I have been trying to find some video lessons for optional amplifiers, But it turns out there is little in my country,, But in here I got so many I need, Thanks, man, :)
I appreciate that you are teaching well prepared and fully illustrated lectures, and I am thanking you for that. I suggest that you establish an independent course for Analogue Electronics. Best regards.
Before you started doing anything, I tried to see if I could come up with how to set the gain myself. And I came up with Gain=R1/R2+1, which is equal to R1+R2/R2. I also did it by simple ratios of the resistors, but added the 1 because the gain value is greater than the ratio of the resistors by 1. A gain of 1 would just be a voltage follower, while a gain of 2 doubles the voltage. So a ratio of 1 would be a gain of 2 because it halves the output voltage, while the voltages on both inputs must be the same, therefore, the voltage on the output must be double that of the input so as to make the halved output voltage equal the input voltage.
Since we are using ideal op-amps, we know that V+ = V-. The fastest way to get to the answer given, is to use the voltage divider and this property. V+ = V_in V- = (R2 / (R1 + R2)) * V_0 By the V+ = V- property, we get: V_in = (R2 / (R2 + R1)) * V_0 Getting V_0 as the lone term: V_0 = V_in * (R1 + R2) / R2.
Aditya Varshney Ok, but try to explain to a novice what a virtual ground is. For that you would need to know about differential amplifiers and feedback circuits. That is why the instructor chose the longer explanation.
I'm visually oriented but this time the math made more sense to me. I haven't seen another basic op amp video that derives the formulas this way 9for beginners) Nice!
Can confirm. My professor's a great teacher in person, but teaching online isn't his forte whatsoever. Videos like this are the only reason I have a clue what's going on.
Wow you are a genius. I am cracking my head with both difference arrangement of formula one is (voltage divider rule menthol) one is (KCL law to solve the equation) then suddenly u make the voltage crystal clear at the point you explain dependent (not really sure wat it means at first but you mention voltage divider rule!! Bingo !!! i got what you mean!!! thousand thanks
Hey, Here is only a problem of definition. In previous video, Vin defined as V+ -V-. But in this specific vedio, Vin was defined as V+ ONLY. I think if you use another name would be easier for understanding. In this video, V+ Only means Vsource , the voltage source you want to measure. (Here used Vin stand for Vsource, and you probably mixed it up). ;)
Yes the explanation in video is a bit confusing because the name of external voltage source is vin. it is much clearer if the name of external voltage source is changed to Vex for example. Since Vex is connected to pin v+ of amp, its clear that Vex = v+ (the point in video at 7:32 till 7:40)
I don't get how gain A can be "really, really big (10⁶)" and at the same time only the ratio of the resistors, which normally isn't a very big number. Can someone please explain this?
I am a first timer at opamps and was trying to get the non inverting amplifier to work. I connected V+ to +5v and V- to gnd( since I will be using it for DC). I am continuously getting a high output voltage of 4.44v. I even tried the offset null circuit but the output voltage remains constant at 4.44v. The feedback resistor is 1Mohm and R1 is 100kohm. Can anyone guide me here as to what could be going wrong? Is the negative power supply absolutely necessary for the opamp to work?
i have a huge question to ask which is bugging me, according to the first equation Vnot = A(V+ - V-), later it was said that V+ = Vinput . again in the last equation it was said Vnot= [(R1+R2) / R2]*Vinput. according to that if we compare the two equations A= [(R1+R2)/R2] . @?!@!@ can someone please explain what im missing. it would be much appreciated if you can help.
Jeff and Gary, Thank you for pointing out an error in the video.I mentioned "vin" twice with different meanings. What a goof. The video makes sense if you completely ignore the "green vin". Go all the way back to 1:58 and pretend I never wrote the green vin. I'm always talking about the blue vin (printed or cursive, its the same vin). I needed the green vin in earlier videos, but not this one. In this video, the only vin is the input to the circuit.
I just posted a repaired video to ruclips.net/video/ZpUZlIjcOzA/видео.html. It's a small change: Just before 2:00 there is no longer the notoriously confusing "green vin".
[] is correct. It might help if you think about the input current not as exactly zero but instead as close to zero (i.e. very very tiny!). Then, we can approximate and just call it zero.
it would be more elegant if you just divied the whole expression at 9:49 with a, then it would be 1/A and that is more logical to be able to just leave out
Can the narrator PLEASE do something where I don't have to hear all the saliva in his mouth? Every flick of the tongue and smack of the lips? It's disgusting.
over 40 years in electronics and this is best explanation of op-amp that a beginner needs...
john james true
I have been trying to find some video lessons for optional amplifiers, But it turns out there is little in my country,, But in here I got so many I need, Thanks, man, :)
Totally!
