Adam I ..... Isn’t that the way learning seems to take place. I too experienced the same feeling from my days (waaaaaay) back in school. I just did not understand why my teachers couldn’t have taught in this way. Even alot of books are hard to learn from because of this......I guess shit happens ....
yea sometimes i wonder , are these college professors really interested in teaching because so many never bother to boil things down into simple form which i thought was the art of teaching. wtf. if they are not capable of doing this then maybe they shouldnt be teachers.
I have watched a lot of instructional videos which use a technique that you also use and I often add my comment " Please use a tripod". My reasoning is mostly to keep me from becoming dizzy or sea sick. I however complement you on a quite steady hand held camera not producing nausea therefore allowing me to focus on your superb, clear, informative content. Thanks for your well done videos. I always learn from you whether or not I am viewing a topic that I think I already know. Another of your 'characteristics' that I appreciate is that you do not waste words using extraneous babble that distracts from the topic. THANKS again. Many of those other videos I can't return to for more information but I come back to yours over and over. Keep them coming!
I don’t usually comment on videos, but I’ve been looking around for a bit for videos to help me make sense of op amps, and this really did it, thank you so much!
Currently an EEE undergraduate student; this video really helped me understand how the op-amp circuit works. Thank you very much for your time and effort!!!
I'm glad this video helped you Ryan. Let me know if there are other topics that you'd like to see a video on. And of course, tell your fellow students too!
I'm glad you like the video. I hope you enjoy the other 200+ that I have on my channel. An index / table of contents can be found here: www.dorkage.com/youtube/W2AEW_video_index.pdf
I have no idea why I found your channel just now.. I have been struggling with OPAMPS and with your clear lectures and actual DESIGN showing what is happening really helped me alot! KEEP UP THE GOOD WORK, YOU, YOU SIR! AND please DO MORE!!!
Hello sir, I am a beginner and I just want to say thanks for the introduction. I have been reading up on op amps and have not gotten it through my head, and most the videos here on youtube make it too complicated and frustrated. this is by far the best video I found on op amps. So thank you and hope you have a wonderful time
Excellent Tutorial showing how to put theory in to practice. I like your choice of circuit as it almost cover everything you need to know on opamps. your results satisfy the formula fully. Vout= -Vin(Rf/Rg)+Vref(R1/R1+R2)(Rf+Rg/Rg) by substituting we get Vout= -4v(20K/10K)+10v(2k2/2k2+2k2)(20k+10k/10k) = 7 Volts which matches your Vout. Thank you and we like to see more tutorials please ! :-)
this is wonderful how you explain, i'm a physics professor too. for fundamental physics and quantum electronics. great job! I loved the experiment you make it after theory solving
This is a difference between a newbie on Monday but expert on Friday,rushing to post a video about something that he slightly licked himself and someone who has a long years indepth knowledge.You have my respect Sir.
I'm amping up (pun intended) for my Biomedical Instrumentation final and this was a helpful refresher on some of the basic stuff. You have a very natural, logical and application-based approach to explaining these circuit configurations - it would be AWESOME if you did a few more. Thanks so much!
Why does everyone have to over-complicate this stuff, its like they make this stuff hard to learn on purpose. Took me so long to find a good explanation online as to why there is a negative voltage at the output of the inverting amplifier. You explained it well by assuming there is a voltage at (-) and that the output has to go negative in order to make that pin into a virtual ground. Other people have tried explaining it to me by FIRST saying there is a virtual ground and it never made sense to me that way. Thank you so much
This Explanation is spot on!! It saved me hours of time. And we know that Time is money. So this dude saved me a lot of money. Please guys fully view the ads in the video and click on them too, this will support the guy who made this video. Thanks
@9:00 we see the date on your scope: 27-Jan-13. This vid was uploaded the day before that: 26-Jan-13. This shows that you're ahead of your time. Thanks for the excellent video.
I'm glad you liked it. Please feel free to use my description with your students, or even share my video (and my channel) with your students. And of course, if there's a topic that you think would make a good video (that would help your students) please let me know.
You're a talented engineer and teacher. I enjoy watching your videos because you offer a wide variety of topics and knowledge in simple and interesting ways. Thanks a lot.
Once again, greatly explained, greatly shown, just great, keep it up!!! Personally I am unable to design analog stuff from scratch, but take plenty of circuit bits and pieces from here and there on opamps and anything I can... and put them together to work. With this simple explanation one can at least understand more about what is happening on circuit when opamps are involved. Thanks once more!!
Thank you so very much for this video (and all the other ones too)! I'm just so glad that knowledge doesn't weigh anything, if not I might have to go on a diet... This really helps a lot! (BTW, thanks to you I calibrated my scope probe, it's really cool to know those kind of things you normally don't look into as an amateur). This is a top youtube channel!
The way you explaining is awesome ...:) and found best of RUclips in these. It's my humble request that, it will be very helpful if you can explain how to find the input impedance of op-amp.
Wow very interesting analogy! I will keep that in mind. What really got me into circuits was the physical analogies that helped explain them. Like the water analogy etc... so this helps. Thanks.
Thanks for posting this, we flew through op-amps when I took circuits two years ago and this made everything make sense again. I might actually finish this controls lab now!
