Mistakes make this learning experience even more real!

Best description on op amps I've found yet... And I've looked around a lot. Subscribed! Thank you sir🙏🏾

great video man, reminded me of how the ideal op-amp calculations work after almost a year of not doing them

amazing explanation and bringing us in your mind to let us learn how to think while solving problems. thank you!

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!

very clear explanation. Thank you for taking the time out of your life to do this.

i am writing on the 9th nov ,you saved me from lot of anxiety

I never understood anything about those things. Seeing those 2 rules being applied made it all click. Thanks!

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.

Sir, thank you for this explaination and including the mistakes that I do.

Excellent and straight forward. Thank you.

Nice video, thanks. In the last example, I think you have 1mA flowing through 2kΩ + 5kΩ, which gives us 1mA * 7kΩ = 7V

All understandable in a easy way ! Thanks good buddy !

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

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.

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

how do u calculate in the third practice problem the 2 volts. cause i dont understand that part

circuit bread #1 fan right herreeee

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.

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.

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?

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

excellent explanation 👍🏽

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

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)

I am confused about the negative voltage, do you have a video that explains what that negative voltage means?

Good job.

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.

Thank you!

Cool video, well done, thank you for sharing it :)

Why in second problem o -vo was taken but in last problem it was opposite?

Thanks 👍

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?

In exercise 1, since this is a non inverting amplifier, couldn’t we use (25/10)0.5 ?

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.

at 9:08 why did he find the current across i1?

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!!!

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 👍

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?

How did you get 1ma across the 5k resistor in problem 3? Shouldn't that be 0.4ma ?

hey! love the vid! I seem to be getting -12.5V for question 1 because my i1 value is 500mu Amps, not 50mu Amps..?

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

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)

For practice problem 1, 5V/10K-ohh is not 50 micro-amps, it will give 5e-4.

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

try to be careful with your solutions, so many errors and we are learning from you.

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

me waiting for the toaster to pop up................... what's the use of the toaster at the end? I didn't get it. 😂

The gain is 2.5 so output voltage is -12.5 v please edit video

I m the 555 th person liked this vid . 555 a great symbole for the great ic 😅 thank sir for the vid

its not 50 micro amps .. its 500 micro amps ... so the output should be -12.5 V I suppose

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

kilo -> k Kelvin -> K

How is Sergei these days?

whats up with the toaster

🤨

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.

9:04 Do some quick Meth? No thanks, it's too early.

His answers were wrong

👁🔺