 So we start with the new topic today oscillators. All of you are aware that unless there is a some kind of a signal generation be it sinusoid, be it triangular view, be it RAM or be it finally the square view or a triangular view. You need to generate signal. Many applications in analog are the sinusoid oscillators are required. There are many applications in digital as well as mixed signals. You need some kind of a square way or a triangular wave generation or pulse waves. Some other applications like synchronization you may require RAMs but all these are not very difficult as far as the periodicity is concerned. So we are more interested in signal repeating at a given period. So that is why it is called oscillator. It changes repeat itself after a particular period or it has certain fixed frequency. A typical feedback system is shown here which has the transfer function of HS open loop gain and if it is given a feedback and assuming the beta network is one unity has a gain of unity then a closed loop gain will be HS upon 1 plus HS and the way it seems here that the output is feedback to the input because that is what beta is one essentially means and it interacts with the input and forms a new output and this process goes on. Now there are of course three possibilities which one of course I have shown here. One is this may be positive, this may be negative, both may be positive, this may be positive, this may be negative and one may be stronger positive positive or stronger minus minus which are possibility exist. One of them shown here that actually input signal some way adds to the feedback signal and therefore if I look at the output of this summer it can be expanded in a geometric series which says Vx is equal to V out Hj omega if at omega 0 at the frequency at which I am interested in into V out Hj omega 0 square V out and go on and so on and so forth. Now one can see from here from this series that if Hj omega 0 magnitude is larger than one positive quantity then what will be the series look like a diverging series because each term will be higher than the last one. So it is a divergent series and therefore in some cases this is called growth condition. So if something keeps increasing and it will finally add to the saturation value whichever it reaches where it is not oscillating it is only going in one direction and maybe saturating the other possibility is Hj omega 0 is less than 1 or minus then of course as the term go ahead they will start reducing essentially means and one can say that this will be a converging series and if there is a convergence in series then one may say that somewhere it may go down at some frequency it may happen that the series sum we can calculate and one knows the geometric series has a say let us say there are p terms so 1 minus 1 upon 1 minus p is the sum of any geometric series so I can evaluate for it such divergent series converging series the net sum value for this. This is what is my next slide is trying to show the first part already I said if Hj1 is a divergent series where if it is less Vx converges series representation and its magnitude is finite and can be given by 1 minus Hj omega 0 omega 0 we are only interested in the frequency at one frequency this is the frequency where system will oscillate that is what we are trying to find out. So what Barkhausen said about this condition that if you have a negative feedback as shown there and if you are this Hj omega 0 is larger than 1 and if the condition second that Hj omega 0 180 degree then the system may actually let us say this is minus sign it should decrease so this is the condition at which which we have already seen this is the condition for stability if it goes up other side system becomes unstable so we are right now trying to see where system starts becoming unstable okay we may see that this instability can be used to generate the signal because instability is something is feedback and it keeps returning back of this Barkhausen criteria which you can hear done in second year I will not repeat it again based on this thinking I can make number of oscillators yeah that you can read in second year book this is just main important part you should look if you have a single transistor and you have a single capacitor okay which may be the CDB of the transistor itself okay and maybe some output internal wire connections then we see it has a single pole it has a single pole which is 1 upon RD CL we know at pole the phase shift goes 45 degree per decade so the maximum phase shift which a capacitor can give is 90 degree J transistor will give you 180 degree out of phase signals input output since I am feeding it back difference phase essentially 180 plus 90 that means 270 but for sustained oscillation signal should return back to add or 0 degrees or 360 degrees which is 360 same as 0 so when the signal returns to same phase then only it will start actually operating as a oscillator so a single transistor with a capacitor does not give any oscillations it is like a system it will die down at the end of the day and no oscillations can be seen we said okay if that is so I had two such systems connected in series this is essential to the principle of ring oscillator you have one inverter is feeding to the next inverter you have one CL here one CL here okay so now you can see 180 plus 180 is 360 this is 90 plus 90 so total 180 out of 180 the return signal is 360 degrees is that correct this is 90 90 and this is 180 180 so essentially now you are saying you are satisfying by our cousin character that the sigma has to be 180 degree and we are satisfying bar cousin criteria however we see this does not oscillate though it looks to that it is satisfying your condition of oscillations but it is still not oscillating this is what is