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 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.
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
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.
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!
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.
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.
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.ua-cam.com/video/GGAt6N-Pz9w/v-deo.html
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.
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.
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.
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.
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 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.
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.
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!
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.
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 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 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!
To make the point, there is no current flowing. Can't prove it without a resistor. Watch my most recent op-amp video ua-cam.com/video/UkfRYfOj3e0/v-deo.html
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.
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!
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!
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).
Chemiprofe asked that earlier. It's never a good idea to connect a device input directly to a power bus, just in case the chip shorts out internally you don't want the whole damn thing burning up. But the the real reason is that there may be a very, very tiny current on those inputs, not enough to worry about when doing first approximation analysis like this. But if you do get a picoamp of current on one input, the other input will probably have the same picoamp of current and so the two inputs will be subjected to the same nanovolt offset. Again, the effect is so tiny you could spend your whole life trying to figure out the circuit if you try to follow each picoamp around the circuit. Better to ballpark it and get really close than spend days trying to figure out every picoamp and nanovolt.
trialen Actually the value on one side should match the equivalent resistance of the feedback network on the other side remembering that the output of the op-amp is virtual ground. In this case I didn't bother with that, I didn't want to complicate things. But when you look at a well designed circuit, let's say a typical op-amp with a gain of ten, 2K on the minus input with a 20K feedback network. The equivalent resistance is (2K*20K)/(2K+20K) = 1.8K. So on the Plus side connect a 1.8K resistor to ground. Now any bias currents on the Minus side will cause a similar voltage drop on the Plus side. But as I've told the other folks, this technique is mostly for doing a first approximation to understand the circuit. Design is a whole other ball of wax. Get Horowitz's The Art of Electronics for info on design, best book out there.
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!!
This is not about "idea" op amps, that is what happens in real life. There are plenty of videos about fantasy devices, just watch almost anyone else's video on Harmonics, all of that is fantasy.
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.
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.
That is kind of an art I guess. You have to think like an electron, "how can I get to ground? What do I have to get through to get there?" It's always been hard for me to do this, so I just think like an electron.
aflyvietguy R9 is on top, R8 is on the right side to ground, R7 and R6 are on the left side to ground. So now you have 1K in series with 2K and 1K in parallel. I guess I should have annotated those. The big point here is that you can ignore R5 totally because there won't be any current flowing into it.
Starting from the output of the op amp, everything is in series with R9. Then, R8 goes straight to ground while R6 and R7 are in series with each other but, as a pair are in parallel with R8, as shown in the equivalent circuit on the far right. Now. R6 and R7 add up to 2K, and 2K in parallel with R8 (1K) is (2K*1K)/(2K+1K) = 0.6666K. And since that is in series with R9, 1K in series with 0.66666K is 1.66666K. Remember that we have to forget about R5, since no current is going to flow into R5. It's like a resistor floating in the air as far as we are concerned. Now, truth be told, you might get a nanoamp flowing into our out of the op amp (or less!) but what difference will a nanoamp make to your calculations? Far less than the tolerances of the REAL resistors.
Could you confirm what ground you would be referencing from if probing the inputs/output given that the rail is in the minus please. Would it be taken from the minus rail to the op amp. or from power supply ground.
Excellent work, though I'm not convinced that NO current flows through the inputs. After all, Voltage = Resistance X Current, and if current is ZERO, then so is voltage. Plus, if no current flows through R4 & R5, then what purpose do they serve?
The current is very tiny, too small to worry about in 99.9% of cases. When the resistors get to be in the Megohm range, then start to worry about them.
Dan Bullard Fair enough. I know that all mathematical calculations are approximate anyway, since one can only compute to a finite number of decimal places. However, as you mention, the resistors have negligible effect in the circuit until they reach the Megohm range, but then, what WILL be their effect?
flurng For the first approximation (i.e. an interview) it's best to ignore those resistors. Once the values get into the Megohm range then those resistors will start making voltage drops which will impact the output. However, most good designs don't put Megohm resistors there. The only reason for them is to prevent massive amounts of current if the op amp input was to blow up and short out. The only reason there are two resistors is that chances are good that the input leakage on one input will be the same on both, and so there will be no differential error, which means no appreciable output difference. See Horowitz Art of Electronics for more info.
