LM3886 - Calculating Fc for low pass NFB filter

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s = j*2*pi*f, though. j is sqrt(-1) (complex numbers). f is frequency in Hz.

Tom

There cannot be "f" in there, because f is on the left side of the formula from the pdf (fc). You cannot calculate f from f.
So this must be some different "s".

TDA2030 from the Workbench is presenting lower bandwidth, like 20kHz instead of 100kHz of LM's. I don't know why. But doubling Cf while Rf2 twice smaller seems good idea (when observing proportions of the TDA graphs and feedback components being changed at the same time)
 
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Yes, but the f is not T(s), see the video mentioned in an earlier post (new link)
(And remember the formula is taken from over 20 year old documentation, I think they used the MIT notation. For those guys Hz was for amatures, they was thinking in radians)
Found a series of MIT lectures:
https://ocw.mit.edu/resources/res-6...os/lecture-1-introduction-and-basic-concepts/
At 19:20 the feedback figure is displayed. There f is the anotation of the feedback transfer function.
So then fc(s) makes sense. Nowadays that is quite confusing🙂
 
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fc = [Rf1 Rf2 (s + 1/Rf2Cf)]/[(Rf1 + Rf2)(s + 1/Cf(Rf1 + Rf2))]
fc = [Rf1 Rf2/(Rf1 + Rf2)] [(s + 1/Rf2Cf)/(s + 1/Cf(Rf1 + Rf2))]
fc = Rf1||Rf2 * [(s + 1/Rf2Cf)/(s + 1/Cf(Rf1 + Rf2))]

fc = Rf1||Rf2 * [s/s] = Rf1||Rf2, if s is much larger than 1/Rf2Cf
fc = [Rf1 Rf2/(Rf1 + Rf2)] [(Rf1 + Rf2)/(Rf2)] = Rf1 Rf2/Rf2 = Rf1 if s is much smaller than 1/Cf(Rf1 + Rf2))

So feedback resistance (or impedace) will go from Rf1 at DC to Rf1||Rf2 at high frequency.
The shift will happen from around CfRf2 to Cf(Rf1 + Rf2) radians.
The slope will be -6dB/oct somewhere between.

That is the old world way of analysis and the formula makes it much easier to do.
 
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