Thank you, it was the best explanation on the integrator circuits that I have seen on UA-cam! What I found really useful is that whenever you were deriving the integral you spent time in properly explaining what is happening, like for example where virtual ground is and what's its purpose. And also it was really convenient that you stated the assumptions that ideal amplifier brings, because that is really important part for actually understanding what is happening. Keep on doing a great work!
This was so helpful - Thanks for posting! It's interesting to see this circuit keeps the charge rate at t-initial, and the time it takes for the output to swing from start to finish is always 1 x the RC time constant regardless of the voltage amplitude.
Good one! Great that you showed the effects the real world can have. I was kind of hoping you would show differentiating the output of the first integrator to make calculus complete. I'll have to try that myself.
Thanks for the video. I don't see you power on the opamp? Also, I copied your circuit parameters but the triangle output waveform got clipped on the top (flatting out). Do you know the cause of it? Thanks
Hello, i loved the clip and hope you're having a great day. When we use ideal integrator with universalOpAmp2 or even simply theoretically, how can we calculate Vcc needed for Op Amp in order for Op Amp not to go into saturation? Thanks!
Your VCC needs to be higher than the maximum you expect the op amp output to be. This is tricky for an integrator if you have a constant input, because you will eventually reach saturation unless the input changes.
@@ElectronXLab Thanks for the quick answer! So basically I have ideal integrator with RC=10ms and square pulse with 1 and - 1 V amplitude with 1 ms period like in the video. That means the maximum output is - 50mV right at the change of polarity. That means that theoretically i need +-50mV VCC in order for op amp to function properly?
Notice in the circuit that the Vin voltage is not being applied across the capacitor, it's applied across the resistor and there is a virtual ground point before the capacitor. The combination of the Vin voltage source and Ri controls the CURRENT through the capacitor, not the voltage across the capacitor. Since current is being controlled, it is what sets the capacitor voltage according to ic=C(dvc/dt)
Thank you, it was the best explanation on the integrator circuits that I have seen on UA-cam! What I found really useful is that whenever you were deriving the integral you spent time in properly explaining what is happening, like for example where virtual ground is and what's its purpose. And also it was really convenient that you stated the assumptions that ideal amplifier brings, because that is really important part for actually understanding what is happening. Keep on doing a great work!
This was so helpful - Thanks for posting! It's interesting to see this circuit keeps the charge rate at t-initial, and the time it takes for the output to swing from start to finish is always 1 x the RC time constant regardless of the voltage amplitude.
Good one!
Great that you showed the effects the real world can have.
I was kind of hoping you would show differentiating the output of the first integrator to make calculus complete. I'll have to try that myself.
Oh, I didn't even think of that, that's a good idea. I did just post a differentiator video though. It'll be out tomorrow.
Thanks for the video. I don't see you power on the opamp? Also, I copied your circuit parameters but the triangle output waveform got clipped on the top (flatting out). Do you know the cause of it? Thanks
These videos are really helpful thank you.
Glad to hear that!
How can I have the same output for vout and vin
I have a lab report that’s due tomorrow 😢
Thanks! It was so helpful!! btw, why did you add R3 and R5 in last circuit? When I did the simulation without adding it, the result was same
how to add ic bro?
Hello, i loved the clip and hope you're having a great day.
When we use ideal integrator with universalOpAmp2 or even simply theoretically, how can we calculate Vcc needed for Op Amp in order for Op Amp not to go into saturation?
Thanks!
Your VCC needs to be higher than the maximum you expect the op amp output to be. This is tricky for an integrator if you have a constant input, because you will eventually reach saturation unless the input changes.
@@ElectronXLab Thanks for the quick answer! So basically I have ideal integrator with RC=10ms and square pulse with 1 and - 1 V amplitude with 1 ms period like in the video. That means the maximum output is - 50mV right at the change of polarity. That means that theoretically i need +-50mV VCC in order for op amp to function properly?
what simulation software is it, plz?
LTspice
One thing I wonder, why will it linear?
A charging/discharging cap isn't linear so why here?
Notice in the circuit that the Vin voltage is not being applied across the capacitor, it's applied across the resistor and there is a virtual ground point before the capacitor. The combination of the Vin voltage source and Ri controls the CURRENT through the capacitor, not the voltage across the capacitor. Since current is being controlled, it is what sets the capacitor voltage according to ic=C(dvc/dt)
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