Alan. You do a perfect job of explaining it. I like to watch your videos, it has helped me very much in understanding how component's work. Keep up the awesome work.
Hey Alan, Watched several of your videos today trying to knock some of the rust out between my ears. Just recently got my technician ticket and getting the bug again to return to electronics a bit as a hobby again some 30+ years later, so many cool things to play with! You do an awesome job clearly explaining, thanks for all your efforts! Best, - Bill KI7QOA
Exactly what I needed to know. I am an amateur radio operator who has used a simple dual op-amp ckt to pre-amplify microphone inputs. The output level was too high and adjusting the second stage of the op-amp feedback to lower the overall gain was the right move. Later, I nearly fried the input dual op-amp ckt in one of my radios. I still have near-perfect operation, but the is now a 2-3khz noise present. I need to understand what component might have been fried in the ckt that needs replacement. This ckt, by design, has a single component that is two diodes in parallel and reversed from each other shorting the inverting input and the output. in parallel with that diode pack, in the feedback line is a resistor and a capacitor (100k & 390pf). I am still trying to figure that out, but this helps (current limiting?). Thank you for the short, practical explanation.
Strictly speaking, the statement at @1:32 should be: "negative feedback keeps any CHANGE in input voltage appproximately equal to zero". In many opamp circuits, V(+) is not grounded and the negative feedback will try to make V(-) - V(+) be zero.
Thank YOU! Using O-SCope to present what You are talking about is the way this topic should be explained. If you ever have time for more videos about Op-Amps, i will be more then happy to watch it, as for me this are the most difficult EE components to understand.
At timeline 7:32 when using a capacitor as negative feedback, the rule you mentioned at the beginning of the video where negative feedback causes the differential input voltage to be near zero, does not apply here, since the non-inverting input is fixed at +5volts and the inverting input swings above and below 5 volts. Am I correct?
Not quite. When the input is brought above 5V, the output voltage gets adjusted to ensure the inverting input stays at 5V - it does this by causing the current through the input capacitor to flow into (or out of) the feedback capacitor. A constant current into (or out of) a capacitor causes a linear change in voltage across that capacitor, which is observed at the output. The junction of the capacitor and input resistor (at the inverting input) remains at 5V.
Sir, I can appreciate that you are trying to make the explanation as simple as possible, and I would not try to explain the operation of the "integrator" in the manner in which I am going to discuss the engineering situation in the op-amp, but it would be interesting to represent the real operation of the op-amp working as an integrator, ( through itself acting as a differentiator from an output point of view!!) At 8:09 you are suggesting that due to the 5 volts at the inverting input there is a constant current flowing in the input resistor which must proceed to flow into the feedback resistor. Then you suggest that this constant current through the capacitor will create a linear ramp voltage across the capacitor. This operation seems to suggest that it is the constant current that creates the ramp output voltage, but in reality, it is not like that. In reality, it is the ramp voltage output which creates the constant current. Here is how it goes. When the input is higher than 5 volts at the inverting input, the output goes negative from a LOW OUTPUT IMPEDANCE VOLTAGE source and so it pulls the output plate of the capacitor quickly down with it. But from the output point of view, the circuit is a differentiator and so to produce a constant voltage of 5 volts on the capacitor plate connected at the input inverting input, the voltage output must have a constant rate of change. In reality, it is even more than that, as there is a very high-frequency oscillation being superimposed on the ramp output which is reflected at the inverting input. So really the practicality of the circuit suggests that it is the constant rate of change of the output voltage that causes the constant current to exist and not the constant current exists because of the 5 volts at the inverting input and this is pushed into the capacitor. The conduction current through the input resistor and the displacement current through the capacitor is created because of the nature of voltages appearing between the two components connected between the output and the input " low impedance voltage sources". Now that in order for a displacement current to flow through the insulation of a capacitor, there must be a rate of change of electric field in the dielectric insulator. Displacement current through an insulator will only flow if there exists a rate of change of the electric field. ( we are in the radio space now and not the DC conduction space) When teaching operational amplifiers, I like to introduce a lever system where the fulcrum is the inverting input at the same potential as the noninverting input. In the case of the integrator circuit, not many books mention the fact that from the output point of view, feeding the same network, the network is a high pass network or a differentiator circuit BEING FED FROM A LOW OUTPUT IMPEDANCE OF A VOLTAGE SOURCE. To really idealize an operational amplifier one needs to ensure that the student fully understands ( in this case voltage sources ) that both the input source and the output of the op-amp have low output impedance and so the junction point at the inverting input is a tug of war between the input source and the output of the op-amp. The junction point of the between the input component and the feedback component will drive the amplifier output so that the tug of war from the output of the op-amp will put the junction point at the same potential as the noninverting input. The real engineering operation of the op-amp from the output point of view is a case of " Depending upon the network connection between the input source and the output of the op-amp, the junction point which controls the inverting input will cause the output to form its nature and shape so as to oscillate the junction point about the potential of the non inverting input. This high-frequency oscillation is fast and of low magnitude due to the huge amplification of the op-amp, which is band limited and hence the high frequency superimposed is not seen. Few people can imagine the practical operation for an inductor replacing the capacitor. To discuss the other permutations where the capacitor and inductor are at the input side is also a good test in both engineering and language logic.
