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Low noise differentiator

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BrownOut

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you might want to ask your self, what does the bode plot of a differentiator look like.... you are really frustrating. look at any standard differentiator circuit or even a high pass filter circuit, there is a cut off frequency at some point.

http://upload.wikimedia.org/wikipedia/commons/6/63/Bode_High-Pass.PNG

it acts as an differentiator in lower frequencies... the cut off frequency is when it starts turning into a flat band.
Why would I ask mysefl that, then look at a bode plot of a high-pass filter? Here is a real bode of a diffeentiator.

Read Differentiator Bode
 

jewdai

New Member
@[email protected]; obviously your not an engineer. Try to implement that in real life. even the slightest resistance will change that bode plot.


at low frequencies my circuit will differenciate, at higher frequencies it is just a flat band.
 

BrownOut

Banned
Obviously, you're not an engineer, else you would understand what I wrote about parasitics. Neither would you have lectured me to review a differentiator plot, the linked a high pass filter. They are not the same thing! It might appear from the bode plot that a HPF is a differentiator up to the corner frequency, but it you were actually the engineer you claim that I'm not, you would know how incorrect that is.
 
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MrAl

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Why would I ask mysefl that, then look at a bode plot of a high-pass filter? Here is a real bode of a diffeentiator.

Read Differentiator Bode

Hello,


The perfect differentiator is simply "s" alone, but we dont always need a perfect "s"
in real life circuits.

The thing that you are missing is not that "s" is different than "s/(s+a)", we know that,
but that "s/(s+a)" can be made to look like "s" over the range of frequencies of interest
simply by making 'a' large enough so that it swamps out the denominator 's' term.
What we end up with is a slope that looks exactly like "s" divided by a constant gain
up until a certain frequency is reached when it flattens out and possibly also rolls off.
The gain can be made up for later.

All you need to do is compare frequency plots of "s" and "s/(s+a)" when 'a' is somewhat
largish and see that they behave almost the same up to a certain point, keeping in mind
that we usually also have control over the gain too.

BTW, for the sake of the OP...
One circuit that is often overlooked is the simple RC passive circuit. That's with C in
series with R and the output is taken across R. It doesnt have the problems that op
amps sometimes bring either, but if gain is needed it can be added in the next stage.
 
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BrownOut

Banned
The thing that you are missing is not that "s" is different than "s/(s+a)", we know that,
but that "s/(s+a)" can be made to look like "s" over the range of frequencies of interest
simply by making 'a' large enough so that it swamps out the denominator 's' term.
Actually, I got that part, after thinking about it for awhile. Was having a real hard time kick-starting my brain yesterday. The only part I didn't understand is where the OP was supposed to show me a differential plot, but linked a HF one instead. I thought it to be more illustrative to link a diff plot if that is the function we are discussing. But oh well...

The thing I was really not getting was why design a bandpass filter if you want a differential amplifier. It's OK because I did a review of differentiator theory, and realized that there are stability issues and increased HF harmonics otherwise. I've never needed to use a circuit like this, and so the last time I went over these issues was in college, way, way back. So I understand why the pole is needed; however, at frequencies near the pole, there will be errors, and so the not entire frequency below the pole is useful. Further, I assume the 2nd pole arises from the amplifier's first order effect, and not in there by design. Then, both poles should be well above the highest frequency of interest, and nyquest criteria satisfied at the crossover frequency.

OK, I've had more coffee, brain is functioning. I'm finally ready to discuss it ( I think ):)
 
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jewdai

New Member
well now that thats settled.


How can i reduce the amount of noise in the system and still differentiate? im kind of opposed to 20 gain stages to get a gain of 1000
 

MrAl

Well-Known Member
Most Helpful Member
Hi again,


If bandwidth limiting isnt good enough, then you can look for other ways of controlling whatever it is you are trying to control. BTW, what are you trying to control? What kind of process?
 

crutschow

Well-Known Member
Most Helpful Member
And what is the gain and signal bandwidth that you need. Without knowing that, a good solution can not be suggested.
 

jewdai

New Member
Audio, Cut off frequency is 40KHz. Its for a microphone controller. gain needed is 1000. I calculated RC = 330 ohms * 10nF
 

crutschow

Well-Known Member
Most Helpful Member
At what frequency do you need this gain?
 

crutschow

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Most Helpful Member
To get a differentiator gain of 1000 at 40kHz, you would need an op amp with a gain-bandwidth product of 40MHz, which is rather high. Not only are such op amps not common but they can readily oscillate unless extreme care is taken with their layout and decoupling.

A better approach would be to divide the gain between two op amps. The differentiator could have a gain of 30.2 at 40kHz with a feedback resistor of 10KΩ and an input capacitor of 12nF (for a standard op amp differentiator circuit). Follow that with another inverting op amp with a gain of 33.2. Each op amp needs a gain-bandwidth product of 2MHz minimum. To reduce high frequency noise you can roll off the second op amp with a LP filter capacitor across the feedback resistor somewhat above 40KHZ.
 

jewdai

New Member
To get a differentiator gain of 1000 at 40kHz, you would need an op amp with a gain-bandwidth product of 40MHz, which is rather high. Not only are such op amps not common but they can readily oscillate unless extreme care is taken with their layout and decoupling.

A better approach would be to divide the gain between two op amps. The differentiator could have a gain of 30.2 at 40kHz with a feedback resistor of 10KΩ and an input capacitor of 12nF (for a standard op amp differentiator circuit). Follow that with another inverting op amp with a gain of 33.2. Each op amp needs a gain-bandwidth product of 2MHz minimum. To reduce high frequency noise you can roll off the second op amp with a LP filter capacitor across the feedback resistor somewhat above 40KHZ.
I am actually quite aware of that. We are planning on using several gain stages to get the system gain to 1000.
 

crutschow

Well-Known Member
Most Helpful Member
Good. Then it sounds like you need no more advice.
 
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