 have very high input impedance. So, we are saying in fact this is what I start doing in the class that what would be the best possible amplifier now that I have done this. So, first it should have very high gain. So, ideally gain should go to infinity, second it should have very low common mode gain. So, ideally that should go to 0, third it should be no burden on its inputs. So, it should not cause any change to the input because you are connected to it. Therefore, its input impedance should be finite and output it should be like you know take any atyachar and not be affected by it. So, it should give you the same voltage whether you take a lot of stuff out of it or not. Therefore, output impedance should be 0. Up to now I have not uttered the word op-amp. We are simply saying what would be an extremely good differential amplifier and once we I convince that these are the desired properties plus of course it should amplify at all frequencies and so on and then we will say voila this is your op-amp. So, an amplifier which has these properties we call an operational amplifier. So, that is how I introduced op-amp to this and then that these are the now consider a configuration like this. Now, this is an ideal op-amp and one of the inputs is grounded. Therefore, the output is some gain times vx and its output impedance is 0, input impedance is infinite. So, as a result the input is not connected to it is not loading anything and it is a voltage dependent voltage source. Its output is a times vx and it is in the inverting configuration here. So, if I have vx at this point which is the input this point. If I have vx at this point then at the output I have minus a times vx and between input and output I have RF and here I have RI. Notice that I have not taken an ideal op-amp. It is actually better to begin with a somewhat more complicated thing do not overwhelm them by saying that you know this is what will happen that the virtual ground will descend on us as if it was wished by the heavens. Now, we just solve that. So, now this is just a network. There is no op-amp, there is no mysteries here. So, you do KCL at x and equate the incoming current without going current. So, the incoming current is vn minus vx divided by RI and the outgoing current is vx plus AVx divided by RF. Equate these two currents and then what you will find is that solve for vx first. Do not tell them that there will be a virtual ground solve for vx and then you will get this and what happens. So, vx has this term inside and what happens as A goes to infinity vx goes to 0. Voila now you say that in case of an op-amp you will have a virtual ground. That means even though this terminal is not connected to ground as the gain goes to infinity, this guy comes down to 0. In fact, the ratio in which this voltage and this voltage changes is such that this point is to be kept at ground. So, they themselves derive the value of vx in the non-ideal case and then take the limit as A goes to infinity and then they themselves will see that vx will come down to 0. In fact, we plot vx as a function of the gain. So, you take gain equal to 100,000 infinity and now you can plot as a function of some input, plot the value of for a given input plot the value of vx and then it will be clear that as the open loop gain of the amplifier reaches infinity, this voltage will come down to 0. In fact, the differential voltage will come down to 0. So, it is not just virtual ground, it is in fact a virtual short. So, which is what we are saying that vx tends to 0 as it tends to infinity. So, the virtual short is actually derived by them. You give them this problem to do that I have a circuit like this, do not even show them the left hand side. So, I have a voltage dependent voltage source and then later say that this is this is my op-amp and therefore they themselves have derived that this voltage will go to 0 as the gain goes to infinity. So, then the virtual short is not a hypothesis, it is not a lemma that you have to assume, it is in fact derived by you, it is not externally forced, it is a result of the network analysis. And then you can calculate the if vx equal to this, then v out equal to a times vx. So, it will be a times and therefore v out by v in will go this. In fact, notice that all the time I am doing the tougher non-ideal case first and then deriving the simpler ideal case as a derivation from that as a goes to infinity. So, then they themselves will say and then you say that look if I already knew that a was infinity, then I could have assumed that this is a virtual short and then the analysis would have been so much simpler. That same result would have arrived in a much simpler way, but let them this is not so difficult that we have to hide it from them. Let them have the pleasure of making this discovery that as a goes to infinity you have much simpler equation because that way when you go back to practical op-amp, then it is simple. They have done that even earlier than the ideal op-amp with limited gain. Exactly the same thing happens with the non-inverting configuration. So, I again take that and use a non-infinite gain here. So, the equivalent circuit is that this is the input voltage. This is the input voltage and because the input impedance is infinite, it is hanging in the air. There is nothing connected to it and then this is a voltage dependent voltage source and what it sees is RF and RI down to ground and this output is the difference of this voltage and this voltage. So, this voltage and this voltage, V in and V in. Because it is a difference amplifier, the output is A times V in minus V in and now you can just derive V x in terms of V out and just collect terms in V in and V out and solve again solve for V x and again you will find that V x is equal