I totally depend on you. Class Mae lecturer Kya padhata hai pata nhi bas topics likh Leta Hun phir aapse padhta Hun and because of you I got very good marks in my mid sem
Can we have a tutorial based on complicated circuits which include a few of the operational amplifier circuits combined with each other? 12hrs away from a crucial basic electronics exam and you're a lifesaver. Thanks!
The timestamps for the different topics covered in the video is given below: 0:52 Non-Inverting O-Amp Configuration 1:51 Derivation of Closed Loop Voltage gain for Non-Inverting Op-Amp Configuration 5:00 Advantage of Non-Inverting Op-Amp configuration over Inverting Op-Amp configuration 6:09 Input Impedance of Inverting Op-Amp 7:25 Input Impedance of Non-Inverting Op-Amp 9:28 Op-Amp as Buffer (or Op-Amp as Voltage Follower)
why you considered negative feedback for both inverting and non inverting configuration and not the positive feedback ?Is virtual ground concept applicable for positive feedback? if no why?
you are one of the best instructors regarding electronics. I am watching your videos but l wish you care a little bit about your sound. you make tired but still, l couldn't find better one and l am listening to you, but your voice is super tiring. Please do something you are a smart person and you can make quadruple your likes only by smoother your voice or anything. The only problem is that.
Easiest explanation ever. I've always hated electronics but now I understand something! I'm in TYBsc graduating through physics. Thank you for helping us. :D
Wow, honestly this is amazing! I truly hope you are proud of the content you put out its just beyond anything given in class at my university. Short, clear and concise! Keep up the good work mate!
It a great videos! Im new with electronic circuit, not really know about the impedance you had mentioned. How is an infinite/high impedance op-amp will benefit a low impedance circuit?
Can I know What are possible variety of questions regarding the whole opam chapter it will be helpful for me Anyway ur videos helped me a lot luv ur videos❤️❤️💯
I have made couple of videos on solved examples on op- amp ( please go through the op-amp playlist). Moreover, for more questions please check the second channel ALL ABOUT ELECTRONICS - Quiz.
Thanks a lottt sir..! We have exam tomorrow and I was so scared because I didn't understand op amp.! But after watching your videos, now I understood everything. Thanks a lot sir.!😍😍 SIR IF POSSIBLE PLEASE UPLOAD A VIDEO ON THE APPLICATIONS OF JEFET AND MOSFET AND ALSO PHOTO ELECTRIC DEVICES..PLESE SIR..WE HAVE EXAM TOMORROW
Hello sir. Thank you very much for these videos, they are extremely helpful. I wanted to ask you, if you have any videos on feedback analysis (shunt-shunt/series-shunt/shunt-series/series-series), on how to find the feedback factor, input feedback resistance etc. Once more, thank you very much.
There is due to virtual ground both the input has same input voltage and we know that opam has a out put voltage is equal to =A×(v2-v1) whare v1=v2 and the out should be zero
I really like your videos! The videos are so helpful, thanks! Now I have a question about the "buffer configuration". I try to imagine the situation with an example: 1st, before switching on Vin, Vin = 0V Vout = 0V 2nd, then I switch on Vin: Vin = 5V => Vout = Vin-Vout = 5V - 0V = 5V 3rd, but then this Vout will be substracted from Vin, so: 5V (= Vin) - 5V (Vout) = 0V Then Vout = 0V. 4, And it runs in a circle like this: Vout is = 0V => Vin - Vout = 5V-0V = 5V Vout: 0V >> 5V >> 0V >> 5V... This example has something wrong in it, because the output is not switching between ZERO and 5Volt in real life. What point do I miss from the example? Thank you for your help! Mark
I suspect the answer to this question is that the voltage source and op-amp don't respond instantaneously. op-amps have a frequency response, which means the op-amp does not respond well to changes that occur faster than some time constant.
In your previous video, you added voltage source at the inverting end with a resistance R1. Now, in this video, while modifying the circuit you shifted the voltage source to the non-inverting side but R1 was kept at the same place. Why is it so?
Its because, using the feedback resistor and R1, the feedback voltage is provided back to the input. At 2:10, as I have mentioned, the feedback voltage is R1*Vo/ (R1 + Rf). So, in short, due to the feedback configuration, there is a Resistor R1 at the inverting terminal. The voltage across R1 is the feedback voltage at the inverting terminal.
Sir you said in your previous video(on inverting opamp) that virtual ground cocept can only be used in inverting configuration but here we have non-inverted configuration than how you applied virtual ground concept at 2:25
Something is missing in your explanation, I think, regarding the input impedance. If you realize, that the 2 grounds in both the inverting and non-inverting op-amps are on the same potential (i.e. connected by a wire), then non-inverting op-amp circuit is exactly the same as the inverting op-amp circuit. I understand, that this is a simplified explanation for students, but I have a feeling, that somebody is cheating on me here. But thanks for your videos anyways, I have learnt a lot from them.
THANK YOU FOR ALL OF YOUR GREAT VIDEOS ON OP AMPS!!! I have a simple requirement that I can't seem to find the answer for: I need an op amp in which I control the GAIN and OFFSET independently!!!!! If I set up a simple inverting amp where I can control the gain with Rf.....then will a variable +/-DC voltage at the "+" input simply move the waveform 'up and down' WITHOUT changing the signal gain??? IOW I need to shift the signal (a 50% duty cycle square wave) without effecting it's shape. THANKS MUCH AGAIN!!!!!
