In 10 minutes, you made me understand what I couldn't in 4 years of engineering! OpAmps no longer look like magic to me!! You are brilliant, Sir. I can never thank you enough.
For some reason, this video I stumbled upon at 3am on (another) sleepless night has just hit the spot. You always just find these videos that either repeat things you already know, or are beyond comprehension complicated and you are lost in the first two minutes. There are always these thoughts where I think „yeah it probably works like that“ but I just never really believe it until someone says it out. This Video cleared a LOT of those up. Thank you so, so much.
This is priceless interview material. In several job interviews I've been asked to analyze the large-signal behavior of different opamps circuits with no equations and this technique has helped me A LOT! Thanks for posting Dan!
If you need something as basic as this on youtube to give you an advantage in a job interview then your training and experience is way below what you need to do that job. Youd be like the aeronautical engineer that thought its ok to cut heat traeted aliminium by laser cutting. Education these days is so poor if it results in qualified people that have that level of knowledge
This is the best intuitive analysis of an Op Amp circuit I've seen. No complex math, just an understanding of a few simple concepts. The rules of an ideal Op Amp, Ohm's Law, and Kirchoff's voltage and current laws.
I usually never write comments on youtube videos but this was a great video. It made so much sense and it made something that looks complicated seem very simple. Thank you so much!!! I have gained back some confidence in my circuit solving skills. You should make more circuit solving videos, a video on how to design op amps according to a specification would be great.
Thank you very much! Up till now I just remembered the formulas for different types of basic opamp circuits, but now I actually understand where they came from and am able to tackle more complecated circuits.
WOW! Absolutely amazing tutorial. I'm currently back in school as a non-trad, and we're working with OpAmp circuits in my upcoming Electronics lab, and when I heard that that would be our second lab, I thought 'ACK! I don't know much about OpAmps, other than that they can be a big stumbling block', and having watched this, I feel much less intimidated. Thank you so much for posting this, this is absolutely amazing, and I'll probably be coming back to your channel as the semester goes on!
Superb explanation! This came at a good time for me, I just finished my Op-Amps class and I'm still wrestling with the circuit analysis. I've been falling back on the algebra formulas, but they don't provide real understanding. The steps 1, 2 and "3 with a twist" are great. Your approach, using the three op amp rules and Kirchoff's Law, provides clear and easy to visualize insight. I've looked at many op-amp tutorials, and this is my favorite, thanks for taking the time to share your knowledge.
This guy helped me understand this in 4 minutes, vs my professor who teaches this in an hour and you are still confused afterward. Thank you sir, will be coming back to see more videos!
I've had several people challenging me to solve this circuit or that circuit. Let's not do that. Use the technique I show you here to solve the circuit you have in mind. Assume a voltage and see what happens. You can always start with the Vout at one of the rails and see where that leads you.
Dan Bullard You assumed there is the same voltage on both inputs in each case (as per your rule 2) and then found a greater voltage on the output in each case? rendering your rule 2 redundant as you stated yourself? This makes no sense? how can you have no differential between inputs(considering firstly that this does NOT use negativr feedback) (ie. 2V on each input and then magically have 10V on the output? again reaching that conclusion by assuming one input is the same as the other "unless the output is greater" then finding that the output IS greater and still taking 10V as the legitimate answer??
Hi Dan, Just to say that I'm 2 minutes into your video and I've paused it just to write this comment. EXCELLENT introduciton, to the point and the 3 rules here, more importantly the first one that I didn't really know about, are just great! I have seen dozens of Opamp videos, but no-one has put rule 1 into plain English: OBVSIOULY this is of GREAT help in understanding the OVERALL CONCEPT. EXCELLENT, I've learned something today, and I'm happy!!!! Keep it up!
You were showing the switch at 2v which corresponded to a Vout of 10v. You referenced back during this time to the initial switch position at 1v with a Vout of 5v.
Simple, yet very helpfull tut. Thanks a lot. I loved that you mentioned that there is no current in the output too. It is very important sometimes for solving the circuit.
This is a great video and you have a talent for conveying challenging concepts. It would be wonderful if you created more videos like this about analog electronic circuit design!
I really liked the video! I'm going to a 2-year tech school for electronics. I'm in my second semester of my first year. The way they do it here is split each semester into 3 5 week modules. This module (as well as the previous one) is all digital stuff, so I'm a bit rusty with calculating these circuits. Again, thanks for the video! It's a nice refresher!
Sir this video you have been taken is just insanely good, is there any chance you could do RL, RC ,RLC and Op-Amp with RL, RC, RLC. I could not believe my eyes, how clear can someone teach something that shot time. Much love.
Not likely, but I appreciate the comments. RL&C are just not that simple, although I did do a video on RC coupling.ruclips.net/video/GGAt6N-Pz9w/видео.html
Dan I am trying to learn electronics and I cant seem to get it. I think the problem lies in my approach or so to say my syllabus (which is me just randomly looking up things in no order in my quest) the question I have for you is.... where would you recommend I start?