to V in now not 0. So, they will notice that if you put A equal to infinity essentially this V x will become equal to V in because this factor will become 1 as A goes to infinity because these terms will become negligible compared to the compared to this one. So, then the virtual short is apparent to them right then and there. It is also apparent to them that you will not have a virtual short if the gain is not infinite that in a practical device if you actually go to measure and the gain is not quite infinite you do expect to see a little bit of the signal there and how much of the signal that you see is dependent on how much is the open loop gain of this. Higher the open loop gain less is the signal which will be visible to you at that particular point then you can do inverting adder because after all the currents were made 0 therefore if you add currents then the output will be proportional to that total current and now you can easily derive the adder you can derive the differentiator by writing the equation because essentially this current and this current have to be equal. So, you now keep on writing this current and that current and then that will give you all the applications of this and finally, if you put a non-linear device here even in that case the same same principle applies that means since this input impedance is infinite the currents here have to sum to 0 and this current happens to be linear, but this current happens to be non-linear. So, you equate the two currents and when you solve eventually what you get is a logarithmic application. Here like we have used one open, but as we know because of virtual grounding we can consider vx to be is equal to grounded. So, sir can we simply say what is the use of actually open because simply connecting v in then r i and diode will also give us the same. No it will not this what will keep this voltage to be what will keep the what will make this voltage equal to ground what will keep it at ground if you actually make it ground then there is no current through this diode because this input will just go into ground and there is no input to that ground. So, the point is that this virtual ground is because the output changes in such a ratio. So, that this voltage divider which is formed by these two results in this being 0. If you anyway force it to be 0 then there will be no voltage higher than the input. In fact, it is very good to give them a qualitative field for this. So, you say look you all agree that this voltage has to be brought to 0. Suppose, this is 1 volt. Now, if this is 10 times this then this voltage has to go 10 times out to bring it to this 0 to 0. As long as there is even the least bit of voltage here this output voltage will because of infinite gain will keep on increasing and how long will it increase till this comes to 0. And when it comes to 0 how much is the output voltage 10. If you did not have in this case if you did not have this op-amp you just grounded this right then this output will just be floating what is the output connected to of this diode right. So, the contribution of the op-amp is not just providing a virtual ground. The virtual ground is just by the way and by assuming it it makes your analysis simple, but the real contribution of this op-amp is to produce such a V output that these two currents are equal and this voltage is there. I think coupling capacitor at the input stage sir. We are using coupling capacitor at the input stage for AAC amplifiers. And we are using capacitors for differentiators also. So, what is the difference between these two I mean when we. They are identical in fact that is the that is exactly the point. Suppose the capacitor is very large which is what you require in case of an AAC amplifier right. So, in that case what it means is that the time constant of this is very large right and the differential at that frequency is the whole signal there is no attenuation of the whole signal. That is also a differentiator actually. So, all the AAC amplifiers are called as differentiator. They are actually differentiator, but they do not show up as differentiator. Essentially what happens is that suppose you have DC plus a high frequency input ok. So, you have DC plus a sinusoid you take the differential. So, how the DC comes sir? Because that is the bias the voltage in general can be thought of some DC component and if there was no DC you do not require a coupling capacitor. Why do you require a coupling capacitor? To eliminate DC. To eliminate. So, DC is there right. So, there is a DC and a sinusoidal signal right. The output may convert sine to cos, but that is equivalent to sine with a phase difference right. And the amount of phase difference will depend on R and C ok. So, therefore at a frequency where your signal is if the capacitance value is very large the phase difference will actually be very very small ok. So, that is the way to say because it will convert sine to cos only if R is not there or you are at a frequency range where the RC time constant is considerable. But if your frequency is at a place where the time is much shorter than the RC time constant then the equivalent phase difference rather than being 90 degrees will be nothing at all and you will get a replica of the input. So, the differential becomes an amplifier you can just derive it from that amount of phase difference between input and output. Thank you. So, in short what we are saying is the following. This is the current in the feedback path right V out by R and this is the current in the input circuit. The current in the input circuit is limited by the capacitance. The current in the feedback circuit is limited by the resistance and these must be equal with the minus sign right. So, what you are saying is that it depends on the