 The next important topic, probably the last one that we can cover is the waveform generation using Schmitt triggers. So, we will talk about the waveform generation using Schmitt triggers and this slide is nothing but just a summary of what we have been doing and instead of drawing the op-amp, the entire op-amp circuit every time, we just show this Schmitt trigger as a black box with these kinds of characteristics and we understand by this that there is a pressure voltage Vth, there is a pressure voltage Vth, Vtl and then the output voltage versus input voltage looks like this. Similarly, we look at the inverting Schmitt trigger and we show that by this particular black box here, just makes things easier for us. How do we characterize the Schmitt trigger? It has got two states, Vo equal to l plus and Vo equal to l minus. In our previous circuit, this was Vsat and this was minus Vsat. How do we make an oscillator Schmitt trigger? With a suitable RC network, it can be made to freely oscillate between l plus and l minus and such a circuit is called an A stable multivibrator or a free running multivibrator. Of course, since this is probably the first time the students are going to hear these terms, it is important to point out that an A stable multivibrator produces oscillations without an input signal. So, it just oscillates freely without any input signal and the frequency is actually controlled by the component values that are there in this RC network. As always, it is good to point out some limits on these frequencies that you can achieve with these circuits. The maximum operating frequency of this oscillator is typically limited to 10 kilohertz because of op-amp speed limitations. If you want to make an oscillator beyond these frequencies, then there are other circuits available that you can use, but it is good to point out this limitation with most op-amps. So, waveform generation using Schmitt trigger, it is best done with an example. So, what we are saying is, what we have is this circuit here. We cannot show all of these at the same time. I am going to go a little back and forth on this one. So, let us say we start with an output voltage of 5 volts. Let us say my plus level that is the V o, high V o is plus 5, low V o is minus 5, my V th is plus 1 and V t l is minus 1. So, in other words, if you look at this figure, this is 5 volts, this is minus 5 volts, this is plus 1 and this is minus 1. That is the situation and this is an inverting Schmitt trigger. And let us just start at some point. Let us say we have V o equal to 5 volts and V i equal to 0. So, if V i is equal to 0, this is 5 volts. This capacitor is going to start charging towards V o and that is what is shown here by this curve. Now, it is important actually to show the students what is happening in the other figure. As the capacitor is charging, the capacitor charge voltage and this V i actually is the same. This V c and V i is actually the same. So, as the capacitor charges, we are moving in this direction. At some point, what will happen is it will cross this V th and that is the time where your output voltage is going to flip like that. So, your output voltage is going to flip like that and now your input voltage, that capacitor is going to start discharging and when the capacitor starts discharging, it is going to go towards minus 5 volts like that. And where are we now on the V o versus V i plane? Now, the capacitor is discharging. So, it has flipped and we are now proceeding in that direction. When we do that, the next crossing is V t l and therefore, when it crosses minus 1 volt, again it is going to change the output voltage. The capacitor is going to charge and then discharge and then charge and so on and it just goes on forever. So, that is what we get for the capacitor voltage and that is the output voltage which will oscillate between l plus which is 5 volts and l minus which is minus 5 volts. Now, it is very important, when you explain a circuit like this in class, it is very important to actually draw this also at the same time and show where you are on this V o versus V i curve, whether you are here or here and in fact, it is very important to point out that you are always in this loop here, in the hysteresis loop. You are not leaving this loop at all. You are not going either in this region of the V o versus V i curve or in this region of the V o versus V i curve and that will happen in all oscillator applications. Now, the next thing that comes to mind is what is the frequency of this particular oscillation. So, that is easy to calculate and this slide shows that calculation. So, let us just take the charging transient first and we know that this from the previous slide, we know that the V c will start off at V t l and it will end up at V t h and that is basically what it oscillates between. So, in the charging transient, let the voltage be described by some a 1 times exponential minus t by tau plus b 1 and what is the time constant in this circuit? It is simply the capacitor is either charging like that through this R or it is discharging like that through this R. So, in both cases, the time constant is R c. So, once we have the equation and the boundary values like V t l here and V t h here, we can proceed and find this time. So, how do we do this? Since we know that at equal to 0, your V t, V c is V t l that immediately gives us a 1 plus b 1. What is the next condition to use to get a 1 and b 1? If this switch did not happen, then this capacitor voltage actually would have gone all the way to l plus. So, that means you can say that if T went to infinity, this