Rank Booster Topic: Operational Amplifiers | Coal India/ GATE 2027 EC/EE | Vipin Mishra

Añadido: 2026-05-27

[00:00:12]Hello everyone welcome to IMSGA Academy.

[00:00:14]Hi myself Bipin Mishan. I welcome you all again in this series the most expected concept for gate 27. Right. So as we promised that we will deliver almost 50 important topics under this series. So I'm back again in this series and today I'm going to cover a topic which is very very important for all EC electrical and I in student and that topic is operational amplifier in this year gate 26 if I talk about just this topic it has contributed almost five marks and I think four questions were there so it is very very important right and it is very easy as well all those questions that this topic has contributed were very very easy right so let's get started. Right? So if we if we talk about operational amplifier, why it is named operational amplifier? So it is a kind of oper amplifier that can perform the mathematical operation as well. Right? Like addition, subtraction, integration, differentiation. That is why it is named operational amplifier. It is a high gain electronic amplifier with differential input and single output and it can perform mathematical operation as well.

[00:01:22]That is why it is named operational amplifier. Right? its gain can be varied to a very high range. So this is an operational amplifier as you can see IC741 is uh there for operational amplifier right now if we talk about operational amplifier it has two inputs right one is known as inverting terminal another is known as non-inverting terminal so input is having two terminal inverting terminal and non-inverting terminal output as having single terminal right output is having single terminal this voltage is plus VCC minus VC CC this is also known as the bicing voltage right this is also known as the bicing or the supply voltage correct and you know output can never be more than this voltage right so your output will always vary in between minus VCC to plus VCC correct and as you can see that we are applying differential input here okay no current will Enter here means it is like the input impedance of operational amplifier is normally taken as infinite and the output impedance of the operational amplifier is normally taken as zero.

[00:02:40]Okay, the operational amplifier output impedance is taken at zero and the difference voltage will appear here. So V difference will be equal to the voltage that we have applied at the inverting terminal. Right? the voltage that we have applied across the non-inverting terminal minus the voltage that we have applied across the inverting terminal. Right? Now this difference of voltage will get amplified and the output voltage will be a * V_sub_2 minus V_sub_1. Right? Or we can also say that V_sub_2 upon V output upon V_sub_2 minus V_sub1 this is nothing but A. And right now because uh we have not provided any kind of feedback. So this gain that we will be getting is an openloop gain. Right? This is known as the open loop gain.

[00:03:33]Right? This gain is open loop gain. And this openloop gain is very very high.

[00:03:41]Right? This openloop gain will be very high. To reduce this gain and to get a significant amount, we will apply the negative feedback. Right? You might have gone through the neg feedback theory. So we have positive feedback, negative feedback, right? We will apply negative feedback so that we can get some finite amount of uh gain. Right? Feedback negative feedback is applied for stable system. Positive feedback is applied to those system which are normally unstable. For example, oxil oxyator circuit, right? Oscillation circuits. So there we used to apply that particular uh uh that particular feedback, right?

[00:04:18]Normally here we will be applying negative feedback is applied for stable systems. Right? So gain which is openloop gain which is very very high to reduce that gain we will apply negative feedback. Okay. So this is what is operational amplifier. IC 741 is a popular operational amplifier.

[00:04:36]Okay. Now as you have seen here that it has two terminals an inverting terminal and non-inverting terminal. So let's suppose that if you have applied the input only at inverting uh non-inverting terminal and let's suppose that uh the inverting terminal is grounded right so we have applied the input voltage here and this terminal is grounded so what will be the difference voltage the difference voltage will be V input minus0 or simply we will be getting V input right then what will be the output so the output will be VA * V difference and that will be simply A * V input right a * B input so output wtage is positive here right so if suppose we have given this small voltage the output wtage will get amplified and you can see that output wtage is positive so phase will not be changing right this type of amplifier is known as non-inverting amplifier right this is non-inverting ing opamp while if we see there here we have applied the input at the inverting terminal and non-inverting terminal is grounded. So if we calculate this difference voltage right right now this difference voltage which is uh which will be equal to 0 minus v input or it will be minus of v input. Now if we calculate V not which will be A * V D so it will be minus of A * V input right so you can see that output voltage is minus A * V input so if you are applying a small voltage at the input it will get amplified but the phase will be shifted right so when you have done something like this this uh configuration is known as inverting amplifier right or inverting operational amplifier.

[00:06:34]So in this way based upon where the input is applied we can have inverting and non-inverting operational amplifiers. I believe this is clear correct. Now before uh getting into the topics it is very important to understand the virtual ground condition.

[00:06:49]What is this condition? So as we have seen that the input impedance as we have seen that the input impedance is very high. It is so input side will be like an open circuit. So normally in OPM what happens? Whatever is the current in it will not go inside the OPM right it will be coming here at the output like this means neither the the current is not entering inside the operational amplifier it is going like this okay because of the input impedance is infinite okay up see in this circuit what is the difference voltage so if we calculate the difference voltage we have applied negative feedback here so difference voltage will be what we we have applied the voltage at the negative terminal or basically difference voltage is always V input 1 V input 2 minus V input one correct this is the difference voltage by the way we have applied different uh V input wtage will be zero here that is another thing now if suppose I'm calculating output so it will be equal to what a * the W D right as we discussed earlier so it will be A * this V input 2 minus V input 1 right or this V input 2 minus V input 1 this will be equal to V upon A what is A is the openloop gain and as I told you that openloop gain is very very high so if suppose I'm keeping it very very high means infinite this will be zero or I can say that V input 2 is equal to V input one means whatever is the voltage here right whatever is the voltage here If V input 2 voltage is because it is grounded, it is zero here. So this voltage will also become zero at this node. This condition is known as virtual ground condition and it is applicable only for negative feedback. Right? This condition is applicable virtual ground condition is applicable only for negative feedback theory. So what you have seen here that if this terminal is grounded, this voltage will be zero. So the voltage of this terminal will also become zero. Right? It means whatever is the voltage at this terminal we you will be having same voltage here. This is known as virtual ground condition and it is very important to understand before proceeding further in operational amplifiers.

[00:09:15]Right? I believe that this is clear.

[00:09:24]Correct? Now after that we can have a lot of configuration. We can design a lot of configuration through this operational amplifier.

[00:09:32]Right? First among them is inverting amplifier. So see because we have applied the input voltage at the inverting terminal and non-inverting terminal is grounded it will be known as inverting amplifier. Right? So this is a circuit and as I know that current will not enter inside the OPM right because of input impedance being infinite. If suppose the current here is I I and current here is if so these currents will be same right the current I I and the current IF will be the same okay I apply the virtual ground condition because this is grounded so we will be having V equal to0 here so the same voltage will come al here also it will be zero right what will be the current I I will be VI minus0 divided by R I right this I can write one R1 let me call it one now what will be this current IF this current IF will be zero minus V upon RF or I can get it VI upon R1 this will be minus of V upon RF or we can write V upon VI this will be equal to minus RF / R1 right or we can also write that output wtage will be minus of RF upon R1 * the input voltage. So in this way we got the output wtage right and this is a very important relation correct this is a very very important relation correct so we are getting V equals to minus RF upon R1 into VI and this is known as inverting amplifier right so you can remember these uh relation that we are getting they are very important okay and you can apply directly to get uh the problems when when we whenever we are solving these problems You can apply them and you can get the result. Now see this this is non-inverting amplifier. So we have applied the input now at non-inverting terminal right and inverting terminal is grounded. That is why it is known as non-inverting amplifier right because of virtual uh virtual ground condition what will be the voltage here it will be VI.