it called essentially this is a latch two inverters connected back to back is essentially a latch okay so if you put one the output of next to 0 if you get that 0 why this does did not oscillate assuming that CL's are same the loads are seen if you have a CMOS kind of inverter what will be the output resistance of them very large RO so the gains will be GM times RO or even here are there will be few tons of kilo ohms so it will be roughly large gains will be available from both these inverters like the larger gains the output which you are looking at okay will be already reaching towards PDT saturation values so essentially what happens that even if you are satisfying the condition the output becomes 1 or 0 depending on the you know if it is 1 year it comes here 0 and it latches itself to 0 1 combination whichever it starts with so it does not oscillate okay so even if the R cousin criteria is made it is not always possible that system will be stable and the condition there is the gain is extremely high if the gain are very small this may be able to and one of the oscillators will make is exactly this that will reduce the gains so low then the system may be oscillate even with two inverter systems okay so this essentially is the feature which we want to utilize so in most ring oscillators how many stages you find always odd number of stages 3 5 7 order numbers and we never put it even because that even will always reach to its saturation values and it will only latch up okay it is like a shift register okay and it will never be able to oscillate now this issue has been taken care let us say I put now instead of two inverters I put it three inverters I have one inverter driving capacitors CL all of them have same CL okay if you look at the these are amplifiers with a j omega 0 as its transfer function and this is AV 0 essentially is the DC gain or low frequency gains super confusion of the first level very very low frequency gain or DC gain so transfer function for individual this will be AV skew upon 1 plus because a triple pole three of them so it is AV 0 and there is no feedback right now it is only open loop system so it is AV cube each has AV AV AV 0 and each have same pole at omega 1 upon RC so we said there are triple pole at omega 0 1 plus S by omega 0 to the power cube this is called a triple pole system and we like to see does this then provides an oscillations somewhere close to omega 0 it may not oscillate at omega 0 but where does it then oscillates we can see and that is what we going to show later okay if we say we can see that oscillation to sustain we can figure out this is the proof will do that the gain has to be at least two for per stage and the then the oscillator frequency will be root three times the pole frequency this is what we should get we will try to prove that we probably get this value as we suggested but however it essentially is trying to tell that three capacitor must give you phase shifts of at least 60 degree per capacitor so that the net is 180 total is 360 plus 360 730 720 okay so you are you must get the oscillations properly now how do I get each RC combination to give me a phase shifts of 60 degree this is what the basic principle of designing a ring oscillator okay but before we go to actually deriving this frequency let us look at this triple pole system with a negative feedback okay is that okay everyone is trivial but this is a proof I want to do it but I will prove later that why I am right or I am wrong let us see whether I am right hopefully I am right is that okay let us say open loop system for a triple pole one is AV 0 and right now I am not making cube I am just saying AV 0 is the net gain for triple system which is AV 0 upon 1 minus SP 1 P 1 is the actual pole remember it is minus side it comes so it is magnitude wise you can like this so it writes 1 minus S by P 1 to the power cube I repeat again AV 0 is the DC gain or if you wish call it a very low frequency gain there is a confusion going on over the end of the course I have restarted rewriting that again and again okay this is your open loop amplifier gain or transfer function and you get a feedback with a feedback of network of which gives you a gain of beta and right now my assumption is beta is independent of frequency so if I close loop gain I look at it ALS upon 1 plus ALS time beta beta could be 1 or beta could be whatever network actually you want to create you may have a capacity network you may have a resistive network to return back to the inputs if you write closed loop gain ALS upon 1 plus beta what I am now going to do is substitute this ALS in this okay and then get the closed loop gain in terms of S by P 1 and AV 0 as we want to write now I say AV 0 is not individual this total gain of the triple pole system later we say it may be cube because each stage will give equivalent of that okay so is that okay everyone if this open loop amplifier gets a feedback with a network gain of beta and beta is not a function of frequency then the closed loop gain is ALS what is the purpose of doing all this I want to find a condition in which the oscillator frequency is decided by omega 0 which is essentially in my hand is that clear because RC is my time constant I can get RC by design and therefore I should be able to relate my oscillator frequency with the design parameter which is only thing which I have maybe I also have a gain but I do not really large again because we have seen if it is too large again maybe I should reduce the gain possibility is one only requirement although I am not very keen about gain so much as I am worried about the conversion of omega 0 times something the oscillator frequency which allows me to actually design an oscillator of a given frequency of my choice okay if I is that okay so