One of the considerations for any op amp circuit is the Slew Rate. Many popular Op-Amps have a Slew Rate 0.5 Volts per microsecond. While this may work with audio circuits, it's not a suitable Op Amp for integrators and other math related circuitry. While this video does touch on the basics, characteristics such as latch up can happen with Bipolar Op Amps. I've encountered Op Amp IC's with their tops completely blown away. So without any further discussion, if your interested in Op Amps, there are many good titles to choose from. While You tube can teach and inform you, it's good idea to read about Op Circuits too.
Great video. I learned a tremendous amount about op amps and integrator circuit. Dont fault author on op amp gain think he got idea that is next needed topic. This is a well scripted and executed tutorial that will last a long time in electronic education, not taught as well in ordinary schools!
+Bosco Khoo It's a simple voltage divider. +10V applied to the top of the series pair of 2.2k resistors. Since the resistors are the same value, the voltage at the intersection of the resistors is 1/2 of the supplied voltage, thus +5V.
I'm using a +10V power supply to the the top of the resistor divider made by the two 2.2k resistors. Since these resistors are equal (and virtually no current is drawn by the non-inverting input of the op amp), the voltage at the junction of the two resistors is +10/2 = +5V.
***** im sorry but im not sure how does ohm law work and to which reference point does V+ take its voltage from, top or bottom resistance and what if we flip the voltage to be at the bottom and ground to be at the top
Great video for a rookie like myself. I'm just wondering (EXACTLY) how you figured out the 5 volts being supplied to the + input of the op amp. You stated it was figured out using OHM's law but with the three values given, (10V, 2.2k and 2.2k) it only gives the current value which is actually zero being pulled from the Op amp if I'm not mistaken. I'm sure I am missing something here!! Thanks.
Alan thank you for making this video. I had one comment to make about the sign convention of the first circuit you used to mark the 10K and 20K resistors. I think they should be labeled opposite of what you did. Starting from the Vout node and going counter clockwise, which is ~7V, there should be a -2V drop across the 20K resistor, which then gets you to the +5V at the inverting input node of the OPAMP, and finally a -1V drop across the 10K resistor to get you +4V at the voltage input.
The sign convention on the 10k and 20k resistors is correct. As drawn, the voltage at the right side of each of these resistors is more positive than the voltage at the left side of each resistor - so the + goes on the right, the - on the left. Using your words, you "drop" down in voltage as you trace from the output to the input, so you go down to a lower voltage as you trace from the output to the input.
Ah, of course that makes sense i didn't look at it from the supply side (aka voltage input). Both are correct and I guess as long as one is consistent in their sign convention, it all works out properly.
can this op amp circuit in timeline 6:58 be considered as a non inverting amplifier? At timeline 9:38, the rule you mentioned about both inputs have to be at the same voltage level when using negative feedback applied only to when a resistor is used as negative feedback and not a capacitor. But here you showed that when a capacitor is used as a negative feedback, the rule still applies.
At 6:58, assuming the signal input is at the upper left, this is an INVERTING configuration. If negative feedback is being properly employed, the two inputs will always be nearly the same voltage.
I know the gain of an inverting opamp is (-) r2/r1, so I was wondering where the +7 volts from +4 volts (-) in came from, then I realized that you are dealing just with the DC, and the 5v at the + input has a gain of r2/r1+1 so that gives 15 out, minus the -8 from the 4v (-) in, leaving +7. It's been a while since I looked at the theory, thanks for the refresher!
👍 Thanks for the great explanation.. Today I understood how it maintain both non-invtng and inverting same voltage after watching many videos video about Op-amp and why we feed back... Simple things always makes great..
Thank you and it did refresh my mind. In theory this can be solved by superposition. For example replace 10V with ground and now you have output = 4x (- r2/r1) = -8V next put back the 10V and replace 4V with ground, now output = 5 x (1+r2/r1) = 15V add -8V and 15V and you get 7V please note r1 =10K r2 = 20K
Hey thanks for answering my last question. Still, there are more to come. Why the 5v at the +In, is it to make the dc pulses to ac (to discharge the cap)? an if I'd feed a sine wave in it,... would it give a 90° phase shift ,... or would it create overtones? Thanks in advance. Greets
Dear Om, I want to thank you, because you helped me a lot to understand op-amps working principals, but there is one issue which is still unclear to me. In the example 5:00, all calculated values are clear to me, but I can't realised how they are not matched with formula which says, Gain=-Rf/Rin=Vout/Vin. We see that -Rf/Rin=-20k/10k=-2, but it's not equal to Vout/Vin=7/4...Maybe I am getting wrong, but I don't know where, so could you help me?
The basic formula assume that the voltage at the op amp inputs is 0V (ground), and it is not at ground in this example. It is at 5V. So the *difference* from the input to the op amp inputs is -1V, and from the op amp inputs to the output is 2V, so the gain is -2.
Very great video sir... I am studying BE-EE now....I am in 2nd Sem. I was bored with OPAMP and its basic.. But after watching this video ..It seems interesting.. Thank You...For such interesting video.
What are some tests you can do for preamp amplifiers circuits and power amplifiers circuits to test for Linearity? Linearity can either mean gain flatness from 20hz to 20Khz or Linearity can also mean how much input peak to peak voltage before the preamps output clips saturates or before the power amplifiers output clips saturates. I'm confused by what EE engineers are considering when a preamp or power are is linear.
