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Old 10th August 2014, 09:03 PM   #11
DF96 is offline DF96  England
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Two cascaded first order filters make a second order filter. However, it will have lowish Q - which may or may not be what is wanted.
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Old 10th August 2014, 10:17 PM   #12
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I'm thinking maybe what you really need is a better calculator. This might help.

http://www.learningaboutelectronics....calculator.php

<< also to attenuate the signal level a bit to help match with the midrange >>

This sounds a little bit strange, at least to me. It sounds like something best done by the amplifier or preamp, not a job for filters.
.

Last edited by bentsnake; 10th August 2014 at 10:26 PM.
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Old 10th August 2014, 10:37 PM   #13
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<< ...a first-order low pass filter can be converted into a second-order type by simply adding an additional RC network to it and the more RC stages we add the higher becomes the order of the filter... >>

This refers to simply a daisy chain of filters, just as you have at the input of the circuit you posted. Its purpose is to get a faster rolloff, but at the expense of more distortion.

But then, "distortion" is a relative kind of thing. Nothing says the distortion from a daisy chain of filters has to be audible, or significant.

However, faster rolloff is not going to attenuate the overall signal. Signal level will remain the same, only that the unwanted frequencies will disappear faster.
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Old 10th August 2014, 11:04 PM   #14
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<< to attenuate the signal level a bit to help match with the midrange (less sensitive)...about how much attenuation will layout this give...? >>

Not very much, since after the however-many-pole filter you're apparently sending the signal to an amplifier.

All considered, it seems to me that what you need is...wait for it...a volume control. Or in this case you'd call it a level control. This was referred to in passing by srenten, but then the conversation went in another direction.

I've sketched one up, it's posted below. The 20k value is arbitrary, it would depend on the rest of your circuit. The Log specification is also arbitrary, I just said that because it's a volume control, but I suspect a linear pot would serve as well.
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File Type: jpg input_filter_a.jpg (60.5 KB, 35 views)
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Old 10th August 2014, 11:22 PM   #15
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PS: I really do doubt that you need a faster rolloff. I say that mainly because speaker crossover circuits are not all that precise, and neither are speakers. Besides, you move a couch or a chair and the whole thing starts all over again.

Last edited by bentsnake; 10th August 2014 at 11:31 PM.
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Old 11th August 2014, 12:01 AM   #16
Pyramid is online now Pyramid  United States
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Quote:
Originally Posted by clm811 View Post
Hello.

I was hoping for some help with a 2 pole low pass filter/attenuator.
With your circuit as drawn, the 3dB cutoff of the filter will be ~145 Hz, not 400Hz. It will also have quite a bit of overshoot (ringing).

If it were me, I would use the circuit in the attached diagram. 400Hz Fc with a Bessel response (no overshoot, constant group delay). If you want an LR response, then make both caps 220n.
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File Type: pdf 400Hz LPF.pdf (20.3 KB, 6 views)
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Old 11th August 2014, 12:33 AM   #17
sreten is offline sreten  United Kingdom
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Quote:
Originally Posted by rayma View Post
Cascaded second order filters like this are common in tube amp and preamp power supplies. The filtering would be very inadequate if they were only first order.
The s domain transfer function is: (1/(1+sRC)) * (1/(1+sRC)), or 1/(1+2sRC+(sRC)^2). This is clearly second order (notice the s^2 term), with a 12dB/octave, or 40dB/decade, asymptote.
Hi,

Techobabble doesn't change the fact is clearly isn't second order,
in any rigourous sense, as its not a perfect cascade in any sense.

Very obvious looking at the circuit, what are you people missing ?

rgds, sreten.
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Last edited by sreten; 11th August 2014 at 12:52 AM.
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Old 11th August 2014, 02:38 AM   #18
Pyramid is online now Pyramid  United States
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Who cares what order it is in any rigorous sense? The OP asked for help with designing a circuit. Do you have anything helpful to post as a response or are you just going to keep flogging a dead horse?
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Old 12th August 2014, 05:19 PM   #19
DF96 is offline DF96  England
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Quote:
Originally Posted by rayma
The s domain transfer function is: (1/(1+sRC)) * (1/(1+sRC)), or 1/(1+2sRC+(sRC)^2).
I think you are ignoring the way the second CR loads the first CR, and the 20k loads the second CR. Your formula would be correct if there were buffers between them and after them.

My algebra (ignoring the 20k load) gives 1/(1 +3sCR + (sCR)^2). This is second order Gaussian?
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Old 12th August 2014, 06:10 PM   #20
rayma is online now rayma  United States
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Quote:
Originally Posted by DF96 View Post
I think you are ignoring the way the second CR loads the first CR, and the 20k loads the second CR. Your formula would be correct if there were buffers between them and after them.

My algebra (ignoring the 20k load) gives 1/(1 +3sCR + (sCR)^2). This is second order Gaussian?
Yes, that was for the RC - buffer - RC version, which gives a coefficient of 2 on the first order term, better than the 3 of the unbuffered version. An active second order filter will be better than either of these near the knee.
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