I appreciate that you are teaching well prepared and fully illustrated lectures, and I am thanking you for that. I suggest that you establish an independent course for Analogue Electronics. Best regards.
FINALLY! someone that explains really how an op amp is modeled and how to get to the gain formula
I've been searching for that explanation for 2 days. All I wanted was the math behind
@@aweirdguy9785 Yes. Unfortunately his algebra is a sloppy mess.
Before you started doing anything, I tried to see if I could come up with how to set the gain myself. And I came up with Gain=R1/R2+1, which is equal to R1+R2/R2. I also did it by simple ratios of the resistors, but added the 1 because the gain value is greater than the ratio of the resistors by 1. A gain of 1 would just be a voltage follower, while a gain of 2 doubles the voltage. So a ratio of 1 would be a gain of 2 because it halves the output voltage, while the voltages on both inputs must be the same, therefore, the voltage on the output must be double that of the input so as to make the halved output voltage equal the input voltage.
Doesn’t that depend on resistors
My god I've found the channel that can teach me this stuff. Everything was frustrating me and this is perfect
Much better than any of the explanations my professor has given
Great video, I couldn't understand half of what my lecturer was trying to get across with their explanation of op-amps but this has really helped
Love Your Videos Sir, Always prefer khan academy videos first . A big fan!
Since we are using ideal op-amps, we know that V+ = V-. The fastest way to get to the answer given, is to use the voltage divider and this property.
V+ = V_in
V- = (R2 / (R1 + R2)) * V_0
By the V+ = V- property, we get:
V_in = (R2 / (R2 + R1)) * V_0
Getting V_0 as the lone term:
V_0 = V_in * (R1 + R2) / R2.
Aditya Varshney Ok, but try to explain to a novice what a virtual ground is. For that you would need to know about differential amplifiers and feedback circuits. That is why the instructor chose the longer explanation.
Thank you for posting such a good lecture on non-inverting opamps.
I'm visually oriented but this time the math made more sense to me. I haven't seen another basic op amp video that derives the formulas this way 9for beginners) Nice!
Your videos can easily replace the best electrical engineering textbooks when it comes to teaching fundamentals.
Can confirm. My professor's a great teacher in person, but teaching online isn't his forte whatsoever. Videos like this are the only reason I have a clue what's going on.
Thank you so much ❤ you saved me
Wow you are a genius. I am cracking my head with both difference arrangement of formula one is (voltage divider rule menthol) one is (KCL law to solve the equation) then suddenly u make the voltage crystal clear at the point you explain dependent (not really sure wat it means at first but you mention voltage divider rule!! Bingo !!! i got what you mean!!! thousand thanks
Great tutorials. I think one of the best source around. Could you please release some tutorial about Oscillators - wave shapers - schmitt's trigger?
how does v+ become vin later in the equation
Sorry for answering 2 years later bro😂
It's simply that since there's no component between Vin and V+,
This makes them equal
This is an excellent video. Thanks for the effort :)
tysm u just fixed my nightmare, cant thank u enough!
in the previous video you said that Vin= V+ - V-
so why here Vin = V+ ??
Osama Hafez V- is not connected to a voltage source. that's why
Hey, Here is only a problem of definition. In previous video, Vin defined as V+ -V-. But in this specific vedio, Vin was defined as V+ ONLY. I think if you use another name would be easier for understanding. In this video, V+ Only means Vsource , the voltage source you want to measure. (Here used Vin stand for Vsource, and you probably mixed it up).
;)
Osama .... you can see that V- in 'grounded'...so V-=0....therefore final expression for Vin becomes Vin=V+ - 0=V+
Yes the explanation in video is a bit confusing because the name of external voltage source is vin. it is much clearer if the name of external voltage source is changed to Vex for example. Since Vex is connected to pin v+ of amp, its clear that Vex = v+ (the point in video at 7:32 till 7:40)
I was confused too, thanks for pointing this ;)
Thanks, explanation super clear.
Thanks a lot for your clear explanation!
Thank you !! Best explantation i found on the internet:)
Amazing explanation Sir. Thanks a lot!
Please give details about inverting and non inverting operational amplifiers in details
Thanks, had a good understanding from lectures but this blows most explanations out the water.
how do you know which resistor to put on top using the voltage divider? (Why R2 instead of R1 in the numerator?)
You put the resistor closest to ground as the top resistor in the voltage divider.
why can’t I have teachers that explain like this at school?
great job khan academy
You make life easy. Thank you very much 😁😁😁
Thanx alot....explained well😊😍
Why is v+ substituted for v-in at 7:46 of the video
I have the same doubt
Because v-in is going directly into the v+ input of the op amp.
Folks I understand this but we get the ratio of resistors. How do we estimate the actual values of the resistors needed?
