DigilentStudio

Question about Real Analog 1.18

Recommended Posts

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!

Share this post


Link to post
Share on other sites

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!

Kaitlyn

Share this post


Link to post
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now