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High resistance components in low frequency cutoff butterworth

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mehrlin
#1
Jul3-13, 10:54 PM
P: 3
I just recently built a low-pass butterworth filter with a cutoff of 50Hz. When I looked to purchase the parts, I noticed that each inductor had a resistance of several hundred ohms. I put this information into my simulator program and the result was something that looks nothing like the expected output.

How would I design a new filter that takes into account the resistances native to each part?
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gneill
#2
Jul4-13, 07:02 AM
Mentor
P: 11,675
Quote Quote by mehrlin View Post
I just recently built a low-pass butterworth filter with a cutoff of 50Hz. When I looked to purchase the parts, I noticed that each inductor had a resistance of several hundred ohms. I put this information into my simulator program and the result was something that looks nothing like the expected output.

How would I design a new filter that takes into account the resistances native to each part?
Does the filter have to be non-powered? If not, consider an active filter version; You can even do away with the inductors! Look up the keywords: "Sallen–Key topology Butterworth".
skeptic2
#3
Jul5-13, 03:13 PM
P: 1,810
If you would be willing to post your circuit we may be able to offer suggestions. For instance, have you considered replacing your resistors with capacitors and your inductors with resistors?

mehrlin
#4
Jul8-13, 11:53 AM
P: 3
High resistance components in low frequency cutoff butterworth

Here's a picture. Unfortunately, going active will not be an option because supplying power to the op-amp will not be feasible.

I'm unfamiliar with switching out components like you mentioned. Can you elaborate?


The resistor at the end of the circuit is simulating the load, and so can't be changed
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Filter Picture.png  
The Electrician
#5
Jul8-13, 02:59 PM
P: 768
Since filters made with real inductors will always have this problem (although typically not as extreme as your problem), a technique is available to deal with it, known as pre-distortion.

A quick search turns up a master's thesis describing a method to pre-distort, and containing many references:

http://etd.auburn.edu/etd/bitstream/...pdf?sequence=2

Your particular inductors have very low Q, and it may not be possible even with pre-distortion to achieve your desired result with those inductors. You may have to get some better inductors.

How large is the signal you are trying to filter? It may be possible to harvest some energy from the signal itself to power a micropower opamp based active filter.
tfr000
#6
Jul9-13, 07:39 PM
tfr000's Avatar
P: 127
A 250 mH inductor? That's a very large inductance - it must be wound with miles of wire, hence the high resistance. This is usually the case with low freq stuff.
Baluncore
#7
Jul9-13, 11:09 PM
Sci Advisor
Thanks
P: 1,888
Increase the capacitance values by a factor of 10 and reduce the inductor values by the same factor.

Capacitors are cheaper than inductors so you might redesign to use a Pi configuration rather than a T. Then you will only need two inductors and three capacitors.
skeptic2
#8
Jul10-13, 11:16 AM
P: 1,810
Why did you choose these inductors?
Does your filter have high current requirements?
Baluncore
#9
Jul10-13, 12:47 PM
Sci Advisor
Thanks
P: 1,888
A low pass filter is a matching device. It matches the impedance of the source to the impedance of the load. Your source impedance is 1 ohm and your load impedance is 50 ohm. It is not just the cut-off frequency that determines the component values needed, it is also the terminal impedance values.
Your LPF transforms 1 ohm into 50 ohm. Is that realistic, or could it be changed to minimise inductor costs.


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