[00:11:51]So the same voltage will appear at this node also. Okay, let this current is I I so same current will be there in the feedback path also. So again if I apply the condition that current I I will be equal to current I uh F and let me call it R1. Okay so what we can write here it will be 0 minus VI upon R1 and we can write it VI minus V upon RF right. So now what we can do here is that if suppose we'll take both the vi at same time. So I can write it v by rf and I can write it vi 1 upon r1 + 1 upon rf. Right? So what will be v upon vi? So v upon vi will be rf r1 plus rf.

[00:12:50]Right? U v upon vi will be what? RF we can send on this side correct and ep 1 upon R1 + 1 upon RF or we can also write it that V upon VI this will be equal to 1 + RF / R1 or V will be 1 + RF / R1 * VI right again we got a very important relation that we will be using while solving many questions. Okay. So this configuration is known as the non-inverting amplifier because we have applied the input at non-inverting terminal. So whenever you are getting these combination you can directly apply these formulas that we have discussed right you can directly apply these formulas that we have discussed. Now phase shifter. So see where we have applied the input here. We have applied the input at inverting terminal and non-inverting terminal is grounded.

[00:13:53]Right? So this is a non-inver this is an inverting operational amplifier. Right?

[00:14:00]So in case of an inverting operational amplifier what is the output upon input?

[00:14:06]This is equals to minus of RF upon R1.

[00:14:11]Right? Now what you can see here is what you can see here is that the feedback resistance and the input path resistance both are same. So what we will be getting is that V upon VI that will be equal to minus1 or V will be equal to minus of VI. Okay. So whatever input you are giving you will be getting the same out in output.

[00:14:36]Whatever input you are giving here you will be getting output exactly equal to this input. But the phase will be shifted. V not will be equal to minus of VI. That is why this circuit is known as phase shifter. So this is again a very important circuit in op. Right? Now voltage follower. This circuit is voltage follower. See we have applied the input at non-inverting terminal. So here the input uh the voltage will be VI at this node. Because of virtual ground condition the same voltage will come here. Now this voltage is coming at the feedback. So output will be what? It will be equal to VI. Right? So when we have output equal to minus of VI, it is phase shifter. But when we have output equals to VI, the input voltage output is equal to input voltage. This will be known as the voltage follower. Means output voltage will follow the input voltage. Right? Output voltage will follow the input voltage. Another very popular combination. And you know this uh la last year they have asked a question based upon this configuration.

[00:15:40]For example, see this circuit of gate 25. And whether what kind of circuit is this? We have to answer here. Okay. So what kind of circuit is this? See if this voltage is V input what will be the voltage at this node? Same V input.

[00:15:55]Right? So if this voltage is V input.

[00:15:59]Okay. Now see uh here also we will get the voltage V input. Right?

[00:16:05]because of uh because this voltage is V output. So here also you will get the voltage V output and V output will be equal to V input here isn't it? We will be getting V output equal to V input. So how this circuit will be acting V input is equals to V output. So isn't it summing amplifier difference logarithmic or buffer output is equal to input. So that can be known as voltage follower or it can also be known as the buffer circuit.

[00:16:37]Right? So this circuit is acting as a buffer circuit.

[00:16:44]Correct? This circuit is acting as a buffer circuit. See, so whatever we have learned, we have applied and we got this problem done. Okay? This kind of question they're asking in gate exam.

[00:16:54]Now see this circuit differential amplifier. Okay, a very popular configuration. So how we will get the output in this circuit right? So see both the terminals are having the input voltage here. So we will apply the superposition theorem here. Okay before applying that can you tell me that if this voltage is V_sub_2 and here this voltage is V what will be this voltage V? To get this voltage V, you will apply the voltage divider. So the voltage V will be V_sub_2 * R4 divided by the total resistance R3 plus R4 right now if this voltage is V this voltage will also be V right by virtual ground condition up what we have to do now we have to apply superposition theorem okay so by superposition right by applying superposition When v_sub_2 is zero, when v_sub_2 is zero, that means v will also be zero. If v_sub_2 is zero, v is also zero, then this vol w voltage will also be zero.

[00:18:01]Okay? Then if suppose you are calculating the output voltage, then you know it will become an inverting amplifier. So output voltage will be what? It will be minus of R2 upon R1 * the input voltage. Right? Similarly, if suppose I'm taking v_sub_1 zero, right?

[00:18:22]Then if I'm taking v_sub_1 zero and only v_sub_2 applied, right? So here what I have done here I have taken v_sub_1 applied and v_sub_2 up here I'm taking v_sub_2 is applied and I'm taking v_sub_0. So if I'm taking v_sub_1 right and v_sub_2 applied then this circuit will be acting as a non-inverting amplifier right. So in case of non-inverting amplifier what will be the output? This will be considered as the input wtage and we will be getting an output wtage V and that output voltage will be 1 + R2 by R1 * the voltage V right and what is this voltage V? You know the value 1 + R2 by R1 right into uh V. We will write it later.

[00:19:13]Now what will be the total output wtage?

[00:19:16]It will be the output wtage 1 that we got here and output wtage 2 that we got here. So it will be V1 + V2 or it will be min - R2 upon R1 into V_sub_1 plus 1 + R2 by R1 into V. If you wish you can put the value of this V. So it will be 1 + R2 by R1 right and what is the value of V? The value of V is R4 upon R3 + R4 * V_sub_2 correct and we had V_sub_1 here - R2 by R1 into V_sub_1. So this is how you will calculate the output voltage in a circuit like this. Right? What we have done? We have just applied the superposition theorem. Right? We have first calculated this voltage by voltage divider. Then we applied virtual ground condition. Then we have applied superposition theorem. First we have taken only the voltage V_sub1. V_sub2 was zero. Then it will be acting as an inverting amplifier. Then we have taken V_sub_10 and V_sub_2 applied. Then it will be acting as a non-inverting amplifier. Right? By applying superposition you can add both the output. Right? And you will be getting the final output in this way.

[00:20:39]Got that? Okay. Now an application of this kind of circuit is if suppose these resistances are also R1 and R2, right?

[00:20:48]So it is exactly similar circuit, isn't it? This is exactly the similar circuit that we have solved here. And what is the output wtage that we calculated? Let me copy it. So see this was the output wtage that we calculated earlier. up here what you can see here what you can see that R3 right what was R3 is R1 here and what was R4 it is R2 here right unless it is the same circuit now if suppose I'm going to put these values here what I will be getting I will be getting minus R2 by R1 into V_sub_1 plus 1 + R2 by R1 + R2 by R1 will be R1 R1 + R2 by R1 right and if we place the same values here R4 is R2 divided by R1 + R2 * V_sub_2 then this will be cancelled and you will be getting - R2 by R1 into V_sub_1 + R2 by R1 into V_sub_2 or you can write it R2 by R1 V_sub_2 minus V_sub_1 right so see an application of the last circuit in which these resistance are R1 R2 and these resistance are also R1 and R2 then you will be getting an output voltage like this they are asking these configuration directly into into the exam and you have to solve those configuration right so I believe it is clear okay this configuration how we have solved it is clear okay up now after that yeah okay so we have done these things. Okay. I believe that you are really uh liking the teaching and learning methodology of IMSGATE Academy. So if you are preparing for GATE 27, GATE 28 or you are preparing for PSU exams like Cole India, DRDO, ISRO, you can join IMSGATE Academy in your preparation journey. Right?

