sallen-key filter design

[pillip]

I am trying to design a 2 pole butterworth, bessel and chebyshev filter
using sallen key design. I am consulting this website
http://www.science.unitn.it/~bassi/Signal/TInotes/sloa049.pdf for
information, but am having trouble calculating the m and n values. Basically
i am limited to 1 microfarad capacitors so i am implementing A1.3 S-K
simplification (see appendix at bottom of webpage). It says " Design should
start with choosing m and K to set the gain and Q of the circuit, and then
choosing C and calculating R to set fc" but since Q is already given for a
specific 2 pole filter, what does it mean by setting Q of the ciruit. Any
help would be appreciated.

 

[John Larkin]

On Sat, 27 Mar 2004 01:22:33 GMT, "pillip" <pillip@pillip.com> wrote:

I am trying to design a 2 pole butterworth, bessel and chebyshev filter
using sallen key design. I am consulting this website
http://www.science.unitn.it/~bassi/Signal/TInotes/sloa049.pdf for
information, but am having trouble calculating the m and n values. Basically
i am limited to 1 microfarad capacitors so i am implementing A1.3 S-K
simplification (see appendix at bottom of webpage). It says " Design should
start with choosing m and K to set the gain and Q of the circuit, and then
choosing C and calculating R to set fc" but since Q is already given for a
specific 2 pole filter, what does it mean by setting Q of the ciruit. Any
help would be appreciated.

Get Don Lancaster's "Active Filter Cookbook."

John

 

[Martin Riddle]

TI has a nice filter designer Filter Pro on their web site.

Cheers

"pillip" <pillip@pillip.com> wrote in message news:1606d6a374ab8b8b8a752f3ad287c6c4@news.teranews.com...

I am trying to design a 2 pole butterworth, bessel and chebyshev filter
using sallen key design. I am consulting this website
http://www.science.unitn.it/~bassi/Signal/TInotes/sloa049.pdf for
information, but am having trouble calculating the m and n values. Basically
i am limited to 1 microfarad capacitors so i am implementing A1.3 S-K
simplification (see appendix at bottom of webpage). It says " Design should
start with choosing m and K to set the gain and Q of the circuit, and then
choosing C and calculating R to set fc" but since Q is already given for a
specific 2 pole filter, what does it mean by setting Q of the ciruit. Any
help would be appreciated.

 

[pillip]

with K = 2, Q = 0.7071 (for 2pole butterworth), m comes out to be 0.5
R = 1/(2Pi*C*f*FSF*sqrt(m))
with FSF =1, C = 1microfarad, f =200Hz, R comes out to 1125.4 ohm
R2 = R = 1125.4 ohm , R1 = m*R = 565.7 ohm
C1 =C2 = C =1microfarad.

Now the problem begins, when i run a simulation in multisim using these
values, the cutoff -3dB frequency comes out to approx 118 Hz as opposed to
200 Hz i used while calculating the values. Any ideas??


"PaulCsouls" <paulcsouls@worldnet.att.net> wrote in message
news:gcv9609bica1qih4ou9uore73j898ru6c3@4ax.com...

On Sat, 27 Mar 2004 01:22:33 GMT, "pillip" <pillip@pillip.com> wrote:

I am trying to design a 2 pole butterworth, bessel and chebyshev filter
using sallen key design. I am consulting this website
http://www.science.unitn.it/~bassi/Signal/TInotes/sloa049.pdf for
information, but am having trouble calculating the m and n values.
Basically
i am limited to 1 microfarad capacitors so i am implementing A1.3 S-K
simplification (see appendix at bottom of webpage). It says " Design
should
start with choosing m and K to set the gain and Q of the circuit, and
then
choosing C and calculating R to set fc" but since Q is already given for
a
specific 2 pole filter, what does it mean by setting Q of the ciruit. Any
help would be appreciated.


You set the equation for Q equal to the value in the table for the
kind of filter you want to build. Either a butterworth, bessel or
chebyshev. You can't make a butterworth, bessel and chebyshev filter.
Then pick a gain. (try 2, its easy) and solve for m. R will equal
1/(2*PI*C*f) .

