3hr 30 min away from exam and u cleared everything.. thanks
@abhishekranjan4170
5 жыл бұрын
Story of every engineer...
@balajidasar396
5 жыл бұрын
@@abhishekranjan4170 soo true.. 😂
@vigneshsambari1733
5 жыл бұрын
The mere truth of any btech guy!!!
4 жыл бұрын
@@abhishekranjan4170 agree
@abhishekranjan4170
4 жыл бұрын
@ 🤗🤗
@Physicsbyshalu
4 жыл бұрын
I'm preparing for NET exam online because my classes are dismissed due to corona virus impacts....but your classes help me a lot.. thanku so much...
@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.
@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!
@gianmarcogriffo3460
4 жыл бұрын
Exam lifesaver, providing a complete description of each operation case and a straightforward understanding method. Thank you so much again!
@pranayverma9984
5 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
@divyagupta2358
2 жыл бұрын
If these teacher teach in school or colleges then there is bright future of india. 👍🏻
@ashishverma8740
5 жыл бұрын
Far better and interesting than NPTEL.. keep doing such work.. :D
@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
@AdityaKushwaha-r5l
4 жыл бұрын
Best playlist for opamp! Loved the video.
@teerthraj_verma
Жыл бұрын
Tomorrow I have semester exam And my last resort is watching you I am watching this for first time and An best part is that i understood most of it
@technicalpsc6441
5 жыл бұрын
best classes i ever watched...
@pierre-olivier7523
11 ай бұрын
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!
@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?
@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?
@tharindumaduwantha1023
2 жыл бұрын
very useful video series for my studies . .thankyou very much for making videos about electronics
@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?
@amanisdreaming3914
Жыл бұрын
tomorrow I have a presentation on Op-amp, yet today Iknow nothing. Thankyou for this.
@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.
@rajeshsubramanyam6669
4 жыл бұрын
Very Simple & Clear explanation I ever seen
@invinciblegirl4386
4 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
@uiticus
10 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
10 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.
@thisisKaushik
4 жыл бұрын
Sir there is a problem due to subtitles. Sometimes it hides the equations
@aiyshafayaz6682
4 жыл бұрын
Concepts get clear in a very little tym . Thanks
@jinusiret
6 жыл бұрын
Expecting more on electronics from this channel
@jinusiret
6 жыл бұрын
Super Video, Now i understand a little, if my faculty following this method then i will pass easily.
@pujithakaricheti8170
5 жыл бұрын
Gud job listening hours and hours is not a matter getting sub with in 15 min really it's very helpful thank you so much
@nhatson3619
11 ай бұрын
thanks a lot for your dedication to free but high-quality education for everyone!
@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 🤩
@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?
@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!!!!!
@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??
@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
how to isolation work if voltage input of one circuit changes then voltage of input of second circuit will also change
@thesilentvampire
5 жыл бұрын
It Helps me with my studies, thanks Yeah just that.
@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...👌👌👌
@SadnanSanim00071
4 жыл бұрын
i didn't understand why we are taking the R1 fraction instead of Rf
@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?
@sahilsingh9661
4 жыл бұрын
How does the 180 phase shift in the output voltage of inverting is a disadvantage?
@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.
@andya8623
2 жыл бұрын
Thank you sir..... Thank you so much Clear all OPM because of you
@divyanshverma4148
3 жыл бұрын
Machadiyaa sir jii... So great video lectures
@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?
@bituphukon6954
5 жыл бұрын
Sir , in input impedance of non inverting OPAM we find Vin= V out. So what is the benefit of Amplification
@ALLABOUTELECTRONICS
5 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
@minakshee4672
5 жыл бұрын
I didn't understand the very last part why we used voltage follower and what was the point behind doing that Please reply in comments thank you
@ALLABOUTELECTRONICS
5 жыл бұрын
In the voltage follower, the op-amp acts like a buffer. It provides high input impedance and low output impedance. And very useful when you want to connect the two circuits. (in impedance matching perspective).