balance between R and 1 by j omega c ok. So, if the time constant is such that then V out equal to. So, indeed the output will increase with frequency. So, if you have a capacitively coupled op-amp amplifier then it will be sensitive to the frequency and the output will in fact increase with frequency unless there is the tailing off with frequency which brings it in and which is the property of a differentiator. That means the output increases with frequency in a differentiator. So, in fact that amplifier is like a differentiator it does cause a phase shift and that phase shift is equivalent to converting a sign to cause and hence a differentiation ok. This is the phase shift that will result from that ok. So, once you do that that basic stuff then we have the idea of negative feedback and again I am not assuming transistor, op-amp or anything. They know that amplifiers exist. So, suppose you have an input, you have an amplifier and you multiply the output by some factor F and feed it back. So, what is V out? V out is A times this voltage and what is this voltage? V in minus F times V out. So, I can collect all the terms in V out on one side and V in on the other side and that shows me that V out by V in equal to A upon 1 plus F times. In fact if you recall in yesterday's talk to talk to a teacher session this was exactly what professor Madhu Belood had answered about that block how to read the block reduction formula. So, if you have a block in a feedback then this is the effect of that that was a formula that he had given it comes exactly from here ok. So, if you have a feedback block this is how it implements the gain. In fact it is important to see what happens if A reaches infinity. If you have very high gain then this will reach 1 by F ok. This 1 will become negligible compared to F into A and therefore it will be A upon F A or 1 by F and it will become independent of the gain itself. So, if the gain is if F times A is very large then the actual gain does not matter. The output will be determined by F and not by A as long as F is very large ok. So, now you can be free of the transistor parameters. You have designed an amplifier and now you have some gain if beta equal to 100 some other gain if beta equal to 150 that is not very good. You sell some product some transistor blows somebody else replaces that transistor by an equivalent transistor, but the beta is different. If the behavior starts changing that is not very good. So, you would like to design a circuit which does not depend on the exact value of the beta. The beta can be in a range and as long as that range is high enough this amplifier will behave exactly the same and that is what the negative feedback has brought for us. Because the negative feedback is set by passive components and they define the gain. Now V out by V in we had made no assumptions about F ok. So, here I subtracted and I got 1 plus F A, but suppose I had added then I will get 1 minus F A. Now this has two regimes of operation. One is that F times A is actually smaller than 1. If it is smaller than 1 what will it do? It will boost the gain right. So, you will get a high gain amplifier. In fact, positive feedback is often used for gain boosting ok. However, it is a dangerous weapon. It has to be done very carefully. Because should F times A become 1 then even the closed loop gain will become infinite. This is the closed loop gain not open loop gain ok. Even the closed loop gain will become infinite and this guy will start oscillating. Indeed that gives us the oscillation condition right there that F times A should be 1 ok. So, then we take oscillators as a special case when F A approaches 1. So, we expect this guy to oscillate and then these are the two feedback factors which result in oscillation by this amount and I actually have a simulation experiment for this. Unfortunately, this computer is refusing to boot with my hard disk otherwise I could have shown you the simulation right here. This algebra is somewhat dirty and you need lots of patience for it ok and be kind to your students if you give it as a problem because it does take a lot of patience to derive it. But eventually you can derive what is the minimum gain and so on in the two cases. So, it takes a lot of time to solve. It takes even more time to format it properly so that it comes out in a document. But since I expected some patience from my students at least I should show the same amount of patience and you can show that the minimum gain that you require in that case is 29 in this case and that that can be shown theoretically that the minimum gain required is 29. And then essentially by using RF and R like this and using an element like this you can now make an oscillator ok. Now, you will need to adjust this RF and R because you are saying that the output impedance of this is 0 it is an ideal op amp. But a real op amp will have some output impedance it will not be 0. So, therefore that will add to this RF and therefore if you just carefully got RF and RI exactly as they were done you will not get an oscillator. So, you need to have a gain which is somewhat higher than 29 to account for this ok. And this you can show very easily in a lab on a breadboard that you just exactly make this circuit notice that this last hour in our RCA amplifier was going to ground. In this case we have connected it to the virtual ground. So, as far as the phase shift oscillator is concerned it is the same rather than going to ground it is going to virtual ground. But now it is possible to have this circuit because the current through this is now being fed and is being amplified. This is actually consider it is. So, there is actually a negative feedback at that point at this point there is negative feedback there is positive feedback with