 We start with the thing which we just left before lunch, a source coupled VCO and I said I can have either N channel all devices or the upper transistors could be P channel to make it CMOS. Let us say since these are two N channel devices, the basically if M5 and M6 are in saturation, these two voltage initially will be VDD minus VT, is that correct? They are in saturation, these are like a pass gates and you put VDD here. So they will give you one VT drop across the source drain and you have VDD minus VT at both ends. If you have VDD minus VT, it should remain permanently that is called metastable position on the cliff, ball is sitting on the top, nothing happens. And if there is a noise, the ball may fall all after right from the cliff, okay. This is called metastable to stable, by stable states. So anyone which may pick up noise little more than the other may bring this little higher than this or this little higher than this and then M3, M4 has been have been connected in latch form. So one will start pulling it up, the other will start pulling it down, is that point clear? Okay, so let us start with this definition what will happen. If you say now at some point of time when you start measuring that this M3 is not conducting, if we say that M3 is not conducting or M4 is conducting, obviously this potential should be such that if this is driven, this potential must be lower to switch it off, is that point clear? If M3 is to be turned off and M4 to be turned on, VJ is minus, let us say initially when V1 must be equal to V2 when it starts, so this VJ is must minus VT must be larger enough for M4 to conduct and this should be less than VT not to conduct. So if that is so, this potential let us say it is VJ is minus VT, then this potential at least should be VJ is minus 2 VT, so that this turns off, that is all you say the levels of out and out bar will be actually seen, is that clear? If you say this is conducting, this potential to be switched off or this has to switch off, out must be something lower than this, so that this conducts and this does not conduct. So if I plot this now, this should be VDD minus 2 VT, this should be VDD minus VT, so this is conducting, VDS minus VT is larger than VDD minus 2 VT, so this will be switched off because this is VDD minus 2 VT, this is VDD minus VT, so this turns on, this turns off by 1 VT shift is sufficient, at the edge but we can say it switches off, V out is VDD minus 2 VT and V out bar is VDD minus VT, so edge off, it will switch off because this minus this and this minus this does not support say let us VT, okay. So if that happens, I have a figure you can draw that, I do not okay, I can keep this, they are the figure I do not have but I can keep like this, so I am interested to know these three potentials here, here and here, okay, V1, V2 and V out, V out bar is a bar of that, so that we can just say the otherwise, okay, think of it, if this potential should be at least and they are equal, if this device has to be switched off, it should be at least 1 VT below, okay, so that is that 2 VT minus 2 VT, I have removed that much voltage equivalently. Actually, one does not know at the start how it works but that is how one thinks that noise would have actually done this kind of a, in a oscillator there is no input signal, so obviously something must start in a way it can, okay, so let us say initially since we said M3 is off and M4 is on, so obviously one may say V out is VDD minus VT M5 minus VT M4 or at least 2 VT N, there is no current goes through M3, okay, please look at your there, there is no current goes through M3, so M4 must provide you 2 IDS current, so that one goes towards the M2 and the other M1 has to receive a current of same IDS assuming M1, M2 are identical, that means capacitor will be charged through what current? 1 IDS current, is that correct? 1 IDS current, this capacitor will be charged, this is what I have shown, this is 4, it provides same current and therefore and there is no current coming from M3, is that clear? M3 is switched off, no current is coming from M3, so if IDS 2, IDS 1 are same as we did here, that means one IDS goes here, the other IDS must supply to charge this capacitance, so that this potential and this potential are plus minus now, is that clear? This potential and this potential are, call it say I have given the name V2 V1, this is V2, this is V1, so if you look at V2, let us say V2 was given by me, this is my V2, this is my V1, so what we see initially, let us say it was VDD minus VT, okay and actually it is, sorry the top portion at this point, it is VDD minus 2 VT, that is why device was switched off and as long as this remains, V2 remains VDD minus 2 VT, so that M3 remains switched off all the time and M4 conducts all the time. During this charge once occur, no changes will occur because capacitor will not then allow any current to flow further. At that instant, since V2 has gone to other higher, this potential and V1, please remember we will not look into V1, we will see it is a complementary to that, V1 and V2 are complementary to each other, so we will right now not assume, you can see from here every cycle, it just follows the V1, okay. So when this occurs, the next stage, case 2 we will see, we will come back and come back to see figure again, if M3 is on and M4 is off, let us say, if M3 is on, so the inverse should occur, is that clear? Now M3 should conduct and M4 should not conduct. Now it is the complementary case of V1, V2 and that is why I said V1, V2 will actually every clock cycle, they will switch over the, they will follow each other. The time taken for time to charge the capacitor, let us say is delta T, so what is