floyd12-25, op-amp input offset voltage

t_n_k

Joined Mar 6, 2009
5,455
Q1. Butterworth - 1st order.

Q2. Diagram 2 is not necessarily 3rd order since there is a feedback path to the RC network shown in each stage. There is no statement of the specific response for each stage. To be a third order filter one would anticipate that each stage has no actual feedback to the RC network and hence each (stage) is a simple single pole filter. I can't comment on the merits of using the Diagram 2 option in relation to a known Butterworth format other than to state that the Butterworth is intended to give a maximally flat response. If maximal flatness is the requirement then one would probably choose Butterworth.

Q3. Yes all in one stage is certainly possible for 3-pole.

Check this link ...http://www.beis.de/Elektronik/Filter/Act3PoleLP.html
 

Thread Starter

PG1995

Joined Apr 15, 2011
832
Thank you very much, t_n_k.

1:
Q1. Butterworth - 1st order.
But why can't it have Chebyshev- or Bessel-1st order response. Perhaps, by adjusting the values of R1 and R2 we can get other responses besides Butterworth?

2:
Q2. Diagram 2 is not necessarily 3rd order since there is a feedback path to the RC network shown in each stage. There is no statement of the specific response for each stage. To be a third order filter one would anticipate that each stage has no actual feedback to the RC network and hence each (stage) is a simple single pole filter.
I think I was wrong about the Diagram 2 in my last post. A single pole filter does not have a feedback path to RC network, right? At least this is how I see in the book I'm using. In diagram we have a three stage filter where each stage has a feedback path to the RC network. A Sallen-Key configuration uses two poles per stage and has a feedback path to RC network. Now please have a look here. Do you think I'm correct? By the way, I'm using this book. Replace the *'s with s-c-r-i-b-d dot com; you would find the material around page #789.

3:
Please help me with it. Thanks.

4:
Please have a look here and kindly help me with the query.

Regards
PG
 

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t_n_k

Joined Mar 6, 2009
5,455
Q1. Yes it can also be a Bessel or Chebyshev type. In essence there will be no difference in response since there will only be one real valued pole. So no distinction exists between the various types for a single pole LPF.

I may respond to the other questions later but another member might be kind enough to oblige.
 

Jony130

Joined Feb 17, 2009
5,487
Q2 - Diagram 2 show 6 order active filter.
Q3 - We also have a second order Multiple Feedback Filter
Q4 - Yes you can build Chebyshev- or Bessel filter with the Sallen–Key topology.
But Butterworth filter is most widely used.
 

Thread Starter

PG1995

Joined Apr 15, 2011
832
Thank you, t_n_k and Jony.

Please help me with the queries below.

1:
Q3 - We also have a second order Multiple Feedback Filter
Do you think this diagram of MFB low pass filter correct? The diagram was taken from here.

2:
Q4 - Yes you can build Chebyshev- or Bessel filter with the Sallen–Key topology.
But Butterworth filter is most widely used.
The response is defined by the damping factor, DF= 2- R1/R2. So, don't we only need to change the values of R1 and R2 in Sallen-Key configuration to get Butterworth, Chebyshev or Bessel responses?

Regards
PG
 

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Jony130

Joined Feb 17, 2009
5,487
Q1 - Yes, the diagram show correct MFB low pass filter.
Q2 - We need to change all components value if we want to change the filter response.
 

Thread Starter

PG1995

Joined Apr 15, 2011
832
Hi

Could you please help me with the query included in the attachment? Thanks a lot for the help.

Regards
PG

PS: Even this page says that closest standard value to use is 680 ohm for 703 ohm. The table used in the attachment can be found here.
 

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PG1995

Joined Apr 15, 2011
832
Hi

To me, it looks like there is an error in the answer given by the solution manual. The manual says that nearest standard value is 720 ohm but 720 ohm isn't a standard resistor value. I was also saying that one could use 680 or 750 ohm resistor instead.

Regards
PG
 

Thread Starter

PG1995

Joined Apr 15, 2011
832
Hi

Would someone please help me with the queries included in the attachment? I would really appreciate your help. I don't think you would be able to see the text clearly in the attachment, therefore please use the given link for high-resolution image: http://img818.imageshack.us/img818/8164/wikipage8x.jpg. If you are not able to open the linked image, then please use this username:imgshack4every1 and password:imgshack4every1. Thank you very much.

Regards
PG
 

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steveb

Joined Jul 3, 2008
2,436
For Q1, it isn't a S-K filter and the text clearly says that. It calls it a VCVS filter (I assume that means voltage controlled voltage sourse). So, it's a different circuit, and you correctly noted that.

For Q2, one can derive these transfer functions by normal circuit analysis. Just write out the linear equations and solve by normal techniques. There are also shortcut methods to derive with, when you have experience.
 