I recall from a discussion with EE that when an op amp is used as a simple DC signal amplifier, the internal resistance of the input device itself (like a sensor coil) has to be included as part of total input resistance at the summing point. In that case the gain would actually be the feedback resistance / total input resistance.
Thanks Dan for the very clear explanation. One suggestion would be to label the appropriate ref designators for the 3rd case equivalent circuit so the “student” can quickly identify which resistor is which.
Good video. You could also mention that rule 2 applies only to closed circuits, where the output connects in some way with the - input, and the impedance therein does not prevent the max output voltage from being too low to match the input.
Dan, this video is so good. I have seen many online tutorials on op amps and this is by far the best. Why don't you upload more like this? You would become very popular very quickly. We need more lectures of this quality there are so few about.
Just an adjustment to your rules. If there is output feedback to the inverting ( - ) input, rule 2 applies unless V(out) is beyond the range of the rails or source.
Yeah, but... It's easier to just let someone do the analysis ignoring what might be feeding back to what. What happens when you have both negative and positive feedback? Who knows, until you analyze it.
Unbelievable how easy you make it seems with this short video. Thank you! I have one question, what is the purpose of R4 and R5 in this case when there isn't any current flowing there?
This circuit is used to test an interviewee's knowledge, so the purpose of R4 and R5 is to see if the interviewee knows that no current ever flow into or out or the inputs.
Thanks, a very helpful video. But could someone explain the purpose of R4 & R5 or are they simply not needed if there's no current flowing through them and no voltage drop? Thanks in advance.
R4 and R5 are just current limiting resistors in case something bad happens to the Op Amo.. That and in this case it allows me to make the point about Rule #1.
This is so elegant. Many thanks for this. Where can I find details on the Jim Harris method? I can't find any references to Jim Harris anywhere? Did he write a text book?
He was an obscure man who built a small school in Silicon Valley called Technical Training Center. He was a genius at simplifying difficult concepts. He's gone now but many engineers and technicians learned everything from Op Amps to uPs just this way.
Hi Dan, I gotta doubt, in fact I think there is something wrong in what you explained. I apologize if I wrong. In your first case, where 1v input is supplied, you said there would be a total of 5v across R8 and R9. Actually it is a potential divider network of equal resistance, it cant divide differently. It should be either 3v ,3v or 2v,2v not as 2v,3v. Please let me know if I am missing something
Ash R8 and R9 are not forming a voltage divider on their own. If you want to look at it like that you must consider R6 and R7 as well. It is then a voltage divider that has 1K (R9) in series with 0.666Ohms (R8 in parallel with R7 and R6 in series).
I might be a little bit too late to join the party 😄 I appreciate the clear explanation and detailed solution you provided. Sometimes we all forget that electronics is not that complicated and that all circuits can be handled by simply adhering to a few basic laws. I have a question, and I would greatly appreciate it if you could help me understand the design decision: what is the purpose of R4 and R5 in this circuit?
R4 and R5 are typically placed there to keep the current down to a minimum if the Op Amp fails, but in this case they are used to trick the observer, to see if they understand that an Op Amp consumes no current on the inputs. It's a standard ploy imposed on Employee Applicants.
@@DanBullard I appreciate your response. They were current-limiting resistors in my mind. Although it's a wonderful concept, I've never utilised them in any designs. In my subsequent design, I'll give it a shot.
Very Good Circuit Analysis ,The Way you say it ,it shows you are a teacher ,,I disagree with some of it ,but bottom line VERY Good Troubleshooting . and thats how we learn from each other .
I don't get how you got the 2V and 1V at the beginning of the video 1:55 ? There is zero current, yes. But how did you find out that it only drops 1 volt over each resistor?
R5 came out while I was calcuating Requivalent because it will never have any current going in or out of it, so for the purpose of calculating Req it's best to just ignore it. We ignore the insulation on the wires because we know that no current flows through the insulators, so we don't need to concern ourselves with it.
It's simple Ohms Law if you ignore the Op Amp, which you can because it draws no current. Around 7:47 you will see how I simplify the circuit. R9 is in series with R8 to ground and R6 and R7 to ground.
Hi Dan, I am delighted with this logic. I don't understand why degrees don't teach this way?! Thank you. I haven't checked the other videos but are there different config which uses L and C?
Thanks, I haven't done anything with L and C. I worked on RADAR and radio, so I do know a fair bit about it, but I'm working on Harmonic Distortion mostly right now.
Hi, I have a question. If there's no current flowing through R4 and thus no voltage drop across it, then what's the purpose of R4 in the first place? Thanks in advance.
I get asked this all the time, I should have addressed it in the video. Most often it's not a good idea to hardwire a power supply into an input, digital or analog. If the pin shorts out internally, something bad could happen. Better to limit the current than letting one faulty transistor set the chip and hence the device on fire!
a very helpful example,,,thank you for releasing this video,,, but how did you figure the gain in the second step 5V/1V?? ,,, while the Vout =10 and the Vin=2 ,,, is the gain supposed to be 10V/2V?? thanks again.