One explanation is the large feedback resistor provides a path for the higher voltage and higher current from the amp's output to go back to the input device or circuit. Without a large value for the feed back resistor, the voltage at the output would easily be quite overwhelming at the input device. If the input device was a moving coil transducer (like in a seismometer), the voltage and current would create a motor effect and the whole system would become unstable and probably oscillate.
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.
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.
Hi, why when doing the equivalent circuit why R5 and OP Amp (-) input is not part of the circuit?? :/ Thx for the explanation I just understood it as I never have before, just that little question confused me.
+Edu "Turupá" Sánchez R5 and the Op Amp don't count as part of the circuit because the input impedance of the Op Amp is infinite. We don't make the insulators on the wire part of the equivalent circuit because their impedance is infinite. R5 is there, but since the input impedance of the op amp is infinite, it won't carry any current, so we can leave both of them out of the equivalent circuit.
+Dan Bullard Oh Thank you very much!!! That makes sense since the current there is 0, I'm still struggling a bit to understand impedance though. I'll keep studying :D
+Edu "Turupá" Sánchez Impedance in this case just means resistance. No need to get XL and XC involved since this is all DC. High impedance just means high resistance in this example.
why are the voltages at the switch 3v, 2v and 1v though, how do we have voltage drop across the resistor with no current flow. is it because the current flows to ground? and if so how is the voltage drop 1 volt if its a 1K resistor?
@@DanBullard yes perfect sense now that I look at it like that, I was doing the math on each separately on each resistor to find the current instead of the whole series to ground. Been a while, just getting back into this stuff so I gotta shake off the cobwebs. Thank you so much.
@@matthewlocke1942 To let you prove you know how Op Amps work. This is a classic interview circuit, it doesn't actually do anything, but employers use it to see if you really understand Op Amps. I developed this video for a friend who was interviewing for a job at General Electric in Reno NV. Long story short, she got the job.
@@DanBullard ok so it’s just there to show there is no actual current flow, and if you wanted to prove you know how it works you would show 0 voltage drop, got it. Sorry I’m trying to learn more about op amp circuits in my own time and I didn’t catch that part in the intro. Thank you for your time, I really appreciate it.
GREAT tutorial. I have to admit, I didn't understand most of it. I've got a probable reverb recovery unit op amp problem in my Marshall guitar amp, and I am hoping if you could walk me through testing the circuit, or at least the op amps, in an effort to identify the defective component? I've got a link to the schematic, and a multimeter. Should I just start at pin #1, measure the voltage, then try to use this video to solve for the other pin, and measure them..? Thanks
Robert McBride I think it would take a bit more than that. One thing you can do is check the voltage difference between the two inputs, pins 2 & 3 if I am not mistaken. They should be the same voltage if everything is OK. Give me the link to the schematic and I can see if I can walk you through testing it.
The Op Amp datasheet is the best place to look for that. Input bias current is spec'ed as well as input offset voltage. Some Op Amps have such high input impedance that Rule 1 is true no mater what. The point is, your first attempt at solving any circuit would be to assume that my rules cover all cases. If you see a 20MOhm input resistor then you should understand that that large an input resistor MAY have some impact on the input voltage, but it is by no means guaranteed.
thanks for the video it is very clear. i want to ask that, how should we think about resistors, in order to calculate the gain as in formula: 1+(Rf/R). what is Rf and R resistors here. first, i thought that R5 is R and rest of the network (6,7,8,9) are parts of Rf but didn't work. could you explain? thank you very much.
Yeah, that resistor to ground really screws everyone up. That's the problem with the way they teach this stuff in college. You become an expert at math and can't do anything else. What's the formula for a circuit that breaks? The math won't tell you what happens, but using this technique will. To calculate gain just do like the video says, change the input voltage, analyze the circuit and see what the output change is. Gain is delta_Vout/delta_Vin.