the current we have? IDS, so what is the charge dumped on the capacitor? IDS into T, delta T. However, what is the potential through which it is going through? VDD minus VTN and VDD minus 3 VTN, that is it shifts through a voltage of 2 VTN. So what is the charge because of 2 VTN on the capacitor? C into 2 VTN, they must be same, so IDS delta T is C times 2 delta VTN or delta T is 2 C VTN by IDS, then the frequency is 2 T delta T repeats again, you can say it is IDS upon 4 C VTN. So frequency of this multivibrator or this VCO, source coupled VCO is decided by what? It decided by IDS, IDS coming from where? Please look at it, IDS is decided by whom? V control, so I have now, I have voltage control I wanted, so adjusted V control, so that I decide my IDS, capacitance I have put it, so this will give me the range, okay, what size range you are looking for, threshold of course is the process technology which you are working at, is that clear to you? So by adjusting the control voltage as well as the capacitance value, I can have a voltage control oscillator which gives me pulses at the output, is that clear? Now the only thing why this, okay, maybe it has advantage, we shall see little later, one of the advantage of this source coupled VCO is, it is comparatively low power because it does not require huge power, depends of course see how much you want to charge, typically it is having a relatively compared to the other one which I discuss is a low power circuit. But it has a very bad feature in the sense, the other one which I am going to show you does not require external capacitor, it uses its own, ring oscillator uses its own capacitances, is that clear to you? So because of this, now please remember capacitance take firstly larger areas and secondly it requires what kind of process, either I will have to create an MIM capacitor between two metal layers or I should create a double poly process, this means I need additional process to have these kind of specialized capacitors made available to you. Let us say the thickness is very small for you or thickness is larger for you, then you have one kind of capacitance and you want to change the capacitance. So you must have more than one capacitor in series or in parallel to adjust the desired value of C, so it requires little larger area to make the capacitance of your charge, is that point clear? You have one unit area you know this much, so you want larger or smaller depends on you make 4 or 5 or 8 and parallel series whatever combinations you do, you can adjust C value and if you do so, you will waste not waste actually, you will use larger area of silicon for one capacitor, okay. This is something problem, it requires technology, different technology and it requires at times larger areas just even to reduce it you will have to put in series of them, so it is not that it is a parallel every time, you can even series them and change the capacitance values, so all VCO source couple VCO that they are called are good because comparatively to the other ring oscillator circuit, one finds that they are low power relatively but requires new process or rather different analog process and also it requires area additionally, this is a weakness the other is maybe some advantage. So if you are looking for relatively lower frequency circuits, okay, then you may have a problem because then you have to actually put large C, if you are very large frequency then your minimum C should be sufficient enough to go there, so there are issues which are not very easy to handle, so it limits your performance in most cases, but it is the easiest way of making a oscillator, this is therefore very popular in many circuits, if you see actual circuits in chips you will find many have their use source coupled VCOs, the second one which I said you is called ring oscillator, now it is also called current starved VCO, this M1, M2 essentially forms a mirror kind of input side, so and that is controlled by V control, so if I have an M1, if I decide my V control I decide the current in M1 and since it is mirrored on that, wherever I connect the same current can be proportionately size if I use, I can pass in the other arms, you can see through this P channel I am going on number of vertical inverters chains, each column is a inverter chain or other one inverter plus additional something in between 4 transistor inverter has been shown here, the method is here is they decide the current, this decides the current, this is decided from M1 and M2 will decide the current in M6, M10 whatever that this side, depending on the sizes I can have different values, I may have same also, if I put same size correctly then I can ask this current and this current may be identical, but to show it right now I am showing you as if this current is 6 and this current is 5 or rather 3 I should say, ideas 3 they are not same, but they may be normally I will make it same, in between these 2 this is an inverter sitting here okay, M5 and M4 forms an inverter this, the output has a capacitance which is essentially the output capacitance of this inverter at this node plus the input capacitance of the next stage which it is driving, you know ring oscillator is inverter driving the next inverter. So obviously I know W into L is the area for N channel and P channel into C ox from the input side and from CDB and this also are functions of areas, so one will be able to calculate the net capacitance which will be area dependent for both omega n I mean N channel area and P channel areas okay, so this capacitance which is self created is that clear to you, it is self created, I do not put external capacitance here, I put natural capacitance which