ALL ABOUT ELECTRONICS Kindly upload video about ADC and DAC. How to convert there are few videos about it and their explaining method is not very attractive.
In example 5, if we apply KCL at inverting terminal then, the answer comes as -R2/R1. Can we directly join the resistances in parallel as one terminal is virtual ground but the other is actual ground? Please clarify this doubt.
If you rearrange the resistors in #5 it will look like an inverting amplifier with R3 as load so you didn't have to make it parallel 😊 I think the gain is -R2/R1.
In example 5 how you took R2//R3 because from my point of view Vout is much greater than Vin so current in R2 will go to R1 which is Vout-0/R2 = 0-Vin/R1 and current in R3 should be equal to Vout-0/R3 and also as R3 is connected to ground and not terminal connecting R2 and R1 so current in it should be independent and current will go from high potential to low potential so there is no way of any current going from R3 to R2 also as R3 is connected to ground and it can extract any amount of current from ground so current in R2 + current in R3 may not be equal to current in R1 so total gain should be Vout-0/R2 = 0-Vin/R1 which is -R2 /R1
As he took R2||R3 in example no.5 if we take 10k||R in example no.4 we get R=-11k ohms whichs not possible.. So I think we CANNOT consider virtual ground and actual ground same and connect them and proceed..
In ques #4 , by virtual ground , Va = 0 volt , now resistor 10k and R both have one terminal at 0v and other terminal is common for both . so why we can not let (10k || R) ? and further taking Rf =(10k || R)+10k and then appling gain formulae for inverting terminal i/p i.e A=-(Rf/R1) ?? just like you did in next ques #5 .
In example 5 @13:21 it is vir tual ground , not actual ground, so current cannot flow between the virtual ground and actual ground, so we can't join those ends
- I have set up the circuit in example 5, unfortunately the answer is not practically the same as your solution. I have all resistors 10k and Vi=1v ==> Vo=-1v ==> Vo=-Vin==> A=-1 that is R3 load resistance and A=R2/R1 - But as far as example 4 is concerned, I have practically the same answer as your solution. Thank you for your effort
In example 5 , R3 is connected parallel to R1 because of same potential difference . Why in example 4 , R cant be paralllely connected to 10k? In example 3 why the circuit is integrator(low pass filter)?
Thanks for all your lovely videos...But I h’ve a doubt...How we can identify directly whether the op amp circuit is Low pass or high pass or band pass/stop filter just by looking at circuit..
I have made the video on active filters using the op-amp. You can check that video for more help. But in general for any circuit by finding the transfer function(Vout/Vin), we can conclude whether it is low pass, high pass or band pass/stop filter.
Sir in first question if we do not consider the complex part that is j....And write XL= wL and XC= 1/wC....Than too the answer is 1 ... So is it correct method?
In this case, although the answer is coming the same, but that's not the correct way. You need to consider the imaginary part as well. (instead of just considering the magnitude of the reactance). I hope it will clear your doubt.
Sir in example 4 ,you have solved the ques by using KCL .Can it also be solved by connecting the ground terminal of resistor R with Va voltage as by virtual ground , Va also equal to 0.But by solving this way, I get my answer wrong. Please help me with this.
Sir why can't we do like.........by virtual shot circuit Va =0 and R is grounded so effectively it becomes an inverting opamp but by that R comes negative...please tell the error sir!!
sir 11:12 lets assume that there is a current I1 through Vin to Node A, and another current I2 through Node B to Node A , did you take applying KCL at node A like this ?
This doubt is based on Practical simulation of Multisim Where I design a circuit of instrumentation amplifier. The problem is : Saturation region is +15/-15 Volt. Now I calculated output voltage is 20V (Theoretical) and according to ur videos it should be exceed the saturation voltage so it should be saturated to +15 V. But in simulation I got only 12V ?? Why??? Same if i change input voltage and let's i got output voltage 120V ,it should be saturated to 15 V but in simulation i got only 7V. Same in case if got let's say -110 V i got only -7 Instead of -15. WHY????
The thing is all op-amps are not rail to rail op-amp. Meaning that all op-amps will not saturate at the biasing voltages. And in fact, for most of the op-amp, it used to be less than the supply voltage. The op-amps which can go up to positive and negative supply rails are known as rail to rail op-amps. That's why you are getting 12V as output. About your second query, just want to know, have you connected the load resistor at the output of the op-amp. And if yes, then what's the value of the load resistance.
to teach a subject it takes a logic, a didactic and it must be formed for that otherwise teaching adds confusion and complicates understanding, everything is easy when we know how to transmit knowledge and everything becomes complicated when we do not know to teach, so I think you need to focus on how to present your classes rather than the content itself. The purpose of teaching is not to show that you know things but to help others understand .