Exactly. Lots of profs teach some mathematical solution to gain, but the best way that I have found is just to do analyze the circuit with a couple of values to see what the DC gain is. No capacitors or inductors, no worries about AC gain, it's the same as DC gain (pretty much). I once got through an interview at Tektronix with this strategy and got the job.
Here's a question about the "ideal" op amp. According to the ideal model, the open loop gain is infinite and any input would also cause the output to saturate or go full conduction. In that case, the op amp would act like a switch -IE- a solid state relay and any input would case it to turn on with full conduction. Even with negative feedback, it seems the op amp would still behave as a switch rather than a linear device and the whole system would be unstable and prone to oscillation. Accordingly, my opinion for modeling the ideal op amp is to assume the it has a very high (but not infinitely high) gain so it still behaves as a linear device, but not have the characteristic of a switch.
I think that negative feedback would keep things in check automatically, no matter what the gain is. Any tiny difference would instantly be corrected. Now, toss a capacitor into the mix and watch it oscillate because the feedback can't get there fast enough especially when the gain is infinite.
Dan Bullard Here's a practical experiment that might verify whether or not a device with an infinite DC gain will be stabile. Get a DC/DC solid state relay (a device that turns full on with a small input signal) like the one in the link below and connect it as an inverting amplifier. That is the + output is tied back to the - input so the output is 180 degrees out of phase with the input. Then apply several different levels as the turn on signal and see if the output is in fact a linear function of the input. www.crouzet-ssr.com/english/products/_gndc.shtml In my opinion, it's a good idea to avoid obscure mathematical constructs that have "infinite" quantities which can lead to confusion in a practical analysis. That's why the concept of a mathematical "limit" was introduced because infinite quantities should not be inserted in equations such as the one for amplifier gain. Accordingly, it would be better just to say the ideal op amp has a gain that approaches - but never reaches- infinity.
I would try to find a school that does not take the classic college approach. See if you can get a sit-in for a day in an early part of the course, like sometime in the first month. If they are teaching mathematics or just Ohms Law, leave and never go back. If they are teaching logic, op-amps, computer programming, and spend at least 20% of the day on a lab exercise then sign up. I wish I could start a school that worked like Jim Harris' school TTC, but no can do Aaron!
THANKYOU SO MUCH DAN...U SAVED MY LIFE...GOT MA FINALS 2MRW...ELCTRINCS...I DONT EVEN KNW THIS CONCEPT OF OP AMPS TILL NOW I HAD SOME LIGHT LOL...CHEERS MAN!! THANKYOU SO MUCH...U'R BEST!!
Dan Bullard Put isn't the voltage divider rule R2/R1+R2 all multiplied by Vin to equal Vout. So from what ive learned in the past is that if both resistors are the same then the voltage will be halved? So 1.5V if you use the equation I use?
Shane Quinn There are three resistors there, so 3V divided by three 1K resistors equals 1mA through each one. One milliamp times 1K = 1V, so each resistor has 1V across it. zero plus one is one, so the voltage at the top of R3 is 1V. One plus one is two, so the top of R2 is 2V. And 2 plus one is three, so the top of R1 is 3V, which all works.
If the differential between V+ and V- is zero, won't Vout become 0? If Vout = 0, won't the differential become maximum again? So the output voltage will just keep oscillating between 0 and maximum A((V+) - (V-))....
Think of it as the reference point (i.e. 0 Volts) for the voltages in the circuit. If you're using "ground" as "the reference point", then, yes. I've learned over the years to use "ground" carefully, because it has different meanings, depending on who you talk to (i.e. big difference if you're an electrician wiring up a house vs. an electronic tech fixing an amp).
Im sorry but i dont understand the first step. I get why R4 has no current flowing through it, but why is VR1 and VR2 equal to 1 volt? How can you determine that the voltage drop between the 3v node and R4 is 2v? Thanks!
R1, R2 and R3 form a simple voltage divider. 3V/3K=1mA, so each resistor drops 1V starting from the 3V source. The junction between R1 and R2 is 3V - (1K * 1mA) = 2V. The junction between R2 and R3 is 3V - (2K * 1mA) = 1V. You couldn't know any of that if there was any current flowing into, or out of the + input of the op amp, which is the whole point of Rule 1. If you had to think about how much current the + input was drawing, you might never figure out the input voltage.
@@DanBullard ohhh i get it now, the circuit "keeps going" after R3, i wasnt seeing that so i wasnt taking R3 into account. Thanks a lot for answering so quickly!
My teacher told me today that you cannot have a higher voltage than your supply voltage without causing clipping, not sure how this applies to your rules or if it does.
+Juiicy Rich I think you need to watch the video again. His explanation of his "rules" are pretty clear. Your teacher is right, which is why when getting a Vout that't higher than the positive rail, he backtracks and solves the circuit again for the voltage at the negative input.