respect to this point because this is the phase shift element. If you look at this feedback this is negative right it is this which has brought the gain down from infinity to 29. The positive feedback is with respect to this ok. So, there is a huge amount of negative feedback otherwise you had infinite gain to begin with. And if you did not do this then what you will get is a square wave because it saturate both ways. So, it is selection of RC component to obtain the phase shift what is whether it is going to be very crucial because if I use low value of the R and C then whether it is going to give the loading to this op amp output. If this is an ideal op amp then it does not matter. But this is not an ideal op amp. So, whether we need to add the output whether we have to add a buffer and give it. The buffer is no different from an op amp. After all what does the buffer do? Means buffer with at the output stage whether we have to put one transistor and then we have to give it. No no no this will oscillate happily with reasonable values of R. But do not make this value of R any smaller than the output impedance which is a few hundred ohms. So, as long as this is in kilo ohms this will oscillate. So, 4 or 5th RC units normally 3 because 60 60 ultimately we have to make that 180. But by changing the values of R and C I can make the phase shift of 1 unit let us say 20 degree then I need such 9 units. So, is it actually possible to build that? That is also possible as long as the total phase shift through this is 180. The more interesting question is why do you need 3? Why are 2 not enough? Is it optimized or there is a. So, it is an excellent question to discuss with the class and the main thing is that we are cheap. We would like to use the minimum number of components possible. So, the question is what is the minimum number of components? Now, if you have just an RC what is the phase shift that you can get? Less than 90. Not 90 less than 90. So, by putting 2 of them together you cannot get full 180 degree phase shift. So, therefore, you require at least 3, but there is no problem you can have 4 5. No, but if we go on adding the number of units then actually you get the attenuation as we go in the next stage. So, the minimum. So, then is that gain has to be increased according to that then that 29 will fail correct. The 29 is only in this case, but you can analyze that and you can you will get a new value for that gain. So, the generation of sinusoidal I am going to come to. So, we have done 2 cases. One in which F times A was less than 1 that was the positive gain boosting. The other that time A was equal to 1 or of the order of 1 and that is the oscillator. What happens if F times A is much larger? That is what happens if your output is huge feedback is huge even compared to 1 you are giving more than necessary feedback. The question is what happens there? So, the other 2 cases we have done we stop there. What happens in the third case which is much more important because you are never going to find that get exactly equal to 1. So, what happens if the feedback is equal to is greater than 1. So, therefore, essentially what happens is that suppose you were to give it an impulse does that impulse infinitely increase or infinitely decrease or remains constant. If you have a stable amplifier then it stabilizes after you suddenly perturb it. If you have an oscillator and oscillator in order to work does require a perturbation. If you had 0 noise in the universe then an oscillator will not work because the oscillator is essentially an amplifier with feedback and it needs to amplify something. So, you assume that the noise is always but suppose the feedback is more than 1. In short, in fact this is the way I introduce oscillators in my class. This is my amplifier black box and this is my input voltage and I am giving it through a switch for a particular reason. So, I have applied the input voltage let us say this switch is now closed. Now this is my output voltage. Here I have some attenuator of a factor F and here I have produced suppose the gain of this is A. So, I have produced A times F times V in. Now let us consider various cases. Suppose I make A times F equal to 1. What does it mean that A times F is 1? It means this is V in. It is a replica of the input. If it is a replica of the input then suppose I was to present this instead of the input then will this amplifier know the difference? It will not know the difference. If I make an exact replica of the input then the amplifier will not know the difference and then that is how we get the oscillator and that is why the condition is A times F should be 1. If A times F is let us say half then suppose you had a 1 volt signal this became half then when you feed half that will become one quarter that one quarter will become one eighth and so on. So, it will steadily diminish. So, that is what I was saying. If it is equal then it will sustain itself, but if it is greater what happens? If it is greater the amplitude will continuously increase of this. How long will this amplitude increase? So, it will increase till there are some non-linearities which come into it in the sense that it starts the output starts let us say to take a very extreme case the output becomes somewhat like this. It starts being cut off. So, instead of being a proper sine wave it gets cut off like that because you have already reached BCC. What does it really mean? It simply means that your total output not only has the original frequency F, but it has its harmonics also it is distorted. So, it has the output contains F 3F and 5F if it is like a square wave. Therefore, the effective gain which matters for feedback that is only this term. Therefore, even though the overall gain is large the gain which