Where is the leftmost (4V) pin/wire connected in the first circuit? To GND? The Voltage comes from the output of the OpAmp and drops 2V over 20k, then it will be "sensed at the non-inverting and another 1V drop over 10k. Is that right?
10V applied across two 2.2k resistors in series. Current through both resistors is the same (10V/4.4k). Thus, the voltage drop across the lower resistor is 2.2K * (10V/4.4k), which can be re-arranged as 2.2k/4.4k * 10V, which is 1/2 * 10V or 5V.
Thank you thank you thank you...., if only for the fact that you made me look the differential input equation again, i thought it was just an over-complicated way of working out the output so i just glanced over it (see initial comment). I put the values used in the video into the said equation and the output came to............7v. You way is much easier and you only need to know basic ohms law. Thank you again.
Would you ever consider doing a brief tutorial on OTA's? There's not nearly as much info on RUclips about them and they pop up in audio circuits a lot. Thanks!
at 9:03, I'm a little confused. Since the charge/discharge is from the capacitor, why is the yellow waveform a ramp, rather the typical curve a charge/discharge cycle of a capacitor?
Hello sir , op.amp.explantion is very nicely and understanding of physics students. Iam physics prof. it is useful for my students .Thank u very much sir.
I don't understand why the slope is negative. As charge accumulates on the capacitor plates, shouldn't the curve first increase linearly from zero? Could you also tell me what the peak values of the linear curve are and how we can determine them?
ok. So what you are saying is that since its connected to a virtual ground, Vout starts decreasing from zero to the neg peak value. Did i get that right?
Thanks for the informative video! Was able to apply this in comparator circuit for a thermistor controlled fan. Much more sensitive than using a single BJT and easier to adjust.
When you apply the pulse, what is the voltage at the inverting terminal? Could you explain it with respect to the fact that it tries to become 5V (like the non-inverting)?
Hey, but how come that it is an almost linear rise? Does the timeconstant of the cap is only partial used wich would result in a more linear curve? Also would the capacitor provide lower amplitude for higher frequency squarewaves because of the linear rise? Thanks in advance.
Same here, more configs, especially explaining where to use them apart from how... differential amplifiers, even voltage follower that seems to simple, but explain the usual pitfalls that make them oscillate...
I have been in the electronics industry for 50 years and that has to be one of the best tutorials I have seen.
Keep up the great work.
You explained what my ECE professor couldn't in over a week of classes in 13 minutes. Thank you.
pass this video along to your classmates!
@@w2aew I definitely will! Thanks again!
2 wks of listening to my professor- didnt understand
6 minutes of watching this- everything clicks in my brain.
Played at the speed of X2, XD!!
Adam I Ii
Adam I ..... Isn’t that the way learning seems to take place. I too experienced the same feeling from my days (waaaaaay) back in school. I just did not understand why my teachers couldn’t have taught in this way. Even alot of books are hard to learn from because of this......I guess shit happens ....
because you are out of focus in class
yea sometimes i wonder , are these college professors really interested in teaching because so many never bother to boil things down into simple form which i thought was the art of teaching. wtf. if they are not capable of doing this then maybe they shouldnt be teachers.
I have watched a lot of instructional videos which use a technique that you also use and I often add my comment " Please use a tripod". My reasoning is mostly to keep me from becoming dizzy or sea sick. I however complement you on a quite steady hand held camera not producing nausea therefore allowing me to focus on your superb, clear, informative content. Thanks for your well done videos. I always learn from you whether or not I am viewing a topic that I think I already know. Another of your 'characteristics' that I appreciate is that you do not waste words using extraneous babble that distracts from the topic. THANKS again. Many of those other videos I can't return to for more information but I come back to yours over and over. Keep them coming!
I don’t usually comment on videos, but I’ve been looking around for a bit for videos to help me make sense of op amps, and this really did it, thank you so much!
Currently an EEE undergraduate student; this video really helped me understand how the op-amp circuit works. Thank you very much for your time and effort!!!
I'm glad this video helped you Ryan. Let me know if there are other topics that you'd like to see a video on. And of course, tell your fellow students too!
***** you're a legend, thanks for putting your time into helping people you don't know. I for one appreciate it greatly.
peter hooper Thanks Peter! I don't think I've ever been called a legend! I'm glad you find my videos useful.
good
You do awesome videos! Finally found a knowledgeable guy on electronics that doesn't talk weird or obnoxious. Your very clear and easy to understand.
I'm glad you like the video. I hope you enjoy the other 200+ that I have on my channel. An index / table of contents can be found here:
www.dorkage.com/youtube/W2AEW_video_index.pdf
Holy Moly!
try great scott
@@AWESOMEEVERYDAY101 GreatScott is amazing,but too advance for a college guy like me..😂
Love his videos though
I have no idea why I found your channel just now.. I have been struggling with OPAMPS and with your clear lectures and actual DESIGN showing what is happening really helped me alot! KEEP UP THE GOOD WORK, YOU, YOU SIR! AND please DO MORE!!!