Since the inputs were not equal when switch was at 3V, there was a difference in voltage between the input terminals. Does that mean that the last calculations would be wrong since current would be flowing out of the inverted input? Or is input impedance so high that there would still be no current flow?
The input impedance is WAY too high to allow any >significant< current flow to make up for the discrepancy in input voltages, but there is usually a small amount, in the nano-amps to micro-amps range when the op amp saturates. Not enough to make any difference across the 1K input resistors, but that will cause some, in some bi-polar op amps. That's why I have the rule of no current flow on the inputs, any current flow that happens by accident from saturation, etc is so infinitesimal it's not worth worrying about.
Hello Dan, Please give me help, I have exam soon and I need to undestand this circuit, please let me know how I can contact you to send my exercice please advice
Should the ground line for the op amp circuit be connected to the chassis ground which is the green wire (or safety earth ground) for the AC power supply? I've seen many permanent installations (such as seismological and meteorological stations) in which the op amp's ground line was also connected to the green wire but some others did not have a connection. However it seems that the amp's ground itself probably should not be connected to the green wire because any current through the green wire would create a voltage and make the amp's ground line subject to erratic ground reference. Furthermore, the voltages in op amp circuits themselves are so low that there's no safety issue and only the chassis of 120 VAC power supply would need to be earth grounded.
Dan Bullard Although the IC itself doesn't have a ground pin, the non-inverting input is grounded (to provide a reference point for the differential input) and the external load at amp's output (a galvanometer for example) also has a return to ground. Seems that connecting the chassis ground wire to the ground line that's also used for the complete circuit might introduce noise.
OK, OK! It was my first one! I will do some more and have already started a positive feedback version. Stay tuned!
Thank you. Respect from india
Thanks alot I really appreciate you from KENYA
Very Good tutorial Dan, hope you can still share some more of your technical knowledge.
In 10 minutes, you made me understand what I couldn't in 4 years of engineering! OpAmps no longer look like magic to me!!
You are brilliant, Sir. I can never thank you enough.
Thank you! I used to teach at a school where we analyzed circuits like that every day. The students got really good at op amps. Glad I could help.
That's concerning.
Ok, this is PURE GOLD, to me at least. This is the simplest explanation ever I heard.
Sir, you helped me overcome 25 years of dread for op-amps... i can never thank you enough.
For some reason, this video I stumbled upon at 3am on (another) sleepless night has just hit the spot. You always just find these videos that either repeat things you already know, or are beyond comprehension complicated and you are lost in the first two minutes.
There are always these thoughts where I think „yeah it probably works like that“ but I just never really believe it until someone says it out.
This Video cleared a LOT of those up.
Thank you so, so much.
This is priceless interview material. In several job interviews I've been asked to analyze the large-signal behavior of different opamps circuits with no equations and this technique has helped me A LOT! Thanks for posting Dan!
If you need something as basic as this on youtube to give you an advantage in a job interview then your training and experience is way below what you need to do that job.
Youd be like the aeronautical engineer that thought its ok to cut heat traeted aliminium by laser cutting.
Education these days is so poor if it results in qualified people that have that level of knowledge
@@davefoord1259 based
This is the best intuitive analysis of an Op Amp circuit I've seen. No complex math, just an understanding of a few simple concepts. The rules of an ideal Op Amp, Ohm's Law, and Kirchoff's voltage and current laws.
I usually never write comments on youtube videos but this was a great video.
It made so much sense and it made something that looks complicated seem very simple. Thank you so much!!!
I have gained back some confidence in my circuit solving skills.
You should make more circuit solving videos, a video on how to design op amps according to a specification would be great.
Ideal op virtual ground (v+=v-)
Find the Circuit vo is not complexe and then v?v? .. ! Great explain Sir 😊
Thank you very much! Up till now I just remembered the formulas for different types of basic opamp circuits, but now I actually understand where they came from and am able to tackle more complecated circuits.
Great tutorial! I went out and bought a HP6235 power supply just so I could recreate this lesson. It makes so much more sense now. Thanks Dan.
WOW! Absolutely amazing tutorial. I'm currently back in school as a non-trad, and we're working with OpAmp circuits in my upcoming Electronics lab, and when I heard that that would be our second lab, I thought 'ACK! I don't know much about OpAmps, other than that they can be a big stumbling block', and having watched this, I feel much less intimidated. Thank you so much for posting this, this is absolutely amazing, and I'll probably be coming back to your channel as the semester goes on!
Thank you!
This is the first time I've fully understand how op amp works. Thanks a lot!
You are very welcome!
Superb explanation! This came at a good time for me, I just finished my Op-Amps class and I'm still wrestling with the circuit analysis. I've been falling back on the algebra formulas, but they don't provide real understanding. The steps 1, 2 and "3 with a twist" are great. Your approach, using the three op amp rules and Kirchoff's Law, provides clear and easy to visualize insight. I've looked at many op-amp tutorials, and this is my favorite, thanks for taking the time to share your knowledge.
