Butterworth Filter Sallen Ket Cut Off Frequency?

[Volt]

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,
Adam

[ericgibbs]

hi, Look at this link: Filter Pro from TI. freeware.

http://focus.ti.com/docs/toolsw/fold...filterpro.html

[Volt]

I'm on a mac i'm afraid

[nubcookie]

here try this link

http://focus.ti.com/lit/ml/sloa088/sloa088.pdf

its quite comprehensive but try page 15 you should be able to change those equations to get the cut off frequency

[Ron H]

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.

[Jony130]

Well, your supervisors is wrong.
Because IF R1=R2 and C1 = C2 the
Fo=1/(2*pi*RC)=0.16/(RC)=44KHz

[msr]

Did you already take a look at wikipedia?
http://en.wikipedia.org/wiki/Sallen%...93Key_topology

I find it very well explained.

[Volt]

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...

[Ron H]

Originally Posted by Jony130

Well, your supervisors is wrong.
Because IF R1=R2 and C1 = C2 the
Fo=1/(2*pi*RC)=0.16/(RC)=44KHz

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.

[Darren Holdstock]

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.