V c of infinity would become l plus. So, if I put T equal to infinity here, I will get b 1 and if I put V c equal to l plus, I will get b 1 equal to l plus. b 1 equal to l plus and a 1, I will get from the first equation and I will get then a complete expression for V c of time, V c of t between 0 and t 1 and this t 1, I can compute by saying that at t 1, my V c is going to be V t h. So, etcetera. So, if you do all that, you will get this particular expression and it is instead of giving this expression as the readymade formula, it is always much better to derive this and show the students that actually follows some logical steps and then you get the final result. Otherwise, the students just remember this and then reproduce it in the exam and then you do not know how it came. Here is an example. It is a simulation example, but you can also do this in lab with actual op-amp. In this one, it is the same circuit, but for this particular Schmidt trigger, we can either use an op-amp 741 or op-amp 411 or some other op-amp with different characteristics and 741 is a little slower than 411. Therefore, you can see that these transitions are actually not sharp. So, this transition here is not as sharp as what you get in 411. So, all these things are good to point out to the students. Otherwise, the students never get an idea of the practical side of these things. All right. Here is another oscillator and this actually is qualitatively sort of different than the first one, because it uses an integrator. So, what is happening here is that this voltage is going to be either plus V sat or L plus or L minus. When it is L plus, this integrator, sorry, this voltage, the output voltage here is going to be either L plus or L minus. When it is L plus, the output of the integrator is going to fall because it has got a negative sign as we have seen yesterday. When the output is negative, when the output of the asymmetrical is negative, this is going to rise and therefore, this is going to continue going up and down, but in a linear fashion as opposed to the last circuit where this was actually exponential because the capacitor is charged through a resistor. So, that is the difference between these two circuits. Here the capacitor, here is the input, here this capacitor voltage Bo1 minus 0 is linear, is varying linearly with time. You can also use some, use Gner diodes to limit this voltage in case you are interested in doing that. For example, the op amp here might give you, so this is your integrator. Same circuit has lost as the last one, except now we have shown the Schmitt tricker as a regular op amp circuit, the actual implementation and that is the same integrator as before. If you do not like plus minus V sat, if you want some other voltage, then you can actually put a pair of Gner diodes like that. So, this here, the light blue is at this end and the dark blue is at this end. So, the Gner diodes will actually help you limit the voltage to whatever value you like. So, that and then that of course, as you know it is given by the forward voltage drop of this plus the reverse breakdown voltage drop of this. So, if this was 4.4 and this is 0.7, then the total is going to be 5.1 volts. Now, this is also available in SQL as an example and maybe in the remaining time, I can show you some of these examples. You can play with the parameters, you can ask students to change some parameters and do some things and so on. That may be worth it. So, let me open some of these examples and let us take a look. So, this is this op amp, this is the circuit that we are just discussing. So, there is an integrator, the R value is 12.5, the C is 1 nanofarad, R here is 10k. I am reading it out because you may not be able to see it. This R is also 10k and then there is a limiting resistance of 1k. So, these are these are the pair of Gner diodes and that serves to limit the voltage to whatever value we want. So, this if you look at this diode, the Gner diode, the breakdown the VZ value is 5 volts. So, the output voltage we should get limited to 5 volts plus 0.7 or 5.7. So, the output should swing between plus minus 5.7. So, if I run this solver, I run this example, look at the output. That is what I get. The output is between something like 5.7 and minus 5.7. What about the frequency? You can actually ask the students to run this example, find the frequency and then check that out with the simulation results, the calculated results. Let us suppose I change a few things. Suppose I change this capacitor. Suppose I make this, instead of 0.001 micro, I make it 0.002 micro or in other words, I increase it by a factor of 2. You can think about what should happen here. Obviously, the output levels are not going to change because they are determined only by the Gner voltage. If I change this, what is going to happen? You think about it but later in the meanwhile, let me just run this and show you the results or let us do that later. Let me just even for the first example, let me even with the previous value, let me plot a few more things. Let me plot this output voltage here, which is the integrator output and this output voltage, which is the circuit output, which is the oscillator output together. Let me plot v out and v 3 together. You can see that this is the capacitor voltage and the integrator output and it is pretty much rising and falling linearly. It is absolutely linear. There is no exponential charge discharge. It is linear. Now, let me come back to that same question. If I change this capacitance value, how will that change things? Let me run it again and let me plot these