Okay?
Paul

 

[Max Hauser]

Linear Technology has had a very handy filter-design tool for years,
FilterCAD. Useful for filter designs with or without LTC's parts, including
much of the basic mathematics. Used in college courses on filters.
http://www.linear.com/software/

"pillip" <pillip@pillip.com> wrote in message
news:1606d6a374ab8b8b8a752f3ad287c6c4@news.teranews.com...

I am trying to design a 2 pole butterworth, bessel and chebyshev filter
using sallen key design. ... what does it mean by setting Q of the ciruit.
Any
help would be appreciated.

 

[Ban]

pillip wrote:

I am trying to design a 2 pole butterworth, bessel and chebyshev
filter using sallen key design. I am consulting this website
http://www.science.unitn.it/~bassi/Signal/TInotes/sloa049.pdf for
information, but am having trouble calculating the m and n values.
Basically i am limited to 1 microfarad capacitors so i am
implementing A1.3 S-K simplification (see appendix at bottom of
webpage). It says " Design should start with choosing m and K to set
the gain and Q of the circuit, and then choosing C and calculating R
to set fc" but since Q is already given for a specific 2 pole filter,
what does it mean by setting Q of the ciruit. Any help would be
appreciated.

When you have limited values for C, it is much better to choose both caps to
be equal. To simplify the calculations you should also choose equal Rs. This
leads to the following circuit. I already calculated most of the parts for
you. The gain of the circuit sets the Q, whereas the pole frequency is
independent.

||
+-------||----------+
| ||C |
| 1/2 TL072|
___ | ___ |\ |
o-|___|-+-|___|-+-----|+\ | ___
R R |C | >--+-|___|-+---o
--- +-|-/ | 100 |
--- | |/ .-. |
| | | | ---
=== | | |Ra --- 47n
GND | '-' |
+-------+ ===
.-. GND
Bessel: Ra=0.268*Rb | | B= 0.786
| |Rb
Butterworth: Ra=0.586*Rb '-' B= 1
|
3dBChebycheff: Ra=1.234*Rb === B= 1.389
GND
RC= B/(2pi*fg)
created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de
set view\ font\fixed

--
ciao Ban
Bordighera, Italy

 

[pillip]

You are suggesting a vcvs filter design, and the values you specify are also
given in the art of electronics. I've tried them as well, but the simulation
still gives an incorrect cut off frequency.


"Ban" <bansuri@web.de> wrote in message
news:uO89c.107448$z23.4547468@news3.tin.it...

pillip wrote:
I am trying to design a 2 pole butterworth, bessel and chebyshev
filter using sallen key design. I am consulting this website
http://www.science.unitn.it/~bassi/Signal/TInotes/sloa049.pdf for
information, but am having trouble calculating the m and n values.
Basically i am limited to 1 microfarad capacitors so i am
implementing A1.3 S-K simplification (see appendix at bottom of
webpage). It says " Design should start with choosing m and K to set
the gain and Q of the circuit, and then choosing C and calculating R
to set fc" but since Q is already given for a specific 2 pole filter,
what does it mean by setting Q of the ciruit. Any help would be
appreciated.

When you have limited values for C, it is much better to choose both caps
to
be equal. To simplify the calculations you should also choose equal Rs.
This
leads to the following circuit. I already calculated most of the parts for
you. The gain of the circuit sets the Q, whereas the pole frequency is
independent.

||
+-------||----------+
| ||C |
| 1/2 TL072|
___ | ___ |\ |
o-|___|-+-|___|-+-----|+\ | ___
R R |C | >--+-|___|-+---o
--- +-|-/ | 100 |
--- | |/ .-. |
| | | | ---
=== | | |Ra --- 47n
GND | '-' |
+-------+ ===
.-. GND
Bessel: Ra=0.268*Rb | | B= 0.786
| |Rb
Butterworth: Ra=0.586*Rb '-' B= 1
|
3dBChebycheff: Ra=1.234*Rb === B= 1.389
GND
RC= B/(2pi*fg)
created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de
set view\ font\fixed

--
ciao Ban
Bordighera, Italy

 

[John Woodgate]

I read in sci.electronics.design that pillip <pillip@pillip.com> wrote
(in <c520694eaba38f63dc386fe380af89a0@news.teranews.com&gt about 'sallen-
key filter design', on Sat, 27 Mar 2004:

You are suggesting a vcvs filter design, and the values you specify are
also given in the art of electronics. I've tried them as well, but the
simulation still gives an incorrect cut off frequency.