@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 ?
@poojaYadav-sv2xq
5 жыл бұрын
Sir there is a mistake in difference between inverting and non inverting opamp regarding to input impotence
@mbukee
2 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
2 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.
@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
Жыл 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.
@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.
@RaviShankar-xc9fu
6 жыл бұрын
Sir greatly explained ... Sir please complete....
@arfatshaikh4669
Жыл бұрын
Sir why u take Voltage divider rule in non inverting amplifier and why not in inverting amplifier
@Conceptuallearning16
Жыл бұрын
11:18 Sir then why Don't we connect both the circuit with just a simple wire.....then?? Why i have to use buffer instead of just a simple wire?? Thank you sir
@uiticus
10 ай бұрын
That is because you want to isolate the input circuit from the output circuit.
@ShreyaRoyCello
2 жыл бұрын
When explaining Non Inverting Op -Amps you said that Vx= (R1/(R1+Rf))* Vout. Why is it (R1/(R1+Rf) and not (Rf/(R1+Rf). I couldn't understand that bit
@ALLABOUTELECTRONICS
2 жыл бұрын
Its voltage divider rule. Since no current is flowing in op-amp terminal (assuming ideal op-amp), the current is flwoing through Rf and R1. Vx is the voltage across R1. Using voltage divider rule, it is R1*Vo/ (R1 + Rf)
@saumadeepsen8029
5 жыл бұрын
Vaiya many many many thnxx😘😘 keep going..... We need full semester 1 topics....
@sangachidam3219
3 жыл бұрын
All your videos are excellent. Good job guys keep it up.
@cadcamdesigntutorials
4 жыл бұрын
What a great explaination it is! Thank you so much 🙏🙏
@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.
@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
@dharunkumar8461
3 жыл бұрын
U have something beautiful in terms of explanation.Develop them
@durgaprasad_S
2 жыл бұрын
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?
@happytulip1451
3 жыл бұрын
No word's sir... I Subscribed your channel....
@rashmibajaj1396
2 жыл бұрын
Exceptional explanation, thank you so much
@AnkitSharma-jf9fg
4 жыл бұрын
you should hold some live question sessions although good videos man I appreciate your work
@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
@somesharunvn
4 жыл бұрын
Input impedance of an inverting Amp you considered Virtual GND. Why don't you consider it for the non-inverting?
@ALLABOUTELECTRONICS
4 жыл бұрын
With the negative feedback, V+ = V-. In case of the non-inverting op-amp, inverting and the non-inverting inputs are at the same potential. (i,e V+ = V- = Vin) And to find the input impedance, we are finding the ratio of Vin and Iin at the non-inverting terminal. So, there is no question of virtual ground here. In case of the inverting op-amp, the input is also at the inverting terminal and the non-inverting terminal is grounded. So, virtual ground concept was applied. I hope it will clear your doubt. If you still have any doubt, let me know here.
@karandev9316
3 жыл бұрын
At 5:44 you mention that non inverting op has high impedance while deriving Acl for inverting opAmp you told us that it has high impedance. Which is why Ir = If...can someone please explain that part?
@ALLABOUTELECTRONICS
3 жыл бұрын
For, non-inverting op-amp, I was referring to the input impedance of the entire non-inverting circuit. (op-amp with external resistors). Of course, the input impedance of the op-amp is very high. But when we connect external resistors, then it changes the characteristics of the circuit. In case of the inverting op-amp the input impedance is equal to R1. (it is input impedance of the entire inverting circuit ). While I was referring to the Ir = If, I was referring to the input impedance of the op-amp. (That's why it can be assumed that, no current is entering into the op-amp terminal) I hope, it will clear your doubt.