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steveb

Joined Jul 3, 2008
2,436
Can you tell some more about this "shortcut".
I'm sure there are many that I can't think of off the top of my head.

However, the one that I see clearly is the non-inverting amplifier that is embedded within the VCVS filter. Rather than solving the whole circuit from an opamp model. Treat the opamp and the Ra/Rb resistors as a gain stage with Vout=(1+Rb/Ra)*Vp, where Vp is the voltage on the positive input to the opamp.

One that might be helpful (I haven't tried it) is to take C1, C2 and R2, which are in a PI or delta configuration and convert it to a T or wye configuration. It seems that one of the components will then be removable, and effectively one component is removed. I'm not sure how much this helps because the circuit is already easy to solve.
 
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Thread Starter

PG1995

Joined Apr 15, 2011
832
For Q1, it isn't a S-K filter and the text clearly says that. It calls it a VCVS filter (I assume that means voltage controlled voltage sourse). So, it's a different circuit, and you correctly noted that.

For Q2, one can derive these transfer functions by normal circuit analysis. Just write out the linear equations and solve by normal techniques. There are also shortcut methods to derive with, when you have experience.
Thank you, Steve.

I have tried to solve the circuit but my solution doesn't look right. Please have a look on the attachment and kindly help me. Thanks a lot.

Regards
PG
 

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Jony130

Joined Feb 17, 2009
5,487
Your DC gain equation looks good.
As for your transfer function, try further simplify the equation.
For example try divide the numerator and denominator by C1*C2*R1*R2*R3*Rb


We can solve this circuit directly




And nodal equation for V1 node

\(\frac{Vin - V1}{R1}=\frac{V1}{\frac{1}{s*C1}}+\frac{V1-Vout}{Rf}+\frac{V1-V2}{\frac{1}{s*C2}}\)

And additional

\(V1 = \frac{V2}{R2}*\frac{1}{s*C2}+V2\)

\(V2 = Vout *\frac{Ra}{Ra+Rb}\)

And after we solve this for Vout/Vin we get this

\( \frac{Vout}{Vin}(s) = \frac{C2 R2 (Ra + Rb) Rf s}{C1 C2 R1 R2 Ra Rf s^2+(-C2 R1 R2 Rb + C1 R1 Ra Rf + C2 R1 Ra Rf + C2 R2 Ra Rf)s+R1 Ra + Ra Rf}\)

Next I divide the numerator and denominator by (C1 C2 R1 R2 Ra Rf) and do some more fractional reduction we get to this

\( \frac{Vout}{Vin}(s) = \frac{\frac{C2 R2 (Ra + Rb) Rf }{C1 C2 R1 R2 Ra Rf}s}{s^2+\left(\frac{-C2 R1 R2 Rb + C1 R1 Ra Rf + C2 R1 Ra Rf+C2 R2 Ra Rf}{C1 C2 R1 R2 Ra Rf}\right)s+\frac{R1 Ra + Ra Rf}{C1 C2 R1 R2 Ra Rf}}\)

Also we can reduce this

\(\frac{C2 R2 (Ra + Rb) Rf }{C1 C2 R1 R2 Ra Rf}s\)

to this form

\(\frac{(Ra + Rb)}{C1 R1 Ra}s\)

And also this

\(\frac{R1 Ra + Ra Rf}{C1 C2 R1 R2 Ra Rf}\)

to this form

\(\frac{(R1 + Rf)}{C1 C2 R1 R2 Rf}\)

And the reduction of this expression

\(\frac{(-C2 R1 R2 Rb + C1 R1 Ra Rf + C2 R1 Ra Rf+C2 R2 Ra Rf)}{(C1 C2 R1 R2 Ra Rf)\)

I leave to you as a homework .
 

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Thread Starter

PG1995

Joined Apr 15, 2011
832
Many. many thanks for the help, Jony.

I have tried to solve the problem using a little different approach but my end solution has one problem. Kindly have a look on the attachment and please help me. Please use the link for high-resolution image: http://img560.imageshack.us/img560/7386/vcvsfilter.jpg. If you can't open the image,please use this username:imgshack4every1 and password:imgshack4every1. Thank you.

Regards
PG
 

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steveb

Joined Jul 3, 2008
2,436
Many. many thanks for the help, Jony.

I have tried to solve the problem using a little different approach but my end solution has one problem. Kindly have a look on the attachment and please help me. Please use the link for high-resolution image: http://img560.imageshack.us/img560/7386/vcvsfilter.jpg. If you can't open the image,please use this username:imgshack4every1 and password:imgshack4every1. Thank you.

Regards
PG
You should have been able to identify your mistake immediately. In every place there is a capacitor, you have 1/C instead of C. What does that tell you? Doesn't that tell you that you defined the admittance of the capacitor wrong?

Admittance of a capacitor is Y=sC, not s/C, right. You had the impedance correct as Z=1/(sC), but somehow when you inverted you got the wrong answer.
 
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