It is just possible that you are identifying the 'cut-off frequency' in
a different way from that assumed in deriving the formula for the
component values.

How are you determining it?
--
Regards, John Woodgate, OOO - Own Opinions Only.
The good news is that nothing is compulsory.
The bad news is that everything is prohibited.
http://www.jmwa.demon.co.uk Also see http://www.isce.org.uk

 

[Ban]

pillip wrote:

You are suggesting a vcvs filter design, and the values you specify
are also given in the art of electronics. I've tried them as well,
but the simulation still gives an incorrect cut off frequency.


Pillip,

these values refer to an ideal opamp, but in reality there is a (shifted)
pole in the transfer function to get the opamp stable at unity gain. This
pole falls exactly into the region in question and will give an additional
phase-shift because of limited gain, so the filter has a (slightly)higher Q
as calculated.
But the difference is really neglegible.
I have simulated a Butterworth filter 2nd order with R=1k, C= 1uF Ra= 586
Ohms, Rb= 1k and the TL072.
My simulation gives 159.8Hz, whereas the calculated value should be
159.15Hz. You can correct that by making Ra a bit smaller, say 582 Ohms.
With a Chebycheff filter the value will be a little more different, but we
talk about less than 1/2%. Will be difficult to find components exact to
this value either.
If your simulations differ more you might have an error somewhere.
--
ciao Ban
Bordighera, Italy

 

[Helmut Sennewald]

"PaulCsouls" <paulcsouls@worldnet.att.net> schrieb im Newsbeitrag
news:dgcb60ta2rotg3rod251odvum1n4b2g14l@4ax.com...

It should work. Are you sure the gain is 2? The feedback resistor and
the resistor to ground from the negative of the op-amp should be the
same. What values are using for these resistors?

Hello Paul,

you have it. If the gain of the opamp is set to 1, then Fc is indeed 118Hz.
With the correct gain of 2, the corner frequency is 200Hz as intended.
A gain of 2 requires for example R3=R4=10k.

Best Regards,
Helmut

 

[pillip]

yes the gain (K) is 2, and the feedback resistor and resistor to ground are
10kohms. With these capacitance and resistance, the curtoff comes out to
324Hz as opposed to 200Hz. Could there be a problem with the values of Q and
FSF given in the tables.

"PaulCsouls" <paulcsouls@worldnet.att.net> wrote in message
news:dgcb60ta2rotg3rod251odvum1n4b2g14l@4ax.com...

It should work. Are you sure the gain is 2? The feedback resistor and
the resistor to ground from the negative of the op-amp should be the
same. What values are using for these resistors?

Paul



On Sat, 27 Mar 2004 05:01:47 GMT, "pillip" <pillip@pillip.com> wrote:

with K = 2, Q = 0.7071 (for 2pole butterworth), m comes out to be 0.5
R = 1/(2Pi*C*f*FSF*sqrt(m))
with FSF =1, C = 1microfarad, f =200Hz, R comes out to 1125.4 ohm
R2 = R = 1125.4 ohm , R1 = m*R = 565.7 ohm
C1 =C2 = C =1microfarad.

Now the problem begins, when i run a simulation in multisim using these
values, the cutoff -3dB frequency comes out to approx 118 Hz as opposed
to
200 Hz i used while calculating the values. Any ideas??