if I substitute here of course this is has been done in second year third year whatever book you have studied I am just repeating it again so that you once for all you get this clear how a triple pole system or maybe our pole system actually gives you the oscillator frequencies so if I substitute here I can get AV 0 upon 1 minus S by P 1 to the power cube into a just this multiply here this multiply here and then you get this term now you can see from here a loop gain can be defined as AV 0 times beta T 0 remember both terms are independent of frequency and therefore is called T 0 is AV 0 beta not any other frequency of so we can find that loop transfer function which is essentially T 0 upon 1 minus S upon P 1 to the power cube this is essentially frequency dependence this is essentially frequency independent term low frequency term so if I write this and I write that S is equal to j omega and then put P 1 as its magnitude value then it becomes T 0 upon 1 plus j upon omega upon root P 1 magnitude P 1 to the power cube this equation you keep we will use a little later but since I was writing T 0 I say we can also write T j omega at any frequency in this fashion okay I will use this later on to prove that Bode's plot also gives the same result as the solution of this equation what are these called which technique I am trying to show you solution of this equation of denominator is will give you roots and the technique to figure out whether system is going to grow or not is decided by root locus technique so I am trying to now bring back something which 30 a student must be doing control course or must have finished no no next semester fourth year of course have done okay and empty of course should say they have done okay they cannot say they are not then except if there are few MSc students then they may say well I have not done it okay so we observe the poles for closed loop gain as the denominator solution equal to 0 so 1 minus S by P 1 okay here everywhere either plus 1 upon j times or put P 1 like this then I get a solution from this 1 minus S by P 1 to the power cube is minus T 0 or 1 minus S by cube root of minus T 0 because equation how do people solve okay this is a trivial maps but I always realize that sometimes reality has forgotten by many so I solve for you this essentially say let us say x is equal to cube root of minus y then I can write x cube plus y is equal to 0 or x cube plus y to the power 1 by 3 to the power 3 so this is like a cube plus b cube equal to 0 okay which one oh yes you are right but that is relevant to write now okay this minus will be taken care when the solution will come plus minus okay okay or maybe put a mod and put a sign plus whichever way you look at it so if I solve this yes oh sorry he is saying plus so maybe I will do this okay no I think that was correct now this 1 plus was correct why did you say so if I bring it 1 plus T 0 was correct in fact okay right now whatever mod will see later but 1 minus this then this is 1 minus sorry if this is 1 minus so 1 minus root 2 0 is correct okay so 1 plus okay you are you are right okay 1 plus now is correct sorry so if I now have an equation which is x cube plus y cube y to the power 1 by 3 cube I can expand this x plus y by a plus b into a square plus a b plus b square this is so known to everyone of us so there are 3 possible solutions out of this 1 is x 1 is minus 1 by y to the power 1 by 3 x 2 is minus y to the power 1 by 3 e j pi by 3 and x 3 is minus 1 by 3 a j pi by 3 pi by 3 is root 3 by 2 or root 60 degree so e to the power j 60 degree so if I now substitute the poles s 1 s 2 s 3 I will get p 1 plus 1 plus root 2 0 p 1 into e to the power this and the other is e to the power minus j pi by so they are called roots of the poles which you have actually wanted to find for the ACL please remember this is for closed loop feedback negative feedback system we are still working on negative feedback system okay and we want to see does that give you oscillation age where it will start oscillating after all what is negative feedback is used for stabilizing something so if I can get to the age of instability I may have growth condition starts and we may say we have oscillation conditions met so if I if you have shown this maybe I can have you drawn written down this expressions so if I plot the root locus that is I plot sigma against j omega for these 3 roots please remember all 3 poles are at at t is equal to that we already initially at the same point for the open loop system now if I change the t values a beta values for initial value 0 t 0 your pole at 1 which is minus of whatever that p 1 value which we got okay however if I get it e to the power j pi by 60 or pi by 2 3 I see this is the curve for that similarly for the minus 60 degree this is the curve for that at this point at that is at where the j is this axis cuts at j omega axis we now see e to the power this divided by this is the tan inverse or tan theta of 60 degree okay this is essentially a frequency omega 0 at which what is the omega 0 definition was we observe that omega 0 is the frequency at which this locus intersects j omega axis so this frequency is omega 0 clearly here the real values must vanish what do you have j omega axis real values must vanish that is why it is called imaginary axis if I find the real value to be 0 from the expression which I have one for this point 1 minus real value of root 3 t 0 e to the power j pi by 3 must be 0 solving this I get t 0 equal to 8 at omega equal to omega 0 okay at omega equal to omega 0 a beta requirement is 8 value is an a beta is 1 and there are 3 stages gains how much should be individual stage gain should be 2 is that point clear so if you can now create an inverse or amplifying system which has a gain of 2 then at the same