Excellent tutorial. Something that confused me was R4. I did the calculation with a Voltage divider with a Load (R4) and I got a different Volatge for the +Input. therefore if I analyze your schematic, Omitting R4 (because there is no Curent flowing through it,) will I get the same result? And if so, why is it necessary to include it in your example? Is R4 a mock component?
+Luis E. Batres Generally, on sensitive input pins we like to protect them from too much current in case the pin shorts internally. The circuit will fail if that happens, but without a resistor, the whole box might catch fire. This is especially true on MOSFET circuits, because the gate is just a capacitor with a very thin insulator to keep the thing from blowing sky high.
+Dan Bullard Thank You for your response. I am still conflicted though; because if as Rule 1 states: there is no Current going into the Inputs of a Op-Amp (for an Ideal Op-Amp that is, because of the High input Z of about 1M ohm for a Real Op Amp.) is your Low Resistance (in R4: 1k) helping in any way to prevent Damage or Over Heating?
+Luis E. Batres Because if the input shorts, the max current will be only in the milliamps. Some engineers bias op amps right from a power supply line capable of 10s of amps. What if the op amp fails and you have no resistor on that line? Even if in normal conditions the input Z is a megohm, if it shorts, (either internally or externally) you'll have grounded a 100W power supply which will probably burn up the runs on the board or start a fire in the component or elsewhere in the circuit. It's easy to use a power supply line to provide a static "1" or "0" to a digital circuit IC for example, but if the device shorts out internally, your power supply can now dump 10s of amps into the chip causing a real safety issue. A nominal resistor, say 1K, will not reduce the voltage much on a high impedance pin, but may prevent a fire, and that alone is a good reason to have a resistor on those pins.
+Dan Bullard Ok. Now I get it. So R4 is not cecessary in this "simple" circuit, but it is of Utmost Importance for the protection of other Circuits in ase this one fails. Thank You for the clarification and your excellent Analysis of an op-amp.
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.
Hi Dan, The gain is also defined as Vout/Vin. Won't that work as well for this problem? You then won't have to find a change in the output for a change in the input.
Hi Geert! Nice to see that you have moved from politics to electronics. Less chance of getting blown up for sure. The point is that the current is so small, it doesn't affect your calculations. Let's say that it has1nA of current flowing into one of the inputs. What's 1nA * 1K? 1uV. Is that going to give you extra points in your job interview, or will they be amazed that you solved their complex circuit in a few minutes like what happened to me at Maxim (twice!) in the interview. By the way, I got the job, and they were blown away that I was able to solve it so quickly.
I get asked that a lot. There may actually be a very tiny current on the inputs, but so small they can usually be ignored. But for absolute accuracy they are included to keep the input impedance the same between the two inputs just to cancel out any voltage drops.
Electrons moving is current, force of electron pressure is voltage. You don't need current for voltage to have effect. Voltage is like the pressure in a hose. Even if there is no water flowing, the hose will still bulge and be stiff because of the pressure.
Alexphal That would be the 'pressure' as Dan described. Another way to think of it is the current through R4 in reality is not absolute zero. There will be some very small current, as in picoamps or nanoamps (will vary with the op amp used). This would mean the voltage on R4 may actually be e.g: 1.0000000V on the left and 0.9999999 on the right.
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.
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.
i have a question: i am using opamp in open loop configaration vcc=12v, vee=0v, v- =2v, v+=(0 to 5v) ac square wave output is=(3 to 12volts)ac sqr???? i am expecting output=(0 to 12 volts) ac sqr. what happened here please explain.i want 0 to 12 instead of 3 to 12
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!
They are there for two reasons. One, to make the point that no current flows, and two, very often you will find resistors on those pins and you have to learn to ignore them. Remember the topic: Solving Op Amps. When you see resistors on the inputs of the Op Amps, you must learn to ignore them. If I didn't show them, you would be wondering what to do and be unable to solve the circuit.
