Truth be known, I'm obsessed with this device (and the transistor, LOL). I was a circuit board designer (mechanical layout of circuits) for years and was always impressed by the many uses I've see with op-amps. This video instills great confidence! Thank you for the time you put into making these videos!
@CircuitBread Жыл бұрын
Nice! Op-amps were so confusing to me at first but once I started to figure them out and saw how incredibly useful they are, I admit I got a bit obsessed as well.
@followthetrawler5 ай бұрын
I was also a PCB layout draughtsman - never ceases to amaze me how we were just seen as monkeys with tape (yes I am that old) when in fact we could quite often have to fully understand the circuit design to lay it out properly - we in fact had the same skills as the circuit designer - and then some. Nice video chap.
@memirandawong Жыл бұрын
Mistakes make this learning experience even more real!
@remcooosten254211 ай бұрын
Nice video, thanks. In the last example, I think you have 1mA flowing through 2kΩ + 5kΩ, which gives us 1mA * 7kΩ = 7V
@gsiqhaiqj11697 ай бұрын
we dont know the voltage between the non inverting input and output, but we know that between the non inverting input and ground we have a 2V voltage drop, therefor we can calculate the curent that goes from node that connects the 2K and 7K resistor to the ground, which is 2V/2k=1mA. If 1mA exists that node, then 1mA needs to entry that node, and the only way is through the 7k resistor (KCL)
@wediadi6788 Жыл бұрын
Another method of calculating Vout using the gain of the opamp. Gain=Vout/Vin, since same current I1 is flowing through both resistors, Gain=R2I1/R1I1=R2/R1=25Kohm/10Kohm=2.5. Vout=Gain*Vin=2.5*0.5=1.25V. The sign of Vout should be negative because the input on the inverting port of the opamp is higher than that on the non-inverting. So Vout=-1.25V
@ufukdeniz30269 ай бұрын
great explanation
@kelvinbyabato8 ай бұрын
try to be careful with your solutions, so many errors and we are learning from you.
@xxbatman69xx984 ай бұрын
People make mistakes as long as he catches them it’s fine
@chuckbenedict7235 Жыл бұрын
I was confused about the sign of example 1 until I realized equation 2 should be (0-V0 ) = (50 microamps)(25K ohms), yielding V0=-1.25V. The voltage drop across the 25K resistor is 0-V0, which is the tricky part. In example 2, its strange to think about negative resistance (IMHO), so I'd suggest R = -V0/i1. It's the same math, just seems to better reflect what is happening in the circuit. In example 3, I believe the equation to compute V0 should be 2-V0 = (1mA)(5K) = -7V.
@mustafanaber Жыл бұрын
thanks for corrections, It made me understand. But In example 3, third equation; where is 5V comes from? Am I missing something?
@andy2a2m Жыл бұрын
from what you said 2-V0=5V, V0 will be -3V not -7V
@100subswovideos211 ай бұрын
the answer is actually 7 volts
@gsiqhaiqj11697 ай бұрын
@@mustafanaber 1mA*5k ohms
@sirhuman56011 ай бұрын
amazing explanation and bringing us in your mind to let us learn how to think while solving problems. thank you!
@zjmullan2 жыл бұрын
great video man, reminded me of how the ideal op-amp calculations work after almost a year of not doing them
@CircuitBread2 жыл бұрын
Glad it was helpful!
@BaviBavii-jw4dn5 ай бұрын
Hi sir!! I'm bavithra from Tamilnadu.I had a Cadence interview on April 28. I watched your video I was very useful sir. Thankyou sir for teaching me in understandable manner.But I need your help sir I didn't have enough materials to practice these type of problems.Can you refer any book for preparing my interview sir? Thankyou sir. I hoped you will reply to my comment sir.
@CircuitBread5 ай бұрын
I'm glad you found the information helpful! For further study, you can explore our textbooks available at www.circuitbread.com/textbooks. Utilize the search bar on the left to quickly locate topics of interest.
@Zzzzzzzzz6786 ай бұрын
Seeing you get a bit lost in the solution made me feel slightly better about myself, lol. But it was well explained, and I've subscribed to your channel.
