Opamp low-pass roll-off question - diyAudio
 Opamp low-pass roll-off question
 User Name Stay logged in? Password
 Home Forums Rules Articles diyAudio Store Gallery Wiki Blogs Register Donations FAQ Calendar Search Today's Posts Mark Forums Read Search

 Solid State Talk all about solid state amplification.

 Please consider donating to help us continue to serve you. Ads on/off / Custom Title / More PMs / More album space / Advanced printing & mass image saving
diyAudio Member

Join Date: Oct 2002
Location: NJ
Opamp low-pass roll-off question

Can someone help me figure out how to use this formula to calculate a low-pass roll-off 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
Attached Images
 question.gif (24.3 KB, 605 views)

 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.
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 low-pass roll-off at 100khz -3b (that should allow for a flat response to 20KHZ?)
Attached Images
 non-inverting-opamp.gif (3.0 KB, 559 views)

 1st December 2002, 06:11 PM #4 On Hiatus   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 diyAudio Member   Join Date: Nov 2002 Location: Auckland, New Zealand If you use a sallen-key circuit, the values are much easier to calculate... just a thought
diyAudio Member

Join Date: Aug 2002
Location: Gaithersburg, MD
Quote:
 Originally posted by ppereira 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 low-pass roll-off at 100khz -3b (that should allow for a flat response to 20KHZ?)
Sorry, it wasn't my intention to lose you, only to
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 Bessel-type 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 diyAudio Member     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 low-pass 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 diyAudio Member   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

 Posting Rules You may not post new threads You may not post replies You may not post attachments You may not edit your posts BB code is On Smilies are On [IMG] code is On HTML code is OffTrackbacks are Off Pingbacks are Off Refbacks are Off Forum Rules

 Similar Threads Thread Thread Starter Forum Replies Last Post EUVL Pass Labs 117 9th May 2017 02:25 PM kiwi_abroad Pass Labs 1 17th September 2006 10:53 AM SSassen Subwoofers 5 17th March 2005 03:02 AM GrahamnDodder Chip Amps 6 23rd September 2003 09:32 PM

 New To Site? Need Help?

All times are GMT. The time now is 07:03 PM.