are available to me, now the way it happens that I have a chain of such inverter of course odd number as I said, we already said odd is to be there to make it satisfy the Barkhausen criteria and the last one for example is fade back here so that it forms the chain, now why it is called starved is the following, the current here and current here, the way it will happen depending on the output here 1 or 0, either N channel will turn on or P channel will turn on, so if P channel turns on this IDA 6 will go through IDA whatever current ID 5 can take, only this much current it can push in, your sizes may be decided by the RC time constant you are looking but the currents are coming from current sources and current sinks, is that clear to you, so it may be able to take larger currents but it supplied lower current from the top for the case of this P channel, so it is called M5 phase starved of current, it can drive larger currents but it cannot because the current is provided from and why it has to be done because this current is controlled through the VCO sorry V control, so this is deciding how much current this arm can have, so this charges the capacitor this and the opposite this capacitor discharges through M4 and M3, this is like a normal inverter having a load of capacitor which charges through P channel and discharges through N channel, the only difference what is happening here is the currents in this are not just decided by W by this but also decided by the control voltage which should be stronger than the capacitance effect coming from W by this, is that clear to you, that is why this is called current starved VCO, if of course this oscillator frequency is IDS upon N is odd number, please take it I not written it but N is odd numbers 1, 2, 3, 4, 5, N is odd, C total whatever please remember what is C total I repeat the capacitance output of this and input of the next stage, that is the total capacitance at every node, even here this is the output this is the input so it is universally true, now this for the secondary students or third year who may do some experiment, normally if you see in our lab we put additional inverter here to monitor the frequency, do you know why, if I do not do this then this will start loading this the last stage will have a different you this you have a probe which probe has a much larger capacitance, so normally the last is a buffer stage through which you actually pick up the outputs in the lab, so please check it, there I have if my sheet old sheets are still running I am told 92 last N log lambda I took and I think I have see the other day the same experiments are still sitting there which is little hard for me, so is that clear so this is another VCO which is like a ring oscillator, please remember this is no different from ring oscillator, the only difference what we did we added a control part of this and by doing this I can now make currents available for inverter of my choice which is proportional to control voltage, generally ideas 6 and 3 will be matched, same size will keep, generally 90% the way it is shown it is correct, this current and this current are same, but you can have a different duty cycles by adjusting different currents in the two arms, charging current will be different discharge current can be different, okay that is what a normal CMOS inverters we keep it equal but it need not be I can have a duty cycle of 30% 40% maybe 50 which is equal, okay so I can always do whatever currents I push I can always adjust its duty cycles, okay so this is also very popular in ring oscillator because it allows you to convert your normal ring oscillator into a VCO, is that clear why it is becoming popular because ring oscillators we use very often, so okay here additional circuit if you do it can convert itself into a voltage control, this is one class of oscillators we saw today, there are another class and I keep saying you this answer I may give at the end maybe or you should think, there are circuits in which we only use RC networks okay RC components to make a oscillations, most of the feedback circuit we looked earlier have only RNC, but there are another class of circuits which requires LC combinations instead of RCs, so firstly think of it why LC can do what RC cannot and if initially if RCs were doing why LCs were not used then, okay either way at certain bandwidths or frequencies we always use RCs and at much higher frequency prefer to go for LC circuits, so why complementary to each other was not tried, let us think of it, this is an issue which is not very true every time but yes there are issues with RCs advantage sometimes LCs have advantage so we look into second class of oscillators which are essentially called LC oscillators inductance capacitance, their basic principle lies in the is called tank circuit, an inductor in parallel to capacitance let us say both inductors and capacitors are in fact actually capacitors also are leaky to some extent, so in real life if the capacitance are used have different dielectric is tan delta has to be figured out and the resistive part even there can occur though 99% we never show that, you have seen inductance we show but capacitance in real life even C can actually create additional resistance there but okay in ideal case if I see this circuit if I monitor the capacitance on this side at a frequency which I call it F is equal to 1 upon 2 pi root LC which is called the resonant frequency of this tank circuit, it is called resonant tank circuit, now if we find out this frequency if I monitor Z what is the impedance do you expect this to happen, how much at this frequency when it resonates, so which I have to be current 0, infinite