Operational Amplifier: Non-Inverting Op-Amp and Op-Amp as Buffer (Op-Amp as Voltage Follower)

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Operational Amplifier: Non-Inverting Op-Amp and Op-Amp as Buffer (Op-Amp as Voltage Follower)

Рет қаралды 773,009

Күн бұрын

In this video, the non-inverting op-amp configuration and how to use Op-Amp as a buffer or as a voltage follower (Unity Gain Amplifier) has been discussed.
In this video, the input impedance of both inverting and non-inverting opamp configuration has also been derived.
And inverting and non-inverting op-amp configurations are compared with respect to input impedance.
Non-inverting Op-Amp:
In non-inverting Op-Amp configuration, the input is applied at the non-inverting terminal of the op-amp and feedback is applied from the output the inverting end of the op-amp.
In this configuration, the input and output voltages are in phase with each other.
The input impedance of this configuration is ideally infinite and practically it is very high.
Op-Amp as a buffer (Op-Amp as Voltage follower):
Op-Amp can be used as a buffer in the non-inverting configuration. In this configuration, output voltage follows the input voltage. Or in another way, the gain of the Op-amp is one (Unity). That's why it is also known as unity gain amplifier.
The input impedance of this configuration is very high and that is why it can be used to isolate the different circuit stages.
The timestamps for the different topics 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)
This video will be helpful to all students of science and engineering in understanding the concept of non-inverting op-amp and understanding how to use the op-amp as a buffer or as a voltage follower.
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Пікірлер: 390

@tirthadas5767 5 жыл бұрын

3hr 30 min away from exam and u cleared everything.. thanks

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4 жыл бұрын

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@Physicsbyshalu 4 жыл бұрын

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@goyaldeekshant 4 жыл бұрын

All the best.

@ALLABOUTELECTRONICS 6 жыл бұрын

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)

@anujkondhalkar9776 6 жыл бұрын

ALL ABOUT ELECTRONICS 🤘 You Just Rocked.

@rajkishornishad1055 6 жыл бұрын

Good

@vinaykumarm9211 5 жыл бұрын

what an explanation!!U have done a great job.Timestamps concept in the description is very useful.

@badcaptain1089 5 жыл бұрын

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?

@Mubarak47 5 жыл бұрын

@@badcaptain1089 Positive feedback makes the circuit unstable that's why it is avoided.

@gianmarcogriffo3460 4 жыл бұрын

Exam lifesaver, providing a complete description of each operation case and a straightforward understanding method. Thank you so much again!

@ashishverma8740 5 жыл бұрын

Far better and interesting than NPTEL.. keep doing such work.. :D

@jamielannister509 5 жыл бұрын

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!

@pranayverma9984 4 жыл бұрын

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

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@divyagupta2358 2 жыл бұрын

If these teacher teach in school or colleges then there is bright future of india. 👍🏻

@user-nq4nt6ex2b 4 жыл бұрын

Best playlist for opamp! Loved the video.

@noelsmaison685 3 жыл бұрын

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

@pierre-olivier7523 9 ай бұрын

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@teerthraj_verma Жыл бұрын

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very useful video series for my studies . .thankyou very much for making videos about electronics

@rajeshsubramanyam6669 3 жыл бұрын

Very Simple & Clear explanation I ever seen

@nhatson3619 9 ай бұрын

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@aiyshafayaz6682 4 жыл бұрын

Concepts get clear in a very little tym . Thanks

@miraada7979 5 жыл бұрын

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.

@shubhamdesai7749 3 жыл бұрын

Voice me Kya rakha he?

@eonang9751 5 жыл бұрын

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?

@jinusiret 6 жыл бұрын

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@RaviShankar-xc9fu 6 жыл бұрын

Sir greatly explained ... Sir please complete....

@andya8623 2 жыл бұрын

Thank you sir..... Thank you so much Clear all OPM because of you

@amanisdreaming3914 Жыл бұрын

tomorrow I have a presentation on Op-amp, yet today Iknow nothing. Thankyou for this.