Hello sir, I am a beginner and I just want to say thanks for the introduction. I have been reading up on op amps and have not gotten it through my head, and most the videos here on youtube make it too complicated and frustrated. this is by far the best video I found on op amps. So thank you and hope you have a wonderful time
Pure gold. Somewhere back in my Mims days I got op amps wrong and they've been unduly confusing ever since. Finally I feel straightened out. Thanks!
Excellent Tutorial showing how to put theory in to practice. I like your choice of circuit as it almost cover everything you need to know on opamps. your results satisfy the formula fully. Vout= -Vin(Rf/Rg)+Vref(R1/R1+R2)(Rf+Rg/Rg) by substituting we get
Vout= -4v(20K/10K)+10v(2k2/2k2+2k2)(20k+10k/10k) = 7 Volts which matches your Vout.
Thank you and we like to see more tutorials please ! :-)
I wish I had been able to see this 40 years ago - it would have made a big difference in how I dealt with op-amp circuits!
Was "big difference" an intentional pun?
😂
this is wonderful how you explain, i'm a physics professor too. for fundamental physics and quantum electronics. great job! I loved the experiment you make it after theory solving
This was one of the best explanations how opamps work. Short and practically useful. The rest shows the data sheet.
This is a difference between a newbie on Monday but expert on Friday,rushing to post a video about something that he slightly licked himself and someone who has a long years indepth knowledge.You have my respect Sir.
I know it's been almost ten years since you posted this video, but it helped me understand things today in 2022. Thank you.
This has been my most popular video - viewed over 700,000 times over the last 9 years...
I'm amping up (pun intended) for my Biomedical Instrumentation final and this was a helpful refresher on some of the basic stuff. You have a very natural, logical and application-based approach to explaining these circuit configurations - it would be AWESOME if you did a few more. Thanks so much!
This made op amps make so much more sense I wish I had watched this in school, I never really understood what they did
Why does everyone have to over-complicate this stuff, its like they make this stuff hard to learn on purpose. Took me so long to find a good explanation online as to why there is a negative voltage at the output of the inverting amplifier. You explained it well by assuming there is a voltage at (-) and that the output has to go negative in order to make that pin into a virtual ground. Other people have tried explaining it to me by FIRST saying there is a virtual ground and it never made sense to me that way. Thank you so much
Maybe the Best easiest explanation i've seen on op amps !
This is my favourite EE tutorial video. Your explanations are crystal clear. Thank you so much for making me feel so smart.
This Explanation is spot on!! It saved me hours of time. And we know that Time is money. So this dude saved me a lot of money. Please guys fully view the ads in the video and click on them too, this will support the guy who made this video. Thanks
"This wasn't meant to be a tutorial on how to design op amp circuits" Please do one- you are a great teacher.
Excellent illustration, one of the best OpAmps videos here ever; Thank you !!
Finally got to understand how op-amps work with a simple explanation. Keep it up sir.
Great tutorial, I feel like I have struck gold here !...cheers.
@9:00 we see the date on your scope: 27-Jan-13. This vid was uploaded the day before that: 26-Jan-13.
This shows that you're ahead of your time. Thanks for the excellent video.
Darn it, where were you 40 years ago when I was trying to figure this stuff out? =:o}
Guess he was either at tektronix or in college 🤣🤣🤣
Your description is better than mine. I have been teaching beginning Op
-Amps for 20 years.
I'm glad you liked it. Please feel free to use my description with your students, or even share my video (and my channel) with your students. And of course, if there's a topic that you think would make a good video (that would help your students) please let me know.
Well done, good refresher, being immersed in digital for the last 10 years.
You're a talented engineer and teacher. I enjoy watching your videos because you offer a wide variety of topics and knowledge in simple and interesting ways. Thanks a lot.
I found this extremely helpful. I needed the physical intuition on how the opamp works and this is exactly what I was looking for. Thank you!
Once again, greatly explained, greatly shown, just great, keep it up!!! Personally I am unable to design analog stuff from scratch, but take plenty of circuit bits and pieces from here and there on opamps and anything I can... and put them together to work. With this simple explanation one can at least understand more about what is happening on circuit when opamps are involved. Thanks once more!!
Thank you so very much for this video (and all the other ones too)! I'm just so glad that knowledge doesn't weigh anything, if not I might have to go on a diet... This really helps a lot! (BTW, thanks to you I calibrated my scope probe, it's really cool to know those kind of things you normally don't look into as an amateur). This is a top youtube channel!
Very very good. You're a great teacher. I really look forward to your uploads. Especially stuff for newbies like me! Keep up the great work!!
Nice. One of the best explanations of basic op amp operation that I've ever heard.
The way you explaining is awesome ...:)
and found best of RUclips in these.
It's my humble request that, it will be very helpful if you can explain how to find the input impedance of op-amp.
Love the way you explained this. Great job. Well done.
Wow very interesting analogy! I will keep that in mind. What really got me into circuits was the physical analogies that helped explain them. Like the water analogy etc... so this helps. Thanks.
sweet this is exactly what I needed. thank you brother, very well explained
Thanks for posting this, we flew through op-amps when I took circuits two years ago and this made everything make sense again. I might actually finish this controls lab now!
Alan. You do a perfect job of explaining it. I like to watch your videos, it has helped me very much in understanding how component's work. Keep up the awesome work.
