The output is proportional to the integral but the voltage is not necessarely directly representing the exact value. One needs to multiply the output voltage with Rin*C/t
well my question is about the arrow 1:12 , my slide book shows it all goes to the left, i assume it goes all to the left to form some formula. can we do that?, and what is the theory behind that? and from what i learn, all the arrow goes to the right. anyway i was trying to simulate a derivative function using op-amp. thank you professor.
If you are referring to the direction of the input current (I_in), this would be for a positive input voltage and using conventional current flow. If the input voltage was negative, conventional current flow would be to the left (and of course electron flow is the reverse of conventional flow). There is some extra detail in my free op amps text book.
Apologies to bother you, but in the case where you have the feedback example, I was told by my professor that we need to solve for the impedance Z of Rf in parallel with the capacitor. So Z = Rf||(1/jwC). Then you use Kirchoff and get -Vin + Iin(Ri) + Iin(Z) + Vout = 0. Then you use the equation for current to solve for Vout, so plug in Iin = Vin/Ri. My doubt is, do we use Xc for the capacitor, or 1/2(pi)fC? Or when dealing with 1/jwC, you simply plug in the value of the capacitor in farads? Every resource and video I've looked at of the op amp integrator seems to take a simplified approach, where they never really take the impedance of the feedback resistor in parallel with the capacitor in consideration for their equations.
Xc *is* 1/(2pifC) so I don't understand the "or" in your question. Like all circuits, there are multiple ways of solving it. In this particular case, the op amp creates the virtual ground at the input, and you can use the current/voltage equation of the cap to solve for the output. The addition of Rf won't make much impact if the input frequency is well above Flow. If it's near Flow, then the accuracy of the result will suffer (and for higher accuracy, use something like your prof suggested). Finally, if the frequency is below Flow, then you don't have an integrator any more, you just have an inverting amplifier. Check of the frequency response graph starting at around 4:30 for a visual.
Hi Sir, for practical integrator op-amp by introducing the feedback resistor, the output equation will be different from one used in the earlier without feedback resistor. So why the same output equation is used?
As long as your signal is sufficiently above the lower limit frequency, the equation is the same. Below that, the circuit is no longer an integrator, it's an amplifier.
If we need to integrate non-ideal signals, although they are similar to sine waves, can we use the integrator circuit? Or do we have to use methods of numerical integration? Thanks!
@@alfonso4998 Overload=clipping. Like any amplifier, if the input signal is too large, the op amp will go into saturation and clip the signal. End result, an inaccurate rendering of the signal. It all depends on the frequenc(ies) of the input, the gain(s) at those points, and the amplitude of the input signal.
@@ElectronicswithProfessorFiore Then, if we use, for example, an input signal of 15V amplitude and we feed the operational with +-12V the output would saturate. Or, in a different case, with also +-12V in the operational, a signal whose amplitude is lower than 12V (5V, for instance), if the gain in the signal's frequency is 5V/V, output would saturate, too. Are these the different situations in which we can have clipping? Thank you!
@@alfonso4998 Pretty much, yes. Whenever the input amplitude multiplied by the gain at that frequency is at or above the supply rails, you've got clipping. The tricky bit with an integrator is that the frequency response falls at a rate of 20 dB/decade, so if you have a complex signal that contains many partials, it is a bit more involved to compute. Integration may change the shape of the waveform, not just increase it like a typical amplifier. For example, integrating a square wave does not yield a square wave; it yields a triangle wave. But if you're looking for a quick approximation, just find the fundamental (lowest) frequency of the input signal and then multiply it by the integrator's gain at that frequency. If it comes within a volt or two of the supply rails, you need to be concerned about potential clipping. That will get you close, assuming that the lowest partial has the highest amplitude.
Ideally, it would hold the level, but in practice not so much. First, no cap is perfect and will eventually leak. Second, a practical integrator will include Rf which creates a lower frequency limit and serves as a discharge path for the cap.
@@kalma4431 There is always some discharge because of leakage paths (the op amp and cap are not perfect). Further, don't forget that the inverting input is at a virtual ground which effectively removes Ri from consideration.
Yes, thank you. You have no idea how much you’ve helped me understand some basics that I’ve struggled with.
Thank you Professor Fiore, Excellent video with math,electronics and graphs .
You are welcome!
The output is proportional to the integral but the voltage is not necessarely directly representing the exact value. One needs to multiply the output voltage with Rin*C/t
-1/(Rin*C)
Thanks very much for these, this and the differentiator vid are both great.
well my question is about the arrow 1:12 , my slide book shows it all goes to the left, i assume it goes all to the left to form some formula. can we do that?, and what is the theory behind that? and from what i learn, all the arrow goes to the right. anyway i was trying to simulate a derivative function using op-amp. thank you professor.