same things again. Now, this is what has happened. Now, in the same time, this is like 0.2 milliseconds. In the same time, I have more cycles here and I have less cycles here. Obviously, the frequency has decreased and you can ask your students to actually play around with these things and see what happens. What will happen if I make this voltage? If I apply a non-zero voltage here, now it is 0. Suppose I make this non-zero, what will that do? What it will do is it will change this mid trigger characteristics. It will push it to either on the positive side or the negative side depending on the value of this DC voltage. Let us say I make it 1 volt or 2 volts and run the program again and then plot this same thing again. This is what has happened now. Now, the capacitor voltage is not symmetric around 0, but it is symmetric around some other point. Now, it is up to you to figure out why this happened and you can discuss these things with each other on modal. Let us take some other example. I am just trying to show that simulation can be sometimes very useful in understanding these things. There are lots of filters examples and there is this bias equalizer circuit which we have already seen. You can run this and you will get the frequency response of that circuit and that is what you are getting. So, it is enhancing higher frequencies and it is attenuating lower frequencies. Now, what this circuit does? It will actually depend on this pot setting. Let me change this pot setting. Right now, it is 0.1. Let me make it 0.8 for example and run it again. If I do that, so this was the previous one and this is the new one. Now, you see that just by changing the pot setting I have completely changed the functioning of this circuit. So, here in this particular case, I was enhancing the higher frequencies. In this case, I am enhancing the lower frequencies. All of these things are, these circuits are actually very useful and if you change the pot settings to some other values, you will get a whole range of these curves. Let us see if there is some other example that would be of interest. Super diode. Let us look at super diode. So, in this example, there is an AM source like we had in the slide and then there is this super diode and then there is this load resistor and capacitance filter. So, let us run this and see what happens. This is not what I wanted to show. Let me just see, show you VS and V out. So, this is the blue one is the input voltage, the AM amplitude modulated voltage and the green one is the demodulated output. Now, suppose I change this capacitance. Right now, the value of this capacitance is 20 nanofarads. Suppose I make this 50 nanofarads and run it again. Let us see what happens. So, and this is the, this is actually a major problem in real life. If your capacitance value is too large, then you will actually miss out some information altogether because this capacitor will not be discharging fast enough and you just miss out all this information. What happens if we make it too low? Let us do that. Suppose instead of 20 nanoseconds, I make it, sorry, 20 nanofarad, I make it 5 nanofarads. Now, let us see what happens. So, this is with 20 nanofarad and this is with 5 nanofarads. So, obviously the ripple now has gone as increased and that is not something that you want and if you increase the capacitance too much, this is what happens. So, there has to be a, the capacitance value actually plays a crucial role in this AM demodulated circuit and this is a good way to illustrate that point. Let us see if there are some other useful demos. This is the op-amp precision rectifier which we actually discussed in the slides and as we said, this is, these resistors, this circuit, this block is behaving as a summer and this block is behaving as a rectifier, half wave rectifier. Let me see what we should do here. That is the input voltage. So, let me look at output voltage versus the input voltage and the output voltage versus time. So, that is what we get and that is what we expect from a full wave rectifier. You can also plot instead of choosing time, here we can choose the input voltage here and then plot this and that is what we get as we expect. So, I think we will stop here. We have covered a lot of examples and we have covered too much material for one day, but I hope you have got something out of it. So, there was a question about the equalizer and what it does. Now, first of all, what is an equalizer? If you have a DVD player or a CD player, you will see that there are some switches there in various frequency ranges, low frequency, medium frequency, high frequency and so on. And those, they are not switches. They are called what are the slide kind of switches. You can slide those up and down to either suppress or enhance, say low frequencies or high frequencies. To give an example, if you have a singer like Lathamangeshkar and there is some tabla playing in the background and you want to enhance the voice and suppress the instrument, then you can enhance the high frequency and suppress the low frequency. So, each one of those sliding devices are connected to a pot like this. There is a potential divider pot. Now, depending on where it is, whether the pot setting is on this side or that side, you will either get this kind of response or you will get this kind of response. Alright, so 0 dB means basically no gain. This one is attenuation and this one is enhancement. Now, this is happening only at a particular frequency. So, that is why you have those five or six different sliding bars. So, that is what it does. I suppose that explains it. So, let