position if you have triple pole equivalence this will give a some kind of a frequency at which system is now at the age of instability and still at the j omega axis this point we want to clear from the root locus figure this divided by this is how much this is P1 this is omega 0 so time 60 is omega 0 by P1 more of that is that okay have you drawn if I substitute time 60 is equal to omega by P1 then I get omega 0 is root 3 P1 1.73 times P1 is that clear to you now at this point what is the frequency of a omega 0 if you exceed little bit omega 0 sigma is slightly positive as you cross t 0 slightly more than that your sigma higher than 0 and imaginary part will only exist okay this is like a positive feedback system which which will show you growth response so at this point ahead there is a growth okay so we should have gain not exactly 2 but slightly more than 2 okay which naturally will allow you to get it so you at that point you say system will start into equivalence of positive feedback is that clear so a negative feedback system has been converted to a positive feedback age system you can say by just adjusting the values of t 0s is that clear and the we can see typically if you look this all that this is e to the power sigma t sin omega t is the growth equation for the waveform to go this is how it actually grows now we see from here if this is so then we can use so we have figured out that the oscillating frequency starts at when a 0 is roughly equal to slightly more but equal to 2 or rather 8 right now t 0 is a beta which is 8 if you keep it then you will be able to have start off growth system okay now we can also figure out the same frequency which we are talking about using a Bode's plot okay Bode's plot can be plotted for the loop gain itself okay so I will have you finished this writing then I will show you I can do the same analysis using Bode's plot also so what is the frequency it will start oscillating root 3 times the 1 upon RC whatever you are fixing for yourself R and C are within your hand and therefore you are fixing the oscillating frequency okay and what is to be guaranteed that t 0 should be just about 8 is that okay so we can use for this side as I said this is not the part of this really course this should have anyway everyone should know this so we can use the Bode's plot for the loop gain T s is t j omega which I have already written a function and then we plot the magnitude of t j omega 0 and we also plot the angle and that is the phase for that which we call the Bode plots please remember each pole gives how many how much degrees per decade 45 degrees so there are 3 poles so it should reach 135 degree at the pole position is that in the Bode plot at P1 frequency of P1 you must get a phase shift of minus 135 degree we also know if t j 180 is positive there is a magnitude at 180 degrees positive is that clear to you and phase has each 180 your growth system is that clear what is the stability or criteria was that if 180 should not reach before the gain goes to 0 dB here gain is positive and you are reached 180 degree now let us see how it looks so what is the gain for how many dB is per decade minus 20 plus minus 20 plus minus minus 60 dB per decade gain starts falling okay so if I do this I find the magnitude which is the loop gain in dB is is minus 60 dB per decade this is my pole 3 dB down point okay and at this point the phase should be 135 if I extend it because there is no other pole it will go straight and then match it saturate by another 135 degree to 0 or minus 270 135 here 135 here and somewhere here we have phase of 180 W1 this minus 180 I extend it on the gain axis so what is the gain margin now positive you have this much gain available at W equal to a W 180 degree phase okay this at 180 degree phase minus 180 we have positive gain there is that clear so what does this mean what does this expression means it means that the stability is disturbed and if you can make that t 0 equal to 8 as I started with this point will get exactly equal to the root 3 times the pole frequency this is what essentially Bode also said so and also the root locus method the third method will be what they go square they are take how many times minus 1 0 actually in circle you get the same point again if you wish we can do that as well but is that okay so just look at it either use Bode's technique or use root locus technique or use request criteria and at the end of the day we figure it out that the oscillator frequency is root 3 times the pole frequency let us say if omega 0 is the pole frequency then omega 0 tan inverse pi by 3 is root 3 by root 3 omega 0 so essentially what is it trying to say at that point the loop gain becomes 1 when okay here is let me redraw it so we now return to ring oscillator okay we now return to a case is that okay everyone has drawn this a polka or 135 laga so Nietzsche 180 that here or they go again positive who I can so this is the condition of oscillations so we return to our case of ring oscillators omega 0 here is out time CL to the power minus 1 please remember our out is normally the output this if it is a Rd kind it is much easier to control omega 0 and a 0 is stage gain so H s is minus a 0 upon 1 plus s omega 0 to the power 3 the net transfer function omega used here is omega 0 each stage should contribute to 60 degree phase is that clear that is what we said now 60 60 60 must appear so that total is 720 or 360 or 0 whichever you call 180 180 180 plus 180 so you return back to original phase okay 0 or 360 degrees so if you do this I write omega sc should be tan inverse pi by 360 degree means pi by 3 so tan inverse pi by 3 is into omega 0 should be the or 3 times omega 0 should be the oscillator frequency and at that time loop gain is unity okay loop gain is unity