Nice video, just a minor issue: there was a mistake in the math at 8:10, which you corrected with another mistake at 12:30. At 8:30, the sign of the second term of the Vo should have been positive. At 12:35, the negative sign in the c2 term turns positive, cancelling the earlier mistake. All in all a good video- I learned something new.
Absolutely correct, my apologize, although, the final answer is correct as noted by a previous comment by Mike Lui "there was a mistake in the math at 8:10, which you corrected with another mistake at 12:30." Sorry for any confusion Thanks for watch man
We use ideal op amps to model the real world (i.e. non ideal), solving problems for non ideal op amps requires first using ideal op amps to model the problem and then, using real components and a breadboard and 5V's to get real world values. Hope this helps
Alex McLeod Oddly as it may sound, placing a resistor as you describe has no effect on the circuit. In other words, the process and solution would be the same. This is due to the fact that for ideal op-amp ZERO current flows into the amp, therefore ip is zero
A seemingly simple question: TL;DR: s being negative represents a stable system, s does not always represent jw (see root locus, control systems, step response, fourier, laplace) Remember that the equation you are looking at is in the frequency domain, to bring it into the time domain take the inverse laplace transform and remember that the inverse laplace transform of 1/(s+a) = e^(-at). Notice as "a" (the input ) in the time domain increases to infinity the equation (i.e., system) trends to zero. This fact/equation/system produces what is called BIBO stable system (bounded input bounded output) Your question touches on control systems, which is an entire field in its own right. What you are asking about is primarily the topic of root locus analysis. Here is a good tutorial on root locus from the University of Michigan: ctms.engin.umich.edu/CTMS/index.php?example=Introduction§ion=ControlRootLocus "No such thing as negative frequency assuming s=jw, right??" Correct, frequency is not negative, but "s" is and you must interpret its impact by using root locus. Don't assume that s=jw. Furthermore, Fourier s=jw, Laplace s= ơ +jw and in this case for the problem in the video s=ơ because jw=0 based on s=-5000 and s=-1000 from the values in the video. That's right s is purely real for the system in the video... which makes sense because it is a DC analysis, or in control system analysis it would be considered a step response analysis. Hope this helps Cheers! Tim from Doers & Thinkers
I hope you get paid a lot because you're like 1,000,000 times more coherent than my prof, thanks.
Opened google typed how to solve op amp circuits saw this video and then solved my homework. All in only 20 minutes! Great explaination!!!
THANK YOU!!!! This stuff is so intimidating, but really it's not that hard. You made it really straight forward!
Thanks for saying that Ellie :)
Thanks for the refresher example. Much appreciated.
Nice video, just a minor issue: there was a mistake in the math at 8:10, which you corrected with another mistake at 12:30. At 8:30, the sign of the second term of the Vo should have been positive. At 12:35, the negative sign in the c2 term turns positive, cancelling the earlier mistake. All in all a good video- I learned something new.
what does it mean to say impedance 1/sC in capacitor, I find it hard to interpret
He explained this at the beginning of the video .
Nice job; thank you!
Thanks my man. Huge Help!
Really like it! Thanks!
You made a mistake at 8:02. you took -Vo out as a common factor but didnt change the sign of the second term. it should be -Vo((1 + 40000SC2)/40000)
Absolutely correct, my apologize, although, the final answer is correct as noted by a previous comment by Mike Lui "there was a mistake in the math at 8:10, which you corrected with another mistake at 12:30."
Sorry for any confusion
Thanks for watch man
pls I want to send u a question to help me solve
absolute legend, thanks so much.
perfect! spot on, thx!
nice job bro :D thanks
hey thanks man this helped alot!
glad you liked it, thanks buddy
If i put vg in the plus , how the answer will change?
At your node where Vn is, why aren't you treating C2 and the resistor below it as a single impedence?
thanks man it was helpful. how do you solve problems for non ideal op amp.
We use ideal op amps to model the real world (i.e. non ideal), solving problems for non ideal op amps requires first using ideal op amps to model the problem and then, using real components and a breadboard and 5V's to get real world values. Hope this helps
Thank you! 😄
thank you.. helped a lot
What is this OP AMP Circuit called? or rather what does it do? is it a filter/amplifier?
Can you please explain to me how you rearranged the equation in the step before you got Vo/Vg at 9:00 ?
What will happen if there was a resistor attached to the Vp lets say 10kohm. Vp=Vn would it become complicated to calculate?
Alex McLeod Oddly as it may sound, placing a resistor as you describe has no effect on the circuit. In other words, the process and solution would be the same.
This is due to the fact that for ideal op-amp ZERO current flows into the amp, therefore ip is zero
Why do the poles occur at a negative number? No such thing as negative frequency assuming s=jw, right??
A seemingly simple question:
TL;DR: s being negative represents a stable system, s does not always represent jw (see root locus, control systems, step response, fourier, laplace)
Remember that the equation you are looking at is in the frequency domain, to bring it into the time domain take the inverse laplace transform and remember that the inverse laplace transform of 1/(s+a) = e^(-at). Notice as "a" (the input ) in the time domain increases to infinity the equation (i.e., system) trends to zero.
This fact/equation/system produces what is called BIBO stable system (bounded input bounded output)
Your question touches on control systems, which is an entire field in its own right. What you are asking about is primarily the topic of root locus analysis. Here is a good tutorial on root locus from the University of Michigan: ctms.engin.umich.edu/CTMS/index.php?example=Introduction§ion=ControlRootLocus
"No such thing as negative frequency assuming s=jw, right??"
Correct, frequency is not negative, but "s" is and you must interpret its impact by using root locus.
Don't assume that s=jw. Furthermore, Fourier s=jw, Laplace s= ơ +jw and in this case for the problem in the video s=ơ because jw=0 based on s=-5000 and s=-1000 from the values in the video. That's right s is purely real for the system in the video... which makes sense because it is a DC analysis, or in control system analysis it would be considered a step response analysis.
Hope this helps
Cheers!
Tim from Doers & Thinkers
You did not explain what it means to have a negative pole
Better if did not take so many shortcuts.
it seems that two wrongs do make a right! xD