[00:22:54]We will be offering you 600 plus hours of live classes, quality material, mentorship session, okay? Test series, right? So mentorship sessions are very important in your entire journey of GATE preparation. We will be helping you out how to prepare, how to strategize, everything we'll be doing for you, right? So if you wish to join IMSGATE Academy, you can call on these numbers and you can book a free counseling session. Right? Many students have written their success story with us. For example, Jata and Rai, they got a 15, a 10 this year. Okay. Similarly, many students have written their success story with IMS GATE Academy in gate 26, right? Previously, gate 25 also many students have written their success story with us. Similarly, in gate 23, 24 and right now there are at very places you can be the next one as well. So, you can also join IMSGATE Academy for the preparation of gate 27 or 28 exam.

[00:23:50]Right? Coming back to the topic now see this one. Right? Okay. If I'll huh now see this one. So in this circuit this is known as inverting adder on the inverting terminal we have applied multiple inputs okay and non-inverting terminal is getting it grounded so this voltage will be 0 volt same voltage will appear here because of virtual ground okay then whatever is this current same will get here so if I write here the current I will be equal to the current I of F right up current I is the summation of current I1, I2 and I3 and that will be equal to I of F.

[00:24:35]Right? Now what will be the current I1?

[00:24:38]So see it will be v_sub_1 - 0 by r plus v_sub_2 -0 by r will be I2 v_ub3 -0 by r will be i3 and that will be equal to I right and what will be if I if will be 0 - v divided by r right we can cancel r from everywhere then it will become v_sub_1 plus v_sub_2 2 V_sub_1 + V_sub_2 + V_sub_3 right that will be equal to minus of V or V not will be minus of V_sub_1 + V_sub_2 + V_sub_3 right because of minus sign it is known as inverting adder so what we have done we have added all the inputs applied to that inverting terminal right this circuit is known as inverting adder I believe it is clear okay we We can have another configuration like this. If suppose the same thing now we have applied at this non uh inverting terminal. Okay. So if we have applied this at non-inverting terminal. Now this voltage the voltage at this node is V.

[00:25:48]Due to the virtual ground the same voltage will come here and that voltage will be V. Okay. That voltage will be V.

[00:25:55]Because you have applied the input at non-inverting terminal. So what is going to be the output?

[00:26:02]Output is going to be 1 + what is there in the feedback resistance? It is R of F divided by R1. Right? This is R of F divided by R1* this voltage V. Right? Now we have to calculate this voltage V. So what will be this voltage V? To calculate this voltage V, we have to apply superposition here. Okay? So apply superposition. So by superposition right by superposition if suppose I'm taking V_sub_1 only applied and I'm taking V2 0 and I'm taking V_sub3 0 then you will be having a circuit like this right then you will be having a circuit like this whose voltage is V here this resistance is R this resistance is R this resistance is R and this plus V_sub_1. Let me call this voltage as V M1. Now can you tell me that what will be this voltage V M1? It will be the voltage across this parall combination.

[00:27:10]The parallel combination of these will be R by 2. Okay. So V M1 we can write that it will be the voltage V_sub_1 into RX2 divided by R + RX2. So when you are going to solve it, it is going to come V_sub1 by 3, right? It is going to come v1 by3. Now similarly right similarly what you can do when you will take v_sub_2 only right when you will take v_sub_2 only and v_sub_1 is zero and v_sub_3 is also set to zero then let me call the voltage v12 and it will be coming v_sub_2 by3. If suppose I'm taking V_UB_3 only and if I'm taking V_sub_10 if I'm taking V_sub_2 0 then you will be getting the voltage V M3 and that also will be equal to V1 by3 right this is V_sub_1 by3 this is V2X3 so this will be V_3x3 correct now what will be this voltage V so this voltage V this voltage V will be actually V M1 1 + V M2 + V M3 right after applying superposition up. Now what you can do you can just write here 1 + RF by R1 and what is voltage V?

[00:28:34]Voltage V is basically V_sub_1 + V_sub_2 + V_sub_3 by 3. This is how you will get the output wtage in this particular configuration. Right? You need to practice all these configuration because you know they are repeatedly asking these configuration right they're regularly asking one of these configuration as a question in gate exam. Okay. Now after this let us discuss this circuit logarithmic amplifier. Right. Logarithmic amplifier or we can call them data compressor.

[00:29:17]Here data they are also known as data compressor. Here data compression will happen right up what will be the output voltage here. Okay. So see because this terminal is grounded this voltage will be 0 volt. So this voltage will also be zero. Okay. Because this voltage is zero. So this will also be zero. Right?

[00:29:38]Suppose again I will write the same thing. Whatever is the current going here same current will be going here.

[00:29:43]Okay. So uh what will be the current here? If I calculate this current here. So this current I will be what? This current I will be vi - 0 by R or this is simply VI / R. Right? The same current is going here. Correct? The same current is going here. So if we will apply the KL here, what I will be getting? I will be getting zero minus the voltage drop in the diode minus the output voltage equal to zero or I will be getting that output voltage is equals to minus of W D right the output wtage is equals to minus of the WD the drop across this voltage the drop across this diode V not is equals to minus of W D okay now we know that the diode current is given by I = I E power V / Nita V_T right and let me write this V voltage as V D E T minus1 whom we can approximate and can write like this E a V_T correct we can write it like this also now if the current across the diode is given by this okay this expression Then what I can do further I can write e ^ v d over na v_t. Okay. And let me take na as 1. Okay. I can take na s1 or let it be. It it depends upon what is given.

[00:31:21]So leave it v d / na v_t. Okay. This will be equal to I over I right. This will be equal to I / I or if I take the log here then ln here then what I will be getting? I will be getting V D equals to NA V_T ln I / I right ln I / I and what is this current I? So this current I is basically na V_T ln. This current I you can write from here and it will become VI upon I into R. Correct. This will be the voltage V D. And to get the output wtage which is our objective output wtage is equals to minus of W D.

[00:32:07]So what will be the output wtage? It is minus of V D or it will be equal to minus of NA V_T ln VI / I R. And the learning that you are getting here is that V is proportional to ln of vi right vot is proportional to ln of vi that is why it is known as logarithmic amplifier right logarithmic amplifier or data compression correct data compression if suppose vi I'm taking 10 ^ 3 okay then what I will be getting I will be getting 3 * ln 10 which will be a very small value. Data is compressed. Correct? So this is also known as data compression.

[00:32:56]If this is known as data compression, we have diode here in the feedback path.

[00:33:00]What if the diode is there in the input path and that is our next configuration that we will discuss. Before that let me solve this question. Okay. So just see this question.

[00:33:11]In the feedback path if suppose you are having diode or you are having BJT.