impedance is infinite, okay. What essentially I am saying that the j omega L at this frequency is equal to opposite 1 upon j omega C for this they are equal and opposite is that correct and because of that the current in this arm is 0 and therefore the impedance seen is infinite but any other frequency they may not be equal and therefore there will be a decrease in there as you move away from the resonant frequency, now this fact is very interesting that infinite whenever a tank occurs it gives you infinite Z, okay this can be understood very easily the problem starts like this if let us say capacitor is charged and there is no dissipation so it discharges through the inductor LDI by DT this see the inductor stores energy that energy is discharged through a capacitor and it will keep happening and times so you will get oscillations charged discharge discharge okay but that is essentially saying that there is no dissipation, so in real life there is no such thing like ideal tank circuit okay or ideal resonant circuit, inductance being always metallic wire it always have some small series resistance associated is that clear to you and as soon as now you say R is there some IR drop will keep coming so one time VC may be full but next time it will go down okay because that is heated out heat out so you will not have infinite impedance essentially equivalently saying it will be some finite impedance and the name which essentially says dissipation factor is called Q okay the Q of the tank circuit which is essentially given by omega L by R okay that means there is a finite Q in normal case ideally what Q should have been infinite okay in most CMOS or NMOS technology the Q which will get may be 3 to 4 or 5 okay our attempt to get higher Q's actually have much more trouble some other day in designs. So since the typical inductance are made out of wires or metal lines printed on the silicon substrate or other silicon dioxide areas it is something like a spiral we create this is how inductance are printed okay the gap that turns they decide net inductance of this spiral which we create why we do not create circular that is ideal but circular this on a graphic terminal is very difficult okay so easiest is Cartesian XY okay so all spirals are normally XY spirals but they could be circular by some tricks but it requires better graphic and longer time to do this which does not help you great anyway in some sense so if you now say it has some finite resistance so let us see what is the Z now available to you where you say Z will be different from the infinite value which we created at that frequency. So I calculated RS plus L1S upon 1 plus L1 C1 S2 plus RS C1 j omega L and RS are in series and parallel is the capacitor okay so if I do this this is what essentially RS plus j omega L1 parallel to 1 upon j omega C if you expand this I get this terms and if I evaluate the value of this at square so I can say magnitude of this can be RS L1 square omega square 1 minus omega square L1 C1 square this is a complex quantity magnitude real magnitude while this value will appear now you can again check this if RS is 0 what will happen this is 0 this is 0 okay so what will happen at omega equal to L1 C1 upon L1 C1 1 minus 1 is 0 and impedance goes to infinite so as long as our basic condition initially ideal situation is still available if RS is 0 okay at the frequency of omega equal to 1 upon root root L1 C1 this still gives you infinite impedance but says RS is present so this term is not becoming 0 so there will be finite resistance even at resonant frequency typically if you plot it will be something like this you can see the denominator parts so it will start decreasing from its peak value okay and if you plot its angle omega 1 there is no imaginary quantity and we say that is that is why it is echoing please remember expand this make imaginary quantity 0 then you say at that point phase is 0 on what is the tank principle that the impedance is matched at that frequency so the imaginary part as it goes away so the net phase this is angle Z is 0 at resonant frequency that is what resonant word means beyond which what is the kind of impedance on this side and what is the impedance on this side this is inductive the other is capacity lead and lag whichever way you look at it so since only at the resonance it has a much larger Z which is essentially replicating trying to say there is a Q finite though it is not infinite but larger Q is certainly available which may be as I say in ICs typically is 4 5 best of times 10 so if we say now that I have this RS is going to be there whether I like and I do not like if I want to use this so I must survive through this Q availability how much Q larger Q is better but larger is how much so that we will see and we will take care of that Q in our design so that the system is always still oscillating at we are not worried very much because the phase goes to 0 anyway at resonant frequency even if Q is not infinite okay that part is this LC circuit tank circuit always provides so the first advantage of tank circuit over any RC circuit with active devices there is no question of Barkhausen criteria has to be satisfied I am independently going anyway do what you do okay and that is the fun part in LC circuits this is one example advantage I told you over RC is that okay okay so in real life as I said there will be but I want to have some better way of representing LC circuit which is called the parallel representation of this resonance circuit with RS as the resistance so I say okay if you are RS series tank this is therefore sometimes called series tank circuit actually it is also parallel but all components I want