is significant for the feedback because you are replacing the original input signal. The gain at the original signal becomes less. I am giving you an example let us say I have a gain of 5 volts supply. Let us say I have a gain of 10. Now, if I give it 0.1 volts the output is 1 volt and 1 volt can be happily produced by this amplifier. Suppose the input is 0.5 volts the output is 5 volts it can barely manage to produce this if it is properly designed with the VCC equal to exactly half and so on. But suppose I give the input to be more than 0.5 volts suppose I give it a 1 volt input the output cannot be 10 volts even though the gain is 10 the output cannot be 10 volts it will reach a value 5 and then saturate. So, the effective gain is 5 the input is 1 volt the output is only 5 volts all the rest of the power is going into harmonics right. So, essentially what it means is that the gain which was constant up to some point has now started reducing alright. That means once I reach into distortion my gain does not remain constant anymore it comes down ok. Therefore, the gain is now non-linear why is the gain non-linear what do you what do we say V out equal to A times V in and if A is constant then this is a linear relation. But even A is not constant as you increase V in A starts coming down right. So, that is a non-linear relationship because A itself is a function of V in. So, in short when you become non-linear you can reduce the gain ok. So, this is what really happens if you do not do anything you cannot make F times A exactly equal to 1 to be safe you will give it a little more gain. So, A times F will become more than 1 if it becomes more than 1 the amplitude will increase once the amplitude increase you will reach this non-linear regime. So, till the amplitude increases and output is within the power supply range the amplitude will continue to increase. But once it starts getting saturated the gain will come down and when the gain comes down the effective value of F times A will become 1 ok. Let us just go through this by step. Let us say that my feedback factor is one tenth ok. Now, when the gain is exactly tenth I will get oscillation correct. But my input was initially some very small noise micro volt my gain was 10. So, 1 micro volt became 10 micro volts 10 micro volts went back ok. Since the gain is more than 1 right if the gain was 1 it will it would it would remain at that at that frequency it will oscillate right. But now what I am saying is that I am feeding back with gain of more than 1. Therefore, the amplitude grows right this amplitude grows so the micro volts becomes milli volts and milli volts might become let us say 100s of milli volts. So, it will keep growing the amplitude will keep growing of this oscillator till till what point will it grow till the output starts distorting. Because the output is now distorting as you know that the effective gain has come down right. So, the output will go into distortion and so much into the as it goes into distortion the gain starts keeps coming down right. So, it will stabilize when A times F becomes equal to 1 in a distorted state right. So, for example, let us take the case that it stabilizes let us say my supply is 5 volts it stabilizes when the input is 3 volts that means the gain is only 5 by 3. Though I had the gain of 10 to begin with now the effective gain is only 5 by 3 right. So, the gain has reduced and it has needed to distort so much in order to reach a voltage where the effective gain is 5 by 3 ok. So, therefore, the amplitude stabilization actually the original differential equation says that the amplitude will be stable only when F times A is actually 1, but F times A cannot be made exactly equal to 1 F times A cannot be made exactly equal to 1. So, therefore, if F times A is greater than 1 the amplitude will grow as the amplitude grows. However, all the amplification does not go into the fundamental some of the amplification goes into you start creating harmonics. Therefore, the part of the gain which is in the fundamental actually reduces right. So, as a result now the effective gain reduces till F times A becomes 1 and at that amplitude it will now stabilize. It is in your interest that your gain is not so high that it stabilizes at a high value of distortion because then you will not get a pure sine wave. Therefore, the idea is that you should be able to adjust the gain such that it is just over 1 ok. Now, how do we adjust that gain? To do that then you actually use some non-linear element sometimes you use a diode to cut the gain. So, if you recall I had this. So, these two resistors had set the gain to 29 that made F times A to be equal to 1, but let us say that I have had to it in order to get any oscillations at all I have had to adjust this gain to 30 ok because there are some parasitic or whatever. Now, the oscillation has started, but I am slightly more than 1 as a result the amplitude will keep growing. The amplitude will keep growing till I start distorting. As soon as I start distorting the gain will come down a little bit and as soon as it reaches 29 I will have steady state. So, in steady state there will be some distortion. If I started out with 30 there would not be too much distortion. If I started out with 100 then I will have a huge amount of distortion. It is also possible that I may convert this from a sinusoidal oscillator to a square wave oscillator. In fact, a relaxation oscillator is nothing but this ok. In a multivibrator what do we have? We have the same thing. We have feedback through a capacitor, but it is producing a much higher gain. It is full gain and it just goes from 0 to 1 and 1 to 0. Otherwise that is also a feedback amplifier, but it does not produce a sinusoidal output, produces a square wave output