@@brockchambers3733 No, if I make them Megohms then nanoamps suddenly matter. Almost every op amp circuit sports these small resistors. Making them Megohm resistors just invites disaster.
Do you know about, how to read the program inside a microchip? I heard the firms can read secret probrams of Pics. What are the keywords of thise topic ?
No, if it's not documented, there is no chance of figuring out what's inside. When I started working on Smart Cards (ISO 80211), a bunch of Koreans I worked with created a microchip to talk to the Smart Card, but because it was a proprietary design that they didn't want to get stolen, they put their chip on the circuit board and then smothered it with potting compound (epoxy) to keep the secret from being stolen. Paranoid, but expected, as Koreans steal everything and when they do invent something they do all they can to protect their secrets.
@@karisikpekmez6850 There is no way to look at the silicon to figure out what the device will do. This is pure fantasy, stop listening to people who tell you such things.
No way! Positve feedback is a completely differnt animal, but if you analyze it the same way you will figure out how positive feedback works. It becomes a Schmitt Trigger.
Hi Dan, Do you know how synchro- servo systems work? I really like your videos they've helped me a lot. Do you think you can do a video on sychro servo systems? I have not seen any videos anywhere.
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.
My fours years at engineering school could not make me understand what you did. Thanks for this wonderfule content.
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.
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.
Ok, this is PURE GOLD, to me at least. This is the simplest explanation ever I heard.
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.
Sir, you helped me overcome 25 years of dread for op-amps... i can never thank you enough.
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
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!
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
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.
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.
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!
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 the first time I've fully understand how op amp works. Thanks a lot!
You are very welcome!
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.
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.ua-cam.com/video/GGAt6N-Pz9w/v-deo.html
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.
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.
Those people who dislike this...why? This explanation is so clear and superb!
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.
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.
Hi Dan , thanks for the video good job ,
can you please explain why do we need R4 and R5 ?
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
The way you did this numerical is fantastic and probably the best
Really great explanation of working through this op-amp circuit. Thank you very much for taking the time to do this. Cheers
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 😊
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.
I wish you had more of these, but then again, this video was a quantum leap in my understanding of op amps
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!
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!
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.
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!
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.
you really know how to teach things. after all research, I think I really understand op amps now. thanks a lot
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 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!
if no current goes in the inputs then what's with the 1k/R4?
thanks
To make the point, there is no current flowing. Can't prove it without a resistor. Watch my most recent op-amp video ua-cam.com/video/UkfRYfOj3e0/v-deo.html
@@DanBullard I see thanks Mr Bullard!
Terrific, but what is the point in recalculating the voltage at the negative input terminal ? Jim
Because I have to prove Rule 3.
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.
Dan! you saved a lot of time ... instead of reading and understanding, i simply understood everything... :D thanks a lot...
Please upload more videos
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.
question: Why have such a complicated feedback loop? What does it accomplish?
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 ! 👌🏻
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 -ua-cam.com/video/UkfRYfOj3e0/v-deo.html
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).
Holy shit! This explanation was the one I was waiting for! So easy to understand
Brilliant, very clear, a good speed, a good amount of repetition/variance. Wish this was around when I was learning.
i never comment on youtube videos, but this was so helpful and articulately explained- THANK YOU!
One of the clearest explanations I have heard so far. Nice Vid Dan
Please explain the function of R4 and R5. If no current is flowing into the inputs why are they required ?
Chemiprofe asked that earlier. It's never a good idea to connect a device input directly to a power bus, just in case the chip shorts out internally you don't want the whole damn thing burning up. But the the real reason is that there may be a very, very tiny current on those inputs, not enough to worry about when doing first approximation analysis like this. But if you do get a picoamp of current on one input, the other input will probably have the same picoamp of current and so the two inputs will be subjected to the same nanovolt offset. Again, the effect is so tiny you could spend your whole life trying to figure out the circuit if you try to follow each picoamp around the circuit. Better to ballpark it and get really close than spend days trying to figure out every picoamp and nanovolt.
