# Butterworth Filter Sallen Ket Cut Off Frequency?

Discussion in 'Homework Help' started by Volt, Feb 24, 2010.

1. ### Volt Thread Starter Member

Nov 28, 2008
24
0
Hello,

I am having a little trouble calculating the upper cut off frequency of this Butterworth Filter in the Sallen and key configuration.

Below you can see the component values in the pic:

http://twitpic.com/152zmr

The upper cut off frequency here (according to my supervisors) is 28kHZ,

however using the equation I found:

fc=1/2pi(root(R1R2C1C2))

,it doesn't equal 28kHz, I am getting 44.209kHz?!

Am I using the wrong equation?

Any help would be much appreciated,
Regards,

Jan 29, 2010
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3. ### Volt Thread Starter Member

Nov 28, 2008
24
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I'm on a mac i'm afraid

Apr 27, 2009
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5. ### Ron H AAC Fanatic!

Apr 14, 2005
7,050
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1. That is a filter, but it is not a Butterworth filter. You can't make a unity gain Butterworth with equal valued resistors and equal valued capacitors.
2. Stray capacitance will cause the actual cutoff frequency to differ from the calculated value. You should use higher capacitance values.

6. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,939
1,089
Because IF R1=R2 and C1 = C2 the
Fo=1/(2*pi*RC)=0.16/(RC)=44KHz

Jul 8, 2008
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8. ### Volt Thread Starter Member

Nov 28, 2008
24
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When I simulate this on multisim it comes out as 28kHz, is there an equation I could use to determine that value?

I have tried v hard to work it out but can't!

Thanks for all your help so far...

9. ### Ron H AAC Fanatic!

Apr 14, 2005
7,050
655
The cutoff frequency is indeed 44kHz, but that equation does not refer to the -3dB frequency. It is the natural frequency of the filter. The gain at that frequency is equal to the Q of the filter. A Butterworth filter has a Q of 1/√2. The OP's filter has a Q of 0.5. Therefore, the Butterworth is 3dB down at the natural frequency. The OP's filter is 6dB down at the natural frequency.
A simulation of the OP's filter and a B'worth shows that the OP's filter is in fact 3dB down at 28kHz, which I'm sure can (and probably will) be calculated by another member of the forum.

A filter with higher Q will have less loss at the natural frequency, and in fact will have peaking as the Q increases.
See the Wikipedia entry for Sallen & Key.

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10. ### Darren Holdstock Active Member

Feb 10, 2009
262
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Just to say, all hail Ron H, this is filter gold. In one clear and concise paragraph Ron H has explained it beautifully.
I blew out a job interview at TAG McLaren Audio many years ago by asking the technical director a similar question regarding damping and the various resonant frequencies, as I couldn't find any textbook that would explain it to my satisfaction. He didn't know either, and took his revenge for my impudence in asking. Still, I learned not to ask awkward questions at interviews, lest a thirst for knowledge be mistaken for being awkward.

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11. ### Ron H AAC Fanatic!

Apr 14, 2005
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Thanks for the kind words, Darren.
I have to admit that I didn't know this until I looked at the simulation, which drove me to the equations in Wikipedia.

12. ### Volt Thread Starter Member

Nov 28, 2008
24
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Thankyou so much, explained it perfectly!

13. ### Jony130 AAC Fanatic!

Feb 17, 2009
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If we have the transfer function

$\frac{Vout}{Vin}=\frac{1}{(1-\omega^2 R_1*R_2*C_1*C_2)^2+\omega^2*((R_1+R_2)*C_2)^2}$

And we want to know cut-of frequency for 3dB=1/√2.
So we have

$\frac{Vout}{Vin}=\frac{1}{(1-\omega^2 R_1*R_2*C_1*C_2)^2+\omega^2*((R_1+R_2)*C_2)^2}= \frac {1} {\sqrt2}$

And for R1=R2=R and C1=C2=C
We get

$(1-(\omega*R*C)^2)^2+4*(\omega*R*C)^2=2$

And if we solve this we get:

$(1+(\omega^2*C^2*R^2))^2=2$ we substitute $X=(\omega^2*C^2*R^2)$

$(1+X)^2=2$

$X=\sqrt{2}-1$

And cut-off frequency is equal

$fg=\frac{\sqrt{\sqrt{2}-1}}{2*\Pi*R*C}$

And this is a special case only for R1=R1, C1=C2

14. ### hgmjr Moderator

Jan 28, 2005
9,030
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jony130,

Shouldn't the equation for the magnitude of Vo/Vi be:

$\frac{Vout}{Vin}=\frac{1}{\sqrt{(1-\omega^2 R_1*R_2*C_1*C_2)^2+\omega^2*((R_1+R_2)*C_2)^2}}$

hgmjr

Oct 9, 2007
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