[00:33:15]Okay. Then what will be the output voltage here? So you have seen that output voltage was equals to minus of V D right and here it will be equal to minus of the base emter voltage right here we will be having it equal to base emter voltage. What is base emter voltage? So you know that base emter voltage this is an NPN transistor base emter voltage is 7. So what will be the output voltage? it will be equal to minus of.7 okay they may have BJT they may have diode here whatever they have you can solve it like this instead of NP and if suppose they have PNP then you know that VB is equals to minus.7 so this will become positive.7 like that question they have asked they can ask it again right I'm now see another configuration where diode is there in the input path last time we had it at the feedback path now what will happen Right. So see again this voltage is zero. So this voltage will be zero.

[00:34:17]Virtual ground condition. And if you will apply KL it will be 0 minus I the drop here minus V =0 or we will be getting V = minus of I into R. Right? So you are having V= minus of I into R.

[00:34:37]Correct? Now what is the current I?

[00:34:39]Current I is there in this diode. Right.

[00:34:42]So how do we write the diode current?

[00:34:46]How do we write write the diode current?

[00:34:49]So this uh diode current last time what we have written is I not e ^ vd / ma v_t approximated equation. Then if I put this value here so I can write it minus i e ^ vd / navy v_t into r.

[00:35:11]Right? Or I can write minus I r e ^ v d / na v_t. Right? If you will apply KVL here, you can see that it is vi minus v d minus0 equal to0 or vi is equals to simply v d. Okay. So if I put the value here I can write it minus of I r e ^ vi / na v_t or what is the learning here is that v is proportional to e ^ vi right e ^ vi right this is that is why it is known as anti-log amplifier or data expansion Right? If suppose VI is 100, E to ^ 100 will also be a very large number. So whatever we are giving it is getting expanded. Correct? This is known as antilog amplifier. Right? Now let's move into very very popular circuit which is known as instrumentational amplifier. A lot of time gate question have been asked through this configuration. Right?

[00:36:20]This type of configuration is known as instrumentational amplifier. How to solve it? Just use the basic configuration that you have done. Let us focus on this operational amplifier. So they have applied input on non-inverting terminal right and negative feedback is there. So can you tell me that what will be the output here? Right? What is going to be the output here?

[00:36:46]What is going to be the output here? Let me call it V_sub1. So what will be V_sub_1 voltage here? It will be equal to 1 + R2 by R1 * this V input 1. Correct?

[00:37:01]Similarly, see this configuration. This is also a non-inverting amplifier.

[00:37:06]Right? And we have negative feedback.

[00:37:08]So, what will be the output here?

[00:37:11]Output here is V_sub_2 and that will be 1 + R2 by R1 into the input applied here which is VI2.

[00:37:21]In this way you will get these two outputs. Okay. Up. Now let us focus on this configuration. After that let's suppose the circuit is a reduced one.

[00:37:32]Right? Let's suppose that the circuit is a reduced one in which now we are having two inputs right minus plus. So now see we are having two inputs. input applied here is V_sub_1.

[00:37:55]We have a feedback resistance, right? This voltage we are calling it V_sub_1. This resistance is R3. This resistance is R4. Okay? And then we have a configuration here also which is R3 and it is R4 here.

[00:38:18]Correct. And the voltage here is V_sub_2. Up. Now you can tell me that after reducing this circuit into this circuit whether we have discussed this one as well already we have discussed this one or not. Okay. So let me take you to this configuration that we have already discussed. Just see say see this circuit if these two resistances are same right and two inputs are applied to two different terminal what is the output? The output is going to be R2 upon R1 V_sub_2 minus V_sub_1. Okay.

[00:38:53]So you can use this learning. Then what will be the output here? The output here will be R2 upon R1. The feedback resistance divided by the input resistance R3. Input side resistance R4 upon R3. And you can get it V_sub_2 minus V_sub_1. All right. Now R4 upon R3.

[00:39:13]what is v_sub_2 and v_sub_1. So if I'm going to calculate v_sub_2 minus v_sub_1 I can get it 1 + r2 upon r1 vi2 minus vi1.

[00:39:28]Okay. So that will be my output wtage.

[00:39:32]In this way I can calculate the output wtage of this circuit.

[00:39:39]Correct? I hope that it is clear. So it was looking to be a very complex circuit but see how easy this solution is going to be clear. So I believe it is clear to you.

[00:39:54]This is a configuration which they have asked most of the time. Okay. A modified instrumentational amplifier can be this.

[00:40:02]It is a modified amplifier. Now what is going to be the output here right? So see how we are going to solve this circuit. If you see they have applied voltage here at non-inverting terminal non-inverting terminal and see the way they have provided feedback. Now here there is no grounding present. So we cannot do it like that. Okay. What you can see here is this current I is coming I is going and same current I will come here also and the same current is getting back. So output is forming a loop with these resistances. Right?

[00:40:34]Output is forming a loop with these resistances. Now if this voltage is V input what will be the voltage here through the virtual ground condition this voltage will be V input 1. If this voltage is V input 2 the through the virtual ground condition this voltage will be V input 2. Okay up you can easily tell me that what is going to be V input 1 minus V input 2. This is going to be equal to I into R. Right? I into R1. What I have done I have applied KL in this path. Okay. Now after that let us apply KVL in this outer path. Right?

[00:41:11]I'm talking about this path. Let us apply KL in this path. If you are going to apply KL in this path, you will be getting V minus I * R2 - I * R1 - I * R2 =0. Or you will be getting V = to I * V = to I * R1 + 2 R2. Correct? This is what you will be getting as V. Now from here what will be I? So I will be V upon R1 + 2 R2. Let me call this equation two. And this was our equation number one. Up. Coming back to this equation, V input minus V input 1 minus V input 2. This is our I I we can put here V upon R1 + 2 R2, right?

[00:42:07]And we have R1 here. So what will be the ratio of V output upon V input 1 minus V input 2? That will be equal to R1 + 2 R2 divided by R1. Or we can write 1 + 2 * R2 by R1 right this will be the output wtage of this configuration a lot of time they have given the same configuration you just have to calculate the output okay so I think by using this relation if suppose you can remember it you can directly apply it otherwise I think there is no rocket science in this circuit you can analyze this circuit as well correct we can also analyze this circuit I believe it is clear Okay, again very popular configuration. Now see this one integrator circuit which will be acting as a lowass filter. Okay, so capacitor it is present at the feedback path. So whenever capacitor is there at the feedback path or output path device will be acting as a low pass filter. If it is there in the input path it will be high pass filter. Right? If it is present in both the way means input and feedback it will be acting as a band pass filter.

[00:43:20]Okay? We'll discuss everything. Don't worry. If suppose you have this configuration then what to do? Okay. So see uh the voltage here is what 0 volt.

[00:43:30]What will be the voltage here? 0 volt.

[00:43:32]Right? If you will apply KVL you will be getting 0 minus VC minus output equal to zero or output wtage will be equal to minus of VC. Okay. Of this this current I is going to go here also. How we can write the voltage VC here? Okay. If suppose I have to write the voltage VC, how we can write this voltage VC here?

[00:44:01]Tell me the voltage VC can be written as minus 1 upon C integral I DT right where I is the current across this capacitor right the same current will be going here. Now if I talk about the value of this current so vi -0 by R will be the value of this current or current I will be VI upon R. Okay. Okay, if I'm going to put the same thing here, I will be getting V equals to minus 1 upon RC integral VI DT, right?

[00:44:37]Because output is equals to output that you are getting you are getting it by integrating the input voltage and that is why it is known as integrator.

[00:44:49]Correct? That is why it is known as an integrator circuit.