so I can how can I design an oscillator all that I have to design an oscillator is to decide my R and C okay that is for R is decided by in transistors by what W by else so CL is also decided by CL because CDB and other capacitances are also area dependent terms is that clear to you so adjust your CL and R such that you get oscillator frequency of your choice that is how oscillators are designed so what is the typical a 0 therefore you should have now for each stage for a root gain equal to 1 at this omega 0 what does that mean a 0 by 8 is cube by it is 1 so a 0 should be 2 ring oscillator thus will oscillate at frequency 2 pi omega 0 into sorry 2 pi not omega 0 by 2 pi so omega 0 root 3 by 2 if the stage gain is equal to 2 each stage gives you a gain of 2 and then at this frequency ring oscillator will start oscillating clearly from earlier discussion that triple pole case says that v 0 is a some constant e to the power a 0 minus 2 by omega this is the growth equations into cause of this is the expression for the growth waveform which you generate see it more on the solutions so obviously a 0 is less than 2 it will start damping a 0 is greater than 2 uncontrolled growth may start and a 0 equal to 2 sustained oscillations will be observed so a number of number of inverter stages should be always odd why because otherwise the net phase can never become 0 so if you are 5 this will be a to the power 5 by something and it will still come to 32 it may come and a 0 may still be 2 okay and you keep on doing this for any number of stages typically in a large ring oscillators which we test how much is typical in CMOS the delay per stage you have any idea what is style per picoseconds so now if you want a frequency figure out how many stages you should have so that n 2n plus 1 times so much will be the frequency of your ring oscillators and to get that W by L for each stages adjusted by you is that clear so the design is essentially only on the W by L and associated please remember CDB which is the output capacitance of the transistor is essentially is governed by area see input the next stage is also governed by area okay and therefore whatever W by L you choose please see to it it satisfies both your current conditions as well as your capacitor conditions is that clear so that the gain is 2 as well as you get the frequencies of whatever your choices this is the most standard square wave generator which we use in the case of digital hardware everywhere you find the easiest method of generating pulses are the ring oscillators just put large number of stages for low frequencies okay if you anyone who is doing an empty project or something you must have seen to test your process we are one of the test area which we have created is the ring oscillator what is it tells I have 125 stage oscillators so that the time which it will require will be large enough frequency low so I can exactly monitor my frequency and by stage law I know how much per stage the delay is essentially coming and that will decide my speed of the digital hardware which I am going to use is that clear so one of the test block on any chip test area is always a ring oscillator and typically it has 125 or even higher number of stages simply because of what because the delay has to be measurable on a CRO okay otherwise the inaccuracy in your measurement may actually cause the problem and since you know that net delay you know how much is per stage delay you have so that is how ring oscillators are used as the test block for all digital signal processing okay then please remember all chips have one test chip itself what is it there may be many transistor areas or many components on a chip each representative of that is brought on these separate area of a chip okay each like you have a D flip flop so you have few D flip flop your adder your small added there every block which I use in real life is actually brought for test individual testing so large pads are used to test individual blocks large pads are normally not available so what do we do you put a multiplexer there to use same pads one after the other okay this is how the test patterns are test parts are created once every chip must have at least how many test areas can you think it may not be a poor a wafer may get that test areas on a chip 5 top bottom left right center okay all variations are taken care if it satisfies everywhere you are well within your design to work at every possible is that clear so please remember chip when you make many of students have been told but they do not want to think that way whatever component you require first must be brought on the test so test you must be designed and put in the center two corners and two bottom and lower corners so that the test is performed before the chip is because chip will have no pad internal to test anything let us say there are number you can prove there so how do you know which part didn't work okay so in real life whether it is analog chip or a digital chip that test chip area must be left for you sometimes even more tests are done but at least these many tests are even for example process test is done whether the wave metal is going through down so a small trenching which I did here put a metal and actually to pad I test whether continuity was okay or not okay so all test things are always provided on a test area please remember this is something no one tells but I figure out my own student do it in spite of being told so it is okay I mean I can't blame anyone but that test is essentially if you are really getting your chip fabricated and want to test you apply input and nothing works then what do you do just hope for the best every time God may help once a while but not every time so test is something and it is required lot of thinking what testing