DansFlix OK, thanks for the explanation. Are there any design considerations for the values of R4/5 or is it safe to just use 1k ?
trialen Actually the value on one side should match the equivalent resistance of the feedback network on the other side remembering that the output of the op-amp is virtual ground. In this case I didn't bother with that, I didn't want to complicate things. But when you look at a well designed circuit, let's say a typical op-amp with a gain of ten, 2K on the minus input with a 20K feedback network. The equivalent resistance is (2K*20K)/(2K+20K) = 1.8K. So on the Plus side connect a 1.8K resistor to ground. Now any bias currents on the Minus side will cause a similar voltage drop on the Plus side. But as I've told the other folks, this technique is mostly for doing a first approximation to understand the circuit. Design is a whole other ball of wax. Get Horowitz's The Art of Electronics for info on design, best book out there.
Dan Bullard Thanks Dan, very interesting.
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!!
Great video , I have a doubt if the potential at the terminals turn out to be different would that not make the op amp non ideal ?
This is not about "idea" op amps, that is what happens in real life. There are plenty of videos about fantasy devices, just watch almost anyone else's video on Harmonics, all of that is fantasy.
hi M Dan I have a problem ,I need 5 volt at the output when the input is 0 volt
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
Why is there no voltage drop across R5? 2:30
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.
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.
how did you come up with the equivalent circuit at 7:40?
That is kind of an art I guess. You have to think like an electron, "how can I get to ground? What do I have to get through to get there?" It's always been hard for me to do this, so I just think like an electron.
Dan Bullard I mean where did those 4 resistors come from? which resistors did you replace from the original circuit?
aflyvietguy R9 is on top, R8 is on the right side to ground, R7 and R6 are on the left side to ground. So now you have 1K in series with 2K and 1K in parallel. I guess I should have annotated those. The big point here is that you can ignore R5 totally because there won't be any current flowing into it.
How di you calculate the equivalent resistance of 1.666k and could you explain the 6mA running through R9 please? (around 8:00)
Starting from the output of the op amp, everything is in series with R9. Then, R8 goes straight to ground while R6 and R7 are in series with each other but, as a pair are in parallel with R8, as shown in the equivalent circuit on the far right. Now. R6 and R7 add up to 2K, and 2K in parallel with R8 (1K) is (2K*1K)/(2K+1K) = 0.6666K. And since that is in series with R9, 1K in series with 0.66666K is 1.66666K. Remember that we have to forget about R5, since no current is going to flow into R5. It's like a resistor floating in the air as far as we are concerned. Now, truth be told, you might get a nanoamp flowing into our out of the op amp (or less!) but what difference will a nanoamp make to your calculations? Far less than the tolerances of the REAL resistors.
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.
Could you confirm what ground you would be referencing from if probing the inputs/output given that the rail is in the minus please. Would it be taken from the minus rail to the op amp. or from power supply ground.
The minus rail is a power supply, I never reference anything relative to a power supply. Ground is ground.
Excellent work, though I'm not convinced that NO current flows through the inputs. After all, Voltage = Resistance X Current, and if current is ZERO, then so is voltage. Plus, if no current flows through R4 & R5, then what purpose do they serve?
The current is very tiny, too small to worry about in 99.9% of cases. When the resistors get to be in the Megohm range, then start to worry about them.
Dan Bullard Fair enough. I know that all mathematical calculations are approximate anyway, since one can only compute to a finite number of decimal places. However, as you mention, the resistors have negligible effect in the circuit until they reach the Megohm range, but then, what WILL be their effect?
flurng For the first approximation (i.e. an interview) it's best to ignore those resistors. Once the values get into the Megohm range then those resistors will start making voltage drops which will impact the output. However, most good designs don't put Megohm resistors there. The only reason for them is to prevent massive amounts of current if the op amp input was to blow up and short out. The only reason there are two resistors is that chances are good that the input leakage on one input will be the same on both, and so there will be no differential error, which means no appreciable output difference. See Horowitz Art of Electronics for more info.