Posted 8 years ago and it is more helpful than other sources i have been read and watched. Thank for this video I learned a lot about op amp
This guy helped me understand this in 4 minutes, vs my professor who teaches this in an hour and you are still confused afterward. Thank you sir, will be coming back to see more videos!
+[Dan Bullard] , you sir were the first one to break it down easy enough for me to understand this, thank you very much
That was extremely clear and helpful. Thanks for posting this video.
I've had several people challenging me to solve this circuit or that circuit. Let's not do that. Use the technique I show you here to solve the circuit you have in mind. Assume a voltage and see what happens. You can always start with the Vout at one of the rails and see where that leads you.
Dan Bullard You assumed there is the same voltage on both inputs in each case (as per your rule 2) and then found a greater voltage on the output in each case? rendering your rule 2 redundant as you stated yourself? This makes no sense? how can you have no differential between inputs(considering firstly that this does NOT use negativr feedback) (ie. 2V on each input and then magically have 10V on the output? again reaching that conclusion by assuming one input is the same as the other "unless the output is greater" then finding that the output IS greater and still taking 10V as the legitimate answer??
I am wondering if by 'voltage supply' he means the plus/minus 10V attached to the op amp
Yes, the power supply
You just save my ass for my tomorrow exam! Thanks
Alma Brew i hope he save my ass too cuz i have an exam tomorrow xD
a bouchra II Good luck!
Really great explanation of working through this op-amp circuit. Thank you very much for taking the time to do this. Cheers
Very logical step by step explanation. It has been a while since I worked with Op Amps and I am boning up for job related testing.
Thanks! I wrote it to help a friend bone up for a test at GE She passed the test and got the job!
Hi Dan,
Just to say that I'm 2 minutes into your video and I've paused it just to write this comment. EXCELLENT introduciton, to the point and the 3 rules here, more importantly the first one that I didn't really know about, are just great! I have seen dozens of Opamp videos, but no-one has put rule 1 into plain English: OBVSIOULY this is of GREAT help in understanding the OVERALL CONCEPT. EXCELLENT, I've learned something today, and I'm happy!!!! Keep it up!
You were showing the switch at 2v which corresponded to a Vout of 10v. You referenced back during this time to the initial switch position at 1v with a Vout of 5v.
Thank you very much! This has made Op Amps infinitely easier!
Like an opamps open loop gain hehe
you really know how to teach things. after all research, I think I really understand op amps now. thanks a lot
Simple, yet very helpfull tut. Thanks a lot.
I loved that you mentioned that there is no current in the output too. It is very important sometimes for solving the circuit.
I wish you had more of these, but then again, this video was a quantum leap in my understanding of op amps
Brilliant, very clear, a good speed, a good amount of repetition/variance. Wish this was around when I was learning.
This is a great video and you have a talent for conveying challenging concepts. It would be wonderful if you created more videos like this about analog electronic circuit design!
My newest -ruclips.net/video/UkfRYfOj3e0/видео.html
We went from 5V out to 10V out. That's 5V. That change happened with a 1V change on the input hence the gain of 5.
I really liked the video! I'm going to a 2-year tech school for electronics. I'm in my second semester of my first year. The way they do it here is split each semester into 3 5 week modules. This module (as well as the previous one) is all digital stuff, so I'm a bit rusty with calculating these circuits. Again, thanks for the video! It's a nice refresher!
Great video Dan! You are really good at teaching these concepts in a simple and elegant manner. Thank you for making this, please make more!
The way you did this numerical is fantastic and probably the best
Sir this video you have been taken is just insanely good, is there any chance you could do RL, RC ,RLC and Op-Amp with RL, RC, RLC. I could not believe my eyes, how clear can someone teach something that shot time. Much love.
Not likely, but I appreciate the comments. RL&C are just not that simple, although I did do a video on RC coupling.ruclips.net/video/GGAt6N-Pz9w/видео.html
Dan I am trying to learn electronics and I cant seem to get it. I think the problem lies in my approach or so to say my syllabus (which is me just randomly looking up things in no order in my quest) the question I have for you is.... where would you recommend I start?
Holy shit! This explanation was the one I was waiting for! So easy to understand
I recall from a discussion with EE that when an op amp is used as a simple DC signal amplifier, the internal resistance of the input device itself (like a sensor coil) has to be included as part of total input resistance at the summing point.
In that case the gain would actually be the feedback resistance / total input resistance.
Yes, but sometimes it's hard to figure it out unless you do a static analysis like this.
Thanks Dan for the very clear explanation.
One suggestion would be to label the appropriate ref designators for the 3rd case equivalent circuit so the “student” can quickly identify which resistor is which.
Good video. You could also mention that rule 2 applies only to closed circuits, where the output connects in some way with the - input, and the impedance therein does not prevent the max output voltage from being too low to match the input.
One of the clearest explanations I have heard so far. Nice Vid Dan
Thank you for comming and sharing with me !
Thank one million !