we should do okay so what is the best technique is go to the earlier chips find what test area they are used implement all of them with you at such a this is the game all of us have been playing and no new things come okay our aim was not to do this theory as we did so long that okay why it should oscillate our interest was only in VCO which is what we are going to use VCOs you all know stands for voltage control oscillators that means by changing the voltages control voltages I should be able to change the frequency but why did I do all this to once for all make it clear that from where everyone is bringing the theory to utilize in design of an VCO the part which I did here is no valid bearing on my VCO designs but as a concept you all should know that how I actually get oscillations in real life and why I just gains to these values okay is that clear so why two inverters things will never work as oscillator because they immediately their gains are so high okay so that is that clear GMR 0 so high that is who is already crossed long time okay so this fact that there is no oscillations can be understood if you know what is the theory behind okay so this is something I repeated what secondary third-day students should do anyway or no there are two kinds of oscillator which I didn't say here one is called which only generates synocytes the other will generate other than synocytes the names are relaxation oscillators or phase shift oscillators the other way but anyway right now I am not separated all of them I just want to use a normal sense VCOs the two VCOs will generate the square roots which are shown here one is called source coupled VCO which is also popularly known as free running multivibrator okay the second is also similar like a ring oscillator but it is called current starved now this world is very interesting current starved it does not have it I can draw more current I want more current but my current source or since I am not providing me that I am limited by someone else okay my we just wants this much current to flow but the lower current source is it seems as it may so these are called starved the starved inverters okay if you look at a VCO all that I am asking is we should have V controlled which will give me output frequency of my choice okay. So if you look at simple expression omega 0 is a frequency when I say there is no control signal or 0 control signal okay is called omega 0 then I say omega out should be this is equal to omega 0 plus K some constant times V W out V controlled so obviously the unit of this should be what something radian per second per volt kind of thing because it should convert into radian per seconds okay is that okay this is the schematic of a VCO this is what I want to design okay is it okay now what are the features of a VCO if I plot the same curve which I just now same expression W out versus V controlled I have linear relationship that means K is treated constant I may vary the key but K is slope which is constant that means I can I can do this but for given this K VCO is the giving a slope for this curve okay Y is equal to MX plus C this is the frequency omega 0 at which typically V control is close to 0 intercept as we call. So we operate our VCO say this center point some ways call it VC center frequency or central omega C is come kind of a center frequency and V1 and V2 are on its left and right so what is the range in which this was VCO will give me outputs for V1 there is a frequency of omega 1 and for V2 there is a frequency of omega 2 so I have a range of omega 1 to omega 2 as my tuning range what is it called tuning range so a VCO I should be able to tune at least in a band of say W1 to W2 or to say the range is W2 minus W1 and on an average this is half of that is roughly the central frequency it is need not be exactly half because there is an intercept going on but magnitude wise it should be roughly half okay so from this expression I can write KVC 0 is 2 pi f max minus f min call it f max call it f min divided by V max minus V min okay is these called VCO gain KVCO is called VCO gain. Typical power dissipated or used here in most VCOs is of the order of 1 to 10 milli watts okay typical that does not mean I cannot make low power or I can never make high power but larger power will require larger currents and therefore larger W by L so you will have a lower frequency down so the given bands will have limited or some maximum minimum powers available generally VCO operation is 1 mega radian per second okay W max as a minimum and maximum be 10 mega radian per second okay omega these are radian per second is omega because it is say not f multiplied by 2 pi and you get or rather divide by 2 pi and you will get frequencies so for this such typical values KVCO varies between 0 to 5 mega radian per second per volt. Let us look at the real characteristics of V control and W out which is extended beyond this okay this we see and this we see they are not same this is central and this is not central so we say prior to this and ahead of the VCO frequency which you have the relationship between V control and W out is not linear only this region to this region KVCO is constant ahead of it goes down and goes saturates okay. So we say as long as KVCO is in linear this that is the range of a VCO maximum range is that clear if this goes in the nonlinear area or this goes beyond on this region then the VCO will not operate the fashion in which I thought it should okay. So in real life the characteristics of this to this relation is not linear everywhere for all frequencies so the minimum frequency from which linearity start and up to the maximum criteria till which the linearity is maintained is the actual VCO range is that clear this is also VCO range fair enough for this actual