"PaulCsouls" <paulcsouls@worldnet.att.net> wrote in message
news:gcv9609bica1qih4ou9uore73j898ru6c3@4ax.com...
On Sat, 27 Mar 2004 01:22:33 GMT, "pillip" <pillip@pillip.com> wrote:

I am trying to design a 2 pole butterworth, bessel and chebyshev
filter
using sallen key design. I am consulting this website
http://www.science.unitn.it/~bassi/Signal/TInotes/sloa049.pdf for
information, but am having trouble calculating the m and n values.
Basically
i am limited to 1 microfarad capacitors so i am implementing A1.3 S-K
simplification (see appendix at bottom of webpage). It says " Design
should
start with choosing m and K to set the gain and Q of the circuit, and
then
choosing C and calculating R to set fc" but since Q is already given
for
a
specific 2 pole filter, what does it mean by setting Q of the ciruit.
Any
help would be appreciated.


You set the equation for Q equal to the value in the table for the
kind of filter you want to build. Either a butterworth, bessel or
chebyshev. You can't make a butterworth, bessel and chebyshev filter.
Then pick a gain. (try 2, its easy) and solve for m. R will equal
1/(2*PI*C*f) .

Okay?
Paul

 

[pillip]

ok, i'll try that.
Thanks alot

Phillips

"Helmut Sennewald" <helmutsennewald@t-online.de> wrote in message
news:c44p3k$6s3$02$1@news.t-online.com...

"pillip" <pillip@pillip.com> schrieb im Newsbeitrag
news:8ab30d46a36a3b2ba5cca4bff44de9f7@news.teranews.com...
yes the gain (K) is 2, and the feedback resistor and resistor to ground
are
10kohms. With these capacitance and resistance, the curtoff comes out to
324Hz as opposed to 200Hz. Could there be a problem with the values of Q
and
FSF given in the tables.

Hello "pillip",
al the formulas and tables are correct. I have simulated it with
LTSPICE and Fg is 200Hz.
R1=563, R2=1125, R3=10k, R4=10k, C1=C2=1uF

PS: You should better use C1=C2=0.1uF.
Then R1 have to be 5630Ohm and R2 11250Ohm. You would choose R1=560Ohm
and R2=10kOhm+1.2kOhm.



Best Regards,
Helmut

This is the schematic file for the simulation with LTSPICE.
Put all the text in a file named "butter.asc" .


Version 4
SHEET 1 880 680
WIRE 64 224 96 224
WIRE 96 224 96 272
WIRE 176 224 96 224
WIRE -48 224 -48 112
WIRE -48 112 0 112
WIRE 64 112 224 112
WIRE 224 112 224 208
WIRE 176 272 176 352
WIRE 224 288 224 304
WIRE 96 336 96 384
WIRE -80 224 -48 224
WIRE -48 224 -16 224
WIRE -160 224 -272 224
WIRE -272 256 -272 224
WIRE -272 336 -272 384
WIRE 352 304 352 352
WIRE 352 224 352 112
WIRE 352 112 224 112
WIRE 352 112 384 112
WIRE 352 352 176 352
WIRE 352 352 352 384
WIRE 352 464 352 496
WIRE -272 224 -288 224
FLAG 224 304 0
FLAG 96 384 0
FLAG -272 384 0
FLAG 384 112 out
IOPIN 384 112 Out
FLAG 352 496 0
FLAG -288 224 in
IOPIN -288 224 In
SYMBOL e 224 192 R0
SYMATTR InstName E1
SYMATTR Value 10000
SYMBOL res -176 240 R270
WINDOW 0 32 56 VTop 0
WINDOW 3 0 56 VBottom 0
SYMATTR InstName R1
SYMATTR Value 563
SYMBOL res -32 240 R270
WINDOW 0 32 56 VTop 0
WINDOW 3 0 56 VBottom 0
SYMATTR InstName R2
SYMATTR Value 1125
SYMBOL cap 80 272 R0
SYMATTR InstName C1
SYMATTR Value 1ľ
SYMBOL cap 0 128 R270
WINDOW 0 32 32 VTop 0
WINDOW 3 0 32 VBottom 0
SYMATTR InstName C2
SYMATTR Value 1ľ
SYMBOL voltage -272 240 R0
WINDOW 123 24 132 Left 0
WINDOW 39 0 0 Left 0
SYMATTR Value2 AC 0.5
SYMATTR InstName V1
SYMATTR Value 0
SYMBOL res 336 208 R0
SYMATTR InstName R3
SYMATTR Value 10k
SYMBOL res 336 368 R0
SYMATTR InstName R4
SYMATTR Value 10k
TEXT -264 56 Left 0 !.ac dec 100 1 10k
TEXT -264 0 Left 0 ;Butterworth Filter 200Hz