[00:44:55]Correct. This is known as integrator circuit. Whenever you have and this circuit is also behaving as a low pass filter that you might have covered in network that is not our topic. So we will leave it. Whenever the capacitor is there in the feedback path always remember that this will be acting as a lowass filter and integrator. Correct?

[00:45:14]What if the capacitor is there in the input path? If suppose we have capacitor here. Okay. Again this will be acting as a differentiator. How suppose that this voltage is zero and this voltage will also be zero. Right? Then if you will apply KL in the output path then you will be getting 0 - IC into R - V=0 or you will be getting V= to minus IC into R. Okay. Now what is this current IC?

[00:45:50]So how you will write this current IC?

[00:45:52]This current IC is basically across the capacitor. So across capacitor how to write the current across capacitor we will write the current like C D VC over DT right into R. Correct. Up what is this voltage VC? So if we apply K well VI minus VC=0 equal to0. So we will be getting VC is equals to VI. So what I can write here is that V will be equal to minus of RC D VI upon DT. Okay. So now what you can see that this time when you are getting output voltage you are getting it by the differentiation of input voltage. Right? And that is why this is known as a differentiator circuit.

[00:46:38]Right? It will always differentiate the input voltage and then the output will come. So this circuit is also known as differentiator circuit. Okay. And because capacitor is present at the input path, so it will be known as a highass filter. Whenever capacitor is at the feedback path, it will be an integrator and low pass circuit.

[00:46:57]Capacitor at the input path, it will be an high uh it will be a differentiator and high pass filter. Right?

[00:47:05]Now see a different kind of configuration. Right? So here what you can see that capacitor is neither present at the feedback path nor it is present at the input path. It is present in this terminal a different terminal right same here it can be present in this sun path it can be present in this path also. Now how how these circuits will act right how these circuits will act.

[00:47:30]So if suppose you have this circuit okay if suppose that you have this circuit then in this circuit what is the ratio of output to input right let's suppose that I have applied input here okay so what will be the ratio of output to input here the ratio of output to input will be 1 minus J 2 pi F RC upon 1 + J 2 pi F RC. Okay. And if you will calculate the magnitude, the magnitude will come out to be one which you can see that it is independent of frequency.

[00:48:16]Right? If it is independent of frequency that means it will be acting as an allp pass filter. It will pass all the frequency.

[00:48:28]Correct? it will pass all the frequencies write that now if suppose I'm calculating the angle here the angle will come as - 2 tan inverse 2 pi f rc right the angle here will be like this this will provide a negative uh negative phase lag okay similarly if suppose we have this configuration okay if suppose we have this configuration and you have applied input Then output to input ratio here will be 1 + j 2 pi frc divided by 1 + j 1 - j 2 pi f rc. Again if you will calculate the magnitude of this gain it will come one right which is independent of of frequency. So again it will be acting as a all pass filter. It will pass all the frequencies. Okay. And if you will calculate the angle of this ratio V upon VI that angle will come positive tan inverse 2 tan inverse 2 pi f RC. Right? That angle will come positive 2 tan inverse 2 pi fc. Okay. So capacitor is present here. Capacitor is not present in these path. It is present here in opposite terminal. If it is suned, it is all pass filter. If it is present here also, it is all pass filter. Right? In this configuration, this is going to be the output. In this configuration, this is going to be the output. If suppose you wish to calculate output, how to calculate? We have already discussed this thing, isn't it?

[00:50:13]What you have to do? You have to calculate this voltage by applying voltage divider. Then after that you have to apply superposition as we have done in our one of the configuration right differential amplifier configuration you can solve these circuits it is going to come like this right so when capacitor is present here it will be acting as a right it will be acting good evening everyone it will be acting as a allp pass filter got that now let us have a question here so see in this question you are having capacitor present here at the feedback back path capacitor present here at the input path as well. Now can you people tell me that how this circuit will be acting? It will be acting as a band pass filter. Right? It will be acting as a band pass filter. I believe it is clear.

[00:51:07]Right? I hope that this question is clear. If capacitor is present at both the places it will be acting as a band pass filter. Okay. By the way what they're asking here all the component in the band pass filter they have given it band pass right. So see if in capacitor is present at both the feedback and input path it will be a band pass filter they have also given it right. If it is only here if it is only a high pass right just see low pass. Sorry if it is only in the feedback path it is low pass. If it is there only in the feedback uh input path high pass. If it is there in the inverting terminal non-inverting terminal right either centered or in series it will be all pass filter but if suppose capacitor is there in the feedback path as well as in the input path it will be like a acting like a band pass filter. So whether it is PSU exam or sometimes in gate exam or many other exam sometimes they will just give give you a configuration where you just have to identify the type of filter.

[00:52:06]This is how you can do if suppose you have inductor entire things will be reversed right high pass will become low pass low pass will become high pass right up now what they're asking the lower 3db frequency of the filter one is filter is 1 mhertz the upper 3dB frequency will be what will be the upper 3dB frequency the lower frequency is given up how to find the frequencies okay so see general expression for frequency is 1 upon upon 2 pi RC you have to find the frequency by using this expression f= to 1 upon 2 pi rc which r and which c we have to use because a lot of r and c's are here okay so let me give you a clarity if suppose you have c in the feedback path okay if you have c in the feedback path then that circuit will be a lowass filter right that circuit will be a lowass filter then frequency will be 1 upon 2 pi right then frequency will be 1 upon 2 pi R C okay then frequency will be 1 upon 2 pi RC right they will give you a feedback resistance they will give you a feedback capacitor RF into CF if suppose capacitor is there in the input path right if capacitor is there in the input path then it will be acting as a highpass filter and then you have to find the frequency by doing 1 upon 2 pi R1 into C1 right by using these two okay and when capacitor is at both when capacitor is at both input and feedback path then you will be having two frequencies lower 3dB frequencies upper 3dB frequency okay then the lower 3dB frequency See you will be calculating by using this expression 1 upon 2 pi R1 C1 R1 C1.

[00:54:11]Okay. And you will be calculating the upper 3dB frequency or higher frequency by using one upon 2 pi RF into CF.

[00:54:22]Everything is summarized here. Right?

[00:54:23]Just remember these things and you will be able to solve a lot of questions in PSU exam, in GATE exam, many other exams. Okay? What they have given here is they have given the lower uh 3dB frequency. Okay. So they have given us lower 3db frequency.

[00:54:40]So given the lower 3db frequency which is 1 mhertz 1 megahertz right now use this expression 1 upon 2 pi.

[00:54:53]What is R1? R1 is basically R here.

[00:54:58]Right? R1 is given as R and what is C1?

[00:55:02]C1 is given as 10 C.

[00:55:05]Correct? So what you can do by solving this you can calculate the value of this 1 upon RC. So 1 upon RC when you will solve it I have the solution it is coming 20 pi into 10 ^ 6. Okay there is no rocket science in solving this up.