@Abhisheksharma-lk4ll
5 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
5 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
5 жыл бұрын
@@ALLABOUTELECTRONICS ok Sir thank you very much :) i got it now
@sriiisriii795
3 жыл бұрын
super...................... no words................... only all about electronics.................
@starrynightskyable
5 жыл бұрын
Hello I am confused with the input impedance of the inverting opamp. The ideal opamp should have an infinite input impedance and zero output impedance, so shouldn't the input impedance for this inverting opamp be very high, why is it equals to R1? Am i misunderstanding a concept here? The input impedance of the non-inverting opamp however, is very high, which relates to the infinite input impedance though. Can you help me out with this?
@ananyasatpathy7694
5 жыл бұрын
No you have understood the concept correctly. Inverting opam has less input impedance I.e only r1 hence it is a disadvantage. While non inverting opamp is more efficient in this respect as it has infinite input impedance
@omwmakingprogress
Ай бұрын
In the non-inverting op-amp, the input impedance is infinite, so if the voltage applied to circuit, it will appear across this circuit. Could anyone pls explain it
@rssings8540
2 жыл бұрын
Listening to this in the exam hall thanks
@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 ❤
@Googleeknowledge1111
2 ай бұрын
Sir,in the buffer circuit, won't the output be out of phase with input?
@ALLABOUTELECTRONICS
2 ай бұрын
No, it will be in the same phase with input.
@rajatdalal8234
5 жыл бұрын
One question, why is the input current in the non inverting amplifier is taken to be zero? Why in ideal amplifier, input current is zero?
@thowhidulislamrifat3369
2 жыл бұрын
sir i have the same doubt..
@panditmanasranjan
3 жыл бұрын
In the video 2 you told that for inverting opamp , the input impedance is infinite. Here you are telling it's depending on R1, can u make things clear
@hm2715
3 жыл бұрын
He told that input impedence of ideal opApm is ♾.... Also, in Inverting Opamp R1 is the input impedence because R1 is connected across Vin... Whereas in non inverting OpAmp the input impedance is not R1 since it is not connected across Vin,, The input impedance of non Inverting OpAmp is not shown here, but it is maximum.... ...Hope it helps
@sais.here.5869
4 жыл бұрын
bhai engg colleges should recommend your lectures
@yeswanthtenali1674
4 жыл бұрын
I have a doubt that why resistance is not connected in non inverting op amp when voltage source is connected??????
@ALLABOUTELECTRONICS
4 жыл бұрын
Generally, we connect the series resistor with the source to limit the current in the circuit. Ideally no current flows in to the op-amp terminal. Actually, it will be very very low. So, you can also connect the voltage source directly the non-inverting terminal. I hope it will clear your doubt.
@damlacanatc2949
4 күн бұрын
simple and perfect explanation
@deepakdaswani7929
6 жыл бұрын
Sir in case of buffer circuit Why we have taken R1 equals to infinity ? Even by putting Rf equals to 0 only we are going to get Vin = Vout.
@ALLABOUTELECTRONICS
5 жыл бұрын
Yes, then also you will get Vo = Vin. But here what I mean to say is, there is no need to connect the R1 to use the op-amp as a buffer.
@inhdungho9509
3 жыл бұрын
thank you very much Sir ! as everyone did comment, I dont know what I should say more, just Thank you
@mnada72
4 жыл бұрын
Are you going to discuss the internal structure of the op-amp
@ALLABOUTELECTRONICS
4 жыл бұрын
Yes, someday I will make a video on it.
@AnkitSharma-jf9fg
4 жыл бұрын
i tried in pspice when I don't give any feedback output voltage range get shifted to negative region of graph can you tell why also I keep input voltage 1uV and Vcc and Vee to 15v and -15v respectively
@ALLABOUTELECTRONICS
4 жыл бұрын
In simulation, if the op-amp which you have selected is ideal then it has infinite gain. So, even for 1uV input signal it will get saturated.