Wow! Thanks for the amazingly prompt ( and coherent ) response! Really helped clear up a few things for me!
One explanation is the large feedback resistor provides a path for the higher voltage and higher current from the amp's output to go back to the input device or circuit.
Without a large value for the feed back resistor, the voltage at the output would easily be quite overwhelming at the input device. If the input device was a moving coil transducer (like in a seismometer), the voltage and current would create a motor effect and the whole system would become unstable and probably oscillate.
Excellent video - well paced, and very well explained.
Sir how can I use it on stabilizer that has 4 different types of voltages?
Please help me out
You will have to design a circuit. That I can't do for you.
This was the one of best video i found,thanks ❤
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?
Great vid, but i cant figure out how you got the equivalent resistance at the end. thanks
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.
Thank you very much! This has made Op Amps infinitely easier!
Like an opamps open loop gain hehe
Thank you for comming and sharing with me !
Thank one million !
thank you so much sir! clean and concise.
Hi, why when doing the equivalent circuit why R5 and OP Amp (-) input is not part of the circuit?? :/
Thx for the explanation I just understood it as I never have before, just that little question confused me.
+Edu "Turupá" Sánchez R5 and the Op Amp don't count as part of the circuit because the input impedance of the Op Amp is infinite. We don't make the insulators on the wire part of the equivalent circuit because their impedance is infinite. R5 is there, but since the input impedance of the op amp is infinite, it won't carry any current, so we can leave both of them out of the equivalent circuit.
+Dan Bullard Oh Thank you very much!!! That makes sense since the current there is 0, I'm still struggling a bit to understand impedance though. I'll keep studying :D
+Edu "Turupá" Sánchez Impedance in this case just means resistance. No need to get XL and XC involved since this is all DC. High impedance just means high resistance in this example.
Great. THX!!!
why are the voltages at the switch 3v, 2v and 1v though, how do we have voltage drop across the resistor with no current flow. is it because the current flows to ground? and if so how is the voltage drop 1 volt if its a 1K resistor?
3V across 3K is 1mA with a clear path to ground, NOT ZERO. That's a simple voltage divider. Does that do it?
@@DanBullard yes perfect sense now that I look at it like that, I was doing the math on each separately on each resistor to find the current instead of the whole series to ground. Been a while, just getting back into this stuff so I gotta shake off the cobwebs. Thank you so much.
@@DanBullard my next question is what is the point of R4?
@@matthewlocke1942 To let you prove you know how Op Amps work. This is a classic interview circuit, it doesn't actually do anything, but employers use it to see if you really understand Op Amps. I developed this video for a friend who was interviewing for a job at General Electric in Reno NV. Long story short, she got the job.
@@DanBullard ok so it’s just there to show there is no actual current flow, and if you wanted to prove you know how it works you would show 0 voltage drop, got it. Sorry I’m trying to learn more about op amp circuits in my own time and I didn’t catch that part in the intro. Thank you for your time, I really appreciate it.
+[Dan Bullard] , you sir were the first one to break it down easy enough for me to understand this, thank you very much
I do not understand how you got 5V for the Delta(Vout)... Anyone care to explain?
What is the op amp is inverted? how do the equations reflect the effect of such change?
No equations here, there are plenty of videos that explain these with equations. Watch those.
GREAT tutorial. I have to admit, I didn't understand most of it. I've got a probable reverb recovery unit op amp problem in my Marshall guitar amp, and I am hoping if you could walk me through testing the circuit, or at least the op amps, in an effort to identify the defective component? I've got a link to the schematic, and a multimeter.
Should I just start at pin #1, measure the voltage, then try to use this video to solve for the other pin, and measure them..?
Thanks
Robert McBride I think it would take a bit more than that. One thing you can do is check the voltage difference between the two inputs, pins 2 & 3 if I am not mistaken. They should be the same voltage if everything is OK. Give me the link to the schematic and I can see if I can walk you through testing it.
