# Determining BETA for an MFB filter

Discussion in 'Homework Help' started by kdillinger, Nov 6, 2009.

1. ### kdillinger Thread Starter Active Member

Jul 26, 2009
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Hello all,

I need some advice on an approach to determine β for a multiple feedback filter. My scanner is broke, but I have pages and pages of lengthy derivations and I am stuck.

My first approach was to consider two β's and them add them together using superposition. β1 would be at the R3/C1/R2 node (see attached) and β2 would be at the classic summing junction.

I was oh so close to Equation 16 from TI's application note (also in the attached).

Rather than ask for someone to help me with the math at this point, I would like to know if my approach makes sense. β1 + β2 = β(total) or as shown in the attached WORD doc.

The more I think about this, I believe I cannot do this but I am lost. Is there a better approach of how I can view the circuit? I can trod through the math, but I need some insight on how I can setup my nodes in these two feedback loops.

One is easy, two is driving me to drink.

-Ken

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2. ### The Electrician AAC Fanatic!

Oct 9, 2007
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Please post in some other format than .doc; .pdf preferably.

I viewed your .doc attachment on another computer. I don't see anything that looks like it came from a TI app note.

Can you provide a link to the TI app note, or the title of the note?

Last edited: Nov 6, 2009
3. ### Ron H AAC Fanatic!

Apr 14, 2005
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I don't see any equations in the attachment.

Jan 28, 2005
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hgmjr

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5. ### t_n_k AAC Fanatic!

Mar 6, 2009
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Re the attachment - capacitors CT & C3 seem pointless. If the op-amp is functioning 'normally' then the negative input should be driven to ground potential - thereby shunting those caps to ground.

6. ### kdillinger Thread Starter Active Member

Jul 26, 2009
141
3
I figured I would post the solution since I have been away due to personal errands. I finished this over the weekend if everyone is still interested.
I am posting only the first and last page of my derivation (total of 6).

And, yes, TNK, the capacitor will be shunted to ground when considering the signal gain, not the closed loop again or noise-gain as it is referred to in literature.

This exercise is to find β and 1/β and the capacitor from the inverting node to ground absolutely comes into play.

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7. ### The Electrician AAC Fanatic!

Oct 9, 2007
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I don't think you explained yourself very well in your very first post.

Are you assuming that the general feedback formula:

$Vout=\frac{A}{1-A\beta}Vin$

is to be taken as applicable to your filter, and then you are to find β as it appears in the general formula?

Or is the β you are trying to find something else?

8. ### kdillinger Thread Starter Active Member

Jul 26, 2009
141
3
Yes, β, as it appears in the general feedback formula. The feedback attenuation at the summing junction.

9. ### The Electrician AAC Fanatic!

Oct 9, 2007
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I had to go dig out one of my old textbooks to be sure of what is meant by the β network in a circuit like this.

It looks to me like this is a shunt-shunt feedback network, in which case the β network is determined by shorting the input, applying a test voltage (1 volt) to the part of the network connected to Vout, and calculating the current into a short to ground at the Vfb node. This is different from what you calculated.

Are you still working on this problem, and do my comments make any sense to you?