us say you want to, this particular frequency suppose you want to enhance. In that case, you will choose this kind of setting. If you want to suppress it, then you will choose this kind of setting. And all that is controlled by simply this pot. So, I hope that explains the question, the equalizer pot. We can go to the next one. Give a numerical example to measure CM or R practically. Alright, now I do not have a circuit to do this in front of me, but these are actually available on the net. So, if we just go to one of these sites like linear or national or one of these things, they will actually give you an application note in which they have a circuit to measure the CM or R. So, and these are fairly simple circuits to understand. Can you please tell us the details of sample and hold circuit using op amp? Please tell us the details of the FET to be used. Alright, now, okay, I will come back to that later. But in your SQL distribution, you do have a simulation example called sample hold. So, we just take a look at that one. But essentially what you need from the switch is that it should, its resistance should be, the on resistance should be small and off resistance should be infinite as large as possible. Otherwise, it will leak the charge. What is the use of capacitor in internal diagram of 741 IC? Okay. So, here is the actual circuit diagram and the function, there is only one capacitor in this entire circuit and that is meant for compensation. Essentially, it compensates, it prevents the circuit from just oscillating. I am not sure how to explain this without a diagram and so on. Maybe I will invite Professor Sharma and see if we can do it better. It is, this is called internal compensation. If you do not have this capacitor because of the very high gain of the op amp, there is a very strong likelihood that the circuit will just oscillate because of any parasitic inductance capacitance and so on that you have. But maybe Professor Sharma can address this better. The basic problem is this that if you have a multi-stage op amp like a 741. So, consider the following situation. You have the first stage, some defamp and this has a second stage. This might be the output driver and then you have the feedback to the inverting input. Now, normally this feedback is supposed to be negative and that stabilizes this op amp. However, the model of this is that this has some RC constants or delay associated with it and this has some RC constants and delay associated with it. In fact, if you look at the 741 circuit, you will find that there are three stages. Now, if the phase difference for all these stages adds up to 180 degrees, then the feedback that you thought is going to the inverting input will actually add up aiding the input rather than suppressing it. As a result, you now have positive feedback. There is 180 degrees difference because of the inversion and an additional 180 degree at some frequency due to these RC elements. So, there is a total of 360 degrees of phase difference and therefore, what you thought was negative feedback can now become positive feedback and this circuit can start oscillating. So, the answer to that is that the very first of these, the time constant must be changed so that it becomes a dominant pole and the output amplitude at the frequency where it would have oscillated. The gain becomes so small that oscillation is not possible and for that you need a large capacitor and this capacitor is the one that you see. It is connected in a miller configuration so that its actual value is multiplied by the gain of that stage and therefore, it is connected to a high impedance point and a high gain stage so that the value of R is large and the value of C is large. So, that this pole is low enough and reduces the gain sufficiently at high frequency so that this circuit cannot oscillate. Details and quantitative details of this we will see in a later lecture that I will take in which we will look at the innards of a 741. However, the internal circuit has been put up by Professor Patil you would have seen that and if you go to Moodle and see the notes that I have put up, there is a tutorial on the insides of 741 and in that you will find many more details about how the 741 works from the inside. I think for now this will have to be. The next question is to operate the opamp in linear region. Do we need to consider open loop gain or closed loop? The question is not too clear but let me try to answer it. So, here you have an opamp and the output here. Now, if you just use open loop without connecting anything between the inputs and outputs, the opamp is as we have seen you need a very, very small voltage to make it go to either minus V sat or plus V sat. And now we consider that there is some kind of feedback whether it goes to the plus terminal or to the minus terminal all these things will actually depend on the details of that particular circuit. And as we have shown in the slides both for the positive the inverting amplifier as well as the non-inverting amplifier, the feedback actually can be shown to be negative, qualitatively. But if you want to actually compute the feedback factor that is something that we should be able to do next week because professor John is going to teach feedback amplifiers and then that will become more clear to you. But just a quick answer to your question is definitely we need to consider what whether the loop is closed first of all. If it is not closed then there is no question of any feedback and then the circuit will not operate in the linear region. So in fact if you take an opamp just go to the lab