At 2:00, how was the voltage divider rule applied to get that formula? I can't seem to derive that... Also, how did we alter the connection of R1 like that?
Sir you have used input impedance = infinity for derivation of inverting input op-amp configuration . But in this video Input impedance is depending on R1 how??
I think you misunderstood that. The input impedance of the ideal op-amp is infinite. In the previous video, when I was saying Rin is infinite, it is from node X. (The input impedance of the op-amp is infinite, but not of inverting amplifier configuration). For the inverting amplifier, if you look from the source side (from where the input is applied), the input impedance is R1. I hope, it will clear your doubt.
Don't understand. If you supply a small voltage at the non inverting input then the op amp will amplify this non inverting input thereby sending this amplified signal to the inverting input. The voltage at v+ can't be the same as output voltage.
According to the equivalent circuit of OPAMP input impedance is the impedance which is present inside the opamp between the two terminals denoted as 'Ri'. But here you said in inverting amplifier input impedance is R1 which is actually the external impedance connect to inverting input. And if input impedance is R1 and not infinity there will be a current flow inside the terminal but we have considered that current flowing through R1 is equal to current flowing through feedback resistance which means no current is flowing into the terminal which is actually contradictory!
For any circuit, the input impedance is the impedance seen the by the source side (voltage/current source). And in fact is it is Vs/Is. When op-amp is operated in the open loop configuration, then no external resistor is connected. And let's say even if you connect any resistor R1 in the open loop configuration, then also Zi is very high. Because current drawn by the op-amp from the source is very low. Hence, Vs/Is is very high. And it can be said that input impedance is very high. In the case of feedback, for inverting configuration, the current drawn from the source is Vs/R1. (Due to virtual Ground) Hence, the impedance seen by the source for the inverting op-amp is Vs/ Is = R1. So, here even though the internal impedance of the op-amp is very high, the impedance seen by the source is R1. And hence, it can be said that the input impedance of the op-amp in inverting configuration is R1. I hope it will clear your doubt.
In video timeline 5:53 you stated that the input impedance of the negative terminal of the op amp is equal to R1. But isn't it more correct to state that the input impedance seen into the negative terminal should be R1//Rf?
RESPECTED ALL ABOUT ELECTRONICS GROUP: AT 1:25 In the blue diagram, when the voltage is supplied to the non-inverting point and to the ground, how can one say that the Vin as a whole is applied to the non-inverting point?? Please do reply as soon as possible ....
Because if you measure the voltage between the non-inverting input terminal and ground it will be equal to Vin. So, entire voltage is applied at the non-inverting input terminal.
If iI get it correctly, the potential difference between the 2 terminals of the circuit is Vs/2. But you're saying that the entire potential (Vs) appears between the terminal. Why?
Here this input impedance is the input impedance of the inverting op-amp configuration. It is the impedance seen by source. But in that video, I was talking about the input impedance of the op-amp. It is the impedance seen by looking into the input terminals of the op-amp. (For the ideal op-amp, the input impedance is infinite) I hope, it will clear your doubt.
Sir in the buffer circuit output of circuit 1 will appear in the vout Sir but what if the input resistance of circuit 2 is low, it would still get fractions of vout. So what is the point of using buffer?
Let's say the input resistance of source signal is 200 ohm. And the input impedance of the second circuit is 50 ohm. Input signal is 1V. In such case, the actual input appearing at the second circuit is 50x 1V /250= 0.2V. But the op-amp has low output impedance. Typically few ohms for some op-amps. Ideal op-amp as zero output impedance. But lets say our op-amp has 10 ohm output impedance. In such case, the input appearing at the second stage is 50*1V / 60 = 5/6V. So, buffer does make a difference. For very low input impedance it won't make difference. But input impedance of most of the practical circuit is more than 10s of ohm. So, buffer does make a difference. I hope it will clear your doubt.
Why the examples are always on the negative feedback meanning the feedback is connected to the negative input. It is because this configuration is more stable therefore largely used in practice than positive feedback?
Actually, I have the doubt that why it draws current at inverting and no current draws at non inverting... 1.U told me the concept of impedance based on this current draw it is good but why current is like that? 2. It is possible for inverting and non inverting with positive feedback?
You can refer op-amp by ramakant gyakwad. Which covers most of the required topics. If you want to go in detail then you can refer the op-amp handbook of analog devices by Walt Jung. But I think for the academic perspective the first one will be enough for most of you.
Really nice explanations on op-amps, but what i dont understand is that basically you made an assumption? in a previous video that A is about 10power6 and biased voltage is say 10v (+ and -) so therefore input is 10microV... so hence the virtual short and 0 potential difference at the 2 inputs (inverting and non inverting). am I correct? So what if this werent the case? if say there is a significant voltage applied at the input like 1-2Volts that you cannot assume is negligible? I mean do we always work with tiny input signals and hence the use of op-amps due to some favorable assumptions in functionality. thx!
The concept of virtual ground is applicable only with negative feedback. In that case, even if we are applying 1V or 2V voltages, the difference between the two terminal adjust the voltage accordingly so that we can get finite output. In open loop configuration, even if we voltages in mV, let's 50 mV at the non-inverting terminal and 10 mV at the inverting terminal, then also output will get saturated to the supply rail (Becuase of the very high gain). But with negative feedback, the gain of the op-amp is finite. Or in other words, we can say that the for the very high gain ,the effective input voltage appearing between the two terminal is almost the same. I hope it will clear your doubt.