Here the op-amp is ideal op-amp. And, the bandwidth of the ideal op-amp is infinite. So, the cut-off frequency which is referred in the example is the cut-off frequency of the circuit.
In ex.2 when we comsider v1 then due to virtual ground concept the o/p resistance R and 3R are in parallel,then the o/p expression due to v1 changes,am i right sir,if wroung could you please explain?
No, it won't act like a positive feedback. Because as both (One terminal of resistor R and non-inverting input terminal) are at ground potential, there is no input as feedabck to the non-inverting terminal.
I noticed a mistake In Example 2. you should have included R (which is connected to Vout) in the calculation of R feedback of the OpAmp, in parallel to 3R.
The problems which have been discussed in this video are GATE examples from past 5-6 years. And in next video too, I am going to cover some more examples.
ok sir,actually in some opam,many tricky problems come into picture like transient concept,phase angle,diode on off etc.hope it will be covered in the upcoming videos,,,,thanks for your wonderful explanation
The book on Op-amp by Ramakant Gayakwad is one of the books where you will find some examples on op-amp. The best source for this kind of examples is the previous year GATE or IES exam papers. Where you will get some good quality examples which will thoroughly test your understanding.
You solved example 5 in the wrong way. virtual ground and real ground are not the same so we cannot take R2 || R3 . And hence the voltage gain will be equal to the gain of inverting amplifier .
Because there is one more node B in this circuit. VB and Vout are at different potentials. So, directly it can't be applied. In example 5, as the parallel combination is directly connected to the output node, it is possible to apply that formula. You can even apply the nodal analysis in example 5 like we did in example 4 and still, you will get the same result. I hope it will clear your doubt.
You have wrongly solve 5th question.. Deliberately you make virtual short circuit into actual short circuit...R3 act as load resistance...even the solution given by GATE is not matching with your answer.
Answer 5th is solved in the wrong way. Solution 5 is completely absurd. You cannot connect the virtual ground point to the actual ground. Even if we assume that your answer 5 is correct, then check your way of solving question 2, since Q2 and Q5 are similar if we keep V2 in Q2 equals to zero. If you solve Q2 as you solved Q5, then you will get different answers for Q2. It signifies that your two answers are contradicting.
Because the signal from the first stage will get amplified by the gain A2. So, let's say if Vin2 is the input to the second stage, then the output would be A2*Vin1. Where Vin1= A1*Vin. So, overall, V0 = A1*A2*Vin. I hope it will clear your doubt.
There is no dedicated book. But what you can do is, you can go through the previous years papers to get more questions on op-amp. For electronics, there is book for GATE called R K Kanodia. There you will get some 40 50 questions on op-amp. If you want you can by that. But for just op-amp questions I don't it's worth it to buy it. But still if you want you can buy it. Also you can check the quiz section for op-amp questions.I used to post it as a community post. For quiz and solution, I have started a dedicated channel, ALL ABOUT ELECTRONICS QUIZ. You can also check that for more questions.
I am sure, you are aware of the complex number. Here j is used to represent the reactance of capacitor and inductor. In general, when the circuit contains a capacitor and inductor, the impedance is a complex number. Which can be represented in the form (A + jB) ( similar to the complex number) I hope it will clear your doubt.
Hello, in example 5, even if the potential at - terminal of opamp is same as gnd, but how can we consider r2, r3 to be parallel??? As the case of current going to ground through r3 is not same as current going to - terminal of opamp through r3.....please clarify it
The timestamps for the different examples covered in the video.
0:31 Example 1
4:06 Example 2
7:21 Example 3
9:01 Example 4
12:45 Example 5
ALL ABOUT ELECTRONICS Kindly upload video about ADC and DAC. How to convert there are few videos about it and their explaining method is not very attractive.
Yes, I will upload it very soon.
why you did not connect 10k ohm resistor in parallel with unknown resistor r for example-4 but you have done in example 5
In example 5, if we apply KCL at inverting terminal then, the answer comes as -R2/R1. Can we directly join the resistances in parallel as one terminal is virtual ground but the other is actual ground? Please clarify this doubt.
If you rearrange the resistors in #5 it will look like an inverting amplifier with R3 as load so you didn't have to make it parallel 😊 I think the gain is -R2/R1.
Agree! The current through R1 is simply going through R2 only, and not parallel combination of R2 and R3.
