Glad they are helpful! I recently made most of my screencasts for my classes public and renamed titles to be more helpful. Hopefully this will allow you to find other videos. I don't edit, so occasionally there are small errors. These videos are made to go with a face-to-face classes where we can follow up in-person. However, please use as fitting!
dumb questition:at the differentiator circiut...why is it not the spike at the Output goes to negative when the squarewave goes to positive,because the right side od the capacitor charge the opposite of Vin?
Instead of making the unhinged assumption, in the case of the integrator, that the output voltage must be very small compared with the input, you could simply write, say for the positive half-cycle of the input, the capacitor voltage as a function of time: Vo (1 - exp(-t/RC)) and make the appoximation for t
I love the ''charge now!...now discharge now!" explanation in 14:24 xD Thank you so much!
How can we access the other videos? Thanks for uploads very neat explanations!
Glad they are helpful! I recently made most of my screencasts for my classes public and renamed titles to be more helpful. Hopefully this will allow you to find other videos. I don't edit, so occasionally there are small errors. These videos are made to go with a face-to-face classes where we can follow up in-person. However, please use as fitting!
awesome
dumb questition:at the differentiator circiut...why is it not the spike at the Output goes to negative when the squarewave goes to positive,because the right side od the capacitor charge the opposite of Vin?
solved through Spiice Labs
Instead of making the unhinged assumption, in the case of the integrator, that the output voltage must be very small compared with the input, you could simply write, say for the positive half-cycle of the input, the capacitor voltage as a function of time: Vo (1 - exp(-t/RC)) and make the appoximation for t
Tq
Hiii
can u help me with something!!
kya chahiye bhai?