@maneki9neko Жыл бұрын
very clear explanation. Thank you for taking the time out of your life to do this.
@CircuitBread Жыл бұрын
You are very welcome
@armisol006 ай бұрын
how do u calculate in the third practice problem the 2 volts. cause i dont understand that part
@bunnie75566 ай бұрын
So because there is a voltage divider you use Vin ( R2 / R1 + R2 ) to get Vout. Which is 3v ( 8k / 4k + 8k ) giving you Vout of 2v.
@WalyB0110 ай бұрын
Putting the formula their gives your brain more power to calculate other stuff. Basically like working memory on a pc. Weirdly enough we are wired quite similarly
@firstlast98134 күн бұрын
8:20 Sorry. To me, this isn't as simple. How did you know to drop 1 volt over the 4k resistor and 2 volts over he 8k resistor?
@Aakanksha.KPoojari-tq5uf16 сағат бұрын
Yessss
@wediadi6788 Жыл бұрын
My question is why do'nt we consider the 25K resitor when we are calculating I1? Isn't the same current I1 flowing through both resistors?
@abrahamjones363511 ай бұрын
That's exactly what I was thinking, if you found an answer please let me know.
@gsiqhaiqj11697 ай бұрын
We have .5V from the source to the non inverting input, not from the source to the output. That means that between the source and non inverting output (where the potential is 0V) we have a .5V voltage drop, therefore we can calculate the current that goes through the 10K resistor. The voltage between the source and output is 0.5V+1.25V=1.75V, and if we apply ohm law we have i=1.75V/(10k+25k)=50uA, I hope this make it more clear@@abrahamjones3635
@magicflour Жыл бұрын
Another thing I would explicitly state for the pedagogy of circuit theory is that you should never do nodal analysis on Vout as it does not have the same assumptions for conventional nodes in that current does not accumulate in any node. I find that some of my tutees did not know this because it was simply implied but never clarified.
@sunrayseducation Жыл бұрын
Sir, thank you for this explaination and including the mistakes that I do.
@andrewpervez3009 Жыл бұрын
Excellent and straight forward. Thank you.
@INONAuto11 ай бұрын
i am writing on the 9th nov ,you saved me from lot of anxiety
@RixtronixLAB Жыл бұрын
Cool video, well done, thank you for sharing it :)
@CircuitBread Жыл бұрын
Thanks for watching!
@kaleevans16923 ай бұрын
Best description on op amps I've found yet... And I've looked around a lot. Subscribed! Thank you sir🙏🏾
@monfry2675 Жыл бұрын
All understandable in a easy way ! Thanks good buddy !
@KyleMalcolm24 Жыл бұрын
I am a bit confused between Example 2 and 3. How come in example 2 it is 0 - Vo but in example 3 it is Vo - 2V? (I am asking about the Vo being subtracted vs not)
@CircuitBread Жыл бұрын
Great question! It is somewhat arbitrary. As long as you state your current flow assumptions correctly, you can do it either way. So, in example 2, I assumed that current was flowing from the inverting input to the output. Whereas in example 3, I assumed that the current was flowing from the output to the inverting input. That's the only reason. I could have assumed the opposite and swapped those, but then the sign of the current (whether it was positive or negative) would have flipped, because my assumption of the direction would be different. Does that help?
@Cookingcousin Жыл бұрын
I’m confused on that too actually, if you assumed current is flowing from high to low when solving for current down the 2kohm resistor wouldn’t it be from positive to negative across the 5kohm resistor as well (from the negative terminal to Vout)
@abiuniverse21 күн бұрын
At 9:36, how do we know that the current through 2KOhM resister (i1) is the same as through the 5KOhM resister. Shouldn't the current be different on the 5Kohm side? Not intuitive there.
@menkveld35659 ай бұрын
I never understood anything about those things. Seeing those 2 rules being applied made it all click. Thanks!
@ramupattu94444 ай бұрын
I1 = 5 /10 k = 0.5 milli amps or 500 micro amps (not 50 micro amps) Voltage drop across 25 k ohms= 0.5 ma* 25 k = 12.5 volt Vout = - 12.5 v
@dawsonbrown39573 ай бұрын
I am confused about the negative voltage, do you have a video that explains what that negative voltage means?