Yes he has done it wrong in example 5 as R3 will make up as the load resistor
Agree, the gain will be -R2/R1
Agree
😥😥Mai ye comment exam ke bad padh rha hu.
same question aaya tha, pura 10 marks kat gya.
Prof se bahas bhi ki, ki mera answer shi h.
In example 5 how you took R2//R3 because from my point of view Vout is much greater than Vin so current in R2 will go to R1 which is Vout-0/R2 = 0-Vin/R1 and current in R3 should be equal to Vout-0/R3 and also as R3 is connected to ground and not terminal connecting R2 and R1 so current in it should be independent and current will go from high potential to low potential so there is no way of any current going from R3 to R2 also as R3 is connected to ground and it can extract any amount of current from ground so current in R2 + current in R3 may not be equal to current in R1 so total gain should be Vout-0/R2 = 0-Vin/R1 which is -R2 /R1
s thirumalai
Yes ex:5 is wrong!!
Ans is -(R2/R1)
As he took R2||R3 in example no.5
if we take 10k||R in example no.4 we get R=-11k ohms whichs not possible..
So I think we CANNOT consider virtual ground and actual ground same and connect them and proceed..
At 13:22 the voltage between R2 and R3 MUST be Vo. Then KCL at the inverting terminal will read (0 - Vin)/R1 + (0 - Vo)/R2 = 0 or Vo = -(R2/R1)*Vin
You are the only person on TH-cam who goes through electronics the most thoroughly and with great math.
Tough problems but good. Makes sure that the principles are understood.
EX 5 IS WRONG ANSWER SHOUDL BE -R2/R1(V1) U CANNOT CONNECT VRITUAL GROUND TO ACTUAL GROUND
Please check this
In ques #4 , by virtual ground , Va = 0 volt , now resistor 10k and R both have one terminal at 0v and other terminal is common for both . so why we can not let (10k || R) ? and further taking Rf =(10k || R)+10k and then appling gain formulae for inverting terminal i/p i.e A=-(Rf/R1) ??
just like you did in next ques #5 .
my attempt also same manner
then R becomes -11K 🙄🙄
@ALLABOUTELECTRONICS
same query
my answer also same... NO CHANGE
In example 5 @13:21 it is vir tual ground , not actual ground, so current cannot flow between the virtual ground and actual ground, so we can't join those ends
- I have set up the circuit in example 5, unfortunately the answer is not practically the same as your solution. I have all resistors 10k and Vi=1v ==> Vo=-1v ==> Vo=-Vin==> A=-1
that is R3 load resistance and A=R2/R1
- But as far as example 4 is concerned, I have practically the same answer as your solution.
Thank you for your effort
In example 5 , R3 is connected parallel to R1 because of same potential difference . Why in example 4 , R cant be paralllely connected to 10k? In example 3 why the circuit is integrator(low pass filter)?
Thanks for all your lovely videos...But I h’ve a doubt...How we can identify directly whether the op amp circuit is Low pass or high pass or band pass/stop filter just by looking at circuit..
I have made the video on active filters using the op-amp. You can check that video for more help. But in general for any circuit by finding the transfer function(Vout/Vin), we can conclude whether it is low pass, high pass or band pass/stop filter.
In example 3, you obtained the cut off frequency, but how do you know that it is at 3db?
The expression is for 3 dB frequency only. For more info, please watch the playlist on the analog filter, where I have derived the expression.
in the 5th problem, when you redraw the circuit, wont R3 act as a load resistance? and the gain be equal to -R2/R1?
same doubt.
Also, same doubt. Confirmed with ltspice simulation. There is a problem here.
i noticed the same way
Applying Kirchhoff's current laws gave same answer
@@noweare1 here comes the ltspice guy
In the Q2 we have to neglect R connected at output terminal or is there any conceptual reason which i am missing? please explain
No voltage drop between ground and outp
Sir in first question if we do not consider the complex part that is j....And write XL= wL and XC= 1/wC....Than too the answer is 1 ... So is it correct method?
In this case, although the answer is coming the same, but that's not the correct way. You need to consider the imaginary part as well. (instead of just considering the magnitude of the reactance).
I hope it will clear your doubt.
Thanks a lot for these solved problems. Helped me a great deal!
What r u studying now ?
Thanks sir. But in #2 time 5:06 I think Rf = 3R || R because R is connected to output terminal and to the ground so as 3R. Thanks.
Sir how would we go about solving the final equation in example 2..provided we were given the values of those resistors?