@divyanshverma4148 3 жыл бұрын

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@cadcamdesigntutorials 4 жыл бұрын

What a great explaination it is! Thank you so much 🙏🙏

@rssings8540 2 жыл бұрын

Listening to this in the exam hall thanks

@sangachidam3219 3 жыл бұрын

All your videos are excellent. Good job guys keep it up.

@sriiisriii795 3 жыл бұрын

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@rashmibajaj1396 2 жыл бұрын

Exceptional explanation, thank you so much

@happytulip1451 3 жыл бұрын

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@invinciblegirl4386 3 жыл бұрын

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

@Tejas.Lipare 6 жыл бұрын

Nice... It is helping me right now while studying Electronics 12th std. Good job.... I will complete all OpAmp videos...👌👌👌

@aparnatrivedi1497 5 жыл бұрын

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

@kontiimanalatit8987 3 жыл бұрын

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.

@ALLABOUTELECTRONICS 3 жыл бұрын

It is yet to be covered on the channel. Soon, it will be covered.

@debasishbhowmik1005 4 жыл бұрын

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❤️❤️💯

@ALLABOUTELECTRONICS 4 жыл бұрын

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.

@AnkitSharma-jf9fg 4 жыл бұрын

you should hold some live question sessions although good videos man I appreciate your work

@saumadeepsen8029 4 жыл бұрын

Vaiya many many many thnxx😘😘 keep going..... We need full semester 1 topics....

@inhdungho9509 3 жыл бұрын

thank you very much Sir ! as everyone did comment, I dont know what I should say more, just Thank you

@curious_enough1288 4 жыл бұрын

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?

@ALLABOUTELECTRONICS 4 жыл бұрын

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.

@dharunkumar8461 3 жыл бұрын

U have something beautiful in terms of explanation.Develop them

@n4v33nkum4r7 2 жыл бұрын

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?

@manogyurka969 3 жыл бұрын

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

@taravanova 11 ай бұрын

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.

@AnkitSharma-jf9fg 4 жыл бұрын

how to isolation work if voltage input of one circuit changes then voltage of input of second circuit will also change

@berenedain8427 9 ай бұрын

Woah what the hell?! That just clicked in my head now, thank you

@tpsicmin 2 жыл бұрын

Amazing Explanation Thank You Sir

@sionmarak1916 5 жыл бұрын

You sir are a god for me.

@sais.here.5869 4 жыл бұрын

bhai engg colleges should recommend your lectures

@binarysaiyan9389 3 жыл бұрын

Your I and Z looks similar. I was having a tough time figuring out which was what

@deepak1952 5 жыл бұрын

I am amazed that i understand this in a minute

@sammetakarthik1476 5 жыл бұрын

tq sir your lectures are very much useful

@rameshmadhuranga7429 3 жыл бұрын

best teacher

@abdullatifomar111 Жыл бұрын

Thanks you very much for your lecture

@Hari-kk7my 3 жыл бұрын

thank you sir ,you are a saviour

@sidcreator2132 3 жыл бұрын

You did it very easily 😍. Nice Explainations. Maja aa gaya. It would be better if you provide notes..✨✨✨✨✨✨

@ALLABOUTELECTRONICS 3 жыл бұрын

For notes, please check the website. www.allaboutelectronics.org

@dalenassar9152 Жыл бұрын

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!!!!!

@chenthanar895 4 жыл бұрын

Super sir if u teach me i will be the topper of my university.

@thisisKaushik 4 жыл бұрын

Sir there is a problem due to subtitles. Sometimes it hides the equations

@ARUNKUMAR-bu6jd 6 жыл бұрын

Very interesting... Sir!!!

@rohan8758 4 жыл бұрын

You are doing good but keep voice speed is less than the present speed..

@hrishie3757 3 жыл бұрын

Then it will be easier to watch them at a faster speed

@amitghosh3938 3 жыл бұрын

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?

@ALLABOUTELECTRONICS 3 жыл бұрын

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.