to be in parallel so I say LP parallel RP parallel CP is equal to this identically equal to this so what is the method we have to do it calculate impedance of this calculate impedance of this real value on the left must be equal to real value on the right imagine a value on the left must be right imagine value on the right so we are two equations will get and those condition relationship between LP L1 RS all can be figured out is that method clear points out for this points out for this equate real part equal imagine part equal and solve this so RS plus j omega L1 is this and on this okay first equation is if I do that RP LPS upon RP LPS is equal to j omega SFI substitute j omega RP LPS equal to RS plus j omega L and then please remember I am saying capacitors are identical they may not be you can choose different and then you have long different difficult situation to get I assume CP the same as C1 then I separate them so whatever is this parallel combination must be equal to this if I solve this this is the equation I get I did lot of equation solving I actually put this equation into one I write RS please look at it the way I am now doing RS into j omega L into RP into this if I get it then that is equal to j omega RP this is one equation I say write Z for left write Z for right equate the real and imaginary terms I get two equations I think the first part I just showed you in case you do not want to use it then you actually write Z for left Z for right equate the real and imaginary so I get LP RS plus L1 RP is RP LP when I equate real values and RS RP minus omega square L1 LP equal to 0 from the imaginary side okay from here I can write omega square L1 LP RS RP from here so LP RS RP upon omega square L1 LP by this so I actually what is the value I want to find RP I want to find the equivalent RP corresponding to L and S on the series side from this equation I get omega square L1 LP is equal to RS RP so RP is equal to omega square L1 LP upon RS so I got one expression for RP in terms of L1 LP and RS that I will use now in the other equation substitute there I replace RP is by the term which I am going to get and then solve again to get what relationship between LP and L1 I am interested in two terms I am interested in LP related to L1 and RP related to RS is that correct these are the equivalents I want to see so if I get RP omega square L1 LP by RS substitute in the earlier equation 3 which is this and then you expand that and at the end of the day you forget this term may directly I am multiplied by RS from here and I get LP square RS square omega square L1 square LP square RS LP RS L1 omega square L1 LP square from here I can directly get this then I realize I need not have done this from here you come here or may be from here you can come here okay what is Q omega 1 L1 by RS is the Q is that correct omega 1 L1 by RS is the Q so if I look at these terms okay omega square cancels this cancels so I only get LP by L1 and on this side what cancels RP square cancels so one plus RS square upon omega square L1 square is LP by L1 so that to get that expression what I was looking for okay is that one clear this square P square and then I get term 1 plus LP square R square upon omega square this is time goes to divide care subject if I do this I get an expression 1 plus RS square omega square L1 square is LP by L1 but Q is nothing but omega 1 L1 by RS so then I can write LP by L1 that is LP is equal to N1 times 1 plus 1 upon Q square is that correct normally Q is much larger than 3 or 4 or 5 so 1 upon 3 will be or 1 upon 4 square or 1 upon 5 square will be 1 by 20 1 by 9 1 by 25 what larger number so typically LP is very close to L1 larger than Q you are as much closer smaller than Q you will be you will have to use this expression is that clear if Q is smaller this term may not be small and then you have to find the actual value normally Q is larger than 3 at least even on silicon so one may say 1.1 I can say 1 and I say LP is L1 I substitute it back here and then I solve for this RP is equal to Q square RS so RP is equal to remember RP is omega square L1 LP by RS I just calculated earlier where you have seen that expression RP is omega square L1 LP by RS I say L1 is equal to LP is that okay L1 is equal to LP so omega square L1 square by RS omega square L1 square by RS square into RS which is Q square into RS so RP is Q square times the RS now the way I did it capacitance I have separated anyway so Cp is C1 so now I know from the series circuit I can go to equivalent parallel circuit why I am doing it because if I see in real life when I make a oscillator I am going to play this game very nicely okay so I have now RS in series now all three are in parallel and they are connected to so Q square RS is RP Cp is C1 and LP is if you want exactly this otherwise LP is close to L1 now again if you have that frequency of omega square is 1 upon LC L1 C1 essentially or LP Cp because L1 and L2 are same so LP are same so again say ZH peak at maximum value occurs at redundant frequency maybe call it omega 1 or omega whatever way and the phase correspondingly is 90-90 so this is the principle of a tank circuit okay I know that when it resonates it gives you large impedance large means this should be much order higher order not infinite but much higher order and at that point phases both L and C phase will cancel and therefore it will be 0 phase so based on this principle of time circuit I can make my first cross coupled oscillators is that clear I now start using this principle I know now how tank circuit works I can use now tank circuit with the transistors to tank circuits and cross couple them is it okay figure Z is same as what I said