Opamp low-pass roll-off question

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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
 

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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.
 
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
 
ppereira said:
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
 
(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.
 
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)
 
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
 
Low frequency roll-off?

The circuit looks like it is drawn wrong. I assume you want to know how big the capacitor should be so that you don't cut off the bass notes, right? I don't think Sallen-Key has anything to do with his problem. Except to introduce him to more confusing math.

Jocko
 
Yes, the circuit is incomplete. doesn't show the output and input. Everything else is Ok, i think. All I want is to trim the frequency to have a flat response from 20HZ - 20KHZ. the 1.2K resistor on the feedback leg in conjunction with the 33uF cap sets the -3db point to 4HZ. The question was how to achieve a low-pass roll-off at 100KHZ -3db....

Thanks
 
Account Closed
Joined 2001
You've got more components than you need to achieve your goal.
Remove C2, R5, R4 and C3. Connect the junction of R3 and R2 to ground.
Lower the value of C1 to set you high-pass roll off to the desired frequency.
Connect C4 directly across R2 and select its value to achieve the low-pass roll off you desire.
I think you'll find C1 ends up about 1uF and C4 ends up about 100pf.

Cheers,

Davey.
 
Well I don't wear a white coat...but I am constantly trying to hide from those who do. ;)

I'm sure Nelson isn't suggesting that the DIY crowd are all ignorant of or incapable of understanding mathematics. Understanding how to calculate poles and zeros of a circuit is one of the core skills needed for audio circuit design and it really isn't all that difficult. Implying this is the preserve of white-coated boffins may lead the amateur to feel excluded and I reject this notion.

Explanation of complex notation can be found all over the internet. If you can understand Pythagoras' theorem (right-angled triangles) you can understand complex number notation. This notation is not intended to make it more difficult for the DIYer ( :rolleyes: ) but rather is intended to make the mathematics simpler for everyone to use.

By the way, "s" is shorthand for j.w (where J is root -1 and w is the signal frequency in radians per second. w is the same as 2.pi.f where f is the signal frequency in hertz).
 
Grab the tool that can do the work for YOU.

traderbam said:
Well I don't wear a white coat...but I am constantly trying to hide from those who do. ;)

I'm sure Nelson isn't suggesting that the DIY crowd are all ignorant of or incapable of understanding mathematics. Understanding how to calculate poles and zeros of a circuit is one of the core skills needed for audio circuit design and it really isn't all that difficult. Implying this is the preserve of white-coated boffins may lead the amateur to feel excluded and I reject this notion.

Explanation of complex notation can be found all over the internet. If you can understand Pythagoras' theorem (right-angled triangles) you can understand complex number notation. This notation is not intended to make it more difficult for the DIYer ( :rolleyes: ) but rather is intended to make the mathematics simpler for everyone to use.

By the way, "s" is shorthand for j.w (where J is root -1 and w is the signal frequency in radians per second. w is the same as 2.pi.f where f is the signal frequency in hertz).

- - Grab the tool, that can do the work for YOU.

Most things is easier to understand
once you take a practical example
and show how you can use it, as a tool,
with real values.
It is a way to make abstract symbols real.
A plain formula with a lot of strange "words"
can not easily be understood, as the context is missing.

So we can present a problem instead.
Where using that tool would be a good way to solve it.
Most problems can be solved in several ways.
One tool might fit one person, while another person
will prefer "his" tool.

The abstract mathematical way, would be the most inaccurate way.
While changing components and take measures
would be the exact way to do it.

Somtimes being somewhat primitive
turns out to more wise
and in tune with the real world out there.

/halojoy
Sweden
måndagen den andra december 2002 anno domino

There might be a formula somewhere, that I can use,
that describe what the perfect woman for me should be like, look like,
and where I with highest probability would find HER.

But we all know that is not how it works, for real.
Neither the time or the place would be correct.
She will not look like the image, and not be like she ought to be.
And above all that
the woman I love the most of them all
will be far from perfect
for me!
:D THAT'S LIFE :D
 
You don't set the "low-pass" point with a RC network.......that is determined by the unity gain point of the op-amp and the gain you select. The capacitor you are thinking about is for stabilty, not setting -3 dB point.

The circuit you have drawn is still wrong. Do it the way National tells you.

Yes, grasshopper.......you must don your white lab coat to achieve the Zen of "S".

Jocko
 
Jocko,

Help me out... What is wrong about the circuit? I've tried many different configurations (including this one) and they all worked fine... I drew the diagram in a hurry to exemplify my question so it doesn't really follow any convention... Forget about the drawing, it really doesn't matter. National's own specsheet states that a high frequency pole can be set by using this combination of cap/res in parallel with the feedback resistor....

The low frequency pole is determined by one of the two: the input cap C in conjunction with Ra (impedance) or by Ri in conjunction with Ci. The low frequency pole would the highest of the two. In my example it is set by Ri||Ci at 4HZ which gives an almost flat response from 20HZ up.

The reason I asked this question in the first place is because, intuitively (no white coats here :) it seems that if you narrow the bandwidth of the amplifier to the audible range this would result in some sort of improvement... Can anyone wearing a white coat tell me if I'm wrong/right?
 
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