# Need help with sallen key bandpass filter

Discussion in 'Homework Help' started by alialiali, Mar 12, 2009.

1. ### alialiali Thread Starter Member

Dec 12, 2008
26
0
Hi every one
I'm working on bandpass secound order filter and I have to have band pass between 23k and 53k using sallen key filters whaen I use the formola to dsign my high pass which should gave me 23k fc its working and I used the folwing calculation

R1 = mR, R2= R, C1=C , C2=nc FSF = 1 from the table
FSF * fc = 1/ 2pi RC squrt mn
m= 0.229 from the table
n= 3.3 from the table
I'm looking for LP for 53k
1*23k=1 / 2pi * (C=0.01u from th table) * (R which we need 2 find) * squrt 0.229*3.3
R = 1 / 2pi * (C=0.01u ) * (23k) * squrt 0.229*3.3 = 707 ohms
Since R = R2 so my R2 value is 707 ohms
R1 = R*m = 0.229 * 707 = 161.9 = 162 ohms
C1 = C which is 0.01u so the value of my C1 is 0.01u
C2 = n * C which is 3.3 * 0.01u = 33n
so my C2= 33n

and my LP filter which should gave me the fc=53k as

R1 = mR, R2= R, C1=C , C2=nc FSF = 1 from the table
FSF * fc = 1/ 2pi RC squrt mn
m= 0.229 from the table
n= 3.3 from the table
I'm looking for LP for 53k
1*53k=1 / 2pi * (C=0.01u from th table) * (R which we need 2 find) * squrt 0.229*3.3
R = 1 / 2pi * (C=0.01u ) * (53k) * squrt 0.229*3.3 = 345
Since R = R2 so my R2 value is 345 ohms
R1 = R*m = 0.229 * 345 = 79.005 = 80 ohms
C1 = C which is 0.01u so the value of my C1 is 0.01u
C2 = n * C which is 3.3 * 0.01u = 33n
so my C2= 33n

the main problem when I contact the output from my HP to the input of my LP filter the over all fc is not maching the fcs I looking for

can any one advice me how to conact the two filters together this can be using athered filter but my quation is do I need to change the holl calcultion and how?

many thanks

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Apr 5, 2008
15,802
2,387
Hello,

Your resistors have a very low value.
What happens when you divide the capacitors by 10 and multiply the resistors by 10 ?
The frequency will be the same but the impedance will be changed.

Greetings,
Bertus

3. ### alialiali Thread Starter Member

Dec 12, 2008
26
0
I dont know can you explin bit more please

Apr 5, 2008
15,802
2,387
Hello,

What I am trying to say is : recalculate the filters with capacitors of 1 nF and n* 1nF in stead of 10 nF and n* 10nF.

Greetings,
Bertus

5. ### alialiali Thread Starter Member

Dec 12, 2008
26
0
Ok I did that divide the capacitors by 10 and multiply the resistors by 10 I can see some difrernt but the problem now is I am getting 57.71kHz for the LP which it should be 53kHz and the HP= 20.84kHz which should be = 23kHz
please can you help me more

Apr 5, 2008
15,802
2,387
Hello,

What kind of filter are you making ?
Where is the table you are talking about ?
Most sallen-key filters I know have equal resistors for the low pass and equal capacitors for the high pass.

Greetings,
Bertus

7. ### alialiali Thread Starter Member

Dec 12, 2008
26
0

I enclod copy of the table so you can have look and gave me some advice

Table 1. Butterworth Filter Table
FILTER FILTER
ORDER
Stage 2
FSF Q
1.000 0.7071
1.000 1.0000 there is full table in the picture below

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• ###### the fsf.JPG
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Last edited: Mar 13, 2009

Apr 5, 2008
15,802
2,387
Hello,

The table does not look familiar to me.
Here is a link to a PDF from TI about reference designs.

focus.ti.com/lit/an/slod006b/slod006b.pdf

From chapter 16 filters are discussed, sallen-key from 16.3.

Greetings,
Bertus

9. ### alialiali Thread Starter Member

Dec 12, 2008
26
0
Thx a lot for the Pdf file but the problem it's very difcult 2 understand it can you please help me bit mor how 2 apply the formlu to my filter as I need band pass betwen 23k and 53k by uding 2 secound order sallen keys filter only

Last edited: Mar 13, 2009

Apr 5, 2008
15,802
2,387
Hello,

Here ia a PDF special on sallen key filters.
Chapter 3 is for low pass
Chapter 4 is for high pass
Each chapter has simplifications in it.

I used the calculator on the web to have the 23 kHz highpass filter
.http://www.pronine.ca/acthpf.htm

R1 = 4.89 k Ohm, R2 = 9.78 k Ohm and C1 = C2 = 1 nF

The same I did with the 53 kHz lowpass filter.
http://www.pronine.ca/actlpf.htm

R1 = R2 = 2.12 k Ohm , C1 = 2 nF , C2 = 1 nF

Greetings,
Bertus

• ###### sloa024b_sallen-key.pdf
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Last edited: Mar 14, 2009
11. ### alialiali Thread Starter Member

Dec 12, 2008
26
0
Many many thx

12. ### alialiali Thread Starter Member

Dec 12, 2008
26
0
This two circuits is great just I’d like to ask how to connect them together so I get my final out put from one port instead to have tow deferent out puts port ?
Do I need to implement a there’d filter to feed this two in it ?

Many thx

13. ### alialiali Thread Starter Member

Dec 12, 2008
26
0
When I contact them together as in the picture it gave me defrnt anwsers

• ###### 23 + 53.JPG
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Apr 5, 2008
15,802
2,387
Hello,

You can connect the highpass filter at the output of the lowpass filter.
This will create the bandpass filter.

Greetings,
Bertus

15. ### alialiali Thread Starter Member

Dec 12, 2008
26
0
Thats great the high pass signal is ok as it's gaving me 23kHz but when I test the output for the 53k it's gaving me 61KHz so please can you gave more advice.

I think we need to add another filter such as adder or first order filter

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Last edited: Mar 15, 2009

Apr 5, 2008
15,802
2,387
Hello,

Perhaps a buffer between the highpass and lowpass will help.

Greetings,
Bertus

Last edited: Mar 15, 2009
17. ### alialiali Thread Starter Member

Dec 12, 2008
26
0
Sorry but what do you mean with buffer as I have no idea a bout it at all

Apr 5, 2008
15,802
2,387
Hello,

A buffer is an opamp with the + as input and the output conected to the - input.

When you place this between the two filters they will not "see" each other.

Greetings,
bertus

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19. ### alialiali Thread Starter Member

Dec 12, 2008
26
0
Do you know what I learned hear from you I wouldnt learn it in one year
I greatly appreciate your kind help

The result of the test now still negative as nothing changed do you think we need to add some resistors to the buffer ?

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Apr 5, 2008
15,802
2,387
Hello,

What happens when you change the TL072 (wich has a bandwidth of 4 Mhz)
to a LF357 ( wich has a bandwidth of 20 Mhz) ?
Perhaps the limited bandwidth is giving some problems.
(you are using rather high frequencies).

Greetings,
Bertus