One of the best op-amp explanation. Hats off
Hey Alan, Watched several of your videos today trying to knock some of the rust out between my ears. Just recently got my technician ticket and getting the bug again to return to electronics a bit as a hobby again some 30+ years later, so many cool things to play with! You do an awesome job clearly explaining, thanks for all your efforts!
Best,
- Bill
KI7QOA
Exactly what I needed to know. I am an amateur radio operator who has used a simple dual op-amp ckt to pre-amplify microphone inputs. The output level was too high and adjusting the second stage of the op-amp feedback to lower the overall gain was the right move.
Later, I nearly fried the input dual op-amp ckt in one of my radios. I still have near-perfect operation, but the is now a 2-3khz noise present. I need to understand what component might have been fried in the ckt that needs replacement. This ckt, by design, has a single component that is two diodes in parallel and reversed from each other shorting the inverting input and the output. in parallel with that diode pack, in the feedback line is a resistor and a capacitor (100k & 390pf). I am still trying to figure that out, but this helps (current limiting?).
Thank you for the short, practical explanation.
Strictly speaking, the statement at @1:32 should be: "negative feedback keeps any CHANGE in input voltage appproximately equal to zero". In many opamp circuits, V(+) is not grounded and the negative feedback will try to make V(-) - V(+) be zero.
Yes, I *wrote* it correctly as the deltaV at the input, I just didn't *say* it accurately.
Thank YOU!
Using O-SCope to present what You are talking about is the way this topic should be explained.
If you ever have time for more videos about Op-Amps, i will be more then happy to watch it, as for me this are the most difficult EE components to understand.
Your videos are always so helpful
At timeline 7:32 when using a capacitor as negative feedback, the rule you mentioned at the beginning of the video where negative feedback causes the differential input voltage to be near zero, does not apply here, since the non-inverting input is fixed at +5volts and the inverting input swings above and below 5 volts. Am I correct?
Not quite. When the input is brought above 5V, the output voltage gets adjusted to ensure the inverting input stays at 5V - it does this by causing the current through the input capacitor to flow into (or out of) the feedback capacitor. A constant current into (or out of) a capacitor causes a linear change in voltage across that capacitor, which is observed at the output. The junction of the capacitor and input resistor (at the inverting input) remains at 5V.
@@w2aew Thank you~ This is interesting!
Sir, I can appreciate that you are trying to make the explanation as simple as possible, and I would not try to explain the operation of the "integrator" in the manner in which I am going to discuss the engineering situation in the op-amp, but it would be interesting to represent the real operation of the op-amp working as an integrator, ( through itself acting as a differentiator from an output point of view!!)
At 8:09 you are suggesting that due to the 5 volts at the inverting input there is a constant current flowing in the input resistor which must proceed to flow into the feedback resistor. Then you suggest that this constant current through the capacitor will create a linear ramp voltage across the capacitor. This operation seems to suggest that it is the constant current that creates the ramp output voltage, but in reality, it is not like that. In reality, it is the ramp voltage output which creates the constant current. Here is how it goes.
When the input is higher than 5 volts at the inverting input, the output goes negative from a LOW OUTPUT IMPEDANCE VOLTAGE source and so it pulls the output plate of the capacitor quickly down with it. But from the output point of view, the circuit is a differentiator and so to produce a constant voltage of 5 volts on the capacitor plate connected at the input inverting input, the voltage output must have a constant rate of change. In reality, it is even more than that, as there is a very high-frequency oscillation being superimposed on the ramp output which is reflected at the inverting input. So really the practicality of the circuit suggests that it is the constant rate of change of the output voltage that causes the constant current to exist and not the constant current exists because of the 5 volts at the inverting input and this is pushed into the capacitor. The conduction current through the input resistor and the displacement current through the capacitor is created because of the nature of voltages appearing between the two components connected between the output and the input " low impedance voltage sources".
Now that in order for a displacement current to flow through the insulation of a capacitor, there must be a rate of change of electric field in the dielectric insulator. Displacement current through an insulator will only flow if there exists a rate of change of the electric field. ( we are in the radio space now and not the DC conduction space)
When teaching operational amplifiers, I like to introduce a lever system where the fulcrum is the inverting input at the same potential as the noninverting input. In the case of the integrator circuit, not many books mention the fact that from the output point of view, feeding the same network, the network is a high pass network or a differentiator circuit BEING FED FROM A LOW OUTPUT IMPEDANCE OF A VOLTAGE SOURCE.
To really idealize an operational amplifier one needs to ensure that the student fully understands ( in this case voltage sources ) that both the input source and the output of the op-amp have low output impedance and so the junction point at the inverting input is a tug of war between the input source and the output of the op-amp. The junction point of the between the input component and the feedback component will drive the amplifier output so that the tug of war from the output of the op-amp will put the junction point at the same potential as the noninverting input. The real engineering operation of the op-amp from the output point of view is a case of " Depending upon the network connection between the input source and the output of the op-amp, the junction point which controls the inverting input will cause the output to form its nature and shape so as to oscillate the junction point about the potential of the non inverting input. This high-frequency oscillation is fast and of low magnitude due to the huge amplification of the op-amp, which is band limited and hence the high frequency superimposed is not seen. Few people can imagine the practical operation for an inductor replacing the capacitor. To discuss the other permutations where the capacitor and inductor are at the input side is also a good test in both engineering and language logic.