Dan, this video is so good. I have seen many online tutorials on op amps and this is by far the best. Why don't you upload more like this? You would become very popular very quickly. We need more lectures of this quality there are so few about.
Just an adjustment to your rules. If there is output feedback to the inverting ( - ) input, rule 2 applies unless V(out) is beyond the range of the rails or source.
Yeah, but... It's easier to just let someone do the analysis ignoring what might be feeding back to what. What happens when you have both negative and positive feedback? Who knows, until you analyze it.
This was the one of best video i found,thanks ❤
Your Rule #2 only applies when there is feedback.
Those people who dislike this...why? This explanation is so clear and superb!
That was a great video. I immediately subscribed after watching this so that I do not miss your videos. That was just great.
Unbelievable how easy you make it seems with this short video. Thank you!
I have one question, what is the purpose of R4 and R5 in this case when there isn't any current flowing there?
This circuit is used to test an interviewee's knowledge, so the purpose of R4 and R5 is to see if the interviewee knows that no current ever flow into or out or the inputs.
@@DanBullard Aha! Thanks for the response. I hope you will keep doing these videos, good teachers with good knowledge are rare.
Awesome vid, clear concepts. Thank you very much. I' would love more circuit analysis videos!
question: Why have such a complicated feedback loop? What does it accomplish?
Thanks, a very helpful video. But could someone explain the purpose of R4 & R5 or are they simply not needed if there's no current flowing through them and no voltage drop? Thanks in advance.
R4 and R5 are just current limiting resistors in case something bad happens to the Op Amo.. That and in this case it allows me to make the point about Rule #1.
Awesome video sir how easily you solve the problems sir.Hats off to you sir.Thanks for uploading this video.Sir please upload some more numericals.
Hi Dan , thanks for the video good job ,
can you please explain why do we need R4 and R5 ?
Verry clear and helpfull video, I have an exam tommorow and some of these rules could be of use thank you
Thank you so much for this video. Keep up the great work, it's super appreciated! Greetings from Sweden.
Dan! you saved a lot of time ... instead of reading and understanding, i simply understood everything... :D thanks a lot...
Please upload more videos
This is so elegant. Many thanks for this. Where can I find details on the Jim Harris method?
I can't find any references to Jim Harris anywhere? Did he write a text book?
He was an obscure man who built a small school in Silicon Valley called Technical Training Center. He was a genius at simplifying difficult concepts. He's gone now but many engineers and technicians learned everything from Op Amps to uPs just this way.
Dan Bullard So, who will train the young Padawans in the secrets of the Force so they become Jedi Knights just like you, Oh-venerable Grand Master?
Hi Dan, I gotta doubt, in fact I think there is something wrong in what you explained. I apologize if I wrong. In your first case, where 1v input is supplied, you said there would be a total of 5v across R8 and R9. Actually it is a potential divider network of equal resistance, it cant divide differently. It should be either 3v ,3v or 2v,2v not as 2v,3v. Please let me know if I am missing something
Ash R8 and R9 are not forming a voltage divider on their own. If you want to look at it like that you must consider R6 and R7 as well. It is then a voltage divider that has 1K (R9) in series with 0.666Ohms (R8 in parallel with R7 and R6 in series).
Your thought me something that my professor couldn't
I might be a little bit too late to join the party 😄 I appreciate the clear explanation and detailed solution you provided. Sometimes we all forget that electronics is not that complicated and that all circuits can be handled by simply adhering to a few basic laws. I have a question, and I would greatly appreciate it if you could help me understand the design decision: what is the purpose of R4 and R5 in this circuit?
R4 and R5 are typically placed there to keep the current down to a minimum if the Op Amp fails, but in this case they are used to trick the observer, to see if they understand that an Op Amp consumes no current on the inputs. It's a standard ploy imposed on Employee Applicants.
@@DanBullard I appreciate your response. They were current-limiting resistors in my mind. Although it's a wonderful concept, I've never utilised them in any designs. In my subsequent design, I'll give it a shot.
Very Good Circuit Analysis ,The Way you say it ,it shows you are a teacher ,,I disagree with some of it ,but bottom line VERY Good
Troubleshooting . and thats how we learn from each other .
I don't get how you got the 2V and 1V at the beginning of the video 1:55 ? There is zero current, yes. But how did you find out that it only drops 1 volt over each resistor?
The resistors on the left side form a voltage divider. 3 volts across 3K means 1mA is flowing, so each 1K resistor drops 1V,
Thank you! Cold you also tell me why you removed the R5 resistor in the final step/task? (When we calculated the Req etc. )
R5 came out while I was calcuating Requivalent because it will never have any current going in or out of it, so for the purpose of calculating Req it's best to just ignore it. We ignore the insulation on the wires because we know that no current flows through the insulators, so we don't need to concern ourselves with it.
It's a classic voltage divider, 3V across a total of 3K resistance, 1mA flows, therefore each resistor drops one volt.
It's simple Ohms Law if you ignore the Op Amp, which you can because it draws no current. Around 7:47 you will see how I simplify the circuit. R9 is in series with R8 to ground and R6 and R7 to ground.