@CircuitBread3 ай бұрын
This confused me when I was in college as well - we discuss it briefly in our intro video and are actually doing final review of another script that will touch on it again (at our pace, it will probably be another 1-2 months before it's published). Here's the one we did awhile ago: kzbin.info/www/bejne/sHmQZp-pZb50f5Isi=hKlGOWLUR6hBOgak
@bilisuma_b8 ай бұрын
me waiting for the toaster to pop up................... what's the use of the toaster at the end? I didn't get it. 😂
@CircuitBread8 ай бұрын
Haha, it’s amazing how many people comment on the toast not popping up. The toast/toaster is just part of our branding and we were originally going to do a bunch of different outros but never got around to it, so for a lot of our videos we just have the toast going down. I guess it adds some suspense. 😅
@liliaoakeley4402 Жыл бұрын
circuit bread #1 fan right herreeee
@StevieOU8128 ай бұрын
I am confused, how can there be a -1,25 volts on the Vout? Shouldn't that be positive? Just need some help. Sorry for the dumb question.
@StevieOU8128 ай бұрын
I think I get it, it is a negative signal compared to ground? Correct?
@CircuitBread8 ай бұрын
Exactly! 0V or ground can be somewhat arbitrary. It's just a reference point and to have a negative voltage just means the voltage is lower than that arbitrarily chosen reference voltage level.
@jameskidd7906 Жыл бұрын
I m the 555 th person liked this vid . 555 a great symbole for the great ic 😅 thank sir for the vid
@Speed-dq2sn9 ай бұрын
🤨
@tensorprodukt9 ай бұрын
its not 50 micro amps .. its 500 micro amps ... so the output should be -12.5 V I suppose
@gsiqhaiqj11697 ай бұрын
we have 0.5 V, not 5V, it's easy to mistake because he put .5
@adarshsk099 Жыл бұрын
Why in second problem o -vo was taken but in last problem it was opposite?
@andy2a2m Жыл бұрын
To anyone that is still confused: In example 1: the 25k ohm is the feedback resistor Rf, the 10k ohm is the input resistor Rin, Because the supply is connecting to negative side, which means the signal must be inverted and amplified, The multiplier is Rf / Rin = 25k ohm / 10k ohm = 2.5, again, it is inverted, so it is -2.5 So the final answer, the output is 0.5V * -2.5 = -1.25V
@andy2a2m Жыл бұрын
In example 3, because the current path is only one way, so, we can use 1mA multiply by 2k + 5k ohms (they are in series, no current going into op amp as the rules stated), the answer is 0.001A*7000ohms, 7V !
@grzesiek1x Жыл бұрын
2:41 a small question as I have already seen similar explanation but one thing is not allowing me to sleep ;) why don't we add those resistances? I knoe that no current enter the inverting input of OpAmp but I see two resistors here 10 k and 25 k but we calculate current for the one of them 10k ? I don't get it. Normally to calculate the current I would add all resistors and divide it by the voltage right? for example 5V / 10 000 ohms + 25 000 ohms to get the total current ..or we just treat that 25k resistor as an "invisible" resistor for the moment to calculate the right output of the op-amp or something like that?
@CircuitBread Жыл бұрын
That's a great question. The reason we only calculate the current for the 10K resistor is because, in this question, we know the voltage across the 10K resistor (because we know the voltage at the inverting input) but NOT the voltage across the 25K resistor (because we don't know Vo). So, given what information we're given, we cannot calculate the current through both of them from the beginning. But we know they have the same current through them so we find the current through the 10K resistance and, with that information, we find the voltage across the 25K resistor, giving us Vo.
@mickhursey48022 жыл бұрын
It would be useful for dunces like me if you explained the 'obvious'🤔 simple maths, I have no idea how you got your 5V from in the last equation
@CircuitBread2 жыл бұрын
Dang! It's always hard to find the balance between boring people with too many details or confusing people with too few details. Sorry that the balance wasn't quite right for you... For that, what happens is that the 1mA on the right gets multiplied by the 5Kohms on the left (to cancel it out of the denominator on the left), so .001A*5000Ω = 5V.