It's was very helpful and easy to understand for someone like me who is really a novice. Great stuff!! Thank you.
Awesome intro. Very simple and to the point, but not in a way that detracts to the quality of information.
great tutorial, one of the best explanations of the basic operation of op-amps I've seen yet on RUclips. A+
One of the considerations for any op amp circuit is the Slew Rate. Many popular Op-Amps have a Slew Rate 0.5 Volts per microsecond. While this may work with audio circuits, it's not a suitable Op Amp for integrators and other math related circuitry. While this video does touch on the basics, characteristics such as latch up can happen with Bipolar Op Amps. I've encountered Op Amp IC's with their tops completely blown away. So without any further discussion, if your interested in Op Amps, there are many good titles to choose from. While You tube can teach and inform you, it's good idea to read about Op Circuits too.
Great video. I learned a tremendous amount about op amps and integrator circuit. Dont fault author on op amp gain think he got idea that is next needed topic.
This is a well scripted and executed tutorial that will last a long time in electronic education, not taught as well in ordinary schools!
At 3:19, how did you get 5V? could you show workings?
+Bosco Khoo It's a simple voltage divider. +10V applied to the top of the series pair of 2.2k resistors. Since the resistors are the same value, the voltage at the intersection of the resistors is 1/2 of the supplied voltage, thus +5V.
use potential divider
I see 40v written!!!!!! Wtf
OhNoitsBadmanMo I did to in the beginning but it’s +10 V
Its the formula for voltage divider vout = (vin x R2)/(R1 + R2) => (10 x 2200) / 4400 = 5.00V
Thanks for another excellent video. Your tutorials are very helpful to me. I would enjoy seeing more about Op-Amps.
Hi, i am unsure at that part in 3:29 where V+ is 5V and how does the ohm law work even with different value of resistance ?
I'm using a +10V power supply to the the top of the resistor divider made by the two 2.2k resistors. Since these resistors are equal (and virtually no current is drawn by the non-inverting input of the op amp), the voltage at the junction of the two resistors is +10/2 = +5V.
***** im sorry but im not sure how does ohm law work and to which reference point does V+ take its voltage from, top or bottom resistance and what if we flip the voltage to be at the bottom and ground to be at the top
***** sorry i got what you means, very sorry for the trouble and thanks for your patience and giving me the answer
Jiejun Huang Glad to help!
Thank you for this teaching.It is so important to understand it before designing with OPAmps
Wow - I finally got the "amp" portion of the device name! Yet another great bit of info - THANKS!
Thank you! It is extremely valuable!
my man you know what your doing, i think you just saved my grades
You are one helluva good teacher. :) I've learned so much watching your videos.
Another well made video Alan. Only wish I had had that basic description when I learnt op-amp theory at TAFE.
Great video for a rookie like myself. I'm just wondering (EXACTLY) how you figured out the 5 volts being supplied to the + input of the op amp. You stated it was figured out using OHM's law but with the three values given, (10V, 2.2k and 2.2k) it only gives the current value which is actually zero being pulled from the Op amp if I'm not mistaken. I'm sure I am missing something here!! Thanks.
Alan thank you for making this video. I had one comment to make about the sign convention of the first circuit you used to mark the 10K and 20K resistors. I think they should be labeled opposite of what you did. Starting from the Vout node and going counter clockwise, which is ~7V, there should be a -2V drop across the 20K resistor, which then gets you to the +5V at the inverting input node of the OPAMP, and finally a -1V drop across the 10K resistor to get you +4V at the voltage input.
The sign convention on the 10k and 20k resistors is correct. As drawn, the voltage at the right side of each of these resistors is more positive than the voltage at the left side of each resistor - so the + goes on the right, the - on the left. Using your words, you "drop" down in voltage as you trace from the output to the input, so you go down to a lower voltage as you trace from the output to the input.
Ah, of course that makes sense i didn't look at it from the supply side (aka voltage input). Both are correct and I guess as long as one is consistent in their sign convention, it all works out properly.
can this op amp circuit in timeline 6:58 be considered as a non inverting amplifier? At timeline 9:38, the rule you mentioned about both inputs have to be at the same voltage level when using negative feedback applied only to when a resistor is used as negative feedback and not a capacitor. But here you showed that when a capacitor is used as a negative feedback, the rule still applies.
At 6:58, assuming the signal input is at the upper left, this is an INVERTING configuration. If negative feedback is being properly employed, the two inputs will always be nearly the same voltage.
@@w2aew Thank you sir
I know the gain of an inverting opamp is (-) r2/r1, so I was wondering where the +7 volts from +4 volts (-) in came from, then I realized that you are dealing just with the DC, and the 5v at the + input has a gain of r2/r1+1 so that gives 15 out, minus the -8 from the 4v (-) in, leaving +7. It's been a while since I looked at the theory, thanks for the refresher!
👍 Thanks for the great explanation.. Today I understood how it maintain both non-invtng and inverting same voltage after watching many videos video about Op-amp and why we feed back... Simple things always makes great..
I'm going to have to come back to this one. Very well explained, but my brain is full. Thanks for sharing!