earlier so even for parallel circuit they are same simply because I have made equivalence of that okay so the first cross coupled oscillator can you now tell me if I want to make a VC out of it from where I can make a VC out of it ISS I can control the ISS from my V control I can also do something else pass it okay so I have two ways I can look into it to convert it into its VCO operation okay you can see here this is essentially trying to say okay at one time let us say it is at fixed value when this is operating the other time and this is operating this is at a fixed value these are two transfer functions one due to this one due to this so what is the net transfer function will be if h1 omega is transfer function of this h2 omega is for this since they are fed back please remember this is the way shown here this output is going to the gate of this this output is going to the gate of this is that point clear to you says these are equivalent like a last situation if this voltage changes the current here will change here and therefore change the voltage across this is that clear this is what the cross coupled word means so we say at when the result close to resonant frequencies are at DC value of that what is the gain of each stage when they are in resonant LNC goes away only what remains there RP so GM RP and GM RP if they are separate GM 1 RP and GM 2 RP are the gains so the net gain is product of the two which is because one is feeding the other so GM square RP square is the actual gain of this stage so I can use this cross couple oscillator and what is the frequency at which it will operate as long as please remember this is like the latch situation but why it is not latching now what is the value I am here looking RP are extremely small RP is the equivalent of a series resistance which is less than ohms okay is that correct so GM times RP is very very small compared to GM times are zero in the case of the latch action we said there a 0 when to be on twin no time okay and therefore at only a DC when the capacitance are no phase two inverters could give 360 phase out and therefore the last way 1010 outputs okay whereas here it is identical to that but now the gains are actually and where this will happen only at frequencies of this interest so only when they resonate at Omega equal to one upon root LC then our root LPR CP then only at that time the gain is much smaller and therefore one can say it will act like a oscillator of course that condition must be that should be slightly greater it should be positive otherwise near a feedback will occur it should be positive but should not be very large positive and this will help you to oscillate at a frequency of 2 pi 1 upon 2 pi root LP CP or L1 C1 which is same is that correct so this is called cross coupled LC time circuit based oscillator which is often used and as some of you said I can either change CP by voltage and we will see later or I can change the ISS value by ISS because ISS will decide GM 1 GM 2 and therefore it will decide the point at which oscillations can have that okay Colpett there are two names very famous Colpett oscillator and the other is Hartley what they did will not solve full of it but maybe quickly will show you I have one transistor here I have one CP across the time circuit and I have another capacitance between output to this and please remember here what is this transistor is working like no it is not negative what is this common gate is that correct and the input which is noise is actually available to at the source of this transistor is that clear source of this transistor which is shown here as Colpett oscillators can you think what will be its oscillating frequency roughly how much you see at this node these are ground so this is also ground these are in series now C1 C2 upon C1 plus C2 is essentially the capacitance which is coming at this node is that clear so L1 into C1 C2 upon LP means LP upon C1 C2 upon C1 plus C2 one upon of course that is the omega square term for this with the Cp plus and that is what you are said Cp plus C1 C2 upon C1 plus C2 will be the new capacitance if I also hold this capacitance is that clear so essentially it will be decided by what values LP and the series combination of C1 and C2 will decide the frequency it is kind of analysis quickly may be cut up I am right now not using Cp but later I will add please remember Cp is only paralleling there so I have additional capacitance if I need that the reason why I removed this Cp from here because the actual Colpett oscillator did not have a Cp there that is why actually it was not used we can add that to C equivalent of our network but since in actual Colpett there was no Cp here okay the only thing is there was no transistor then it was a vacuum try out which was used in those cases okay the first Colpett oscillator was used with vacuum device at least 4 to 5 inches long at least 2 inches in dia quads okay or other condition okay this is a common gate amplifier so this is the noise voltage because oscillations cannot start unless there is some signal comes so we assume I was the noise voltage and please remember this is the source end of the transistor so source cable many current source like I am a transistor the VGS do hey please remember the gate is grounded in common gate gates are grounded VB and a fixed value so that is grounded and between source and the gate let us say the potential is V1 which gives me output current of GM V1 then there is a capacitance C1 and C2 so part of the current goes through C2 there is also current in C1 then this is your LPRP and this is your output voltage so how much current actually how many at a given node at this how much