[00:55:21]Now what we have to calculate is the upper 3dB frequency. Okay. So to calculate the upper 3dB frequency use the formula 1 upon 2 pi RF into CF okay or 1 upon 2 pi what is RF value it is.1 C sorry RF value is 2 R it is 2 R and what is CF value it is 1 C so what we will get here we will get 2 into 2 4 RC C right what is 1 upon RC 1 upon RC is 20<unk>i into 10 ^ 6 divided by4 right you have to just solve it 4 pi correct you are having4 pi into RC so now when you will solve this I have the solution it is coming 50 MHz okay don't waste the time in solution I have already done that so this is how you can solve this question right this is how you can solve this question okay hi Bhavani I believe it is clear okay so remember this how to calc how how to judge the how to judge the filter type by looking into the position of C where it is connected and how to calculate the frequency everything is summarized in this one question whether it is PSU exam GA exam whatever exam you are going to give you are going to get a question based upon this topic this particular circuit Okay. So, uh high pass, low pass, band pass are discussed. All pass we have discussed separately. So, I believe this topic is clear now. Okay.

[00:57:02]Now, see this circuit.

[00:57:04]Sometimes you have been given the open loop gain and you have to calculate the closed loop gain. Right? So, what is the relation between the open loop gain and closed loop gain? Correct? What is the relation between the open loop gain and closed loop gain? Suppose that here they have given the difference voltage, right? And here you are not using any feedback. Okay. So because you are not using any feedback here you will be getting open loop gain. What is open loop gain? Open loop gain here will be equals to V upon V D. Right? And this will be very very high. Right? This will be very very high.

[00:57:42]Right? The open loop gain is very very high. To reduce this gain and to get a significant value we will apply negative feedback. Right? negative feedback is applied to stable system. We will be able to reduce the openloop gain which is already very high to a significant value. Right? So what we will be doing?

[00:58:00]We'll be using the negative feedback.

[00:58:02]Now after using the negative feedback you know that in this circuit the gain or the ratio of output to input will be equal to what minus RF upon R1. Right?

[00:58:14]This is R1. Correct?

[00:58:17]This is open loop gain. This is closed loop gain. Do we have any relation between them? Yes, we have a relation between them. Closed loop gain is equals to minus RF upon R1 1 + 1 upon open loop gain and RF by R1. Remember this relation because you know this relation is very important. By knowing this relation only you can solve gate questions directly by putting directly by putting the values in this formula. Okay. Directly by putting the values in the formula. Correct. So now see okay I'm from diploma sir lateral entry student. Okay I believe you are able to understand all these things right.

[00:59:05]Very good. Up. Now see we have a question directly asked based upon this particular formula only in this circuit they are using negative feedback. They have given you open loop gain. Input impedance is infinite. Output impedance is zero which is normally there in opamp. Right? No need of anything specific given here. Now openloop gain is 105. They are asking that what is going to be the voltage gain. Means they are asking about the closed loop gain.

[00:59:32]Right? This is your RF and this is your R1. Okay. Now use the relation that the closed loop gain is equal to what? The closed loop gain is equals to minus RF upon R1 upon 1 + 1 upon the open loop gain into RF upon R1.

[00:59:57]Okay. Just calculate this direct formula based question we are having here and I believe you can solve it correct just put the values what is the value of RF it is - 100 upon R1 it is 5 1 + 1 upon what is open loop gain given 105 again 100 / 5 okay now when you will solve it it is going to give you -6.6 66 and that will be the answer. This will be the voltage gain by using the feedback and that will be known as the closed loop gain. Okay. So see just knowing one formula you could have scored two marks in gate 26. Okay. A lot of question have been asked through this topic OPM and it is very very important topic whether it is GATE exam, whether it is any PSU exam, whether it is semester exam, whatever exam you talk about, OPM is a very very important topic for all EC and electrical student.

[01:00:56]It has contributed a lot of gate questions right.

[01:01:01]So with this we have discussed almost all important configuration. Now let us apply whatever we have learned through these configuration to solve these gate questions. Okay. For example have a look over this gate question of 25. The opamp in the circuit following circuit are ideal. The voltage gain of the circuit will be what is going to be the voltage gain is what we have to calculate.

[01:01:23]Right? If I'm talking about voltage gain actually what I have to calculate if I'm talking about the voltage gain I have to calculate the ratio of the output voltage to the input voltage.

[01:01:42]Okay. Now see this circuit in this circuit it is V input which is applied here. Right? So we have applied the input voltage to which terminal? We have applied the input voltage to non-inverting terminal. Right? Then what will be the voltage here?

[01:02:00]Tell me what is the voltage here? What voltage you will be getting here?

[01:02:07]What voltage you'll be getting here? For the simplicity, let me write this voltage to be V_sub_1 and this voltage to be V_sub_2. Then our voltage V input is basically nothing but V_sub_2 minus V_sub_1. Okay. Up. Now suppose this voltage is V_sub1. So this becomes a non-inverting amplifier. So what will be the output voltage that you'll be getting here. Okay. Let me call this voltage to be V of X. Okay, that let me call this voltage to be V of X. So V of X will be what? 1 + RF upon R1 * V_sub_1 or 1 + RF is 10 R1 is 10 * V_sub_1 or we will be getting it 2 * V_sub_1 right 2 * V_sub_1 and what is this voltage this voltage is V_sub_2 right this voltage is V_sub_2 if this voltage is V_sub_2 this voltage will be V_sub_2 right and we will be having this voltage as equal to V_sub_2. Yes or no? Why? By virtual ground condition, we will be having this voltage equals to V_sub_2.

[01:03:17]Correct? We will be having this voltage equals to V_sub_2. After that see now focus should be on this second OPM.

[01:03:25]Let's suppose whatever is the current flowing here, let me call it I, right?

[01:03:30]Nothing will go inside the OPM. So, the same current you will be having here as well. Okay. Now what will be this current? Let me call it I1 and let me call it IF. But we know that I of 1 will be equal to I of F. How you will calculate I of 1? I of 1 will be Vx minus V_sub_2 divided by 10.

[01:03:56]Correct?

[01:03:57]And I of F will be V_sub_2 minus V output divided by again 10. Correct? Now then what we will be getting is that V of X - V of 2. This will be equal to V_sub_2 minus V. What is Vx? VX is equals to 2 * V_sub_1. So 2 * V_sub_1 minus V_sub_2.

[01:04:22]2 * V1 minus V_sub_2. Let me call this V_sub_2 also here is equal to V or we will be getting 2 * V_sub_1 minus V_sub_2 this is equals to V or right or V_sub_1 minus V_sub_2 can be written as 2 * the input voltage equal to output voltage or you will be getting the output upon input voltage this will be equal to two right you got your answer you got your answer what I have Op is used in multiple application in industry. There are multiple configuration.

[01:05:01]Uh there are multiple configuration X sign. Yes, you are right. Okay. We have discussed almost all those configuration. Okay. So here we are only discussing the linear application. There are many other nonlinear applications of oper O op also that we I will cover in sometime cells. Right? I will cover sometimes else in other lecture. Here my focus is bhani only discussing the linear applications.

[01:05:28]Correct? I believe it is clear. Okay.

[01:05:30]Linear nonlinear application. I'm discussing only linear application.

[01:05:34]Nonlinear application like uh it will be used in our stable multiviibrator.

[01:05:38]It can be used as a voltage comparator circuit. Okay. Many things can be done.

[01:05:43]It can be used as a triple 5 timer also.

[01:05:46]Okay. It can be used as a timer circuit also. R table multi viibrator is midmit trigger a lot of configuration is there those are considered as nonlinear that we will discuss sometime else right now I'm discussing some basic configuration in this circuit right OPM is a very uh uh large topic it cannot be covered only in one lecture so here I'm just discussing the basic concepts of or basic circuits of OPM which are mostly asked in most of the exams right but you are right okay what you are saying now have a look over this circuit this Question was asked in gate 26. An ideal OPM circuit is shown in the figure. The output wtage V not is X when the switch is open. Which of the following option represent when the switch is closed?