@mickhursey48022 жыл бұрын
@@CircuitBread Thanks for that. Not a complaint, I loved the pace of the video just my math isn't up to scratch so ended up rather confused😮💨
@yahyaalhashmy1313 Жыл бұрын
can you explain how you dropped volts using simple math in 3rd question, when 3 volts passes 4R it drops one volt and when it passes thro 8R it drops 2V, how?... I ain't no math genius
@CircuitBread Жыл бұрын
The 3V are dropped over 12K ohms total. I'm not sure if it helps to think about it this way but, with 3V, it's easy to divide that 12K ohms into three parts and you will drop 1V per 4K ohms. So you'll drop one volt over that 4K resistor - giving you 2V at the non-inverting node. This is a semi-intuitive way to look at it but may not be the best when approaching it for the first time. The more rigorous way to do it would be to look at it like a voltage divider. I recommend checking out our tool on this that can help you calculate it, shows the actual equation, and can let you play with it until you get a better intuitive feel for what's going on: www.circuitbread.com/tools/voltage-divider
@paulwood3460 Жыл бұрын
Nice one. Suggest always to put a zero in front of a real number that only has a fractional part I.e 0.5 not .5 I was scratching my head because I thought it was 5 👍
@CircuitBread Жыл бұрын
That has caught a couple of people - I'll try to remember to do that in the future. Thanks for the feedback!
@phionadesitter12032 жыл бұрын
excellent explanation 👍🏽
@Jonhalloum Жыл бұрын
In exercise 1, since this is a non inverting amplifier, couldn’t we use (25/10)0.5 ?
@CircuitBread Жыл бұрын
Kinda! This is actually an inverting amplifier and the equation for Vo/Vi is -R1/R2, so we could do -(25/10)*.5 to get the final output a lot faster. For most situations, you'd just use this equation to make your life simpler, but the idea was to show how to solve for op-amp problems utilizing the characteristics of an op-amp. For this, the learning is in the process more than the result. Hope that helps!
@astrophotographysometimes2303 Жыл бұрын
at 9:08 why did he find the current across i1?
@boonedockjourneyman7979 Жыл бұрын
Good job.
@bendtrom7200 Жыл бұрын
I don’t want to be the wise guy because I am not. Perhaps there is a small math error in the above video in the beginning 2:25 minutes in the video? U (Volt= V) / R (Ohm = Ω) = I (Amperes = A) therefor 5V/10KΩ i.e., 10000 Ω = 0,5 milli amperes or 0,0005 Ampere or 500µAmpers (Micro Amperes) not 50µAmpers. So 0.0005A x 25KΩ = 12,5 V not 1,25V. Ω=ohm symbol
@CircuitBread Жыл бұрын
Hey Bendt, I had no idea this was going to cause so many problems when putting this problem together but there have been many comments on this. The period in front of the .5V seems to be missed very easily. So, your math is totally right but with the wrong input. It's .5/10K not 5/10K, so the answer in the video is actually correct. I appreciate you taking the time to provide the feedback!
@gsiqhaiqj11697 ай бұрын
@@CircuitBread yeah, putting .5 instead of 0.5V was a mistake, easy to overlook the point.
@thomaslacey10012 жыл бұрын
Thanks 👍
@thespencerowen Жыл бұрын
How did you get 1ma across the 5k resistor in problem 3? Shouldn't that be 0.4ma ?
@CircuitBread Жыл бұрын
Hi Spencer! Since we know the inverting input is at 2V and the resistor to ground is 2K ohms, we know that the current through that resistor is 1mA. As the 2K and the 5K resistors are in series, the current through the 5K resistor has to also be 2mA. And that's how we find the output voltage of the op-amp. I hope that helps!
@thespencerowen Жыл бұрын
@@CircuitBread Ah thank you. I was following the current from the wrong pin so I didn't notice that they were in series.
