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Question about Real Analog 1.18


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The Following Question was posted on one of our Real Analog videos:

I liked your lectures on the electric circuits.Indeed these are very informative and add a lot to one's knowledge.However I got confused in watching the lecture( 1.18) on "Active First Order System Example". You presented two active circuits.1.RC circuit connected to the +ve terminal of the Op-Amp.For this you told that there flows no current into the Op-Amp and this means there is no current in the feed back resistor and hence no voltage drop acros the resistor and so Vo=Vc(time15:25). But when you took the 2nd circuit ie the integrator then in that you told again that there is no current into the Op-Amp but now you changed your statement.You applied KCL there and told that it's equal to the current flowing into the capacitor(time 18:45).Now these two statements are contradictory.So which one is true?

Please help clear this up.

Thank you!

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Hi there Digilent Studio, 

I can certainly see the confusion there. In the earlier section the feedback loop has only one possible loop. Through the +ve terminal of the opamp through the resistor and connected to the output of the opamp. However, in the second circuit you discussed there are two possible loops for the feedback loop. Yes it does connect to the positive terminal of the opamp, but it also connects to the loop going through the resistor and Vin. So, although there is still no current into the opamp, the capacitor has current flowing through it, because of the other path.

I hope this clears that up!


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