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2nd Order low pass filter - Why a topology?
BartK:
Hi all,
Although I am new to this forum, I am familiar with electrical designs and such. I am designing a second order low pass filter to filter a 0-10V signal from a sensor. The theory states that there are two possibilities for an active low-pass filter; the Sallen-Key toplogie and the Multiple Feedback topology.
However, I wonder why I should use one of these topologies. Can't I just take a second order passive low pass filter and place an opamp buffer behind it?
I used LT spice to compare the two and this is what the simulation shows:
Can someone explain to me why this is "not" possible? I can't find anything about this. For many of the articles found, an active filter is always made with one of the two topologies mentioned.
Thanks in advance for your great help!
iMo:
Because they are not identical?
AndyC_772:
You can use whichever topology you prefer; there's no definitively "right" or "wrong" option.
Tip: do always take into account the tolerances of your components, and the parasitic elements that you can't eliminate. You might be surprised just how much of an effect they can have on the behaviour of an analogue filter.
rs20:
As hinted at by imo's somewhat cryptic post, the two filters have a different response. What is generally considered ideal for a 2nd-order filter is a Butterworth filter (often implemented using the Sallen-Key topology), which is sharper and generally considered superior to two standard 1st-order filters in series. (Although I suppose it depends on the application as suggested by AndyC)
I think your error was assuming that the ideal output of a 2nd-order filter of cutoff frequency f is the same as two 1st order filters in series. But we can actually do better.
Notice, in imo's post, how the Butterworth filter (Sallen Key) filter's response stays nice and flat real close to the cutoff frequency, and then quite sharply transitions into the roll-off region? Whereas your proposed topology takes a much more lazy route. More numerically, the amplitude response of a 1st-order filter at the cutoff frequency is -3dB, which means putting two in series would yield a 6dB attenuation at the cutoff frequency. By contrast, the Butterworth filter only attenuates by 3dB at the cutoff.
BartK:
--- Quote from: imo on May 07, 2019, 08:11:29 am ---Because they are not identical?
--- End quote ---
I now see that I have accidentally taken the butterworth response for the Sallen-Key topology. I meant to use one of my other models. This was indeed my mistake.
The difference that can be seen comes from the other response, namely the quality factor of 0.7071(butterworth). However, if I take a quality factor of .5 for the Sallen-Key, the graph is almost the same.
--- Quote ---Quality factors of common systems
A unity-gain Sallen–Key lowpass filter topology with equal capacitors and equal resistors is critically damped (i.e., Q = 1⁄2).
A second-order Butterworth filter (i.e., continuous-time filter with the flattest passband frequency response) has an underdamped Q = 1⁄√2.[11]
A Bessel filter (i.e., continuous-time filter with flattest group delay) has an underdamped Q = 1⁄√3.[citation needed]
--- End quote ---
See: https://en.wikipedia.org/wiki/Q_factor
Stil though, this is debating about the response.
My question is more about the topologie. In the case of a second-order active low-pass filter, no mention is made of the option of taking a second-order passive filter with a buffer, why not?
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