If you are referring to the direction of the input current (I_in), this would be for a positive input voltage and using conventional current flow. If the input voltage was negative, conventional current flow would be to the left (and of course electron flow is the reverse of conventional flow). There is some extra detail in my free op amps text book.
Apologies to bother you, but in the case where you have the feedback example, I was told by my professor that we need to solve for the impedance Z of Rf in parallel with the capacitor. So Z = Rf||(1/jwC). Then you use Kirchoff and get -Vin + Iin(Ri) + Iin(Z) + Vout = 0. Then you use the equation for current to solve for Vout, so plug in Iin = Vin/Ri.
My doubt is, do we use Xc for the capacitor, or 1/2(pi)fC? Or when dealing with 1/jwC, you simply plug in the value of the capacitor in farads? Every resource and video I've looked at of the op amp integrator seems to take a simplified approach, where they never really take the impedance of the feedback resistor in parallel with the capacitor in consideration for their equations.
Xc *is* 1/(2pifC) so I don't understand the "or" in your question. Like all circuits, there are multiple ways of solving it. In this particular case, the op amp creates the virtual ground at the input, and you can use the current/voltage equation of the cap to solve for the output. The addition of Rf won't make much impact if the input frequency is well above Flow. If it's near Flow, then the accuracy of the result will suffer (and for higher accuracy, use something like your prof suggested). Finally, if the frequency is below Flow, then you don't have an integrator any more, you just have an inverting amplifier. Check of the frequency response graph starting at around 4:30 for a visual.
Hi Sir, for practical integrator op-amp by introducing the feedback resistor, the output equation will be different from one used in the earlier without feedback resistor. So why the same output equation is used?
As long as your signal is sufficiently above the lower limit frequency, the equation is the same. Below that, the circuit is no longer an integrator, it's an amplifier.
If we need to integrate non-ideal signals, although they are similar to sine waves, can we use the integrator circuit? Or do we have to use methods of numerical integration? Thanks!
As long as the signal is within the integration frequency range and doesn't cause overload, you can integrate it.
@@ElectronicswithProfessorFiore Excuse me, but what do you mean by overload? What causes output saturation?
@@alfonso4998 Overload=clipping. Like any amplifier, if the input signal is too large, the op amp will go into saturation and clip the signal. End result, an inaccurate rendering of the signal. It all depends on the frequenc(ies) of the input, the gain(s) at those points, and the amplitude of the input signal.
@@ElectronicswithProfessorFiore Then, if we use, for example, an input signal of 15V amplitude and we feed the operational with +-12V the output would saturate. Or, in a different case, with also +-12V in the operational, a signal whose amplitude is lower than 12V (5V, for instance), if the gain in the signal's frequency is 5V/V, output would saturate, too.
Are these the different situations in which we can have clipping? Thank you!
@@alfonso4998 Pretty much, yes. Whenever the input amplitude multiplied by the gain at that frequency is at or above the supply rails, you've got clipping. The tricky bit with an integrator is that the frequency response falls at a rate of 20 dB/decade, so if you have a complex signal that contains many partials, it is a bit more involved to compute. Integration may change the shape of the waveform, not just increase it like a typical amplifier. For example, integrating a square wave does not yield a square wave; it yields a triangle wave. But if you're looking for a quick approximation, just find the fundamental (lowest) frequency of the input signal and then multiply it by the integrator's gain at that frequency. If it comes within a volt or two of the supply rails, you need to be concerned about potential clipping. That will get you close, assuming that the lowest partial has the highest amplitude.
its lost again, i think we can't put image link in this comment. i'm trying to screen capture the problem and i upload it in imgbb.
What is output if input drops to zero. Will output remains const or output discharges?
Ideally, it would hold the level, but in practice not so much. First, no cap is perfect and will eventually leak. Second, a practical integrator will include Rf which creates a lower frequency limit and serves as a discharge path for the cap.
@@ElectronicswithProfessorFiore thank you. Sir. But why capacitor cannot discharge through R1 when feedback resistance is not connected ?
@@kalma4431 There is always some discharge because of leakage paths (the op amp and cap are not perfect). Further, don't forget that the inverting input is at a virtual ground which effectively removes Ri from consideration.
Thanks Professor🙏
Thank you professor
why you deleting my earlier question.
I didn't.
@@ElectronicswithProfessorFiore i'm really sorry.. i'll try asking again. maybe the network error. forgive me professor.
Pleas write V as V and not like a sin lol.