current you can see this current this current this current and this current is that clear output per current so we will do that so V1 so hey if I want to find this voltage this is ground so I am trying to find this voltage so I in a ground either I minus this current minus this current okay minus because opposite sign a minus by sign I in minus V out by LPS minus V out by RP into this capacitance 1 because this is current into 1 upon C1S is V1 drop across C1 is V1 minus because it is the opposite polarity there so I can calculate this voltage V1 so GM times V1H known to me is that clear I want to calculate current in this arm how much is that this is V out minus of minus V1 into C2S is the current in this TV or V out plus V1 into C2S so I see 2 is and then I substitute V1 from this here so I got V out minus I in minus V out by this into C2S so I got IC2 as the current in this arm I know GM V1 because GM times this current is also known to me I know this current I know this current so some total current should be 0 at node okay so I is that okay 4 current 8 node per 0 some of them are 0 so I calculate this current I know this current I know this current I add 4 currents and equate them to 0 is that written down so to say GM V1 plus IC2 V out by LPS V out by RP is 0 or any other substitute K you have to come in V out by I in it number of expression out I mean a rough may calculate the fear finally is fairly good okay what is the condition of oscillation we say the impedance should tend to infinite okay so we are buying should tend to infinite if oscillation at occur at frequency of omega R which is called resonant frequency so he's got the up he's got zero current a denominator code so I'm going to get a omega square R is 1 upon LP times C1 C2 upon C1 plus see what I'm going to be able to take it now and the second part from the this GM RP real and imaginary part so GM RP C1 C2 upon 1 plus C2 by C1 square if I use C2 by C1 1 that is C1 same as this now GM RP 4 so at least the minimum DC gain you expect is 4 slightly more because it should sustain it okay and if you you know C1 C2 here this is C1 square upon 2 C1 so just LP C1 by 2 is the frequency available for you at the resonance for this and I mean I said earlier that you want to put it in this tank circuit so a CP capacitance so I can calculate the resonant frequency of a Colpitt oscillator by deciding and GM is decided about the bias current I will put RP of course is not in my hand it is inductance whatever I have I will decide on that q square RS is RP okay So, this is also an oscillator, Colpitt, now instead of capacitors and inductance, you do the opposite. There is a capacitor and here we will put inductance, so this is the circuit, what will it say? Hartley oscillator. Where C A is there, L is there, C A is there, C A is there, C A is there, that will become Hartley oscillator. Please remember, Hartley came later, though in many old books, they always start with Hartley oscillator and they show Colpitt later, but I may tell you again Colpitt oscillator was made first and Hartley was a modification. Do you know what kind of capacitance is we have, we used to use in those cases, variable capacitance, we have no diodes as such with variable, what do kind of capacitors we used? C A is, what can I say, you must know, C A, C A, C A, C A is there, C A is there, C A is So, here is another kind of oscillators before we quit for the day this may be this one. So, here is the other kind of oscillators this is only fun part in that there are other kinds of oscillators before we quit for the day this may be the last this is very interesting someone earlier in between said that okay though now I will show you that a typical noise source is shown here is fed to a tank circuit a C P R P L P because if you give noise it will die down after some time because all people dissipate out a bar the paragraph or a bond may die. Okay let us see if this RP patient make a minus RP then it will act like a ideal tank circuit and then it should oscillate even with the small noise signal it should start oscillating. So, the game was that you are a tank circuit and you connect it with a some active circuit such that the input of that is equivalent of minus RP so it is called negative impedance converter or generator if I can create minus RP out of this active circuit then I will be able to cancel this RP RP and minus RP and therefore we may say that I have a ideal tank circuit to proceed it and it does not require then any other this only you have an active device but in fact actually are using only one time one start it starts however this circuit should maintain minus RP all that time so you need that you know that it is not something like that I am okay is that point clear one simple negative impedance in the generator circuit is shown here now okay and based on this we will show you a oscillator which is also as I say GM oscillator as it is called transconductance oscillator is figure clear RP cancelled by minus RP this minus RP kya jagam in a active circuit lagaya ki jo uska input impedance is minus RP and the fun part there is it is independent of frequency SNHK resonant frequency by minus RPA how much I minus RP okay that is why it will once you provide that it will always oscillate there is nothing to worry about is that okay a case of a head this is negative impedance generator I have to transistor M1 M2 this is a common gate circuit this is common base common source circuit and you have a risk a input terminal hey okay this is biased through IB1 this is biased through IB2 or call it IB1 also it is immaterial it is biased same or different if it is different then GM's are different