 1st section is just f=1/2piRC. The 2nd section is also the same freq calc IF the gain is correct. I can't remember what it should be because of atrophy of 59 year old synapses, but its something like 1.414. Search for "equal RC voltage controlled voltage source filters" and I bet you find the chapter in the active filter book by Huelsman from around 1970. Make sure the source impedance driving the input is << (way less) than the input R, or that freq will be off.

Imagecraft compiler user

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Without R10 this would be a classic Sallen Key LPF, adding R10 changes this. I agree with Bob, this circuit is missing an input resistor, and the impedance of the input RC, without its own op amp buffer, will impact the remainder of the filter.

Wiki Sallen Key Filters http://en.wikipedia.org/wiki/Sallen–Key_topology might be a reasonable starting point, but there are many filter design web sites out there.

If you are not comfortable calculating the filter's transfer function then start with a known, defined, filter design, such as the above.

Be careful, also, about filters designed for bi-polar power supplies, (e.g. +/- 12 V), vs those designed for uni-polar supplies, (e.g. 0/5V). The "Mid-Point", and "Gnd", are very fifferent between the two.

JC

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DocJC wrote:

Be careful, also, about filters designed for bi-polar power supplies, (e.g. +/- 12 V), vs those designed for uni-polar supplies, (e.g. 0/5V). The "Mid-Point", and "Gnd", are very fifferent between the two.

JC

Are you basically saying that the design of a filter will change if I use DC volage? How?

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Everything is possible. Do you want to use +-supplies, or just a positive supply? With bipolar supply, its a lo pass filter. It will go right down to DC. No DC blocking coupling caps needed. With a single supply, you need to bias the + inputs of the amps (that are grounded in the bipolar design) to vcc/2, and probably need coupling caps to keep the DC in the filter from banging into whats in front of and behind the filter.

Imagecraft compiler user