I don't get how gain A can be "really, really big (10⁶)" and at the same time only the ratio of the resistors, which normally isn't a very big number. Can someone please explain this?
Ahh.. Is the gain of the opamp near infinite while the gain of the circuit is the ratio of resistors?
this is the neat thing about op amps you buy one and you choose the gain you want by choosing the values of those 2 resistors
and why is that? how is that possible??
1/0.0001 = 10000. Here 1 and 0.0001 are not so huge but their ratio is
Why it is V- = Vo * ( R2/(R1+R2), not V- = Vo * ( R1/(R1+R2))?
If you want to measure the voltage at V- .You need voltmeter to measure across R2 and ground right? So that's why the voltage at V- = Vr2.
what about ideal non-inverting? would they just be 0 since v- and v+ seem to be important here
Why is he able to say (R1+R2)/R2 is 2? That would only work if they have the same values of resistance right?
awesome explaination
Very nice video!
What if there's a resistor between the voltage source and V+? How would the equations change, then?
There's no current flowing so it won't do anything (is my guess...I'm just learning this stuff myself).
op-amps are still cool... thanks..:)
I am a first timer at opamps and was trying to get the non inverting amplifier to work.
I connected V+ to +5v and V- to gnd( since I will be using it for DC). I am continuously getting a high output voltage of 4.44v. I even tried the offset null circuit but the output voltage remains constant at 4.44v. The feedback resistor is 1Mohm and R1 is 100kohm.
Can anyone guide me here as to what could be going wrong?
Is the negative power supply absolutely necessary for the opamp to work?
if you feed opamp around 5V to 0. The opamp will compare 2.5V Dc best. Your signal Vp must be around 2.5 V for best . Otherwise you will always get 5V
I would to have a video with the digital circuits of opamp please
@7.42 v+ is vin but there is resister so no voltage drop ?
Why do we substitute V+ with Vin.
If V- = Vo * ( R2/(R1+R2)), shouldn't Vo = V- * ((R1+R2)/R2) ?
Yes its correct. V+=V-=Vin(source) in this op amp
Vin comes twice? one that is supplied as V+ and the other one as difeerence of V+ and V- . Whats the differnce ?
How come you changed v+ to vin?
There's a way to contact the professor? Great work!
Thank you very much
can someone tell me why gain will be a very very big number?
i have a huge question to ask which is bugging me, according to the first equation Vnot = A(V+ - V-), later it was said that V+ = Vinput . again in the last equation it was said Vnot= [(R1+R2) / R2]*Vinput. according to that if we compare the two equations A= [(R1+R2)/R2] . @?!@!@ can someone please explain what im missing. it would be much appreciated if you can help.
Jeff and Gary, Thank you for pointing out an error in the video.I mentioned "vin" twice with different meanings. What a goof. The video makes sense if you completely ignore the "green vin". Go all the way back to 1:58 and pretend I never wrote the green vin. I'm always talking about the blue vin (printed or cursive, its the same vin). I needed the green vin in earlier videos, but not this one. In this video, the only vin is the input to the circuit.
Thank you soo much Gary :)
I just posted a repaired video to ruclips.net/video/ZpUZlIjcOzA/видео.html. It's a small change: Just before 2:00 there is no longer the notoriously confusing "green vin".
thank you Willy McAllister :) your teaching is awesome
Sir, How did you the equation that v- =v°R2/R1+ R2 ?
Ronny Official it's a voltage divider equation. search on RUclips for voltage divider circuit, you'll find your answer there!😃
I dont understand how you can have i+ = i- = 0 you put a voltage/signal?
pebre79 it's a characteristic of the op amp, high input inpedance meaning there is a high opposition to current
[] is correct. It might help if you think about the input current not as exactly zero but instead as close to zero (i.e. very very tiny!). Then, we can approximate and just call it zero.
not=ex) I'm not a US citizen.
Variable in electrical engineering is naught...
Thanks pal this helped me a lot! :)
thank you sir.
How did he get 1
What is the use of R2 here
Pull down resistor.
Awesome
Amazing
sir can we consider (r1+r2)/r2 as total gain=A+(r1+r2)/r2
I was with you till about 6:20 and I have no idea where and how you manipulated the equation.
it would be more elegant if you just divied the whole expression at 9:49 with a, then it would be 1/A and that is more logical to be able to just leave out
explanation is not so good
check out Michel Van Biezen
much better
Can anyone test this out and give feedback? Search: 'Circuit Solver' on Google Play.
wtf
Can the narrator PLEASE do something where I don't have to hear all the saliva in his mouth? Every flick of the tongue and smack of the lips? It's disgusting.
mate its free stuff. if you don't like it the don't watch. that simple really
I agree.