@amitghosh3938 3 жыл бұрын

@@ALLABOUTELECTRONICS thankyou sir very much it has cleared my all doubts 🤩

@KotanaPhaniSriRahulAP 5 жыл бұрын

Thanks borther! your videos have been helping me a lot

@sahilsingh9661 4 жыл бұрын

How does the 180 phase shift in the output voltage of inverting is a disadvantage?

@prateekmishra4661 6 жыл бұрын

How the log amplifier can be tuned around to provide antilog function?.

@ALLABOUTELECTRONICS 6 жыл бұрын

If you know the log amplifier using the Op-Amp, the diode is connected in the feedback between the output and the inverting terminal. The same circuit can be turned into the anti-log amplifier by interchanging the position of resistor and diode. So, in anti-log amplifier resistor will be in the feedback and diode will be connected between the input and the inverting input terminal.

@PavanKumar-lo1nr 5 жыл бұрын

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 ....

@ALLABOUTELECTRONICS 5 жыл бұрын

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.

@binarysaiyan9389 3 жыл бұрын

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?

@ALLABOUTELECTRONICS 3 жыл бұрын

Would you please mention, where you are referring to in the video?

@Abhisheksharma-lk4ll 4 жыл бұрын

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

@ALLABOUTELECTRONICS 4 жыл бұрын

What I mean to say was, it can only be applied with the negative feedback. ( When the op-amp is operating in the linear region).

@Abhisheksharma-lk4ll 4 жыл бұрын

@@ALLABOUTELECTRONICS ok Sir thank you very much :) i got it now

@islamicfact3010 2 жыл бұрын

Video is very good and clearing many doubt but I didn't understood one point, how Vx=(R1/R1+Rf)Vo ?

@tusharbehl199 4 жыл бұрын

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??

@ALLABOUTELECTRONICS 4 жыл бұрын

Would you please tell me where exactly in the video you are referring to ( timestamp). So, it will be easy for me to answer.

@SACHINKUMAR-qo5cg 4 жыл бұрын

sir !! u re best...Thanks

@ogbuddha7835 3 жыл бұрын

At 1:23 : Inverting input (-) is grounded and non inverting input have input voltage = Vin Doubt : since its a negative feedback opamp and one terminal is grounded then why other terminal i.e Vin isn't 0 ? Both the terminals have equal voltage isn't it?

@sejalvinodwasule8868 4 жыл бұрын

Sir please upload videos about transducers, Integrated circuits, U will cover 12th syllabus

@ojaspatil2084 4 жыл бұрын

At 4:25 you are giving the sinusoidal input of 1V to opam and getting the output of 3V sinusoidal signal. So, her 1 and 3 V are Vmax or Vrms??

@nandindwandwe1821 4 жыл бұрын

This helps thanks. You explain this well but you talk too fast sometimes.

@sejalvinodwasule8868 4 жыл бұрын

Then decrease speed

@poojaYadav-sv2xq 4 жыл бұрын

Sir there is a mistake in difference between inverting and non inverting opamp regarding to input impotence

@uiticus 8 ай бұрын

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?

@ALLABOUTELECTRONICS 8 ай бұрын

The more appropriate would be, for the inverting configuration, the input impedance seen through the input (applied at the inverting terminal) is R1.

@samarpreetsingh5318 4 жыл бұрын

Excellent video!👌

@shresthshukla3061 4 күн бұрын

00:18 Non-inverting op-amp configuration and using op-amp as a voltage follower. 01:39 Op-Amp in Negative Feedback Configuration 03:10 The non-inverting op-amp configuration allows control over the gain by adjusting the values of RF and R1. 04:43 Non-inverting op-amp configuration provides amplified AC signal with same phase as input. 05:59 The input impedance of the inverting op-amp configuration is equal to R1. 07:24 In the non-inverting op-amp configuration, the input impedance is infinite. 08:55 The non-inverting op-amp configuration has a high input impedance, allowing the entire applied voltage to appear across the circuit. 10:27 The non-inverting op-amp circuit can be used as a voltage follower or buffer circuit.

4 жыл бұрын

Sir, thank you so much. 😊

@cristianbretfelean2805 2 жыл бұрын

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?