@fernandogallardo34772 жыл бұрын
For practice problem 1, 5V/10K-ohh is not 50 micro-amps, it will give 5e-4.
@CircuitBread2 жыл бұрын
Phew, you scared me! I hate it when I mess up my math. In this case, you are correct with your math but you're reading it wrong. You're using 5V, not .5V. In the question, it is .5V/10K, which does give us 50 microamps. Thanks for the scrutiny and feedback, I *do* like to be corrected when there are mistakes, so don't hesitate if you see anything else that looks out of place.
@TomVanGaever Жыл бұрын
In the first problem, the output should be -12.5V, not -1.25V. The closed loop amplification of the inverting amp is -2.5 not -0.25 (Acl = 25k/10k)
@CircuitBread Жыл бұрын
The output should be -1.25 but you are correct about the amplification. I think the confusion comes from the input voltage, which is .5, not 5. .5 * -2.5 = -1.25
@test89267 Жыл бұрын
I also felt in this trap. Did calculations with 5v instead of .5v
@ChrisLee-yr7tz2 жыл бұрын
9:04 Do some quick Meth? No thanks, it's too early.
@CircuitBread2 жыл бұрын
😂 And here I get on my kid's case for speaking clearly...
@panospanos81408 ай бұрын
whats up with the toaster
@CircuitBread8 ай бұрын
The fact that we show a toaster or that the toast doesn’t pop up?
@klevisimeri607 Жыл бұрын
Thank you!
@CircuitBread Жыл бұрын
You're welcome!
@Motivational_women13 ай бұрын
His answers were wrong
@CircuitBread3 ай бұрын
All of them? Any of them? Could you let me know what you're referring to? We've had questions in the past about these but it seems like the biggest mistakes he made was not investing in handwriting practice.
@ravichandransrinivasan5453 Жыл бұрын
The gain is 2.5 so output voltage is -12.5 v please edit video
@CircuitBread Жыл бұрын
The input voltage is .5, not 5. So -2.5 *.5 = -1.25 not -12.5V - those decimal points can be easy to miss!
@georgeifarouq1082 Жыл бұрын
Hi for example 1 could you explain why the current going through the 10k and 25k resistors was defined by 0.5V/10k Ohms instead of 0.5V/(10k + 25k) Ohms?
@CircuitBread Жыл бұрын
Hi Georgei! The important thing here is to remember that the inverting and non-inverting inputs to the op-amp are the same voltage level. Since the non-inverting input is tied to ground, you can assume that the inverting input is tied to ground. Thus, the current going through the 10K ohm resistor is found by (0.5-0)/10K. And, since you know no current goes into the input, you know that same current has to go through the 25K ohm resistor. I hope that helps!
@Hamza__E Жыл бұрын
@@CircuitBread I had the same question and you cleared it up perfectly! Thank you so much! Also, I just want to say how thankful I am that you answered this. Even though the video was uploaded 11 months ago.
@mcpoogle9058 Жыл бұрын
Im actually still confused on this; if no current flows into the inputs then the only current path would be through both the 10k and 25k resistor right? So wouldnt the current flow be defined as .5 / (10k+25k) ?
@gsiqhaiqj11697 ай бұрын
@@mcpoogle9058 hi, I know this is an old comment but i will explain anyway. We have .5V from the source to the non inverting input, not from the source to the output. That means that between the source and non inverting output (where the potential is 0V) we have a .5V voltage drop, therefore we can calculate the current that goes through the 10K resistor. The voltage between the source and output si 0.5V+1.25V=1.75V, and if we apply ohm law we have i=1.75V/(10k+25k)=50uA, I hope this make it more clear
@monoc4d Жыл бұрын
Great video... I was struggling with op amps, your video help me enormously. I have a question. How at 6:41 you get negative R? I rewatch that part many many times and still don't understand. Thanks!!!
@CircuitBread Жыл бұрын
Hi there! This is a problem that I've seen a lot and have definitely done myself. It goes back to first step where we show the equation as (0 - Vo)/R = i1 It is very easy to drop the 0 and forget that the negative sign stays! So that's where it's simplified to -Vo/R=i1 and then that negative sign switches over to the other side. Does that help?