and just connect both of these two ground and measure the output voltage you will see that you are either stuck at this point or at this point because even if the input voltage is exactly 0 there is an offset voltage and so therefore your output is likely to be at plus V sat and or minus V sat that is open loop. So the opamps are except for the purpose of comparison opamps are rarely used in that configuration. Yeah so more than that I guess we cannot say right now quantitatively but this topic will be taken up later by professor John. Maybe if I will ask professor Sharma if he wants to add think of it as a black box. This black box could be with or without feedback. Now let us say that I have applied some input to this and some output to this and if the closed loop gain is G closed loop then we can say that V 0 is G closed loop times V i. Now there is a maximum value of V 0 which you can have from saturation condition. Therefore V i max is then this V 0 max divided by G closed loop. So the maximum input voltage that you can apply will then be constrained by the closed loop gain. However as professor Patil has just told you suppose you are operating it open loop in that case this G is extremely large and therefore the voltage that you can apply without saturation is close to 0 and that is why what he told you that if you take an open loop op amp and connect nothing to it just the offset voltage which is close to 0 is enough to saturate it that is because of this because the highest voltage that you can connect to it in that case approaches 0. However if you are using negative feedback then this loop is finite and if you take the maximum swing voltage which is the output saturation plus minus output saturation minus this is the maximum saturation voltage and if you divide it by the closed loop gain then that is the maximum incursion allowed at the input voltage. If it exceeds it then your output will get saturated I hope we are actually second guessing your question I hope this has answered your question that was your question. So there is a question from Warangal which says what is the significance of negative power supply in op amp for example minus 15 volts. First of all this plus minus 15 volts is not holy because 741 uses I have seen many questions on the moodle which seem to assume that an op amp must be operated with the bipolar supply and with plus minus 15. This is only one particular op amp one particular case the op amp is a differential amplifier that means it amplifies the difference of two quantities a and b. Now it is possible that a is greater than b or b is greater than a therefore a minus b can be both positive and negative and so the output can be both positive and negative. Well if the output can be both positive and negative then you must have both a positive supply and a negative supply if you do not have a negative supply how is the output going to go negative. So as a result if your reference voltage is 0 then the output can go to positive or negative voltages and therefore you require a positive and a negative supply without this you will not be able to produce a negative voltage the output is anyway constrained between the power supply voltages. So therefore if you have a single can you use an op amp with a single supply and the answer is indeed yes. What you can do is that you can use the analog reference as V dd by 2 that means all voltages are now measured with respect to V dd by 2 your signal input and your signal output must now be referenced to this V dd by 2 and this is equivalent to connecting two power supply one of plus V dd by 2 and the other of minus V dd by 2. So physically if you have one power supply you can still manage notice however that that is equivalent to connecting two power supplies each of half the magnitude and plus and minus sign and now your signals must be referenced to this half way point. But in general we do not bother for most lab kind of experiments at least we use two power supplies and this is the reason that a negative as well as a positive supply is in fact required. I hope that answers your question there is a question on how many differential amplifiers are there in a 741. In fact 741 has only one differential stage it is a very complicated circuit and if you really want to understand it please look at this specific tutorial that I have put up on 741 it is there on your model side. However since this question has been raised I would just like to point out the inner structure of this the inputs to 741 are with respect to these two terminals notice that Q1 and Q2 form emitter followers the input is connected to the base and the output is taken from the emitter. So it is an emitter follower common collector configuration as you would know this is a high input impedance configuration and that configuration is in fact required for an op amp because we want the input impedance to approach infinity. The output of this emitter follower then goes to this common base stage this particular stage is also part of the same differential amplifier and forms a common base amplifier the input is at the emitter the output is taken at the collector. So now you have a differential but common base amplifier. So you have a combination of common emitter feeding a differential common emitter feeding a differential common base. The output from these two differential arms is combined in this circuit and the final output is now single ended from this point onwards all signals inside the 741 are single ended. This is a common emitter amplifier output then drives this push pull output stage. So this is the