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If this is true, why would anyone bother with an op amp integrator?

The only difference that I can see is that an RC circuit would be more sensitive to drawn current than an Op amp integrator would be. Am I missing something?

- Thread starter TheDude_
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- #1

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If this is true, why would anyone bother with an op amp integrator?

The only difference that I can see is that an RC circuit would be more sensitive to drawn current than an Op amp integrator would be. Am I missing something?

- #2

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Also op-amp integrator can be faster. If you have a high speed current pulse, you can see the op-amp step up much faster and settle faster. Also, op-amp integrator can be design to hold the output level much better by having a very high impedance discharging path. Where as the RC LP filter get discharged by the R. It is not unusual to have op-amp integrator that can react to nS pulses and hold for mS.

- #3

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It is true that an unloaded RC circuit has an integral transfer characteristic.

It is also true that it can be at least as fast as an op amp, it just depends upon realising suitable circuit values.

But connect suitable circuit values for integrting the frequencies of interest to an input and an output and what happens?

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Filters are fun.

- #5

jim hardy

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If this is true, why would anyone bother with an op amp integrator?

there are times you need a true integration not the approximate one you get from a simple RC circuit .

An example would be a process controller with integral term.

Sometimes an approximate integral is close enough for what you're trying to measure and that's where you might get away with the simple RC.

For example if you need the integral of a sinewave a simple RC can get you pretty close.

But if you need the integral of a step function, whhich would be a straight line of constant slope, observe that the RC only approximates that for a fraction of a time constant.

It'd be a 'Judgement Call" by the fellow doing the work..

- #6

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Use of integrators as basic building blocks of filters was a significant advance in analog integrated circuit technology because these "state variable filters" can be made very insensitive to component tolerances. Discrete high order filters required precise component values, often requiring manual calibration. Components inside of integrated circuits are much poorer in quality than what can be achieved with discrete components, so implementing anything other than a very basic low order filter was out of the question. Use of integrator based state variable filters allowed precise, high order, filters to be implemented in silicon thanks to the insensitivity to component tolerance inherent in this topology. These used to be called KHN filters back in the day. Kerwin, Huelsman & Newcomb developed this specifically for the analog IC use. I have a copy of their original paper somewhere. If I find it I will post it for those interested.

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- #7

psparky

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Like said above, an op amp can give you gain. Like when you turn up the volume on your stereo. This gain translates into + dB.

An even better example would be an equalizer on your stereo. When you adjust the different frequencies, you will either have a rise in dB or a fall in dB at that specific frequency. This can not be done with an RC circuit...but can certainly be done with an op amp on each frequency. Set your break frequency for that amp and then put a variable resistor (knob or adjustable button on EQ) in the feedback loop.

An RC circuit will only give a gain of 1 or 0 dB.

An even better example would be an equalizer on your stereo. When you adjust the different frequencies, you will either have a rise in dB or a fall in dB at that specific frequency. This can not be done with an RC circuit...but can certainly be done with an op amp on each frequency. Set your break frequency for that amp and then put a variable resistor (knob or adjustable button on EQ) in the feedback loop.

An RC circuit will only give a gain of 1 or 0 dB.

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- #8

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Try different values in the scope app.

http://www.st-andrews.ac.uk/~jcgl/Scots_Guide/experiment/integ/int.html

- #9

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yungman: I could be off base here, but doesn't the value of R*C determine how fast integration takes place? I.e. if the value of RC determines the 3db cutoff frequency that is passed in the low pass filter, why would a op amp be 'faster' than a simple RC circuit?

Also: I have a rudimentary understanding of impedance, but I'm not sure how the RC Low Pass filter 'would get discharged by the R.'

jim hardy: Along the same lines, if both circuits integrate, then I'm confused why one would integrate 'faster' than the other - they both would integrate the step function in the same way, and thus would only be correct for a part of the signal, right?

- #10

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Consider an RC integrator.

The governing equations are

[tex]\begin{array}{l}

{V_{in}} = IR + \frac{1}{C}\int {Idt} \\

{V_{out}} = \frac{1}{C}\int {Idt} \\

\end{array}[/tex]

Now we can only make this an integrator of input voltage if we make the time constant RC very long compared to the period of the input wave, resulting in a small voltage across the capacitor.

Under those circumstances we can write

[tex]\begin{array}{l}

{V_{in}} \approx IR \\

hence \\

I = \frac{{{V_{in}}}}{R} \\

hence \\

{V_{out}} = \frac{1}{{RC}}\int {{V_{in}}dt} \\

\end{array}[/tex]

Which is the equation of the integrator we seek.

But the condition on RC places severe practical constraints on the values and frequencies we can adopt.

Further, as I have already mentioned, the input and output are also affected by the impedances of the preceding and following circuit blocks.

- #11

jim hardy

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Her's the "words" to go with it.

If you build both an RC and an opamp integrator and hand them a step input,

their outputs will both start to change with same slope because initial current flow into both capacitors is the same.

After a short while the RC integrator's slope will decrease -

that's because current into the RC's capacitor = (Vin - Vcapacitor)/R

but the current into opamp's capacitor = Vin/R

so the RC briefly approximates integration but the opamp does it exactly.

That's what Studiot was telling you when he said

and that's the significance of Yungman's commentNow we can only make this an integrator of input voltage if we make the time constant RC very long compared to the period of the input wave, resulting in a small voltage across the capacitor.

opamp holds one end of resistor at zero so current into cap = Vin/R not (Vin-Vcap)/R. With the latter, as Vcap goes up the slope gets smaller - less current into cap.For one, op-amp integrator can hold virtual ground at the integrator input,

So the latter gives you an exponential approach to Vin not the straight line you get from true integration.

Further the RC integrator won't "hold" when input voltage returns to zero, the cap discharges back into signal source through input resistor.

I once built an opamp analog integrator that'd hold last value within 0.1% overnight.

hope you have enough offerings here topiece together a mental model that works. When it agrees with Studiot's math then it's probably right.

old jim

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Thank you both! I understand much more clearly now. I promise I read each response several times...

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