Thank you very much sir, it helped me a lot
Hi Dan, I am delighted with this logic. I don't understand why degrees don't teach this way?! Thank you. I haven't checked the other videos but are there different config which uses L and C?
Thanks, I haven't done anything with L and C. I worked on RADAR and radio, so I do know a fair bit about it, but I'm working on Harmonic Distortion mostly right now.
@@DanBullard Thank you again. I have always been fascinated by OPAMPs, quite a powerful component.
Excellent video - well paced, and very well explained.
Nice! What a nice way of explaining op-amps! Loved it.
Hi, I have a question. If there's no current flowing through R4 and thus no voltage drop across it, then what's the purpose of R4 in the first place? Thanks in advance.
I get asked this all the time, I should have addressed it in the video. Most often it's not a good idea to hardwire a power supply into an input, digital or analog. If the pin shorts out internally, something bad could happen. Better to limit the current than letting one faulty transistor set the chip and hence the device on fire!
Thanks for your quick response!!
I was looking for this comment ! 👌🏻
a very helpful example,,,thank you for releasing this video,,, but how did you figure the gain in the second step 5V/1V?? ,,, while the Vout =10 and the Vin=2 ,,, is the gain supposed to be 10V/2V??
thanks again.
Exactly. Lots of profs teach some mathematical solution to gain, but the best way that I have found is just to do analyze the circuit with a couple of values to see what the DC gain is. No capacitors or inductors, no worries about AC gain, it's the same as DC gain (pretty much). I once got through an interview at Tektronix with this strategy and got the job.
Dan Bullard yes indeed ,,, its a very good strategy.
thank you.
I need some clarification on that. Did you open the capacitors and short the inductors ans analyze as DC?
i never comment on youtube videos, but this was so helpful and articulately explained- THANK YOU!
Very good explanation. Thank you.
Thanks for the help. I hope the rules you said has got no error in it.
Here's a question about the "ideal" op amp.
According to the ideal model, the open loop gain is infinite and any input would also cause the output to saturate or go full conduction. In that case, the op amp would act like a switch -IE- a solid state relay and any input would case it to turn on with full conduction.
Even with negative feedback, it seems the op amp would still behave as a switch rather than a linear device and the whole system would be unstable and prone to oscillation.
Accordingly, my opinion for modeling the ideal op amp is to assume the it has a very high (but not infinitely high) gain so it still behaves as a linear device, but not have the characteristic of a switch.
I think that negative feedback would keep things in check automatically, no matter what the gain is. Any tiny difference would instantly be corrected. Now, toss a capacitor into the mix and watch it oscillate because the feedback can't get there fast enough especially when the gain is infinite.
Dan Bullard Here's a practical experiment that might verify whether or not a device with an infinite DC gain will be stabile.
Get a DC/DC solid state relay (a device that turns full on with a small input signal) like the one in the link below and connect it as an inverting amplifier. That is the + output is tied back to the - input so the output is 180 degrees out of phase with the input. Then apply several different levels as the turn on signal and see if the output is in fact a linear function of the input.
www.crouzet-ssr.com/english/products/_gndc.shtml
In my opinion, it's a good idea to avoid obscure mathematical constructs that have "infinite" quantities which can lead to confusion in a practical analysis. That's why the concept of a mathematical "limit" was introduced because infinite quantities should not be inserted in equations such as the one for amplifier gain.
Accordingly, it would be better just to say the ideal op amp has a gain that approaches - but never reaches- infinity.
I would try to find a school that does not take the classic college approach. See if you can get a sit-in for a day in an early part of the course, like sometime in the first month. If they are teaching mathematics or just Ohms Law, leave and never go back. If they are teaching logic, op-amps, computer programming, and spend at least 20% of the day on a lab exercise then sign up. I wish I could start a school that worked like Jim Harris' school TTC, but no can do Aaron!
THANKYOU SO MUCH DAN...U SAVED MY LIFE...GOT MA FINALS 2MRW...ELCTRINCS...I DONT EVEN KNW THIS CONCEPT OF OP AMPS TILL NOW I HAD SOME LIGHT LOL...CHEERS MAN!! THANKYOU SO MUCH...U'R BEST!!
Thanks! I love how clearly you explained the method (it's exactly what my prof seems incapable of doing)
Dan, this is an excellent op amp tutorial. Love it. Thanks
Terrific, but what is the point in recalculating the voltage at the negative input terminal ? Jim
Because I have to prove Rule 3.
holly crap!! compair this to the video by the indian proffesor at MIT This is now too easy. Thank you sir
This is definitely a simplified look at Op Amps. Would you be able to create a tutorial on bjt transistor biasing?
thank you. you are a life savior! I ' ve got an exam in two days and things begin to clear up. Can't wait for your next video :D
Regards, Vlad
Thanks for posting this problem.
How do you get that 2V and 1V on the far left at the beginning of the video? Where is the math for that?
faceinthegrass It's a voltage divider. 3/3 = 1. 1+1=2. That's the math.