push pull output stage and the output is therefore single ended. This is the architecture of 741 just to quickly go over it differential input differential input emitter follower feeding a differential common base amplifier outputs combined the combined single ended output goes to a common emitter amplifier the common emitter emitter amplifier through phase splitter feeds a push pull amplifier. Now all of this requires biasing and this biasing is then provided by these current mirrors. So you have this as the current source and there is a Vidlar current source current mirror here which has this resistor included and exactly how it is calculated is given in my notes and then this current is then mirrored into various sources. So this current is then mirrored into various sources which bias all the stages. So all the stages are bias through current mirroring and therefore as the current through this changes the current through various stages will change and how exactly it is done please read through the notes that I have put on the movie. There is a question saying in op-up integrated circuit integrator circuit we get negative output why the answer is that to reduce error the other side of the RC circuit I will just draw it for your reference again. If we want no error in this integrator then we want the current the current through this resistor should represent the input voltage which is being integrated the integration is being done in this capacitor. So if the current is to depend only on the input voltage then this point must be virtual ground and therefore it is convenient to use the inverting configuration. Because we use the inverting configuration for the op-amp the output is obviously negative with respect to the input. So that is not a big deal you can for example just put a buffer here with a gain of minus 1 and now you will get the positive output. So it is not a big deal you can very easily get a positive output as well but to understand this consider the traditional RC integrator which is this and this will be a true integrator only if the output is very very small compared to the input otherwise it will have error the output will not be exactly proportional to the integral of the input and that is precisely what we are doing in this circuit by returning the resistor not to the output but to the virtual ground. If we reduce it to the virtual ground then the output is guaranteed to be directly proportional to the integral of the input. This passive integrator has errors and the output is indeed as you can work out from the differential equation the output is the integral of the input as long as the output voltage is negligible compared to the input. So to avoid this error we use this configuration in which the input the current is between input and virtual ground which is guaranteed to be negligible with respect to the input and then this current is integrated over this capacitor which charges it up and therefore this output is the integral of that current and hence the integral of this voltage. So that is why we choose the inverting configuration and the inverting configuration will then give you minus times the integral. If you want positive you can easily put a stage of gain minus 1 here. It is also possible to make integrators using non-inverting configuration but that is much harder we would not have time to do it right here. This there is a question can we make a buffer using inverting and non or non-inverting configurations. This I had taken actually when I first introduced the inverting and non-inverting configuration. Let us look at both the configurations. This is the inverting configuration with a gain of minus 1. This is the non-inverting configuration with a gain of plus 1. Notice the gain was 1 plus r 1 by r 2. So therefore to make it truly 1 the feedback resistance should be 0 and the other can be infinite. So you get this simple configuration. You can also of course put a resistor here. It will make not make a difference. Now the point is a good buffer should not draw any current from its input whereas this one draws a finite amount of current from its input because this point is a common ground v i by r flows from the source. So apart from inverting the input this has the disadvantage of pulling some current from the source. This one is a good buffer because the input is connected to the non-inverting input and here the input impedance is supposed to be infinite and therefore no current is drawn from the input. Therefore this is in that sense a better buffer. The gain of this with plus 1 whereas the gain of that will be minus 1. Therefore here the output is an exact replica of the input. However this can drive any load whereas this one is not loaded by the op amp at all and which is what a buffer is supposed to be. So therefore this is a preferred configuration for buffer. I had answered this question about inductor feedback also on Moodle. There is no difference between inductor feedback and capacitor feedback and so on. The basic methodology of calculating input output remains the same. However since you have raised this question let us just look at the case where the feedback is inducted. Consider this configuration. In that case the current flowing in is v i by r and the voltage drop from this is 0 and the voltage drop v o is such that l d i by d t that is the total voltage drop across and that is equal to v o. This is the current in this arm that should be v i by n, v i by r. This is the output voltage and this d i by d t is minus l d by d t of this. So it is minus l by r d v i by d t. So this will differentiate