 

[Helmut Sennewald]

"pillip" <pillip@pillip.com> schrieb im Newsbeitrag
news:8ab30d46a36a3b2ba5cca4bff44de9f7@news.teranews.com...

yes the gain (K) is 2, and the feedback resistor and resistor to ground
are
10kohms. With these capacitance and resistance, the curtoff comes out to
324Hz as opposed to 200Hz. Could there be a problem with the values of Q
and
FSF given in the tables.

Hello "pillip",
al the formulas and tables are correct. I have simulated it with
LTSPICE and Fg is 200Hz.
R1=563, R2=1125, R3=10k, R4=10k, C1=C2=1uF

PS: You should better use C1=C2=0.1uF.
Then R1 have to be 5630Ohm and R2 11250Ohm. You would choose R1=560Ohm
and R2=10kOhm+1.2kOhm.



Best Regards,
Helmut

This is the schematic file for the simulation with LTSPICE.
Put all the text in a file named "butter.asc" .


Version 4
SHEET 1 880 680
WIRE 64 224 96 224
WIRE 96 224 96 272
WIRE 176 224 96 224
WIRE -48 224 -48 112
WIRE -48 112 0 112
WIRE 64 112 224 112
WIRE 224 112 224 208
WIRE 176 272 176 352
WIRE 224 288 224 304
WIRE 96 336 96 384
WIRE -80 224 -48 224
WIRE -48 224 -16 224
WIRE -160 224 -272 224
WIRE -272 256 -272 224
WIRE -272 336 -272 384
WIRE 352 304 352 352
WIRE 352 224 352 112
WIRE 352 112 224 112
WIRE 352 112 384 112
WIRE 352 352 176 352
WIRE 352 352 352 384
WIRE 352 464 352 496
WIRE -272 224 -288 224
FLAG 224 304 0
FLAG 96 384 0
FLAG -272 384 0
FLAG 384 112 out
IOPIN 384 112 Out
FLAG 352 496 0
FLAG -288 224 in
IOPIN -288 224 In
SYMBOL e 224 192 R0
SYMATTR InstName E1
SYMATTR Value 10000
SYMBOL res -176 240 R270
WINDOW 0 32 56 VTop 0
WINDOW 3 0 56 VBottom 0
SYMATTR InstName R1
SYMATTR Value 563
SYMBOL res -32 240 R270
WINDOW 0 32 56 VTop 0
WINDOW 3 0 56 VBottom 0
SYMATTR InstName R2
SYMATTR Value 1125
SYMBOL cap 80 272 R0
SYMATTR InstName C1
SYMATTR Value 1ľ
SYMBOL cap 0 128 R270
WINDOW 0 32 32 VTop 0
WINDOW 3 0 32 VBottom 0
SYMATTR InstName C2
SYMATTR Value 1ľ
SYMBOL voltage -272 240 R0
WINDOW 123 24 132 Left 0
WINDOW 39 0 0 Left 0
SYMATTR Value2 AC 0.5
SYMATTR InstName V1
SYMATTR Value 0
SYMBOL res 336 208 R0
SYMATTR InstName R3
SYMATTR Value 10k
SYMBOL res 336 368 R0
SYMATTR InstName R4
SYMATTR Value 10k
TEXT -264 56 Left 0 !.ac dec 100 1 10k
TEXT -264 0 Left 0 ;Butterworth Filter 200Hz