Dan Bullard Put isn't the voltage divider rule R2/R1+R2 all multiplied by Vin to equal Vout. So from what ive learned in the past is that if both resistors are the same then the voltage will be halved? So 1.5V if you use the equation I use?
Shane Quinn There are three resistors there, so 3V divided by three 1K resistors equals 1mA through each one. One milliamp times 1K = 1V, so each resistor has 1V across it. zero plus one is one, so the voltage at the top of R3 is 1V. One plus one is two, so the top of R2 is 2V. And 2 plus one is three, so the top of R1 is 3V, which all works.
If the differential between V+ and V- is zero, won't Vout become 0? If Vout = 0, won't the differential become maximum again? So the output voltage will just keep oscillating between 0 and maximum A((V+) - (V-))....
Are the black arrows the same thing as ground?
Thanks for the tutorial by the way.
Think of it as the reference point (i.e. 0 Volts) for the voltages in the circuit.
If you're using "ground" as "the reference point", then, yes.
I've learned over the years to use "ground" carefully, because it has different meanings, depending on who you talk to (i.e. big difference if you're an electrician wiring up a house vs. an electronic tech fixing an amp).
Im sorry but i dont understand the first step. I get why R4 has no current flowing through it, but why is VR1 and VR2 equal to 1 volt? How can you determine that the voltage drop between the 3v node and R4 is 2v? Thanks!
R1, R2 and R3 form a simple voltage divider. 3V/3K=1mA, so each resistor drops 1V starting from the 3V source. The junction between R1 and R2 is 3V - (1K * 1mA) = 2V. The junction between R2 and R3 is 3V - (2K * 1mA) = 1V. You couldn't know any of that if there was any current flowing into, or out of the + input of the op amp, which is the whole point of Rule 1. If you had to think about how much current the + input was drawing, you might never figure out the input voltage.
@@DanBullard ohhh i get it now, the circuit "keeps going" after R3, i wasnt seeing that so i wasnt taking R3 into account. Thanks a lot for answering so quickly!
My my it's really a good insight to use opamps for many different things. Thank you.
Wow! This was explained extremely bold.
Thank you really much!!
My teacher told me today that you cannot have a higher voltage than your supply voltage without causing clipping, not sure how this applies to your rules or if it does.
+Juiicy Rich I think you need to watch the video again. His explanation of his "rules" are pretty clear. Your teacher is right, which is why when getting a Vout that't higher than the positive rail, he backtracks and solves the circuit again for the voltage at the negative input.
You forgot to mention, that the OP Amp rules you talked about, only apply in negative feedback!
Rule 3 takes care of that for you, otherwise you have to be able to tell the difference.
Since the inputs were not equal when switch was at 3V, there was a difference in voltage between the input terminals. Does that mean that the last calculations would be wrong since current would be flowing out of the inverted input? Or is input impedance so high that there would still be no current flow?
The input impedance is WAY too high to allow any >significant< current flow to make up for the discrepancy in input voltages, but there is usually a small amount, in the nano-amps to micro-amps range when the op amp saturates. Not enough to make any difference across the 1K input resistors, but that will cause some, in some bi-polar op amps. That's why I have the rule of no current flow on the inputs, any current flow that happens by accident from saturation, etc is so infinitesimal it's not worth worrying about.
Very nice tutorial, Dan! thank you! I hope you have more tutorials on Op-Amp application, you did really good job in explaining!
superb !! I've never got this much clear explanation even in my university classes..
can we also state that rule 2 no longer applies if there is no feed back or positive feedback
Yep, could do that, but then than leaves you no place to start.
thank you you're a smart man the third rule was what I need💛
hi M Dan I have a problem ,I need 5 volt at the output when the input is 0 volt
Hello Dan, Please give me help, I have exam soon and I need to undestand this circuit, please let me know how I can contact you to send my exercice please advice
Thank you very much! Very much appreciated! Clear, simple and easy to use :D
Should the ground line for the op amp circuit be connected to the chassis ground which is the green wire (or safety earth ground) for the AC power supply?
I've seen many permanent installations (such as seismological and meteorological stations) in which the op amp's ground line was also connected to the green wire but some others did not have a connection.
However it seems that the amp's ground itself probably should not be connected to the green wire because any current through the green wire would create a voltage and make the amp's ground line subject to erratic ground reference.
Furthermore, the voltages in op amp circuits themselves are so low that there's no safety issue and only the chassis of 120 VAC power supply would need to be earth grounded.
The op amp has no ground wire. Power+ Power-, In+, In- and that's pretty much it.
Dan Bullard Although the IC itself doesn't have a ground pin, the non-inverting input is grounded (to provide a reference point for the differential input) and the external load at amp's output (a galvanometer for example) also has a return to ground.
Seems that connecting the chassis ground wire to the ground line that's also used for the complete circuit might introduce noise.
Excellent Sir. So far I haven't seen someone mentioning rule 3.