please remember if you bias it differently their GM's are different but if you assume same then GM's will be same a viscar equivalent what is the equivalent of this how do I calculate impedance I apply a voltage VX at that terminal and find the current there so VX by IX is the R there or Z there okay so I have a source VX at this terminal please remember where I am calculating all in is that okay so this VX is supplied by me and I am monitoring this current IX at this node is input make it not current JRI AC part so if you look at it since it is going to this is M1 and this is M2 from VX yes cut rain M2 cut M2 cut rain and to make a current flow go GM2 V2 what is V2 we will see but GM2 V2 a drain has cut a yes cut M1 cut gate be a M1 cut gate okay I repeat this is D2 and this is G1 okay so now from G1 there must be some VGS V1 which is creating current in M1 which is GM V1 but this is going to VDD for AC it is ground so AC it is ground so this is the drain of M1 okay and this is the source of M1 from source of M1 to G1 must be some potential V1 which is causing this current to flow okay but this source is connected to the source of so do no case or same or say here for voltage VGS because otherwise there is no M2 conduction so there should be a potential V2 and this is ground so this is ground is that equivalent clear to you all of you I start with this go supply source of VX and I want to monitor IX okay the D2 of this there is a GM2 V2 and since this is same as G1 I have a separate voltage which is VGS for M1 so that gives me V1 which gives me a current source of GM V1 in this but the other end is grounded so that is grounded and the source of this and source of these are connected so source of these and source of these are connected okay now we see if I see IX this is the current IX but if you see opposite of this this is going to the ground this side which essentially means IX is also equal to minus GM1 V1 but what is VX you can see from here this is VX and this is V1 so V1 minus V2 is nothing but VX is V1 V2 so V1 V2 that is the VX so VX is V1 minus V2 but IX is GM2 V2 minus or equal to minus GM1 so VX is equal to IX upon minus GM1 minus IX upon GM2 let us say IVs are same so you may say one upon GM1 same as 2 by GM so minus 2 by GM so input seen from this terminal is minus 2 by the negative impedance meaning up go now adjust this GM value by what is current or even sizes to some extent you can adjust W by L as well and get the value of Rn which is equal to RP okay and as you tune it you will actually get oscillations sustained oscillations this is very interesting and the actual use of this negative impedance terminal last oscillator for this side is that okay is that non clear so trivial it looks but very powerful negative impedance generator do you know a similar negative impedance generator can be used generation has been tried in opams or the fans what is it called second number I tell you need to move only capital and the other minute garator a negative impedance generator with opams is called garator okay so garator both popular circuit a equivalent garator is shown here with two transistors connected in this fashion you can create negative impedances and GM decides the value for your RP you are choosing at time this please remember RP are very small RP are very small 10 RS 20 RS RS is ohm less than a womb this means GM should be very high and this higher GM means power so to create this we will pump power for a circuit person first thing power kya gaya so abhij frequency they their power is quite equal and sustainable circuit do you have a scope of the cross couple oscillator but abhi kya nani abhi scope out are such that he you can see from here this and these two transistors are so connected they actually are providing negative impedance at these two terminals one by one think of it so as GM of their you will get RP and minus RP across this and they will be sustained oscillations from this generators this is essentially negative impedance based LC oscillators is that clear that is very there are no this essentially requirements these RP's are cancelled please look at it is cool about we be like here is cool we be like here or what do not circuit up us may a dressing okay all that is a I SS a message are a new server is a little bit but essentially I SS by 2 is provided to each of them and therefore it will act like a same circuit which I showed at this note it was minus RP for this at this it is minus RP for this and you can you can now say this is sustainable oscillator which is used this is negative GM oscillators is that clear negative GM oscillators okay and these are also very popular up is car a core modified version Friday go the card it may have nearly not key abhi to be trailer CP code just that will be the one which we see we are going to often use in real life circuits is that okay the problem I said in this morning that there are issues of deaths available irrespective to what okay and you want to retain that change in frequency or change in time which is called jitter and phase noise how to get rid of them and the circuit we use is PLL PLL will not be part of this exam course but I will tell you what is PLL because that is something you all should know because I don't know your mixed signal circuit people if at all you take that course whether they believe that is actually PLL is a pseudo it can be a mixed signal or it can be analog whichever way you look at it like phase detectors are essentially digital parts okay XOR an offer. Maybe if time permit I will also show you very interesting what is the switch circuits are okay which is again mixed signal should talk about it. So see you till Friday.