[01:06:30]Right?

[01:06:32]What type of configuration is this? The input is applied at non-inverting terminal. Right? Thanks for understanding. So input is applied at non-inverting terminal. Okay? So it is non-inverting amplifier. Now when the switch is open, when the switch is open then what will happen?

[01:06:52]So when the switch is open right when the switch is open what how much will be the feedback resistance? So when the switch is open we will be having 2 ohm plus 1 ohm right 2 kiloohm plus 1 kiloohm and total feedback resistance will be 3 kiloohm. Okay. And then what will be the output wtage? The output wtage will be 1 + RF upon R1 into the input voltage. Right? Or what we will get? We will get it 1 + 3 upon R1 is 1 into input voltage and how much is the input voltage? Input wtage is given one.

[01:07:33]So in this way you will be getting V not equals to 4. And they are saying it that you call it X. Okay, we are getting 4 volt. Then they are saying that call it X volt. So let me call it X. No problem.

[01:07:47]Now after that they are saying that when the switch is open, sorry, when the switch is closed. So if the switch is closed, so when you will close the switch this 1 kiloohm will be shorted.

[01:07:58]So you'll be having only 2 kiloohm in the feedback part. At that time RF will be equal to only 2 kiloohm. Right? RF will be equal to what?

[01:08:10]It will be equal to 2 kiloohm and that will be shorted to zero. So 2 kiloohm only. Then what will be the output wtage? Again you will apply the formula same formula RF upon R1. Right? And into V_sub_1. So you will get 1 + 2 upon 1 into 1 or you will get 3 volt. Let me call it V1. Let me call it V2. Right? So they are asking that how much is the output wtage 2 in terms of X. Okay.

[01:08:43]Output wtage V output 2 upon V1. Right?

[01:08:47]Numerically this is equals to what?

[01:08:49]Numerically this is equals to 4 upon 3.

[01:08:54]Isn't it? Numerically this is equals to 4 upon 3.

[01:09:02]Got that? Sorry 3 upon 4 not 4 upon 3 it will be 3 upon 4. Okay. Now from here you can write that output wtage 2 will be 3x4 * output wtage 1. And they are saying that output wtage 1 you can also call x. Let me call it x. Okay. And I will not put the value of x which is four. So what relation I'm getting?

[01:09:26]Output wtage is equals to 3x 4 * x. This question was asked in gate 26. I believe you understood how easy this question was, isn't it? So if you know the basic configuration, you know you can solve many gate questions, right? And you can score maximum marks as well. Then have a look over this circuit. This was also asked in gate 26 and was carrying two marks. Okay, what we have to do here is the idle OPM circuit is shown in the figure. Which of the following option gives the correct value of IX? What will be the value of IX? Okay. So what type of configuration it is? See this terminal is grounded. The voltage is 0 volt. Same voltage will come here which is 0 volt. Okay. Input is applied at the inverting terminal. So it is inverting amplifier. Okay. If I'm calling this voltage to be output voltage. So because it is inverting amplifier what will be the output voltage? Output wtage will be - RF upon R1 * the input voltage or it will be - 1 upon 1 into 2 right or simply we will be getting minus 2 volt right so you'll be getting V not equals to 2 volt okay see the current is coming at this terminal okay and let's suppose that a current is also coming at this terminal okay so If we assume some arbitrary direction of currents, you can consider it going downward. You can consider is coming upward. Whatever you wish, you can assume it. Right? If your assumption is right, current will be having the same same sign. Right? As per the sign of the current, our assumption may be right or wrong. Okay?

[01:11:11]Let me call this current I1 and let me call this current I2. Okay? Apply KCL.

[01:11:17]So if you will apply KCL then see incoming currents are I1 and I2 and outgoing current is IX that is why I have assumed them like this. So I will be equal to I1 + I2 correct I of X will be equal to I1 + I2.

[01:11:38]Now what will be the current I1? To calculate current I1 see this voltage is 0 volt 0 minus V. So 0 - V divided by the resistance which is 1 kiloohm plus the current I2 current I2 will be this voltage is 0 volt 0 - V 0 - V divided by 2 kiloohm so what we will get we will get minus of V minus of V by 2 right now how much is V not v is minus2 so minus of -2 minus of -2 2 divided by 2. So it will be 2 + 1 or it will be 3 ampere or milliampere. The resistances are given in kilm. Voltage is given in one in volt. So current will be in milliampere. So we will be getting 3 milliampere and that will be the answer. Correct. What we have done?

[01:12:37]Nothing. We know that it is a inverting amplifier. We calculated the output voltage by applying the inverting amplifier formula. After that I have just applied KCL that you will that you will cover during the networks basic KCL right so through that you could have solved this question and you could have easily scored two marks now see again the same type of question okay is there any difference between the last question and this question yes only one difference right now it is an non-inverting amplifier okay if this voltage is VI this voltage will also be VI okay now if this voltage is VI this voltage is VI and they have given that VI is 2 volt. So if they have given it let me write it 2 volt. Okay. Now they're asking that what will be the current I. So see there will be a current going here. There will be a current going here and let me assume a current here also. And let me call this voltage to be V. Now what will be the voltage V? Because it is a non-inverting amplifier. It will be 1 + RF upon R1 * the input voltage. Okay. Now 1 + RF RF is 1, R1 is also 1 and input wtage is 2.

[01:13:50]So we will be getting output voltage equals to 4 volt 1 + 1 2 into 4. Right?

[01:13:57]Now after that let us apply KCL. Right?

[01:14:00]So if you will apply KCL see this is the incoming current. This is the incoming current and we have an outgoing current.

[01:14:07]Okay. So what will be the current? So I uh we will have let me call this current as I. So this current I plus this current I not they are equal to the outgoing current and let me call this current I1 and this current as I2.

[01:14:25]So I will be having I1 + I. The incoming currents are equal to outgoing current. You have to calculate I so it will be equal to I2 minus I1. How to calculate I2? I2 will be V upon 1 kiloohm right and minus what will be I1 I1 will be 2 - V upon 1 kiloohm okay or I not will be what is V it is 4 - 2 - 4 correct so how much we will get we will get 4 - -2 or we will get it 6 miampere just check if I have not done any uh calculation wrong otherwise this is going to be the answer right otherwise this is going to be the answer up see how easily we have solved this question right correct voltage is in volt resistance are in kiloohm so answer will be milliamp they're asking in milliamp only correct I believe it is clear now have a look over This question, this is also a gate 26 question. A lot of resistances are given but they all are same. Right? Now what they're asking is that uh the closed loop gain will be closed loop gain means you have to calculate the ratio of V to VI and they are asking about the magnitude. What is the magnitude of the closed loop gain?

[01:15:58]Okay. So how to solve this circuit?

[01:16:01]This voltage is going to be 0 volt. So this voltage will also be 0 volt. Okay.

[01:16:08]Up. See it is not a normal feedback path. They have given a a combination of resistance. Let us suppose that this voltage is V of X. Okay. Let's suppose that this voltage is V of X. Now what we can do? There will be a current like this. Then there will be a current like this. Okay. And both these currents will be same.

