# What is this Low pass filter?

Discussion in 'General Electronics Chat' started by Tera-Scale, Jun 7, 2012.

1. ### Tera-Scale Thread Starter Active Member

Jan 1, 2011
164
5
Does this filter fall under the sallen key topology? If not what is it's toplogy and how can I calculate the cut-off frequency? I know it has a linear gain of 30 that is adjustable. thanks

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2. ### #12 Expert

Nov 30, 2010
18,076
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Not Sallen-Key. C11 || R13 gives a pole when Xc = R13
It is merely a frequency limited amp.

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3. ### Tera-Scale Thread Starter Active Member

Jan 1, 2011
164
5
Can you give me a reliable source where I can read more on this please? So I can get the formula as well.

4. ### #12 Expert

Nov 30, 2010
18,076
9,676
Xc = 1/(2pi F C)
392k = 1/ (2 pi f 1e-7)

at that frequency the gain of the amp will decrease at the rate of a 1 pole filter.
Maybe that is 3db/octave. I forgot, but you can do the math and find out.
Crutschow or AG will probably pop in and fill in the info on this point.
Don't know a "source" for capacitave reactance at "x" frequency.

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5. ### Tera-Scale Thread Starter Active Member

Jan 1, 2011
164
5
Oki, I understand now. thanks a lot.

6. ### mlog Member

Feb 11, 2012
276
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Isn't there also a pole with R12 and C10?

7. ### #12 Expert

Nov 30, 2010
18,076
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Oh yeah...I missed that, but it's at a different frequency by a factor of 4. Too far apart for me to recognize this as a Sallen-key filter. Same formula. 2 break points on the diminishing gain graph.

8. ### Ron H AAC Fanatic!

Apr 14, 2005
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There is a pole at Fp1=1/(2*pi*R13*C11)≈4Hz.
Ther is a zero at Fz1=1/(2*pi*(R13||R14)*C11)≈17Hz.
There is another pole at Fp2=1/(2*pi*R12*C10)≈16Hz.
Fp2 approximately cancels Fz1, so the output rolls off at -6dB/octave, beginning at 4Hz.
It is basically a single pole filter with a low frequency gain of G=1+R13/R14=4.24=12.55dB.
I ignored RV2, which has very little effect.
Ignoring noise gain, the same filter could be realized by omitting C11, and changing C10 to 390nF.
Before anyone asks, you could leave C10=100nF, and change R12 to 392k, but 100k does a better job of cancelling input offset current.

Last edited: Jun 8, 2012
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9. ### mlog Member

Feb 11, 2012
276
36
Those are the same values I got for the poles and zero, except you transposed the R12 and C10, but I know what you mean.

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10. ### Tera-Scale Thread Starter Active Member

Jan 1, 2011
164
5
Thanks a lot for the workings. I understand the cancelling of the pole and zero 'gradients' in the bode plots but the -6dB per octave is not clear to me. can you converted to decades please?

11. ### Ron H AAC Fanatic!

Apr 14, 2005
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Oops! Maybe I'll edit my post to fix that.

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12. ### Ron H AAC Fanatic!

Apr 14, 2005
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For frequencies much greater than the corner frequency, the output of a single RC lowpass drops by half each time the frequency doubles (octave).
20*log(0.5)=-6.02dB.
By the same token, it drops to 10% of its previous amplitude each time the frequency increases by a factor of 10 (decade).
20*log(0.1)=-20dB.

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13. ### Audioguru Expert

Dec 20, 2007
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A resistor feeding a capacitor drops the signal voltage level 6dB (to half) per octave which is a drop of 20dB (1/10th) in a decade of frequency.

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