Sir but the impedance Z1 in the second part of te first circuit should be (r2*r2+Xc*Xc)^1/2
Sir in example 4 ,you have solved the ques by using KCL .Can it also be solved by connecting the ground terminal of resistor R with Va voltage as by virtual ground , Va also equal to 0.But by solving this way, I get my answer wrong. Please help me with this.
same query
applying kcl at node A it should be Vin - VA/10 + Va-Vb /10 right?
Sir why can't we do like.........by virtual shot circuit Va =0 and R is grounded so effectively it becomes an inverting opamp but by that R comes negative...please tell the error sir!!
Thank you sir,I got that example what I am searching
In the first question why is the second circuit not a differentiator circuit because of the capacitor it should differentiate the input signal, right?
Yes, that's true. But we are intersted in finding the overall gain of the circuit. So, just impedance of each circuit component is considered.
ALL ABOUT ELECTRONICS So if I were to find the output signal, will I consider it as a differentiator or an inverting amplifier?
For the given example, just consider it as an inverting amplifier.
@@ALLABOUTELECTRONICS sir why did we consider j terms in the reactance earlier we didn't put j terms in calculation of impedance
sir 11:12 lets assume that there is a current I1 through Vin to Node A, and another current I2 through Node B to Node A , did you take applying KCL at node A like this ?
The KCL is applied at node B. That means the summation of all the incoming and outgoing current from node B is assumed as zero.
sir, what about the presence of resistor 'R' between Vout and and ground in question no 2.
Please clarify @ALL ABOUT ELECTRONICS
This doubt is based on Practical simulation of Multisim Where I design a circuit of instrumentation amplifier. The problem is : Saturation region is +15/-15 Volt. Now I calculated output voltage is 20V (Theoretical) and according to ur videos it should be exceed the saturation voltage so it should be saturated to +15 V. But in simulation I got only 12V ?? Why???
Same if i change input voltage and let's i got output voltage 120V ,it should be saturated to 15 V but in simulation i got only 7V.
Same in case if got let's say -110 V i got only -7 Instead of -15. WHY????
The thing is all op-amps are not rail to rail op-amp. Meaning that all op-amps will not saturate at the biasing voltages. And in fact, for most of the op-amp, it used to be less than the supply voltage. The op-amps which can go up to positive and negative supply rails are known as rail to rail op-amps. That's why you are getting 12V as output.
About your second query, just want to know, have you connected the load resistor at the output of the op-amp. And if yes, then what's the value of the load resistance.
I got 11V not 7V (There was a small error in my circuit and just now i rectified i got 11V).
to teach a subject it takes a logic, a didactic and it must be formed for that otherwise teaching adds confusion and complicates understanding, everything is easy when we know how to transmit knowledge and everything becomes complicated when we do not know to teach, so I think you need to focus on how to present your classes rather than the content itself. The purpose of teaching is not to show that you know things but to help others understand .
Bhai tu he bachaega mujhe fail honey se
In reference to example 3- Is the cut off frequency of the circuit different from cut off frequency of the OP AMP?
Here the op-amp is ideal op-amp. And, the bandwidth of the ideal op-amp is infinite. So, the cut-off frequency which is referred in the example is the cut-off frequency of the circuit.
Gracias por compartir estos ejercicios. 🙂
In example 2 why 2R doesn't play a role in voltage division?
Thank you so muh sir . Your video are helping me a lot
sir can u mention from which year's question paper these questions are taken from?
In ex.2 when we comsider v1 then due to virtual ground concept the o/p resistance R and 3R are in parallel,then the o/p expression due to v1 changes,am i right sir,if wroung could you please explain?
Bb
In example 2 when we applying superposition principle,if v1 is on then output resistor acts as positive feedback ,is it right? Could u plz clarify..
No, it won't act like a positive feedback. Because as both (One terminal of resistor R and non-inverting input terminal) are at ground potential, there is no input as feedabck to the non-inverting terminal.
I noticed a mistake In Example 2. you should have included R (which is connected to Vout) in the calculation of R feedback of the OpAmp, in parallel to 3R.
SAME DOUBT
Great videos I would like to get the ppt or images for your excellent videos so that I can refer at any time I need
can you provide gate problems for ec,ee and instrumentation?
The problems which have been discussed in this video are GATE examples from past 5-6 years. And in next video too, I am going to cover some more examples.
ok sir,actually in some opam,many tricky problems come into picture like transient concept,phase angle,diode on off etc.hope it will be covered in the upcoming videos,,,,thanks for your wonderful explanation
Still confused why VA is equal to zero.. What is ground potential?