[01:16:31]both these currents will be same. Okay, considering this just apply KVL here and again you will get the answer.

[01:16:41]Again you will get the answer. So you know that both these currents will be same. Let me call this I1. This will be also I1. Okay, correct. Let me call this I1 and let me give it another name so that we can apply KCL. I1 is equals to I2.

[01:17:02]So see we have I1 equals to I2. How to calculate I1? It will be VI -0 divided by R R1. Okay. I2 will be 0 - V of X divided by R2. Okay. Now because R1 and R2 both are same. So what I can write? I can write that VI upon R. This will be equal to minus of vx upon r or simply I will be getting v of x = to minus of v of i. Right? I got this vx voltage is now known to me. Okay. Now once vx voltage is known to me. Then what I can do? I can do nothing else but applying KCL is an option. So let me correct calculate this current as I3. Let me call this current as I4. Okay. Now apply KCL.

[01:17:55]KCL at node uh Vx. So what will be KCL? Incoming current is I2 and outgoing currents are I3 and I4. Now I2 current what will be I2 current? I2 current will be zero minus Vx divided by R2. And I3 current will be simply Vx -0 divided by R4. and I4 current will be Vx - V divided by R3 right up then what I can do because all these are same R1 R2 R3 R4 they are all same I can just cut them and I will be writing here minus of Vx okay equals to V of X plus V of X - V or I will be getting minus of VX = to 2 of VX - V or from here I will be getting that V is equals to 3 * Vx right that V not will be equal to 3 * VX and what is VX VX is equals to minus of VI so I will be getting it something like this okay then from here if I will calculate V upon VI I will be getting it minus3 okay and then if I'll calculate the magnitude of it V upon VI this will come out to 3 you got your answer right. So what rocket science was there in this question that if suppose it was the part of gate exam which it was in 26 you could not have done this there is nothing like that isn't it simple configuration you should know inverting non-inverting amplifier and then after that you are just applying KVL or KCL isn't it there is nothing like analog electronics specific analog electronics that we are doing right I believe it is clear okay then see this question this was asked in gate 21 Okay, isn't it looking like something called as instrumentational amplifiers combination that we have discussed? Okay, so the way we have solved that particular question we will try to solve it as well.

[01:20:04]Right? So see in this circuit in this circuit if you will focus here okay basically this was present here this was grounded it's a typing mistake okay now see if you see this circuit what type of circuit is this input is applied at inverting amplifier right it is inverting terminal inverting amplifier what will be the V1 so the output wtage V1 will be simply minus R2 upon on R1 * V input right okay then if you see this configuration the input is applied at the non-inverting terminal so what will be the output wtage 2 the output wtage 2 will be 1 + R2 by R1 * V input right this is what you will get and after that this circuit will be reduced to a circuit like this.

[01:21:10]Just have a look. We can have a reduced circuit in this way that this is minus terminal. This is positive terminal.

[01:21:18]This is grounded. Here you are taking the output. Right? You have a feedback resistance whom they are saying R4 right. And now input is connected like this.

[01:21:35]Just see at the at the inverting terminal you are getting this voltage V1 and V2 and these resistances are R3 and this resistance is also R3.

[01:21:51]Okay. What is V1? What is V2? That you have already calculated. It is there with you. Okay. Now how to solve this question? So how to solve this question?

[01:22:00]What we have to do? We have to apply superposition theorem. Isn't it? We have to apply superposition theorem. So by super position theorem, right? If suppose I'm considering only V_sub_1, if I'm only considering V1, what will be my output voltage? If suppose I'm only considering V1, what will be my output voltage? My output voltage, it is an inverting amplifier. it will be - R4 upon R3 * V1 correct plus if suppose now only if suppose I'm considering V_sub_2 only then what it will be then again it will be minus of R4 upon R3 right with V2 correct the way I have applied superposition everywhere else I have done the same thing here this is what I will be getting okay up see what I can write it I can write it that minus R4 by R3 I can take common then it will be V1 + V2 right up V1 and V2 both are there with you so R4 upon R3 now what is V1 it is minus R2 upon R1 V input and what is V2 v2 is 1 + R2 upon R1 and V input Right? So finally what you will get? You will get minus R4 upon R3. This will be cancelled and you will be getting V input only. Correct? You will be getting V input only. So if you are going to calculate V output upon V input this is going to be minus R4 upon R3. Right? It is going to be minus R4 upon R3. This will be the correct answer. Correct?

[01:23:55]Again see what I have done inverting non-inverting amplifier what is their output voltage I should know and after that you should know how to apply superposition theorem that we have learned during different configuration right in this way you can solve questions like this correct I believe it is clear let us solve this last question now see this question of gate 21 consider the circuit with an ideal opamp shown in the figure assuming V input is less than VC V reference is less than VC ignore all those things. They're asking that output voltage will be equal to what? Okay, what will be the output voltage? Again, see what is the difference between this circuit and this circuit? There is no difference at all. Right? There is no difference at all. This is what this is an inverting amplifier. By superposition, right? By superposition, what will be the output wtage? The output voltage will be let's suppose that I'm ignoring this and I'm only taking input voltage then it will be minus of RF divided by R time V input. Okay. Plus if suppose I'm ignoring this and I'm only taking this input then it will be again minus of RF by R. Right? And this time this input just see the polarity it is connected you have to take it minus V reference right do not forget to do that after that what we will get right after that what we will get the condition at which the output is equal to zero so they are saying that output is equal to zero let me put it equal to0 so minus RF upon R V input plus RF upon R reference. Okay. So what we we will be getting? We will be getting V reference equal to V input. Yes or no? We will be getting it V reference equals to V input. Yes or no guys? Give me a reply.

[01:26:05]Okay. So V input is equals to V reference.

[01:26:09]What I have done? Nothing. Right? I should know inverting non-inverting amplifier. I should know applying superp position and this problem is done. Okay.

[01:26:17]So in this way you can solve many complex problem.

[01:26:21]Now opamp is not just this one. There are other lot of configuration that we can discuss like voltage comparator circuit right like smid trigger a stable multi viibrator triple five timer using OPM a lot of things are there but I have discussed only those configuration here those who have produced a lot of gate questions. So if you are watching this lecture you know your OPM will be revised. If you are suppose first time watching it I think the concepts of OPM will be very very clear to you. You'll be able to solve many GATE questions also whether it is coal India, whether it is GATE exam or any other PSU exam that you are giving. Correct. Now so this is the end of this lecture. Okay.

[01:27:03]Uh like this video, share this video among your group so that everybody will be benefited. And uh do not um if suppose you have not subscribed our channel do subscribe it so that whatever we are going to take whatever lectures we are going to take in future you'll be getting regular updates as well okay and join our telegram also for the same whatever we are doing here whatever we are going to do you'll be getting a regular updates in our telegram channel okay also the class PDF also we will share there okay so I hope that you liked this lecture you people are giving a very good response to this series most expected concept series in which I have already covered 4year transform row and stability root locus and today I've taken a lecture on OPM right the next lecture is going to be on another very very very important topic which has highest probability of producing gate question and it has produced a lot of gate question BJT bicing circuit so next lecture is going to be on this topic right till then stay connected so that whatever we are doing you'll be getting regular update Okay, like, subscribe and share this video also. Okay, thank you.

[01:28:14]Thank you so much for watching this video. Best of luck.