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!
Thank you for the great explanation. What are the rules for non ideal Op-amp?
The Op Amp datasheet is the best place to look for that. Input bias current is spec'ed as well as input offset voltage. Some Op Amps have such high input impedance that Rule 1 is true no mater what. The point is, your first attempt at solving any circuit would be to assume that my rules cover all cases. If you see a 20MOhm input resistor then you should understand that that large an input resistor MAY have some impact on the input voltage, but it is by no means guaranteed.
thanks for the video it is very clear. i want to ask that, how should we think about resistors, in order to calculate the gain as in formula: 1+(Rf/R). what is Rf and R resistors here. first, i thought that R5 is R and rest of the network (6,7,8,9) are parts of Rf but didn't work. could you explain? thank you very much.
Yeah, that resistor to ground really screws everyone up. That's the problem with the way they teach this stuff in college. You become an expert at math and can't do anything else. What's the formula for a circuit that breaks? The math won't tell you what happens, but using this technique will. To calculate gain just do like the video says, change the input voltage, analyze the circuit and see what the output change is. Gain is delta_Vout/delta_Vin.
ok then i will use this :D thank you very much.
Excellent tutorial. Something that confused me was R4. I did the calculation with a Voltage divider with a Load (R4) and I got a different Volatge for the +Input. therefore if I analyze your schematic, Omitting R4 (because there is no Curent flowing through it,) will I get the same result? And if so, why is it necessary to include it in your example? Is R4 a mock component?
+Luis E. Batres Generally, on sensitive input pins we like to protect them from too much current in case the pin shorts internally. The circuit will fail if that happens, but without a resistor, the whole box might catch fire. This is especially true on MOSFET circuits, because the gate is just a capacitor with a very thin insulator to keep the thing from blowing sky high.
+Dan Bullard Thank You for your response. I am still conflicted though; because if as Rule 1 states: there is no Current going into the Inputs of a Op-Amp (for an Ideal Op-Amp that is, because of the High input Z of about 1M ohm for a Real Op Amp.) is your Low Resistance (in R4: 1k) helping in any way to prevent Damage or Over Heating?
+Luis E. Batres Because if the input shorts, the max current will be only in the milliamps. Some engineers bias op amps right from a power supply line capable of 10s of amps. What if the op amp fails and you have no resistor on that line? Even if in normal conditions the input Z is a megohm, if it shorts, (either internally or externally) you'll have grounded a 100W power supply which will probably burn up the runs on the board or start a fire in the component or elsewhere in the circuit. It's easy to use a power supply line to provide a static "1" or "0" to a digital circuit IC for example, but if the device shorts out internally, your power supply can now dump 10s of amps into the chip causing a real safety issue. A nominal resistor, say 1K, will not reduce the voltage much on a high impedance pin, but may prevent a fire, and that alone is a good reason to have a resistor on those pins.
+Dan Bullard Ok. Now I get it. So R4 is not cecessary in this "simple" circuit, but it is of Utmost Importance for the protection of other Circuits in ase this one fails. Thank You for the clarification and your excellent Analysis of an op-amp.
+Dan Bullard I had the same question too. I am assuming the same explanation holds for R5.
in a simple inverting opamp....what if there's no resistor in the feedback?
Make R7 and R9 0 ohms and follow my procedure. You'll get the answer yourself, which is the point of the video.
Very good explanation. Thank you.
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.
Hi Dan,
The gain is also defined as Vout/Vin. Won't that work as well for this problem? You then won't have to find a change in the output for a change in the input.
but how can there be no current in the input?
Hi Geert! Nice to see that you have moved from politics to electronics. Less chance of getting blown up for sure. The point is that the current is so small, it doesn't affect your calculations. Let's say that it has1nA of current flowing into one of the inputs. What's 1nA * 1K? 1uV. Is that going to give you extra points in your job interview, or will they be amazed that you solved their complex circuit in a few minutes like what happened to me at Maxim (twice!) in the interview. By the way, I got the job, and they were blown away that I was able to solve it so quickly.
if there is no current on inputs then it doesn't matter what are the values for R4 and R5? I even can remove them from circuit?