@AnkitSharma-jf9fg
4 жыл бұрын
@@ALLABOUTELECTRONICS it is not gaetting saturated its range become something from -1.22v to -1.51v
@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 :-)
@siddharthnevagi8370
3 жыл бұрын
can somebody tell me how voltage Vx = R1 / R1 + Rf * Vout ??? shouldnt it be Rf / R1 + Rf * Vin. Correct me if Im wrong
@sharetechpathshala
Жыл бұрын
sir how do you teach...using PowerPoint and pen tab....or something else
@jigarpatel6520
4 жыл бұрын
why (in both the cases: say inverting and non-inverting op-amp) FEEDBACK is only given to the inverting terminal [pin#02 of op-amp] ??
@ALLABOUTELECTRONICS
4 жыл бұрын
Because in both configuration, the negative feedback is given to the op-amp.
@jigarpatel6520
4 жыл бұрын
@@ALLABOUTELECTRONICS what would happen if we give the positive feedback to op-amp ?
@ALLABOUTELECTRONICS
4 жыл бұрын
With positive feedback, the output of the op-amp will go into the saturation. Ideally, the output of the op-amp should increase continuously with the positive feedback. But it can't go beyond the supply voltage, hence the output of the op-amp will get saturated near the supply voltage.
@farhanupaul
4 жыл бұрын
Shouldn't the input impedance for inverting op amp be R1|| rin [rin = very high], so input impedance =R1?
@ALLABOUTELECTRONICS
4 жыл бұрын
Yes,but here since the op-amp is considered as the ideal op-amp, so Rin is infinite.
@farhanupaul
4 жыл бұрын
@@ALLABOUTELECTRONICS After finishing your BJT series everything is coming as small signal analysis. Thank you for that.
@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.
@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.
@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.
@tpsicmin
2 жыл бұрын
Amazing Explanation Thank You Sir
@apekshaphadte4443
Жыл бұрын
6:12 isn't input impedance present inside the triangle symbol of the op amp instead of being at Vin?
@ALLABOUTELECTRONICS
Жыл бұрын
Here we are finding the input impedance of inverting and non inverting configuration. ( instead of input impedance of the op-amp alone). In this configuration we are also connecting external resistors to the opamp. In general for any circuit, the input impedance of the circuit is the impedance seen by the input voltage source. That’s why here we are looking at the impedance seen by Vin. If you want to know more about input and output impedance, you can refer the following video: kzitem.info/news/bejne/mKCtmJOucZaqhG0
@apekshaphadte4443
Жыл бұрын
@@ALLABOUTELECTRONICS thank you so much! Ur reply means a lot! Gbu
@706bhaskar
4 жыл бұрын
Hi what is the theoretical formula for Closed loop non inverting gain?
@ravisanju1629
Жыл бұрын
In previous vedio the input impedance of inverting op amp is infinite and in this vedio you said input impedance is equal to R1.
@ALLABOUTELECTRONICS
Жыл бұрын
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.
@AnkitSharma-jf9fg
4 жыл бұрын
I tried the voltage follower circuit in pspice and it is not showing any output across a resistor although it giving right feed back
@ALLABOUTELECTRONICS
4 жыл бұрын
It would be better if you send me a diagram. May be you can message me on instagram your simulation.
@AnkitSharma-jf9fg
4 жыл бұрын
@@ALLABOUTELECTRONICS on what account i have to instagram you
@AnkitSharma-jf9fg
4 жыл бұрын
@@ALLABOUTELECTRONICS i send you pics please check it
@chenthanar895
4 жыл бұрын
Super sir if u teach me i will be the topper of my university.
@binarysaiyan9389
3 жыл бұрын
Your I and Z looks similar. I was having a tough time figuring out which was what
@basantasubedi1960
3 жыл бұрын
What application is he using to present this, can anybody tell me?
@berenedain8427
11 ай бұрын
Woah what the hell?! That just clicked in my head now, thank you
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