the input. In fact r l does exactly what c r would do. Just interchange the position of c and r and change l for c that is the behavior that you expect. So for an integrator you would put the inductor where in series with the input and put a resistor in the output. So in short there is no new technique required just put in the components as they are and the current should flow like this assuming if there is negative feedback assuming a virtual shot. So therefore you already know that this voltage is 0. Therefore the output voltage is the drop across this inductor which is l d i by d t and you know that what is the value of i because i is given by v i by r. Just substitute and now you will be able to find out v 0 in terms of v i. So in fact there was a question on Moodle which says what happens if you have inductors and current sources. The point is it does not matter bring them on just connect the whole thing and you must make sure that the sum of currents at this point goes to 0 which is what we are doing by saying that as much current which comes in is the current which goes out. So that is essentially nothing but k c l at this node and it is for this reason that this node is also called the summing node. So the total the sum of current is to be calculated at this point the voltage here is given by the virtual shot and the sum of current should be 0 and that gives you a sufficient set of conditions to calculate the v 0 as a function of v i. So whether you have inductors capacitors resistors non-linear diodes non-linear circuits like diodes or current sources it does not matter just connect them up some the current at the summing node put that current equal to 0 and made the voltage here equal to this and that will give you a sufficient set of conditions to evaluate them. Yeah in normally we do not use inductors because they are inconvenient to use they are bulkier you can make very small capacitors and they require currents to flow through them all the time. So therefore, you can make low power and compact circuit using capacitors. So if you have a choice of using an inductor or a capacitor then typically you tend to use an R C circuit rather than an L R. There is a question from N i T k saying that when a bond is broken the free electron moves, but the movement of hole is also due to movement of electrons. So why is the mobility different? The mobility of something is not dependent on what it is the mobility is dependent on what surrounds it. For example, when I walk through the C S T station in a big crowd my mobility is very small because I am scattered very often. When I move on an empty road I tend to walk very fast. So therefore, mobility is not a property of what is moving. Mobility is the property of the surrounding through which I am moving. So in case of hole movement it is a bound electron which moves and therefore, moving of an electron from one atom to the other atom involves detachment from the binding force of one nucleus and then attachment to the next. This obviously is a different mechanism from the mechanism of the same electron moving as a free electron. In short electrons move through the bound electron cloud under a field when hole transport takes place and they move through the interstitial space between the electron clouds when they move as free electrons because they do not get bound they do not get inside the orbital of an atom. So therefore, it depends on which one is easier to move. In general it is obvious from this qualitatively that it is easier for a free electron to move because it does not have to bother about the binding force of the previous atom and then going to attaching to the next atom. Therefore, in general holes have lower mobility than electrons though in both cases it is electrons which are moving. However, notice that this is in general true but there are cases where hole depending on crystal structure and so on where hole mobility could be higher than electrons, but it is true that it that is rare. So, there is a question from H G S I T Indore which says that if we need a gain of 10 then we can select resistors any resistors in a ratio of 10, we could use 10 k and 1 k, we could use 1 meg and 100 k. So, what choices do we have for selecting these resistors? There are two things which limit the value of the resistor. So, let me just go back to this diagram again. If the value of r is too small what is the problem? What it means is that a large current has to be supplied by the input source and a large feedback is to be feedback current is to be supplied by the output. That means the output is loaded to drive so much current through the feedback resistor and the input is loaded because we are pulling a huge amount of current from it. Therefore, we should not make r too small at the same time we should not make r too large because remember the input impedance of this is not truly infinite. It may be of the order of meg. So, if you start using multiple mega ohms in these resistors then you cannot make the assumption that there is no current flowing into this. After all it is a real op-amp not an ideal op-amp and in that case even though the current drawn is very small the parasitic leakages and resistances will become important. So, in general multiple mega ohms is too large and just ohms is too small. You must use in between values and a typical value might be 1 k and 10 k or 10 k and 100 k. Typically values in case are convenient to use too high a value of resistor will lead to errors because of leakages to low a value will lead to errors because of loading of the op-amp as well as the input.