I get asked that a lot. There may actually be a very tiny current on the inputs, but so small they can usually be ignored. But for absolute accuracy they are included to keep the input impedance the same between the two inputs just to cancel out any voltage drops.
If there's no current through R4, how can be there are 2 v across R4 if Ohm's law said V=IR (I=0 ==> V=0)?
R4 never "Drops" 2V. when there is no current, a resistor transmits the voltage unhindered through it.
Dan Bullard Thank you Mr. Dan bullard. I still confuse. Why there's voltage at input if there's no current moving across R4 (no electrons moving)
Electrons moving is current, force of electron pressure is voltage. You don't need current for voltage to have effect. Voltage is like the pressure in a hose. Even if there is no water flowing, the hose will still bulge and be stiff because of the pressure.
Good explanation! Thank you again. One more question. With R4=1kOhm as a resistance, why is there still 1V on the right side of R4 (Op Amp input)?
Alexphal That would be the 'pressure' as Dan described. Another way to think of it is the current through R4 in reality is not absolute zero. There will be some very small current, as in picoamps or nanoamps (will vary with the op amp used). This would mean the voltage on R4 may actually be e.g: 1.0000000V on the left and 0.9999999 on the right.
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.
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.
i have a question:
i am using opamp in open loop configaration
vcc=12v, vee=0v, v- =2v, v+=(0 to 5v) ac square wave
output is=(3 to 12volts)ac sqr????
i am expecting output=(0 to 12 volts) ac sqr.
what happened here please explain.i want 0 to 12 instead of 3 to 12
Not all Op Amps are "rail-to-rail" op amps. Check the specs and see what the specs are for rail to rail operation
Thanks! I love how clearly you explained the method (it's exactly what my prof seems incapable of doing)
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!
Why do you have R4 and R5 resistors. If the op amp input impedance is essentially infinite why not connect direct without those two resistors?
They are there for two reasons. One, to make the point that no current flows, and two, very often you will find resistors on those pins and you have to learn to ignore them. Remember the topic: Solving Op Amps. When you see resistors on the inputs of the Op Amps, you must learn to ignore them. If I didn't show them, you would be wondering what to do and be unable to solve the circuit.
@@DanBullard thanks. You added them for modeling purposes. To show internal resistance (just thought you would pick 1megohm instead to show
@@brockchambers3733 No, if I make them Megohms then nanoamps suddenly matter. Almost every op amp circuit sports these small resistors. Making them Megohm resistors just invites disaster.
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.
Do you know about, how to read the program inside a microchip? I heard the firms can read secret probrams of Pics. What are the keywords of thise topic ?
No, if it's not documented, there is no chance of figuring out what's inside. When I started working on Smart Cards (ISO 80211), a bunch of Koreans I worked with created a microchip to talk to the Smart Card, but because it was a proprietary design that they didn't want to get stolen, they put their chip on the circuit board and then smothered it with potting compound (epoxy) to keep the secret from being stolen. Paranoid, but expected, as Koreans steal everything and when they do invent something they do all they can to protect their secrets.
@@DanBullard Thanks for answer. I herard that, with an acid the upper layers are solved then the inner region is detected.
@@karisikpekmez6850 There is no way to look at the silicon to figure out what the device will do. This is pure fantasy, stop listening to people who tell you such things.
@@DanBullard but we wonder how... If could be :?
same procedure is applicable if feedback on positive input..?
No way! Positve feedback is a completely differnt animal, but if you analyze it the same way you will figure out how positive feedback works. It becomes a Schmitt Trigger.
Can u make videos on solving cmos/pmos configurations
Hi Dan,
Do you know how synchro- servo systems work? I really like your videos they've helped me a lot. Do you think you can do a video on sychro servo systems? I have not seen any videos anywhere.