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I have an RC low pass filter connected to the non inverting terminal of the opamp. There is a feedback resistor from output of opamp to its inverting pin.Then an RC low pass filter is placed at the output of the opamp. So while designing the filter,should I consider these two low pass filter together as a second order filter and design or consider each RC low pass filter separately and design the filter.Which is correct to get the desired response?enter image description here

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  • \$\begingroup\$ You should learn that filters have group delay , the derivative of phase shift and most are non-linear except some. So start with band pass characteristics -3dB , Phase response and Gain and available GBW limits . ALWAYS start with specs, then add more like Bandstop rejection, Q, component tolerance, gain error, linear phase, maximally flat, zero overshoot \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Dec 13 '19 at 6:46
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You have your sallen-key topology wrong. Even if it was correct, I don't like the circuit because attaching a load will mess with the transfer function. Ideally your load attaches to the output of the op-amp or a buffered node, as then attached loads will not draw current from the filter but the op-amp itself, preserving the filter's characteristics.

schematic

simulate this circuit – Schematic created using CircuitLab

^ Sallen-Key two pole low-pass.


To answer your question, for the sallen-key you need to analyze as a whole (or look up TF, it's somewhat complex). If you used buffered RC/Active Lowpass, it would be each TF multiplied, as the op-amp will provide the current.

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U1 completely buffers the two RC stages from each other. This means you can calculate their frequency responses independently, and simply multiply the responses together.

This is not the ideal way to make a 2nd order filter, it will have a very 'soggy' response around the cuttoff frequency, which most people don't want. While technically it is a 2nd order filter, most people would refer to it as 'two cascaded first order filters'.

For a better approximation to a flat passband, followed by a steep rolloff, design a proper 2nd order filter, using the opamp as part of the filter.

Sallen Key is a reasonable topology, there are many cookbooks available on the net to help you design one. Bear in mind that a SK lowpass filter fails in the stopband at high frequencies when the opamp output impedance rises. Use a 3rd order Sallen Key, which uses the same single opamp, to mitigate (doesn't cure, but makes it easier to live with) this fault, and gives you more design freedom on the filter transition as well.

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Deepa - it is correct that your circuit is a second-order lowpass. However, as mentioned already, it is a rather bad one.

Two reasons: It works like a passive filter only with a "bad" amplitude response (two real poles) within the passband and it does not provide a low-resistive output which is insensible to any load.

I have mentioned the "poles". Note that we have various different second-order lowpass responses - and the most useful response is the so called "Butterworth" response which has a maximally flat amplitude response within the passband.

This filter type has not two real poles but two conjugate-complex poles and a quality factor Qp=0.7071 (in contrast, the quality factor of your circuit cannot be larger than 0.5). Such a Butterworth filter can be realized only as an "active filter" - for example based on the mentioned Sallen-Key topology (other filter structures are aslo possible).

In any "active" filter the amplifier is not used as a buffer only (as in your case) but it has a feedback network which is part of the filter function. However, calculation of the parts values based on a certain filter specuification is more involved...in most cases, we make use of formulas readily available in textbooks or other sources (internet).

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