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Just think of it this way your not necessarily paying for the education what you are paying for is the chance to earn that fancy piece of paper that serves as proof that you know what your doing.

you're paying to take the tests.

And also paying to see those chicks around your campus...

I totally agree with you, bro. My professor just talks following the PPT without any useful information. She only knows to give us a lot of hard tutorial sheets...

I learn only topics of the lessons from my lecturers.

gootno

the video was 6 mins

Thirukumara racimo

Thirukumara racimo extremo

Thirukumara

It’s not all that bad. If you’re still having troubles RSD Academy has a simple and straight forward 9 part opamp playlist.

HollowHandPuppet it would amplify the difference if there was no feedback. If there’s feedback then output and inverting input monitor and adjust each other constantly

That must be a mistake he made right? Im pretty sure that a non-inverting opamp set up this way where R1=R2 would generate some kind of sinus-wave(ac) at the v0 node because of this "paradox". v0 would go from 2v to 0v to 2v and then back again, oscillating?

@@eyeballs97 assuming v+ was 1v and v- was 1v-6uV (there is a very very small difference between v+ and v-), then the amplified output would be 6v (assuming A=1,000,000). In the example, if Vo increased, it caused v- to increase, which will then cause Vo to decrease back to 6V (Vo's original value), this then makes v- decrease back to 1-6uV. At this point all voltages are back to its initial condition, so there wont be any oscillations. Only when Vo is once affected, then will it cause v- to change accordingly.

The resistors have the same value, so (1 + R2/R1) = (1 + 1) = 2 Vo/Vi = 2 => Vo = 2*Vi

Vout will decrease only when the input voltage decreases, they are not inversely proportional to each other, they are just in opposite phase.

You have the idea right. Any change to Vout causes V- to change, which pushes Vout in the opposite direction of its change. That is the essence of feedback. In real life, Vout will settle at exactly the voltage you want and will stay there. (It may waver a tiny bit, but too small to notice.)

Sexcyvidoes

by its definition : inverting input

v+ and v- are APPROXIMATELY 0 due to the virtual ground theory but they are actually a very very small number... but when the difference between the small number is multiplied by A then we get Vout.

+Hillary Lastname 0967777302

Hillary Lastname but then that does not explain why the output stays at 2v

ATM Jahid Hasan it's not exactly, remember that the Vo=2Vm was obtained by approximating A as large (see the previous video) so that we ignored a small corrective term the actual v- differes from 1 by something of order 1/A which is very small.

No it wont be zero , this would be true for open loop configuration . As soon as we add feedback this formula Vout = gain*(V+ - V-) no longer applies

Saptaki Debnath , yes that is besically making feedback.If you want to know more , you can check this video ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-nFD8O1Lp_1k.html

Yes Sir Khan please bring forth deeper lessons for us please... Multiplexer would be nice. Great tutorial we love it!

www.khanacademy.org/science/electrical-engineering/ee-amplifiers/ee-opamp/v/ee-feedback

Cindy Robinsin m

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