Second example answer is wrong , correct answer is V output is -3V1+2V2
bro is there any use of R present at the output?
can you please suggest some books where this kind of examples is given ?
The book on Op-amp by Ramakant Gayakwad is one of the books where you will find some examples on op-amp. The best source for this kind of examples is the previous year GATE or IES exam papers. Where you will get some good quality examples which will thoroughly test your understanding.
thanks
Where you get this type questions sir..I want more questions for preparation
There is a seperate channel ALL ABOUT ELECTRONICS- QUIZ. You will find more solved examples on op-amp and other topics.
Thanks a lot Sir ❤
sir apne fail hote hote bacha liya
bhaii....
Which book you follow
love you bro
very nice explanation
Sir, in ex 4 where did Va go?
Good 👍
Thank you sir😍
Kindly upload video about ADC and DAC. How to convert there are few videos about it and their explaining method is not very attractive.
You solved example 5 in the wrong way. virtual ground and real ground are not the same so we cannot take R2 || R3 . And hence the voltage gain will be equal to the gain of inverting amplifier .
You could not pick easier problems to solve.
What is that j in jwL
J is like imaginary part i in complex no. Since I is reserved for current we use J
Why are we not applying inverting amplifier formula directly in example 4
The direct formula would have been applied if there was no R in the circuit. But becuase of Resistor R, which is grounded, it can't be applied.
ALL ABOUT ELECTRONICS
But why are we not considering "R" resistance parallel to 10k resistance like we have done in example 5
Because there is one more node B in this circuit. VB and Vout are at different potentials. So, directly it can't be applied. In example 5, as the parallel combination is directly connected to the output node, it is possible to apply that formula. You can even apply the nodal analysis in example 5 like we did in example 4 and still, you will get the same result.
I hope it will clear your doubt.
Thanks a lot👌❤️
You have wrongly solve 5th question.. Deliberately you make virtual short circuit into actual short circuit...R3 act as load resistance...even the solution given by GATE is not matching with your answer.
Answer 5th is solved in the wrong way. Solution 5 is completely absurd. You cannot connect the virtual ground point to the actual ground. Even if we assume that your answer 5 is correct, then check your way of solving question 2, since Q2 and Q5 are similar if we keep V2 in Q2 equals to zero. If you solve Q2 as you solved Q5, then you will get different answers for Q2. It signifies that your two answers are contradicting.
Good video. Best wishes
V. Nice......
5:04 why?
11:28 how can you assume Vb=-Vin ????
At 10:20, by applying KCL at node A.
Bhai video me watermark lgane ki jrurt nhi h
Lectures notes pdf link broo
❤️
i never understand op-amp without example. i think this formula not example.
I think he messed in example 5 th like they are not parallel
how w=10^6?
Cos(wt) is general form.
Gr8
but why A=A1*A2 AND NOT A=A1+A2
Because the signal from the first stage will get amplified by the gain A2. So, let's say if Vin2 is the input to the second stage, then the output would be A2*Vin1.
Where Vin1= A1*Vin.
So, overall, V0 = A1*A2*Vin.
I hope it will clear your doubt.
You could explain it more clearly. Sometimes things you say doesn't make sense to me
Thanku sir
Sir please tell me best book to do questions related to opamps for iit jam
There is no dedicated book. But what you can do is, you can go through the previous years papers to get more questions on op-amp. For electronics, there is book for GATE called R K Kanodia. There you will get some 40 50 questions on op-amp. If you want you can by that. But for just op-amp questions I don't it's worth it to buy it. But still if you want you can buy it.
Also you can check the quiz section for op-amp questions.I used to post it as a community post.
For quiz and solution, I have started a dedicated channel, ALL ABOUT ELECTRONICS QUIZ. You can also check that for more questions.
@@ALLABOUTELECTRONICS sir can you pls give pdf of rk kanodia gate ece.... I have downloaded online but it has missing pages
What is j
I am sure, you are aware of the complex number. Here j is used to represent the reactance of capacitor and inductor.
In general, when the circuit contains a capacitor and inductor, the impedance is a complex number. Which can be represented in the form (A + jB) ( similar to the complex number)
I hope it will clear your doubt.
Pisot
Hello, in example 5, even if the potential at - terminal of opamp is same as gnd, but how can we consider r2, r3 to be parallel??? As the case of current going to ground through r3 is not same as current going to - terminal of opamp through r3.....please clarify it
@ALLABOUTELECTRONICS please respond sir! 🥲
In the Q2 we have to neglect R connected at output terminal or is there any conceptual reason which i am missing? please explain
bro thats already at ground potential and we dont need that to measure output voltage