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17th December 2008, 05:29 #1
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op amp integrator differentiator
Plz explain me physically
How the capacitor when connected at the output (feedback) of opamp acts as integrator & when connected at the input of opamp acts as differentiator?

17th December 2008, 05:29

17th December 2008, 08:56 #2
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op amp differentiator
Originally Posted by mess123
An opamp with neg. feedback has a voltage at the neg. terminal which is app. zero (virtual ground principle).
1.) Integrator: With this in mind you simply write the input current Iin=Vin/R and the output voltage Vout=1/C(∫Iout•dt). Then, replace Iout=Iin and you get the desired result.
2.) Differentiator: Do the same for Iin and Vout.
For my opinion, that is the most simple explanation. But keep in mind that not the capacitor ALONE acts as an integrator. It is the whole circuitry !!

17th December 2008, 08:56

17th December 2008, 18:49 #3
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function of differentiator in opamp
for an ideal opamp in negative feedback the voltage at the virtual ground node on the positive terminal is equal to the voltage at the virtual ground node on the negative terminal. assume virtual ground node is at 0V. therefore, voltage between output node to ground is equal to voltage between output node to virtual ground = voltage across the feedback capacitor(Vc). Vc is integral of the current through it (as a consequence of electrostatic theory). the current throught the capacitor is = current through resistor = Vin/R. hence the Vc = Vout is an integral of Vin with some proportionality constant.
if the capacitor is in the input and resistor in the feedback, the input current = Ic = CdVin/dt. this flows through resistor, hence a differentiator.

17th December 2008, 18:49

16th August 2010, 12:11 #4
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op amp integrator & differentiator
check the link, it would lead you to the concept,
why it is called integrator?
and, how, it is working?
http://www.circuitfantasia.com/circ...integrator.htm

16th August 2010, 12:58 #5
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Re: op amp integrator & differentiator
Perhaps it is helpful to someone, however I would clearly contradict this statement about the basic inverting OP integrator circuit:
"It will be simple, small, easy remembering but... nonunderstandable!"
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