max to mean what you will do but this is the ultimate limits beyond this or beyond this VCO will not operate in standard W0 into K times omega K times V control. So we must for a given technology first to these markers are figured out how much is the range in which my VCO can operate is that clear to you so this is something physical range you know up to which you can really operate if I plot W out by V control and this may have a frequency dependence so this is what it says KVCO is constant and then it start decreasing beyond those frequencies okay. We say this is the range up to which VCO should be used in your application corresponding to this frequency whatever voltages are possible only those voltages should be allowed to swing on okay is that clear so otherwise you know you may swing from 0 to 15 volt what do we do okay you may do it but the system may not respond it because you may reach somewhere here or poles may start actually get reducing the KVCO terms okay is that point clear to these are something limits which many a times a priority should be known for a given technology and once these known then you can actually go to this curve and you know what is the value of this this possibly is normally measured value these are not derived values these are normally measured value for an VCO and this data is many times what in your NL of model models okay please remember VCO design is very very crucial in some other circuits in RF where do you think we need in RF circuits I have shown you first few days my many slides in a receiver when the signal arrives from antenna and you actually boost it through amplifier that frequency may be very very high and your processing has to be at digital low frequencies so what do we do we mix it with some other frequency which is close to this frequency and you create beat frequencies okay or you scale down that frequency by VCO tuning so this VCO has to be our variable frequency so that I can tune out okay this is called the mixer has one end as VCO other is the signal and that together will give me the output which is omega 1 minus omega 2 which is small frequencies okay for the image which needs to be rejected okay so please remember VCO is a essential part in any mobile system or any such system in which signals is received from outside okay so this is why I thought that you should know you need VCO irrespective the problem is on which probably I do not know whether we will be able to cover by Friday if VCO changes its frequency because of the drifts or I want this range to be different then I should be able to synthesize another frequency out of same VCO okay I want one frequency F okay range is over but I want another range like in bands in the case of mini this you will have bands you want to tune different bands what do you do they are not a circuit which is used is called phase lock loops which is also called fringe frequency synthesizers okay normally will be lower than the VCO used so you first design highest frequency VCO keep dividing as much as you wish okay this is required so VCO is the kind of VCO I am talking is more useful for digital applications but so is true for any analog applications if needed before we start working ahead let us have some terminology given in a database book nothing much from my side a mid-range value of the VCO is called center frequency I already said I just a minute in the range somewhere in the mid band is we call it center frequency mid-range value then the second terminal second term which is of relevance for a VCO is decided by a skull tuning range and is decided by variation of VCO center frequency with process and temperature now this is another issue which I was just talking you need stability lock you should need so you need to have a variation of VCO center frequency with process and temperature variation is limited by that and the frequency range needed for an application how much which band you are working at and how much process variation and temperature will decide the range up to which circuit can give you proper performance now there is an issue which I do not know whether I have written but I think I remember maybe he has written he is we said that after all the faith and frequency of any circuit will also be decided by the noise sitting on the control because if we control itself changes other things will proportionately change anyway is it not and that may be aside essentially how much noise it picks up so we must be able to figure out frequency noise at the output and which is proportional to KVC this noise output noise is proportion to the VCO gain larger the noise or give larger is the noise output you get so what should I do to reduce this noise at the output I should reduce the VCO gain KVCO because proportion to that is okay I reduce that value itself but the tuning range was proportional to what KVCO so if you want a larger tuning range I want larger KVCO I want a low noise outputs then I should have lower KVCO this is the design is that here the two cases one you need higher VCO this KVCO for larger tuning range and you need lower KVCO for lower noise outputs is that okay this so therefore there is a design issue which how much noise you can tolerate and how much gain see frequencies limit you have the third parameter of interest is the tuning linearity now we will see this latter since the capacitance the tuning is normally done by change in capacitance so we can say KVCO output by dVDD how much is by power supply variation output changes which can be written as dA poro DC disk where capacitance is a function of power supply voltage what is this means that if I change the VDD my KVCO is not a constant I am also worried how much it varies actually the fourth term of my interest is I need larger output amplitudes