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1st December 2002, 03:28 PM  #1 
diyAudio Member
Join Date: Oct 2002
Location: NJ

Opamp lowpass rolloff question
Can someone help me figure out how to use this formula to calculate a lowpass rolloff for an opamp. This was taken from the datasheet for the National LM3875. I just can't figure out where the "s" variable comes from....
Help! Thanks 
1st December 2002, 04:49 PM  #2 
diyAudio Member
Join Date: Aug 2002
Location: Gaithersburg, MD

I think they are using the "s" in the sense of the
Laplace transform. The impedance of the capacitor can be expressed 1 / (sC) The formula is the overall impedance of the parallel combination of the two branches. There was some fancy factoring to make the resulting expression look like that of a couple of parallel resistors: ab / (a + b). I'm not sure why they left the s in there like that and then set the whole thing equal to "f", though. I would take s itself to mean "j*2pi*f" where j is the square root of 1. E. 
1st December 2002, 05:41 PM  #3 
diyAudio Member
Join Date: Oct 2002
Location: NJ

You lost me I can add, subtract and, if I make an effort, multiply and divide...
Can you please look at this schematic and tell me which values I should use to set the lowpass rolloff at 100khz 3b (that should allow for a flat response to 20KHZ?) 
1st December 2002, 06:11 PM  #4 
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Join Date: Nov 2002

Refering to first diagram
first take the values at 1 kHz
at that f the resistance of C (also called reactance Cx) is very large, almost infinite, it blocks currents with low freq almost totally so gain is like (20k/1k)+1=21 at 1 kHz the Cx of C gets smaller the higher f at 100kHz it is 100 times smaller than at 1 kHz say it is 10 kohm resistance of Cf+Rf2= 30kohm Rf1=20kohm the resulting resistance is for the signal currents these 2 resistances in parallell, 20k//30k gain at 100kHz is (resulting resistance/1k)+1 somebody else will have to fill in here with more exact info and formulas 
1st December 2002, 07:11 PM  #5 
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Join Date: Nov 2002
Location: Auckland, New Zealand

If you use a sallenkey circuit, the values are much easier to calculate... just a thought

1st December 2002, 07:41 PM  #6  
diyAudio Member
Join Date: Aug 2002
Location: Gaithersburg, MD

Quote:
try to explain where they got the "s" from. Optical brings up a good point: why not try a different topology with the same part? If you set up a Besseltype of response you can get pretty much linear phase (constant group delay) and you won't need to shoot for so much bandwidth. (More bandwidth is not necessarily better.) Erik 

1st December 2002, 07:50 PM  #7 
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Join Date: Aug 2002
Location: Gaithersburg, MD

(Ooops... ran out of editting time. )
R3 should go to ground, by the way. If you set the equation they give in the app note equal to the value of R3, you will get a value of "s" that satisfies the condition. It would make it easier if you set one of C or R to a value before hand. Pick a value of part that you have laying around or make the two Rs the same. E. 
1st December 2002, 09:28 PM  #8 
diyAudio Member
Join Date: Jan 2002
Location: Earth

In the datasheet formula quoted there is a zero (s+1/(Rf2.Cf)) and a pole (s+1/[Cf.(Rf1+Rf2)]
These take the form of (S+a) and the 3dB points will occur when f=a/(2.pi) The pole or lowpass corner frequency will be f = 1/[2.pi.Cf.(Rf1+Rf2)] and the zero at f = 1/(2.pi.Cf.Rf2) 
1st December 2002, 10:35 PM  #9 
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Join Date: Oct 2002
Location: NJ

Thanks everybody. It seems that if I apply traderbam's formula  f = 1/[2.pi.Cf.(Rf1+Rf2)]  and use a 1.2K resistor for Rf2, leave Rf1 at 30.1K and a 50pf Cap for Cf I should get around 100KHZ.... Did I get this right?

1st December 2002, 10:52 PM  #10 
diyAudio Member
Join Date: Oct 2002
Location: NJ

Forgot to ask. What does this do to the high frequency pole? It was set to (almost) exactly 4KHZ by Ri and Ci (20KHZ at .25db) fc = 1/(2p Ri Ci). Does it stay the same after I add Rf2 and Cf? I should have stated my goal when I started this thread: a high frequency pole at 4KHZ and a low frequency pole at 100KHZ....
Thanks again. Paul 
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