because the feedback is not controlled. If you want sinusoidal output then you must be able to control the gain based on the amplitude and very often what they do is that they put a thermistor here which senses the temperature of some resistor through which the oscillations are going. And if the, so actually the thermistor will have to come here, but essentially what it means is that as the amplitude increases it decreases the gain. As the amplitude decreases it increases the gain. So, as a result this thermistor automatically adjusts to a value of resistance so that f time a becomes exactly equal to 1 and now you will not have too much distortion. So, very often you have to use some non-linear element which will make sure that you remain in the vicinity of f by a being just fractionally greater than 1. Doing it manually is very difficult therefore you use non-linear elements like diode or thermistor which will then control the amplitude of a. You could, but then the trouble is that as the temperature changes the actual value of resistance will change because of the thermocode. And it is possible that what was stable in October is unstable in December or in January. So, therefore manual controls are in theory you could adjust it every time, but manual control will not be able to adjust it very nicely for all times to come. Therefore you need some non-linear element whose value changes with amplitude. So, suppose you have something whose resistance changes with amplitude because it heats up. So, now if the amplitude becomes too much the value of r will become high the value of gain will become less. So, for example let us say that this I make this one amplitude dependent that thermistor which should have a negative tempcode. This is an ordinary resistor with a high value of thermo. Normally the thermistor is kept here because its resistance comes down with temperature. So, as the amplitude increases the resistance value comes down and that reduces the gain. If you put a positive term tempcode here for example a nichrome wire. So, now as the amplitude increases this resistor will become hotter. As it becomes hotter its resistance will slightly increase. As the resistance increases the gain which is the ratio of these two will come down. So, there are many adjustments like this which of course first you should set it almost just over one. But then that fine adjustment rather than being done by hand is done by these non-linear elements like thermistors, diodes and nichrome wires. It is going to give you the amplitude nearer to the plus minus V cc. We cannot control in between. No you can control in between. Suppose if I want to have. The amplitude will stop growing when f times a is 1. But adjustment of such type of f times a critically to be equal to 1 will be difficult in this case. Why? Because adjustment because controlling the amplitude because whenever the gain a f if I adjust more than 1 it will try to go towards the plus minus V cc. But on the other hand if whatever you are using for adjusting the gain kixing at a lower amplitude then that is fine. It will not reach till the amplitude reaches equal to V cc. Let us say that this guy heats up when the amplitude is only 1 volt that depends on the value of this. So, suppose it starts heating up when the amplitude becomes 1 volt then this guy will equalize the gain pull down the gain long before you have reached full V cc. So, that just depends on the component that you have chosen. But the idea is that there must be some controlling element in an oscillator if you want to control the amplitude which will take you either you should be willing to live with distortion in which case distortion itself limits the gain. The gain at that frequency becomes less at the harmonics becomes more. So, if you are living to willing to live with distortion then there is not too much worry. If you want to have a distortion free pure sine wave then you must have these amplitude control devices which will reduce the gain as soon as the amplitude reaches some preset value. So, in short what are you replenishing? There are some losses in the circuit and what you are doing is that by amplifying you are making up for those losses. Now, for steady state you should just about make up for that loss. If you do not make up sufficiently for that loss the amplitude will diminish continuously. If you more than make up for that loss then it will grow. But the point is that if you have some control element which will balance them at a particular value of loss because the amount of loss also increases as the amplitude increases. Therefore, the amount of feedback required also goes up. Therefore, the equilibrium is maintained when the amount of loss becomes equal to the amount of feedback. So, if you have a lossy element which is calculated to have its loss change significantly at this value then you will have the proper adjustment at that value. In short that 29 is to 1 was calculated by this loss how much is the loss in that phase shift element. Now, if you say that as the amplitude increases I will have a non-linear resistor whose resistance increases then as the loss increases it will come back into balance. If the amplitude increases the loss will become more and now it will come back into balance. If the amplitude decreases the loss will become less and the amplitude will be forced to increase again. So, elements like that have to be used in an oscillator to stabilize the amplitude of it. We will stop here. I just did not get time to come to practical current sources but in fact the same characteristic of transistors that I talked about they are very often used to make current sources and you generally have a fixed gate bias and in only in this regime of voltages it will act like a current source.