because you want a larger oscillating signals okay but that may limit the gain part from that because if that increases you may get instabilities so how much maximum amplitude I can reach before stability criteria is violated stringently I mean excessively typical power that I have said for VCO is around 10 milliverts 1 to 10 milliverts and you have to trade off between power frequency and noise so whatever we started the same this is still valid for even in VCO we need to have trade off between noise power and frequency so you cannot have large tuning range low power and low noise there is nothing kind like it is like a triangle system you stretch one the other will reduce anyway okay the last part for the VCO property this is I repeat first words we talked about what center frequency tuning range tuning linearity output amplitude power and last but not the least last property of VCO which is of relevance is signal purity even with a constant KVCO and V control that is not no noise and V control VCO is KVCO is constant output signal is not always periodic there is still noise sitting over output frequency output signals both in phase and in frequency these are called phase noise and jitter the pillar essentially controls this phase lock loops essentially controls phase noise and jitter please remember phase and frequency is time so frequency domain and time domain they seem to be different but they are always connected so one can figure out not very straight forward way if you are given a jitter what will be equivalent phase noise or given a phase noise what will be equivalent jitter jitter is always in time frames phase noises like between the frequency domain so is that okay to you so these are the properties so please remember so we need an PLL at the end because otherwise this will take care of this your noise part which you thought is not there even if you maintain KVCO and you maintain the control voltage even then okay so this is very essential in any VCO design which is the most common VCO okay also popularly known as multi vibrator because you have worked mostly on the bipolar circuits you will not immediately guess this is same circuit but it is essentially a multi vibrator and implemented in CMOS technology or NMOS technology if it is a this is all NMOS if I wanted to make a CMOS the two transistors M5 and M6 will be P channel and that is the only difference between a CMOS and fully NMOS I repeat if I want to make a CMOS out of that this M5 and M6 are P channel devices and their gates are connected to ground essentially making them saturated either way that okay you can see from here what is I am trying to show you that this M1 M2 are essentially controlled by the V control okay for given W by L and XS voltage available to me for a given technology they may give me current sources okay IDS1 and IDS2 if I make sizes VT same these two currents will be also equal right now I showed a different but in reality I will make them equal I have connected a capacitor here why I need a capacitor because unless I create omega somewhere I cannot create omega oscillations okay root three time omega 0 is okay but omega pole has to be created so here is one capacitor of course right now in my analysis because this is what generally everyone does I have not taken up parasitics but if you put all parasitics this may go little heavier and you need a simulation to perform okay right now to make analytical thinking we assume it could be one C constant in reality you will have to figure out net capacitance at these nodes and that is the values you will have to figure it out okay but theory wise what is it happening see these are two current sources okay for some reason we will see we will come back after this all the time going if one of them is conducting let us say M4 conducting and M3 not conducting okay so there is no current coming from M3 because you said M3 is not conducting this partly may occur because of the values which I get on V out V out bars okay if there is no current here and M4 is providing you a current this requires ideas because this on this current source require this now this is connected through a capacitor so when I switch over from state not only one current which it has to supply to maintain this but to maintain this the other equivalent current must flow through the capacitor down okay so how much current this should supply twice so that is the theory we are looking into inverse happen if this is on off and this is on this must provide two ideas current so that one ideas goes there and one ideas goes through capacitor to create let us say for reasons this is digital circuit and therefore M3 is on and M3 is off and M4 is on fully on no current goes through M3 if this current is coming from M4 from power supply side like this this must maintain one ideas now this current source is connected to the other end and requires an ideas current so when I am switching from this off means some value here and this as which this immediately at switch point must provide twice the current one to go through this and one to go through down okay so that means it must supply twice ideas currents by same logic if your inverse is true that M3 is on and M4 is this M3 must provide twice ideas currents to go through these two and then the capacitor which are like this or charge like this is that clear to you this is essentially the feature of source by stable element and that is what by stable is it is not by stable by stable is essentially it does not come out of it it remains stable latch this is not a latch it is a stable or rather it is called free running okay so we will come in the evening and will now study the waveform for this also for the start one okay and we will also look into other oscillators time permitting.