@magicflour Жыл бұрын
Another way to state it is that (conventional) current always flows from a higher voltage to a lower voltage. So the zero volts at V+ *has* to be higher than what Vo is. Thus Vo must be negative valued.
@prabhjotsingh69415 ай бұрын
@@CircuitBread if we subtracted 2v from Vo in equation that would give -3v. is this correct?
@artysanmobileАй бұрын
You are teaching some very bad information here. I’ve taken my time to see if it’s just interpretation, but it isn’t. 1. The voltages at the two input pins are NOT always the same! If they were, the OpAmp would have no use. An OpAmp is an ideal differential amplifier. That’s all it is. The voltage difference at the two opposite polarity inputs is the input signal of any differential amplifier. If there is no difference, as you say explicitly over and over, the output will always be 0v. 2. The output pin should be considered a perfect current sink. What’s a perfect current sink? Well, ground is. Treat it as ground for the purpose of feedback R value computations.
@sejlahadzic9712 Жыл бұрын
kilo -> k Kelvin -> K
@neverhavingfun Жыл бұрын
👁🔺
@michaeldrake25122 жыл бұрын
This doesn’t have nearly enough views! I would literally lose sleep over how op-amps worked and circuit analysis in general. Thanks to this I’ll sleep peacefully now. I really feel that the theory for electronic circuits is made overly complicated by those that are teaching this but the way you’ve explained this was straight to the point! Perhaps that’s something to do with limited application of their skills outside of an academic course. I think it’s brilliant you leave your mistakes in because, even though it’s simple algebraic operation still so easy to make a mistake. Seeing the reasoning is good
@CircuitBread2 жыл бұрын
Thanks! I'm still self conscious of my mistakes but I hope they help people somehow. And, particularly, I'm glad this was helpful to you!
@sebastiandelbanorollin54822 жыл бұрын
How is Sergei these days?
@CircuitBread2 жыл бұрын
It's complicated and I don't know how much of his personal life he'd want me to share but he's alive, healthy, and unbelievably stressed and sad. When I met him he was living in Mariupol and there'd be times we were talking during the 2014 Crimean invasion that there'd be explosions in the background, so that's why he moved to more central Ukraine. But Mariupol is where he grew up and he still has family there, so it's heartbreaking to see it as an active warzone.
@sebastiandelbanorollin54822 жыл бұрын
@@CircuitBread sorry to hear that. send him best wishes and thanks for his great work.
@HS-cu6iw2 жыл бұрын
hey! love the vid! I seem to be getting -12.5V for question 1 because my i1 value is 500mu Amps, not 50mu Amps..?
@CircuitBread2 жыл бұрын
Thanks! Did you happen to use 5V instead of .5V? That seems to be the most likely cause making your current 10x what it should be.
@HS-cu6iw2 жыл бұрын
@@CircuitBread omg i cant believe i missed the "." 😂..reminder for me to pay close attention! Thanks again, btw perfect release time; I am prepping for my exams and opamps a big chunk of it!
@danielkioko53787 ай бұрын
Had missed it too.
@gkdresden7 ай бұрын
These were more the theoretical problems with op amps. The practice problems are much different, like: - the input current is not really zero, especially not in the case of bjt op amps - in certain cases you need to take care of the input offset voltage, especially in the case if you use op amps as comparators. - when you operate with pulse voltages at the inputs you have to make sure, that the operating voltage stays stable, even if you need to operate with a certain capacitive load at the output. So you have to use stabilization capacitors between the Vcc and GND pin of your op amp or two of them if you operate with a symmetrical operating voltage. Otherwise your circuit will fall into oscillations - the nonlinearities of the op amp input circuit can demodulate rf. So if you want to build up an audio amplifier it is sometimes necessary to use a low pass filter at the signal input to suppress rf interferences to avoid that you have some closely located radio station music in the background of your audio amplifier output. - if you need higher voltage amplifications (for microphone or phono signals), you need to make sure, that the gain bandwidth product doesn't compromize your amplifier at the higher frequencies. Also the input referred voltage offset and noise voltage become important in such application cases. etc.