@mayurshah9131 6 жыл бұрын

Just Superb

@mbukee Жыл бұрын

I want to ask something. At 11:11, there is a figure which shows the connections between circuit 1, opamp and circuit 2. In this configuration, we use opamp to transfer the signal from circuit 1 to the circuit 2 with gain=1. Why don't we connect the output of the circuit 1 to the input of the circuit 2? Why do we need to buffer the signal? What is the logic behind the opamp? Thanks

@ALLABOUTELECTRONICS Жыл бұрын

When there is impedance mismatch and when the second circuit has low input impedance then to avoid loading, it's advisable to use buffer. That's why the output of circuit is not directly connected to second Circuit. I hope, it will clear your doubt.

@bituphukon6954 4 жыл бұрын

Sir , in input impedance of non inverting OPAM we find Vin= V out. So what is the benefit of Amplification

@ALLABOUTELECTRONICS 4 жыл бұрын

I think you mean to say buffer circuit right. Well, the input impedance of the non-inverting amplifier is very high. So, whenever you want to avoid the loading effect, it is very useful. Basically, it is useful in driving the low impedance circuits. Also buffer circuit can provide a decent current to the other circuit. Some times the normal source (not necessarily a voltage source often derived from the power supply) won't able to supply that much current. In such scenario it is useful

@AnkitSharma-jf9fg 4 жыл бұрын

and why the wave at output is not clipped at every alternative time period as input at both terminals of op amp is same at some point

@lukaskunc2869 2 жыл бұрын

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.

@johanburger4454 4 жыл бұрын

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.

@Matter743 5 жыл бұрын

How input impedence of inverting depends on R1??!!!!!!

@anshumannegi2409 4 жыл бұрын

it doesnt bro. This guy has just somehow made a lecture with concepts that suit the situation without telling the real concept. all his previous lectures are like that only.

@Matter743 4 жыл бұрын

@@anshumannegi2409 that's the problem ...I wanted to learn concept ..but most video I see are like...they define some set of rules and around these rules whole video concepts will be based🤷🏻‍♂️🤷🏻‍♂️

@AJ-et3vf 2 жыл бұрын

Great video sir! Thank you!

@christophersekaran3848 6 жыл бұрын

Excellent video! Was really helpful!

@SadnanSanim00071 3 жыл бұрын

i didn't understand why we are taking the R1 fraction instead of Rf

@rimjhimghawry1344 2 жыл бұрын

Thank you for your help ✨💜

@GSPmediaO28 3 жыл бұрын

Asum explanation sir👍 tqsm

@huyhq_official1601 9 ай бұрын

Amazing ! Thanks so muchhhh

@AshwinGopa 4 жыл бұрын

Hi, quick question. For the input impedance comparison, if we apply the concept of virtual ground; then we get the same voltage at Vx as Vin right? Then in that case there will be current passing to Vx due to which Zin would be 1? Can you please explain this part to me, not able to understand as to why is Zin infinite here and not 1. Also, thank you so much for the videos; really helpful :-)

@mdapurbo8230 5 жыл бұрын

In the Negative feedback circuit,it is essential to have a phase shift of 180 degree between the Vin & Vf. Now Vout is feded back as Vf, multiplied Vout with a factor B. Now here,my question is,if I am giving input signal to the non inverting terminal of the op amp,Vout will surely be in phase with the input signal,then the feedback voltage Vf also be in the same phase of the input voltage i.e. between them 0 degree phase shift is presented.But the condition of negative feedback is : phase shift between Vin and Vf = 180 degree.Please help me to clear my confusion.

@ALLABOUTELECTRONICS 5 жыл бұрын

In non-inverting op-amp configuration, the feedback voltage is applied at the inverting input terminal. So, it always opposes the input signal. Or in other words, it is 180 degree out of phase with the input signal. And that's how the condition of the 180-degree phase shift is getting satisfied. I hope it will clear your doubt.

@mdapurbo8230 5 жыл бұрын

@@ALLABOUTELECTRONICS Thanks ❤

@iliyanbahchevanski862 5 жыл бұрын

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!

@ALLABOUTELECTRONICS 5 жыл бұрын

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.