Thank you and it did refresh my mind. In theory this can be solved by superposition. For example replace 10V with ground and now you have output = 4x (- r2/r1) = -8V next put back the 10V and replace 4V with ground, now output = 5 x (1+r2/r1) = 15V
add -8V and 15V and you get 7V please note r1 =10K r2 = 20K
Correct, thats the way I learned.
Hey thanks for answering my last question.
Still, there are more to come.
Why the 5v at the +In, is it to make the dc pulses to ac (to discharge the cap)?
an if I'd feed a sine wave in it,... would it give a 90° phase shift ,... or would it create overtones?
Thanks in advance.
Greets
Dear Om, I want to thank you, because you helped me a lot to understand op-amps working principals, but there is one issue which is still unclear to me. In the example 5:00, all calculated values are clear to me, but I can't realised how they are not matched with formula which says, Gain=-Rf/Rin=Vout/Vin. We see that -Rf/Rin=-20k/10k=-2, but it's not equal to Vout/Vin=7/4...Maybe I am getting wrong, but I don't know where, so could you help me?
The basic formula assume that the voltage at the op amp inputs is 0V (ground), and it is not at ground in this example. It is at 5V. So the *difference* from the input to the op amp inputs is -1V, and from the op amp inputs to the output is 2V, so the gain is -2.
Thanks a lot, it's clear now.
Awesome video! Very interesting. Hope you keep the op-amp videos coming- Can't wait to see more!
thank you very much for up loading a great video by which the op amp student can gain value able basic knowledge of op amps
wow you are grate cos you actually show the logical thinking and the physical circuit working as well love it . thanks for this .
I try to do this with all of my videos. Thank you for noticing.
w2aew cool please don't change it. cos a lot of channels will either just show you the practical or the theory only . keep it up .
One of the clearest op amp lesson on youtube! THANK YOU :D
Very great video sir...
I am studying BE-EE now....I am in 2nd Sem.
I was bored with OPAMP and its basic..
But after watching this video ..It seems interesting..
Thank You...For such interesting video.
What are some tests you can do for preamp amplifiers circuits and power amplifiers circuits to test for Linearity? Linearity can either mean gain flatness from 20hz to 20Khz or Linearity can also mean how much input peak to peak voltage before the preamps output clips saturates or before the power amplifiers output clips saturates. I'm confused by what EE engineers are considering when a preamp or power are is linear.
Holy shit!!! you explain much better that university teachers!!! Thank you for the great video!
Where is the leftmost (4V) pin/wire connected in the first circuit? To GND? The Voltage comes from the output of the OpAmp and drops 2V over 20k, then it will be "sensed at the non-inverting and another 1V drop over 10k. Is that right?
Are three power sources needed? The Vin +4v, The 10v, and power to pin 8(V+) and pin 4 (Ground)?
Thank you.
I am so lucky to find this explaination
I loved the real life breadboard examples. Great vid
3:15 Can someone help explain to me how he got that voltage? Im unclear how Ohm's law was applied
10V applied across two 2.2k resistors in series. Current through both resistors is the same (10V/4.4k). Thus, the voltage drop across the lower resistor is 2.2K * (10V/4.4k), which can be re-arranged as 2.2k/4.4k * 10V, which is 1/2 * 10V or 5V.
@@w2aew wow thank you that makes sense
Thank you thank you thank you...., if only for the fact that you made me look the differential input equation again, i thought it was just an over-complicated way of working out the output so i just glanced over it (see initial comment). I put the values used in the video into the said equation and the output came to............7v. You way is much easier and you only need to know basic ohms law. Thank you again.
And your explanation really helped me. Week before exams. :/ Thank you :)
You are a gentleman and a scholar.
not an acrobat?
Would you ever consider doing a brief tutorial on OTA's? There's not nearly as much info on RUclips about them and they pop up in audio circuits a lot. Thanks!
Excellent explanation of opamp mechanics
at 9:03, I'm a little confused. Since the charge/discharge is from the capacitor, why is the yellow waveform a ramp, rather the typical curve a charge/discharge cycle of a capacitor?
Because the op amp + feedback results in charging/discharging the capacitor via a constant current.
Hello sir , op.amp.explantion is very nicely and understanding of physics students. Iam physics prof. it is useful for my students .Thank u very much sir.
I don't understand why the slope is negative. As charge accumulates on the capacitor plates, shouldn't the curve first increase linearly from zero?
Could you also tell me what the peak values of the linear curve are and how we can determine them?
Great tutorial for learning how to step through an op amp circuit! Thanks for the video!
ok. So what you are saying is that since its connected to a virtual ground, Vout starts decreasing from zero to the neg peak value. Did i get that right?
Thanks for the informative video! Was able to apply this in comparator circuit for a thermistor controlled fan. Much more sensitive than using a single BJT and easier to adjust.
When you apply the pulse, what is the voltage at the inverting terminal? Could you explain it with respect to the fact that it tries to become 5V (like the non-inverting)?
Hey, but how come that it is an almost linear rise? Does the timeconstant of the cap is only partial used wich would result in a more linear curve?
Also would the capacitor provide lower amplitude for higher frequency squarewaves because of the linear rise?
Thanks in advance.
brilliantly explained, learnt to see the simplicity of the operation
Same here, more configs, especially explaining where to use them apart from how... differential amplifiers, even voltage follower that seems to simple, but explain the usual pitfalls that make them oscillate...